326603890-control-engineering-marcel-dekker-inc-8-brown-forbes-t-engineering-system-dynamics-a-unified-graph-centered-approach-marcel-dekker-2001-pdf.pdf

ENGINEERING SYSTEM OYNAMICS

CONTROL ENGINEERING .4 Series of Reference Booksand Textbooks Editor NEIL MUNRO,PH.D., D.Sc. Professor AppliedControl Engineering University of ManchesterInstitute of Scienceand Technology Manchester, United Kingdom

1. NonlinearControl of Electric Machinery,Darren M. Dawson,Jun Hu, and Timothy C. Burg Intelligence in Control Engineering,RobertE. King 2. Computational 3. Quantitative FeedbackTheory: Fundamentalsand Applications, Constantine H. Houpisand StevenJ. Rasmussen 4. Self-LearningControl of Finite MarkovChains,A. S. Poznyak,K. Najim, and E. G6mez-Ramirez 5. RobustControl and Filtering for Time-DelaySystems,MagdiS. Mahmoud 6. Classical FeedbackControl: With MATLAB, Boris J. Lurie and Paul J. Enright and Applications: 7. OptimalControl of Singularly PerturbedLinear Systems High-AccuracyTechniques,Zoran Gajid and Myo-TaegLim 8. Engineering SystemDynamics: A Unified Graph-CenteredApproach, Forbes T. Brown Additional Volumesin Preparation AdvancedProcessIdentification Najim

and Control, Ensolkonen and Kaddour

Modem Control Engineering, P. N. Paraskevopoulos Sliding ModeControl in Engineering,Wilfrid Perruquetti and JeanPierre Barbot

ENGINEERING SYSTEM OYNAMICS A Unified Graph-CenteredApproach

Forbes T. Brown Lehigh University Bethlehem, Pennsylvania

MARCELDEKKER, INC. DECKER

NEWYORK¯ BASEL

ISBN: 0-8247-0616-1 This bookis printed on acid-free paper. Headquarters MarcelDekker, Inc. 270 Madison Avenue, NewYork, NY10016 tel: 212-696-9000;fax: 212-685-4540 Eastern HemisphereDistribution Marcel Dekker AG Hutgasse 4, Postfach 812, CH-4001Basel, Switzerland tel: 41-61-261-8482;fax: 41-61-261-8896 World Wide Web http://www.dekker.com The publisher offers discounts on this book whenordered in bulk quantities. For more information, write to Special Sales/Professional Marketing at the headquarters address above. Copyright © 2001 by Marcel Dekker,Inc. All Rights Reserved. Neither this book nor any part maybe reproduced or transmitted in any form or by any means,electronic or mechanical, including photocopying,microfilming, and recording, or by any information storage and retrieval system, without permission in writing from the publisher. Currentprinting (last digit): 10987654321 PRINTED IN THE UNITED STATES OF AMERICA

This book is dedicated to Henry M. Paynter Marjorie H. Brown who in their distinct ways madeit possible.

Series

Introduction

Manytextbooks have been written on control engineering, describing new techniques for controlling systems, or new and better ways of mathematically formulating existing methods to solve the ever-increasing complex problems faced by practicing engineers. However, few of these books fully address the applications aspects of control engineering. It is the intention of this newseries to redress this situation. The series will stress applications issues, and not just the mathematics of control engineering. It will provide texts that present not only both new and well-established techniques, but also detailed examplesof the application of these methods to the solution of real-world problems. The authors will be drawn from both the academic world and the relevant applications sectors. There are already manyexciting examples of the application of control techniques in the established fields of electrical, mechanical(including aerospace), and chemical engineering. Wehave only to look around in today’s highly automated society to see the use of advanced robotics techniques in the manufacturing industries; the use of automated control and navigation systems in air and surface transport systems; the increasing use of intelhgent control systems in the manyartifacts available to the domestic consumer market; and the reliable supply of water, gas, and electrical power to the domestic consumer and to industry. However, there are currently many challenging problems that could benefit from wider exposure to the applicability of control methodologies, and the systematic systems-oriented basis inherent in the application of control techniques. This series presents books that draw on expertise from both the academic world and the applications domains, and will be useful not only as academically recommendedcourse texts but also as handbooks for practitioners in many apphcations domains. Engineering System Dynamics is another outstanding entry to Dekker’s Control Engineering series. Nell Munro

Preface Chapters 1 - 6 of this textbook/reference constitute a basic course in system dynamicsprimarily for juniors in mechanicalengineering and allied fields. It can be read at more than one level, and includes problemsof varying sophistication, so it also workswell for graduate students whohave not been exposedto its unifying methodology.By avoiding someof the topics relating to nonlinear systems, material from Chapter 7 or 8 or AppendixB maybe added to emphasize comprehensivelinear methods, automatic control, or classical vibrations. A senior-level course emphasizinglinear methodscan be centered on Chapters 7, 8 and 11, and a senior-level course emphasizing modelingcan be contered on Chapters 9, 10 and 12, where nonlinear systems predominate. These latter chapters work even better with graduate students. Details are given in the section entitled To The Instructor that follows thi~ Preface. Aninstructor’s manualgives further suggestions, as well as solutions to the problems. Modelingis the inherently pivotal and difficult element in the process of analyzing real physical systems. Therefore, the text is organized around the behavior of physical systems rather than around linear mathematics. It starts with graphical descriptions of the steady-state characteristics of engineeringdevices and the concepts of stable and unstable equilibrium. Nonlinearity is not avoided in modelingor simulation, since it impedes only hand analysis. The modelingof mechanical,electrical, and fluid systems is treated simultaneously, in order to develop a sense of analogy. Emphasisis placed on real engineering componentsand hybrid systems. This is made possible through the use of the bond graph language. The traditional modelinglanguages of linear mathematics, linear block diagrams, and medium-specificdiagrams such as circuit diagrams, free-body diagrams and fluid-power diagrams also are used, when appropriate. Linear methodsare developedfully despite their subservient role. The powerof bond graphs is widely appreciated, but they have been taught more to graduate students than undergraduates, missing the majority of potential users. The author set out explicitly to develop a textbook appropriate for undergraduates. Throughthe crucible of several years of trial and error, including manypreliminary editions, he gradually discovered the special impediments encountered by undergraduates, and howto address these di~culties. Students learn the bond graph language easily enough, but the modeling concepts that the language so concisely express are new to most undergraduates, and need to be developed gradually and carefully. These concepts exist independent of language, and are essential to an ability to modela range of physical systems without over-particularization. Textbooksthat do not employbond graphs tend to address modelingin a relatively superficial or ad hoc manner. Viewedthis way, the use of bond graphs is the simplest wayto achieve competencein the modelingof systems based on power, energy and forces. Most topics are introduced through case studies. Exampleproblemshelp get the student started, and the "guided problem" is designed to encourage vital active learning. A problemis posed, and its relevance to the student’s learning vii

viii

PREFACE

noted. A list of suggested steps or hints are then given. A solution is given a fewpages later. MATLAB is employedextensively. Nonlinear models are treated with standard numerical integrators. Linear modelsare treated with standard operatorbased routines. It is important that students understand what these routines mean, but it is no longer critical that they becomehighly proficient in hand analysis and calculations. The advent of the information age has impacted the cognitive processes of young people in both positive and negative ways. Certain social and communicative skills have been enhanced.On the other hand, critical thinking likely has declined, including its logical, visual and kinesthetic components.A decrease in the time children are given for self-directed creative play contributes to this phenomenon.Faced with a problem, students today are apt to ask where they can find a solution, not howthey can solve it themselves. Theyoften view the sample problem, rather than fundamental concepts, as the repository of knowledge. The change is partly justified by moderncommunicationnetworks, but it hinders the learning of basic modelingand analysis skills. The remedyincludes better integration of the engineering curriculum. Industry had taken a back seat to governmentin shaping engineeering departments, particularly because of Cold Warfunding. An increasing specialization resulted, including teaching assignments and the developmentof a new traditional style that could be called "prescription learning." A. rebellion emerged in the 1980s, in the form of an insistence on the primacy of "design." More recently, most remainingcourses at major universities in the relatively specialized area of mechanical vibrations have been replaced by more general courses in system dynamics,often with an introduction of control. It is ironic that the resulting plethora of textbooks in systemdynamics,while nominallyintegrative, itself tends to employthe prescription learning that is antithetical to integrative learning. This text is intended to be moreintegrative, both in its subject and in its cognitivestyle. This booknever wouldhave been written had I not had the privilege of contact with HenryPaynter, the creator of bondgraphs, largely from the late 1950s and through 1970. It also benefited from contributions by myson, Gordon,Professor KennethSawyers, Professor RamChandranat Kettering University, especially Professor Timothy Cameronwho is presently at Virginia Commonwealth University, and most especially myLehigh colleague and co-teacher Professor No DukePerreira. Forbes T. Brown ¯ .. well-connected representations let you turn ideas aroundin your mind, envision things ~rommanyperspectives ¯ .. that is what we meanby thinking. Marvin Minsky ¯ .. there is somethingessential in humanunderstandingthat is not possible to simulate by any computational means. Roger Penrose

To the Instructor

The structure of the first part of the book favors a basic course in modeling and simulation, with more emphasison nonlinear systems than is conventional. By reducing this emphasis, you can fashion an alternative course that emphasizes comprehensivelinear methods, an introduction to automatic control, or linear classical vibrations. Suggeste d sequencesof sections are given in the diagrambelow; the detailed comments that follow identify certain further possible variations. Suggestedsequencesof chapters and sections for first undergraduatecourses, four different emphases 3.6-3.7’ 4’

4.1-4.3

5

5.1(skip5.1.4),5.2.1-5.2.3,5.3,5.4.1-5.4.6

6.1-6.2.5 6.2.6-6.2.8~ 6.3

6.1-6.2.5 optional )6.2.6-6.2.8 topics 6.3

optional 8 ~ topics modelingand linear automatic simulation methods control

classical vibrations

Modelingultimately is a nondeterministic process, and consequently this text is not written in the minimalistic style of elementary textbooks in engi: neering science. Rather, someengineering color or practical asides are.inserted ix

x

TO THE INSTRUCTOR

from time-to-time. The objective is two-fold: to conveyinteresting practical information, and to developthe students’ critical ability to distinguish core essentials from peripheral details. The students need to understand at least in general terms what is expected of them, however, so you maywish to discuss this issue with them at an appropriate time. The learning objectives should depend on the maturity and level of the students, and you have considerable latitude in the sophistication of the problemsyou choose to assign as homework or discuss in class. The first six chapters represent about three credit hours of workfor a typical undergraduate. The remaining six chapters represent about five semester hours more, assumingfew topics are skipped. The later chapters have less redundancy than the earlier chapters. Graduate students typically not only can skip some of the more routine mathematical sections, but.can proceed at a faster pace, covering the entire unfamiliar material in about five semester hours. There are manyoptional topics that can be deleted to form coherent courses with fewer credits, depending on the objectives and the mathematical preparation of the students. The author believes it is important to have all the topics under one cover, however,to provide flexibility in course design and continuity between successive courses, to allow the workingengineer to extend his/her capabilities, and to demonstrate that the basic approach extends to sophisticated modeling and analysis, unlike someother methodologies. Chapter 1 introduces beginning students to system modelingand analysis, discusses contrasting learning styles and addresses the treatment of dimensions and units. Advancedstudents could skip someor all of this chapter. Chapters 2-3 represent a first pass at modelingand analysis, with about 1 credit hour of content for the beginning undergraduate. Chapters 4-6 continue with a second pass, significantly extending the domainof the modelsby allowing greater structural complexity for the same energy types. Chapter 2 starts by developing the seminal idea of steady-state source-load synthesis, which is strangely missing from manyengineering curriculums. This focuses attention on the constitutive characteristic, which is represented graphically more than analytically in order to develop the student’s basic grasp of its meaning.The chapter proceeds to develop the special cases of static couplers known(in bondgraph language) as transformers and gyrators, and applies these, particularly, to maximizethe power transmitted from a source to its load. Although there is little dynamicsin the development, the student begins to address dynamics qualitatively, including the determinationof stability for certain cases.. The first three sections of Chapter 3 completethe introduction of the fundamental modelingelements: the compliance, the inertance and the junction. Examplesare limited to models with a single junction, because this element requires the student to identify variables and the constraints between them, whichlies at the heart of the modelingprocess and most challenges the student. The .chapter proceeds to deduceand solve the linear differential equations for these models. Sections 3.6 and 3.7 introduce nonlinear elements and the use of numerical simulation particularly for nonlinear models. These sections maybe skipped if other matters are assigned a higher priority.

TO THE INSTRUCTOR

xi

The heart of the basic modelinginstruction lies in Sections 4.1 and 4.2. Here, as elsewhere, you can adjust the level of sophistication through your selection of classroom examples and homeworkproblems. Someof Section 4.3 on model equivalences and all of Section 4.4 on equilibrium can be skipped if little emphasis is intended on nonlinear models. Sections markedwith an asterisk in the table of contents and the text in this and other chapters are optional, even if nonlinear modelsare a priority. Modelsare converted to differential equations in Sections 5.1 and 5.2. The treatment of under-causal modelsthrough the use of virtual compliancesor inertances is unique to this book, and seems to be readily applied by students. Youmayskip Sections 5.2.4-5.2.6 if difficult nonlinear modelshave lowpriority. The introduction of operator methodsin Section 5.3 assumes no prior knowledge. The discussion of linearization in Sections 5.4.1-5.4.6 probably should be included in any course, but the subsequent subsections are optional. The classical solution of linear differential equations in Section 6.1 assumes no formal training. The use of the phasor methodfor frequency response in Section 6.2 is similar to that given in almost any course in automatic control, including the use of Bode plots. If you emphasize the use of MATLAB rather than hand calculations, and place a low priority on the inverse problem of systemidentification through frequencyresponse, you could skip subsections 6.2.6-6.2.8, saving considerable time. There is considerable latitude in whether, or when, to introduce Laplace transforms formally. As ordered in the text, the formal study of Laplace transforms is deferred until after their mathematicalbases in Fourier transforms and in convolution are established. The earlier use of the operator (Heaviside) notation allows transfer function expansion to be given in Section 6.3, so that problemswith zero initial conditions can be treated, including full use of the Laplace transform tables. Chapter 8 on automatic control is accessible on this basis. Should you nevertheless wish to introduce Laplace transforms early and explicitly, particularly so that initial-value problemscan be treated, you may insert Section 7.2 (less subsections 7.1.1 and 7.1.2) right after Section 6.1; the details of the partial fraction expansioncan be introduced as needed. A full presentation of linear methodsshould include Section 6.4 on convolution, but the topic is otherwise optional. The Fourier story in Section.7.1 is important should you wish to focus on vibration, but otherwise is also optional. Section 7.3 is a rather classic exposition of matrix methodsin the study of linear models. Section 7.4 is a modernexposition on the use of the loop rule to expedite the determination of transfer functions for linear modelsrepresented by bond graphs. The material also is part of the basis of somemodernbondgraph software. As indicated by the asterisks, however,Sections 7.3 and 7.4 are optional for most courses of study. At least one-half of a first coursedevotedexclusivelyto control, not including modeling, is accommodated by the text. This includes material in Chapters 5 and 6 as well as Chapter8. A briefer treatment can end after the first or second of the three sections in Chapter8, and still represent a balancedcoverageof the subject.

xii

TO THE INSTRUCTOR

Chapters 9 and 10 return to the subject of modeling, greatly extending the domainof applicability. These chapters can be studied independently of Chapters 7 and 8. Chapter 9 focuses on modelswith variable couplers, such as modulated transformers and gyrators, activated bonds, nonconservativecouplers and irreversible couplers. The irreversible coupler introduces thermal systems with heat conduction, and connections to other energy media. Section 10.1 elaborates on the meaning and methodology of lumping. The balance of Chapter 10 focuses on coupled complianceand inertance fields, treating a variety of energy transducers and providing a unique insight into the distinction between holonomic and nonholonomicconstraints. Chapter 11 introduces the seminal idea of distributed-parameter modelsand analysis, with restriction to one-dimensionalmodelsand linear media. Sections 7.1 and 7.2 are appropriate prerequisites. Emphasisis placed on wave-like behavior and on boundary-value problems. Someof the boundary conditions are nonlinear. Models are categorized according to the numberof distinct powers propagated, and organized in such a fashion as to minimizecomputational effort. Chapter 12 addresses the modeling and analysis of thermodynamicsystems, particularly with a flowing compressiblepure substance. The material in Chapters 7, 8 and 11 is not prerequisite. It is natural to adapt bQndgraphs to thermodynamicsystems, since the graphs deal inherently with energy flux and can be extended to deal with the distinctions described by entropy. Muchof the material has not been published elsewhere, and maybe of special interest to those whoare already expert at traditional bond-graph methods. Practical codes are developed for the simulation of unsteady or nonequilibrium models, including multiphase flows, without any need for iteration in most cases. Accurate analytical modelsare used for the properties of selected gasses and liquids. Codesfor evaluating the properties are given in AppendixD and, together with the simulation codes given in the chapter, can be downloadedfrpm the Lehigh University website. Chemicalkinetics is introduced in the last section of the chapter, employinga further extension of bond graphs. Appendix A is a primer on MATLAB, which together with the "help" windowshould suffice for the tasks given in the book. Youcan assign this material early in the first course, or as needed. AppendixB places special emphasison the particular advantages of the classical vibrations approach, whichis based on second-orderdifferential equations rather than the first-order or state-space e}tuations used in the rest of the text. The second-order approachis particularly useful in the treatment of high-order models without damping, or in’ which dampingis small enough to treat in an approximatemanner.~ A course that emphsizes mechanical vibrations also ought to makespecial use of manyof the vibrations examplesand problems scattered throughout Chapters 2-6. The material in Chapter 11 on distributed-parameter modelsalso is highly relevant to mechanicalvibrations. Forbes T. Brown

Contents

Table of Bond Graph Elements ..............................

inside front cover

Series Introduction by Neil Munro...........................................

v

Preface ....................................................................

vii

To the Instructor

ix

..........................................................

Chapter 1 INTRODUCTION ........................................... Example; Modeling and Engineering Science; Modeling Languages; Modeling for Control; A Word to the Wise About Learning; Treatment of Dimensions; Treatment of Units; References Chapter

2 SOURCE-LOADSYNTHESIS ............................

2.1 System Reticulation .............................................. 2.1.1 Case Study: Induction Motor as a Source; 2.1.2 Case Study: Water Sprinkler System as a Load; 2.1.3 The Source-Load Synthesis; Case Study; 2.1.4 Summary 2.2 Generalized Forces and Velocities ............................... 2.2.1 Efforts and Flows; 2.2.2 Electric Conductors; 2.2.3 Longitudinal Mechanical Motion; 2.2.4 Incompressible Fluid Flow; 2.2.5 Rotational Motion; 2.2.6 Lateral Mechanical Motion; 2.2.7 Microbonds; 2.2.8 Analogies; 2.2.9 Summary 2.3 Generalized Sources~ Sinks and Resistances. : ................. 2.3.1 Independent-Effort and Independent-Flow Sources and Sinks; 2.3.2 General Sources and Sinks; 2.3.3 Linear Resistances; 2.3.4 Nonlinear Resistances; 2.3.5 Source-Load Synthesis; 2.3.6 Power Considerations; 2.3.7 Summary 2.4 Ideal Machines: Transformers and Gyrators .................. 2.4.1 Ideal Machines; 2.4.2 Transformers; 2.4.3 Gyrators; 2.4.4 Mechanical Devices Modeled as Transformers; 2.4.5 Electric Transformers; 2.4.6 Transducers Modeled as Transformers; 2.4.7 Mechanical Devices Modeled as Gyrators; 2.4.8 Transducers Modeled as Gyrators; 2.4.9 Summary xiii

1

17 18

30

40

57

xiv

CONTENTS

2.5 Systems with Transformers and Gyrators ...................... 2.5.1 CascadedTransformers; 2.5.2 CascadedGyrators; 2.5.3 Case Study of a TransformerConnectinga Source to a Load; 2.5.4 Second Case Study of a Transformer Connecting a Source to a Load; 2.5.5 Case Study of a Gyrator Connectinga Sourceto a Load; 2.5.6 Transmission Matrices*; 2.5.7 Summary

72

Chapter 3 SIMPLEDYNAMIC MODELS ........................... 3.1 Compliance Energy Storage ..................................... 3.1.1 Linear Springs and Energy; 3.1.2 The Generalized Linear Compliance;3.1.3 Electric Circuit Compliance;3.1.4 Linear Fluid Compliance Due to Gravity; 3.1.5 Fluid Compliance Due to Compressibility; 3.1.6 Summary 3.2 Inertance Energy Storage ....................................... 3.2.1 Mass, Momentum and Kinetic Energy; 3.2.2 The Generalized Linear Inertance; 3.2.3 Common Engineering Elements Modeledby Constant Inertances; 3.2.4 Tetrahedron of State; 3.2.5 Summary 3.3 Junctions ......................................................... 3.3.1 Junction Types; 3.3.2 Mechanical Constraints Modeled by 1-Junctions; 3.3.3 Electrical Circuit Constraints Modeledby 1Junctions; 3.3.4 Fluid Circuit Constraints Modeledby 1-junctions; 3.3.5 MechanicalConstraints Modeledby 0-Junctions; 3.3.6 Electric and Fluid Circuit Constraints Modeledby 0-Junctions; 3.3.7 Simple IRC Models; 3.3.8 Summary 3.4 Causality and Differential Equations .......................... 3.4.1 Causalities of Effort and Flow Sources; 3.4.2 Junctions With Elements Having UncoupledBehavior; 3.4.3 Junctions with Elements Having CoupledBehavior; 3.4.4 Writing Diffferential Equations; 3.4.5 Summary 3.5 Solutions of Linear Differential Equations ..................... 3.5.1 The First-Order Differential Equation; 3.5.2 Responsesto MoreComplexExcitations Using Superposition; 3.5.3 The Second= Order Differential Equations; 3.5.4 Solution of the Second-Order Differential Equations; 3.5.5 Summary 3.6 Nonlinear Compliances and Inertances ........................ 3.6.1 Nonlinear Compliances; 3.6.2 Nonlinear Fluid Compliance Dueto Gravity; 3.6.3 Nonlinear Compressibility Compliance;3.6.4 Junctions with Multiple BondedCompliances; 3.6.5 Nonlinear Inertances; 3.6.6 Kinetic and Potential Energies; 3.6.7 Summary 3.7 Numerical Simulation ........................................... 3.7.1 State-Variable Differential Equations; 3.7.2 Simulation with MATLAB; 3.7.3 Integration Algorithms; 3.7.4 Second-Order Runge-Kutta; 3.7.5 Fourth-Order Runge-Kutta; 3.7.6 Summary

91 92

*- optional topic (see ToTheInstructor)

102

110

134

145

162

174

CONTENTS Chapter 4 INTERMEDIATE MODELING ......................... 4.1 Simple Circuits .................................................. 4.1.1 SimpleElectric Circuits; 4.1.2 Fluid Circuits; 4.1.3 Mechanical Circuits; 4.1.4 Summary 4.2 System Models with Ideal Machines ...........................

xv 191 191

205

4.2.1 Electric Circuits; 4.2.2 Fluid/MechanicalCircuits with Positive DisplacementMachines;4.2.3 Losses in Positive Displacement Machines*; 4.2.4 Losses with DCMotor/Generators*; 4.2.5 Case Study with Source and Load*; 4.2.6 Twoand Three-Dimensional Geometric Constraints; 4.2.7 Case Study: Pulley System; 4.2.8 Modeling Guidelines; 4.2.9 Tutorial Case Study; 4.2.10 Common Misconceptions; 4.2.11 Summary 4.3 ModelEquivalences ............................................. 242 4.3.1 Thevenin and Norton Equivalent Sources and Loads; 4.3.2 Passivity with Respect to a Point on a Characteristic*; 4.3.3 Truncation of Transformers and Gyrators Bondedto R, C, or I Elements; 4.3.4 Reduction of Two-Pair Meshes; 4.3.5 Transmission Matrix Reduction of Steady-State Models; 4.3.6 Summary 4.4 Equilibrium ....... ............................................... 258 4.4.1 Reductionof Steady-State Modelswith a Single Source; Case Study; 4.4.2 Alternative Approaches to Reducing Steady-State Models; 4.4.3 Removalof Elements for Equilibrium; 4.4.4 Case Study with a Steady-Velocity Equilibrium; 4.4.5 Case Study with Stable and Unstable Static Equilibria; 4.4.6 Case Study with LimitCycleBehavior; 4.4.7 NecessaryConditionfor Instability or LimitCycle Oscillation*; 4.4.8 Summary Chapter

5 MATHEMATICAL FORMULATION ..................

5.1 Causality and Differential Equations .......................... 5.1.1 ApplyingCausal Strokes; 5.1.2 Differential Equations for Causal Models; 5.1.3 Case Study: A Linear Circuit; 5.1.4 Case Study: Nonlinear Stick-Slip; 5.1.5 Case Study with Transformers and Gyrators; 5.1.6 ModelsReducible to Causal Form; Order of a Model; 5.1.7 Summary 5.2 Over-Causal and Under-Causal Models ....................... 5.2.1 Treatment of Over-Causal Models; Case Study; 5.2.2 Equations for Under-CausalModels; 5.2.3 Algebraic Reduction Method; Case Study; 5.2.4 Differentiation Method; Case Study*; 5.2.5 Method of Non:Zero Virtual Energy-Storages; Case Study Continued*; 5.2.6 CommercialSoftware for DAE’s;5.2.7 Case Study With Meshes*; 5.2.8 Summary

281 281

309

xvi

CONTENTS

5.3 Linear Modelsand Simulation .................................. 5.3.1 Superposition and Linearity; 5.3.2 Linearity and Differential EqUations;5.3.3 Operator Notation; 5.3.4 Transformationfrom State-Space to Scalar Form; 5.3.5 Transformation FromScalar to State-Space Form*; 5.3.6 Transformations Using MATLAB; 5.3.7 Simulation of Linear Models Using MATLAB*; 5.3.8 Summary 5.4 Linearization ..................................................... 5.4.1 Case Study with Linearization of a Resistance; 5.4.2 Linearization of a Function of OneVariable; 5.4.3 Essential Nonlinearities; 5.4.4 Linearization of a Functionof TwoVariables; 5.4.5 Linearization of a First-Order Differential Equation; 5.4.6 Linearization of State-Variable Differential Equations;5.4.7 Case Study with Three Different Types of Equilibria; 5.4.8 Case Study: Stick-Slip System with Inertia*; 5.4.9 Summary Chapter 6 ANALYSIS OF LINEARMODELS,PART 1 .........

341

358

389

6.1 Direct Solutions ’of Linear Differential Equations ............ 6.1.1 The Homogeneous Solution; 6.1.2 Singularity SystemInputs; 6.1.3 Exponential and S~nusoidal Inputs; 6.1.4 Power-LawInputs; 6.1.5 The Methodof UndeterminedCoefficients; 6.1.6 Application of Initial Conditions; 6.1.7 Solutions to ImpulseInputs; 6.1.8 Differentiation and Integration Properties; 6.1.9 Step and ImpulseResponses Using MATLAB; 6.1.10 Summary 6.2 Sinusoidal FrequencyResponse................................. 6.2.1 The Phasor Method;6.2.2 Bodeplots; 6.2.3 ModelsWithout Damping;6.2.4 Models Comprising a Single Pole or Zero; 6.2.5 ModelsComprisinga Pair of ComplexPoles or Zeros; 6.2.6 Factorization of Higher-OrderModels;6.2.7 ,Bode Plots for Highei-Order Models; 6.2.8 The Pure Delay Operator*; 6.2.9 Summary. 6.3 Transfer Function Expansion................................... 6.3.1 Impulse Responses; 6.3.2 Step and Other Transient Responses; 6.3.3 Determinationof Partial Fraction Expansions;6.3.4 Summary 6.4 Convolution*..................................................... 6.4.1 Decomposing Signals into a Sumof Steps; 6.4.2 Discrete Convolution; 6.4.3 Discrete Convolution by MATLAB; 6.4.4 Convolution Integrals; 6.4.5 Summary

389

Chapter 7 ANALYSIS OF LINEAR MODELS,PART 2 ......... 7.1 FourierAnalysis .................................................. 7.1.1 Fourier Series; 7.1.2 Responseof a LinearSystemto a Periodic Excitation; 7.1.3 Fourier Transform;7.1.4 Digital Spectral Analysis*; 7.1.5 Fourier Analysis Using MATLAB*; 7.1.6 Summary

485 485

408

459

470

xvii

CONTENTS 7.2 The Laplace Transform ..........

’ ...............................

507

7.2.1 Development from the Fourier Transform*; 7.2.2 Development from the Convolution Integral*; 7.2.3 Definition and Inverse; 7.2.4 The Derivative Relations; 7.2.5 Singularity Functions and Discontinuities; 7.2.6 Other Key Relations; 7.2.7 Finding Laplace Transforms of Output Variables; 7.2.8 Partial Fraction Expansions; 7.2.9 Initial and Final Value Theorems; 7.2.10 Summary 7.3 Matrix Representation

of Dynamic Behavior*

......

: .........

526

7.3.1 The Matrix Exponential; 7.3.2 Response to a Linearly Varying Excitation; 7.3.3 Eigenvalues, Eigenvectors and Modes; 7.3.4 Case Study: Three Fluid Tanks; 7.3.5 Case Study with Complex Roots; 7.3.6 Modified Method for Complex Eigenvalues*; 7.3.7 Case Study: Vehicle Dynamics; 7.3.8 Application of MATLAB; 7.3.9 Response to Exponential and Frequency Excitations; 7.3.10 Representation in the s-Plane; 7.3.11 Summary 7.4 The Loop Rule* .................................................

561

7.4.1 Signal Flow Graphs; 7.4.2 The Loop Rule for Signal Flow Graphs; 7.4.3 Converting Bond Graphs to Signal Flow Graphs; 7.4.4 Direct Application of the Loop Rule to Bond Graphs Without Meshes; 7.4.5 Bond Graphs with Meshes; 7.4.6 Determination of State Differential Equations; 7.4.7 Summary Chapter

8 INTRODUCTION

TO AUTOMATIC

CONTROL ....

583

8.1 Open and Closed-Loop Control ................................ 8.1.1 Example Plant; 8.1.2 0pen-Loop and Optimal Control; 8.113 Feedback Control; 8.1.4 Response to Disturbances; 8.1.5 Root Locus Basics; 8.1.6 Use of MATLAB; 8.1.7 Criteria for Stability; 8.1.8 Summary

583

8.2 Dynamic Compensation .........................................

605

8.2.1 Proportional-Plus-Integral Control; 8.2.2 Proportional-PlusDerivative Control; 8.2.3 Proportional-Plus-Integral-Plus-Derivative Control; 8.2.4 Phase Lead Controllers; 8.2.5 Phase Lag Controllers; 8.2.6 Phase Lead-Lag Controllers; 8.2.7 Digital Control Systems; 8.2.8 Summary 8.3 Frequency Response Methods .................................. 8.3.1 Polar or Nyquist Frequency Response Plots; 8.3.2 The Nyquist Stability Criterion; 8.3.3 Measures of Relative Stability; 8.3.4 Nichols Charts; 8.3.5 Dynamic Compensation Using Nichols Charts; 8.3.6 Approximate Correction for Digital Sampling; 8.3.7 Summary

629

xviii

CONTENTS

Chapter.

9 MODELSWITH STATIC COUPLERS ................

657

9.1 Modulated Transformers ........................................ 9.1.1 Remotely Modulated Transformers; 9.1.2 Locally Modulated Transformers; 9.1.3 Increase in the Order of a Model Due to Modulation; 9.1.4 Dependent Inertance with a Locally Modulated Transformer; 9.1.5 Inertance Dependent on Local Displacement; Case Study; 9.1.6 Summary 9.2 Activated Bonds .................................................

657

9.2.1 Definition

682

and Application; 9.2.2 Causality; 9.2.3 Summary

9.3 Nonconservative Couplers ...................................... 9.3.1 Causal Relations; 9.3.2 Equilibrium; 9.3.3 Dimensional Analysis; 9.3.4 DynamicSimulation; 9.3.5 Linear Couplers; 9.3.6 Summary 9.4 Irreversible Couplers and Thermal Systems ..................

697

718

9.4.1 Effort and Flow Variables; 9.4.2 Heat Conduction; 9.4.3 The Irreversibility Coupler; 9.4.4 Application to Diction; 9.4.5 Use of 0- and 1-Junctions; 9.4.6 Thermal Compliance; 9.4.7 Pseudo Bond Graphs for Heat Conduction; 9.4.8 Summary Chapter

10 ENERGYSTORAGEFIELDS .........................

735

10.1 Field Lumping ....... i .......................................... 10.1.1 Scalar Fields; 10.1.2 Rigid-Body Vector Fields; 10.1.3 The Role of Approximations; 10.1.4 Nodic Fields; 10.1.5 Planar Vector Nodic Fields; 10.1.6 Estimating Upper and Lower Bounds for Field Transmittances*; 10.1.7 Three-Dimensional Vector Nodic Fields; 10.1.8 End Corrections for the Inertance of Channels; 10.1.9 Multiport Vector Nodic Fields; 10.1.10 Summary

735

10.2 Discrete Compliance Fields ................................... 10.2.1 Generic Relationships; 10.2.2 Examples with Geometrically Varied Capacitance; 10.2.3 Thermodynamic Coupling Between Mechanical and Thermal Energies; 10.2.4 Summary 10.3 Displacement Modulated Inertances ......................... 10.3.1 Inertances Dependent on Holonomic Constraints; 10.3.2 Inertances Dependent on Nonholonomic Constraints; 10.3.3 Summary 10.4 Linear Multiport Fields ........................................

752

10.4.1 Mutual Inertances and Resistances: the Real Electrical Transformer; 10.4.2 The Rigid Inertive Floating Link; 10.4.3 MultiCoil Transformer with Perfect Flux Linkage; 10.4.4 The Piezoelectric Transducer; 10.4.5 The Thermoelastic Rod; 10.4.6 The Piezomagnetic (Magnetostrictive) Transducer; 10.4.7 Generalized Linear Media; 10.4.8 Reticulation of General Multiport Fields; 10.4.9 Further Equivalences; 10.4.10 Summary

760

771

CONTENTS

~x

Chapter 11 INTRODUCTION TO DISTRIBUTED-PARAMETER MODELS ............................................................... 11.1 Wave Models with Simple Boundary Conditions

.................

795 796

11.1.1 Comparison of Lumped and Distributed Models; 11.1.2 The Pure Bilateral-Wave-Delay Model; 11.1.3 Analysis of the Pure Bilateral-Wave-Delay Model; 11.1.4 Fixed and F~ee Boundary Conditions; 11.1.5 Fourier Analysis with Fixed or F~ee Boundary Conditions; 11.1.6 The Hodograph Plane; 11.6.7 Summary 817

11.2 One-Dimensional Models ...................................... 11.2.1 General Formulation; 11.2.2 One-Power Models; 11.2.3 Symmetric One-Power Models; 11.2.4 Multiple-Power Models; 11.2.5 Summary

825

11.3 WavePropagation .............................................. 11.3.1 Simplest Model: Pure Transport; 11.3.2 First Modification: Thermal Leakage to a Constant-Temperature Environment; 11.3.3 Second Modification; Thermal Compliance in the Walls; 11.3.4 Dispersion and Absorption; 11.3.5 Group Velocity*; 11.3.6 Summary 11.4 One-Power Symmetric Models ................................

837

11.4.1 WaveBehavior; 11.4.2 Energy Velocity in Conservative Media*; 11.4.3 Boundary-Value Problems; Transmission Matrices; 11.4.4 Exponentially Tapered Systems*; 11.4.5 Summary 11.5 Multiple-Power Models .........................................

858

11.5.1 Symmetric and Anti-Symmetric Variables and Models; 11.5.2 Case Study of a Degenerate System: A Counter-flow Heat Exchanger; 11.5.3 Case Study of a Symmetric Model: The Bernoulli-Euler Beam; 11.5.4 Summary Chapter 12.1

12 THERMODYNAMIC SYSTEMS ....................... The Convection

Bond and Compressible

867

Flow .............

867

12.1.1 Flow Through a Port; 12.1.2 The Convection Bond; 12.1.3 The RS Element for Fluid Flow; 12.1.4 Summary 12.2 Heat Interaction and Junctions ............................... 12.2.1 The Reversible Heat Interaction Element; 12.2.2 The General Heat Interaction Element; 12.2.3 The HRS Macro Element; 12.2.4 The 0-Junction for Convection Bonds; 12.2.5 Merging Streams: the 0S Junction; 12.2.6 The 1-junction and IS-Junction with Convection; 12.2.7 Summary

876

12.3 Case Study with Quasi-Steady

891

Flow .........................

12.3.1 Bond Graph Model; 12.3.2 Irreversibilities; tation of Results

12.3.3 Compu-

xx

CONTENTS

12.4 ThermodynamicCompliance .................................. 12.4.1 Application of Simple ThermalCompliance;12.4.2 Causality; 12.4.3 General ThermodynamicCompliance; 12.4.4 The CS Macro Element; 12.4.5 Computations for the Ideal Gas; 12.4.6 Case Study: A Piston-Cylinder Compressor; 12.4.7 Treatment of a Quasi-Steady-State Fluid Machine; 12.4.8 Fluid Inertance with Compressible Fluid Flow; 12.4.9 Pseudo Bond Graphs for Compressible Thermofluid Systems; 12.4.10 Summary 12.5 Evaluation of ThermodynamicProperties ................... 12.5.1 The Most Commonly Available Analytical Form for State Properties; 12.5.2 HelmholtzAnalytical Formfor State Properties; 12.5.3 Application to Gases; 12.5.4 Application to Refrigerants; 12.5.5 Application to Water; 12.5.6 Case Study: a Refrigeration Cycle; 12.5.7 Application to the Liquid Region; 12.5.8 Considerations of Reversing Flows; 12.5.9. Summary 12.6 Systems with Chemical Reaction* ............................ 12.6.1 Energy of a Pure Substance; 12.6.2 Energy of Multiple Species; 12.6.3 Stoichoimetric Coefficients and Reaction Forces; 12.6.4 ChemicalEquilibrium; 12.6.5 Reaction Rates; 12.6.6 Models of Reactions Without Mass Flows; 12.6.7 Modelsof Reactions With Mass Flows; 12.6.8 Summary

897

APPENDIX A INTRODUCTIONTO MATLAB.................. Scalar Calculations; Variables; ComplexNumbers;Arrays and Matrices; Evaluating and Plotting Functions; Fitting Curves to Data; Control Flow Commands; Script Files; Data Files; Function Files; CommunicationBetween Files

953

APPENDIXB CLASSICALVIBRATIONS........................ B.1 Models With Two Degrees of Freedom ....................... B.I.1 Normal ModeVibration; B.1.2.Forced HarmonicMotion B.2 Higher-Order Models ........................................... B.2.1 Modal Motions; B.2.2 The Initial Value Problem; B.2.3 Forced Response; B.2.4 Modal Damping; B.2.5 Example Using MATLAB; B.2.6 Mode Reduction

965 965

APPENDIX C LAPLACETRANSFORM PAIRS .................

979

919

943

968

APPENDIX D THERMODYNAMIC DATA AND COMPUTER CODE ........ . ......................................................... 983 D.1 Programs and Data for Air and Components; D.2 Programs and Data for Refriferant R-12; D.3 Data for Refrigerant R-134A; D.4 Data for Refrigerant R-22; D.5 Programs and Data for Water Index ....................................................................

997

ENGINEERING SYSTEM DYNAMICS

Chapter 1

Introduction This Chapter is directed primarily at undergraduates taking a first course in system dynamics and related areas. It addresses three questions: Whyare you taking this course? What style of study works best? Howcan you reliably treat dimensions and units? Advanced students may proceed directly to Chapter 2. Your job title is Project Engineeer, the quintessential assignment for a mechanical engineer or an engineer in an allied field such as aeronautics or agriculture. Your responsibility is to shepherd a new product idea from the conception stage to the production stage. It may be a component such as a pump or an automotive strut, it may be a machine tool, or it may be an entire production machine. You might have a small team under your direction, or you might work alone, but in any case you have access to technicians, machinists and production specialists. You are directed in turn by a managementthat has a limited understanding of the technical possibilities, so you also play a role in choosing the product ideas to pursue and the marketing strategy. Part of the success of the company rests on your shoulders. Howdo you do your job? You discover a need, or one is presented to you. Next, you interpret this need in terms of specifications, that is a list of the specific functions the product should exhibit, including quantitative performance objectives. Then, as a good designer, you conceive possible solutions, preferably several of them. The design process you are engaged in is summarized, albeit in a very cryptic fashion, in Fig. 1.1. At this point it is unclear which of the solutions are feasible, if any, and what values of their various parameters would be necessary or desirable. If you follow the experimental option shown in the diagram, you construct and test physical models of the proposed solutions. If instead you follow the analytical option, you model the designs abstractly, and likely use a computer to help determine the behavior. In either case, you display the results of the experiment or the analysis in an intelligible manner, so that you can compare them with the desired result, namely the original specifications. As designer, you must then decide what to do next; surely the discrepancies between the desired and the actual behavior require further iterative passes around the loop. At some point you decide either that .the design has succeeded or failed, and

2

CHAPTER 1. need

~1 problem -1 lbrmulation ]

I specification ~

~

INTRODUCTION

conceptualization solution ]

experimental option t~s~ data reduction collection, L I physical l- I modeling analytical option analysis and I~ I conceptual L._ computation1I modeling["Figure 1.1: The engineering design process

you go on to another design concept or another project. A successful conclusion is rewarding to you and your company, even if it is the abandonment of a poor idea that has been intriguing your boss. To be successful it must not take too long to accomplish, however, even if it is a revolutionary advance in its field; time costs money, and your competitors are not standing still. You will spend more time constructing and executing analytical or computer models of your designs than you will conceiving them. Efficient design implies quick analysis. Howcan you carry out the analysis process rapidly as well as effectively? The answer is that you learn how to construct an abstract model of nearly any system you may design, and how to predict performance with this ~nodel far quicker and cheaper than you could get a working physical model built and tested. Further, you should learn how to make the abstract model appropriately simple. An unnecessarily complex model, even if constructed readily, produces unnessesarily complex results, and can be expensive to execute computationally. You should keep in mind the decision toward which the model is created. Whenyou first go around the design loop, the intended decision likely is simple: whether or not to pursue the particular concept further. A crude model would be appropriate. Later, as a final design is approached, details of the system becomeof interest, perhaps requiring a rather fancy model. The purpose of this text is to empoweryou to analyze real engineering systems, whether they exist already or are in the conceptual stage. Emphasis is placed on the process of modeling, that is, the construction of conceptual abstractions with well-defined meanings that behave in an appropriately approximate way to the real thing. Considerable attention also is needed on the analytical methods available to reveal this behavior. It is the modeling that is inherently difficult, however, for it is not a "deterministic process. It involves intent, recognition and judgement. It is an art based on science, physical insight and experience. Together with the genius required to conceive new products, it is the "existential pleasure," the spice, that makes engineering an intensely human activity. This text deals largely with systems that can be characterized by how they

1.1.

3

EXAMPLE

employenergy or power, either in large or small quantities. A wide variety of mechanical, fluid, electrical, thermodynamicand hybrid systems are included. Someof the systems are in a time-invariant state, particularly in Chapter 2. Most of them are time-varying, justifying the title system dynamics. Systems that do not operate because of energy balances, such as the informational considerations of computersystems, economicor political systems, traffic systems, etc. are not addressed.

1.1

Example

As an exampleof the process, imaginethat you are a project engineer developing the basic concept of a vehicle poweredby batteries and a DCelectric motor. Youquickly discover that a motor with a maximum powerof 30kWcan drive the vehicle fast enough, at a steady 45 m/s (about 100 mph). It cannot accelerate the vehicle from rest to 27 m/s (about 60 mph)in under 18 seconds, however, even with a perfectly variable transmission. To reduce this time, you consider the electrical-hydraulic drive represented diagramatically as: e RI t fixed field

"i~¢, shaft

~ accumulat°r/""x

flow, Qp flow, Qm[ motor 1 velocity,

v ~

The motor has a fixed field and a 12 volt armature voltage, e. Its shaft rotates at a variable angular velocity $, driving a hydraulic pump,which has inertia and resistance. Fluid is stored in a hydraulic accumulator(a tank containing a compressedgas and hydraulic oil separated by a bladder or piston) at a high pressure, P. The fluid drives the vehicle at a variable speed v through the use of a hydraulic motor with a volumetric displacement (volume of fluid passed per revolution of the shaft) that is varied by the driver. The driver then can release the energy stored in the accumulatorto achieve rapid acceleration. Further, he can also makethe displacementnegative, braking the vehicle by having the hydraulic motor pumpfluid back into the accumulator. This feature, called regenerative braking, recovers the energy of motionrather than dissipating it, unlike conventionalbrakes. The vehicle has mass, frictional drag and an additional hill-climbing load. Youwouldlike to have answers to such questions as: Howlarge (and heavy) a hydraulic accumulatoris necessary to accelerate the vehicle to 27 m/s in less than 6.5 seconds, without requiring a larger motor? Whatpressures should be used?- What maximum volumetric displacements (sizes) should the pumpand motor have to accomplish this? Howshould the volumetric displacement of the pumpbe automatically controlled? Whatwouldthe energy efficiency of various schedules of speed be? Ultimately, is this concept worth pursuing? As a first step in answeringthese questions, you makea simplified modelof

4

CHAPTER 1.

INTRODUCTION

the system in the form of a bond graph:

R,

R.

RL

A bondgraph is a minimalist representation of the implied model. Its structure also directs in routine fashion the assemblyof the corresponding differential equations. Someof the componentscan be described by parameters (constants): Ri, G, Ip, Rp, and IL. Others require more complexconstitutive relations: C and RL. The values of TM and Tp are varied to control the flow of energy to or from the accumulator. The modelleads directly to a computersimulation for the dynamicperformanceof the systemfor whateversituation is of interest. Since each run takes only a couple seconds, you can simulate hundredsof different situations, if you like, adjusting the various parametersuntil you are satisfied with the result. Here are typical results after someof these changes have been made: 5O [

40~ 30

f motor speed ~/10,

rad/s

/SP~P-’~’~o=7MPa speedv, m/s

2O 10 0

0

50

100 150 time, seconds A speed of 27 m/s is indeed achieved in 6.5 seconds, by setting the volumetric displacement of the motor at a particular value. The displacement is then changedto a certain larger value, resulting in the samefinal steady speed. The pressure and energy in the accumulatorfalls during the acceleration, but recovers afterward, due partly to the automatic control of the pumpdisplacement that can be designed partly through analysis of the bond graph. Thus a subsequent burst of acceleration also wouldbe robust. This control also affects

1.2.

MODELING AND ENGINEERING

5

SCIENCE

the speed of the electric motor. The motor is 507o efficient when its power is maximized (at 250 rad/s) and 100%efficient when its power approaches zero (at 500 rad/s). The control successfully prevents inefficient speeds (below tad/s) and achieves the maximumpossible 947o efficiency at the steady speed. In another simulation, the displacement of the motor after 6.5 seconds is set to the value which is found to produce the maximumpossible steady speed, which is the desired 45 m/s: 5O speed v, m~ 40

30 20

speed ~/10, rad/s t~~~pressur

e

P/IOPo Po =7 MPa

oo

150 time, seconds

The motor speed becomes the 250 rad/s for which it produces the necessary maximumpower, at 507o efficiency. The pressure drops to its minimum, which is the charging pressure of the accumulator (the pressure when it is virtually empty of hydraulic oil). As project engineer, you are now in a position to render some judgments, which of course also involve factors not discussed here, such as cost, weight and reliability. As a student, you should be able to address some simpler problems in this manner after studying Chapters two and three, and address some comparable problem as a project after completing Chapters four and five. Most of the problems you will address in the course will be considerably shorter, however.

1.2

Modeling and Engineering

Science

Real engineering systems, like the example above, do not closely resemble the pristine systems presented in introductory textbooks in mechanics, thermodynamics, fluid mechanics, etc. These textbook systems are in fact highly abstracted mode~s, presented in the form of some language, usuMly largely pictoral, that convey a unique meaning. You hopefully absorbed some of the modeling idea, but what you really practiced was the application of physical principles to translate the given model as expressed in one language into another language,

6

CHAPTER 1.

INTRODUCTION

which usually was mathematical. The new expression was intended to reveal certain features or behaviors of the model which at first were hidden. This ap-. proach helped you develop a working understanding of physical principles that must be acquired before you can advance to considering real systems. But the gap between real engineering systems and approximate abstract models must be bridged. Most non-structural engineering in the nineteenth century evaluated design concepts largely through experiment. Slowly it became apparent that the analytical option can substitute, saving time and money and sometimes reducing danger. The science of thermodynamics, for example, was born of the need to understand steam engines better. Structural science, so critical to safety, became particularly well developed. The analysis of dynamic systems lagged, because of the difficulty in solving the describing differential equations. Nevertheless, the goad of increasingly complex military hardware in World War II spawned a formal systern science, based on differential equation models. Linear equations were emphasized, because only they could be solved routinely, albeit often with considerable tedium. Alternatively, components and systems were designed explicitly to act nearly linearly, sometimes just so that their behavior could be predicted. The advent of the computer removed the computational roadblock. Today an engineer can construct a computer model of nearly any proposed system, and predict its performance far quicker and cheaper than a prototype could be built and tested. The engineer then can adjust its parameters until the result is satisfactory, or until it appears that some different approach to the problem should be sought. Not all engineers today are competent to do this with muchgenerality, but those who can are in a different league from those who can’t. Experiment need be employed only sparingly, largely either to understand phenomenawhich defy analysis (often using relatively simple apparatus) or verify a final design. The Boeing company, for example, now designs and builds its most complex aircraft without a single prototype being constructed and tested (although key component~are tested separately).

1.3

Modeling

Languages

A limited number of generic building blocks will be identified from which you will construct models. The language chosen to represent these elements should be general enough to encompass virtually all systems which centrally involve the management of pow.er and energy, and narrow enough to impose the conservation of energy as the default option. Such a language allows you to take full advantage of the analogies that exist between seemingly different engineering systems. The language also should be chosen for the ease with which its objects can be reduced to mathematical form. Perhaps the .most serious hazard in modeling is the threat of making an outright mistake. Everyone is an occasional victim; the expert keeps a commonsense vigil, routinely comparing the structure of the model or its consequences

1.4.

MODELING FOR CONTROL

7

with expectations. The modeling language or languages you use to represent models also greatly affects your propensity to make and discover mistakes. Mathematics and derivative computercode are the final language for most models, but these languages tend to obscure mistakes. Mathematics jokingly has been called "the universal solvent," since everything dissolves in it, becoming invisible. This book employs a graph-based approach to modeling. Graphical languages comunicate better to your senses than mathematics. Considerable use is made of sketches, diagrams and plots. Where practicable, the system behavior implicit in a graphically expressed model is also determined graphically. The models for most dynamic systems are converted to differential equations, which are then solved. This conversion is made as automatic as. possible. Since the subject is systems that control power and energy, a principal modeling language is desired that automatically conserves energy, unless it is directed explicitly to do otherwise. It is helpful also if the language identifies the critical geometrical and analagous constraints of such systems. Finally, it is desired that the language be both graphical and precise in its meaning, and allow routine conversion to equations or equivalent computer code. Such a language should be optimally efficient, and minimize mistakes. Several languages apply to specific types of systems. The circuit diagram is prototypical. Similar languages apply to classes of mechanical systems and to hydraulic systems. They fail to address interdisciplinary systems, however, which today means most systems. Actually, the electric circuit diagram often is applied by analogy to represent non-electrical phenomena,such as heat transfer or acoustics. It lacks adequate generality to represent all systems, however, and is awkwardeven when it is usable. A more generic language is sought. Such a language exists, fortunately. Bond graphs were devised in their basic form in 1959 by Henry M. Paynter at MIT. Bond graphs are used by a significant fraction of mechanical and electrical engineers, and even by some life scientists. The language has been extended to model systems with heat fluxes, compressible fluid flow and chemical reaction. It has been promulgated by thousands of professional papers and, since 1991, by the biennial International Conference on Bond Graph Modeling (ICBGM). Although the advantages of a knowledge of the bond graph language are well known, it has not been taught widely in undergraduate core courses, and is not to be dabbled in lightly. By making its mastery a manageable and natural task, this book aims to increase its popularity for the benefit of engineering.

1.4

Modeling for Control

A model or its direct consequence is itself built into a control system, in the form of the controller with its control algorithms or alternatively its hardware configuration. Regardless of whether you study the introduction to control in Chapter 8, you should knowthat the control engineer typically succeeds or fails according to the workability of the model or models that he devises to represent

C H A P T E R 1. INTRODUCTION

8

the behavior of the system to be controlled. Control systems combine a muscle part, or “plant,” with a brain part. The brain part is directed to act by some command signal, which often is a measure of the desired performance of some plant variable such as a speed, an angle or a temperature. The brain usually measures this “output” variable, and compares it t o the command signal. The difference between these signals is an “error” which is used as the input to the controller part of the brain. The output of the controller then is sent to the plant for action. This feedback scheme also potentially helps correct for unwanted effects due to external disturbances on the system. The controller must be tailored to the system it controls. Its design depends on a model of the system created by the engineer. Its performance hinges on the accuracy of the model. Its cost depends on the complexity of the model. Typically, the most practicable design results from a compromise between accuracy and complexity.

1.5

A Word to the Wise About Learning

Your primary challenge as a student is to learn how t o approach problems that you, and maybe everyone else also, have never seen before. The detailed minimalistic procedures that are the focus of many elementary courses address only specially conditioned and presented problems that could be solved by a clerk with a fancy computer program. They represent only a first step. The celebrated mathematician Roger Penrose’ claims that human understanding and even consciousness itself are non-computational, that is, cannot be simulated by a Turing machine (digital computer). This book complements its direct presentation of topics with four different types of problems: case studies, numbered and highlighted examples, guided problems and assigned problems. Case studies typically introduce and illustrate new ideas, and should be viewed as integral parts of the text. The numbered examples, on the other hand, are sample problems that merely demonstrate application of the basic concepts and procedures. They logically may be skipped. A person engaged in the problem-specific or training mode of learning, suggested in part (a) of Fig. 1.2, develops problem-solving prescriptions by examining the solutions t o sample problems. He or she views these detailed and often voluminous patterns, rather than the relative simple underlying concepts, as the information needed for success. Real-world engineering, however, covers a thousand times more situations than possibly could be addressed by sample problems. The breadth of the problems in this text may help convince you of the futility of reliance on superficial procedures, despite your earlier experience. Excessively detailed focus on solved problems would distract from your developing the essential “physical thinking,” the picturing in your mind’s eye of what really is going on. This insight, together with a firm grasp of the grand conservation principles, comprise the essential skills. Without them, recall of Roger Penrose, Shadows of the Mind, Oxford University Press, 1994.

1.5. A W O R D T O T H E WISE ABOUT LEARNING

sample problem

sample problem

sample problem

sample problem

9

(b) problem-general or education mode (a) problem-specific or training mode Figure 1.2: Contrasting modes of learning all existing engineering formulas is useless. The problem-general learning mode, suggested in part (b) of the figure, stores the information needed for success in an integrated network of basic concepts and procedures. Case studies, solved problems, guided problems and assigned problems are not definitive patterns t o be imitated, but are suggestive illustrations that help you develop your integrated network. Enriching physical demonstrations also can help greatly. There is no substitute for direct observation and contemplation of basic concepts. A small set of concepts casts myriad shadows in printed solutions t o relevant problems. It is inefficient to infer reality solely from its shadows. You must, rather, recognize what basic concepts apply t o your real-life problems. In this information age there are many ways you can find the details. The variety of situations addressable through the learning mode, on the other hand, grows exponentially with the number of the core concepts and procedures understood. Fbrther, the structure becomes an increasingly redundant network of ideas. If some element is forgotten, a detour around the flaw usually can be found, and the flaw itself often can be repaired without recourse t o outside information. It may seem that the amount of specific information you need t o recall actually decreases with experience. Engineering becomes heady stuff. Is your pleasure growing with each successive course? To learn the true meaning of the core concepts and procedures you must actively solve relevant problems without reliance on highly specific patterns. The “guided problems” in this text start with a problem statement, and proceed with a list of “suggested steps” that is less than a unique prescription for a solution but is intended t o relate the solution t o the core concepts and procedures. You may then compare your solution with one given at the end of the section. Each guided problem exists for a purpose, which is briefly suggested at the outset.

10

CHAPTER

1.

INTRODUCTION

Figure 1.3: Journal bearing for case study (Some guided problems are indicated as advanced, or optional.) You defeat their primary purpose by looking at the given solution before attempting to create your own. Even when you fail to produce a solution, you will usually benefit from the attempt.

1.6

Treatment of Dimensions

Engineers and mathematicians write algebraic equation~, in which numbers are represented by symbols, usually letters of the Romanor Greek alphabets. These equations are solved for unknowns, or dependent variables, which are written on the left sides of the equations, as functions of the knowns, or independent variables, which are grouped on the right sides of the equations. Substitution of the numbers for the symbols then produce a second type of equation which gives numerical answers. For only the engineers, however, do the symbols carry added meaning: physical dimensions. These dimensions, which themselves can be represented by words or symbols, comprise the terms of a third type of equation. The experienced engineer always keeps the dimensional equations in mind, for they give him or her a tremendous advantage over the mathematician in the use of equations. A case study illustrates this advantage. Consider that you are asked to find the torque required to rotate a lightly-loaded shaft in a well-lubricated journal bearing at the angular velocity u~. The shaft has diameter d, rotates concentrically in a sleeve or cylinder of length L and diameter d + 2e, so that e << d is the radial clearance between the shaft and the sleeve, as pictured in Fig. 1.3.. This space is filled with oil of absolute viscosity #. (This is a classical problem, which you may address in a first course in fluid mechanics, if you have not done so already.) Perhaps you deduce the following relation to approximate the torque or moment, M0: (1.1) Mo - ~r#d2 Lu~ 2e No simple model can be exact, but is this one correct, within the several assumptions upon which it is based? You know that most potential mistakes violate dimensional consistency, so you proceed, before substituting numbers to get a specific value for the torque, to substitute the dimensions of each of the terms. You know the following dimensions:

1.6. TREATMENT OF DIMENSIONS

11

~r has no dimensions L, d, e have dimensionof length a~ has dimensionsof l/time (giving units such as rad/second) # has dimensionsof forcex time/length squared Therefore, you substitute to get force x time length ~- x length 1 [M0] - ~ length length time’

(1.2)

where the square brackets around Mdesignate "dimension of." Cancellations give (1.3) [M0] = forcelength x time length 3 = force. 3 xx time You know, however, that the torque or momentM0has the dimensions of force x length, not force. Somethingis wrong! The equation for Mo~nust be changedto increase its dimensionby one length. Youtherefore re-visit your derivation. Or, perhaps, you reason as follows: the shear stress on the surface of the shaft is proportional to the velocity of its surface, whichin turn is proportional to the diameterd. The shear force also is proportional to the wetted surface area of the shaft, whichalso is proportional to d. Finally, the torque equals the shear force times the radius, which again is proportional to d. Therefore, the term d2 in the numeratorof equation (1.1) should be d3. Withthis correction, equation (1.3) gives force x length, which you knowis the correct dimensionsfor torque. Youthen return to your derivation, and find that you forgot to multiply the shear force by the radius, which is d/2. The proper torque therefore is Mo- 7r pd3 Lw (1.4) 4e Youare finally ready to substitute the numericalvalues of the parameters. This scenario is typical of a goodengineer. Goodengineers are humanand make mistakes, but have gotten into the habit of catching them by checking dimensions. In order to deal with dimensionseffectively, it is necessary to work with a minimal consistent or primary set thereof. The most commonand recommended primary set of dimensions in use by engineers today for treating mechanical, electric circuit and thermal systems is force (F), length (L),time (t), electric charge(Q) and temperature(0). Other dimensionsare called derived, since they can be expressed in terms of the consistent set. The namesof someof these are listed in the first columnof Table 1.1, and symbolstherefore are listed in the second column. (The other columnsare discussed later.) A second primary set of dimensions replaces the role of force with mass, M. Specifically, wheneverforce F appears in Table 1.1, this set of dimensions substitues M.L/t"-, which follows from Newton’s second law: F = rna. This so-called "absolute" consistent set of dimensionssometimesis featured in ele-

12

CHAPTER Table 1.1 Dimensions

mass (m) angle velocity angular velocity acceleration angular accel. energy power pressure entropy electric current electric potential

F L t 0

newton, N meter, m second, s degrees Kelvin, K coulombs, C

Q

INTRODUCTION

and Units

name I symbol I SI units primary set of consistent dimensions and units: force length time temperature electric charge derived dimension

1.

I Am. Std.

units

pound force, lb foot, ft second, s deg. Rankine, R

and units: F.t2/L Lit

1/t ~ lit F.L F.L/t ")F/L F.L/O

Q/t F.L/Q

N.s2/m = kg (radian) m/s 1/s (rad/s) ~ 1/s Nm= J (joules) N-m/s= W (watts) N/m’-> -- Pa (pascals) N.m/K- = J/K C/s= A (amperes) N-m/C= V (volts)

lb.se/ft= (radian)

slugs

ft/s 1/s (rad/s) ~ ft/s 21/t

ft.lb ft.lb/s "~ lb/ft ft.lb/R

mentary books on dynamics. Nevertheless, it is relatively awkward for most engineering systems, and in practice is rarely used. For example, engineers do not often think of pressure as having the dimensions M/t2L, or energy as having the dimensions ~-. M. L2 /t Beginners often try to use a mixed system of diinensions, with both mass and force. This attempt prevents critical cancellations from being made. It abandons much of the benefit derived from using dimensions.

1.7

Treatment

of Units

Most implementation of dimensions is carried out through the use of a specific system of units. Three systems are most common, and are addressed here: SI units, American Standard or British Gravitational units, and the old English Engineering units. The first two can be treated as either force-based or mass-based. Old English Engineering units are by tradition mixed force- andmass-based, and as a result are not listed in Table 1 but are treated separately below. Fluid mechanics texts tend to use slugs, which if not converted represent a mass-based American Standard system; dynamics texts tend to use the recommendedforce-based American Standard system; thermodynamics texts tend to use pounds mass as part of a mixed old English Engineering system. These texts often fail to adequately delineate force-based and mass-based use of the SI system, also. A generation of engineers have been confused, particularly when interdisciplinary systems are involved.

1.7.

TREATMENT

13

OF UNITS

The confusion regards the treatment of mass. As advocated above, the role of mass should be replaced by force through the use of Newton’s second law. Force units do not necessarily equal mass units times acceleration units, however, unlike the fundamental dimensions. Rather, force units are, in general, merely proportional to the units of mass times acceleration. In general, m F = --gc a, (1.5) where gc is a constant of proportionality that may or may not have the numerical value 1.0000. The standard weight W is therefore related to the mass m by m W = -- g, gc

m W -- = --. gc g

or

(1.6)

Whenever you face a mass m in an equation, you may substitute either m/gc or Wig (and p/g¢ or 7/g for p, where 7 is the weight density), except in the cases of the more careful thermodynamics texts in which g~ appears already. The result is twofold: first, the units of m~s will be converted automatically to equivalent units using force, as desired and as listed in Table 1.1; second, the numerical values will be properly represented, regardless of the system of units being employed. Values and units ~re given in the Table 1.2. The old English Engineering system of units by tradition is a mixed system, containing both units of mass and force, which can complicate their treatment. Use of the method above automatically replaces its pounds mass (lbm) with equivalent untis using pounds force (lbf, or just lb). The result is virtual conversion to the force-based American Standard system. Note that the value of g¢ in the old English Engineering system is 32.174, not 1.0000, so use of the recommended method produces a symbol for every number (other than true conversion factors, which are dimensionless though not unitless), which is a good general policy in analysis. The system also uses British Thermal Units (Btu) for energy, Table 1.2 Units and Values for Conversion Equivalent units of

values and units of

m

L

F

kg

m

N

9.81 ~5

kg.m 1.0000 2 N.s

slug

ft

lb

ft 32.174 ~

1.0000 slug- ft 2lb.s

"sluglet"

in.

lb

in. 386.09 -~.

lbm

ft

lb

ft 32.174 ~

gc

m

1.0000 sluglet.in. lb-s" 32.174 lbm-ft 2 lb.s

of Mass to Force units of m W gc g

2lb.s ft 2 lb.s in 2lb.s ft

14

CHAPTER I.

INTRODUCTION

rather than ft-lb, but this poses no real problem since the ratio of ft.lb to Btu is a dimensionless conversion factor (with value 778.16). Engineers probably use inches more often than feet, such as in "psi" for pressure. The units of mass then are lb.s"/in. The rather whimsical name "sluglet" is given in the table for this unit; it equals the weight Win pounds force divided by g in inches per second squared, or one-twelfth of a slug. This unit of mass has no official or recognized name. This namelessness underscores the preference of engineers for force rather than mass units. Manymore physical phenomena directly involve force than involve mass. The SI system of units has d particular advantage for electromechanical systems: a volt-coulomb equals a Newton-meter (equals a Joule). This book employsSI units for all such systems, as well as all electrical circuits. Therefore, Table 1.1 gives electrical units only in $I. EXAMPLE 1.1 Evaluate the classical formula for the natural frequency, w, = v/~-/m, for the following particular values of the mass m that is attached to ground by alinear, spring with rate k: (a) ra = 2 kg, k = 8 N/m; (b) m = 2 slugs, k = 8 lb/ft; (c) m = 2 lbm, k = 12 lb/ft; (d) W= 2 lb, k = 1.0 lb/in. Solution: (a)

symbol equation : w~ = ~/-~-~ units equation : number equation :

(b)

symbol equation : units equation : number equation :

(c)

[w~] = a~n =

~/N

1 kg.m ~ kg N-s"

1 s

- ~ ¯ 1 = 2 rad/s

Wn = [w~] = w~ =

~/lb

1 slug.ft 2~- ’ slug lb.s

1

¯ ~ - 1 = 2 rad/s

symbol equation : units equation : number equation :

~ [Wn] =~/lb ~-’lbm 1 lbm.ft lb.s

_ 1s

~z. = 12- ~ - 32.2 = 13.9 rad/s

1.7.

15

TREATMENT OF UNITS

(d)

symbol equation units equation

1 ft ~-in = 1~ [~n] =~/lb ~n’l-~’sZ’

1i

number equation : a~ = 1.0. ~ ¯ a2.2.12 = la.9 rad/s The las~ case aboveillustrates the most common style of Americanengineers, except they likely woulddirectly substitute the value and units a86 in/s ~ for 9. In all the c~es the result h~ the units s-~. This ~ is a circular frequency; a dimensionless angle is implied, whichis the radian. To’convert from radians per secondto cycles per second(Hz), division by 2n is necessary. Guided Problem 1.1 This first guided problem illustrates the role of gc even whenthe mass does not appear explicitly in a formula being evaluated. It aiso showsthe role of true conversion factors. Unless you already are quite competent with treating dimensionsand units, you are strongly advised to at least attempt the problem before looking at the solution on the next page. Find the velocity v of a perfect gas in the throat of a nozzle, given the formula v = x/~-~--~, and (a) Ah = 10 kJ/kg and (b)Ah = 5 Btu/lbm. Suggested Steps: 1. Note that the Ah given is the enthalpy per unit mass, or the specific enthalpy. Therefore, replace this implied mass by m/gc to modify the formula. 2. Write the units equation for the SI example. The k3 should be expressed in terms of the primary consistent units. Are the dimensionscorrect? Is there a need for a conversionfactor? (If so, write the units for this factor in your equation.) 3. Write the numbersequation for the SI example, evaluate the velocity and note the units. 4. Write the units equation for the Btu example. The conversion factor between Btu and foot-pounds should be included. Are the dimensions and units of the answer correct? 5. Write the numbersequation for the Btu example, evaluate the velocity and note the units. There are 778.16 ft.lb per Btu. References Textbooks that can serve as references for this text have been authored by D.C. Karnopp, D. L. Margolis and R.C. Rosenberg, 2 P. Gawthrop and L. 2SystemDynamics: a unified approach,2nded., Wiley,NewYork,1990.

16

CHAPTER

1.

INTRODUCTION

Smith, 3 J.U. Thoma,4 F.E. Cellier, 5 and A. Mukherjee and R. Karmaka~ Notable collections and reviews of papers are given in a special issue of the Journal of the Franklin Institute 7 and a book edited by P.C. Breedveld and G. DauphinTanguy. s For conference proceedings and more see www.BondGraph.com. PROBLEMS 1.1 State the standard weight of a mass of I lbm. 1.2 The energy of a compressed spring accelerates a mass. Find the maximum velocity of the mass assuming a spring rate of 20 lbs/in:, an initial spring compression of 5 inches from its free length and a mass weighing 10 lb. Neglect the mass of the spring. 1.3 A DCmotor operates at 70% efficiency with a power supply of 24 volts and 2 amps. Find the velocity at which it can raise a 10 kg object. 1.4 Find the velocity of the flow of water (weight density 62.4 lb/ft 3) in the throat of a nozzle, given the pressure drop from the rest upstream state, Ap = 10 lb/in. 2, and the formula v = ~. 1.5 A hydraulic fluid suffers a pressure drop of 2000 psi in flowing through a valve. Neglecting any effect of compressibility or heat transfer, the dissipated energy simply raises the temperature of the fluid. Find the temperature rise 3. assuming a specific heat of 0.5 Btu/lbm.°F and a density of 1.6 slugs/ft

Solution

to Guided Problem 1.1

N m 2. [vl--v /kN-m ~gg " kg.m Ns2 ’k--~= ~3. v = x/2.10.1.

1000 = 141.4

~/Btu lbm.ft ft.lb ft 4. Iv] = Vl--~-~m. lb.s 2 " Btu - s 5. v = ~/2.5.32.174 ¯ 778.16 = 500 ft/s a MetaModelling:Bondgraphs and dynamicsystems, Prentice-Hall, 1996 4Simulationby Bondgraphs:introduction to a graphical method,Springer Verlag, Berlin, NewYork, 1990. 5 ContinuousSystemModeling,Springer-Verlag, NewYork,1991. 6Modelingand Simulation of Engineering Systems ThroughBondgraphs,CRCPress, Boca Raton, 1999. 7j. of th~ FranklinInstitute, guest editor P.C. Breedveld,v 328 n5/6, 1991. S BondGraphs.for Engineers, North-Holland,Amsterdam,1992.

Chapter 2

Source-Load

Synthesis

An engineer usually starts to model an engineering system by subdividing it, conceptually, into components with explicit boundaries. In the simplest case only two components are defined: a source and a load. In somewhat more complex instances the source and the load may be interconnected by a third component. This chapter considers sources and loads which either are joined directly, or have one or more components in between to form a chain. Each component is characterized in isolation, and these characterizations are combinedto reveal the behavior of the assembled system. The engineer also wants to know how to configure a particular system to achieve the best possible behavior. Variables, such as force, velocity, voltage and current, generally change over time, justifying the name. This chapter, however, deals with components and systems that can be characterized in terms of variables that remain constant over time. These components or systems are said to reside in steady state, or in equilibrium. Sometimes more than one equilibrium exists for a given synthesis of components. Sometimes, further, a small disturbance on the system produces a radical change in the state of a system; such an equilibrium is said to be unstable. The challenge of modeling a system in equilibrium might seem vastly simpler than the modeling of a dynamic system, in which the variables change over time. This is usually not the case, however. Most of the mistakes that modelers of dynamic systems make can be traced to the geometric and quasi-geometric constraints of the system, which apply to the equilibrium states as well. Thus you are urged to pay serious attention, even if your primary interest is dynamics. 17

18

CHAPTER 2.

SOURCE-LOAD

SYNTHESIS

surroundings~

syste~

~

)

contro~ Figure 2.1: Partitioning

2.1

of the universe into a system and its surroundings

System Reticulation

In the study of thermodynamics, the universe is divided customarily into a system and its surroundings, as pictured in Fig. 2.1. The system might be closed, represe.nting a fixed quantity of matter or a control mass. On the other hand it might be open, representing a region in space or control volume into which and/or from Which matter can flow. In either case, the boundaries of the system, often called the control surface, might be fixed or moveable, giving a total of four different possible combinations. Conceptual partitions of space or bodies of matter also are basic to the study of mechanics. In the study of statics and dynamics one draws a free body diagram of the system in question. In fluid mechanics one chooses a control volume. It should be no surprise, therefore, that in the general modeling of physical systems one starts by carefully defining a system, or perhaps five or twenty different sub-systems. The process of endowing a system with structure often is called reticulation. The Latin stem reti means (fish) net; to reticulate ineans to make into or like a net. This image helps you focus not only on the sub-systems, but also on the meshpoints or bonds of the net. Reticulation is not as straightforward as sometimes it may seem in practice. It is easy to overlook the simplifying assumptions that are implicit in this act, with resultant trouble. The discussion here starts simply, as you did in the study of statics and thermodynamics, by identifying a single system and its surroundings. 2.1.1

Case

Study:

Induction

Motor

as

a Source

Consider an induction motor as pictured in Fig. 2.2. This is a complicated device. The g0al is merely to characterize its behavior as it applies to users, not to understand the details of its construction or operation. The entire machine and its source of power is defined as the system, letting only the shaft penetrate the system boundary or control surface. The system is characterized in terms

2.1.

SYSTEM RETICULATION control surface F ~ I r---"’--~

,’

I I L

19 surroundings ~ ~ I shaft

oC, oi:

’

~ ~ ¢ M,,, I energy port (intersection motor system I of shaft and control I surface)

Figure 2.2: Representatioh of an induction motor as a one-port system

Figure 2.3: Shaft and crank of appropriate variables that can be observed at the point of interconnection between it and the environment, namely the shaft. The shaft, or more precisely its intersection with the control surface, is said to be the energy port or just the port of the system. The description "one-port model" distinguishes this representation from potentially more complex models having two or more defined ports. The first of two variables that most engineers would identify is angular velocity, ~ (¢ is the rotational angle of the shaft), usually measured in radians per second. The second variable is the t.orque or moment, M~, in the shaft. The product of the two variables, Mine, is the power, P, generated .by the motor and propagated through the port (shaft) to the environment, assuming ~ and M,~ are both defined as positive in the same sense a~ seen by an observer stationed near the motor: T~ = -~¢m~. (2.1) To establish this result, a normal force F acting on pictured in Fig. 2.3. If the the work done is F58. But (is

consider that the torque and the power arise from an imaginary attached crank of radius arm r, as crank is displaced by an infinitesimal distance equals r 5¢, where 5~ is the corresponding angular

20

CHAPTER 2.

SOURCE-LOAD

SYNTHESIS

200 2.5

150

f.o

1.5 Nm 1.0

100 rad/s

0.5 0

5O 0

50

I00 150 ~, rad/s

200 I

00

Figure 2.4: Torque-speed characteristic

1.0

2.0

of the induction motor system

rotation of the crank and shaft, so the work becomes Fr 5¢. The power is the work per unit time, or 7 ~ = Fr de/dr = Fr~. Finally, the product Fr is the applied moment, Mm, establishing equation (2.1). Imagine an experiment in which the environment is adjusted until ~ reaches some desired value, and then Mmis measured. This experiment is then repeated for many other values of ~ until a complete characteristic Mm = Mm(~)

" (2.2)

can be plotted. A typical characteristic for an induction motor is shown in part (a) of Fig. 2.4. Alternatively, the environment is adjusted until J¥/m reaches some desired value, at which point ~ is measured. The consequence of repeating this experiment for many different values of Mmcan be written ~ = ~(M,~).

(2.3)

The corresponding characteristic for the induction motor is shown in part (b) of the figure. Care must be taken because there are two values of ~ for some values of Mm. These two plots foll(~w the convention that the horizontal axis (abscissas) represents the independent variable on the right side of the equation, and the vertical axis (ordinate) represents the dependent variable on the left side the equation. This convention actually is quite arbitrary, however; the two plots represent the same characteristic, with the axes merely reversed. The general idea of a characteristic overlooks any considerations of independence and dependence, that is of cause and effect. This perspective could be called existential, as opposed to causal. Thus, the two plots are completely equivalent. (L/~ter, ~ you will employ a causal perspective of a characteristic plot, however, for specific computational purposes.)

2.1.

SYSTEM RETICULATION

21 control surface h~ sprinkler __~

surroundings load system

-~Q shaft

pump

[

Me ~ ]_water

supply

~

Figure 2.5: Representation of the load as a one-port system

Either plot completely describes the motor as long as the motion is steady (no acceleration or inertial moments), the machine does not run backwards, you do not care about heat and noise, etc., and you are not worried about failure of the power or the motor. This model of the motor is described as having only one port, with the conjugate variables ~ and M,~. It is represented by the word bond graph Mm INDUCTION MOTOR ~ The port is penetrated by a power bond or, for short, bond, indicated by the horizontal line. The outward-directed half-arrow defines the direction of power flow when P = Mm~is positive. 2.1.2

Case

Study:

Water

Sprinkler

System

as

a Load

Consider now that the motor drives a pump which pumps water from a basement tank through a pipe to some upper floor of building where it passes through sprinkler heads and extinguishes a fire. This new equipment, collectively, now can be considered as the system; the motor becomes the new environment. The role of the system and the environment thus are inverted. The new system can be represented summarily by a second word bond graph: Mp LOAD ~ SYSTEM Like the motor, this model of the system has only one port, which is a shaft with an angular velocity ~ and moment Mp. Note one difference: the power

22

CHAPTER 2.

SOURCE-LOAD thirtieth floor

2.5 2.0 1.5 N.m 1.0

SYNTHESIS

~

twentieth floor tenth floor

0.5 I

50

I

I

100 150 ~, rad/s

Figure 2.6: Torque-speed characteristics

I

200

for the load system

product ~Mpis positive when power flows into the load, not out. This is the only difference, in fact, between a "source" and a "load." It is now possible to run an experiment in which the environment of this load (i.e. the motor) is adjusted to set any desired speed, at which the torque Mp is measured. Repeating this experiment for the whole range ~ > 0 of interest produces a load characteristic. Three such characteristics are plotted in Fig. 2.6, for the sprinklers being on the tenth, twentieth or thirtieth floors, respectively. These characteristics also could be found by inverted experiments in which Mpis set and ~ is measured, as with the motor. This description alone does not tell you how much water flow is associated with any particular combination ~, Mp; it doesn’t even recognize the existence of water. For now, just recognize that the faster the shaft turns, the greater the volume flow rate of water will be. 2.1.3

The

Source-Load

Synthesis;

Case

Finally, the particular motor is joined to the particular the following word bond graph: ]l~l INDUCTION LOAD MOTOR SYSTEM . ~

Study load, as suggested by

These two sub-systems now share a commonspeed (~) and a common moment, so that M,~equals .hip." Therefore, if their respective characteristics are plotted on commoncoordinates, as in Fig. 2.7, the resulting equilibrium must be represented by their intersection. For the tenth-floor case, there is only one intersection of the characteristics, and therefore only one possible steady operating speed, flow and moment. Such an equilibrium is called a steady state, ilnplying that the state variables are unchanging over time. For the thirtieth floor case, there are no intersections at all. Moreover, the load torque exceeds the motor torque for all speeds ~ > 0. The motor simply will not run. It might even burn itself out trying.

2.1.

23

SYSTEMRETICULATION

2.5

load characteristics, Mp." thirtieth floor

~

twentiethfloor tenth floor

1.0 0.5 0

I

0

50

I

I

II

100 150 ~, rad/s

200

Figure 2.7: Synthesis of the source (motor) and load characteristics

M

(h~

(a)

,K>K

,Mp>Mm

(d)’ I, change~

i

Figure 2.8: Stable and unstable equilibria for the twentieth floor sprinkler The case of the twentieth floor sprinkler is moreinteresting. There are two intersections of the source and load characteristics at the speeds labeled ~ and ~2 in Fig. 2.8. Both of these intersections represent equilibrium states, which meansthat the source speed equals the load speed and the source torque equals the load torque. Equilibrium, however,is not a sufficient condition to insure actual operation. This is because someequilibria are unstable, that is fail to persist. Prediction of whetheror not a particular equilibrium or steady state is stable requires consideration of a nearby non-equilibrium or unsteady state. For the motor-pump-sprinkler system this means a departure from the equilibrium angular velocity, so that the torques of the source and the load are different. Onesuch pair of source and load states near the equilibrium state ~2 is indicated in the figure as points (a). Since the shaft is assumedto be rigid, the source (motor) and the load have the same angular velocity, (~. A straight lin6 drawnbetween the two points therefore must be vertical. The source and the load do not have the same torque, M. The electromagnetically induced torque of the motor can be different from the fluid-induced torque of the pump. A

24

CHAPTER 2.

SOURCE-LOAD

SYNTHESIS

difference between the two torques causes the shaft, with its attached rotors in the motor and the pump, to accelerate or decelerate. Acceleration would drive the speed further above the equilibrium, indicating instability. On the other hand, deceleration would cause the speed to approach the equilibrium. You need to know, therefore, whether the shaft accelerates or decelerates. For states (a), you can see that the motor torque, ~im, is less than the pump torque, ~.lp, producing a net negative torque and a decrease in the angular velocity ¢. As the speed decreases, the torque slowing it downalso de.creases, so the deceleration decreases. Whenthe speed reaches the equilibrium, ¢2, the net torque and the deceleration vanishes. The equilibrium appears to be stable, but to make sure you ought to check what happens if the starting speed is somewhat below ¢2. This check can be done by considering the non-equilibrium states (b). Here, the source (motor) torque exceeds the load torque, so the speed of the common shaft accelerates, as indicated by the arrow. Again, the acceleration vanishes as the speed aproaches ~)2. This observation confirms the stability of t.he equilibium speed ¢2. At the same time it indicates that the equilibrium speed ¢1 is unstable; no matter how close states (b) are to this equilibrium, an acceleration will drive the speed away from it. This conclusion is reinforced by consideration of states (c), which produce a deceleration, as indicated by the associated arrow. Finally, consider the rest case (d). As with case (c), Mm< Mp. This means that the motor will not start, and therefore the system is unacceptable, despite the stability of the speed ~2. (It could be made practicable, however, in the event that some auxiliary means were available to start the shaft rotating faster than ~. One such means is suggested in Guided Problem 2.2.) As a more familiar example of an unstable equilibrium, consider a perfectly symmetrical sharpened pencil placed perfectly vertically on its tip in a perfectly still room. It is in equilibrium because there is no force to make it fall over. The slightest deviation from the vertical orientation, however, or the slightest puff of air, produces a small force in a direction awayfrom the vertical equilibrium. As the departure from equilibrium grows, the destabilizing force grows also, and the pencil soon falls over. In practice, tiny errors in the initial state and tiny disturbances are impossible to avoid. Real pencils in real rooms never stand on sharpened tips for more than a brief moment. 2.1.4

Summary

A system is identified by its control surface. A shaft penetrating this surface defines an energy or power port of the system. The behavior of this power port can be described by the conjugate variables torque and angular velocity, the product of which is the power passing into or out from it. If the system is not influenced by any other external variables, it is said to have a single port, or be a one-port system. Its environment, similarly, could be described as another one-port system. Which is the system and which is the environment is quite arbitrary; you could simply call them system A and system B. Further, the steady-state behaviors of each such system, for any and all environments,

2.1.

SYSTEM RETICULATION

25

can be represented by the relation it imposes between its conjugate variables, as determinable by experiment and as representable by an equation or a plot. Such a relation is called the system characteristic. Two one-port systems connected at a commonport experience an equilibrium state or states where they have commonvalues of both of the conjugate variables. If the two characteristics are superimposed on a commonplot, these equilibria are represented by their intersections. An equilibrium can be stable or unstable, depending on whether an imposed departure from equilibrium spontaneously grows or shrinks. This determination requires information not contained in the characteristics themselves. For the case of the motor-pump-sprinkler this information is two-fold: the shaft is rigid, so both source and load have a commonangular velocity, but the torques can be different. If the source torque exceeds the load torque, the combined system accelerates. This situation can be represented by a vertical line segment connecting the two characteristics at the particular speed, and an arrow pointing to higher speeds. The pattern of such arrows for different speeds shows which equilibria are stable and which are not. The purpose of presenting this case study far transcends systems with shafts. Similar considerations apply to systems with interconnecting rods, fluid pipes and electrical conductors, as you shall see. Guided

Problem

2.1

This first guided problem involves a basic source-load synthesis. You are strongly advised to carry it out, as least roughly, before examining the solution at the end of the section. An internal combustion (IC) engine drives a propeller that powers a boat which weighs effectively 4000 lbs. The thrust produced by the propeller depends on the throttle position and the speed of the boat, as plotted in Fig. 2.9. The drag force on the hull also depends on the speed of the boat, as shown. (a) Plot the steady-state position.

speed of the boat as a function of the throttle

(b) Assume that the throttle has been at 30% long enough for the boat to reach equilibrium speed. The throttle then is suddenly advanced to 100%. Find the instantaneous acceleration. Also, determine the acceleration when the speed of the boat has increased half way to its final equilibrium. Suggested

Steps:

1. The thrust and drag are forces that are analogous (similar) to torque, and the linear speed is analogous to angular velocity. To answer part (a), determine the intersections of the thrust curves with the drag curves, cross-plot the speeds of these points against the throttle positions, and connect the resulting points with a smooth curve.

26

CHAPTER 2.

800

¯ % throttle

600

60--

SOURCE-LOAD

100 90- ~thrust 80 ~_~..~ ~

~"><

SYNTHESIS

drag ~

~

force, F 400 lbf 200

0

0

10

20

30 40 speed, ~, ff/s

Figure 2.9: Thrust and drag for boat of Guided Problem 2.1 2. Get the net force which accelerates the craft by subtracting the drag of the hull at the equilibrium speed for 30% throttle from the thrust at the same speed corresponding to 100%throttle. 3. Use F = ma to determine the acceleration. 4. To answer the final question, find the average of the initial and final speeds, and repeat steps 2 and 3.

Guided

Problem

2.2

This problem illustrates stable and unstable equilibria as well as basic sourceload synthesis. Its value depends on your effort. Clearly the water-sprinkler system you have examined is unsatisfactory for the twentieth floor and above. One solution is to replace the motor, which is of the standard shaded-pole design commonfor small induction motors. The proposed motor, characterized in Fig. 2.10, is of the capacitor-start induction type. It has two windings; the one with a series capacitance is switched out of the circuit by a centrifugal switch at 75%of synchronous speed. Ignore potential problexns with overheating. (a) Estimate the maximumpossible powers below and above the switching speed for the proposed motor, and compare to the original motor. (b) Find the equilibrium speeds for the sprinklers on the tenth, twentieth and thirtieth floors, respectively, using the proposed motor. (c) Discuss the stability

and any peculiar operation of these equilibria.

2.1.

27

SYSTEM RETICULATION mainand auxiliary winding ~~’--~,,switching 2.0 /’//~

0

-~1

0

-’~ main windi~ oNy~ .~.~/~ I I I I I I 50 150 200 100~,rad/s

Figure 2.10: The motor of Guided Problem 2.2 (d) As project engineer for the system, recommend a course of action. Suggested Steps: 1. Power is the product M6. The maximumpowers can be estimated by trial-and-error guesses of points on the characteristics. Later, youwill see soIne graphical meansthat will help you zoomin on the maximaquickly. 2. All you really need to find the equilibrium points is the motorcharacteristic, represented by the solid line in Fig. 2.10, and the three load characteristics given in Fig. 2.7. Superimposethe load characteristics on the motorcharacteristics, and find the intersections. 3. Carry out a stability analysis as in Fig. 2.8. Notethat the vertical segment of the load charateristic merely connects its two end points; operation cannot occur at an intermediate point. Instead, switching (rapid?) may Occur.

4. Consider whether the nominalsolutions seem practical; use your common sense to make a suggestion beyond the formal scope of the analysis as presented. PROBLEMS 2.1 Whenthe boat of GuidedProblem2.1 and Fig. 2.9 is traveling at a steady 20 ft/s the throttle is abruptly re-set at 80%. (a) Find the thrust and drag forces before the velocity has time to change. (b) Find the instantaneous acceleration of the boat.

28

CHAPTER 2.

SOURCE-LOAD SYNTHESIS

2.2 Estimate the ma~ximumpower 7)maz the induction motor characterized in Fig. 2.4 can deliver, and the associated torque Mand angular velocity ~. Also, find the ratio of this power to the power 7)ma~ M.delivered when the torque is maximized. 2.3 Consider the motor of Guided Problem 2.2 and Fig. 2.10 driving the twentieth’~. floor sprinkler system. The load has an effective inertia of 0.015 kg.m (a) Estimate the. average net torque available for acceleration speed range 0 < ¢ < 100 rad/s.

over the

(b) Repeat (a) for the speed ranges 100 < ~ < 130 rad/s and 130 < 150 rad/s. (c) Estimate the time required for the system to accelerate the switching speed of 150 rad/s.

from rest

2.4 An induction motor and load have the characteristics shown below. The shah and the load are rigidly connected, but have some inertia. Determine the stabilities of the three equilibrium points.

M~

~::~

~ m --

load

otor

SOLUTIONS

TO

GUIDED

PROBLEMS

~ 30 ft/s20

10

0

0

60 0 throttle, %

1 O0

2.1.

29

SYSTEM RETICULATION

The 100% thrust curve must be extrapolated to estimate the thrust at 13 ft/s: about 870 lb. Subtracting 270 lb for the drag force at the same speed gives a net force of 600 lbs. Fg 600 x 32.2 a .... 4000 mg

3.

~. 4.83ft/s

13 + 35 The average of the initial and final speeds is 2 - 24 ft/s. At this speed, the difference between the maximumthrust and the drag is 770 - 490 = 280 lbs, which gives an acceleration of 280 × 32.__2 ~. _ 2.25 ft/s 4000

Guided

Problem

2.2

main and auxiliary winding The three maximum powers are at the points indicated by dots, and have the powers, Me, in watts, indicated by the numbers. They can be found by trialand-error. Note that the points must lie on the descendingportions of the respective characteristics.

! 327

2.0 M, N-m 1.0

original motor ~.

.-main winding only

~. 50

0

100

~

150 ~, rad/s

200

2~ 3.

2.0 M, N.m 1.0 "

10

Points a and d are seen to be stable equilibria. The intersection of the 20thfloor characteristic with the dotted switching line wouldbe a stable equilibrium if the dotted line represented a real characteristic. In reality, only states b and c exist, so the motor switches back and forth rapidly between these twostates.

e d

50

100

150 ~, rad/s

200

Something must be done to eliminate the rapid switching for the 20th-floor case, which would burn out the switch quickly. The simplest solution employs a more powerful motor, and likely would not be any more expensive than fancier solutions ~vhich retain one of the existing motors. Better matching between one of these motors and the existing pumpnevertheless presents an alternative class of solution, as discussed in Section 2.4.

30

2.2

CHAPTER 2. SOURCE-LOAD SYNTHESIS

Generalized Forces and Velocities

The power that is transferred through the bond between the two sub-systems in the case study of Section 2.1. equals the product of the torque Mon the shaft and the angular velocity ¢ of the shaft. This idea of factoring power into a product of two conjugatevariables also applies to mechanicaltranslation, mechanicalshear, electrical power, fluid power, heat transfer and other energy domains. Observation of the scalar or vector nature of the resulting conjugate variables establishes analogies(special similarities) betweenvariable types across these different domains.As a result, analogiescan be drawnalso between the various types of componentsthat the bonds interconnect. The variables and componentsin an electric circuit, for example, behavefunctionally in the same wayas the correspondingvariables in an analogous mechanical"circuit." These analogies enable you to transfer your knowledgeof one type of physical or engineering systemto another. Theyalso reduce the variety of functionally different componentsin a modelof an interdisciplinary system.

2.2.1

Efi‘orts

and Flows

The powerfactor variables for a generic (or general abstract) bond are called its effort or generalized force, labeled as "e," and the flow or generalized velocity, labeled as "0".~ Thus the power becomes ~

(2.4)

Equation(2.4) represents the first of the two. parts of the definition of efforts and flows. Thus, for example, since P = Me, Mand ~ qualify as effort and flow. Whichis the effort and which is the flow are determined by the second part of the definition. Powerflows along a bond in one direction or the other. Flows are normally defined as directed, also, while efforts are normallyeither true scalars or are treated as scalars. (There are infrequent instances in which someadvancedbond graphers mayconsciously invert the roles of effort and flow, but you should overlook this possibility at this time.) Scalars such as voltage, pressure and temperatureare efforts. Directed quantities such as electric current and volume fluid flow are flows. Conventionally, the symbolchosento represent an effort variable is written abovea horizontal bondor to the left of a vertical bond, and the symbolchosen to represent the powerconjugate flow variable is written below the horizontal bondor to the right of the vertical bond:

1The traditional bond graph symbol for flow is f. This book substitutes 0 to emphasize its relation to the displacement, q, and to reflect the practice in analytical dynamics.

2.2.

31

GENERALIZED FORCES AND VELOCITIES

(a)

(b)

Figure 2.11: Convention distinguishing generalized force (effort) ized velocity (flow)

from general-

e

et

i:

Figure 2.12: Power propagated by a pair of electric

conductors

The half-arrows on the bonds indicate the direction that power flows when T’ > 0. If e is a true scalar, this is also the direction of positive flow, 0- One detail: according to a standard set by a committee during the llth IMACS World Congress (1985), the half-arrows should be placed on the flow side of the bond. Consider the examples of Fig. 2.11. In case (a) the positioning of u above and to the left of the bond, and v below and to the right, signals that u is the effort (generalized force) while v is the flow (generalized velocity). In case however, only the location of the half-arrow resolves an ambiguity to show the same association.

2.2.2

Electric

Conductors

Each power flow in an electric e, and current, i:

circuit is represented by the product of voltage,

(2.5) Voltage is a symmetric or scalar variable with respect to the conductor; it is identified therefore as a generalized force or effort, represented in this book also by the symbol e (the sy~nbol v also could be used). Current, on the other hand, is anti-symmetric with respect to the conductor; it flows one way or the other. It is identified therefore as a generalized velocity or flow, represented by either the symbol 0 or the symbol i. A pair of electric conductors, one grounded, is shown in Fig. 2.12. Note that the voltage e is the difference in electric potentials of the two conductors, while the current 0 = i is equal in magnitude and opposite in direction. In practice one usually does not bother to draw the grounded conductor in a circuit diagram, but nevertheless tacitly assumes its existence. The corresponding bond graph symbol also is shown, including the power convention half-arrow.

32

CHAPTER 2. SOURCE-LOAD SYNTHESIS control surface subsystem A I subsystem B

velocity~ push-rod: F veloci SUBSYSTEM A

SUBSYSTEM B

Figure 2.13: Subsystemsinterconnected by a cable or push-rod The time integral of the electric current,

~

"~dt= q.~

(2.6)

- ql,

equals the net electric charge or electric displacement, which happily is traditionally represented by the samesymbol, q, as mechanicaldisplacement.

2.2.3

Longitudinal

Mechanical

Motion

Consider an inextensible cable or a push-rod that connects two sub-systems, as illustrated in Fig. 2.13. The cable has a tension, Ft, with dimensionsof force. The push-rod can sustain either tension or compression;the choice of variable shownis compression,Ft. If the velocity of the cable or rod, 5, is defined as positive in the rightwarddirection, the powerconveyedfrom the left to the right becomes 17)= -Ftgc

or 7)= Fc~:.

[

(2.7)

Note that if the cable or the rod is in tension (Ft > 0 or F~ < 0) and k > 0, the poweris negative, that is being propagatedto the left. The "powerconvention" (7) > 0 for powerpropagatedto the right) is indicated by the half-arrow on word bond graph. The variable 2 is a real velocity aligned with the bond. The generalized velocity or flow is directed similarly. Therefore, Ft or F~ mustbe the generalized force, or effort. Forceis defined classically as an action on a system describable by a vector. The tension F~ or compressionFc, on the other hand, refers to both the action on the systemand the reaction on its environment.In this symmetric view the tension or the compressionacts between two systems like a scalar. A

2.2.

33

GENERALIZED FORCES AND VELOCITIES control

observer A

tension

axis of bond observer B

Figure 2.14: The perspectives of two observers at opposite sides of a bond "generalized force" or effort is a generalized concept of such a variable; voltage is another example. A powerful way to determine whether a variable has the directedness of a flow or generalized velocity or the symmetry of an effort or generalized force is to consider two observers facing each other head-to-head and toe-to-toe on either side of the control surface or either end of the bond, as suggested in Fig. 2.14. They both perceive the variable in question at the control surface through which the bond penetrates. If their perceptions are identical, the variable can be said to be symmetric with respect to the observers, and the variable is an effort or generalized force. In our immediate example, if one observer senses tension, so does the other; if instead he senses compression, so does the other. If the variable is i, however, one sees a motion directed away from himself, while at the same time the other sees motion directed toward himself. Anti-symmetry implies a flow or generalized velocity.

2.2.4

Incompressible

Fluid

Flow

The flow of a fluid through a pipe or tube can be represented at any cross-section or control surface as the sumof the flows through a bundle of infinitesiInal stream tubes, each with cross-sectional, area dA, as suggested in Fig. 2.15. Each stream tube acts much like a push-rod with compression force dFc = P dA, where P is the static pressure, and velocity ~ = v. Thus its energy flow rate or power is d7) = ~ dFc = Pv dA. Assuming P is uniform over the section, -~ section gives the total power

integration

(2.8) of this power over the

2Theapproximationof equations (2.8) and (2.9) omits the transport of both internal kinetic energies, whichcan be important particularly whenP is small. Nevertheless, they

34

CHAPTER 2.

,

S~treamlines

. t" . ~’""

~~

SOURCE-LOAD SYNTHESIS

stream tube

micro-bond:

~’ control surface

integrated:

P vdA P

Q

Figure 2.15: Channel with elemental stream tubes of area dA

7 ) = PQ;

Q=. f vdA,

(2.9)

The symlJol Q stands for volume flow rate. This representation is particularly useful when the density of the fluid can be assumed to be constant, for then Q is uniform along the tube or pipe and across any throttling restrictions. (The analysis of compressible flow in Chapter 12 employs mass flow rate instead.) The pressure P is the same whether the observer is facing upstream or downstream; this implies symmetry and identifies P as an effort or generalized force. It is representable by the generic symbol e, like simple force, momentor voltage. Conversely, Q has a direction, and thus is a flow or generalized velocity, representable by the generic symbol 0, l~ke simple velocity, angular velocity and electric current. Note that the time integral of Q (representable by the generic symbol q) is the net volume of fluid which has penetrated the control surface. Sometimes, as a result, the symbol V is preferred for this volume or generalized displacement, and the symbol ~’z is preferred fbr the volume flow rate.

2.2.5

Rotational

Motion

The power transmitted along a rotating shaft was seen in Section 2.1 to be the product of the moment, M, and angular velocity, ~: T’ =/~.

(2.10)

Whichof the two power factor variables, _~,I or ~, is the effort and which is the flow? To answer this question, imagine the two observers facing each other at the two ends of the shaft, as shown in Fig. 2.16. If one sees clockwise rotation, the other sees counterclockwise rotation. (Check this out by rotating your pencil apply with considerable accuracy to most hydraulic "fluid power" systems. More complete representations for compressible flow systems are presented in Chapter 12, where it is shown that it is more accurate to substitute the stagnation pressure for the static pressure.

2.2.

35

GENERALIZEDFORCESAND VELOCITIES control

observer A

as seen by both observers

shaft Mas seen by observer A

Mas seen by observer B

observer B

Figure 2.16: Observersviewing rotating shaft I

or I I

controlsurface Figure 2.17: Power propagated by lateral motion in one direction and observing it from both ends.) Thus, ~ is a flow variable. Now,let the two observers sense the torque or moment.Both will feel the same direction, either clockwise or counterclockwise.(Checkthis out, also, by placing a clockwise momenton one end of your pencil with your left hand, as seen by your left wrist, and balancing that momentwith your right hand. Your right hand then also exerts a clockwise moment,as seen from its wrist.) Thus, momentMis an effort or generalized force.

2.2.6

Lateral

Mechanical

Motion

Transverse or lateral motion of a membertransmits power through a control surface if a corresponding shear force is present. As suggested in Fig. 2.17, whichrepresents a top view, the poweris the product of the shear force, Fs, and the lateral velocity, vs: (2.11) Observers facing each other at the two ends of the bondsee the shear force in the samesense, to their respective right sides or respective left sides. If one observer sees the motion as rightward, on the other hand, the other observer sees it as leftward, and vice-versa..Thus, the force is the effort and the velocity is the flow.

36

CHAPTER2. observer A

control

SOURCE-LOAD SIWTHESIS vector examples: symmetric:(a) and (b) asymmetric:(c) and (d)

axis of bond observer B Figure 2.18: Transversevectors as seen by two observers at opposites sides of a bond Should the shear motion be vertical rather than horizontal, however, the roles of the two variables becomereversed: the velocity is seen as the s~ne by both observers, while the shear force has opposite sense. Howthe observers see transverse vectors is pictured in Fig. 2.18. But do you want the distinction betweeneffort and flow to hinge on whetherthe motionis lined up with the axes of the observers? In exceptional cases like this, whenneither variable is a true scalar, you are permitted to choose whichis the effort and which is the flow. (Most~nodelerschoosethe force as the effort, and the velocity as the flow.) The convention of the powerhalf-arrow is inviolate, on the other hand: the arrow points in the direction that the powerflows whenthe product of the effort and flow variables happensto be positive.

2.2.7

Microbonds

The bonds addressed so far can be called macrobonds, in contrast to microbonds which describe the power propagated through a microport of infinitesimal cross:sectional area, dA. A bond or macrobondis thus the sum or integral of an infinite numberof microbonds;its existence implies one or more simplifying assumptions. For the incompressible fluid flow above (Fig. 2.15), the powerof a microbondis given directly by equation (2.8). Thus, P is the effort and v dA is the flow. This equation can be integrated to form the macrobond only if either P or v is uniformover the channel. If P is uniformor is assumedto be uniform (the more commonassumption), the equation integrates to give the product PQ, where Q = f v dA, as noted above. To the extent that neither P nor v is uniform, however, the macrobonddoes not properly, model the situation.

2.2.

GENERALIZED FORCES AND VELOCITIES

~

,,

37

f

flywheel

-~flywheel

¯

I

control surface

control surface

Figure 2.19: The effect of power factoring control surface or port location

on accuracy, due to the choice of

Microbonds for longitudinal motion share the flow variable a? with the microbond. Therefore, the effort on the microbond is the compressive force -a dA, where -or is the normal compressive stress. If ~ is uniform throughout the cross section, the effort of the macrobondbecomes the integral of the stress, namely the force. If 2 is not uniform, the macrobond cannot model the situation precisely. The rotating shaft also can be considered to be the sum or integration of an infinite set of microports. At the microport level, the rotation is a lateral motion, with the power density being the product of the shear stress ~- and the shear velocity vs. Neither of these quantities is commonto all area elements dA. The moment M is r times the moment arm integrated over the area, 3I = vr[27rr

dr],

(2.12)

where r is the radius variable and a is the radius of the shaft. Mdoesn’t exist at a point. The power propagated is the integral of the shear stress times the shear velocity, vs, over the area, or 7) = 7vs[27rrdr];

vs = r~.

(2.13)

A comparison of equ.ations (2.12) and (2.13) shows that ) =M~ is correct onl y if you assume that ¢ is uniform, and thus can be taken forward of the integral. This happens when the cross-section of the shaft rotates as a rigid uuit. To the extent that ~ is not uniform, the very concept of the macroport is in error. Only in extremely rare cases would the assumption of uniform (~ not be entirely appropriate. The point is, however, that whenever you employ a macroscopic variable (such as Mor ~) you are inevitably introducing some measure of approximation, of inexactness. Sometimest-he error is not small. As an example, consider a "shaft" that is a thin but large-diameter flywheel, as shown in Fig. 2.19 part (a). The boundary between the system and the environment is assumed to lie right in the middle of the flywheel. The center and the periphery of this flywheel are muchmore likely to experience different

38

CHAPTER 2.

Table 2.1 Effort

SOURCE-LOAD

SYNTHESIS

and Flow Analogies generalized force generalized velocity or effort, e or flow,

Fluid, incompressible: approximate t micro-bond Mechanical, longitudinal: tapproxi~nate micro-bond Mechanical, transverse: rotation translation (shear) micro-bond Electric conductor:

P Fc

M Fs ~-dA e

v dA v =

vs vs i

t The powersassociated with the transport of kinetic and potential energies are added in the Chapter 12. For most mechanicaland incompressible fluid cases they are small. angles of rotation than would the center and the periphery of a small shaft. If the periphery were suddenly braked, for example, the core might continue to rotate through an additional small angle. Or, if the core is oscillated rotationally at a high frequency, the periphery might not follow in lock-step; in fact, a "natural frequency" might well exist at which resonance occurs. Should this phenomenon possibly exist, you would be well advised to place the system boundary to the left or to the right of the entire flywheel, as shownin Fig. 2.19 part (b). It is a good idea, then, to partition a system from its environment at a location where the interaction can be described simply (in the present case by the variables q~ and M). There are no inviolate rules for doing this; consequently, modeling ultimately becomes an art. (You should not be discouraged by this fact, for any subject that can be reduced to rules can becomevery dull, and its practitioner is subject to being replaced by software.) Cutting a motor through its middle, for example, likely would cause you grief, even though it would not be incorrect.

2.2.8

Analogies

The effort and flow symbols for the various media that have been examined are summarized in Table 2.1. The various effort variables are said to be analogous to one another, as are the various flow variables, due to their commondefinitions with respect to power and directivity. The implications will gradually become clearer, as you identify commontypes of elements that represent various classes of interactions between the efforts and flows or their time integrals or time derivatives.

2.2.

GENERALIZED

FORCES AND VELOCITIES

39

The direct analogy between mechanical and electric circuit variables associates mechanical force and torque with electric voltage, and linear or rotary mechanical velocity with electric current. 3 This analogy can help the student well grounded in mechanical systems transfer his knowledge to electric circuits. Similarly, it can help the student well grounded in electric circuits to extend his understanding into mechanical "circuits." Similar conceptual and computational advantages in the transference of knowledge accrue from the other analogies. A few physical interactions are not best ~epresented by simple bonds or product conjugate variables. The example of compressible fluid flow is treated in Chapter 12.

2.2.9

Summary

The power flowing through a simple port or simple bond customarily is factored into a product of two variables, e0. The "effort" or "generalized force" e normally is symmetric, and the "flow" or "generalized velocity" 0 normally is anti-symmetrlc with respect to two observers facing each other from opposite ends of the bond. This simple factoring applies to longitudinal, lateral and rotational motions of a rod, cable or shaft, to electric conductors and to the flow of an incompressible fluid, establishing important analogies l?etween variables used in these different domains. The direction of positive power flow is indicated on a bond by a half-arrow. A standard port or macro-port and its bond can be considered to be the sum or integration of a bundle of micro-ports and their micro-bonds, each representing an infinitesimal cross-sectional area. This summingor integration inevitably introduces some degree of approximation. All use of (macro) bonds, and indeed the very use of the word "modeling," implies approximation. Only pure mathematics is exact. The practical question is not the presence of error, but rather its extent and significance. PROBLEMS 2.5 A cable attached to the top of the machine shown on the left at the top of the next page is pulled upward, thereby producing some energetic effect. Represent the interaction with a bond, annotated with a power half-arrow and effort and flow variables, which should be defined. 3Electrical engineers sometimesdrawthe "Firestone analogy"betweenforce and current on one hand and velocity and voltage on the other. This gives a greater apparent similarity betweencircuit diagramsand somecorresponding mechanical"circuits." For basic physical reasonsincludingits extensibility to non-circuit situations, however,the author strongly prefers the direct analogy.

4O

CHAPTER 2.

SOURCE-LOAD

SYNTHESIS

[cable

~ machine

I

!machine[ 2.6 A lever attached to the machine shown above right is rotated upward. Represent the interaction by an energy bond, annotated by symbols for effort and flow which you should define and by a power convention half-arrow. 2.7 Heat conduction is a special kind of power that can be treated as the product of an effort and a flow, like the other power types above. The effort variable is absolute temperature, which you can label as T. The power itself is the rate of heat transfer, which you can label as Q, where Q is a quantity of thermal energy called heat. Determine the flow variable, 0, in terms of these variables, and express its integral as the displacement variable, q. Relate this displacement variable to a commonvariable used in thermodynamics. (This question presupposes that you have been introduced to thermodynamics. Further consideration of heat transfer and more general thermodynamics is postponed until Chapters 9 and 12.)

2.3

Generalized

Sources,

Sinks and Resistances

The case study of Section 2.1 was represented INDUCTION MOTOR

M ~9

by the word bond graph LOAD SYSTEM

Both the induction motor and the load system were modeled by particular static relations between the commonshaft moment, M(or Mmor Mp), and the shaft angular velocity, ~. In Section 2.2 such static relationships were generalized, so that the form of this system becomes

in which S stands for "source" and R stands for "resistance." Sources normally emanate power, whereas resistances normally absorb or dissipate power. The effort e and flow ~ in this graph could represent any of the energy types discussed in the last section. The graph could represent an electrical system, a system with incompressible fluid flow, or a mechanical system with either longitudinal, lateral or rotational motion. The source and resistance elements in each case might be characterized in one of several special ways, which will now be discussed.

2.3.

GENERALIZED SOURCES, SINKS AND RESISTANCES Se--.-~-

or

eo .~...-~e

41

S

s° --g2or . qo Figure 2.20: Characteristics of effort and flow sources and sinks 2.3.1

Independent-Effort Sources and Sinks

and Independent-Flow

An independent-effort source, usually called simply an effort source, and a independent-effort sink, usually called simply an effort sink, are defined to haveefforts whichare independentof their flows. This meanseither that e is a constant, or in general a function of time, e = e(t). The elements provide or absorb any flow that the attached system mayrequire in response to the imposed effort. This behavior is represented in the plot of Fig. 2.20 by a horizontal line in the e-0 plane. Both elements are designated by the symbol Se. The only difference betweenthem is that the source is designated with an outwarddirected powerarrow, and the sink with an inward-directed powerarrow: effort source: effort sink :

Se ~

~ Se

They are really the same element, since there is no absolute requirement that the powerflows in the direction of the powerarrow, that is e0 > 0; it is just nice to have different wordsto express the different functions normallyserved by the element. Physical examples approximated by the Se element include a force source as from a weight in a gravity field, a torque source, a voltage source as from an ideal battery, and a pressure source or sink from a large body of water at some elevation. These examplesare suggested in Fig. 2.21. Coulombor "dry" friction also can be modeledas a force source, as long as the motion never reverses direction, as discussed below. The independent-flow source and the independent-flow sink, more simply called the flow source and the flow sink, are similar to the effort source and the effort sink except that the flow is independentof the effort rather than the effort being independentof the flow. As this reversal or roles suggests, the flow maybe either constant or an independent function of time. The elements provide whatever effort that the systemrequires in response to the independent

42

CHAPTER2.

SOURCE-LOADS-YNTHESIS

mechanicaltranslational:

mechanicalrotational:

fluid (gravity):

electrical (ideal battery):

Figure 2.21: Exampleeffort sources and sinks

2.3.

GENERALIZED SOURCES, SINKS AND RESISTANCES

43

flow. Flow sources and sinks are designated by the symbol S~: flow source:

S,t

flow sink:

~

Sf

The generalized characteristic is plotted in Fig. 2.20 as a vertical line. A synchronous motor turns with a constant speed, determined by the fl’equency of the driving ACvoltage (usually 50 or 60 Hz), as long as the load torque does not exceed some limit. A source SI element models this behavior, but does not recognize the torque limits. Any other constant velocity source, constant current source or constant fluid flow source also can be represented by the element. Notice again that a current sink of 5 amps is the same as a current source of -5 amps, as a pressure source of 5 psi is the same as a pressure sink of -5 psi. Thus, you are permitted to avoid effort and flow sinks in favor of effort and flow sources, or vice-versa, although most modelers prefer to use both types. 2.3.2

General

Sources

and

Sinks

A general source (S ~) can represent any prescribed static (or algebraic) relationship between its effort and its flow. Thus, as noted above, the induction motor of Section 2.1 can be represented as a general source. Other examples include a model of a battery for which the voltage depends on the current, a model of a velocity-dependent force applied to a mechanical system and a model of a pressure source that depends on the fluid flow. In the most general case, the relationship itself could be an explicit function of time. The general sink element (~S) is not employed per se in this book, in favor of the equivalent term resistance and its symbol R, as discussed below. You will see in Chapter 4 that any general source can be represented well by an effort source or a flow source in combination with a simple resistance. Ultimately, the general source is employed, if at all, only as a convenient shorthand. 2.3.3

Linear

Resistances

The generalized resistance,

indicated as e

can have any functional dependencybetween its effort and its flow. It is helpful to distinguish different classes of these elements. The simplest class is the linear resistance which, in a plot of effort vs. flow as illustrated in part (a) of Fig. 2.22, is represented by a straight line drawn through the origin. The slope of this characteristic is defined as the modulus of the resistance, or for short its "resistance," which in this book is given the same symbol, R. Thus, a symbol Ri in a bond graph indicates that a resistance relation applies between the associated effort ei and flow ~i. The algebraic

44

CHAPTER 2.

ei

(a) linear

SOURCE-LOAD

SYNTHESIS

// -R

(b) nonlinear

qi

(c) power as area

(d) linear biased

(e) multiple valued

Figure 2.22: Types of resistance characteristics expression ei = Ri~)i, where R~ is a constant, details this relationship. This or any other algebraic expression is written separately from the bond graph, however; algebra is never included within a bond graph. The two meanings of the resistance are distinct. The electrical resistor may be the most familiar approximation to a linear resistance. The effort is the voltage drop e across the resistor, the flow is the current i which passes through it, and e = Ri. This and other linear resistances are pictured in Fig. 2.23. A commonlinear fluid resistance results from the assumption of viscous laminar fully-developed steady flow through a circular tube. In your fluid mechanics course you will derive the relation that the pressure drop P across the tube is related to the volume flow rate Q through it by P = RQ; R =- 8p_~_L

(2.14)

where a is the inner radius of the tube, L is its length, and # is the absolute viscosity of the assumed incompressible fluid. More generally, whenever the viscous forces overwhelm the inertial forces in an incompressible flow through a passage, that is when the Reynold’s number is low, the resistance can be approximated as linear, and its modulus as a constant. Examples include flow through a thin slit such as a leakage path in a machine, flow through a porous plug and "creeping" flow through an orifice. Mechanical devices that can be approximated by a linear resistance are hard to make, but are very co~nmonly assumed in analysis. They are called dashpots, and are given the special symbol shown in part (c) of Fig. 2.23. The ratio

2.3.

GENERALIZED SOURCES, SINKS

AND RESISTANCES

45

(a) electrical resistance ---~i

R te---’~R

e~--~~ (b) fluid resistance P "~

e =Ri

-~~~

Q ~’R

L

P=RQ

(c) translational dashpot linear fluid resistance

x

F=tUc,

R=b

M=R~,

R=b

X

(d) rotational dashpot

classical symbol

thin film of fluid

physical example

bond graph

relationship

Figure 2.23: Devices approximated by linear resistances

46

CHAPTER 2.

SOURCE-LOAD SYNTHESIS

of the force to the velocity traditionally is designated by the symbol b, so the modulus of the bond graph element is R = b. One way to construct a linear translational dashpot is to convert the relative velocity of the two translating membersto the flow of ~: liquid, which can be done by employing a piston and cylinder. Then, as shown in the example, this liquid is forced through a passage that behaves as a linear fluid resistance according to equation (2.14). A rotational dashpot is designated by a similar symbol, as shown in part (d) of the figure. Translational and rotational dashpots often are described as "viscous dampers" because it is hard to approximate linearity mechanically without using a viscous fluid. The shear flow of a viscous liquid in a very thin annulus between a rotating cylinder and a cylindrical shell, for example, can produce a torque proportional to angular velocity, that is a rotational linear resistance. As a practical matter, however, the small amount of fluid trapped in the annulus is apt to overheat, causing its viscosity to drop, or worse, causing the fluid to oxidize and turn black, like the motor oil in your car. 2.3.4

Nonlinear

Resistances

Any component with a single effort or generalized force that is a function of an associated flow or generalized velocity, but is not proportional to it, can be described as a nonlinear resistance. The general resistance is the same as the general sink, and is used in its stead. Flow through an orifice at anything higher than creeping Reynold’s numbers is an example. If Bernoulli’s equation is used to model the flow between an upstream state with zero velocity and the point in the throat where it achieves maximumvelocity, and complete dissipation of the kinetic energy is assumed to occur downstream, the result is P=-~pv

~=~p

~

=

~

.

(2.15)

Here, p is the fluid density and c4 is a flow coefficient which represents the fact that the flow separates from the walls of the orifice to form a narrower vena contracta or effective orifice area CdAo. (For a sharp-edged orifice, Cd = 0.611; in most other cases it is larger but less than 1.0.) The equation often is used even when the assumptions above are recogniZed as inaccurate; in this case, Cd is adjusted to accomodate the error, and likely is determined experimentally. The characteristic given by equation (2.15) is plotted in Fig. 2.24. Replacing the linear fluid resistance in the translational viscous damper in Fig. 2.23 with an orifice flow produces a nonlinear mechanical damper, as suggested in Fig. 2.24. The shock absorbers in your car are essentially of this construction. Not only is the nonlinear behavior easier to obtain, it is desirable; for small bumps you want the shock absorber to act gently, whereas for large bumps you prefer that it act more vigorously so as not to bottom out (reach its mechanical limit). A mechanical brake is another kind of nonlinear resistance. As shown in Fig. 2.25, the classical model for such a resistance comprises a constant or coulomb

2.3.

GENERALIZED SOURCES, SINKS

nozzle:

--~---

47

-----

p .~ ~ plug: "--~~

valve or porous

AND RESISTANCES

P or F

large hole with small orifice shock absorber:

F 0o

Figure 2.24: Nonlinear fluid resistance

and application

isk

~o

in a mechanical damper

MI

e

Figure 2.25: A mechanical brake and its ideal resistance characteristic friction torque for forward motion, associated with the kinetic coefficent of friction, and an equal but opposite torque for reverse motion. This assumes application of a fixed force to the brake calipers. Whenthere is no motion, the resistance torque balances whatever torque is applied, as long as it does not exceed the "breaknway"torque which is associated with the static coefficient of friction. If the disk never rotates backwards, however, this model gives nothing but a constant torque, and can be modeled better as an effort sink. You also will see more exotic friction models in Chapter 4, such as the one that makes a violin sing or a dry door hinge squeel. The resistance characteristic shown in part (b) of Fig. 2.22 is nonlinear two ways: it is curved, and it does not pass through the origin, or is biased. The modulus of the resistance, R, is defined as the slope of the chord from the origin to the state point of interest, as shown. This chordal resistance therefore is not constant, but rather is a function of t) or e: e = R~);

R = R(c)) or R = R(e).

(2.16)

These equations also apply to the linear case, except that R becomes constant. The power dissipated in the resistance, for either the linear or the nonlinear case, equals the area. of the rectangle defined by the origin and the operating point, as indicated in part (c) of the figure. Algebraically,

P = eO= R(~)O". A linear biased characteristic

(2.17)

is shown in part (d) of Fig. 2.22. In Chapter

48

CHAPTER2.

SOURCE-LOADSYNTHESIS

4, biased characteristics are represented as combinationsof an effort or flow source and an unbiased linear or nonlinear (curved) characteristic. Finally, multiplezvaluedcharacteristics are shownin part (e). The characteristics illustrated, nevertheless, are globally passive, unlike the biased curves (b) and (d), since they reside only in the first and third quadrants where the powerproduct e4 is positive everywhere.Anexampleof a multiple-valued characteristic is given in GuidedProblem2.3. Note that any bias destroys passivity, but a lack of bias does not assure passivity.

2.3.5

Source-Load

Synthesis

The interconnectionof a rotary source to a rotary load, by a shaft, was treated in Section 2.1.3 by finding the intersection of the two torque-speedcharacteristics. The same type of synthesis applies to any simple interconnection in which the poweris represented by the product of a pair of conjugate generalized forces and velocities.

EXAMPLE 2.1 The output pressure of an impeller pumpreduces the net flow below that of the geometrically slip-free flow. The result is a nonlinear source characteristic, as plotted below. The particular load characteristic plotted in the figure also is nonlinear, typical of the flows through sharp restrictions as discussed above and as employedin the water-sprinkler nozzles of Section 2.1. Determinethe equilibrium state for the combinationof these elements, andits stability. 8O

P, psi

60

~ ~

40

~

~750

rpm

IMPELLERPUMP~ LOAD

20 0

load

I

0

I

100

I

I

200

I

~ I

I

I

300 Q, in3/s400

I

2.3.

GENERALIZED SOURCES, SINKS

49

AND RESISTANCES

Solution: The equilibrium state is given by the intersection of the characterisitics. This equilibrium is stable, since {f operation is attempted at a larger flow, the pressure of the load would exceed the pressure of the pump, ibrcing the flow back toward equilibrium. If operation is attemped at a smaller flow, the pressure of the pumpwould exceed that of the load, again forcing the flow toward equilibrium.

2.3.6

Power

Considerations

The power th’at is transmitted from a source to a load often is of interest. Loci of constant power can be constructed on an e - ~ plot. One such locus is shown in part (a) of Fig. 2.26. The power equals the cross-hatched area of the rectangle of height e and width ~. All points on the locus must have the same area, as suggested by the two other rectangles shown. The curve is a rectilinear hyperbola. The slope of a chord drawn from the origin to any point on a locus of constant power equals the magnitude of the slope of the tangent to the locus drawn at the same point. This fact, illustrated in part (b) of the figure, aids the sketching of constant power loci. Note also that the chord and the tangent form isosceles triangles with the ~ axis.

P=e~=4

3 e

3 e

2

0

0

1

2 0 3 (a) rectangles

4

2

0

0

i

2 0

3

(b) chords and tangents

Figure 2.26: Loci of constant power as rectilinear

hyperbolas.

4

50

CHAPTER 2.

SOURCE-LOAD SYNTHESIS

EXAMPLE 2.2 Locate the point of maximum power on the torque-speed characteristic the induction motoras plotted below.

of

2.5

1.0 0.5 0

0

50

100 150 ~, rad/s

200

Solution: A family of loci for various constant powers is superimposed on the plot below. The locus representing the maximum power, shownby a dashedrectilinear hyperbola, is tangent to the cha[acteristic at the operating point for that power. All other points on the characteristic have less power, including the point of maximulnmoment.A simpler way to estimate the location of the maximum power equates the magnitUdeof the slope of the chord from the origin to the point, and the slope of the tangent to the characteristic at that point. Thesestraight lines are drawndashed.

2.5

loci for constant powers,W 50 100 200 300 400

1.0 0.5 0

0

50

100 150 ~, rad/s

200

The source-load synthesis of a linear electrical source and linear electrical load is given in Fig. 2.27. The magnitudeof the slope of the source characteristic is knownas the source impedance.The slope of the load characteristic, which is also the resistance of the load, is similarly knownas the load impedance. Maximum poweris transferred from source to load at the midpoint of the source characteristic. Not only does this point maximizethe area of the rectangle betweenit and the origin, but the chord from the origin to this point has the same magnitudeof slope as the tangent to the load characteristic, which is

2.3.

51

GENERALIZED SOURCES, SINKS AND RESISTANCES

~ ~

d impedance e

~,~rce

impedance i

Figure 2.27: Synthesis of linear source and load to maximize power transfer

the load impedance itself. Thus, maximumpower is transmitted when the load impedance equals the source impedance. Such "impedance matching" is standard between audio amplifiers and audio speakers, for example. 2.3.7

Summary

One-port elements with constant effort independent of flow, but possibly dependent explicitly on time, are designated by the bond-graph symbol Se. They are called constant-effort sources if their power arrow is drawn outward, and ¯ constant-effort sinks if it is drawn inward. One-port elements with constant flow independent of effort, but possibly dependent explicitly on time, are designated by the bond-graph symbol Sy. They are called constant-flow sources if their power arrow is drawn outward, and constant-flow sinks if it is drawn inward. General sources have other relationships between the effort and flow, and are designated by the bond-graph symbol S. General sinks normally are called resistances, and are designated by the bond graph symbol R. Resistances occur whenever electrical, mechanical or fluid energy is dissipated into heat. The slope of a chord drawn from the origin to a point on the characteristic curve of a resistance is called the chordal resistance or just the resistance, and is designated by the same symbol, R. Resistance characteristics that pass through the origin of the effort-flow coordinates are called unbiased; others are called biased. An unbiased resistance is called passive if its characteristic remains in the first and third quadrants of the coordinates, since then the power 7) =e~ is always positive into the element, regardless of the operating point. It will be shown in Chapter 4 that both general sources and resistances can be represented by a combination of an effort or flow source or sink and an unbiased resistance. The simplest resistance is linear and unbiased; its value of R is constant. Linear fluid resistances depend of the viscosity of the fluid. Nonlinear fluid resistances depend rather on the density of the fluid. Linear mechanical resistances or dashpots often are called "viscous dampers," and often depend on the viscosity of a fluid. Mechanical friction, as in a brake, is an unbiased nonlinear resistance, but maybe approximated as~ an effort source or sink if its motion is always in one direction.

52

CHAPTER 2. SOURCE-LOAD SYNTHESIS

,

f 6o 800

force, F lbf 400

~

.......

90..~O....~hrustfor indicated percentthrottle" 80

~ 50

~/dral

~

2OO ~

0

I I

0

10

20

30 40 speed,k, ft/s

Figure 2.28: Thrust and drag characteristics 2.3

50

I

60

for the boat in Guided Problem

The source-load synthesis illustrated in Section 2.1 and again in Example 2.1 can be generalized for any of the powertypes considered. The synthesis maximizesthe powertransmitted if the chordal resistance of the load equals the slope of the source characteristic at the equilibrium point: Guided

Problem

2.3

This problemoffers you needed experience in source-load synthesis, and reveals an important phenomenoncalled hysteresis, or history-dependent behavior including sudden jumpsin state. The IC engine and propeller drive from Guided Problem 2.1 is placed in a small boat with a planing-type hull. Unlike the displacement-type hull considered before, the drag characteristic of this hull, plotted in Fig. 2.28, has a pronouncedlocal maximum (at about 24 feet per second). This peak is due the wavegenerated by the motion. At the speed of its maximum,the bow is considerablyelevated over the stern, so. in effect the boat is continually trying to climb a hill of water. At higher speeds the boat "planes," riding high and nearly level, with the generated wavecrest being amidships. Find and plot the steady-state relationship betweenthe throttle position and the speed of the boat. Clearly indicate any hysteretic behavior. Suggested Steps: 1. Draw a word bond graph for the system, naming a source, a load, a ggneralized velocity and a generalized force.

2.3.

GENERALIZED SOURCES, SINKS 10

~5.5V

io, mA

6 4

AND RESISTANCES

/

[ ]

~..------~

~/

53

e~=5.0 V

4..._,_.5 V

/

4.0 V 3.5V

3.ov

¯

"~’~IDo 0,_, 1.5k 12

2

VT

4

6

8

10

12 eD,

~f~ ~-~ ~

14

16

V

@)e~

Figure 2.29: Characteristics

of a MOSFET for Guided Problem 2.4

Were it not already done, you would superimpose plots of the source characteristics and the load characteristic, making sure the dimensions, units and scales are compatible. Cross-plot all intersections of the source and load characteristics onto coordinates with speed on the ordinate and throttle position on the abcissas. If and where it is not clear how to connect the points on your crossplot with a continuous curve, generate more points by interpolating more source characteristics. Identify any jumps in behavior with arrows on the cross-plot. Note explicitly the range of throttle positions for which there are multiple speeds. In this range the actual speed depends on the history of the throttle position as well as its current state.

Guided

Problem

2.4

This type of problem is familiar to electrical engineers. It requires determination of a source characteristic. Your instructor may advise you on its relevance to your situation. An n-channel enhancement MOSFETis driven by a 12 v battery and has a load resistance of 1.5 kfL Given the characteristics plotted in Fig. 2.29, plot the drain current iD and voltage eD as a function of the gate voltage

54

CHAPTER 2.

Suggested

SOURCE-LOAD

SYNTHESIS

Steps:

1. Consider the battery and resistor as the power "source," and the given characteristics as the "load." Superimposethe source characteristic on the given plot. Hint: It is a straight line with negative slope; its intersections with the axes are readily found. 2. Cross-plot the intersections of the source line and the given load characteristics onto plots of iD vs. eG’ and eD vs. eG. PROBLEMS 2.8 A shaft drives a load that resists with a constant torque regardless of speed, assuming that the shaft never reverses direction. Indicate what type of bondgraph element most simply represents this frictional behavior. Label the effort and flow of its bond with symbols that indicate their meanings, and show the power-convention half-arrow. 2..9 As noted in the text and pictured in Fig. 2.23, a viscous fluid damper can be made by placing a cylindrical hole through the piston of a fixed double-rodend cylinder. Fluid leakage around the piston and friction between the piston and the cylinder walls may be neglected. (These seemingly contradictory assumptions could be reasonable if low-friction polymer seals are used between the piston and the cylinder.) You will need to define relevant physical parameters (constants) in order to carry out the following: (a) Find the relationships between the force and the pressure difference across the piston, and between the velocity and the flow. (b) Sketch-plot the force vs. the velocity, and express the resistance relative to the velocity as a function of fixed parameters to the extent possible. Is this approximate resistance a constant? 2.10 Answer the above problem when the cylindrical hole is replaced by a short orifice of area Ao, to create a standard type of shock absorber. 2.11 A shaft of radius a, shown below, rotates concentrically journal with radial clearance e < < a and length L.

in a fluid-filled

2.3.

GENERALIZED

SOURCES,

SINKS

55

AND RESISTANCES

(a) Find the steady-state resistance relative to the angular velocity of the shaft. (b) Discuss any major pract’ical

deviation from the assumed behavior.

2.12 Write the chordal resistance of the flow through an orifice, equation (2.15), in the form R -- R(Q).

as modeled by

2.13 The impeller pump of Example 2.1 (p. 48) drives a load for which P 20 + 0.10 Q psi, where Q has units of in 3/s, in place of the plotted load. Estimate the equilibriura pressure and flow. 2.14 Estimate the maxitnum power that the impeller pump of Example 2.1 (p. 48) could deliver if the load could be changed arbitrarily. Give the corresponding pressure and flow rate, also. 2.15 The hydraulic load plotted in Example2.1 (p. 48) is driven by a hydraulic source described by P = 80 - 0.2Q psi, where Q has units of in3/s, in place of the impeller pump. Estimate the equilibrium pressure and flow. 2.16 Estimate the maximumpower that the internal combustion engine characterized in Fig. 2.39 (p. 75) can produce. (This figure appears in Section 2.5, but you should disregard everything in the figure except for the engine characteristic.) 2.17 An npn transistor in the commonemitter configuration has the characteristics plotted below for various values of the base current, ib.

transistor

ec

~ load

8

R-- 1500 4 50’ ~[ --

~" ~ ’

~" ic’mA " --e°=12V-

~

~

10

~’~~ 00

(a) Superimposeon the plot the characteristic the voltage source eo and the resistor R.

4

8

12 e,., V

for the load which comprises

(b) Plot the load (emitter) current, ic, as a function of the base current, and approximate this relation algebraically.

56

CHAPTER 2.

SOURCE-LOAD

SYNTHESIS

2.18 The boat of Guided Problem 2.3 and Fig. 2.28 pulls one or more water skiers. The force in the tow line for one skier is shown plotted below as a function of speed; note that a planing phenomenonexists which is similar to that of the boat hull, but more pronounced.

200[-

..........

-~

,~ , , , ,-"------~~,,, ~ 0 ~//, 10 20 30 40 0 50 speed, ft/s (a) Carefully estimate the maximum steady speed for towing a single skier. (b) Estimate the maximumnumber of identical skiers that can be accelerated from rest simultaneously, and estimate their maximumspeed. (c) In a fancy maneuver, skiers are transferred to the boat from another boat, at full speed. Estimate the maximumnumber of skiers that can be towed successfully after the transfer, and estimate their maximumspeed.

SOLUTIONS

TO GUIDED

PROBLEMS

Guided Problem 2.3

velocity,.f 40 f~/s 3O 2O l0

¯

jump~

~~

I~/I

PROPELLER F DRIVE

A~ ~ hysteresis band

20

40

60 80 throttle, %

100

2.4

IDEAL

MACHINES:

TRANSFORMERS

AND GYRATORS

57

Guided Problem 2.4 l0 i~,

e:

mA 8

¢o 0

2.

2

4

6

10

12

14

eo, V

16

12

I

2.4

8

~ "1

I 4

I e~, V

Ideal Machines: Transformers Gyrators

I 6

and

An ideal machine transmits work at one of its two "ports" to work at the other "port." Although the effort and the flow usually are changed, energy is not stored, generated or dissipated. Entropy also is not generated. Levers, gears, electric motors and piston pumps are examples of engineering components that may be approximated as ideal machines. In modeling physical or engineering systems one often neglects the dissipation of energy, or more properly the conversion of energy to heat. This simplification is done despite the fact that leakage or slippage in the presence of friction or analogous generalized forces inevitably has at least a small effect. For example, a gear train might be assumed to be frictionless, an electric transformer might be assumed to have perfect coupling and zero resistance, an electric motor might be assumed to be 100%efficient in converting electrical energy to

58

CHAPTER 2.

SOURCE-LOAD

SYNTHESIS

mechanical form, and a fluid pump might similarly be approximated as converting mechanical energy to fluid energy completely. Even if energy dissipation is not neglected, often it is represented separately. For example, a real electric motor can be represented as a frictionless or ideal motor plus an external resistance to represent electrical losses and another external resistance to represent mechanical friction. 2.4.1

Ideal

Machines

All of these devices can be described as "machines," in a generalized sense, and models of them that neglect energy loss are called ideal machines. There are two particular generic (or generalized) types of ideal machine that command special attention. One is called a transformer, and the other a gyrator. (The fatniliar electrical transformer is an electrical approximation of the transformer, but only if it is idealized so as to work perfectly at all frequencies, including DC, unlike a real electrical transformer.) An ideal machine can be represented by the word bond graph IDEAL MA CHINE It is defined to conserve energy, that is to consume or produce zero net power at all times. Thus the input power equals the output power. With the power convention of this graph, e101 = e_~02.

(2.19)

The ideal machine can be run in either direction. Although a positive power flows from left to right according to the power convention arrows chosen arbitrarily above, the power could be negative, and therefore flow from right to left. As well as being able to run forwards or backwards, the ideal machine is thermodynamically reversible, which means that it generates no entropy. As a result, neither power can represent the flow of heat. 2.4.2

Transformers

Equation (2.19) allows for many different types of ideal machines. The most commontype, called the transformer, satisfies the additional relation

=r0,.]

(2.20)

Thus T is defined as the ratio o/the generalized velocity or flow on the bond with the outward power arrow to the generalized velocity or flow on the bond with the inward power convention arrow. This ratio, called the modulus of the transformer, may be constant, or.in general may be a function of the displacement q~ or some other displacement. Until Chapter 9, attention is restricted to tranformers with constant moduli. Constant modulus or not, .the symbol T within a bond graph designates this type of relationship. Other authorg usually

2.4

IDEAL MACHINES:

TRANSFORMERS

59

AND GYRATORS

elaborate with the designation TF, or, if the modulus is not a constant, with MTF(for modulated transformer): ~.1

ql

el._~_T ¯ ql

T ~.2

q’2

el

F~ ~o

q~

MTF

~

~

e2 O~

Subscripts can be added to the symbol T to distinguish different transformers that might appear in the same model. (Most other authors do not-designate ¯ commonsymbol for the modulus of transformers.) ~bu must not confuse the bond-graph symbol T with the modulus T; only the modulus can participate in an equation, such as T = 5. Substitution of equation (2.20) into the conservation-of-energy equation (2.19) gives a simple but profound result: [G1 = T~2 . ]

(2.21)

In words, the ratio of the 9e~evalized forces of an ideal transformer equals the inverse of the ratio of the respective ~eneralized velocities. 2.4.3

Gyrators

The second major possibility definition satisfies

for the ideal machine is the gyrator,

which by

Thus, the modulus of the gyrator, G, is defined as the ratio of the effort on one of the bonds -- either one -- to the flow on the other bond. Substitution of equation (2.22) into (2.19) gives (2.23)

el = G02,

The gyrator in a bond graph is designated by the symbol G; other authors more commonly employ GYor, if the modulus is not a constant, MGY: el

G

e2

e.1

Like T, the modulus G need not be a constant, constant moduli is deferred until Chapter 9. 2.4.4

Mechanical

Devices

e_._L__~ MG Y

GY e.2

Modeled

although consideration

as

of non-

Transformers

Most mechanical drives can be represented as transformers with constant moduli, as long as frictional losses are neglected. Such drives can be based either on non-sliding friction (shear forces) or positive action (normal forces). Friction drives include belt drives, as suggested in Fig. 2.30, and rolling drives, as suggested in Fig. 2.31. Friction drives are inexpensive, and can damp vibrations. On the other hand, the transmitted forces are limited by the frictional properties of the materials.

60

CHAPTER 2.

SOURCE-LOAD SYNTHESIS

Figure 2.30: Examples of pulley drives

cylindrical

conical

Figure 2.31: Examples of roiling contact drives

2.4

IDEAL

MACHINES:

TRANSFORMERS

AND GYRATORS

61

(CourtesySeitz Corporation) (a) timing belt

(b) chain and sprocket

, Bostb-nGearWorks) (C) spur gear

(Courte~.GleasonWorks) (d) bevel gear

Figure 2.32: Examples of toothed drives

Teeth added to a belt gives a timing-belt drive, and a chain of links and pins substituted for the belt gives a chain-and-sprocket drive. Similarly, teeth added to the wheels of a rolling drive produce a gear drive. These positive-action drives can carry heavy loads and high power efficiently. Examples are shown in Fig. 2.32. All of them can be designed to operate very s~noothly, for example by using gear teeth with involute shape. Nevertheless, slight manufacturing errors or deflections due to heavy loads occasionally produce significant vibrations not included in a constant-modulus transformer model. The nominal kinematics of most toothed drives can be analyzed as though the teeth or sprockets weren’t there. The virtual surface of contact for a rotating member is known as the pitch surface. A cylindrical pitch surface possesses a pitch radius, r, and a pitch diameter. For a pulley, wheel or gear, the commonvelocity of the two couple.d me~nbers 1 and 2 at the point of contact is v = r~l = ro.q~_9, where ~ and ¢.., are the respective angular velocities. The

62

CHAPTER 2.

(ideal) transformer

SOURCE-LOAD SYNTHESIS

electrical transformer

(ideal) gyrator

Figure 2.33: Standard electric

circuit

symbols for transformers and gyrators

ratio of the angular velocities of the two pulleys or two gears etc. therefore is the inverse of the ratio of their pitch radii or diameters, i.e. T = ~/~ = rl/r.,..

2.4.5

Electrical

Transformers

An electrical transformer acts like an ideal transformer only for frequencies of ex~citation that are neither too small nor too large, and only if resistance losses are neglected. Whenunqualified, the term transformer refers to the ideal element asdiscussed above, and therfore does not include the real electrical transformer., Circuit diagrams, however, sometimes refer to idealized behavior that is equivalent to the (ideal) transformer. Electric circuit symbols for both the real electric transformer and the (ideal) transformer are given in Fig. 2.33. The non-ideal behavior of an electrical transformer is modeled in Chapter 10 by combining the (ideal) transformer with other elements. Purely fluid devices (fluid-in and fluid-out) that act like transformers without any moving mechanical parts are almost nonexistent. Nevertheless, devices with internal moving parts that behave like fluid transformers can be synthesized from two component devices that act like transducing transfomers or gyrators. An example is given in Section 2.5, after the idea of transducing transformers and gyrators is introduced. 2.4.6

Transducers

Modeled

as

Transformers

A two-port device that converts (or transduces) energy directly from one energy domain to another, without employing a thermal engine, is called a transducer. Electrical and fluid motors, actuators, generators, pumps, compressors and turbines are power transducers that convert substantial power from one domain to another. Some people reserve the word "transducer" for instrument transducers, such as microphones, tachometers, strain gages and instruments that measure temperature, pressure, acceleration, etc. These convert relatively little power into electrical form, since their function is to provide information. Sometransducers are transformational, while others are gyrational.

2.4

IDEAL

MACHINES:

TRANSFORMERS

AND GYRATORS

63

EXAMPLE 2.3 Show that the positive-displacement inechanical-to-fluid power transducer in the ibrm of a piston-and-cylinder or rain, as shown below, can be modeled as a transformer if leakage and friction are neglected. Find the modulus of the transformer in terms of physical constant(s).

\\\\\\ Solution: The volume flow rate of ~h~ ~uid, Q, ~quals ~h~ velocity shaft, ~, times the area of the piston, A:

of i~

Q= ~A. Further, side,

the input power on one side equals the output power on the other

F~ = PQ. Substitution gives F~ = (Q/~)P = AP. If the device is modeled as a transformer, its bond graph must be Fc

T~

The transformer modulus T is defined as the ratio Q]~, and therefore equals the area A. The transformer modulus also must equal the ratio F~/P, which has been verified. Therefore, the transformer model is proper. In the example above, the relation Fc = AP is derived from a velocity constraint combined with the conservation of energy. A simple force balance on the piston would ~lso give this result. Students who have t~ken a course in statics often tend to recognize this result more readily, in fact, than they recognize the velocity constraint. It is important to get in the habit of identifying such geometric constraints directly, however, for in more difficult situations, particularly when dynamics is involved, the force balances can be relatively awkward. In any case, use of the principle of the conservation of energy such as in this example precludes the need to perform both a force balance and a geometric constraint analysis. You can choose, or do both as a check. The power in a trausformer can flow in either direction, regardless of the orientation of the power convention half-arrow. The piston-cylinder device can be operated either as a pump or as an actuator (its most commonrole)~ Rotary limited-angle actuators, such as drawn in Fig. 2.34, also are co~nmon.They, for example, articulate the revolute joints of hydraulic robots. The term pump usually refers to machines that permit continuous rotation,

64

CHAPTER 2. PI

SOURCE-LOAD SYNTHESIS

AP = PI - P2

~

~u

iOnaryabutment M

and vanes

Figure 2.34: Rotary actuator

such as the gear pumpof Fig. 2.35 part (a), the vane pumpof part (b), or piston pumpof part (c). This type of machineis characterized geometrically by the volumeof fluid which passes through it per radian of shaft rotation, knownas the radian displacement of the machine, D. Its definition assumes no leakage flow between the inlet and outlet chambers. (The effects of modest leakage and friction on nominal positive displacement machines is considered in Chapter 4.) It also ignores the unsteadiness of the flow that exists in most pumps. With these idealizations,

Q = D~

(2.24)

A pumpapproximatedas an ideal machinealso has no friction, so powerout equals the powerin. Thus, if Apis the pressure rise from the fluid inlet to the fluid outlet, and Mis the torque on the shaft, then

M~ = Q AP.

(2.25)

2.4

IDEAL

MACHINES:

TRANSFORMERS

65

AND GYRATORS

internal ~eal formed here Unbalancedforces causedby pressureat the outlet

pump or (a) gear CourtesySauer-Danfoss Inc.

motor

~

igh pressure

LOwpressure inlet .......................

~

internal seal formed here

Rotations Shaf

(b) vane pump or motor ,n~_~.t

~

--

- 0.0u.~tlutlet

Casing. Vanes

Above armbelow:Reprintedwi¢h permissionand cou~esyof E~onCorporalion

Cylinder block

Spring force transmitting pin (3 spaced evenly)

(c) axial pis[on spring pump or motor

Shoeretainer plate (retracts pistons)

P ~’::::::~::::~:~ ~

Ports are .~ rotated 900/ flow

Valve~ plate / Cylinder hlnnk

~

\ Piston

\ Sho~

Figure 2.35: Common types of positive-displacement

~ Spherical washer

pumpsand motors

66

~ CHAPTER 2.

SOURCE-LOAD SYNTHESIS

EXAMPLE 2.4 Show that the idealized positive-displacement pump is a transformer. Determine the modulus of this transformer, and find the relation between the momenton the shaft and the pressure rise. Solution: Substituting equation (2.24) into equation (2.25) gives the sired relation between the momenton the shaft and the pressure rise: M = D AP. This result together with equation (2.24) defines the transformer 5I

~ T AP

The modulus of the transformer is the ratio Q/~, which equals the ratio M/AP, so that T = D, the volumetric displacement per radian. What happens to the purap if AP is negative, corresponding to a pressure drop instead of a rise? The equations above and the bond graph still apply, so a negative value of the torque Mresults. This implies that the fluid power is flowing into the machine, whereas the mechanical power is flowing out. A device operating this way is not usually called a pump; rather, it is called a hydraulic motor. Thus, barring any practical problems such as rubber seals being over-stressed and failing, a positive displacement pump run in reverse becomes a motor, and vice-versa, including the machines shown in Fig. 2.35. The transformer representation nicely represents this fact, for it is neutral on the question of which way the power flows. In practice, however, one might prefer to reverse the power arrows when use as a motor is intended, so that both AP and M become positive and T = lID. The pump/motors being considered are called positive displacement machines, since packets of fluid with fixed volume are conveyed mechanically from one port to the other. In the gear machine the packets are the fluid trapped between adjacent teeth. In the vane pump they are the fluid trapped between succesive vanes; the van~s slide in and out of slots to prevent leakage. In the piston pump the packets are the fluid volumes that enter and subsequently are expelled from each cylinder. The axial design pictured has several cylinders bored around a common barrel, shown cross-hatched, which rotate with the drive shaft like the barrel of a six-shooter. Pivotable piston shoes rotate with the pistons, and are also given a sinusoidal axial motion by sliding on a stationary angled cam pla~e. Whenthe volume of a cylinder is expanding, it is connected through a kidney-shaped slot in a fixed val~ving plate to the inlet port. Whenthe volume is contracting, it is connected through another kidneyshaped slot to the outlet port. (These inlet and output ports or lines actually are centered at angles rotated 90° about the machine axis from those pictured.) Positive-displacement hydraulic machines are widely used in industry and transportation because they handle vastly larger forces and powers than their

2.4

IDEAL

MACHINES:

TRANSFORMERS

F2 2

AND GYRATORS

x~

67

x

axes for xI andx,. are normal

Figure 2.36: Gyroscope with idealized

model

electromechanical counterparts (e.g. motors and solenoids) for a given size and weight. Their use comprises the major part of the area of engineering known as fluid power. The remainder of fluid power encompases pneumatics, which uses compressed air or other gases in similar linear and rotary machines. Dynamic hydraulic machines do not have fixed displacements, and are not represented as two-port elements until Chapter 9, since their behavior is more complex. Dynamic machines typically are used in relatively low pressure applications, varying from tiny impeller pumps and fans to enormous hydraulic turbines. In manycases, however, it suffices to represent a dynamic machine as a one-port fluid device, presuming a fixed speed or some other fixed condition for the rotor. Thus, an impeller pump was modeled in the last section as a one-port general source. 2.4.7

Mechanical

Devices

Modeled

as

Gyrators

The word "gyrator" may conjure up in your mind the word "gyroscope." This association is appropriate, for when stripped of its complexities, which are beyond our present interest, a gyroscope indeed exhibits gyrational coupling between two axes of rotation that are both normal to each other and normal to the principal axis of rotation. As Fig. 2.36 attempts to indicate, a torque FIL on one of these axes, where L is the length of the shaft, produces a proportional angular velocity ~.2/L on the other axis, and vice versa. As an example, a gyroscope with a horizontal shaft supported only at one end will not fall downif it is allowed to precess, that is rotate slowly about a vertical axis through the support point (see Problem 2.25, p 71). The strength of the gyrational coupling, that is the magnitude of the tnodulus G, equals the angular momentumof the rotation about the principal axis. As a result, any mechanical system with a significant angular ~nomentum has a potential for gyrational coupling. Such coupling can be enormous if the angular momentumis large. Nevertheless, most mechanical systems do not combine

68

CHAPTER 2.

SOURCE-LOAD SYNTHESIS

field circuit e...G

Figure 2.37: Idealized

DC motor and gyrator

M

model

large angular momentumwith a rotation of its axis, justifying the neglect of gyrational coupling. The electric circuit symbol for a gyrator is included in Fig. 2.33 (p. 62). No simple passive electrical element for low-frequency circuit use acts like a gyrator, however; the symbol is used mostly to represent idealized electric circuit analogies for mechanical systems. 2.4.8

Transducers

Modeled

as

Gyrators

The force on an electrically charged particle moving in a magnetic field is proportional to the charge, to the field and to the particle velocity. The force on an electrical conductor therefore is proportional to the electrical current and the field. In a simple DC electric motor, the armature wi)e is wound and the commutation so arranged that the force produces a torque on a shaft. This torque therefore is proportional both to the current in the armature circuit and to the magnetic field. If, further, the magnetic field is constant, as in a permanent magnet motor or a motor with a field circuit driven by a fixed voltage or current, the torque is simply proportional to the armature current. Thus, as suggested in Fig. 2.37, M = Gi, e = G~,

(2.26a) (2.26b)

where G is a constant dependent on the geometry of the motor. Equation (2.02b) follows from the assumed conservation of energy. It directly suggests the idea of the tachometer or speedometer, an instrument in which the angular velocity of a shaft produces a proportional voltage which can be displayed, with an appropriate conversion factor, by a voltmeter. A large tachmeter qualifies as an electric generator. Thus, the electric motor and the electric generator are the same machine simply operated backwards, much like the hydraulic motor and the hydraulic pump. A translating coil device, such as that examined in Guided Problem 2.6, is a motor with translational rather than rotational motion..It also can be modeled as a gyrator. There are many similar electromechanical devices based on magnetic fields, such as solenoids, although most are complicated by varying energy storage and so are addressed later (in Chapter 10). On the other hand,

2.4

IDEAL

MACHINES:

TRANSFORMERS

AND GYRATORS

69

the class of electromechanical devices based on electric fields rather than magnetic fields can be shown to give transformational coupling. Examples include capacitor microphones and piezoelectric transducers; detailed consideration is again deferred to Chapter 10 because of complications due to energy storage. Transducing gyrators also are useful in modeling dynamic fluid machines. Their modeling in this manner goes beyond the scope of this text. Simpler representations are given in Chapter 9. 2.4.9

Summary

Ideal two-port machines are by definition conservative, reversible and without stored energy. The two simplest forms are the transformer and the gyrator. Many actual machines can be approximately represented by these forms, and rnore complicated models of these and other devices often have transformers or gyrators imbedded within them. You will examine how these elements can be combined with one another and with other elements, such as the sources and loads already introduced, beginning in the following section. The output flow of a transformer equals the product of a modulus, T, and the input flow. The input effort equals the product of the same T and the output effort. The effort on each of the two ports of a gyrator equals the product of a commonmodulus G and the flow of the opposite port. Engineering approximations to the transformer include pulley and gear pairs, electrical transformers, hydraulic rams and hydraulic pump/motors. Engineering approximations to the gyrator include gyroscopes, DC motors and other moving coil devices. Guided

Problem

2.5

This problem and the next should enhance your understanding of the nature and application of transformers and gyrators by having you evaluate the moduli of particular devices modeled as ideal machines, and examine their bilateral behavior. Evaluate the transformer modulus for the rotary actuator shown in Fig. 2.34. The depth of the working chamber may be taken as w, and other parameters may be defined as appropriate. Suggested

Steps:

1. Define with symbols the radius of the shaft and the radius of the vanes/ inner housing, and the depth of the vanes and chamber. 2. Choose whether to relate ¢~ to Q (easier) or 7~/ to Ap, and evaluate T as their ratio in terms of (fixed) parameters. 3. Check to see that the relation 2.

not chosen agrees with the modulus of step

70

CHAPTER 2.

Figure 2.38: The translating

Guided

Problem

SOURCE-LOAD

SYNTHESIS

coil device of Guided Problem 2.6

2.6

A translating coil device comprises a coil wrapped on a tube and positioned in a radial magnetic field. The example illustrated in Fig. 2.38 employs X permanent magnet, as in a loudspeaker. The axial force on the coil is F = 27rrNBi where 2~rr is the circumference of the coil, N is the number of turns, B is the magnetic field strength and i is the electric current. Model this device as a form of an ideal transducer, neglecting losses. Give its modulus and the relation between the voltage e and the velocity Suggested

Steps:

1. Identify the power conjugate variables on the mechanical and electrical bonds which are related by this device. 2. Identify the type of ideal machine and its modulus from the given equation; connect the two bonds with the associated standard bond-graph symbol. 3. Find the relation between e and 2 from this bond graph. Note that it follows from the conservation of energy. PROBLEMS 2.19 Identify a set of variables for a transformer model of the pulley drive systems of Fig. 2.30 (p. 60), define the transformer modulus, and express this modulus in terms of physical constants that also should be defined. 2.20 Answer the question of the preceding problem for the cylindrical drive shownon the left side of Fig. 2.31 (p. 60).

roller

2.4

IDEAL

MACHINES:

TRANSFORMERS

AND GYRATORS

71

2.21 Answerthe question of the preceding two problems for the bevel or conical roller drive system shownon the right side of Fig. 2.31 (p. 60). 2.22 You are given a hydraulic rotary actuator such as pictured in Fig. 2.34 (p. 64), but are not given its specifications. Propose and describe an experiment that could determine the modulus of the transformer that models the actuator. 2.23 Find the transformer ~nodulus for a steady-flow model of the piston pump/ motor shown in Fig. 2.35 part (c) (p. 65). You Inay define the radius of piston as r~, the distance between the machine centerline and a piston centerline as vb, the number of pistons as n a~d the angle of the cam plate from the normal to the axis as 8. 2.24 A tachometer produces 1 volt for each 10 revolutions per second. If the machine is used instead as a motor, find the current needed to produce a torque of 0.02 N.m. Assumethe behavior is consistent with that of an ideal machine, and note whether it is a transformer or a gyrator. 2.25 The shaft of a rapidly spinning rotor (or gyroscope) is horizontal and supported at a distance L from its center of mass. The axis of the shaft is observed to precess about a vertical axis at a steady rate ~. The axis remains horizontal. The rotor has mass m and spins at w rad/s. Knowing that this device can be represented by a gyrator with modulus Iw, where I = mr~ is the mass momentof inertia and r~ is the radius of gyration, determine the rate of precession. 2.26 A bicyclist negotiates a turn with a 20-foot radius at 24 feet per second. Each of the wheels weighs five pounds, has a radius of 1.1 feet and has a radius of gyration of 1.0 feet.

(a) The gyrator modulus.connecting the vertical axis of the turn with the axis of tilt equals the angular momentumof the two wheels. Determine this modulus. Neglect the angle of tilt.

(b) Determine the moment produced by the rotation of the wheels that partly counteracts the centrifugally-induced tilting moment.

72

CHAPTER 2.

SOURCE-LOAD

SYNTHESIS

SOLUTIONS TO GUIDED PROBLEMS Guided

Problem

2.5

1. Radius of shaft: r~; radius of vanes or inner housing: r~,; depth of the vanes and chamber: w. 2. The volumeof one chamberis ~r(r~ -r~)(¢/2~r)w. Therefore, the rate of change of the voluine of one chamberis w(r~ -r~)@2. The flow Q enters two of the four chambers (and leaves the other two), so Q = (1/T)~ with lIT = w(r~ - r’2s) w(r~, - r~)(r,, + r~). 3. The relation M = (1/T)AP can be checked most easily in the limit when r,, - r~ = Ar is muchsmaller than either r,, or r~. The result is M = 2rF where r = (r~ + r,)/2 and F = ApAr, which can be seen to be correct. Guided

Problem

2.6

1. Mechanicalforce and velocity: F and ~?, respectively. Electrical generalized force and velocity: e and i, respectively. e F 2. ~ G ~ From the

given

equation,

G -= 2~rrNB.

Therefore,

e = 2rrrNB k.

2.5

Systems Gyrators

with

Transformers

and

Sections 2.1 and 2.3 describe how to predict the equilibria of sources attached to loads. Section 2.4 introduces the transformer and the gyrator. Howtransformers and gyrators act when cascaded head-to-tail, and how they can be placed between a source and load to improve the behavior, is now investigated.

2.5.1

Cascaded Transformers

A pair of transformers can be cascaded: e2 e 1 ~ T1 .----2-’-~-

T

e~

2

The definition of a transformer requires 03 = T2g12 = ~T~o~,

(2.27a)

el = T~e: = T~e3,

(2.27b)

which says simply that the cascade reduces to a single effective transformer with modulus equal to the product of the component moduli:

e~ T O~

~ q3

T = TIT~

2.5.

SYSTEMS

WITH TRANSFORMERS

AND GYRATORS

73

Thus, if n transformers are cascaded, the result is a single effective transformer with modulus equal to the product of M1 the component moduli: 0n+I = T(~I,

(2.28a)

el = Te~+l,

(2.28b)

T=T1T,., ... T,.

(2.28c)

EXAMPLE 2.5 Showthat gear trains, such as those pictured in below, can be represented as a cascade of transformers if friction is neglected, and that they act as if there were only a single gear pair.

Solution: The transformer modulus for each pair of meshing gears equals the ratio of their numbersof teeth or of their diameters, as noted in Section 2.4.4. The overall train acts like a single gear pair, or single transformer, with a transformer modulus or gear ratio equal to the product of all the individual transformer moduli, that is all the component gear ratios. The inverse torque ratio has the same value only if friction is neglected; the model is invalid otherwise. 2.5.2

Cascaded

Gyrators

Cascading two gyrators produces a transformer with a modulus that equals the ratio of the two gyrator moduli. This result can be deduced step-by-step as follows: e2 ~1

q2

~3

74

CHAPTER 2.

SOURCE-LOAD SYNTHESIS

Cascading a gyrator with a transformer, however, gives a gyrator. (You should take a few momentsto demonstrate this fact for yourself.) As a result, the cascading of any even number of gyrators produces a transformer, while the cascading of any odd numberof gyrators produces a gyrator. EXAMPLE 2.6 Showthat if the two shafts of ideal DCelectric motorswith constant fields are connectedtogether (belowleft), the voltages and currents of the twopairs of electrical terminalsare related like those of an ideal electrical transformer. In addition, showthat if the twopairs of electrical leads are interconnected instead (below right), the two shafts have torques and speeds related like those of a transformer.

Solution: Anideal DCelectric motorwith constant magnetic field has been represented as a gyrator. Equation (2.29) showsthat connecting together the shafts of two such motors produces a transformer: el

i~

M ~G~ ~ ~G2

e~ .-~--’~ ~"

e2 e~ T--.---~--~:---~. II

:2

12

T=~G~ G2

This device has an important advantage over the usual electrical transformer: it works at DC. For the second case, the result is a mechanical transmission knownas a Ward-Leonardsystem: M1

e

~

¢,~ G,-7-~ a2 ~

~

M1

-~

r~

G~ r=

Energylosses are addressedin Section 4.2.4. 2.5.3

Case Study to a Load

of a Transformer

Connecting

a Source

Transformersroutinely interconnect sources and loads. Transducingtransformers do this intrinsincally. Non-transducingtransformers often are chosenexplicitly to improve the matching of the source to the load. The key ideas perhaps are best learned by a familiar example. Consider an automobileIC engine with a fixed throttle position whichdrives a vehicle through a fixed transmission along a level road. The engine can be modeledas a source; the torque-speedcharacteristics are plotted on the left side of Fig. 2.39. The force-speed characteristics of the load are plotted on the right

2.5.

75

SYSTEMS WITH TRANSFORMERS AND GYRATORS

300

//

150 engine torque, 100 M

~ ~

¢

ft-lbs 50

R,’N \

vehicle drag,

/.~ °

200

~, F ~/~ \ equilibriumS{

/

equilibriumfl~ lbs.

/\

~/

/ \ lOO .------.--z

..--" .... -" ..........

0

"1 i I I

0

100 200 300 400 speedof drive shaft, &, rad/s

0

50 100 150 vehiclespeed,k, ft/s

Figure 2.39: Automobilesource and load characteristics with synthesis side of the figure. (This force is the sumof the air drag and the rolling drag.) The drive train between the engine and drive axle comprises a transmission, drive shaft and gear differential. This drive train converts engine torque and speed to vehicle thrust force and speed. Losses in this system are neglected here, so these componentscan be modeled,collectively, as a transformer, shown as Td below: F M ENGINE~ DRIVETRAIN ~ WHEELS -~-=7--.--.--.--.-~-. VEHICLE X

M S~M

F T

F~.

R

The drive axles in turn rotate the wheels. This action is modeledby the transformer Tw.The modulusof this transformer represents the ratio of the linear motionof the vehicle to the rotary motionof the wheels, whichis the radius of the wheels. The two cascaded transformers can be telescoped into one, labeled as T. The modulusof this combinedtransformer equals the ratio of the velocity of the vehicle to the angular velocity of the engine shaft, or the distance moved per radian of rotation of the shaft. Assumethe vehicle is knownto movetwo feet for every rotation of the shaft; therefore, T = 2/2rr = 1/~r ft/rad. To determine the equilibirum speeds of the vehicle and the engine, you need either to transformthe load curve into the torque-speedcoordinates of the engine plot, or to transform the engine characteristics into the force-velocity characteristics of the vehicle. Both of these transformations are shownin the figure,

76

CHAPTER 2.

~

500 vehicle drag,

\\\locus -. \maximum

F 400 lbs. /

equilib~F2~ "\locus.of

341 lbs

ft-lbs 5O / 0

~

q P (but opposite sign) 100 200 300 400 speed of drive shaft, ~b, rad/s

/

300

M 100

0

load for 10%grad~

0.1

3500 lbs

150 engine torque,

SOURCE-LOAD SYNTHESIS

0

equiliy~f/ J load for /level road

50 100 150 vehicle speed, ~, ft/s

Figure 2.40: Calculation of the transformer modulus for maximumpower transmission using dashed lines. In the first case, the transformer and the load are effectively combined into an equivalent load, labeled as R’. In the second case, the engine and the transformer are effectively combined into an equivalent source, labeled as S’. The transformations are carried out by choosi.ng individual points along the characteristic to be transformed. The relations ¢ = (1/T)Sc and M = (first case) or :b = Tq~ and F (1/T)M (s econd case) ar e us ed. The equilibrium state is given by the intersections of the characteristics. The two approaches give the same answer, as they should: & = 123 if/s, F = 226 lbs, ~ = 386 tad/s, M= 72 ft-lb. The transmission ratio of a real car changes when you shift gears to address a changed loading condition. Presume you wish to find the drive train ratio Td which maximizes the steady-state speed of the vehicle for a given load characteristic and a given radius of the wheels. You can reason as follows: maximum load speed means maximumload power. Since the transmission is assumed to be lossless, this means maximumengine power. Thus, you start by finding the point on the engine characteristic which gives the greatest power. The power is the product of the torque and the speed; you can simply use trial-and-error. Better yet, you can use one of the methods described in Section 2.3.6, which are implemented on the left-hand side of Fig. 2.40 by dashed lines. Note specifically that the rectangular hyperbola representing maximumpower is tangent to the

2.5.

77

SYSTEMS WITH TRANSFORMERS AND GYRATORS

characteristic at the maximum powerpoint, and that the slope of the chord from the origin to this point has the sameslope as the tangent to the characteristic at this point. The maximum power is approximately T) = M,n~m= 111.5 ft.lb x304rad/s = 33,900 ft.lb/s. The locus for maximum power can be transformed over into the right-hand plot, whereits intersection with the load characteristic gives the corresponding maximumspeed and force: approximately ~ = 132.4 ft/s and F1 = 256 lbs. The transformer ratio which permits this to occur is therefore T = k~/~l = Mm/FI= 0.4355 ft/rad. (Clearly, you’ve madea mistake if both calculations don’t give you the sameanswer.) Finally, the required transmission ratio Td is the ratio of T/T,., where Twis the radius of the wheel. For example, if this radius is 1.1 feet, there follows Td = 0.396. (This drive train ratio in turn is factored into the ratio for the transmissionand the ratio for the gear differential) Achieving maximum speed on an incline requires a lower transmission ratio (down-shifting). Assumethat the vehicle and its payload weighs 3500lbs., and is climbing a long 10%grade, as shownin the figure. The weight times the sine of the angle gives an additional force of 348 lb of drag. This is added to the air drag and wheel drag to give a newload characteristic and a new solution4: T = :~2/~,,,~ = 73.5/a04 = 0.242 ft/rad or T = lt.lm/F~ = 111.5/46~ = 0.242 ft/rad. 2.5.4

Second Source

Case Study to a Load

of a Transformer

Connecting

a

The matching of a source and a load often is improvedthrough the purposeful introduction of a coupling that acts like a transformer, as the gears in the transmission of the case-study vehicle aboveillustrate. As a secondcase study, consider inserting an ideal pulley or gear drive betweenthe induction motorand the pumpof the water sprinkler system discussed earlier in Section 2.1. The newsystem, shownin Fig. 2.41, can be represented by the bond graph S

M_~_~_~T A/d R

Theobjective is acceptableoperation for the twentieth and thirtieth-floor sprinklers, for whichthe direct drive failed, without using a larger motor. The angular velocity and the torque of the ideal pulley drive are q~ = T~,,, 1 -Aid = ~Mra,

(2.303) (2.305)

4The solution as presented overlooks a small correction for the case of the 10%slope that results frown the fact that the normal load on the pavement is reduced by about one-half percent below the weight. Since about 40 pounds of the load force is rolling friction which would be approximately proportional to this force, the drag should be reduced by about 0.2 lbs.

78

CHAPTER 2.

SOURCE-LOAD SYNTHESIS

Figure 2.41: Addition of pulley drive to the induction motor system ~ locus

for maximum power S’

~

3.0

M~, N’mF t

o

2.0I 0

\~,,~ f’\

,

50

~ locus for ~ maximum

100

one-half power

,io.8200

150

250

~e, rad/s Figure 2.42: Characteristics the pulley drive

of the source combining the induction motor and

where T is the ratio of the diameter of the motor pulley to the diameter of the pump pulley. The motor and the pulley system can be considered as a single equivalent source, S’~Md

R

/An acceptable alternative combines the pulley system and the pump system into an equivalent one-port load.) The torque-speed characteristic of this new source depends on the value of T. Characteristics for four different values are plotted in Fig. 2.42. This is muchlike the choice you have in shifting gears of an automobile with a ~nanual transmission. A large value of T (as ~vith "fourth gear") gives a high speed at low torque, whereas a small value of T (as with "first gear") gives low speed at high torque, all for the same motor speed and torque. The characteristic for T = 1 is identical to that of the motor itself; the ideal pulley acts like a direct drive. The other characteristics are transformations of this basic characteristic. Each point on the basic characteristic is mappedonto

2.5.

SYSTEMS

WITH TRANSFORMERS _x locus

t 3.0~ Md’

|

N.m

/ /

79

AND GYRATORS

for maximumpower

~~//circles (.) show practical operating points ~\ for systems permitting reliable start-up jthirtieth floor ~ I xx Ma

1.3 0

0

50

100

150

200

250

rad/s Figure 2.43: Matching of motor/pulley system to pump/sprinkler

load

a correspond!ng point on the transformed characteristic according to equation (2.34)..The power.associated with the point is unchanged by the transformation; i.e. MdCa= M,,¢,~. The transformed point lies on a locus of constant power that passes through the original point. This locus is a hyperbola with vertical and horizontal asymptotes, the same rectilinear hyperbola discussed in the prior example. Twosuch loci are drawn in Fig. 2.42 as dashed lines. Rapid sketching is aided by the fact, as noted above, that the rnagnitude of the negative slope at each point on the hyperbola equals the positive slope of the chord from the origin to that point. The locus for maximumpower just graces each characteristic. All of these points of tangency represent the same maximumpower the motor can deliver. The various speed ratios and torque ratios equal the transformer moduli or their reciprocals. The other locus drawn in the figure represents one-half the maximumpower. This locus intersects each source characteristic twice. The respective lower-speed intersections have the same speed and torque ratios as both the upper-speed intersections and the maximumpower points, It is strongly suggested that you choose some arbitrary point or points on one of the characteristics and, using only knowledge of the values of T, find and validate the corresponding point or points on another characteristic. The largest possible load flow corresponds to the largest possible load power, which in turn equals the largest possible source power. The maximumpower that can be delivered by the source is indicated in Fig. 2.42 by the locus of maximumpower. In Fig. 2.43 this locus is seen to intersect the load characteristic for the thirtieth floor whenT = 0.81 For the twentieth floor, the intersection is close to T = 1, and for the tenth floor it is about T = 1.1 (interpolating). Are these solutions acceptable? The candidate design solution with T = 1 for the twentieth floor already

80

CHAPTER 2.

SOURCE-LOAD SYNTHESIS

has been eliminated, since unfortunately the pump won’t start. The solutions T = 0.8 and 0.5 for the thirtieth floor suffer the same problem. For T = 1.1, the starting torques for the motor and pump are about equal; starting is too problematical to justify the choice. Thus, unless you were to replace the motor, for example using the "capacitor-start". feature highlighted in Guided Problem 2.2, you must settle for a lower output flow and power in all these cases. Allowing about a 10%margin to cover errors inherent in the modeling, including friction which is being neglected, the solution for T = 0.5 is seen to be acceptable for the twentieth floor. For the tenth floor, both T -- 0.8 and 1.0 are acceptable; the latter gives virtually maximumpossible flow. The acceptable solutions are marked by heavy dots in the figure.

EXAMPLE 2.7 A DCmotor drives a frictionless and leak-free positive displacement pump which forces a liquid through a flow restriction to a load: fluid line restriction The motor has the torque-speed characteristics plotted below left, and the pressure drop across the restriction is plotted below right as a function of the flow through it.

moment,

~mr

1

8 4

pressure

I’ ~ ’ ’ ’ ’ ’ ~l~ ’

drop, lb/in2~

1°°° f /’... ~

0

500 I000 speed, rad/s

~

~.~ restriction-

0 1 2 3 4 5 6 flow, in3/s

Assuming thepressure of theloadis zero,findthedisplacement of thepump that(approximately) maxilnizes theflowto theload,~d theassociated fl0w. Then,repeatthe solution whentheloadpressure is knownto be 1000psi. Solution:Maximizing the flow to the load impliesm~imizing the power transfer. Sincethe pumpis 100%efficient, thismeansmaximizing the power from the motor.M~imumpowerfrom the motor occursat the mid-point of thestraight-line characteristic, as discussed in Section 2.3.6andshown specifically in Fig.2.27.Thusthemotorshouldbe run at 500rad/switha torqueof 7 in..ib, producing thepower500x 7 = 3500in..Ib/s. A locuswith thispoweris superimposed on the pressure dropplotbelowleft,revealing an intersection ~t a pressure of 1000psiandflowof 3.5 in3/s.Thisis the solution whentheloadpressure is zero,so thatthe pressure seenby the pumpequalsthepressure dropacrossthefluidrestriction.

2.5.

SYSTEMS WITH TRANSFORMERS AND GYRATORS

81

~ocus of maximumpower pressure ~, ,, ,\. , . ,~..~ drop, lb/in 2 ~ ~ /" ~ ~ .~ restriction~ 0

1

234 56 flow,in3/s

The pumpacts as a transformer, as shownabove right. Its volumetric displacement is the modulusof the transformer, which can be deduced from the ratio 3.5/500 = 0.007 in3/rad or the ratio 7/1000 = 0.007 in3/rad. The pressure generally seen by the load is the sumof the pressure drop across the restriction and the load pressure. This sumis plotted belowfor the secondcase of a 1000psi load pressure. \ / solution pressure, ~’t~t~l’~’~ ’4 lb/in21000 lO0~O~psi~., t ~ ..~restriction~ 0123456 flow,ina/s The intersection with the same power locus as before is nowat 1400 psi and 2.5 in3/s. The modulusof the transformer that produces this result is the volumetric displacement 2.5/500 -- 0.005 in3/rad or 7/1400 = 0.005 in3/rad. 2.5.5

Case Study of a Gyrator Load

Connecting

a Source

to a

Consider a DCmotor with constant field and momentM = ai, a = 0.15 N m/amp,which drives a load. The electric power source for the motor has the voltage-current characteristic shownon the left side of Fig. 2.44, and the mechanical load has the torque-speed characterisitic shownon the right side. You wish to find the equilibrium torque and speed, neglecting losses. The system can be modeled by the bond graph S

e~G ~

M ----U

The given information M= ai establishes the gyrator modulusas a, so you also knowthat e = a~. Youcould combinethe source and the gyrator into an equivalent source, so

82

CHAPTER 2. SOURCE-LOAD SYNTHESIS S

~

G

~R

0.~kS’ dfor G =0.15 N.m/am N.m

, .~M 2O

k,,,,,,

volts " 10 0

0

~

1

0.4

s

[

\x~\

S

"~quilibrium

2 3 i, ~ps

0

4

40

80 120 ~, rad/s

160

Figure 2.44: Source and load characteristics for DCmotor system that the source characteristic is mappedinto the load plot: S’

M

R

¢

Alternatively, you could combinethe gyrator and the load into an equivalent load, so that the load characteristic is mappedinto the source plot: S

e

~

i

R’

The source characteristic is a straight line in the present case, so the first approach is easier. The mappingis shownin the figure by the dashedline labeled S~. Note that the left end of the source characteristic becomesthe right end of its transformation, and vice-versa. Younowsimply identify the intersection of the two characteristics as the equilibrium point.

2.5.6

Transmission Matrices*

The use of matrix notation expedites the telescoping of a chain of linear twoport elements into a single equivalent two-port element. This idea is introduced here for the special cases involving only transformers and gyrators. Further use of the methodis given in Chapter 4 and beyond. The state of a bondis treated as a two-elementvector, with its effort first and its flow second. The equations describing a transformer of modulusT and a gyrator of modulus G become

2.5.

SYSTEMS WITH TRANSFORMERS AND GYRATORS

83

The cascading of two transformers, as in Section 2.5.1, -el ~T1 ..~__~ 2 ~.’--’-7 -

e2 T q2

__ 01

ez q3

gives

l/TIT2 (t3 01 = T1T2 03 Denotingthe transmission matrix of the ith element in the cascade of n transformers as Ti, there results 01 =T~T~ ...T, Jell

[O~+~J’ 0, =[~ 0 ]lIT[en+l] [en] . T:TIT9 """ rn,

(2.33)

whichagrees with equation (2.28). The cascade of two gyrators, qe

01

q3

gives

° oll

G2

l/G2

which agrees with equation (2.29). 2.5.7

Summary

Transformerscascaded head-to-tail can be represented by a single transformer with modulusequal to the product of the individual moduli. A pair of gyrators in series, on the other hand, is equivalent to a transformer with a modulusequal to the ratio of the individual moduli. A gyrator connected to a transformer gives a gyrator. This meansthat an even numberof gyrators in cascade gives an equivalent transformer, while an odd numberof gyrators in cascade gives an equivalent gyrator. Modelsoften interconnect a source to a load by a transformer or a gyrator. The source then can be combinedwith the transformer or gyrator to give an equivalent source, which can be directly synthesized with the load. Alternatively, the transformer or gyrator can be combinedwith the load to give an equivalent load, which can be directly synthesized with the source. Graphical representations of the characteristics enables you to visualize the solutions of a

84

CHAPTER 2.

SOURCE-LOAD SYNTHESIS

variety of problems, such as determination of the moduli which give maximum powertransfer from the source to the load. Matrix relations can be used to represent the characteristics of linear twoport elements such as transformers and gyrators. This approach is helpful but not necessary. Guided

Problem

2.7

Solving this problem and the next should .help you understand howto choose the modulusof a coupler so as to maximizethe performanceof an electrical, mechanicalor fluid system. Choosethe modulusof an electric transformer which connects an amplifier with a 24-ohmresistive output impedanceto a speaker with a 6-ohmresistive impedanceso as to maximizethe powertransfer. Suggested Steps: 1. Represent the source, transformer and load with a bond graph. Find a relation to characterize the source. 2. Combinethe transfor~ner and the load and draw a new bond graph. Evaluate the modulusof the new load. 3. Express the powerdelivered in terms of the bond Current and the parameters. 4. Find the current that maximizesthis power. 5. Find the transformer modulusthat gives this maximizedpower. Guided Problem 2.8 A hydraulic power supply drives a hydraulic ram (piston/cylinder) against mechanical load. The characteristics of the power supply and the load are plotted in Fig. 2.45. Find the area of the piston which maximizesthe power delivered to the load. Neglectfriction. Suggested Steps: 1. Drawa bond graph of the system. powerpoint on the source characteristic. 2. Find the maximum 3. Find the correspondingpoint on the load characteristic. 4. The transformer modulus sought is the ratio ~/Q of these two points. Relate this ratio to the area of the piston. 5. Checkthat the transformer modulusis also the ratio P/F for these two points.

2.5.

85

SYSTEMS WITH TRANSFORMERS AND GYRATORS

[ nower I tube "40013 ¯ P, psi 30013

80,000 F, lbs 50,000

2000

40,000

1000

20,000

0

I

0

I

0

20

30 40 Q, in3/s

I

0

I

I

0.5

I

I

1.0 1.5 .i:, in./s

I

I

2.0

Figure 2.45: The system of Guided Problem 2.8

PROBLEMS 2.27 Modelthe gear train below as a cascade of transformers. Find the moduli of the componenttransformers and the overall equivalent single transformer. out

i

201

28

15i

numbersof gear teeth

]] !

I

12 i

! I

.1

2,4 I

q~in

2.28 Find the relationships betweenthe boundaryvariables for the following cascade. Canthis systembe represented by an equivalent single transformer or gyrator, and, if so, what wouldits modulusbe? el~T

~2 q2

G ~3 q3

2.29 Carry out the previous problem using the methodof transmission matrices.

86

CHAPTER 2.

SOURCE-LOAD SYNTHESIS

2.30 The vehicle considered in Section 2.5.3 (which has T = 1/~r ft/rad and mg = 3500 lbs) is traveling at 100 ft/s on a level road. Using only the data given by the solid lines in Fig. 2.39 (p. 75), (a) find the drag force. (b) find the engine speed, moment and power. (c) find the thrust force (from the engine power and vehicle speed). (d) find the acceleration of the vehicle.

2.31 The capacitor-start motor described in Guided Problem 2.2 (p. 26) substituted for the basic induction motor used in the system with the motor, pulley drive and pump load analyzed in Section 2.5.4. Determine which of the candidate solutions T = 0.5, 0.8, 1.0 and 1.3 are acceptable for the tenth, twentieth and thirtieth floor sprinklers. 2.32 The analysis of the system with the motor, pulley drive and pump load given in Section 2.5.4 employs an equivalent source bonded to the actual load: St

Md

Draw an annotated sketch of the replacement for Fig. 2.43 (p. 79) that corresponds to the bonding of the actual motor to an equivalent load: S M,.

~ R’

2.33 An ideal hydraulic power supply (comprising a motor, pump and relief valve) has the pressure-flow characteristic as plotted below left. It drives piston-cylinder with the characteristic shown below right. Find the maximum possible speed of the piston, and the area of the piston that results in this speed. Neglect friction. 6000 force, lbs 40O0

2500 pressure, psi

2000 0

0

30 in3/s flow

0

0

6 12 18 velocity, in./s

2.5.

SYSTEMS WITH TRANSFORMERS AND GYRATORS

87

2.34 A motor drives a fan through a belt drive. The torque-speed characteristics of the motorand the fan are plotted below.The pulley ratio is not specified, but the motorshould not be run with a torque higher than its continuousrating of 4.0 in.-lb (to prevent overheating).

12 torque, in.. 1

0

400

800 1200 1600 speed, rpm

(a) Drawa bond graph model of the system, neglecting energy storage elements. (b) Determinethe speed of the motor that maximizesits Mlowablepower, and find this power. (c) Determine the corresponding speed and torque of the fan, and the pulley ratio that achieves them. Neglectbelt losses, so poweris conserved. 2.35 An electrical power supply drives a DCmotor that in turn drives a rotational load. The powersupply is regulated to give constant voltage up to a limiting current, as plotted below. The motor has a constant magnetic field. The load has the torque-speed characteristic, also plotted below. 3.01 e,

i

,

,

i

i

24~~--~~~

i, amps

o

0

100

200 ~, rad/s

300

(a) Drawa bond graph modelfor the system. Neglect losses in the motor. (b) Determinethe maxirnumspeed that the load can achieve.

88

CHAPTER 2.

SOURCE-LOAD

SYNTHESIS

(c) Evaluate the modulus that describes the basic property of the motor.

2.36 A pumpwith the characteristics given in Fig. 2.45 (p. 85) drives a rotary hydraulic motor with volumetric displacement D per radian, in place of the piston load shown in the figure. The load torque on the shaft of the motor is M = a~ where a = 100 in..lb s. Draw a bond graph model, and find the value of D that maximizes the speed of the load. 2.37 Consider the system studied in Section 2.5.5 which comprises an electrical source, an ideal DC motor with a constant magnetic field and a particular mechanical load with characteristics given in Fig. 2.44 (p. 82). In the present case, the modulus a is not specified. (a) Determine the maximumpower that can be transmitted

to the load.

(b) Determine the value of a that accomplishes this transmission.

2.38 The armature circuit of an ideal DC machine with constant magnetic field is completed with an electrical resistance Re, as shown below. The machine is characterized by the relation M= ai, and is otherwise ideal. Show that the system acts like like a linear rotary mechanical dashpot, and determine its mechanical resistance, Rm.

2.39 Repeat Problein 2.9 (p. 54) using the reticulated F ~T

P~R

~

Q

2.40 Repeat Problem 2.10 (p. 54) using the reticulated above. SOLUTIONS

TO

Guided Problem

2.7

1. S e = RLT’~ir T i 2.

S e~.

R~;

GUIDED

RLTi rt~L Ti 2R~ = RLT

model

model of the problem

PROBLEMS

e

=eo--R,~i;

.

R~ = 24 Q

2.5.

SYSTEMS

WITH TRANSFORMERS

AND GYRATORS

3. 79 = ei = (eo - R,~i)i d79 eo 4. ~ = 0 = eo - 2R,~i. Therefore, i = ~R-’~’~’ e = eo/2. 5. e = RLli = RLT2i. Solving for T and using the above, = (eo/2R,~)RL Guided

Problem

1.

S--~ 2,3. 4000 7 P, psi 3000

=

2.8

T

---~--.-R F

~’’~imum power’~ max

80,000 F, lbs 60,000 locus

2000

40,0013

1000

20,000

00

-- 2.0.

10

20

30 40 Q,in-’/s

of maximum~

00 0:5 l:0

-r-- .¢~ = 1.69 4. *--~= 0.050 in2. The piston area is l/T= 20.1 in-’. 5.

1/(pistonarea) = P/F, also.

’ 1:5 ~ 210 Y¢,in./s

89

Chapter 3

Simple

Dynamic Models

The systems considered thus far respond instantaneously with a new state whenever their environment or boundary conditions change. Whena real physical system is subjected to a sudden change in boundary conditions, however, it might in fact respond noticeably slowly. The variables which define its state would approach their new equilibrium either monatonically (non-oscillatorily) or oscillatorily, assuming stability. Such systems, or the models that represent them, are called dynamic. Dynamic physical systems contain mechanisms that store energy temporarily, for later release. The dynamics can be thought of as a sloshing of energy between different energy storage mechanisms, and/or a gradual dissipation of energy in resistances of the type you have examined already. Section 3.1 starts the subject by introducing a major class of energy storage mechanisms called compliances. Section 3.2 continues by introducing the other major category of energy storage mechanisms, called inertances. This chapter makes a first pass through the major ideas in the modeling and analysis of dynamic systems. To accomplish this, attention is restricted to simple models comprising, at most, a source, a resistance, a compliance and an inertance joined together by a junction. Junctions are virtually the only elements that can interconnect three or more bonds, and therefore become the very heart of the structure of a bond graph. They are introduced in Section 3.3. Dynamic models are represented mathematically by differential equations, as contrasted to static or steady-state models that are represented mathematically by algebraic equations. The conversion of the bond graph models to differential equations in presented in Section 3.4. The behavior of the models, as represented by the solutions to the differential equations, is the subject of Section 3.5. Nonlinear compliances and inertances, previously omitted, are introduced in Section 3.6. This leads, finally, to numerical simulation in Section 3.7. 91

92

CHAPTER 3. SIMPLE DYNAMIC MODELS

F

Figure 3.1: Linear spring and spring characteristic

3.1

Compliance

Energy

Storage

Elementsthat store energy by virtue of a generalized displacement are called compliances. Physical approximations include mechanical compliances (e.g. springs), fluid compliances(e.g. gravity tanks) and electrical compliances(e.g. capacitors), amongothers. Compliancesare designated in bond graphs by the capital letter C. 3.1.1

Linear

Springs

and Energy

The mechanical spring is a familiar example of a one-port element that is totally different from the sources, sinks and resistances consideredin Chapter2. As illustrated in Fig. 3.1, the spring is characterized by a relationship between its generalizedforce and its generalized displacement,rather than its velocity: F

~ SPRING

F = kx. (3.1) The coefficient k is knownvariously as the spring rate, the spring constant or the spring stiffness. Note that in equation (3.1), x is defined so as to equal zero whenthe spring force is zero. Springs do not dissipate energy, as resistances do; they store energy. This can be shownexplicitly by integrating the powerflowing into a spring over time, so as to computethe input work, WI~.~, required to compress or stretch the spring from x = Xl to x = x~: Wl-~2 = F~ df

F dx =

kx dx = -~k(x2 - xl~’). (3.2)

As suggested in Fig. 3.2, this workis the area under the spring characteristic between the two values of x. If the process is subsequently reversed and the spring is returned to its original position, the additional workis exactly the negative of the original work:

W~-+l= ~k(xl

=

3.1.

93

COMPLIANCE ENERGY STORAGE

XI

-~’2

X

Figure 3.2: Work as area under characteristic It is customary to say that energy is conserved by the spring; the energy it stores is _-1 ~=11F_~ 2 kx" 2k

(3.4)

so that, in general, Wl-~2

= ~22

-

~1,

(3.5)

A rotary spring acts like a compression or tension spring, except its effort is a moment, M, and its. displacment is an angle, ¢, which is the time integral of the angular velocity, ¢: M = k¢. (3.6) 3.1.2

The

Generalized

Linear

Compliance

A "generalized linear spring" results from replacing the force F of the compression or tension spring in equation (3.1) by the generalized force e, and replacing the displacement x by the generalized displacement q. The constitutive relation is e = ~q.

(3.7)

The same result applies when the M and ¢ of the rotary spring characterized by equation (3.6) are replaced by e and q, respectively. In bond graph form this is written

and is called a compliance element or just a compliance. "Compliance" means the inverse of "stiffness," so C = l/k, a constant, for both translational and rotational mechanical springs. The result is summarizedgraphically in Fig. 3.3. Note that the modulus of a compliance is defined in a manner similar to the definition of the modulus of a resistance. Further, just as the symbol R is used for both the resistance and its modulus, the symbol C is used for both the compliance and its modulus.

94

CHAPTER 3.

SIMPLE DYNAMIC MODELS

e

q Figure 3.3: The generalized linear compliance ea

C

(a) circuit

(b) bond graph

i ~-C

Figure 3.4: Capacitor as a compliance The energy stored in the linear generalized compliancefollows from substitution of the generalized variables into equation (3.4): "1?=-~ ~ q =-~ Ce " .

(3.8)

This is knownas a potential energy. The traditional symbol is capital V, whichin this text is given as script Y (to distinguish it from volume). 3.1.3

Electric

Circuit

Compliance

An electric capacitor is indicated by its traditional symbolin Fig. 3.4. Its applied voltage e and charge or electrical displacement q are modeledas q = Ce.

(3.9)

This equation agrees precisely with equation (3.7)~ revealing that, the capacitor is modeledas an electric circuit compliance, and that the capacitance C is analogous to the compliance C. The coincidence that "capacitance" and "compliance" both begin with the letter "c" contributed to the selection of that letter to designate the generic ratio of generalized displacementto generalized effort. A capacitance usually is considered to be constant; the modelis linear. 3.1.4

Linear

Fluid

Compliance

Due to

Gravity

The mechanicalspring and the electrical capacitor have a generalized force which is proportional to a generalized displacement. A fluid complianceshould have the same thing: a pressure proportional to a volumetric displacement. Think for a momentwhat this could be.

3.1.

95

COMPLIANCE ENERGY STORAGE surface area A P~C

Figure 3.5: Liquid gravity tank as a compliance The most commonanswer is a fluid tank of constant area independent of the depth of liquid in it, as shown in Fig. 3.5. A fluid port enters at or near the bottom of the tank. The depth of the liquid equals the fluid displacement V, which is the volume of liquid in the tank above the port divided by the cross-sectional area of the tank: z = VIA.

(3.10)

The pressure in the tank at the level of the port equals the weight density of the liquid times the depth: P = pgz. (3.11) The compliance of the tank, C, is the ratio of tile volume to the pressure: [c=- e q-VP pgV/AV I

- pg"A

(3.12)

Note that the time derivative of the volume of liquid in the tank is the volume flow rate into the tank, which can be designated either as ~’z or Q. 3.1.5

Fluid

Compliance

Due

to

Compressibility

A second type of fluid compliance is associated with the compressibility of the fluid. Assuming small changes in density only, such as found in considerations of air acoustics and nearly all problems involving the compressiblity of liquids, it is conventional to assume dP = ~--,dP (3.13) P in which the bulk modulus, fl, is treated as a constant. This equation says, for example, that a one percent change in density is associated with a pressure change of virtually one percent of fl; for water with no gas bubbles this gives about one percent of 300,000 psi, or 3000 psi. The equation of state for a perfect gas is P = pRO, where 0 is absolute temperature. Differentiation of this equation and comparison with equation (3.13) reveals that fl = P0 for small isothermal changes; P0 is the mean absolute pressure. In the absense of heat transfer, the pressure and density of a perfect gas with constant specific heats are related by p/pk =constant. Upon

96

CHAPTER 3.

SIMPLE

DYNAMIC MODELS

v (<
rigid chamber as a compliance

differentiation, this gives/3 = kPo, where k is the ratio of specific heats. A one percent change in volume of air at atmospheric conditions therefore, in the absence of heat transfer, would be associated with a pressure change of about .01 * 1.4 * 14.7 = 0.206 psi. A rigid chamber of volume V0 filled with a slightly compressible fluid is shown in Fig. 3.6. The fluid compliance of this system is C = q dq _ Vo dp/p Vo I e - de dP = -~-"

(3.14)

Should the volume of the chamber expand with increasing pressure, the effective compliance would be larger. Note that this compliance is approximated as a constant. EXAMPLE 3.1 A uniform rectangular block has half the mass density as that of the water in which it floats. Find the compliance associated with small rotations induced by an applied moment about an axis normal to the page.

i,

w "’x applied cm+~)moment

Solution: the block triangular side, and

"l ]~

~

The center of mass is on the waterline, so it remains fixed when is rotated. For a clockwise rotation through a small angle ¢, a area shown shaded below emerges above the waterline on the left an equal triangular area is submerged on the right side:

3.1. COMPLIANCE

ENERGY

STORAGE

97

The buoyancy forces on the two sides change by the area of this triangle, (w¢/2)(w/4) = w2¢/8, times the weight density of the water, pg. These forces act through the centroids of the triangles to give a pure moment or couple. The couple equals the force times the separation between the centroids, or pgw3 , This is the momentper unit length of the block in the direction normal to the page. The corresponding compliance, by definition, is the ratio of the angular deflection to the applied moment, or C--- 3l~I¢ - pgw 12

3.1.6

Summary

The reversible storage of potential energy can be represented by one-port elements for which the effort is a function of the generalized displacement. The generalized displacement is the time integral of the generalized velocity or flow. Such elements are designated by the bond graph symbol C, which stands for compliance. This symbol doubles as the modulus of the compliance, that is C -- q/e. For the linear or constant compliance this modulus is the inverse slope of the characteristic. Examples modeled by compliances include the mechanical spring with its force or momentand linear or angular displacement, the electrical capacitor with its voltage and charge, the open fluid tank with its pressure and liquid volume, and a container confining a compressible fluid. Nonlinear compliances are discussed in Section 3.6. Guided

Problem

3.1

It is essential that you grasp the idea that the definition and value of a compliance depends on the displacement variables to which it is referenced. This problem is basic in this regard, while the second problem is more sophisticated. The rotation of a lever is resisted by a spring with rate k, as shown in Fig. 3.7. Find the compliance of the system with respect to the displacement x and force F.

F

Figure 3.7: Guided Problem 3.1

98

CHAPTER 3.

Suggested

SIMPLE

DYNAMIC MODELS

Steps:

1. Find the displacement of the spring, which you could call xs, as a function of the parameters a and b and the displacement variable x. 2. Write the energy stored in the spring as a function of xs. 3. Combinethe results function of x. 4. Equate the result for C.

of steps 1 and 2 to get the energy of the spring as a of step 3 to the known form, ]2 = (1/2C)x ~-, and solve

5. You might also wish to solve for C using a more familiar force-based analysis, although presently it is more important that you learn the method above. Find the force on the spring, which you could call Fs, as a function of the parameters a and b and the effort variable F. Then, solve this equation for F. Solve the equation from step 1 for x. Finally, compute the ratio C. = x/F. The answer, of course, should contain only parameters, and no variables. Guided

Problem

3.2

A long slender-walled cylinder expands under the influence of an internal pressure, creating an effective compliance relative to an incompressible fluid within. The internal radius is r, and wall thickness is t << r and the length is L >> r. The wall material has Young’s modulus E and Poisson’s ratio #. (a) Estimate the compliance of the chamber. (b) Estimate the effective overall compliance relative itself compressible, having a bulk modulus Suggested

to a fluid which

Steps:

end view of one-half of the cylinder. Place the 1. Draw a cross-sectional pressure forces across the diameter and the resisting peripheral forces ta on the shell. Balance the forces, and solve for the peripheral normal stress. 2. Draw a side view, and find the normal axial stress. 3. Use the results of steps 1 and 2 and the elastic parameters of the material to find the peripheral and axial strains. 4. Find the ratio of the change in volume to the original volume, using the results of step 3 and the assumption of small strains. Neglect the effects of the ends of the cylinder.

3.1.

COMPLL4NCE ENERG’Y

STORAGE

99

5. Evaluate the compliance, which is the ratio of the change in volume to the applied pressure.

6. Find the compliance, assuming the shell is rigid and the fluid has a finite bulk modulus,/~.

7. If both the shell and the fluid are elastic, then there are two compliances bonded to a single junction, which also has a third bond to represent the flow that can enter or leave the chamber through a port. Determine whether this is a 0-junction or a 1-junction, and find the combined compliance.

PROBLEMS 3.1 Derive the compliance of a solid body floating on the surface of a liquid with respect to small vertical displacements. Define parameters as needed. 3.2 Find the compliance of a shaft in torsion with respect to its overall angle of twist. Define parameters as needed. (You may refer to your textbook on the mechanics of materials.) 3.3 A pendulum comprises a point mass attached to the end of a massless rigid rod which is pivoted about its upper end. Interest is restricted to small angles of deflection from the vertical equilibrium.

(a) Find the gravity-induced compliance by considering momentfor a deflection from the vertical.

the correcting

(b) Find the compliance by considering the gravity energy associated with 1 2 the deflection. Hint: Use the approximation cos 0 ~_ 1 - .~0 .

3.4 Solve the preceding problem substituting a uniform slender rod of mass m and length L for the massless rod and point mass.

100

CHAPTER 3.

SIMPLE DYNAMIC MODELS

3.5 For the rolling disk of mass m and radius r shown below:

(a) Define a variable to describe the displacement of the disk from its equilibrium position. Find the potential energy of the disk in terms of this displacement. (b) Find the compliance of the disk with respect assuming it has only small values.

to the displacement,

3.6 Find the compliance of the piston/cylinder shown below with respect to F, x. Assume x << L, neglect friction and leakage, and presume (a) isothermal behavior (b) adiabatic behavior.

3.7 Verify that the bulk modulus for acoustic (small) and adiabatic compression of a perfect gas is ~ = kPo, where k is the ratio of specific heats and Po is the mean absolute pressure.

SOLUTIONS TO GUIDED PROBLEMS Guided 1.

Problem

3.1

Xs -~- a._.~_ xa~-b °

1 2 1 2-4. 12 = =kx~ = =k 2

2

a ~ = lx2; ( )~ x’~ --2C .theref°re,

21 C=

(-~)

3.1.

COMPLIANCE

101

ENERGY STORAGE that F ----

x~.

Fs =

From step 1, x ----a+b~xs. a

Finally,

Guided

C -= F = ~

Problem

, which is the same as above.

3.2 2tap = 2rP, therefore

ap = Pr/t

2~rrtaa---- ~rr2 p, therefore aa = Pr/2t 2

nrt

= ap/2

a~~

3. peripheral

strain:

axial strain:

(

ep = ~(ap - #aa) = 1

ea = ~ (aa - #a~)

E--~

- # ~-~

4. volume: V = ~r[r(1 + ep)]2L(1 + e~) ~rr2L(1 + 2ep+ ea ) 7¢r3Lp change in volume: AV ~_ ~rr2L(2ep + ~a) = (2.5 ~- , z#)

5. For the chamber alone,

neglecting

V 6. For the fluid alone, C~ = /~

Q = _~p

o ~’¢" "’’~Q

C~

end effects,

AV = (2.5 Dc = --

p

~rr’2L ~ ~ P Q=~/~

C=Cc÷Ci

C =~rr2L

(2.5-

#)~-~

¯ ~rr3L - 2~)

102

CHAPTER 3.

SIMPLE DYNAMIC MODELS

P

(a) mass

(b) characteristic

Figure 3.8: Accelerating mass and characteristic

3.2

Inertance

Energy Storage

Inertance energy storage, also knownas generalized kinetic energy, is the second of the two categories of energy storage. Most models of this storage are representable as one-port elements, designated by the symbol I. (More complicated models are considered in Chapter 10.) 3.2.1

Mass,

Momentum

and

Kinetic

Energy

Kinetic energy, as its name suggests, is associated with motion, unlike potential energy, which is associated with displacement. Like potential energy, however, kinetic energy is potentially recoverable. Consider the mass m of Fig. 3.8, which is accelerated by a force F(t) from rest in an inertial reference frame to a velocity, 5, and a corresponding linear momentum,p = m2. Newton’s law gives d2 F -m’ dt which can be integrated

(3.15)

to give ~:-

fFdt m

_ p m

(3.16)

The kinetic energy is designated traditionally by a capital T, and is given in this text by script T (to distinguish it from transformer moduli). It equals the work done by the force: T= F/ dx / = m-~2 dt d~ /

= = -~m~" = 2m .(3.17) 1 m ,)2 d2 1._~p.~

This energy is represented by the area under the characteristic, as shown in part (b) of the figure. The same type of development that gives equation (3.5) for the work associated with a change of displacement of a compliance gives the work required to change the velocity or momentumfrom state 1 to state 2: W~.~ = T~ - %.

(3.18)

3.2. 3.2.2

103

INERTANCE ENERGYSTORAGE The Generalized

Linear

Inertance

The generalized compliance has been described by

e = q/c,

(3.19)

which is a linear relation betweenthe generalized displacementand the effort. The corresponding"constitutive relation" for the generalized inertance is (3.20) which is a linear relation betweenthe generalized momentum and the flow. This is precisely equation (3.16) if 0 is replaced by 2 and the generalized inertance, I, is replaced by m. The generalized momentumis defined by lp=/edt

or

e-=

(3.21)

P’I

The relationship of equation (3.20) is represented in part (b) of Fig. 3.8 changing the labels ~ to c~ and mto I. The bond graph symbolfor inertance uses the letter I in the samewaythat the complianceuses the letter C: e=$

I

The momentum of the mass in Fig. 3.8 is proportional to the velocity, so that I = mis a constant; the inertance is said to be linear. The generalized kinetic energy of a constant inertance is

which can be comparedwith equation (3.17).

3.2.3

Common Engineering stant Inertances

Elements

Modeled

by Con-

Arigid bodywhichrotates about a fixed axis, such as a flywheel, resists angular acceleration because of its massmomentof inertia, I, as suggested in Fig. 3.9 part (a). (Some authors use the symbol J to represent the mass moment inertia, in which case you can write I = ~/ for the modulus. Youshould not write J in the bondgraph, however,since it is not a defined bondgraph symbol.) In this application, p is the angular momentum. The kinetic energy is 31¢ or ½P2/I, consistent with equation (3.22).

104

CHAPTER 3.

~

(a) rotational inertia

SIMPLE

l=J

DYNAMIC MODELS

M

I=L (b) electrical inductor

(c) fluid tube

P~ t’IIIQ

IIII/1 I=pL/A

"

Q

Figure 3.9: Examples of constant inertances

EXAMPLE

3.2

Deduce a formula for the value of I for a rigid mass rotating about a fixed axis. Use a momentummethod and then an energy method; the results should agree. Solution:

The angular

momentum is

and the kinetic energy is

where r is the radial and the integrations the first equation to with O = ~, the result

distance of the element dm from the axis, of rotation, are carried out over the entire body. Either comparing equation(a.20) or the second equation to equation(~.22), is I = f r2dm.

This is the mass moment of inertia, familiar from introductory dynamics courses. It gives, for example, mR2 for a slender hoop of radius R, ~mRfor a uniform disk of radius R and (1/12)m(2R) 2 for a rod of length 2R, all about their respective normal centroidM axes. An inductor in an electric circuit impedes the rate of change of current. It also can be modeled in bond graph language by an inertance, as shown in part (b) of Fig. 3.9. In this application, p is still fe dt, the modulusI equals

3.2.

105

INERTANCE ENERGY STORAGE 1 ""

1 "2

the inductance, L, and the generalized kinetic energy is _~Lwor ~p /L. Many engineers use the symbol I for inductance anyway. The mass of a fluid in a tube impedes the rate of change of the volume flow rate. Its use in a fluid circuit also can be represented in a bond graph by an inertance, as shown in part (c) of Fig. 3.9. The standard variables e = P and 1. ~ = Q can be used EXAMPLE 3.3 Deduce the inertance of an incompressible fluid of density p in a uniform channel of area A and length L. Use both a momentumapproach and a separate energy approach. Solution: The momentumapproach can be represented steps in the multiple equation below: I-- p-- dp/dt Q dQ/dt

by the successive

_ P1- P~_ p__~_L A dv/dt A ’

The final step above directly uses Newton’slaw; P1 - P2 equals the net force divided by A times the mass pAL and the acceleration dv/dt. The resulting formula is approximate, and in fact somewhat underestimates the inertance, since it assumes that the velocity is uniformly equal to Q/A whereas the velocity is more apt to be nonuniform over the cross section. Approaching this result from the perspective of kinetic energy, T= 5mv"=

(pAL)

=5

Here, m is the total mass of fluid in the tube and v is its commonvelocity. The term on the right within the parentheses is the desired inertance, and agrees with the right-end term of the previous equation. The inertances Fig. 3.10.

of two cascaded constant-area

channels sum, as suggested in

iUse of e = P neglects the dyr~atnic pressure, as notec~ before. Therefore, the inertance relation presented here will not fully represent the behavior when the dynamic pressure is significant relative to the static pressure; in particular, entrance and exit losses will need to be added.

Q --~

~A~ If-

pL, A1

12 = pL2 -~2

Figure 3.10: Inertance of two cascaded fluid channels

106

CHAPTER3.

SIMPLE DYNAMICMODELS

EXAMPLE 3.4 Find an appproximate expression for the inertance of a fluid channel of varying area A(s), wheres is the distance along the channel. Solution: A channel of continuously varying area can be considered as a cascade of infinitesimally long segments, each of which has the inertance p ds/A, as suggested below:

The overall inertance is the sumof these, or =p

(3.)

--~ds.

The result of Example3.2 is the special case for A = constant.

3.2.4 Tetrahedron of State* The four essential variables of state for physical systems are represented by the vertices of the tetrahedron of state, picture in Fig. 3.11. Manynew bond graph modelers like to .use this diagramto help them rememberthese variables and the R,C and I elements that are defined by relationships betweenthem. It is a conception of HenryM. Paynter. The edge of the tetrahedron between the vertices for the generalized displacement q and the flow or generalized velocity ~ represents a simple time differentiation/integration relationship. The edge betweenthe vertices for generalized momentum p and the effort or generalized force e -- ib also represents a simple time differentialtion/integration relationsh~ip. The. edge betweenthe vertices for e and 0 suggests the relationship that characterizes a resistance,

pd~O~

/( )%

q q dt

Figure 3.11: Tetrahedron of state

3.2.

INERTANCE

ENERGY STORAGE

107

~R, or a general source or sink~ S~or~S. The edge between the vertices for e and q suggests the relationship that characterizes a compliance, ~ C. Finally, the edge between the vertices for p and 0 suggests the relationship that characterizes an inertance, ~ I. (The hidden remaining edge, between p and q, has no special significance.)

3.2.5

Summary

A linear one-port inertance or generalized kinetic energy can be characterized by the proportionality between its flow or generalized velocity and its generalized momentum, where the generalized momentumis the time integral of the effort. Thus, the generalized momentum,p, is to the effort, e, as the generalized displacement, q, is to the generalized velocity (or flow) 0. The inertance relationship is represented in a bond graph by the symbol I. The proportionality constant is the modulus of the inertance, and also is called the inertance and is designated by the symbol I. The generalized potential and kinetic energies equal the areas under their respective characteristics, which for the linear cases considered are, respectively, ~ =. ½ (1/C)q=’~~Ce ~ 2and ~-. 7"= ~1102 = ½(1/I)p Mass, rotational inertia, fluid inertia and electrical inductance are examples of a generalized inertance or generalized kinetic energy. General formulas have been developed for the rotational and fluid cases. Table 3.1 summarizes some of the more primitive physical elements commonlymodeled as compliances and inertances. Guided

Problem

3.3

It is esssential that you grasp the idea that the definition and value of an inertance depends on the generalized velocity to which it is referred. This problem offers a basic example. A disk-shaped pinion of radius r and mass m is engaged to a fixed straight rack, as shown in Fig. 3.12. A slender rod of length L andmass Mis welded to the pinion, as shown. Find the inertance of the unit relative to the horizontal velocity of its center, 5:.

II/11///I//1111 Figure 3.12: Guided problem 3.3

108

CHAPTER 3.

Table 3.1 Primitive

~

~

M

Compliances

compliance gravity

strain

C=l/k

Physical

SIMPLE DYNAMIC MODELS

C=L3/3EI

and Inertances inertance

C= l/Apg

M

C= 1/k

C=Vo/~6

Suggested

C=2L/zraG

P IQ

C 1/mgr

C=A/,og

I=pL/A

Steps:

1. Write the kinetic energy of the unit due to the horizontal velocity of its center of mass as a funtion of 5. 2. Find the angular velocity of the unit as a function of 5. 3. Find the kinetic energy of the unit due to its angular velocity as a function of this angular velocity, and then use the result of step 2 to convert this to a function of ~. 1 "2 4. Add the energies of steps 1 and 3, and set equal to Fix . Solve for I, which should be the desired answer.

PROBLEMS 3.8 Define the symbols L, E, I, G, r, A, p, /3 and l~b in Table 3.1 as they pertain to the (a) cantilever beam, (b) floating block, (c) torsion rod, (d) dulum, (e) compliance volume, (f) gas accumulator, (g) gravity tank and inertance channel.

3.2.

INERTANCE ENERGY STORAGE

109

3.9 A disk of mass mand radius r rolls without slipping downan incline with angle ~ relative to the horizontal. Represent the situation with a bond graph, using the velocity of the center of the disk as the flow. Determinethe inertance, and find the acceleration of the center of the disk. Suggestion: Use the concept of kinetic energy to find the inertance, as in GuidedProblem3.3. 3.10 A tube of length L and internal area. A is filled with and its entrance surroundedby an inviscid liquid of density p. The pressure at its entrance is a constant P = pgh and is zero at the other end. Represent the situation with a bondgraph, using the volumeflow rate Q as the flow variable but neglecting entrance and exit losses. Find the rate of changeof the flow. 3.11 A pendulum comprises a point mass m attached to the lower end of massless rod of length L pivoted at its upper end, as in Problem3.3 (p. 99). (a) Find the momentrequired to producean angular acceleration, neglecting the gravity compliance property, and thereby deduce the inertance with respect to angular velocity. (b) Checkto makesure that 5I¢ "2 , with your I frompart (a), is the correct kinetic energy. 3.12 Solve the preceding problem substituting a uniform slender rod of mass mand length L for the massless rod and point mass(as in Problem3.4, p. 99). 3.13 For the rolling disk of Problem3.5 (p. 100): (a) Find the kinetic .energy of the disk, first in terms of any velocity but finally in terms of ¢, where ¢ is the angle betweenvertical and a line drawnfrom the center of curvature to the center of the cylinder. (.b) Usethe result of step (a) to find the inertance of the disk relative

¢. SOLUTION TO GUIDED PROBLEM Guided Problem 3.3 1 2. 1. T~= ~(m+M)2

2. ~ =Sir.

4.

%+~=~

~ =~

~m+

1+~

M ~.

110

CHAPTER 3.

Therefore,

3.3

SIMPLE

DYNAMIC MODELS

I= ~m+ 1+~ AI.

Junctions

The models presented thus far exhibit no branching. Most models of engineering systems require branching, which is accomplished through the use of simple junctions. The conventional bond-graph abstractions for junctions are given below first, showing that there are precisely two types. Mechanical, electrical and fluid constraints are then modeled by these junctions, and a single junction is used at the heart of models for very simple systems. Application to a variety of more complex systems is made in Chapter 4 and subsequent chapters. 3.3.1

Junction

Types

An arbitrary number of bonds can be joined by a junction, the temporary symbol J:

identified

here by

e3

The power convention half-arrows in this example are chosen arbitrarily, the first two being inward and all others being outward. The following rules are imposed to define two types of junctions:

with

Rule (i): For both types, the junction is ideal, neither storing, creating nor dissipating energy. Rule (ii): For the 0-junction type, the efforts on all the bonds are equal. For the 1-junction type, the flows on all bonds are equal. The first

rule implies that the input power equals the output power, requiring el(~l + e2~2 ’: e3~3+ ... q- en~n.

For the 0-junction

(spoken zero-junction),

r eo

represented as

(3.23)

3.3.

111

JUNCTIONS

the second rule requires el ~ @2=- e3 = .-- ~ en ~- e,

(3.24)

which when introduced into equation (3.23) gives (3,25) The labels el .. ¯ e,~ have been removed from the respective bonds, not of necessity (they could be left on) but because the effort e, written above and to the left of the junction symbol, is recognized as being commonto all the bonds. The 0-junction is the common-effort junction. It is often more useful to recognize its property described by equation (3.25), so it is also the flowsumming junction. For the 1-junction, represented as

e3 q" e.[ the second rule requires 01 = 0u = 93 ..... which through equation (a.2a)

0n = 9,

(3.26)

gives

el -+- e2 = e3 q- ... + en.

(3.27)

The labels 01. ¯ - 9,~ have been removed from the respective bonds, again because they would be redundant. The commonflow or generalized velocity is indicated by the 9 (or a specialized symbol, if one prefers) written below and to the right of the 1. The 1-junction is called the common flow junction. It also is called the effort-summing junction. It is the dual of the 0-junction, since the constraints imposed on the efforts in one case are the same as the constraints imposed on the flows in the other, and vice-versa. The concept and notation for these junctions was introduced by H.M. Paynter 2, and was the most crucial step in his creation of bond graphs. Two junctions of the same type connected by a bond can be collapsed into a single junction, with the bond vanishing, as shown in Fig. a.13. The only difference between the expanded and contracted versions of the graphs shown is the presence or absence of the variable ea or 93, respectively, which may not be of interest. 2H.M.Paynter, Analysis and Design of EngineeringSystems, The MITPress, Cambridge, Mass., 1961.

112

CHAPTER 3.

SIMPLE DYNAMIC MODELS

e5

e2

~4 l~2 q5

Figure 3.13: Bond-graphjunction equivalences

Figure 3.14: Junction for mechanical translation 3.3.2

Mechanical

Constraints

Modeled

by 1-Junctions

Common-flow junctions abound in models of engineering and natural systems. A mechanical examplefor translation is shownin Fig. 3.14. A push-rod with compression force Ft drives two push-rods with compression forces FA and FB. All three push rods share the commonvelocity k, as implementedin the bond graph by the 1-junction. The conservation of powerrequires Ft~ -~ FAS: + FB~,

(3.28)

F, = FA + FB.

(3.29)

so that Equation (3.29) can be found alternatively from fr ee-body analysis which is traditionally taught in courses entitled "statics" and "dynamics." Generalized force balances represent the "conservation of momentum" which is one of three great principles used to analyze combinationsof mechanicalelements, The other two are the geometric constraints betweenthe displacements or velocities (sometimes called "continuity"), and the conservation of energy. Anytwo the three are sufficient; the third is redundant. It is extremelyadvantageousfor

3.3.

113

JUNCTIONS

you to be able to employ any one of the three consequent approaches. When the modeling becomes complex or confusing, the use of the energy balance and the geo~netric constraints tends to be clearer and simpler than either co~nbination using the force balances. It also is more generalizable to non-mechanical phenomena. EXAMPLE 3.5 Model the parallel combination of two masses, a spring and a dashpot shown below with a bond graph. Reduce the graph, if possible, so only one element of each type remains.

F

Solution: The 1-junction of the bond graph model below represents the fact that all four elements share a commonvelocity. The two inertances (masses) sum to give an equivalent inertance (mass). Constant inertances arrayed around a 1-junction always can be summedin this manner. Elements of different types cannot be merged.

F

EXAMPLE 3.6 Model the parallel combination of three dashpots below by a bond graph. Also, show how the individual characteristics, as plotted by solid lines, can be combinedinto a single resistance characteristic.

force

114

CHAPTER 3.

SIMPLE DYNAMIC MODELS

Solution: All three dashpots experience a commonvelocity (5:), whereas the force F is the sum of the three individual forces. Therefore, as shownby the dashed line, the three forces sum for each value of 5: to give the overall characteristic.

b~ 1.i.~--~R

force R~

reduces to:

Mathematically, F = Fa + Fb + Fc = Ra~ + RbiC + Rc~ = RS:, It follows that the equivalent overall resistance R equals the sum Ra + Rb + Re. This equation is most easily understood if the resistances are constants and the characteristics are straight lines through the origin. It is quite correct, nevertheless, if the resistances are functions of 2 and the characteristics are curved lines. EXAMPLE 3.7 Each of the three springs below has a constant spring rate. Determine the modulus of a single compliance element, C, that can represent the combination. kl

Solution: The three individual forces sum to give the total force, as with the dashpot example above. These forces are proportional to the three springrate constants, which are inversely proportional to the three compliances. The combination rule therefore is 1

1

1

1

¯

3.3.

115

JUNCTIONS

In general, the 1-junction represents the geometric constraint in which three or more bonds share a common generalized velocity or flow. The presence of a 1-junction maynot be always obvious, as the following examplesuggests. EXAMPLE 3.8 Modelthe fire-extinguishing system of Chapter 2 with an additional load driven from the sameshaft, as shown. The "pump"includes its fluid load.

~

pulley drive ._M.,~_M. ~_~-_~. ~ auxiliary l°a d~a~ "~

This auxiliary load might be a siren, for example, that is operated at the same time as the pump, or it might be a separate pump. Also, show how the individual characteristics of the pumpand the auxiliary load, as plotted below, combineto give an equivalent overall resistive load.

M’I. 0

O;

5;

’ 100

’ . ’200 250 ’ 150 ¢,1, rad/s

Solution: To rePresent the combination of the drive and the two one-port devices, first answerthis: do these three elementsshare one of their respective conju.gate variables? The answer is yes; they have the same angular velocity, Cd, because they have a common shaft. This fact is recognized by a 1-junction: MOTORM._~.~.~r BELT Md M.~., PUMP ~,,, DRIVE T ~d"~AUXILIARY LOAD For emphasis, all three bonds are annotated with "~a." This is redundant; "~a" need only be noted below and to the right of the "1" symbol, also as shown.

The powers delivered to the load A, load B and their combination, respectively, are

p~ = Mp~; ~Dtotal

~-

p~ = M~,

Mv~d+ Ma[~ = (Mp + Ma)~a= Md~d.

116

CHAPTER 3.

SIMPLE DYNAMIC MODELS

Thus, the total torque provided by the belt drive is

M,~= Mp+ The summationof the last equation above mayimplementedgraphically by "vertical" addition of the two componentcharacteristics, as suggested below,to give the characteristic drawnwith a dashedline. This curve is the effective one-port characteristic of the equivalent combinedload. It can be used as before to carry out a source-load analysis or optimization. 2.5

~..-- M~=M~+Mo

1.5 Nm 1.~3

(same),,

0.5 0o

50

100 150 . 200 250 Ca, rad/s

Mm BELT -- Ma COMBINED MOTOR .----~-~ ~a ~" LOAD Cm DRIVE 3.3.3

Electrical 1-Junctions

Circuit

Constraints

Modeled by

Common-flow implies common-currentfor electric circuits, and common-current implies series interconnection of components.Suppose that three components arbitrarily labelled Z1, Z2 and Z3 are connected in series, as shownin Fig. 3.15. The bond graph follows directly. The total voltage of the source equals the sum of the voltages (or voltage drops) of the three components;the sum

Figure 3.15: Electrical exampleof a 1-junction: series circuit

3.3.

117

JUNCTIONS

of the voltage drops around a loop equals zero. Thus if the elements happen to be resistances, the total resistance of the circuit is the sumR1 + R2 -I- R3. Electrical resistances in series combinelike mechanicaldashpots in parallel. If the elements are inductors, the total inductance of the circuit is the sum I1 + I2 + Ia. The same is analogous to any other type of inertances bondedto a common 1-junction, such as the masses discussed earlier. If the elements are capacitors, the sununation of the voltages implies summingthe reciprocals of the capacitances, or 1/C1 + 1/C.2 + l/Ca. This is like the springs bondedto a common 1-junction as discussed earlier. 3.3.4

Fluid Circuit 1-Junctions

Constraints

Modeled

by

Fluid flow is analogousto electric current. A series interconnection of components such as pumps,valves, hydraulic motors and reservoirs can be represented by a 1-junction with a bond radiating outwardto the appropriate 1-port element for each individual component.The order of these bonds is immaterial. Sometimesthe effect of an element is small enoughto justify its removal from a model. This action can leave a 1-junction with only two bonds attached. In this case, if the two remaining powerarrows are directed unailaterally, it is permissible to removethe junction altogether and connect the two dangling bonds. Example 3.9 Model the hydraulic system pictured below. You mayrepresent the four elements in the bond graph by the words PUMP,VALVE,MOTOR and RESERVOIR. Also, describe how the model can be simplified if the pressure drop across the valve is negligible or the reservoir pressure is virtually atmospheric.

~hydraulic

~ motor I t~l

reservoir

Solution: The first step is to recognizethat all four elements havethe same flow, Q. Thus, their bondsare joined by a 1-junction: VALVE Pm PUMP ~P !Q’~-~ !

MOTOR

RESERVOIR

118

CHAPTER 3.

SIMPLE DYNAMIC MODELS

Note that the pressure on the valve bond, P~,, is the pressure drop across the valve. The pressure Pmis similarly the pressure drop across the motor: The sumof these two pressure drops plus the reservoir pressure equals the pumppressure. This means that the resistances of the components also sum, despite the nonlinearities. If the valve is wide open, Pv -~ 0, and the powerdissipated in the valve is virtually zero. In this case the valve and its bondsimply can be removed from the bond graph model. Similarly, the reservoir pressure P,. maybe considered to be zero gage, so if atmospheric pressure is employedas the reference (to give gage pressure), the reservoir bond carries no power. this case, the bond and the reservoir element maybe removedsimilarly. Finally, if only twobondsare left on the junction, the junction itself can be removedand the two bonds simply connected together, assuming the two bonds have powerconvention half-arrows in the same direction.

3.3.5

Mechanical

Constraints

Modeled

by 0-Junctions

A dashpot is the idealization of a damperwith zero mass; it therefore always has the same force at its two ends. Thus, a dashpot can be represented by a three-ported 0-junction and a resistance, as shownin Fig. 3.16. The force is a function of the velocity difference :b~ - :~2, whichtherefore is the flow on the resistance bond. EXAMPLE 3.10 Showhow a cascade of dashpots can be modeled by a bond graph, and howthat graph can be simplified by combiningthe characteristics of the individual dashpots.

X1

X~

Figure 3.16: Mechanical exampleof a 0-junction; two-endeddashpot

3.3.

119

JUNCTIONS

Solution: The bonds of the successive elements can be joined as shown in (a) below, because of the commonvelocities at the connections. Then, the three O-junctions can be coalesced as shown in (b), since their bonds all have the same effort (force). Finally, since each dashpot has the same force, the bonds of their respective resistances can be joined together with a O-junction, as in (c). R~ ¢R2 (b)

(a)

¯ ---fo---.-. Fo.-----F0

R

1-22~

(c)

l-X4

x, I

~

X3 X4

R~

R~

R~

R~

The generalizied velocities of the individual resistances sum~o give the overall generalized velocity. Therefore, the dashed line representing the overall resistance characteristic is found by a horizontal summationof the individual characteristics:

x

Mathematically,

=F + F +F F R Thus, the overall reciprocal ofthe resistance, 1/R~ called the overall conductance, equals the sum l/R1 + l/R2 + l/R3 of the individual conductances. The same statement applies if the resistances are nonlinear, as long as the individual resistances are expressed as functions of the commoneffort e rather than the non-commonvelocities. An idealized spring has zero mass, so like a dashpot it has the same force at each end. It therefore can be represented similarly, by a three-ported 0-jur/ction and a compliance, as shown in Fig. 3.17. The deflection.which causes the force (or torque, in the case of a rotary spring) is proportional to the difference between the displacements at the two ends, ql - q’_,. Should one end of the spring be motionless, there is no power on the associ-

120

CHAPTER 3.

SIMPLE DYNAMIC MODELS M

F’~’~"~x~

F M~ (b) rotary spring

(a) translational spring ¯

~eo

~eo ~:~>-~--~.

~

C (c) bond graph

C

C (d) right side i~obil~ed

Figure 3.17: Double-endedsprings

ated bond, whichtherefore maybe erased. Part (d) of Fig. 17 showssuch a case. The 0-junction is left with only two bonds, and therefore serves no purpose. It can be removed, and its two bonds joined (assuming the power conventions of the two bonds are compatible). The result is a single bond and a compliance, identical to the case whichintroduced springs in Section 3.1. EXAMPLE 3.11 Determine the combination rule for combining a cascade of springs with individual compliancesC1, C2, C3 etc. into a single compliance, C. Solution: Since the forces are common and the relative displacements sum, and a relative displacementis proportional to a compliancefor a given force, the compliances sum: C=C~+C2+C3+....

3.3.6

Electric and Fluid by 0-Junctions

Circuit

Constraints

Modeled

A common-voltagecurrent-summingjunction is typified by a soldered joint or a parallel combination of elements, as suggested in Fig. 3.18. Conductances sum in the same way as they do for mechanical 0-junctions. Capacitances sum similarly, satisf~ving the sameequation as the springs of Example3.10, which applies to any type of compliance.A0-junction for a fluid circuit is typified by the pipe tee, or interconnection of fluid lines at a point of common pressure. Anexampleis shownin Fig. 3.19. Finally, you need to consider howinertances connected to a common0-junction can be combined.

3.3.

JUNCTIONS

121

t e

Zl

..eo .---~eo---~. eo ~ Zl

Z~

23

Figure 3.18: Electrical exampleof a 0-junction: parallel connection

P

P

Figure 3.19: Fluid examplesof a 0-junction: parallel connection

122

CHAPTER 3.

SIMPLE

DYNAMIC MODELS

EXAMPLE 3.12 Determine the combination rule for multiple inertances 0-junction.

bonded to a common

Solution: The effort is commonto all the inertances, while the flow (and its rate of change) sums. The rate of change for the flow of each inertance is inversely proportional to that inertance. Therefore, the combination rule is 1 1 1 1

=

This means that the overall inertance is less than any individual inertance. Should any of the inertances be zero, for example, the overall inertance is zero also, causing the effort on the 0-junction to be zero. The "geometric constraints" of mechanical circuits are analogous to the "continuity" or "conservation of mass" constraints of fluid circuits. (The reader aware of the difference between the stagnation and the static pressures should recall that this difference is being neglected for the time being.) Since flows sum, conductances, compliances and reciprocal inertances (called susceptances) sum.

3.3.7

Simple

IRC

Models

Mechanical, electrical and fluid examples of simple IRC systems and bond graph models thereof are given in Fig. 3.20. All of these models include a source, a resistance, a compliance and an inertance arrayed around a commonjunction. Consideration of more complex models is deferred to Chapter 4, although the R, C and I elements could be the result of combining two or more elements of the same type. Special RCand IR models result from deletion of the I and C elements, respectively. These examples are the major vehicle used in this chapter to introduce the modeling and analysis of dynamic systems. You are urged to cover up the bond graphs in the figure and practice modeling these systems yourself. 3.3.8

Summary

0 and 1-junctions normally are employed as the only means of constructing a branched model, and therefore play a major role in bond-graph modeling. The junction structure of a model comprises its bonds, junctions, and (as will be illustrated later) transformers and gyrators. The complete model merely adds one-port elements to his junction structure: sources, resistances and energy storage elements. Newmodelers often are uncertain which junction to use in a particular situation. This dilema is reliably resolved only by careful identification of the key variables in the physical system, which is a required step no matter what type of modeling language is used. The bonds on a 0-junction have a commoneffort and flows that sum. The bonds on a 1-junction have a commonflow and efforts

3.3.

123

JUNCTIONS

F k~~

l=m

R=b C=l/k

e

I=L R=R C=C I=pL/Ap

~A~_~ C= At /Pg ~"

QI

(a) O-junction type

orous plug ~’Ap R

C=l/k R=b

C=C

e

R=R

~~

Q

~

Ap

A~ I=pL/A~ C=A,/pg R=8~/~

(b) 1-junction type

Figure 3.20: Simple IRC models

124

CHAPTER 3.

SIMPLE

DYNAMIC MODELS

that sum. Trying instead to remember which type of junction applies to series connections and which applies to parallel connections usually leads to mistakes, since the answer depends in too complicated a way on whether the system is mechanical, electrical or fluid. Multiple resistances bonded to a common1-junction can be combined to give an overall resistance greater than any of its components. Multiple resistances bonded to a common0-junction combine to give an overall resistance smaller than any of its components. The same statements apply to inertances. On the other hand, two or more compliances bonded to a common0-junction produce an equivalent overall compliance larger than any of its components. Twoor more compliances bonded to a common1-junction, conversely, produce an equivalent compliance smaller than any of its components. The difference results from the fact that resistance and inertance are proportional to impedance, whereas compliance is proportional to admittance, the reciprocal of impedance. You are expected at this point to be able to find appropriate bond graph models for si~nple systems such as those pictured in Fig. 3.20. Extracting the behavior of these models comes next. Guided

Problem

3.4

Needed practice dealing with compliances as well as junctions is provided by this problem. Part (b) should become routine to you. Three constant compliances are combined two different ways: C1

(i)

~0~C2

C1

(~i)

~1~C2

l C3 (a) illustrate mechanical, electrical and fluid implementations, and (b) find compliance of the equivalent integrated compliance, ~.C, in terms of the moduli C~, C2 and C3. Suggested

Steps:

1. For part (a) focus on the type of variable (e or q) which sums and the type of variable which is commonto all three compliances. 2. For part (b) sumthe variables of the appropriate type to get a total value, and compare the result with the desired" form.

3.3.

125

JUNCTIONS

Figure 3.21: Guided Problem 3.5

Guided

Problem

3.5

This problem requires the consolidation of distinct inertances into a single inertance. Water pumpedfrom a river to cool the steam in a power plant passes through the circular channels shown in Fig. 3.21. Estimate the overall inertance relative to the total volume flow. Suggested

Steps:

1. Find the inertances of the four separate channels relative to their individual flows. 2. Combine the three inertances

which share a commonpressure.

3. The inertance resulting from step 2 shares a com~nonflow with the fourth channel. Combinethe inertances to get the overall inertance.

Guided

Problem

3.6

This is the first of three short, basic ~nodeling problems which you should be able to do at this point. A drum viscometer comprises a cylinder with a rotating drum submerged in the fluid being tested, as pictured in Fig. 3.22. The rate at which the drum coasts to a stop is observed. Model the system with a bond graph, and evaluate the moduli of its elements. The mass moment of inertia of the drum is known to be 0.12 in lb s ’~, m~dthe viscous drag on it produces the moment M- 27rLr3# (b W

where L = 6 in, r -- 3 in, w = 0.02 in and # = 1.4 x 10-7 ~. lb s/in Suggested

Steps:

1. Identify the key flow variable, and label this on a 1-junction.

126

CHAPTER 3.

SIMPLE

DYNAIVIIC

MODELS

Figure 3.22: Drum viscometer for Guided Problem 3.6

d

~d ¯ density p

Figure 3.23: U-tube for Guided Problem 3.7

2. Identify the important physical mechanisms that perpetuate and those that dissipate energy. Represent these as bond graph elements appended to the 1-junction. 3. State or find the moduli of the elements of part 2, using the given information.

Guided

Problem

3.7

A U-tube is filled with water, as shown in Fig. ter following a non-equilibrium initial condition the system with a bond graph, and evaluate the preparation for an analysis. Neglect the effects of

3.23. The motion of the wais of ultimate interest. Model parameters of your model in viscosity.

3.3.

127

JUNCTIONS

r

r

-7

’,11 ’, _1_, Ili/¢.)l IR~ -.-C source

filter

load

Figure 3.24: Guided Problem 3.8

Suggested

Steps:

Compliance and inertance elements are present. Identify the variable which is commonto them, and deduce which kind of junction is proper. Bond the C and I elements to the junction. Express the moduli of the I and C elements in terms of the parameters given in the drawing. You may have more than one energy storage element of the same type in your model. If so, combine these elements into a single element of the same type, determine its modulus, and draw the reduced bond graph.

Guided

Problem

3.8

A current source drives an essentially inductive load. Small perturbations in the current source are inevitable, and produce undesirable variations in the voltage. A shunt resistor and capacitor are added to the circuit, as shownin Fig. 3.24, to moderate these pulsations. You are asked to model the system, in preparation for an analysis directed at determining appropriate values for the resistance and capacitance. Suggested

Steps:

1. Determine the effort or flow variable that is commonto the four elements, and the effort or flow variables that sum. Draw and label a junction accordingly. 2. Bond the four elements to this junction.

128

CHAPTER 3.

SIMPLE D}%TAMIC MODELS

PROBLEMS 3.14 An engine drives both a virtually ideal DCgenerator with constant magnetic field and a Inechanical load on the same shaft. Modeleach of the components by an appropriate bond graph element and assemble a corresponding complete bond graph model.

~ I engine

~- generator

electrical load ~ mechanical

load I

I. 3.15 The bond graphs below can be reduced to a single resistance, Find the modulusof R’ assumingconstant moduli as given.

(a)

R 1

R 3

R3

~0-~--~1

--1---~,--0 R2

RI

R4

R2

R4

~R

bothcases: R~ = 4 R2=3 R3= 2 R4=l

3.16 Drawa bond graph to represent the electric circuit below. Reducethe load seen by the voltage source to a single resistaace, R, and find its modulus.

3.17 Anapplied force F is resisted by the bending of a slender cantilevered beamof length L, thickness t and width w as well as a spring with rate k, as shownbelow. F

(a) Find the moduli of the two componentcompliances in terms of fixed parameters. (Youmayuse references in your personal library.)

3.3.

129

JUNCTIONS

(b) Represent with a bond graph how the two compliances combine, labeling F and 2 appropriately. (Ask yourself: which of these variables is commonto the two compliances, and which is the result of a summation?) (c) Find the combined compliance.

3.18 Twolinear springs with rates kl and k.~ are connected in (a) series (b) parallel. Find the compliance of the combinations. 3.19 Find the capacitance of two capacitors, (b) parallel.

C1 and C_~, connected in (a) series

3.20 A physical damper has significant mass, which can be assumed to be divided equally between its two ends. Give a bond graph model for this damper~ and relate the moduli of its elements to physical parameters. 3.21 A spring with stiffness k connects a mass Mto ground, as shown below. The mass of the spring, ms, is relatively small, but not insignificant. Investigate its effect on the behavior as suggested by the steps below.

(a) Assumethat the mass of the spring, ms, is not great enough to significantly upset the uniformity of the strain from one end of the spring to the other under dynamic conditions. Define a position variable y such that the left end of the spring represents y = 0, and the right end, y = L. Each infinitesimal segment dy therefore has mass m~(dy/L). Similarly, the velocity of the segment is (y/L)2, where 2 is the velocity of the large rigid mass. Evaluate, by integration, the kinetic energy of the entire spring as a function of ms and ~. (b) Interpret the kinetic energy of the spring as an inertance relative the velocity ~. Then, express the total inertance relative to ~, which you may define as I, as a function of M and ms. (c) The natural frequency w~, of an ideal mass-spring system is shown in Section 3.5 to equal V/~/IC. Find the w,~ predicted by your model, which has the same compliance as if the spring had zero mass. (d) Comment,qualitatively, may entail.

on any limitations

you suspect this analysis

130

CHAPTER

3.

SIMPLE

DYNAMIC MODELS

3.22 A vertical tube of diameter d and length L is capped with a closed chamber of volume ~/b, as shown below. Water enters from the bottom, and air is trapped above. Estimate the effective compliance of the system with respect to the entering water; the following steps are suggested.

water

(a) Evaluate separately the compliances due to gravity and the compression of the air. Assume that the water never enters the upper chamber, and that the compression of the air is (i) isothermal (i.e. slow) (ii) adiabatic (i.e. fast).

(b) Represent how the compliances combine with a bond graph, labeling P and ~ appropriately. (Ask yourself: which of these variables is common to the two compliances, and which is the result of a summation?)

(c) Find the overall compliance.

3.23 Define variables,

3.24 Define variables, graphs.

(a)~

and model th~ electric

circuits

below with bond graphs.

and model the mechanical circuits

below with bond

3.3.

131

JUNCTIONS

3.25 Model the tank and tube shown below left, neglecting the inertance of the fluid. Specify the moduli of your bond graph elements. ’

tank area: 10 in2 3pg=62.4 lb/ft ~-’~

#=2x10~ lb’s/in2

~w~,"

,fluid ~I

3.26 A steel plunger (Ps9 = 0.28a lb/in 3) is released from rest in a fluid-filled cylinder, as shown above right. The fluid can flow unimpeded through an auxiliary channel. Model the system in preparation for an analysis to determine its motion. Neglect fluid inertia, and assume the plunger remains centered. Specify the values of the moduli of your bond graph elements. 3.27 A water tank of area 2.0 m2 drains through a porous plug in its bottom with a fluid resistance of 98.7 kN s/m5. It also drains through a tube of length 10 m and diameter 2.0 cm. Model the system with a bond graph, and evaluate its parameters. Neglect the effect of viscosity.

lporous

~

I plug

tube

~

3.28 Often it is impractical to reduce the vibrational forces of a machine to an acceptable level, .but it is practical to isolate the machine from its foundation, and thereby prevent large oscillations from shaking the floor, creating noise, etc. A standard scheme, valid for y-motion only, is shown below. Model this system in preparation for an analysis which will be used to choose the springs and the dashpot. If your model has separate elements to represent the separate springs, combine these elements and give the modulus of the new element in terms of k.

mass m k~ ~b k ~ support

(conceptual)

~ floor

132

CHAPTER 3.

SIMPLE

DYNAMIC MODELS

3.29 Presume that the LRC filter shown below drives an instrument with a virtually infinite input impedance. The relation betweenthe voltages ei and eo is of interest, since the latter is intended to be a smoothed but not smothered version of the former. Model the system with a bond graph in preparation for au analysis. L

SOLUTIONS

TO GUIDED

Guided

3.4

Problem

(a) (i) common effort

R

PROBLEMS

mechanical: electrical:

_1..

_l_ _[_

fluid: (ii) common displacement mechanical:

electrical: fluid: (b) Themodulineednot be constantsif chordaldefinitionsare used. (i)

q ’~ qi = .~ C,e

~ C = CI

3+ C~_ + C

(ii) e = ~’ei = .S(1/C~)q~ I/C = 1/C~+ 1/C~. + 1/C3 Guided

Problem

3.5

1. Large channel: IL = pL _ 4pL 4 x 1.94 x 100 5. = 6.86lb s’~/ft 7r(6), A 7rd2 = ~ Each small channel: ls = 4 × 1.94 × 20 5. - 5.49 lb s2/ft 7r(3)2 2. 5. Combinationof three small channels: I = ~5.49 = 1.83 lb s~/ft 3. 5. Combinationof all four channels: 1 = 6.86 + 1.83 = 8.69 lb s2/ft

3.3.

133

JUNCTIONS

Guided

Problem

3.6

1,2. The key flow variable is the angular velocity of the drum, ~. There is kinetic energy ½I~2 associated with this velocity, where I is the mass momentof inertia of the drum. Energydissipation takes place because of the viscous drag in the narrowannulus. Theshear force is a function of angular velocity, so this gives a resistance, R. The bond graph that represents these observations is I~ 1.

M-~-~’~R

The mass moment of inertia is the integral 1 = fr2dm 2. = 0.12 in lb s The resistance is the ratio of the torque Mdue to viscous drag to the angular velocity ~, or R = 27rLr31~/w= (2zr×6×33×1.4×10-7/0.02) = 0.00713 in-lb.s. Guided

Problem

3.7

The potential or compliance energy is a function of the displacement of the water, and the kinetic or inertance energy is a function of the velocity of the water. One could choose as the.generalized velocity either the commonvertical velocity of the surfaces of the water or the volumeflow of the water. The associated generalized displacements are the common vertical displacements of the surfaces of the water and the volumedisplacement of the water, respectively. The latter choice is madehere, with symbolsQ = g" and V, since equations (3.12) (p. 95) the compliances and Example3.3 (p. 105) gives the inertance with reference to these variables. Note there are two compliances,one for each free surface, as shownin the bond graph. From equation (3.12), each compliance is C A/p9 = ~rd2/4p9. Fr om Example 3.3, the inertance is ~. I = pL/A = 4pL/Trd The two compliances share a common flow or displacement, so their efforts or pressures sum. This means that the reciprocal of the combined compliance equals the sum of the reciprocals of the individual compliances, so that the cmnbined compliance is C’ = C/2 = 7rd2/8pg. Note that half the compliance meanstwice the stiffness; the two stiffnesses sum. Guided

Problem

3.8

1,2. All four elements share the same voltage difference. Therefore, all four associated bond graph elements share a common0-junction. The commonvoltage at the bottom ends of the elements is taken as ground, or zero.

R

s<--77--o o ----.- z c

134

3.4

CHAPTER 3.

Causality

SIMPLE

and Differential

DYNAMIC MODELS

Equations

The dynamic behavior of a well-posed model with energy storage elements is found by expressing it as a differential equation or set of equations, and solving the equations numerically -- called simulation -- or analytically. The procedures are largely deterministic, unlike in modeling, requiring care but no special insight. They can be expedited by software. A middle course is chosen in this text between blind reliance on software and total reliance on mind and hand, so that you can benefit from modern methods and understand their origin. This section concerns the determination of the differential equations, and the following section deals with their solutions. It is possible to define all the efforts and flows in a bond graph with symbols, and write one equation for each one-port element, two equations for each transformer and gyrator, and one equation for the variables which sum about each junction. This usually large number of equations then can be combined, eliminating the variables of minor interest, to give a solvable set of differential equations. This procedure is excessively complex in practice, however, unless an effective over-arching plan is applied. Such a comprehensive plan best develops one first-order differential equation for each independent energy storage element. This state-space formulation is ideally suited for both numerical simulation techniques and, in the case of linear models, analytical solution. The plan also ought to be so automatic that it can be programmed into software. Such programs are indeed available for bond graph models3. Users of such software (sometimes coupled with separate numerical simulators) do not have to write the differential equations themselves or even find their solutions, which are directly plotted. This rapidly evolving software is not discussed further in this book, however, in favor of the development of a more general and basic understanding of the analytic procedures and the use of more general-purpose and widely used software. 3.4.1

Causalities

of

Effort

and Flow

Sources

The effort source Se imparts the effort e as an independent variable onto the system to which it is attached. The associated flow ~ becomes dependent. You can say that Se is the cause of e, whereas (~ is caused by the reaction to this by whatever system (unshown below) is attached to the right-hand end of the bond. This bilateral causality can be indicated with either of the notations Se

e--~ ~ ~-q

or

Se

e ~ q

Similarly, the flow source S.~ forces the flow ~ into the system, which responds with the effort, e. You can say that Sf is the cause of 9, whereas e is caused by 3See R. Rosenberg, A Users Guide to ENPORT-4, Wiley, N.Y. 1974. Also, Controllab Products B.V. (Netherlands) has a programcalled 20-sim; informa’~ioncan be found http://www.rt.el.utwente.nl/20sim/clp.htm. Lorenz Simulation SA(Belgium)has a package called MS1;informationcan be found at http://www.lorsim.be.CadsimEngineering(California) has a product called CAMP-G; information is available at http://www.bondgraph.com.

3.4. CAUSALITY ’(a)

AND DIFFERENTIAL

135

EQUATIONS

e01~ (b)

l.

q2 eo~O4 q~ C-~ eo~eo /R po = f eo dt R Figure 3.25: Effort and flow sources bonded to a 0-junction behavior

to give uncoupled

the reaction to this (~ by whatever system (unshown below) is attached to right-hand end of the bond. This causality can be indicated as S!

or

The short causal stroke drawn transverse nationally:

The causal orientation entation.

3.4.2

Junctions ior

Sy

I~

to the bond is recognized inter-

is completely independent of the power convention ori-

with Elements Having Uncoupled Behav-

Consider an effort source bonded to a 0-junction, as shown in part (a) of Fig. 3.25. All bonds on the 0-junction have the same effort, by the definition of the junction. This effort, e, is specified or caused by the effort source. The flows ~.), ~3 and ~4, on the other hand, must be separately caused by whatever (unshown) elements are attached to the outer ends of these bonds. This causal pattern is recognized by placing causal strokes at the outer ends of the bonds for these three flows. The flow ~, then, is the sum of 02, ~3 and 04, with signs determined by whatever power convention half-arrows are part of the model. Consider now the specific situation shown in part (b) of the figure, in which the three bonds are attached, respectively, to a C, an I and an R element. The flow on the R bond then can be written as eo/R, the fiow on the C bond as C deo/dt, and the flow on the I bond as f eo dt/I =po/I, assuming constant moduli R, C and I. These three elements are said to be given admittance causality. This means that their flows respond to an induced effort, and are expressed as functions of that effort or its time integral, p0. The behavior of the three ele~nents is said to be uncoupled, which means that the flow on one of them is independent of the size or even the presence of the others. The flow on the source bond completes the picture; it is the sum (~ = eo/R + C deo/dt +po/I.

136

CHAPTER3.

SIMPLE DYNAMICMODELS

The C element also is said to have differential causality, since its causal output is a function of the time derivative of its causal input. TheI element is said to have integral causality, since its causal output is a function of the time integral of its causal input. EXAMPLE 3.13 By analogy to the exampleabove, explain whythe causal strokes in the bond graph below left are drawnwhere they are. Then explain the annotations for the efforts and flows given on the bondgraph belowright, including giving appropriate namesto the causalities of the R, C and I elements. Are the behaviors of these elements coupled or uncoupled?

Solution: A flow source is bonded to a 1-junction. All the bonds and the elements are forced to have, in common,the flow 00 imposedby the source. This fact is underscored by the placements of the causal strokes on the bonds. In general, all the bonds on any 1-junction must have their causal strokes at their junction ends, except for precisely one whichhas its causal stroke at the remote end. Similarly, all the bonds on any 0-junction must have their causal strokes at the remote end, except for precisely one which has its causal stroke adjacent to the junction. Youwill find that these facts makecausal strokes very useful. Three of the bonds are attached to C, I and R elements with the respective efforts expressed as ROo, (1/C) f ~odt = qo/C, I d(lo/dt and R(?o + qo/C + I d~o/dt. Theseelements are said to have impedancecausality, since their efforts are functions of the flow or its integral or its derivative. They are said to be uncoupled, as in the O-junction example, because each effort is independentof the presence of the other two elements. The C elementalso is said to haveintegral causality, and the I elementis said to have differential causality. 3.4:3

Junctions

with

Elements

Having

Coupled

Behavior

The system represented in Fig. 3.26 differs from that in Fig. 3.25 in that the source does not directly determine either the efforts or the flows on any bonds other than the source bonditself. Oneof the bondsfor the R, C and I elements must have its causal stroke placed adjacent to the junction; since there is only one effort, the other two causal strokes must be placed at the outer ends of their bonds. But which of the three possible patterns do you use?

3.4.

CAUSALITY

C~

0

AND DIFFERENTIAL

"---~’I

EQUATIONS

C’~---’~

137

0 ~I

R (b) integral causalites added

R (a) model

" R

R (d) annotation completed

(c) annotation of causal bonds

Figure 3.26: Flow source bonded to a 0-junction

to give coupled behavior

The short answer to this question is that, given a choice, you use integral causality. The reason should become clear later, but for now simply recognize that differential causalities are defined only for C and I elements (the energy storage elements), and under the following causal conditions: compliance :

---~k-~Ci O~

inertance: Integral causalities conditions: compliance:

~.i

qi

Ii

Oi =_C’~-~-, dei

(3.30a)

ei = Ii .

(3.30b)

apply only to these same elements under the inverse causal

~.i

q~

ci

ei = ei(q~)

q~- 1 J C~ gt~ dr, Ci

(3.31a)

1 [ ei dr. (3.31b) Ii For the present model, integral causality is applied to both the C and I elements in part (b) of Fig. 3.26. This forces impedance causality on the R element. The next thing you do is annotate the effort and flow sides of the C and I bonds as follows, depending in general on the directions of the power convention inertance

:

ei .----~I~ qi

0i = O(Pi)

Pi I~

138

CHAPTER3.

SIMPLE DYNAMICMODELS

Ci

~Ci

half-arrows:

~

or

Thecircles drawnaround(li and/5i are a logically optional device that will prove to be helpful later. Note that ibi is a wayof writing ei. Note also howthe sign changes compensatefor the reversal in the powerconvention half-arrow. These annotations, are madein part (c) of Fig. 3.26. The next step is to annotate the efforts and flows for the remaining bonds, as shownin part (d) of the figure. The causal stroke on the C bond, combined with its effort, dictates that the effort on the R bondbe q/C. The R-element responds, with its admittancecausality, by giving the flow q/CR. The only remaining unannotated effort or flow is the effort on the source bond. This is strictly a causally output variable, and need not be labeled unless it is of interest. Here, it is simply labeled with a large cross, signifying disinterest. EXAMPLE 3.14 Apply causal strokes to the bond graph below, which substitues an effort source for the flow source in Fig. 3.26, and substitues a 1-junction for the 0junction, so as to achieve integral causality on both energy-storageelements. Is this causality compatible?Then, annotate all the efforts and flows except for the flow on the source bond, which maybe labeled with a large X to signify disinterest.

s~ C’~---~- 1

R Solution: All but one of the four bondscan and must have its causal stroke placed adjacent to the 1-junction, since all the flows are the sameand its specification should comeonly from one place. Since this reasoning allows a choice, you can choose integral causality for the C bond, and then integral causality for the I bond. The R bondis forced to have impedancecausality.

3.4.

CAUSALITY

AND DIFFERENTIAL

EQUATIONS

139

Se

The causal strokes are shown above left. Next, you place the designations for the effort and flo~v on the bonds that have integral causality, as shown above right. Finally, to complete the annotation of all the efforts and flows in the graph, you observe that the causal input to the R element is its flow, and this flow comes, causally, from the I element, and equals p/I. The effort on the R element is the causal output of the element, and is written as R times the flow. The final result is as follows:

se

Rp/l~p/l R 3.4.4

Writing

Differential

Equations

Once the bonds in a bond graph have all their efforts and flows annotated according to the the rules given above, the writing of the associated differential equations is a simple matter. There is one differential equation for each C or I element with integral causality. The nmnber of such elements or equations is called the order of the model. The left side of each respective equation comprises the term which is circled in the annotation of its bond. The right side comprises what this term is equal to, as dictated by the bond graph with its annotations. For the 0-junction case of Fig. 3.26, the flow ~ = dq/dt is a causal output of the junction, and equals the sumof the causal inputs S0 (with plus sign due to its power convention half-arrow), p/I (with minus sign due to its power convention half-arrow), and q/CR (with minus sign). The rate of change of momentum [9 = dp/dt is another causal output of the junction, which because there is only one effort on the junction equals the causal input, q/C. Thus, the equations become 0 -junction case :

dq 1 1 d--~ = 40 - ~P - ~-~q,

(3.32a)

dt-

(3.325)

dp 1

~q"

140

CHAPTER 3.

SIMPLE

DYNAMIC MODELS

EXAMPLE 3.15 W¥ite the state differential equations for the bond graph of Example 3.14, using the method described and illustrated above. Solution: The effort D = dp/dt is a causal output of the junction, and equals the suru of the causal inputs e0 (with plus sign), q/C (with minus sign) and Rp/I (with minus sign). The rate of change of flow 0 = dq/dt is another causal output of the junction, which because there is only one flow on the junction equals the causal input, q/C. The equations therefore are 1 -junction

case:

dp

1 R d--~- = e0 - ~q - ~p, dq 1 dt - ]P"

The differential equations above represent the mathematical ~nodels of their systems. The variables p and q are called the state variables, since they are the only dependent variables in the equations. The only other variables are the independent or causal input variables 00 and e0, which could either be constants or specified functions of time. The only other letters in the equations represent parameters, or constants: R, C and I. Each case has therefore two equations and two unknowns, and the pairs of equations are solvable. Their solutions are considered in the next section. Should you be interested in some variable other than one of the state variables, you first solve for the state variables, and then express the variable of interest in terms of them. For example, if you are interested in the flow on the 1-junction, you use 0 = p/I; should you want the effort on the zero junction, you use e = q/C. The annotated bond graph provides adequate information. Do you get differential equations for the uncoupled cases of Fig. 3.25 and Example3.13? The answer is yes, you get one for the single instance of integral causality in each case. For the 0-junction case you get dq d-~- = 00,

(3.33)

dp dt

(3.34)

and for the 1-junction case you get eo.

The fact of uncoupled behavior vastly simplifies these cases; they can be considered to be degenerate. Most problems of interest do not involve such uncoupled behavior. Should one of the C or I elements in Fig. 3.26 or Example 3.14-3.15 be zero, or missing, the number of differential equations reduces to one, and the model is said to be of first-order. First-order models without coupling between the C

3.4.

CAUSALITY AND DIFFERENTIAL EQUATIONS

after step I: eolX

after step II:

141

ealX

p/I (a) IR model

after step I: XTOo

after step II: X-~o

(b) RCmodel Figure 3.27: First-order models or I element and the R element are presented in Fig. 3.27. The methodologyis the same, and so are the results, except for the missing terms. Modelsoften are called by the elements they contain: RCmodels, IR models and ICR models. This description is not unique, however,unless you specify whetheror not their elements are coupled, or which junction type joins them. The RCmodelrepresented in Fig. 3.27 gives, on inspection, the single differential equation dq 1 (3.35) d-~ = 00 - ~--~q, whichrepresents the only variable that is encircled. EXAMPLE 3.16 Find the single differential equation for the IR modelgiven in Fig. 3.27. Solution: The only variable that is encircled is ,5. This is placed on the left side of the equation, and terms on the right side are dictated by the efforts annotated on the other bonds, which according to the causal stokes are causal inputs to the junction. Their signs are dictated by the power convention arrows. The result is

@ R d-7= eo- ~-p. Theconceptof causality will proveto be particularly crucial in the writing of differential equations whenmore complexmodelsare considered, including the use of transformers, gyrators and multiple junctions. At that point (Chapter 5) the logical underpinning of the procedures introduced here should becomemore apparent.

142 3.4.5

CHAPTER3.

SIMPLE DYNAMICMODELS

Summary

Eachbond in a bond graph maybe assigned a bilateral causality, as indicated by a transverse markcalled ~ causal stroke. The effort on the bondis causal in one direction, and the flow in the other. This refers only to computation, that is the determination of independent and dependent variables, and mayor may not be related to physical causes and consequences. The use of integral causality for the energy storage elements -- the C’s and I’s -- expedites the writing of the state differential equations for a system. The causal inputs to these elementsequal the time derivatives of the state variables, that is the generalized velocity dqi/dt for each C element and the generalized force dpi/dt for each I element. Each causal output becomesa function of its state variable, specifically qi/Ci or pi/~i. The causal strokes on the other bonds in the system usually allows these state functions and any input variables to be propagated to the other bonds in the model; each effort and flow thus can be indicated as a function of the state variables and/or the input variables. (Exceptions will be noted in Chapter 5.) It is usually desirable to annotate all the bonds before attempting to write the differential equations. Youshould never annotate an effort or flow in violation of the causal strokes, even if what you wouldwrite is functionally correct; such an action frustrates the purpose of the causal method. Onedifferential equation of first order results from equating the derivative dpi/dt or dq~/dt, for each energy storage element with integral causality, to the causal input of its bond. The signs are determinedby the powerconvention halfarrows. The numberof first-order differential equations that model an entire system therefore equals the numberof energy storage elements with integral causality, and is called the order of the model. Differential causality sometimesis forced. Its treatment for uncoupledmodels has been addressed. Its treatment for other cases is deferred. The use and meaning of the causal method is expanded upon in Chapter 5. The present applications involve modelswith a single junction, whichis a very special case. Guided

Problem

3.9

Solving this problemand the next can be considered simple and essential first steps in your writing of differential equations from bond graph models. A simple RCmodel is excited by an indepedent effort source, as shown in Fig. 3.28. Define the state variables, and write a correspondingfirst-order differential equation(s). Suggested Steps: 1. Apply the mandatory causal stoke to the effort source bond, and then (since it is allowed)apply integral causality to the energy-storageelement.

3.4.

CAUSALITY

AND DIFFERENTIAL

EQUATIONS

143

Figure 3.28: Guided Problem 3.9 2. Annotate the efforts integral causality.

and flows on the source bond and the bond with

3. Annotate the effort and flow on the remaining bond according to the mandates of tile causal strokes. 4. Write the differential equation by equating the circled term to the sum of terms dictated by the causal strokes.

Guided

Problem

3.10

Write differential equation(s) (pp. 126-127, 133). Suggested

for the U-tube modeled in Guided Problem 3.7

Steps:

1. Apply the causal strokes, using integral causality if possible. Identify the order of the model. 2. Annotate the energy storage bonds in the standard way. 3. Complete the annotation of any other bonds, if necessary. 4. Write the differential equations by equating the circled terms to whatever the bond graph dictates. PROBLEMS 3.30 Determine whether or not the behaviors of the R and I elements shown below left are coupled. Determine their behaviors for ~0 -- constant.

3.31 Determine whether or not the behaviors of the R and C elements shown above right are coupled. Determine their behaviors for e0 = constant.

144

CHAPTER 3.

SIMPLE

DYNAMIC MODELS

3.32 I and C elements are bonded to a commonjunction which is excited by an effort or flow source. The behaviors or the two energy storage elements are uncoupled. Identify the possible combinations of junction and source types, and describe mathematically the relations between the behaviors of the sources and the I and C elements. 3.33 Define state variable(s) and write the corresponding differential for the model represented by the bond graph below.

3.34 Write appropriate differential lem 3.23 (p. 130).

equations

equation(s) for the electric circuits of Prob-

3.35 Write appropriate differential Problem 3.24 (p. 130).

equation(s) for the mechanical circuits

3.36 Write appropriate differential Problem 3.25 (p. 131).

equation(s)

3.37 Write appropriate differential (p. 131).

equation(s) for the plunger of Problem 3.26

3.38 Write appropriate differential 3.27 (p. 131).

equation(s) for the fluid system of Problem

3.39 Write appropriate differential Problem 3.28 (p. 131). 3.40 Write appropriate differential 3.29 (p. 132).

SOLUTIONS

TO

Guided

3.9

Problem

R~-------- 1 ~-----~C

GUIDED

equation(s)

for the fluid tank and pipe

for the vibration isolator

equation(s) for the filter

circuit

of Problem

PROBLEMS

The model has one energy storage (the compliance); this can be given integral causality, and therefore must be given this causality. ASa result, the order of the modelis one.

3.5.

SOLUTIONS OF LINEAR

dq

1

DIFFERENTIAL

145

EQUATIONS

1

4. -~ =-~eo- -h-dq Guided

Problem

3.10

1. C ’~ 1 --~-~I 2. C ’-~y~

This bond graph assumes the two original compliances have been combined into one, with one-half the compliance of the individual compliances. 1 p-~//I

3. This step is not necessary. dV

1

dPdt - ~V1

3.5

(Makesure you have the minus sign.)

Solutions

of Linear Differential

Equations

The forms of the linear first and second-order differential equations found above are repeated over and over in the modeling of systems with constant parameters. Even much higher-order differential equations are decomposed into first and second-order components. In this section, solutions to first and second-order differential equations are found in the absence of excitations and for stepwise excitations. These solutions are characterized by certain values called time constants, natural frequencies and damping ratios. Once found, they can be used as components in the responses to more complex excitations. 3.5.1

The First-Order

Differential

Equation

The first-order version of the general linear differential equation with constant coefficients, such as equations derived above for the RCand IR models given in Fig. 3.27, can be written in the form

~-~-[ + x = u(t).

]

The constant r has the dimensions of time, and is known as the time of the model. For many IR models such as that given by Example 3.13 considered in Problem 3.31, and for many RC models such as that equation (3.35) and that addressed in Guided Problem 3.9, the time is either

(3.36) constant and that given by constant

146

CHAPTER 3.

SIMPLE

DYNAMIC MODELS

Figure 3.29: Bond graphs for basic RC and IR models

T = RC or

T = I/R.

(3.37)

The four classic cases cited are reproduced in Fig. 3.29. Only the details of the excitation term on the right side of the equation vary. The fact that ~- has the dimensions of time can be seen by comparing the dimensions of the two terms on the left side of equation (3.36) above. This is the only parameter that characterizes the models in the autonomous case of u(t) = Whenthe right side of equation (3.36) is set equal to zero, describing autonomous behavior, the equation is known as the homogeneous equation, and its solution is known as the homogeneous solution, xh(t):

[ xh(t) = xoe-t/L ]

(3.38)

This decaying exponential is plotted in Fig. 3.30. As suggested in the plot, the tangent to the curve at any time t intersects the axis Xh = 0 at the time t + T. During the same time interval the value of the function is reduced by the factor e = 2.718. These facts expedite the sketching of the curve, as well as underscoring its nature. The solution to a non-homogeneous linear differential equation with constant coefficients, called a forced response, equals the homogeneoussolution plus any particular solution.

3.5.

147

SOLUTIONSOF LINEAR DIFFERENTIAL EQUATIONS

|o0

x ~

0"8~x

p(-t/r)

r = RC

or

r = I/R

0.6 0.4 0.2

0

0

0.5

1.0

1.5

2.0

2.5

3.0 t/r 3.5

4.0

Figure 3.30: Homogeneous solution of first-order models

EXAMPLE 3.17 Find the solution to the differential equation whenthe right side is a constant, that is when dx ~-~- + x = xf = constant. Solution: The simplest particular solution is x = X f, because with this solution the derivative term vanishes. The completesolution is therefore x(t) = xoe-t/’~ + xy. The undetermined coefficient x0 depends on the initial condition. To be general, let the initial time be called T, and the corresponding value of x be x(T). Substitution of this value into the solution above gives Xo =

Ix(T)z~]e~/~,or x = Ix(T) z~]e-(’-v)/~ + x~t >T, whichis plotted in Fig. 3.31. In the special case.with T = 0 and x(0) = 0 this is often called the "step response" of the first-order model. Note that the behavior is essentially the sameas the homogeneous response; a final state xf is approachedexponentially from an initial state. The only difference is that the final state is not zero.

148

CHAPTER 3.

X0 ......

SIMPLE DYNAMIC MODELS

-~

x(r)( I I I I

O0

T

T+ r

t

Figure 3.31: Responseof the first-order modelto a step excitation 3.5.2

Responses perposition

to

More Complex

Excitations

Using

Su-

A continuous excitation u(t) maybe approximated as a sum of two or more steps. Anillustration is given in part (a) of Fig. 3.32. The solution of a linear-differential equationfor an excitation whichequals the sumo] two or morecomponentexcitations equals the sumof the responses to the individual excitations. This is knownas the property of superposition. (It does not extend to nonlinear, differential equations.) Since the excitation in the case of present interest has been decomposedinto a sum of steps, the response to this sumof steps equals the sumof the responses to the individual steps. Theindividual responsesare given by dashedlines in part (b) of the figure. Each of these is of the form of the solution to Example3.17 with Xi(T) = since they are independentof each other. Their sumis given by the solid line. The time constant in this exampleis rather long comparedto the intervals between the steps. The response consequently is considerably smoothed. The response to the sumof steps therefore is not muchdifferent from the response to the actual continuousexcitation. If the approximationstill is unacceptable, the individual steps could be madesmaller and closer together. The property of superposition is the basis of muchof the analysis of linear models,as you will see later on. Thesimplicity and predictability of the behavior of linear modelsis the principal justification engineers havefor designingsystems that behavelinearly. In other cases, the simplicity of linear analysis also tempts the assumptionof linearity, justified or not. 3.5.3

The Second-Order

Differential

Equations

Second-ordermodelscomprise a set of two coupled first-order differential equations. Examplesare given by equations (3.32) (p. 139) and the answer to Example 3.15 (p. 140). The two equations can be combinedinto a single second-order

3.5.

SOLUTIONS OF LINEAR DIFFERENTIAL EQUATIONS

149

Au(t) ~

I

~

AU2

~ /lu3

~

(negau

"v e)

t (a) excitation and step approximation note: x(to)=Oassumed sum of step responses=x(t)

toT~

T2

T3 T4 T~ t T6 (b) step responses and sum~ereof

Figure 3.32: Approximationof a response of a first-order modelusing steps and superposition differential equation, with one of the dependentvariables sacrificed in favor of the other. This second-orderequation is of the form d’~x

dx

a2-~ + al -~ + aox = bl

~t )

+

bou(t).

(3.39)

Dividing this equation by a0 and makingthe following definitions natural frequency : w. = x/-~o/a2, dampingratio : ( = al/2 ax/-d~-~,

(3.40a) (3.40b)

gives the conventional form 1 d2x 2( dx bl du(t) w-~n ~ + -~, -~ + x -ao dt

bo u(t) + --ao -- f(t),

(3.41)

The cases given in Fig. 3.26 and Example3.14 (p 137-139) are perhaps the most commonbond-graph models of second-order systems. The first-order equations for the 0-junction case are given by equations (3.32). Taking the derivative of the first equation and substituting the first gives a second-order equation with the dependentvariable q: dt---T~ ~ ~-~ + ~--~q = -~-. dqo ar~- q" + 1 dq 1

(3.42)

150

CHAPTER 3.

Alternatively, both the first variable p:

SIMPLE

DYNAMIC MODELS

taking the derivative of the second equation and substituting and the second gives a second-order equation with the dependent d2p 1 dp 1 dt ~ + ~-~ + -i-~P

1 = -~qo.

(3.43)

EXAMPLE 3.18 Combinethe two first-order differential equations found in Example3.14 for the 1-junction case into a single second-order differential equation in terms of (i) the dependent variable p and (ii) the dependent variable Solution: Taking the derivative of the first equation and substituting second gives a second-order equation with the dependent variable p:

the

d2p R dp 1 deo dt --~" + 7-d~ + -]-~P = ~-" Alternatively, taking the derivative of the second equation and substituting both the first and-the second gives a second-order equation with the dependent variable q: d2 q - +-~-~ dt--~ R dq + -i--~q 1 = ~eo. 1

EXAMPLE 3.19 Find the natural frequencies and damping ratios corresponding to equations (3.42), (3.43) and the two equations in the solution to Example3.18 above. Solution: When the four equations natural frequency is seen to be

are compared to equation (3.41),

in both cases and all four equations. hand, are different for the two cases: " O-junction 1-junction:

The damping ratios,

the

on the other

: ~ - 2wnRC - 21-R ’ I, 1 R RV~_~ ~- 2wn I - 2

Note also that the right sides of the two differential are different.

equations for each case

3.5.

SOLUTIONS OF LINEAR DIFFERENTIAL EQUATIONS

151

Youshould memorizethe left side of equation (3.41), whichapplies generally for second-order models. It is also useful to rememberthe boxed equation for the natural frequency, since it applies broadly to the second-order modelsof the simple types considered in this chapter, but not all second-order models. You are not urged to me~norize the equations for the dampingratios or the excitation terms on the right sides of the differential equations, howeve2,since they apply only to special cases. In general, you should be able to write firstorder differential equations directly from the annotated bond graph, combine theln to get a single second-orderequation, and by comparisonof its coefficients to those on the left side of equation (3.41), determinethe natural frequency,the dampingratio and the forcing function. 3.5.4

Solution of the Second-Order Equations

Differential

The solution to the homogeneous equation is Xh :

(Xhl

COS ~2dt

~- Xh2 sin03dr )

~

<1 or

02 d

>

(3.44)

in which the dampednatural frequency, ~.dd, is lWd---- V~--~: Wn.]

(3.45)

This is such an important result that you are urged to verify it by substituting equation (3.44) and its first and secondderivatives into the differential equation (3.41). Note that WdiS smaller than wn,althoughthe difference is very small for most commonmodest dampingratios ~ < 0.2. To underscore the distinction, w~often is called the undampednatural frequency. Equation(3.44) is plotted in part (a) of Fig. 3.33 for the commonly encountered special case in whichthe initial conditions are xh(0) = 1 and ~h(0) --~ 0. This special case requires Xhl :

1;

Xh2 --

(3.46)

-02d

~

~

Youare urgedto verify this result also. It is found by substituting the solution into the differential equation, and then substituting the initial conditions. Dampingoften is neglected to simplfy the analysis of vibrating systems. The special case with no damping(~ = 0) producesa non-decaying sinusoidal oscillation. The solution for the special initial conditions becomessimply Xh = Xhl COSW~t, with the dampednatural frequency equaling precisely the natural frequency. Real passive systems normally have at least a pinch of damping, that is a positive value of ~, regardless of whetherit is recognizedin a model. Damping causes the homogeneous solution to decay toward zero over ti~ne, even if ~ is very small. Homogeneous solutions are often called transient solutions as a result. (Recall that a transient is someonewhodoesn’t stick around for long.)

152

CHAPTER 3.

SIMPLE DYNAMIC MODELS

4

8

1.0 X/Xo 0.5

0

-0.5

-1.0 0

2

6

10

12

(a) homogeneous response for x(O)=l, x(O)=O 2.0

1.5

1.0

0.5

0

0

2

4

6

8

10

12

(b) responseto a step

Figure 3.33: Responseof the second-ordermodelto initial conditions and a step excitation

3.5.

SOLUTIONS

OF LINEAR

DIFFERENTIAL

EQUATIONS

153

For 0 < ~ < 1 the behavior is oscillatory. As ~ is increased within this range the oscillations decay faster. They vanish altogether for { >_ 1, which is known as critical damping. The solution for this special case is xh = (Xhl +xh~.w,t)e -~n*

({ = 1 or wd = 0).

(3.47)

For ~ > 1, Wd becomes an imaginary number. Rather than using imaginary coefficients, one writes the solution in terms of real numbers as follows: Xh---- Xhle -t/r1

-t/r -{- 2Xh2e

~ > 1.

(3.4s) The same initial

conditions as above (xh(O) = 0, kh(0) = 0) give V/~--1 ± ~ Xhl,2 -- 2V/~-- I

(3.49)

The particular solution forced response of the sec0nd-order equation (3.41) for the excitation f(t) constant = x), is the sameas th e p arti cular solut ion of the first-order model to the same disturbance, namely, for t > 0, Xp = xI. The general solution, then, is the sum of this constant and the homogeneoussolution. It has two unspecified coefficients, regardless of whether equation (3.44), (3.47) or (3.48) is used for the homogeneous part. The values of these coefficients depend on the specific situation of interest. Twoconditions must be specified. Note that any degree of damping causes the homogeneous or transient part of the solution to decay with time, leaving only the particular or forced response. Perhaps the most commonconditions of interest specify that the values of x and its time derivative 2 are zero at the initial time, t = 0. This situation is com,nonly referred to as a step response. After some algebra, the result for the case of ~ < 1 is

x(t)=xl

II--e-;~nt

(COSWdt--~d

sinwdt)]

~<1.

(3.50)

This response is plotted in part (b) of Fig. 3.33. Cases with { _> 1 also are plotted. Tile response of a linear second-order model to a stepwise excitation is, according to the property of superposition, the sum of the responses of the model to the individual steps. Thus the response to any excitation, like that of the linear first-order model as illustrated in Section 3.5.4 above or of any order linear model, can be estilnated by approximating the excitation by a sum of steps. This idea is refined in Chapter 5.

154

CHAPTER 3.

SIMPLE

DYNAMIC MODELS

EXAMPLE 3.20 The floating, pitching block of Example 3.1 has a mass moment of inertia about its center of mass I -- m(w"~ ÷ h~)/12, where m is its mass. Characterize the behavior of the block after it is released from rest at an angle of ~b0. Neglect damping and the effect of any virtual inertia due to the water. Solution: The model comprises a compliance with the value found in Example 1, and an inertance with the value above, bonded to a common1-junction with the angular velocity ~:

pi/Ii Since the mass equals the mass of the displaced water, or m -- ½whp, the inertance is I = hw(w"~ + h~)p/24. The differential equations are

dp 1 ~ = -~¢, de

1

To combine these equations and eliminate the variable p, you can take the derivative of the second equation and substitute the first: d2¢_ ldp dt "~ I dt

1

This is in the .form of equation (3.41) (p. 149) with the forcing function input f(t) = O, the damping ratio ~ = 0 and the natural frequency

~n - ~ - hi1 + (h/w)~] Since there is no forcing and consequently a zero particular solution, the total solution is the homogeneoussolution, which from equations (3.44) and (3.45) (p. 151)is ¢ = ¢1 cosw~t + ¢~ sinwnt. The angle ~ is specified as ~0 at t = 0, so therefore ~1 = ¢o- Also, ~ = 0 at t = 0, which forces ¢: = 0. The final result is ¢(t) = ¢o coswnt. The motion is a sinusoidal oscillation

at the natural frequency, w,~.

3.5.

SOLUTIONS OF LINEAR DIFFERENTIAL EQUATIONS

155

EXAMPLE3.21 A system is described by the second-order differential equation d~ x 6 dx dt --~ + dt +9x=18’ in which the units of time are seconds. Find the natural frequency and dampingratio of the system, and find x(t) assumingx(0) = 1, 2(0) Solution: Dividing by 9 places the equation in the form of equation (3.41) (p. 149): 1 d~" 2 dx

6dt-- +5-Yi+x =2.

Thecoefficient of the first term gives the natural frequency: 1 w~ = --~ = 3 rad/s The coefficient of the secondterm equals 2~/wn, so that 23 The particular solution is the const~t Xp = 18/9 = 2. Since ~ = 1, the homogeneous solution is given by equation (3.47) (p. 153). The total solution, therefore, is X = Xh + Xp = (Xhl

+ Xh2Wnt)e -w"t

+ 2.

Initial conditions nowcan be recognized.The first initial condition is x(0) = 1 = x~a + from whichXh~= --1. The secondinitial condition is 2(0) = 0 = --Xh~Wn+ Xh~Wn= 3(-- -- 1+ Xh~), from which xhz = --1. The complete solution is therefore x(t) = -(1 + 3t)e -3t + 2.

156 3.5.5

CHAPTER 3.

SIMPLE

DYNAMIC MODELS

Summary

Whena constant resistance and either a constant compliance or a constant inertance are coupled, the result is a linear first-order model that can be characterized by a single time constant, v. The homogeneous solution and the response to a step excitation approach equilibrium exponentially, that is with terms of -~/r the .form e Whenconstant I, C and R elements are coupled, the result is a linear secondorder model that, if the effect of the R element is not too large, can be characterized by a natural frequency, wn, and a damping ratio, ~. The homogeneous solution and the response to a step disturbance feature an oscillation at a damped natural frequency, Wd, that is very close to w~ unless the damping is so large that oscillation barely can be discerned. The oscillations decay within -;~the t. envelope e Whenthe effect of the R element becomes so large that ~ > 1, Wdno longer exists as a real number. A large effect sometimes results from a large value of R, and other times results from a small value. The time responses then are represented by the sum of two exponential terms with distinct time constants. The borderline case of ~ -- 1 produces a special analytical form. The response of any model described by a linear differential equation to virtually any excitation can be estimated by approximating the excitation by a sum of steps. The response, according to the property of superposition, equals the sum of the responses to the individual steps. Guided

Problem

3.11

The problems below give practice in solving for the behavior of first-order and second-order dynamic models. The drum viscometer of Guided Problem 3.6 coasts to a halt from an initial velocity of 1.0 radian per second. Find and sketch-plot the speed and position as a function of time. Suggested

Steps:

1. Apply the causal strokes to the bond graph and annotate all the efforts and flows according to the procedure developed, in Section 3.4. 2. Write the differential

equation.

3. Evaluate the time constant by comparing the differential tion (3.36) or through the use of equation (3.37).

equation to equa-

4. The solution is given by equation (3.38). Evaluate the initial sketch-plot the result. 5. The position of the drum is the time integral sketch-plot.

velocity, and

of the speed. Integrate and

3.5. Guided

SOLUTIONS Problem

OF LINEAR DIFFERENTIAL

EQUATIONS

157

3.12

The current source of the electric circuit of Guided Problem 3.8 (pp. 127, 133) undergoes a step change in time, from 0 amps to 1 amp. The load inductance is 0.01 Henry. The voltage should not exceed 15 V. Choose values for the resistance, R, and capacitance, C which satisfy the constraint and provide a good response. Suggested Steps-" 1. Apply the causal strokes to the bond graph and annotate all the efforts and flows according to the procedure developed in Section 3.4. 2. Write the two coupled first-order

differential

equations.

3. Youare interested in the load current. The limiting voltage is proportional to the rate of change of this current. Therefore, choose as the variable for your combined differential equations the state variable proportional to current. 4. Combine the equations so the variable pendent variable.

chosen in step 3 becomes the de-

5. Compareyour differential equation to equation (3.41) (p. 149). Identify the values of the natural frequency and the damping ratio. 6. Find the homogeneoussolution of the differential ence of two undetermined coefficients.

equation. Note the pres-

7. Find the simplest particular solution possible (with no undeterinined coefficient). Sumthis with the homogeneoussolution to get the total solution. 8. Use the initial conditions to determine the undetermined coefficients. semble the solution.

As-

9. The limiting voltage effectively sets the maximumrate of change of the current. Examine the nature of the solution with this constraint in mind, perhaps using plots given in the book for various dampingratios, to choose an effective damping ratio. Only one undetermined parameter should remain at this point. 10. Use the limitation of the maximumallowed voltage to find an equation that determines the remaining parameter. You should now be able to give values for both C and R, and plot the current as a function of time. PROBLEMS 3.41 The U-tube of Guided Problem 3.7 (p. 126) has a diameter d = 0.25 inches and a length L = 10 inches; the fluid is water (pg = 62.4 lb/ft3). Determine the natural frequency of the motion following a non-equilibrium initial condition.

158

CHAPTER 3.

3.42 The capacitors of the electric lem 3.34 (p. 144) have an initial inductors as a function of time.

SIMPLE

DYNAMIC MODELS

circuits of Problems 3.23 (p. 130) and Probcharge q0. Find the currents through the

3.43 The masses of Problems 3.24 (p. 130) and 3.35 (p. 144) are given an initiM velocity 50, but the springs are initially unstressed. Find the position of the mass as a function of time for (i) system (a), (ii) system (b). Assume 3.44 The fluid tank of Problems 3.25 (p. 131) and 3.36 (p. 144) has the Q0 = 0.05 in3/s begin abruptly. Find the subsequent depth of the water in the tank. 3.45 The plunger of Problems 3.26 (p. 131) and 3.37 (p. 144) is released rest. Find its subsequent velocity. 3.46 Tile water tank of Problems 3.27 (p. 131) and 3.38 (p. 144) has an initial depth of 1.0 m. The tube is full of water that has a zero initial velocity. Find the subsequent depth as a function of time. 3.47 The vibration isolator of Problems 3.28 (p. 131) and 3.39 (p. 144) is leased from rest 0.02 m from its equilibrium position, but suffers no other excitation. Find and sketch-plot the position of the mass as a function of time. Assume k = 900 N/m, m = 2 kg and b = 12 N s/m. 3.48 The filter circuit of Problems 3.29 (p. 132) and 3.40 (p. 144) experiences a step change in the voltage ei, from 0 to 1 volt. Find the output voltage eo, assuming L = 1.0 mh, C = 10 #f and R = 2.0 ohms. 3.49 An underdamped second-order model oscillates with decaying amplitude in the absence of a continuing excitation. Show ~hat the logarithmic decrement, A, defined as the natural logarithm of the amplitude of successive peaks, equals the constant -2~r4/x/~ - 42. Next, invert the relationship to give a function 4 -- 4(A)¯ Finally, relate A to the log of the amplitude ratio for n cycles. 3.50 The mass-spring-dashpot system below exhibits the response plotted at the top of the next page when the mass is struck by a hammer. (The blow effectively imparts a velocity to the mass instantaneously.)

(a) Estimate the natural frequency (w~) and the damping ratio (4) system. Hint: The damping ratio is so small that there is an indistin-

3.5.

SOLUTIONS

OF LINEAR

DIFFERENTIAL

159

EQUATIONS

guishably small difference between the natural frequency and the dainped natural frequency. (b) Estimate the spring constant, given that the mass is 4 kg.

position, inches °.’l[

II II //

a

0.2

^

1

0

-0.4

-0.8 -’1 0

9

4

6

8

’10

’~9

14

"16

18

90

time, seconds 3.51 The cylinder shown below has a radius and height r = h = 10 cm and density equal to 750 kg/m3. It bobs vertically in a tub of water of density 1000 kg/m3 with an observed frequency of 1.35 Hz. Damping is to be neglected.

I

motion

(a) Find the change in the bouyancy force resulting from a vertical displacement to the disk from its equilibrium position. Interpret this result to give a compliance. (b) Solve for the effective inertance, using the observed natural frequency and the value of the compliance determined in part (a). (c) By comparing the result of part (b) with the mass of the cylinder, determine the "virtuM mass" of the water that effectively moves with the cylinder. (d) Find the length of a cylinder of water of the same radius r that has mass equal to the virtual mass. (Note: In estimating the inertance of the

160

CHAPTER 3.

SIMPLE

DYNAMIC MODELS

fluid inside a circular tube with thin-walled ends surrounded by fluid, this length also should be added to each end. Such a correction is widely used in acoustics, and is especially important if the tube is short. If the end of the tube is surrounded by a flange, the end correction should be about 39%greater. These cases are discussed further in Section 10.1.8.) 3.52 A long tunnel or penstock of area Av -- 100 ft 2 and length L = 3000 ft carries water from a large lake to a turbine. In an emergency shutdown the flow through the turbine, initially 1000 ft 3/s, ceases abruptly. A large surge chamber of area As = 1200 ft -~ relieves the pressure surge that occurs. ~ .

12e_.

-

.

surge

~///~chamber penstoc~

_-

turbine

~

(a) Define variables and parameters and draw a bond graph model for the system. Neglect all damping and the inertia of the water in the surge chamber. (b) Write differential

equations of motion.

(c) Solve these equations analytically and sketch-plot the results. high does the water in the surge tank rise, and when does it peak? SOLUTIONS

TO

Guided

3.11

Problem

-- = -~--p dt

GUIDED

PROBLEMS

(Makesure you have the minus sign.)

3.

~" = R 0.00713 = 16.84 seconds 4. ~) = ~)oe-t/~ = 1.Oe-t/16.s4 5.

o

~o

:o

~o ,t,

seconds

40

0

~o

:o

~0

t, seconds

4o

How

3.5.

SOLUTIONS OF LINEAR DIFFERENTIAL EQUATIONS

Guided

Problem

1.

161

3.12

R q/C~q/RC ~o q/C~ p

C dq 1 1 2. -~ = io - -~-~q - yp

dp 1 d-~ = -~q 3. The load current i equals p/I. Therefore, q will be eliminated from the equations abovein favor of p. 1 dq 1. 1 q 1 1. 1 dp 1 4. d2p 2dt C dt RC C -~-~P = -~o RC dt ~P 5. Multiplying this equation by IC ~d substituting p = Ii gives i~d2i I di ~+~+i=io Comp~ing~his with equagion (a.41) (p. 149) gives

~=~;

~=

7~-~v~

(These results can be seen to agree wi~h those in Example3.19, p. 150.) 6. Assuming~ < 1, the homogeneoussolution is given by equation (3.44) (p. 151): sinwat); wa = ~ ~w= 7. Since the excitation is a step of amplitude i0 = 1.0 amps, the p~ticul~ solution is i~ = io = 1.0. Thetotal solution is i = ih + io. 8. Theinitial current is zero, so ihl = --io. Therate of changeof this current also is zero. To determine ih2 it is necess~y ~o take ~he derivative of i(t) at t = 0 and set this equal to zero: di It=o= 0 = [(e )(zowasmwdt+ih2wacoswat)-~w,~(e-¢~"t)(-iocoswat dt +ih~ sinwdt)]lt=o = in~wa + ~w~io {h

: e-{w=t(ihl

Therefore,

i=i0

ih2

1-e -~

COSNdt + ih2

= --~Wn/Wd : --iO~/

coswet

~--

if2,

~_~sin~et

whichagrees wi~h equation (a.g0) (p.

~d

,

laa).

9. ~he voltage which must no~ exceed 1~ V equals q/C = dp/dt = Idi/dt. The plots in p~g (b) of Fig. a.aa (p. 1~2) reveal ghag the l~ger ~ is chosen, ghe l~ger wn can be wi~houg ghe slope of ~he curve exceeding ~y p~ticul~ m~imum;a l~ge ~ is desired. When~ is ~eater than 1, however, ~he curves appro~h their final value with an unnecess~ily long ~ail. ~herefore, ~ = 1 is a good solution. Unforguna~ely, ghe equations above fN1 for ghis critical damping. ~he solution becomes

162.

CHAPTER 3.

SIMPLE

DYNAMIC MODELS

-~t. i = io + (ihl + ih’2wnt)e The initial conditions render ih2 = ihl = --i0 = --1.00 amps, from which di ~. -,o~t ¯ d-~ =~" r e d’2i dt ~ = io(w~ - w~t)e The m~imum voltage occurs whendZi/dt ~ is zero, or whenwrit = 1. At this ti~ dt]m~

= Iw~e = 15V =

e

from which w~ = 15eli = 15 x 2.718/.01 = 4077 rad/s = 649 Hz, and 1 C = w~l - (4077)e 1x .01 - 6.015 pf;

o.1

-~ 6.015 x 10 1.0

0

.3.6

Nonlinear

0.4

0.8

1.2

1.6 t, ms

Compliances

2.0

and Inertances

The compliances and inertances considered thus far are constant, and therefore give linear relations between effort and displacement or momentum and velocity. The differential equations considered thus far also are linear. On the other hand, manyof the resistances considered in Chapter 2 are nonlinear. The concept of nonlinearity is now extended to compliances and inertances. The presence of a nonlinear R, C or I element in a dynamic model produces a nonlinear differential equation. Few nonlinear differential equations possess analytic solutions. Numerical simulation, using MATLAB, is employed in Section 3.7 as a practical alternative. 3.6.1

Nonlinear

Compliances

The spring shown in Fig. 3.34 comprises a uniform strip or beam of elastic material, cantilevered at one end, that wraps around a fixed curved memberas it deflects. Without the curved memberit would have a linear characteristic for small displacements. This member,h.owever, foreshortens the free length of the beam as x increases. Therefore the spring becomes progressively stiffer as it deflects. This type of characteristic is represented by a curved line as in part

3.6.

NONLINEAR COMPLL4NCES AND INERTANCES

163

F

(a) system

(b) characteristic

Figure 3.34: Exampleof a nonlinear spring

ei

qi (a) linear

(b) nonlinear

Figure 3.35: Definitions of compliance (b) of the figure. The energy stored in the spring equals the area under the curve, or V(x) = J g dx.

(3.51)

whichalso is true of the linear special case. A generalized nonlinear compliancecan be described by a nonlinear characteristic, whichcan be expressed algebraically either in a form appropriate for integral causality or in a form appropriate for differential causality. For the preferred integral causality,

Thus, the causal input 0 is integrated to give the state variable q, and the causal output is expressed as the appropriate (nonlinear) function of The relation ei = qi/Ci, which is used in the linear case, can be retained for nonlinear cases, if you so prefer. As can be seen from part (b) of Fig. 3.35, this approachcorrespondsto defining the complianceas the inverse slope of the chord drawnfrom the origin to the point of interest on the characteristic. The term chordal compliance sometimesis used to underscore this definition. It allows the writing of differential equations as thoughthe element were linear.

164

CHAPTER 3.

SIMPLE DYNAMIC MODELS

P

fQdt= V Figure 3.36: Nonlinear fluid compliancedue to gravity If C is not constant, however,the co~nputational requirement is that Ci be a funtion of the state variable qi: [ Ci : Ci(qi),

]

(3.53)

ComputingCi involves about the same effort as the direct computation of ei using equation (3.52). As a practical matter, therefore, there is usually advantage to the approach using equation (3.53). The two approaches are analogousto the use of ei = Ri0i as compared to ei = ei (Oi) for a nonlinear resistance with the same (impedance)causality. Placing differential causality on a nonlinear complianceshould be avoided, if possible. Nevertheless,it can be treated directly; applications of differential causality for both linear and nonlinear compliancesand inertances are given in Chapter 4. The differential relation implied for the general complianceis qi(ei) Ci

(ti = ~qi;

qi = qi(ei).

(3.54)

The constant-effort source, Se ~, could be thought of as a very special type of nonlinear compliance, since it serves to store and dispense potential energy and can be represented by an effort-displacement characteristic. Froma practical viewpoint, however,it is useful to consider the constant-effort source as a unique element. 3.6.2

Nonlinear

Fluid

Compliance

Due to Gravity

The gravity complianceof a tank of liquid is nonlinear if the pressure at the port near the bottomof the tank is not proportional to the volumeof the liquid stored abovethis level. This happens, as illustrated in Fig. 3.36, if the surface area of the tank varies with the depth of the liquid; the pressure is proportional to the depth, but the volumeis not. The general relation is P = P(V). For arbitrary nonlinear compliancewith generalized displacementq and effort e, the relation is e = e(q), as given by equation (3.52). The correspondingparticular relation for the tank is found by using the relations P = pgh, where h is the depth of the water, and V = f2~=: A(z) dz, where z is a running variable for the depth.

3.6.

165

NONLINEAR COMPLL4NCES AND INERTANCES

EXAMPLE 3,22 A conical-shapedtank ends in a small orifice of area A0. Determineits fluid complianceas a function of the volumeV of liquid in the tank. Then, give a fully annotated bond-graphmodel, and write a solvable differential equation in terms of V, with Qi, treated as an input.

Solution: The compliancerelation can be found as follows: V = ~rr ~" dz = 7~z 2 tan 2 adz =

3

3

"

This result is inverted to get the desired form P = P(V):

P=pg ~z

3V

~/~

¯

The expression P = V/C still applies, although at this point finding C is an unnecessaryextra step: C(V)=

V_ 1 P(V)

1/3 [7~tan2c~V2~

3

The fully annotated bond graph below includes the effort P(V) above, and the orifice flow resistance relation Q(P).

.~i,I

c P(V) ’_ P(v) _ UsingBernoulli’s equation for the nonlinear resistance that gives Q(P),

d--~=Qi’-Q(P)=Qi’~-A°

=Qin-Ao 2

2g ~t~n

This equation is solvable, since the only unknowns on its right side are the given excitation Qin and the state variable V.

166 3.6.3

CHAPTER 3. Nonlinear

Compressibility

SIMPLE DYNAMIC MODELS Compliance

A hydraulicaccumulator,shownin Fig. 3.37 part (a), is a rigid chamberinto whicha liquid such as water or oil enters or leaves, and whichalso contains a fixed amountof a gas such as air or nitrogen (the latter to reduce the fire hazard with mineral oils). The gas maybe separated from the liquid by gravity or, more effectively, by a piston or a flexible barrier such as a diaphragmor a bladder. In any case the pressures of the two fluids are virtually equal. Accumulatorsserve the purposes of storing significant amountsof liquid and energy, and reducing pressure and flow surges in hydraulic systems. An accumulator can be modeled at least crudely over a wide variation in pressure by a nonlinear compliance. The flow on the compliancebond refers to the flow of the liquid, whichis relatively incompressible.It is positive for flow directed inward. The integral of this flow equals minus the changeof volumeof the gas. The resulting pressure can be estimated from an appropriate equation of state for the gas. Adiabatic conditions for an ideal gas give p/pk =constant, from whichthe characteristic in the desired form is P0 (3.55) k P - (1 - V/Vo) Here, Po is the "charging pressure" that exists whenthe accumulator is empty of liquid but contains the full charge of gas, and Vois the total volumeof the chamber.Isothermal conditions give the same equation except that k is replaced by 1. Thegeneral shape of the characteristic is shownin part (b) of the figure. Compressingor expanding a gas by a large proportion changes its temperature significantly, potentially causing heat transfer with the surroundings.Heat transfer over a significant temperaturedifference is irreversible. Adiabaticconditions are present only. to the extent that there is too little time for significant heat transfer to occur. Isothermal conditions, for which heat transfer is reversible, are present only whenthe charging and discharging is so slow that significant temperature differences do not develop. Actual charging and discharging rates usually lie somewherebetween these extremes. The system no longer conserves mechanicalenergy, and cannot be modeledaccurately as a pure compliance. For the present this complication is neglected. It is addressed in Chapter 10. 3.6.4

Junctions

With Multiple

Bonded Compliances

The characteristics of three arbitrary compliancesare given in part (a) of Fig. 3.38. Whenthese compliances are bonded to a common1-junction, the three efforts sum for a given commondisplacement; the combinedcharacteristic is the vertical sumof the three characteristics, as shownin part (b). Whenthe compliances are bonded to a common0-junction, on the other hand, the three displacementssum for a given common effort; the combinedcharacteristic is the horizontal sumof the three characteristics, as also shown. The chordal definitions of complianceallow the combinationrules for linear compliances to be retained for nonlinear compliances, assumingdependencies

3.6.

NONLINEAR COMPLIANCES AND INERTANCES

valve for adjusting _~~s

167

~1 valvefor adjusting charge

charge

gas

~e,.Vo-V p~ston liquid

standard symbol

P,V x<~ valve that closes "l~:~]’whenV=Oto prevent neck ~]bladder ext~edfrom through neck~~ ~l ~] being Q neck by gas pressure piston type

bladder type (a) majorconstructiontypes

(b) compliancecharacteristic Figure 3.37: Hydraulic accumulators

168

CHAPTER3.

SIMPLE DYNAMIC MODELS

for commonq C~

/ add ¯ q’s for commone C~ e

\

qi

_.

q

(a) individualcharacteristics

(b) combinedcharacteristics

Figure 3.38: Graphical reduction of multiple compliancesbondedto a junction

P relativistic mechanics

inductancewith saturation

Figure 3.39: Nonlinear inertances are expressed as functions of the common variable. Therefore, for the 0-junction, C = ~-~i C~(e), and for the 1-junction, 1/C = ~ 1/Ci(q). 3.6.5

Nonlinear

Inertances

Aninertance is defined as nonlinear if p is not proportional to 0. This is analogous to a nonlinear compliancein which q is not proportional to e. A happy fact: in classical mechanics,including fluid mechanics,massesare not functions of their velocities, and therefore inertances are linear. Youwill find nonlinear inertances only in relativistic mechanics,and (more importantly to engineers) in the electromagneticdomain.The shapes of the characteristics for these cases are shownin Fig. 3.39. (The axes have been reversed to makethem compatible with the next figure.) The electrical inductor has a limiting generalized momentum because the magnetic flux of real materials reaches a limit; the phenomenon

3.6.

169

NONLINEAR COMPLIANCES AND INERTANCES

s called magnetic saturation. (The weight of many electronic packages such ~s stereos often resides mostly in their transformers, which contain considerable tron in order to minimize magnetic saturation.) The preferred integral causality and the associated computation are similar to those for the compliance: (~)’~1~i O~(p) Eor differential

(~i = (ti(Pi);

Pi = f ~ dt.

(3.56)

causality they are dpi

~,P(~]i)d

(3.57)

3.6.6 Kinetic and Potential Energies The power flowing into a simple compliance or inertance, linear or nonlinear, is the product of the effort and the flow, or P = eo = ~0.

(3.5S)

Since this power is flowing into a reversible energy-storing one-port element, the energy stored is its ti~ne integral: /~90dt=/edq-p.

f Od

(3.59)

The integral f e dq above is integrable, as you have seen, if e is a function of q; this is called potential energy, and is given the traditional symbol Y:

In similar fashion, the integral f c)dp above also is integrable if 4 is a function of p; this is called generalized kinetic energy and is given the traditional symbol,/-4:

[T= f O(p)dp. I

(3.61)

The dual nature of the functions e = e(q) and c~ = O(p) and the energies ~; and 7- are suggested in Fig. 3.40. Note that linearity is just a special case. (The axes of the inertances are inverted from those given before in order to emphasize that momentump plays the same role for kinetic energy that displacement q plays for potential energy. Both are state variables, as you knowalready.) 4Asnoted before, roman(non-script) Vand T traditionally also are used for potential and kinetic energy,respectively.

170

CHAPTER3.

SIMPLE DYNAMICMODELS

e=p

(a) constitutive relations andenergy

(b) definitions of C and

(c) linear case

Figure 3.40: Generalized compliancesand inertances

3.6.

NONLINEAR

COMPLIANCES AND INERTANCES

171

The definition of the chordal compliance, C = C(q), is recalled in part (b) of this figure. The definition of the chordal inertance, I = I(p), is given in a similar fashion as the reciprocal of the slope of the chord: q- q

e =- P lc -= ~ e ’ -: "q

(3.62)

Compliance and inertance moduli are assumed to be chordal unless otherwise described. The slopes of the chords, that is the reciprocals of the compliance and the inertance, are called generalized stiffness and generalized suseeptance, respectively. Constant inertances are linear. It is not necessary for an inertance to be constant for its relation between 0 and p to be linear, however; a change in something else, not infrequently a displacement, might also cause an inertance to change. For example, the fluid inertance of a channel that is empty toward the downstream end depends on the displacement of the fluid interface. In another example, the inductance (electrical inertance) of the coil in a solenoid depends on the displacement of the moving mechanical member. Consideration of displacement-dependent inertances is deferred to Chapters 9 and 10. Constant compliances are linear. Similarly to inertances, however, some linear compliances vary because of changes in a variable other than the two used to define the linearity. For example, the capacitance (electrical compliance) of a capacitance microphone varies with the mechanical displacement of its diaphragm. Potential energies dependent on two displacments are considered in Section 10.2, where they are treated as compliances with two ports or bonds. 3.6.7

Summary

Whenan effort and its associated displacement are proportional to one another, the associated C is constant, and the compliance is said to be linear. When proportionality does not exist, C = C(q) and the compliance is said to be nonlinear. In nonlinear cases it is usually more convenient mathematically to use the particular relation in the form e = e(q) rather than e q/ C(q). Wh en the generalized force is independent of the displacement, you should substitute an effort source, Se, for the compliance. Two or more compliances bonded to a common0-junction produce an equivalent overall coinpliance larger than any of its components. Two or more compliances bonded to a commonI-junction, conversely, produce an equivalent compliance smaller than any of its components In classical mechanics, including fluid mechanics, inertances are linear. In engineering practice, most nonlinear inertances occur because of magnetic saturation in electrical inductances. Analagous to the compliances, these cases usually are most readily treated computationally using 0 = 4(p) rather than gl = p/I(p), although either of these forms imply the favored integral causality. The use of differential causality is treated in Chapter 4. Sometimes a generalized kinetic energy or potential energy depends on more than one independent variable. These cases are deferred to Chapters 9 and 10.

172 Guided

CHAPTER 3. Problem

SIMPLE DYNAMIC MODELS

3.13

This problem gives needed practice in determining a mechanical compliance relation from a force balance and also from the use of potential energy. A simple pendulum comprises a slender uniform rod of mass m and length L pivoted about its upper end. Find the compliance relation between the moment and the angle without restricting the size of the angle. Also, find the chordal compliance, and the state differential equations that describe the oscillation. Thes~ equations are solved in Section 3.7.3. Suggested

Steps:

1. Draw a bond graph model for the system. It is suggested that you start with a 1-junction labeled with the angular velocity. 2. Perform a force-moment balance to get the relation induced moment and the angle.

between the gravity-

3. Get the relation between the moment and the angle a different way: express the gravity energy as a function of the angle, and take its derivative with respect to the angle to get the moment. Make sure the two answers agree. 4. Find the chordal compliance as a function of the angle (although this is not needed in practice). 5. Place causal strokes on the bond graph, and annotate the efforts and flows using the results above. equations. 6. Write the nonlinear state differential introduced thus far cannot solve these equations.

Note that the methods

PROBLEMS 3.53 A well-advertised mattress gets abruptly stiffer above a certain deflection. Suggest how this can be accomplished with special shaping and mounting of coil springs, and briefly discuss its desirability. 3.54 Evaluate the nonlinear compliance relation for the two-dimensional paraboliC tank shown below, assuming integral causality. The shape of the tank is described by y = ax2 and the width, w.

3.6.

NONLINEAR

COMPLIANCES AND INERTANCES

173

3.55 Solve the problem above for the tank being axisymmetric about the y axis. 3.56 The fluid tank of Example 3.22 (p. 165) is now formed in the shape of two-dimensional "V" rather than an axysymmetrical cone. Repeat the analysis for this case. 3.57 The conical tank of Example 3.22 (p. 165), but with no supply tube, drains to atmosphere through a vertical tube of area A and length L. Write a set of state differential equations to model the system. Viscous effects may be neglected, and the resistance assumed to correspond to a single head loss in the tube (giving a pressure drop of ~_pv’). 1 Include the inertance effects of the tube. 3.58 The rolling disk of Problem 3.5 (p. 100) and Problem 3.13 (p. 109) is given a large initial displacement, so the compliance is not constant. (a) Determine the compliance relation. (b) Write a complete set of first-order

state differential

SOLUTION TO GUIDED PROBLEM Guided

Problem

3.13

M -- (mgL/2) sin ¢

L

V = m9-~(1

c= ¢--= M mgL

cos¢); sin

dV L M=a¢)-~’---- mg-~sine

¢

C (~mgL~)sin ¢1 1 p’~//t dp

~=

-(mgL/2) sin ¢

vd-~ = i

I = mL’2/3

equations.

174

3.7

CHAPTER 3.

SIMPLE

DYNAMIC MODELS

Numerical Simulation

Only linear differential equations and rather simple nonlinear differential equations allow analytical solution. Until the 1940s other differential equations could be tackled only through limited and laborious graphical or numerical procedures, often using grindingly slow mechanical calculators. This impacted the modeling of dynamic engineering systems and, indeed, the very design of those systems. In particular, great emphasis was placed on linear or linearized models, differential equations and the systems they represent. Linear systems indeed possess some intrinsic virtue that commendsthem for engineering design. Nevertheless, slavish focusing on them is no longer warranted or necessary. The first major tool developed to solve time-based differential equations automatically was the analog computer. In the 1950s these "simulators" evolved away from their mechanical origins to become largely electronic circuits based on the operational amplifier in which vaxious voltages behaved in time as did the associated variables in the equations. A maze of patch cords gave structure to a particular model, and banks of adjustable resistors allowed parameters to be set. Analog computers were used into the 1970s because of their potentially high speed, which was realized most fully when they were hybridized with digital computers. The convenience, flexibility, accuracy and increasing speed of the digital computer finally pushed the analog computer into obsolescence. Even if you have little facility for solving differential equations analytically, you can utilize them effectively in the analysis of dynamic systems. You don’t even need to understand the numerical methods that the software employs, including the MATLAB integrators described below and elsewhere. On the other hand, such integrators can behave erratically; some knowledge of their algorithms is worth the modest effort required.

3.7.1

State-Variable

Differential

Equations

A model first should be reduced to a set of n independent first-order differential equations. This form results naturally from the procedure described in Section 3.5 and used throughout this book. The use of vector notation is helpful, partly for its economy when a model is of high order, and partly because the MATLAB simulator that is used below is based on it. This "state-space" notation is written as dx = f(x, t).

(3.63)

Boldface is used to indicate vectors; when written by hand, the typesetters’ symbol for boldface is recommended: a wavy underline. The state vector, x, is a vector of the state variables, or Ix1 x2 ... x~]T. The number of state variables, n, is called the order of the system.

3.7.

175

NUMERICAL SIMULATION

3.7.2

Simulation

With

MATLAB

MATLAB is a commercial software package that competes with and can be used in conjunction with C, C++ and Fortran. It is especially easy to use, and has features that commendit particularly for the analysis of dynamic systems. It is employed frequently is this book. Before you proceed further it is desirable that you study Appendix A. The MATLAB simulators presented below "integrate" any set of linear or nonlinear differential equations of the form given by equation (3.63). MATLAB also has a simulator for linear models, presented in Section 5.3.7, that is much more efficient computationally because it employs analytic solutions. The first step in carrying out a simulation is to code the coupled first-order state differential equations (equations (3.63)) themselves. This is done special subprogramcalled a function M-file. The first active line of this file is as follows: £unc~ion

f = ~name~(~;,x)

Unless you make special provision, the only arguments that are communicated to this subprogram from the main program are the time t and the state vector x, which comprises the state variables x(1), x(2) .... ,x(n). The derivatives dxl/dt," dx:/dt,... ,dxn/dt are represented by the vector f, which has components f(1), f(2) ..... f(n). The subprogram normally ends with differential equations in the form f(1) =
> Xn

f (n) = f=f, The last statement above converts a row vector to a column vector. Unless you make special provision, only the elements of this vector, f(1), f(2),..., f(n), are communicated back to the main program. With this basic scheme, any parameters that are needed to evaluate the derivatives must be defined within the function file. This file may be executed thousands of times during a single simulation, so the procedure can be quite inefficient. To avoid the repetition, values can be communicatedback and forth between the main program and the function M-file through the use of a "global" declaration. You also can write your main program as a special "script M-file," again using the global declarations, rather than merely entering it into the commandwindow. These optional elements of good programming practice are presented in Appendix A under the heading Communication Between Files. The function M-file is called with its file name, not the name given on its first line, but you should make them the same. All M-files must be given the extension .m, and be properly opened, in order to be recognized. They may reside on your floppy disk, or better yet on the hard drive. Their directory

176

CHAPTER 3.

SIMPLE DYNAMIC MODELS

location should be entered into the path that MATLAB uses. Procedural details are given in Appendix A under the heading Script Files. Before the simulation can be executed, the initial conditions must be specified as a vector. For example, the initial state vector Xl(0) = 1, x~(0) 0, x3(0) = 2 can be entered x0 = [1 0 2]; Next, the commandto carry out a simulation using a particular for the differential equations can be issued directly: [t,x]=ode23(’’,

function file

[0 10] ,x0)

The command"ode" stands for ordinary differential equation; the "23" is a descriptor of the algorithm used (several alternative available integrator algorithims are described under "help: Function functions and ODEsolvers"). The second argument on the right is a two-element vector comprising the initial time and the time the simulation should stop. The third argument, x0, is the initial condition noted above. Further optional arguments can be added to control the accuracy of the simulation. Accuracy is increased, at the expense of execution time, by decreasing the size of the time steps. These, in turn, are set automatically with respect to error indices that you can specify. Default values are set, otherwise. Details are found by clicking on " ODESET"within the help window. Mistakes in the differential equations often result in instabilites in the solution which take almost forever to execute. Should you wish to abort a simulation that seems endless, type control-C. Often it is wise to try a very small value for the final time, until you are sure the behavior is reasonable. Securing a plot of the results

is probably your final step. The command

plot(t,x) will plot all the variables, with a commonscale, as a function of time. The color code (which depends on the version of MATLAB being used) gives you the state vectors in the order they have been defined. Likely different variables have different magnitudes, and some may be of little interest. Should you wish to plot x(1) and 100 times x(3), for example, enter plot(t,x(:,l),t,x(:,3)*lO0) The state vector x is stored as a three-column matrix; the colon (:) is a wild ¯ card. The time is stored as a vector of the same length. The plot command will execute only if the number of rows in t equals the number of rows in x. For information on howto control the colors of individual plots, or to designate individual data points by various symbols, use the "help" window.

3.7.

NUMERICAL SIMULATION

177

EXAMPLE 3.23 The pendulum of Guided Problem 3.13 exhibits interesting behavior because of the nonlinearity of the gravity-induced compliance. Simulate the system, starting from rest at various angles. Plot the angle as a function of time for a period of three seconds. The differential equations are dp L ~ = -mg-~ sin ¢, dt de 1 d--[~=-~P"

Solution: Letting the state vector be x = [p q]’ and the parameters be L = 1 ft, mg=1 lb and g = 32.2 ft/s 2, the function m-file becomes functionf = pendulum(t,x) L=I; W=I; g=32.2; I=L^2*W/(3*g); f(1) = -W*(L/2)*sin(x(2)); f(2) = x(1)/I; f=f,; The final statement above converts a row vector into the necessary column vector. The main program, entered into the command window with its prompt >>, could comprise only two lines: >> [t,x] = ode23(’pendulum’,[O 3],[0 pi/18]); >> plot(t,x(:,2).180/pi) This produces a simulation and plot of the angle, in degrees, from 0 to 3 seconds, with initial conditions comprising zero velocity and an angle of ~r/18 = 10°. Should you wish to superimpose plots for simulations with different initial conditions, enter the command >>hold on Someresultsareplottedat the top of the nextpage.The frequency of the oscillation for the 20-degree swings can be seen to be almost the same as the frequency for the 10-degree swings. This shows that the assumption of linearity, with its natural frequency of w,, = 1//x/~ = v/3 * 32.2/2 = 6.9498 rad/s, or one cycle in 0.904 seconds, is approximately valid for angles as large as 20° . The frequency for an initial angle of 90° , on the other hand, is noticeably smaller. As the angle approaches 180° , the frequency approaches zero. Notice the hesitation as the pendumlumpasses through its maximum angle. Does this agree with your intuitive sense?

178

CHAPTER 3.

SIMPLE D~NAMIC MODELS

° 180 °120 °60 °0

-60° °-120 °-180

3.7.3

0

1.0

Integration

Algorithms

2"Otime,seconds 3.0

The numericalintegration of differential equations is a major subject in itself. There are manydifferent methodsavailable, even for the ordinary differential equations of concern here. Rather than survey these methods, the presentation belowis limited to the simple class of Runge-Kuttasingle-step algorithms. This choice is based on their simplicity and robustness, plus their reasonable if not optimumefficiency. Should you at sometime do a great deal of numerical simulation, you ought to avail yourself of current literature and software packages. The methodpresented below is a basic l~unge-Kutta integration scheme. The fourth-order version with automatic setting of the time increment is a satisfactory work-horsefor most problemsthe engineer is likely to ,encounter. Runge-Kuttaalgorithms estimate the state x(~o + h) based solely on knowledge of f(x, t) and x(to), whereh is the incrementbetweensuccessive times at which the state is coraputed. Oncex(t0 + h) is found, to is reset to equal the previous to + h, and the "single-step" process is repeated. Throwingaway the known history in this mannerreduces the potential efficiency of the process, but makes the methodself-starting, robust and easy to program. The ease with which the step size, h, can be varied at any stage in the solution is a special advantage. This allows the time step to be madeas large as possible, reducing computer time, without the error exceedingset limits. The fourth-order Runge-Kuttamethodwill be approachedinductively, start-

179

3. 7. NUMERICALSIMULATION estimate of X(to+h) I ~,~f, ,~ actual solution area is estimated ] ~tual / =f (area change in x ~ curve is the actual Io t to+h t

underneath the change in x)

(a) classical Eulerintegration actual k =f improvedestimate [ of the change in ~..~ ~l/estimated (trapezoidal ~ integration) I to to +h t (b) second-orderRunge-Kutta

~.(~o+~

Figure 3.41: First and second-order Runge-Kuttaintegration ing with the lowest order, whichis x(to + h) - xo + hf(xo, to),

(3.64)

wherexo --= x(to). This is the classical Euler formula; the estimated state is direct extrapolation of the slope it(to), as shownin part (a) of Fig. 3.41. 3.7.4

Second-Order

Runge-Kutta

Afar moreaccurate estimate of x(to + h) results if the term f(xo, to) in equation (3.64) is replaced by the average of f evaluated at to and f evaluated at to h. The latter is not knownin advance, but can be estimated using a linear extrapolation of ± = f, as shownin part (b) of Fig. 3.41. Hence, the estimate of f at time to + h is (3.65) f+(to) = f[x+(to), to + in which x+(to) = xo + hf(xo, (3.66) is the sameextrapolation as in the lowest-order method(equation (3.64)). the second-order symmetrical Runge-Kutta algorithm becomes h x(to + h) -~ x(to) + ~ [f(xo, to) + f+(to)], (3.67) where f+(to) is found from equations (3.65) and (3.66). A moreformal methodof derivation, whichcan be extendedto higher orders, employsa Taylor’s series expansionfor x(to + h): Of f(to) x(to + h) = x(to) + f(to)h + ~xx to

+ ~-~ -~- + ....

(3.68)

180

CHAPTER 3.

SIMPLE DYNAMIC MODELS

Here, (Of/OX)to is an n × n Jacobian matrix of derivatives. Equation(3.68) can be approximated by the general second-order Runge-Kutta formula x(to + h) ~ x(to) + A,f(to)h + A2f[(x(to) + #,f(to)h, to + #.~h]h, (3.69) in which A,, A2, #, and #2 are constants to be determined. A second Taylor’s series expansion, this time for the rightmost term of equation (3.67), permits the equation to become x(to+h) ~ x(to)+(Al+A2)f(to)h+A2

#1

Of f(to)

"~X

to

+

which can be compared directly with equation (3.68). through second order gives A2 = 1 - A,, 1 #, = #2 = 2(1- A,)’

to

Equating the terms (3.71a) (3.71b)

where A1# 1 but otherwise is arbitrary. Infinitely manychoices exist, but a highly stable and commonchoice is the symmetrical form AI = A2 - ’

(3.72a)

It, = #2 = 1

(3.72b)

whichgives precisely the result of equations (3.65) to (3.67). 3.7.5

Fourth-Order

Runge-Kutta

The symmetrical fourth-order Runge-Kuttaformulae probably are in most commonuse. Twoequivalent forms of this are 1 x(to + h) -~ x(to) + ~(zl + 2z2 + + Z4),

(3.73)

x(to + h) ~- x(to) + z2 1 e = ~(-zi + 4z2 - 2z~ - z4),

(3.74a)

z, = hf(xo, to),

(3.75a)

(3.74b)

in which 1

1

(3.75b)

1

(3.75c)

z2 = hf[x(t0) + ~Zl, to + ~h], Z3=hf[x(to) + ½z2, to + ~h],

(3.754) z4 = hf[x(t0) + z2, to + h]. The error e in equations (3.74) is proportional to the fifth powerof the time interval, h, and the cumulativeerror is proportional to the fourth powerof h.

3.7.

NUMERICAL SIMULATION

181

Thus, for equivalent accuracy to the second-order formulae, much longer time intervals can be used and the process usually is more efficient. The magnitude of the vector e or its square eTe, where the superscript T means transpose, is a very useful gage for the expected error. Thus if eTe is greater than some preset maximum e~o, the entire calculation for the time step can be thrown away, the time interval halved, and the calculation repeated. Similarly, if eTe is less than some small preset fraction of e~, the time interval for the succeeding ti~ne step can be doubled. There is no point in this latter case for throwing away the previous calculation, which is merely more accurate than necessary. A good integrator such as the fourth-order Runge-Kutta often uses very small time steps near the beginning of a simulation, which subsequently increase by a large factor. A fixed-step integrator would either be relatively inaccurate at first, or would eat up excessive ruu time. The results should be insensitive to large changes in the error gage, although an excessively large error gage will produce excessive time steps and resulting error. The problem is more acute for systems demonstrating sudden changes such as occur when a variable is abruptly limited, as in a mechanical impact. The MATLAB integration algorithms are largely hidden from the user. The algorithm ode23 has some of the features of a second-order Runge-Kutta, and some of the features of a third-order. The algorithm ode45 is a hybrid fourthto-fifth order Runge Kutta. MATLAB also offers other integrators with special virtues. MATLAB also offers other integrators with significant particular virtues, which are described in the "help" window. Somefurther discussion of the use of these programs is given in Section 5.2 and Appendix A. 3.7.6

Summary

Computersimulation is an alternative to analytical solution of the differential equations which describe the behavior of dynamic models. It is especially useful for nonlinear models, where analytic solution is typically extremely difficult or impossible. Sometimes it is even used merely to find the equilibrium of a complex system in order to avoid having to. solve difficult algebraic equations, which itself might require iterative numerical methods with uncertain success. Basic fourth-order Runge-Kutta is the most practical algorithm presented. Although it is not the most efficient scheme available, it is simple and robust, and possesses reasonable efficiency for a very broad range of systems and conditions. The simulators ode23 and ode45 are Runge-Kutta algorithms of between second and third-order and between fourth and fifth-order respectively; they are amongst several MATLAB integrators that are simple to use but not fully documented. Guided

Problem

3.14

This first simulation problem for you to do is about-as simple as a nonlinear system can get, and allows comparison of simulated and analytical solutions.

182

CHAPTER 3.

SIMPLE

DYNAMIC MODELS

A water tank has an area of 6 ft 2 and an initial depth of 2 ft. A drain orifice has an effective area of 1.0 in 2. You are asked to simulate the emptying of the tank, using various error indices, and to compare the result to .an analytical solution. Suggested

Steps:

Model the system with a bond graph. The relation between the orifice flow and the tank pressure is nonlinear. Model it assuming Bernoulli’s equation. The "effective" area is meant to include the effect of the vena contracta, or flow coefficient. o

Annotate the graph in the usual way, noting the nonlinear resistance. Find a differential equation relating the volumeof water left in the tank during draining to time. Write the function m-file, which can be done in as few as two lines of code. Simulate the system from the given initial volume until it is empty. Since you do not knowa priori how long this takes, choose a short time to start with, and extend it with subsequent runs. What does your model say if the volume were to become negative? . Plot your result. Solve the differential equation analytically. This is a rare example of a nonlinear equation that is simple enough to be solved: the method of separation of variables applies. Comparethe result to your simulation.

8o

Investigate errors that result from choosing a large error index. You could change the index by factors of ten. Note that the default index for ode23 is .001, and for ode45 is le-6 (see MATLAB help, "ODESET").

Guided

Problem

3.15

This problem gives you practice simulating a nonlinear second-order model, without having to d(~ muchcoding. The resulting behavior is surprisingly exotic. A vibration absorber comprises a flexible disk with a fused rigid hub and fused outer ring, as pictured in Fig. 3.42. The hub is vibrated vertically with the applied sinusoidal force, as shown. The disk acts like a spring, and the rim acts like a mass at the end of the spring. You may neglect the masses of the hub and the disk in your analysis. The rim weighs 3.0 pounds. The force-deflection characteristic of the disk is nonlinear: 3, F = kx + cx

k --- 3001b/in,

3. c = 30, 0001b/in

3.7.

183

NUMERICAL SIMULATION

top view

section view

~induced vertical motion

Figure 3.42: Disk-and-rim vibration

absorber for Guided Problem 3.15

The linear coefficient k results from the bending stiffness of the disk. The nonlinear coefficient c results from radial or membranestresses set up in the disk because the rather rigid rim resists being pulled inward as the disk bends. The effect of non-zero initial conditions does not decay to reveal the forced part of the behavior unless some damping is introduced. Add a damping term to the force-balance equation with a damping coefficient b = 0.02 lb s/in. To explore a little of the behavior of the disk, you are asked first to simulate its behavior, starting from rest and proceeding for 0.5 seconds, for a small hub displacement equal to hosinw~t with the small amplitude ho = 0.0001 inch at the calculated natural frequency. Then, repeat the simulation for an amplitude ten times higher and a frequency 15% higher~ for two seconds or longer. Next, carry out a third simulation that is the same as the second, except for introduction of the initial values h = x -- 0.058 inch (compression, with rim position nqutral). Finally, repeat the last case, increasing the 0.058 inches to 0.059 inches. Plot the position of the rim as a function of time in each case. Commenton your observations. Suggested

Steps:

1. Model the system with a bond graph, omitting the damping. Note particularly that while the force on the disk spring and the rim is common, the absolute deflection of the rim is not equal to the relative deflection of the disk spring. 2. Apply causal strokes to the bond graph in the standard annotate the bonds according to the standard procedure.

manner, and

3. Write the differential equations. Add the damping term. It is necessary to introduce the rim position as a third state variable simply to have it available for plotting.

184

CHAPTER 3.

SIMPLE DYNAMIC MODELS

4. For very small deflections, the cubic term with the coefficient c in the force characteristic can be neglected. Calculate the resulting resonant frequency, which is virtually unaffected by the damping:wn = 1/v~-~. function file. Set the fre5. Codethe differential equations in a MATLAB quency at the frequency computedin step 4, and the displacement associated with the excitation velocity at 0.0001 inch. 6. Runthe simulation with zero initial conditions for the specified 0.5 seconds. Use integrator ode45with the default error indices. Plot the resulting deflection as a function of time. The amplitude of the oscillation Of a linear oscillator excited with a sinusoidal signal at its natural frequency growslinearly in time, except for the effect of damping.Doesyour result seemto agree with this description? 7. Increase the amplitude and the frequency as indicated, and carry out the second simulation for two, three or four seconds. Plot the result. The amplitude of motion for a sinusoidal signal at any frequency other than the natural or resonant frequency should ultimately reach an equilibrium value. Doesit? 8. Repeatthe secondsimulation, but specify the rather large initial deflections indicated. Plot the result. The response of a stable linear model approachesthe samesteady-state oscillation regardless of the initial condition. Doesit in this case? 9. Repeatstep 8 with the almost infinitesimally larger indicated initial conditions. The results should be drastically different, however. Can you explain this? PROBLEMS 3.59 Consider the vehicle of Section 2.5.3 (pp. 74-77) accelerating on level groundfroman initial velocity of 20 ft/s until equilibrium is reached. It weighs 3500lbs. (a) Approximatethe characteristics given in Fig. 2.39 by simple polynomial expressions. (b) Write the differential equation for the speed: (c) Carry out a numerical simulation. 3.60 Carry out the same steps as in the previous problemfor the acceleration from rest of the motor of GuidedProblem2.2 (pp. 26-27). The load comprises an inertia of 2.0 kg.m2 and a linear resistance of 0.005 N.m-s.

3.7.

NUMERICAL SIMULATION

185

3.61 Carry out the same steps as in the previous two problems for the acceleration of the boat of Guided Problem 2.3 (p. 52) from an initial velocity of ft/s. Assume50%throttle and a total effective weight of 3000 lbs. 3.62 The differential equation for the conical tank of Example 3.22 (p. 165), with zero input flow, can be transformed to the tbrm dV~/dt ’ = -(W) 1/6 by defining a nondimensional "volume" V~ = V/Vo, where ~b is the initial fluid volume in the tank, and an appropriate nondimensional "time" t ~. Simulate the emptying of the tank numerically, and compare the result with the analytic solution (which is readily found despite the nonlinearity). 3.63 Simulate the response of the electric curcuit of Guided Problems 3.8 (p. 127) and 3.12 (p. 157) to the step change in the excitation current. Compare result with the analytic solution which was found because the model is linear. 3.64 Simulate the response of the fluid tank system of Problems 3.25 (p. 131) and 3.36 (p. 144) for the initial conditions specified in Problem 3.44 (p. 158), and (optional) compare the result to the analytical solution.

SOLUTIONS

TO GUIDED

Guided

3.14

Problem

I.

PROBLEMS

-~ c-~-----1 .~---~---~¢ ~=.~ ~

2. Q = Ao ~/-~- where Ao = i ~ .in

~.

3. C

WC I Ao 2V~gy. ~- R

A dV

-Ao

2V

_Ao%/2~g

c=~;-~= (Note that the density p does not affect the result.)

4. function f = tank(t,Y) A0=1/144; A=6; g=32.2; f=-A0*sqrt(2*V*g/h); 5,6. Enter into the commandWindow: [t,VJ =ode23(’ tank’, [0 100], 12) plot(t,V) The result shows that the final time needs to be extended. The plot for 350 seconds (rather than the 100 seconds above) is as follows:

186

CHAPTER 3.

SIMPLE

DYNAMIC MODELS

12 ~ V,ft 8

0

0

100

200

300 t, seconds

If the volumeis started negative, or becomesslightly negative due to numerical errors, the square root of the negative value is treated as zero. Normally, on the other hand, MATLAB treats the square root of a negative number as an imaginary number.

= .Ao

dt

2~lWo = -ao~C-~tl’o 2(v/V- v’-~o) -Aow~t 2

A

g

= 12 1 - 1

32.2 2x12x

e12(1 - 0.003284t) The tank empties at t = 1/.003284 seconds. The plot of this equation is indistinguishable from that above found using ode23. The equation can be plotted with the MATLAB instructions t=[0:0.5:304.5]; V=12*(1-0.003284*t).’2;

Note the array power (the dot before the ")

plot(t,V) Increasing the error index by a factor of 1000produces a negligible difference. This result applies to first-order modelsonly; the behavior of high-order models is apt to be more sensitive to the eror index. The most commoncause of a significant error is an abrupt changein a forcing function, such as at the instant of a mechanical impact.

3.7.

NUMERICAL

Guided

187

SIMULATION

Problem

3.15

Y*-- ~k "[ compressionX~__ q =x-y

C kq+cq~ 0 ~---~-

I

s~ dp -~ = kq + cq 3 + damping force = kq + cq 3 + do dx ~c=--d~

dq _ & _ 1 ~p; --dr -

d

=-~(xosinwt)= xowcos~t

dy 1 cl-~ = ~P dx ~ dt

~ ~0~

COS~t

~. = 1/x/7-C= x/gT-J7= x/K~7~ = x/a00x as6/a= 196.47rad/s 5. Let the state vector be x = [p, q, y, x]’. Then, f=funct±on

d±sk(t,x)

k = 300; c = 30000;

b = 0.02;

I = 3/386;

x0 = 0.0001;

om = 196.47;

~(4) = x0*om*cos(om*t); f(3)

= x(1)/I;

f(2)

= ~(4)-f(3);

f(1)

= k*x(2)+c*x(2)’3+b*f(2);

6. Enter into

the command window:

[t,x]=ode45(’disk’,[0

0.5],[0

0 0 0]);

plot(t,x(:,4),t,x(:,2)) The resulting response:

plot (with some notational

refinements)

indeed shows a growing

188

CHAPTER

3.

SIMPLE

D}%rAMIC

MODELS

0.004 inches 0.002

0

-0.002

-0.0040

0. l

0.2

~ 0.3 0.4 t, seconds

0.5

Changing xo to 0.001 in, increasing w by the factor 1.15 and increasing the final time to 4 seconds gives the following result. The ratio of the arnplitudes of the rim to the hub approaches a constant value of 4.0. 0.008 solid line: rim, y(t)

inches 0.004

0

1

2

3 t, seconds

4

Introducing the initial condition [0 0. 058 0 0. 058] gives the following result. Although the amplitude starts large, it ultimately decays to the same =t=0.004 inches as when zero initial conditions are specified. 0.08 inches

rim motion for hub h(t)= 0.001 sin(1.15%t) and initial compressiondisplacementof 0.058 inches

0.~

-0.04

-0.08

0

1

2

3 t, seconds

4

3.7.

189

NUMERICAL SIMULATION

Increasing the number 0.058 inches slightly, to 0.059 inches, causes a major change in behavior: the rim amplitude does not decay, but approaches a value about 17 times larger than before. rim motion for hub h(t)= 0.001 sin(1.15w, t) and initial compressiondisplacememof 0.059 inches 0.08 inches 0.04

-0.04

-0.08

0

1

2

t, seconds

3

4

There are in fact two grosssly different resulting riln amplitudes for the same hub amplitude, depending on the initial conditions. Should the rim motion be momentarily damped, for example, the large amplitude would vanish leaving only the smaller motion, a phenomenon known as the jump effect. The short explanation is that increasing amplitude means increasing stiffness, due to the nonlinear force term with the coefficient c. Increasing stiffness in turn means increasing natural frequency.

Therefore, if the amplitude is large enough at first, the natural frequency will rise to match the applied frequency, 15% larger than the small-amplitude natural frequency, and the disk resonates with a rimto-hub amplitude ratio of about 67. The behavior depends on the history of the state, or demonstrates hysteresis, as the resonance diagram opposite suggests.

resonant amr;l~mde]

~[~ 0

frequ_~ ~ ’,jumps and

~eresis 1 w/w.

Chapter 4

Intermediate

Modeling

This chapter represents a second pass at the modeling of dynamic systems. The physical systems considered require bond graph models with more than the single junction assumed in the previous chapter. Models for the simple circuits considered in Section 4.1 below employ junction structures comprising any number of bonds and junctions. Transformers and gyrators are added in Section 4.2, with special attention on hydraulic-mechanical, electromechanical and mechanism-based systems. Models often can be simplified or transformed to make them more amenable to analysis, without losing meaning. The most commonmodel equivalences that accomplishthis objective are presented in Section 4.3. This leads to a consideration of the equilibrium state of dynamic models in Section 4.4. Stability, instability and limit cycle behavior are critical properties that emerge. The translation of the bond graphs to differential equations, and the finding of numerical or analytic solutions for imposed initial or boundary conditions, are deferred until Chapters 5 and 6.

4.1

Simple Circuits

Circuits comprise components, each of which has two sides, interconnected at simple junctions. Either the generalized forces or the generalized velocities are commonat each junction. The structure of a curcuit is represented by its pattern of interconnections, that is its topology. The geometric orientations of the components, on the other hand, either are i~nmaterial or are constrained in a simple universal way. Electric, fluid and mechanical circuits are addressed below. Someof the systems considered in Section 4.2 are circuits, also. 191

192 4.1.1

CHAPTER 4. Simple

Electric

INTERMEDIATE MODELING

Circuits

Simpleelectric circuits compriseinterconnectionsof resistors, capacitors, inductors, voltage and current sources. Three-port and higher ported elements such as transistors are excluded, and transformers are deferred to the following section. The junction structures.of the bondgraphs are independent of the nature of its .one-port elements. As a result, one-port elements sometimesare designated in this bookwith the generic symbolZ, in both the circuit diagrams and the bond graphs. The following routine procedure can be employedfor simple circuits without transformers: 1. Represent each node by a zero-junction, since the voltage is commonto all joined conductors. It is helpful to label the junction with a symbolfor its voltage. 2. Represent each branch i by the combination

whereZ~ represents the ith element (source, resistor, capacitor or inductor). This says that the current entering the element on one side emerges unchangedon the other side. It is helpful to label the junction with a symbolfor this current. Discard all bondsfor whicheither e = 0 (ground or reference voltages) i -- 0, since they represent zero power. (An exception is the rare case in which capacitors prevent all of the conductors from having any current, and you still are interested in the distribution of voltages.) Eliminate all junctions from which only two bonds emanate, and join the bonds. Coalesce all directly bonded zero-junctions and directly bonded one-junctionsinto single junctions of the respective types. Steps 3 and 4 do not changethe meaningof the bondgraph, but serve to simplify its appearance. Recall that the electrical and bond-graphsymbols for resistance and for capacitance are R and C, respectively, whereasthe electrical symbolfor inductance is commonlyL and the bond-graph symbol is I. Do not write L in the bondgraph;it is undefined. Instead, write I -- L off to one side.

4.1.

193

SIMPLF~CIRCUITS EXAMPLE 4.1 Applythe four steps aboveto get a bondgraph for the electric circuit

Solution: The bond graph after step 2:

After step 3:

and finally after step 4:

After somepractice it is possible to draw the final bond graph directly, without separating the four steps. EXAMPLE 4.2 Drawa bond graph for the Wheatstonebridge with arbitrary elements Z~. The bond graph elments also may be labeled Z~ (as word bond graph elements). This circuit could comprise any combinationof types of elements, althoughin usual practice the four legs are of the sametype. e2

e’~~[~ ~]~6.

,~

194

CHAPTER 4.

INTER_MEDIATE

MODELING

Solution: The bond graph after step 2 is

eb Z’\l./ \ 1./z2 e~o ~ 1~--~------1 Z4~ i~,~x

0

Zs/1/S~Zs

and after step 3 i~ is Z~

O~

Z~

z

x

Z~

1i~0

¯

Z~

Finally after step 4 it is

z~

You are urged to try this problem using a different node as the reference or ground; although the benzene-ring-like form of the resulting bond graph will remain, the order of the elements will change. This change corresponds to the different choices of variables, and does not imply different physical behavior.

4.1.2 Fluid Circuits Fluid circuits include gravity tanks, compression chambers and fluid restrictions such as orifices or valyes interconnected by pipes or channels that may exhibit significant resistance and/or inertance. A very large gravity tank might be approximated as a constant-pressure source, called a reservoir or surap.

4.1.

195

SIMPLE CIRCUITS

Themodelingsteps for fluid circuits are analogousto those for electric circuits. Afluid restriction is modeledlike an electrical resistor, with an R element bondedto a 1-junction, since the input and output flows are the same. A junction of three pipes (a pipe "tee") or moreis represented by a 0-junction, like electric circuit junction, since the pressures are common and the flows sum. EXAMPLE 4.3 Modelthe hybrid mechanical/fluid sytem pictured below in the form of a bond graph:

r~-~

.~

~

~

surge tank area A ~~~--~toload,

¢

P,

i .--.-.~-

area)

Solution: The gravity tank is analogous to a shunt capacitor in an electrical circuit. (Series fluid compliances cannot be madewithout energy transduction to mechanicalform, whichyou will see later.) The pressure in the reservoir is considered in the modelbelowto be negligible comparedto other pressures. C S ~ T ~ lrl---.--PO O[

R

~ ln---2A,

- LOAD

T = displacementper radian since S~ = O, its bondc~ be dropped

4.1.3

Mechanical

Circuits

Anassemblageof masses or inertances, springs and dashpots that moveeither in simple translation or simple rotation is called a mechanicalcircuit. A translational exampleis given in Example4.4 below. Twoor ~nore componentsjoin at a "mechanical node," where they have a common linear or angular velocity. This is different froman electrical node, at whichthe efforts (voltages) rather than the flows or generalized velocities (currents) are common.The two sides of a spring or dashpot have the sameforce but different velocities, whichalso is unlike electrical capacitors or resistors that have common flows rather than efforts. Onthe other hand, the two sides of a masshave the samevelocities but different forces, like the two sides of an inductor that have the samecurrents

196

CHAPTER 4.

INTERMEDIATE MODELING

1but different voltages. Steps similar to those for electric circuits are as follows: 1. Representeach mechanicaljunction with a 1-junction; if inertia is associated with the generalized velocity of a junction, add its inertance:

I It is usually desirable to label the linear or angular velocity with a customized symbol. 2. Represent each spring and each dashpot, respectively, with a 0-junction bonded to the C Or R element: ~0~

0

Again, it is usually desirable to label the common force or torque with particularized symbol. 3. Coalesce bonded junctions of the same type into a commonjunction. 4. Representeach input by S, S)~ or S~, as appropriate. 5. Delete any bondsfor whicheither the generalized force or velocity is zero. 6. Eliminate all junctions from which only two bonds emanate, and join the bonds. The exampleproduces a loop of bonds, called a mesh. This and similar fourbond meshes can be reduced to tree-like bond graphs, as is shownin Section 4.3. 1Somemodelers use the Firestone analogy between mechanical and electrical variables in order to make the modeling diagrams of similar assembledges of electrical and mechanical components look alike. In this analogy, velocity and voltage are knownas "potential" or "across" variables, and force and current are knownas "flow" or "through" variables. Inertial elements are treated as having zero velocity on one side. Fluid flow is treated like a current or a force, and pressure is treated like a voltage or a velocity. This analogy turns a hydraulic cylinder into a gyrator, and a DCmotor into a transformer. "Line graphs" (originally known as "linear graphs") represent the models. These graphs are more complicated than bond graphs, particularly when transformers or gyrators are employed. This author considers the identification of force as a fiow and velocity as a potential to be unnatural, and prefers to define variables by their nature at a single cut in space rather than with respect to a circuit element" with its two sides. Furthermore, the effort-flow concept which emerges from the single-cut idea extends better to thermodynamic systems and distributed-parameter models.

4.1.

197

SIMPLE CIRCUITS

EXAMPLE 4.4 Translate the system drawn below in the language of mechanical circuits into the language of bond graphs: R!

F~

m,

XI

F2

m.

~’ .C

X2

X3

Solution: After step 2: Rl

x3 l x4

After step 3:

Steps 4, 5 and 6 do not apply to the model as pictured. If the right end were fixed to ground, however, the bond with i4 would be deleted (step 5), leaving a two-port 0-junction that could be eliminated by bonding C2 directly to the 1-junction for i3 (step 6). 4.1.4

Summary

Simple electric, fluid and mechanical circuits comprise assemblies of simple components each having two ports with a commoneffort or a commonflow; The components thus can be modeled with a bond graph having two boundary bonds, a 0-junction or a 1-junction and an R, C or I element. Interconnections also occur at 0 or 1-junctions. It is strongly recommendedthat you define symbols for all commonvariables, and place these symbols on the system sketch first and

198

CHAPTER 4.

INTERMEDIATE L

k!

R~

MODELING R 2

~

(a) mechanical circuit e(t~~

P~ (b) fluid circuit h!

accumulator

tank

(c) electric circuit reservoir i

Figure 4.1: Guided Problem 4.1 the corresponding bond graph junctions second. Bonds with either zero effort or zero flow, and therefore zero power, may be erased from the graph. Guided

Problem

4.1

The three circuit diagrams in Fig. 4.1 imply unique models. Translate the language of these diagrams into the language of bond graphs. It is critical that you attempt these simple cases before you examine the solutions. Suggested

Steps:

1. Define symbols for whichever variable is commonfor the two ports of each element, and place them on the diagrams. 2. Carry out the steps for electric circuits (which apply by analogy to fluid circuits also) or mechanical circuits, as given in this section. Label each junction with the proper symbols from step 1. Do not attempt more than a crude .modeling of the accumulator.

Guided

Problem

4.2

Part (a) and step 1 of this problem are a simple example of the use of junctions. The balance of the problem gives insight into some of the virtues and liabilities of major conventional hydraulic circuits. Hydraulic motors often are driven from simple fixed-displacement pumps run at constant speed so they produce virtually constant flow. See Fig. 4.2, which employs~ standard fluid power symbols. The speed of the motor usually

4.1.

199

SIMPLE CIRCUITS

bleed-off: pump

control valve

~

valve ~co:~’ntrol-’~~load

meter-in: ~

pump

relief valve

pump

relief valve

motor

motor

meter-out:

Figure 4.2: Hydraulic system of GuidedProblem4.2

lve

200

CHAPTER

4.

INTERMEDIATE

MODELING

is controlled by an adjustable valve in one of three ways. In bleed-off control, excess pump flow is diverted through the valve. In meter-in control, the valve admits only the desired load flow, while the excess, if any is diverted through a relief valve. The relief valve is constructed so as to limit the pressure of the pump, /~p. Meter-out control is similar, but the control valve is placed downstream of the hydraulic motor. The small symbol at the bottom of each vertical line represents the vented reservoir or tank, which is common.(This is like the ground symbol in electrical circuits.) (a) Draw bond graphs to model the systems, neglecting leakage and friction and assuming constant pump flow. The terms "RELIEF VALVE," "CONTROLVALVE" and "MOTOR/LOAD"may be used for the relevent components. (b) Find the overall efficiency of these systems when operation is at the load-to-relief pressure ratio (PL/PR) of 0.1 and 0.9, and the load-to-pump flow ratio (QL/QP) is 0.2 and 0.8 (for a total of four combinations), neglecting leakage and friction.

Suggested

Steps:

Draw bond graphs for the systems, focusing on 0-junction and 1-junction for series interconnections.

for pipe tees

Plot the pumpcharacteristics, as modified by the relief valve (if any), a map of flow vs. pressure. Also, plot the four operating load points. Find and plot on the map the corresponding operates. 4°

points at which the pump

Computethe et~iciencies ~ as the ratios of the motor power to the pump power, power being the area of the rectangles defined by the operating points and the origin.

PROBLEMS 4.1 Find the bond graph for the circuit of Example 4.1, moving the ground from the lower end of the inductor to the uppper end. Is the structure of the graph changed? Are the meanings of the effort and flow variables changed? (Pay particular attention to signs.)

4.1.

SIMPLE CIRCUITS

201

4.2 Model the systems below with bond graphs. Identify variables and parameters where appropriate on copies of the diagrams and on the bond graphs.

(a) porous plug

]’ tube

4.3 A hydraulic circuit includes a positive-displacement pump, rigid fluid lines, a rotary hydraulic motor and a load with a rotational inertia. Give the structure for a model of the system in bond-graph form. Describe in words what kinetic energies are represented by the inertance(s) in your model. 4.4 Model the translational

mechanical system below with a bond graph.

F

4.5 Model the rotational

mechanical system below with a bond graph.

flywheel

4.6 Draw a bond graph for the circuit

flywheel

shown below. L

coupling

202

CHAPTER 4.

INTERMEDIATE MODELING

4.7 Infections of the middleear frequently are stubborn and potentially serious, particularly for small children. It is common to insert a small tube surgically through the eardrumto act as as a drain for the infectious material. (The tube is rejected by the body several months later.) The direct effect of the tube on the hearing, presumingthe tube and middle ear are free of liquid, is the present concern. The mechanical loading of the eardrum by the tube appears to be negligible; the question is the effect on the vibration of the eardrumby the acoustic bypass hole through the tube.

Assumethat a pressure drop of 1 lb/ft 2 across the eardrum produces a volumetric displacementof 0.5 x 10-6 ft 3. Also, assumethat the effect of the small boneswhichtrans~nit the vibration is s!mplyto increase the effective mass of the eardrum by 50%.Other parameters are listed below.

2( area = O. 10 in eardrum~ thickness = 0.04 in. 3\ (,. density = 2.0 slugs/ft ~ ~~~J4~~/

~bones

that

vibration ~’~er

to ear

ddle ~’~-//////~~.~/eVaC~U__m~.~ \~,~

transmit inner

ear

(neglect) irnni3

~ area of hole i~.0 ~0~1 in ,~~ (diameter ~ 0.035 in.) density of air = 2.4 x 103 3slug/ft

Describethe physical significance of each of the elementsin the simple bondgraph modelshownbelow. Relate the moduli of its elements to the parameters of the physical system, and evaluate those parameters. The motion of the upper 1-junction is transmitted to the inner ear by an "activated" transformer which you need not consider. (Problem 6.21 (p. 447) requests the response of systemto sinusoidal excitations, both with and without the drain tube.)

4.1.

203

SIMPLE CIRCUITS

.1

~

~ T ~ to inner

ear

o ---.-c,,

o 1

SOLUTIONS

Guided

TO

Problem

GUIDED

PROBLEMS

4.1

(a) mechanical circuit 1.

~x2

[.~x~,

forcesin springs:Fk,, Fl~2 force in dashpot:Fb

~ 2.

after step2:

l,

~ i

~ R

C~

12

C 2 (note: further reductionis possible usingFig. 4.18)

I~

~

12

204

CHAPTER

4.

INTERMEDIATE

MODELING

(b) fluid circuit

11

CI

R~

(c) electric

I2

C,

Rs

R~

circuit

2. after step 2:1 L

~

~ R3

R,

I n°te: /=L I1; ~ 0 ~ li~"~

2.

Rf-~----

"1

s, e(.-~t) li

e,_

R,

li--~C

ls i-..~-- S

R,. after step 3:

after step 4: I

R~-.~----- ~

S~

li~

li~C li~Sy

liaR3

i

R.,’ --R2+R 3

Se e(’~ t) 1, 2’ .~,--~-~-elo

--,~-~-R

R~ Sf~ li’---"~’C

4.2.

205

SYSTEM MODELS WITH IDEAL MACHINES

Guided

Problem

4.2

1. bleed-off:

Pe a~

~ 0

PLQL = -P¢ MOTOR/LOAD

CONTROLVALVE meter-in and S meter-out: ~

2,3.

Pe Q~,

o . 1^ Pt MOTOR~LOAD L RELIEF VALVE CONTROL VALVE

_~. pump, bleed-off-~ .~ pump, meter-in or meter-out Q~’[~. pump characteristic ~

o

0.8o~-~,-~............ .:.,,;2 I t four combinations I ’; .of~,o.L~, . 1 I’ 00.1 PR

0.9 P~PR(relief valve setting) P

Bleed-off: ~1 = ~12 = 0.8; ~/a = ~14 = 0.2. Meter-in or out: r/1 = 0.1 x 0.8 = 0.08; r~2 = 0.9 × 0.8 = 0.72; r/3 = 0.1 ×0.2 = 0.02; r/4 = 0.9 x0.2 = 0.18. (Bleed-off control gives higher efficiency, but meter-in or meter-out control is moredirect and less susceptible to errors in the pumpflow, etc.)

4.2

System Models With Ideal

Machines

The models of Section 4.1 comprise one-port elements (S, Se, Sf, R, C, I) interconnected by junction structures of bonds, 0-junctions and 1-junctions. Constant-parameter transformers and gyrators (T,G) are now added to the junction structures to permit modeling a much broader set of systems. Circuitlike systems are considered first. Certain commonlyoccuring two- and-threedimensional geometric constraints are then recognized.

4.2.1

Electric

Circuits

An idealized electric transformer is shown in Fig. 4.3. The modulus of the transformer, T, equals the turns ratio of the two coils. The presence of four wires but only two different currents is represented by the use of two 1-junctions. The four-port model reduces to the two-port model if the voltage differences across the terminai pairs are adopted as the effort variables. This simpler model also applies directly if the voltages on the lower terminals are considered as ground. The bonds with zero voltage carry no power, and may be excised from the graph. This model implies that the transformer behaves the same at all frequencies, including DC. In fact, a real transformer fails completely at DC, and exhibits

206

CHAPTER 4.

INTERMEDL4TE MODELING

eI = 2 Te

i2 = Til (a) circuit representation 1.-~-~- T -------r

1/2~

or

el .-~-T II

.~I1

- e2 .---~-12

(b) bond graph representation Figure 4.3: Ideal electrical

transformer

deteriorated performance for very rapid changes or high frequencies. These limitations are explored in Section 10.4; for the present they are neglected. Step 2 of the procedure for electric circuits given in the preceding section is expanded to include placement of the four-ported combination of the transformers and 1-junctions. EXAMPLE 4.5 Model the following circuit

with a bond graph:

Solution: The bond graph after

step 2 is

4.2.

207

SYSTEM MODELS WITH IDEAL MACHINES The final graph with the grounds accounted for reduces to

4.2.2

Fluid/Mechanical Circuits with Positive Displacement Machines

The piston-and-cylinder idealized in Example2.3 (p. 63) has a potentially serious practical problem:if the piston fits too tightly, friction is apt to be severe; if it fits too loosely, leakage is boundto result. Further, high pressure expands the cylinder walls, opening or wideningthe leakage path. Someleal~ge across the piston usually is tolerated, becauseit is "internal," producingno puddles on the floor. Fluid is ported to both sides of the piston. The symmetricdoublerod variation shownin Fig. 4.4, can be represented by a combination of two 1-junctions (to equate the inlet and outlet flows and the motions of the two rod ends) and an ideal transformer, muchlike the electical transformer. The modulusof this transformer is the annulus area of the cylinder. If the power conventions arrows across T are reversed, the modulusbecomesthe reciprocal of the annulus area.

...........

P~|Q

T

T=A

P~tQ

Figure 4.4: Double-rod-endcylinder as an ideal machine

208

CHAPTER 4.

INTERMEDIATE MODELING

EXAMPLE 4.6 Three symmetricdouble-rod cylinders are placed in series, as shown. Draw a bondgraphfor this circuit. Pl

three cylindersin series

3

Solution: Since all three cylinders have the same flow rate Q, their fluid bonds should emanate from a common-flow1-junction. Also, no mechanical forces are placed on the left sides of any of the piston rods, eliminating the need for any 1-junctions for the mechanicalbonds.

r, PI

~

T 2 r~ 1£

2f

EXAMPLE 4.7 The single-rod cylinder is cheaper and more commonthan the double-rod cylinder. The two ports of this cylinder sometimesare interconnected, as shown, to give a regenerative circuit with a single external fluid port. Modelthis system with a bond graph, neglecting friction and leakage.

4.2.

SYSTEM MODELS WITH IDEAL

209

MACHINES

Solution: The two faces of the cylinder have the same velocity, which is recognized in the bond graph below by a 1-junction. The flows Qa and Qc are different in magnitude (as well as direction), however, because the areas of the two faces are different. Two transformers represent these geometric relations, one having its modulusequal to the total area of the piston, while the other modulus equals minus the area of the annulus. Finally, the two flows are joined at a pipe tee; the flows sum and the pressure is common,as represented by the 0-juction. F ~ ~- 1 -----~ T¢ [’

F ~’P

oo

10_~ 0’ -

P

T

The regenerative circuit also can be viewed as an integrated unit with a single fluid port and a single mechanical port. The flows Q~ and Qd sum to give Q, which means that the single modulus T on the reduced bond graph, drawn on the right side above, equals the sum Tc + T~, which is the area of the shaft. This result can be deduced directly. The sum of the volumes of the fluid and the rod within a fixed control volume drawn around the unit is a constant. Therefore, every net cubic inch of fluid pushed into the cylinder pushes out one cubic inch of rod, and vice-versa, A rotary hydraulic motor/pump with a shaft that emerges at both ends can be modeled as a four-port machine, like the electric transformer and the doublerod-end cylinder. The now familiar structure has two 1-junctions and an ideal transformer, as shown in Fig. 4.5 part (a). If the shaft does not penetrate through the machine, the three-port version of part (b) of the figure applies. If one of the pressures and one of the torques are zero, the model reduces to a two-port ideal transformer, as shown in part (c). 4.2.3

Losses

in

Positive

Displacement

Machines*

The models of positive displacement machines presented thus far neglect the effects of friction and leakage. Mechanical friction and fluid friction (pressure drops due to orifice flow, etc.) can be modeled by bonding the resistances Re

(a) Figure 4.5: Ports of idealized

(b)

’/

(c)

model of a hydraulic motor/pump

210

CHAPTER

4.

INTERMEDIATE

MODELING

(b) leakage

Figure 4.6: Models for friction chanical machines

and leakage in positive displacement fluid/me-

and RQ to the respective terminal 1-junctions, as shown in Fig. 4.6 part (a). Twotypes of leakage flows are represented in i)art (b) of the figure. Qea Qe2 represent external leakage; the 0-junctions indicate the literal diversion of fluid, which either drips on the ground or floor or preferably is carried back to the reservoir by a special drain line at essentially zero (gage) pressure. The leakage resistances are Rea and R~2. The internal leakage Qi occurs as backflow across the piston or the vanes or the gear teeth, etc, from the high pressure side of the machine to the low pressure side. The resistance to this leakage is Ri. The displacement flow of the machine, labeled QD, is the theoretical flow which would exist were there no leakage. Any of the five resistances R~, RQ, Re~, R~2 and Ri can be included or excluded from the complete model, depending on the magnitudes of the phenomena and how accurate you wish to be. Note that the frictional resistances R~ and RQ can be removed if their moduli are sufficiently small, whereas the leakage resistances R~, Re2 and Ri can be removed if their moduli are sufficiently large. 4.2.4

Losses

With

DC Motor/Generators*

The resistance of the armature circuit of a DC motor/generator, given as R in the model of Fig. 4.7, almost always is important. The brushes of a conventional machine impose a nearly constant frictional torque on the shaft; the bearings contribute some additional torque. The bond graph of part (b) of the figure therefore shows an effort-source element Se with its constant torque labeled as Mo. Sometimesa viscous resistance is also significant enough to be considered; although this is not included in the bond graph or the following analysis, it can be appended readily as an R element. Note the familiar four-port structure of the bond graph. As with the other

4.2.

SYSTEM MODELSWITH IDEAL IVlACHINES

M1 l~

M 0 (friction) el

;~

~/]J R’-~-----

1 e-Re

K --schematic e2[i:(M+M°)/~G e~ (a) 1

RMo

Ge

0

1

0

M2l¢: (b) bond graph (e-Ri)/G

1.0

0.5

G(~/e

(c) typical torque-speed characteristic

0

0.2 0.4

0.6 0.8 1.0 ~o / Ge (d) maximum powersand efficiencies

Figure 4.7: Modelof electrical/mechanical machinewith losses

211

212

CHAPTER 4.

INTERMEDIATE MODELING

machines, one or two of these ports may be removed if its effort is zero. The net applied voltage and torque are taken below as e and M, respectively. The bond graph reveals that i = (M + Mo)/G and ~ = (e - Ri)/G. These two relations can be combined to eliminate the current i, giving the nondimensionalized torque-speed relationship RM -l-Q---, Ge

G~. Q= RMo/Ge, e

(4.1)

in which the parameter Q characterizes the quality of the machine. This equation plots as a straight line in part (c) of the figure. As with other such straightline source characteristics, maximumoutput power occurs at its mid-point, for which RM/Ge = ~G/e = (1.- Q)/2, so that 2e

P,na~ = (1 - Q)2 ~-~’ The energy efficiency

(4.2)

of this operating point, defined as M~/ei, is 1 (1 - Q)2 (1 + Q)-I ~max

The maximumenergy efficiency lower power:

~ ~ ~

is higher,

(4.3)

¯

and occurs at a higher speed and

. a~m~ v _ 1 - ~,

=

(4.4b)

~.

(4.4~)

(These results follow from setting the derivative of an expression for the efficiency equal to zero.) The m~imumpowers and efficiencies are plotted in part (c) of the figure as functions of the quality index Most electric motors will overheat if run continuously at high torque and low speed. Complete specifications include a coefficient of proportionality between the power being dissipated and temperature. Forced air convection increases the range of allowable operation. Specifications for permanent magnet DC motors ideally include values for e, R, Mo and G. The modulus G is typically described as a "torque constant" or "torque sensitivity" with the units oz-in/amp, or ~ a "back EMFconstant" with units of volts/rpm. You should convert these to SI units to take advantage of the fact that one Newton-meter/second equals one Watt and one volt-amp; they should then give the same number. Oftentimes, the torque, speed and current are specified instead at two (or more) conditions, perhaps including stall, the no-load condition, or a "rated" condition, all at some voltage, e. This information is sufficient for deducing the values of R, M0and G.

4.2. 4.2.5 ¯

SYSTEM MODELS WITH IDEAL Case

Study

With

Source

MACHINES and

213

Load*

The DCmotor of Fig. 4.8 is driven by an electrical source with characteristic plotted as S in part (c). It drives a mechanical load, with characteristic plotted as R3 in part (d), through a viscous coupler. The motor is characterized by torque constant of G = 0.15 N m/amp, armature resistance of R1 -- 1.67 ohms as plotted in part (c), and friction torque M0= 0.08 N m. The viscous coupler has a slip velocity proportional to the torque, with coefficient R.) -- 0.025 N s/rad as plotted in part (d). A bond graph model is shown in part (b) of the figure. The viscous coupler has three bonds, one for its input power, one for its output power and one for the the difference or dissipated power. The torque (effort) is commonto the three bonds, while the angular velocities (flows) sum. Therefore, the bonds are joined by a 0-junction. To find the equiSbrium, the source S and the armature resistance R~ are combined in part (c) of the figure to give an equivalent source, ~. I t i s f ound by subtracting the effort for R1 from the effort for S, since e3 -~ el - e2 and e~. = R~i. Next, the load characteristic R3 and the viscous coupler characteristic R2 are combined to give the lower dashed line plotted in part (d) of the figure. This characteristic is found by a horizontal addition of the characterisitics R2 and R3, since ~1 = ~2 + ~3 and ~ = M/R2. The frictional torque M0 then is added to this characteristic, vertically because of the 1-junction, to give the higher dashed characteristic labelled R. This represents the overall load, and is transcribed into part (e). The result at this point is a source characteristic S’ bonded through a gyrator to a load characteristic R. Next, the combination S~ ~ G ~ is reduced to S" ~, as indicated and i.mplemented in part (e) of the figure. The definition of the gyrator (e2 = G¢~ and M = Gi) is used to establish the end points (or any other points) of the S" characteristic, as shown by the dashed arrows from part (c) to part (e). The equilibrium operating state is given, finally, by the intersection of the charactertistics R and S": Mm-= 0.24 N m (which gives .M = 0.16 N m) and q~ = 67 rad/s (which gives ~3 = 60 rad/s). 4.2.6

Two-

and-Three-Dimensional

Geometric

Constraints The complexity of two- and-three-dimensional geometric constraints is not shared by electrical or other circuit-like systems, and is one of the fascinations of mechanical engineering. It is not the purpose of this book to present a course in kinematics, but certain key ideas are reviewed. One approach develops direct relations between the displacements of the inputs and the outputs of the mechanisms. The relations between velocities are found by differentiating the resulting equations with respect to time. A different approach deals directly with velocities rather than the displacements. Sometimes this is easier, particularly when there are multiple independent ve-

214

CHAPTER4.

~~

(a) system

(b) bond graph

S ~e~1.’~-

INTERMEDL4TE MODELING

t.~o°upler. load

M

e3 G’~’- Mm 1.’~"

0

~.R 3

Mo ~

e2

R1

Se

R2

(c) given characterisitics S and RI 20~.¢~e,2 (e’s shownfor volts I~ ~ particular i) and reduced S"

s ---~-G

R 0

Mm ~ 0.4

.

....... ¯ S R s~ ~’~=~_

I I ~I

~.-- ~o~ 0

~ ~"

40

80 120 ~, raa/s (d) given R~, R~, M0~d reduced R

] J

0 \\1 k~

N.m~ 0 4~ "’,

2

-.s"

~ 4 // ~, ~ps

~x

,"

S I ~ s ~ equilibrium ..

~ J

~

~ or,

160

0

".. , , 40

80 12( ~,, raa/s (e) ~rther reduction S"andsynthesis

Figure 4.8: Exarnple of DCmotor with a particular source and load

4.2.

215

SYSTEM MODELS WITH IDEAL MACHINES

(b) pinned joint between two bodies (a) two points on a body

parallel to surfaces at contact

(c) rolling contact between two bodies

(d) sliding contact between two bodies

Figure 4.9: Kinematical constraints

for rigid bodies

locities or displacements. The approach is based largely on the following key kinematical relations, as illustrated in Fig. 4.9: The vector velocities related by

for arbitrary

points A and B on a rigid body are

VB = VA + ~) × tAB,

(4.5)

where tAB is a geometric vector from point A to point B, and ~ is the angular velocity vector for the body. 2. Point A on one member and point B on another member have the same velocities if both points are located coextensively at a pinned or swivel joint between the members, i.e. VA = VB.

(4.6)

3. Two instantaneously contacting points A and B which belong to separate membersin rolling contact also satisfy equation (4.6). (The accelerations of these two points are different, however, unlike the corresponding points for pinned members.) 4. Two instantaneously contacting points A and B which belong to separate members in sliding contact have zero relative velocity in the direction

216

CHAPTER

4.

INTERMEDIATE

MODELING

normal to the surfaces in contact. That is, if n is a vector normal to the surfaces of contact, (VA-- Vt~) ¯ n = 0 (4.7) The development of a bond graph model normally starts with identification of the critical velocity variables, including their vector orientations. These should first be noted directly on a drawing of the physical system. In the bond graph, then, a 1-junction is introduced to represent the magnitude and sense of each such velocity, and the corresponding label is attached to the junction. If the orientation of a velocity is not constrained by the geometry of the system, separate 1-junctions may represent its orthogonal components. Next, the constraints interrelating the various 1-junction velocities are sought and represented by proper junction structures. Finally, any inputs or outputs are represented by bonds or sources, and any dashpots, springs and inertias, etc., are modeled by attached R, C or I elements. The procedhre may be understood best by examples. This chapter focuses only on geometric constraints that can be modeled with junctions and constantmoduli transformers..More complex mechanisms are addressed in Chapter 9. 4.2.7

Case

Study:

.Pulley

System

Consider the pulley and parallel cables shown in Fig. 4.10. Points A and B are commonto both the cable and the pulley. The velocities of points A, B and C are assuraed to be horizontal. The velocities of points along a vertical diameter of the pulley are parallel. They form a trapezoidal vector diagram as shown in part (b) of the figure, where ~A and ~B are chosen arbitrarily. The velocities ~A and kB are related to the velocity of the center of the pulley, ~c, and the angular velocity, ~, by kA = 2C -- r~,

(4.8a)

2~B = ~C "~ r~.

(4.85)

The constraints represented by equations, (4.8) are incorporated into the bond graph shownin part (c) of the figure. Specifically, equation (4.8a) is represented by the upper 0-junction, and equation (4.8b) is represented by the lower junction. Study this graph carefully; its construction is the critical step. The applied forces FA, FB and Fc and the moment Mare added in part (d) of the figure. An alternative perspective results from adding equations (4.8a) and (4.8b) and solving for 2c, and subtracting one from the other and solving for r~b: , 1 kc ~(2A + 2,),

(4.9a)

r~= 1~(~, - 2A)-

(4.9b)

These equations are represented directly by the bond graph in part (e) of the figure. Specifically, the first equation is represented by the left-hand 0-junction,

4.2.

217

SYSTEM MODELS WITH IDEAL MACHINES

I (a) physical configuration

(b) velocity diagram 2./A =2r~

k~X, x / re 0-~-~---.

\0~

XB

T=l/r

(c) bondgraphwith kinematicalconstraints (d) bondgraph with forces added

"~’. Xc TI ~ T~ = I/2 Tz= I/2r (e) alternateto (c)

-T2 Ohx ~0 2r--’~ \I~ (f) alternateto (d)

Figure 4.10: Pulley systein

218

CHAPTER 4.

INTERMEDIATE

MODELING

and the second by the right-hand 0-junction. The forces and moments are added in part (f) of the figure. Both bond graph models must represent correct relationships between the forces and the moments, since the kinematical relationships and the boundary powers are proper. Which ought to be preferred in practice depends on additional circumstances. For example, if the moment is zero, the resulting erasures from the bond graphs of parts (e) and (f) lead simpler interpretation.

EXAMPLE 4.8 The floating lever below represents a minor modification of the pulley system, as long as the lever remains nearly vertical. The only substantive difference is the addition of a force applied at point D. Model the system with a bond graph.

A:XA ]~c

: FA r~ I

(small deflections assumed) T~ =ros/r~s

Tc= 1/2

rs=r,,~/r~,

73=1/r~

Solution: The bond graph of part (f) of Fig. 4.10 is adapted below represent the entire system except for the presence of a force at’ point D. The key geometric constraint regarding point D is ~gD : TArA + TBJ:B;

This constraint

~Tc~O

TA --

rOB tAB ’

TB

rDA : ’FAB --.

is implemented by a 0-junction in the bond graph:

~O~x

0r~o¢

T~ ¢

4.2.

219

SYSTEM MODELS WITH IDEAL MACHINES EXAMPLE 4.9

An epicyclic gear train comprises a sun gear, three kinematically redundant planet gears, a ring gear and a spider which contains the bearings for the planet gears. All these members can rotate about their centers, and the centers of the planet gears also move with the spider. Model the syste~n with a bond graph. A~.

3 planet~ ~\ ~ .rp gears "~~~."-~t-.i-.-~.-.-~

ring

gear

~_’~"1~’~

~spider

~onAA

Solution: The key velocities are defined in the diagram below left. The planet gears share a comnlon velocity with the sun gear at their point of mesh, ~s, and a commonvelocity with the ring gear at their point of mesh, 2~. The kinematical relations are

i. s

"~

= ~(z~ 1+" ~s). ~p

These four equations are represented in the bond graph by four transformers. The last equation also requires a 0-junction to represent its summation of velocities.

note: The 1-junctions are written and labeled first to aid in drawing

iCp T.,,

the graph. They may be removed later since they have only two ports.

~ 1.

Tp = ll& T~= 1/2 T.~= T~= r~

(~s

"

Xs

X.

In usual practice, one of the three input shafts is held motionless at any moment in time, by some external means, reducing the system to a net two-port transmission.

220

4.2.8

CHAPTER 4.

INTERMEDIATE

MODELING

Modeling Guidelines

There is no universal modeling procedure. Nevertheless, as a useful guideline for most cases.

the steps below serve

1. Identify any input and output variables, define symbols for these variables, and place these symbols appropriately on a sketch of the system being modeled. If any of these variables are efforts, also define and label the corresponding flow variables. Use arrows to show directionality. 2. Define and label on the sketch all other variables that appear to be important in relating the input to the output variables. 3. Identify whatever generalized geometric constraints exist between the various flow variables you have defined. This can be done in the form of equations, and often involves defining parameters such as the lengths of levers or the areas of pistons. This step may utilize knowledge gained in your engineeering science courses or in some of the examples given in this book. 4. Start to form a bond graph by drawing any boundary bonds for the input variables. Then draw a 1-junction without bonds for each other flow variable identified in step 2, and annotate it with your symbol for that variable. These ~nay be placed on the sheet, in isolation, wherever you imagine will leave enough room for the missing bonds and elements. 5. Implement the constraint equations of step 3 by placing appropriate junctions, bonds and transformers. Identify the meaning or values of any transformer moduli. At this point you may have a completed junction structure, that is an interconnection between all the flows that you have defined and are needed to relate the input variables to the output variables. 6. If one or more of the key variables are not yet interconnected by bonds, the system may possess one or more gyrational couplings, that is proportionalities between efforts and flows. Identify these, and represent them by gyrators. Note their moduli. 7. Should there still be a key dangling variable or two, examine each of the steps above to identify what is missing. 8. Identify the one-port effort and flow sources, resistances, compliances and inertances that you believe are important. Represent each by a bond graph element Se, SI, R, C or I, using distinguishing subscripts if necessary, and place these where they belong on the bond graph. This step and the next may draw heavily from your knowledge of engineeering science. Whether to include a possible element depends on the relative significance of its power or energy compared with the power o~ energy of the other bonds or elements.

4.2.

SYSTEM MODELS WITH IDEAL

MACHINES

221

9. Identify the constitutive relations for each 1-port element introduced in step 8, that is, give the values of constant moduli and equations or plots for nonlinear characteritstics. Consideration of power or energy may help here, also. 10. If the display of your bond graph is messy or awkward, re-draw it to suit your taste. Eliminate any unnecessary 2-port junctions. You, may wish also to simplify the graph in preparation for analysis. (Some simplifications are presented in Section 4.3.) It is important to complete steps 1 - 3 before starting to draw the bond graph. Most of the critical modeling decisions are made here, and have nothing to do with bond graphs or whatever other modeling language you plan to use. The essential virtue of the bond graph is that its junction structure imposes the conservation of energy, saving you from having to find the relationships between the various efforts independently. You can substitute such relationships in the modeling process for the generalized geometric constraints between the associated flows, but this approach often is harder, and is not recommended as a general procedure. Examination of force relationships is most helpful as a check; this becomes an optional step 11. 4.2.9

Tutorial

Case

Study

The steps above are applied to the hydraulic/mechanical system pictured in part (a) of Fig. 4.11. Hydraulic fluid drives a single-rod cylinder which is connected to a weight through a cable and a pulley. The flow emerging from the rod-end of the cylinde~ drives a rotary hydraulic actuator which in turn drives a flywheel. Friction and leakage are to be neglected in the model you are asked to create. Step 1 asks for the key variables to be defined. The input volumetric flow rate Q and pressure P are given already. The velocity of the piston and cable is labeled in part (b) of the figure a.s ~. The angular velocity of the output shaft the rotary actuator is labeled as ¢. These are the key input and output variables. The flow rate between the cylinder and the rotary actuator also appears to be a useful internal generalized velocity, so it also is labeled (Q~) as part of Step Step 3 asks for the key geometric constraints. The velocity ~ is related to the flow Q by Ap&= Q, where Ap is the area of the piston. The flow Q~ is in turn related to & by Q~ = Aa&, where Aa is the area of the annulus (the area Ap minus the area of the shaft.) This flow also is related to the angular velocity ~) by Q~ = Dq~, where D is the volmnetric displacement of the rotary actuator per radian of rotation. This displacement can be related more specifically to the diameters and width of the actuator chamber (Problem 2.22), if desired. Only now that the key variables and constraints are identified ought one start to draw a bond graph. The input boundary bond and 1-junctions to allow the key generalized velocities to be registered are shown in part (c) the figure, following the instructions of Step 4. Their placement on the sheet is arbitrary, but enough room for likely further development should be allowed.

222

CHAPTER 4.

INTERMEDL4TE MODELING

hydraulic ram

!(Q~.[J....~~]

.|

(~

pulley

cable

rotary motor / with attached ~ I"~--~--~.~ I flywheel (not shown) reserv°ir (a) given information

(b) labeling of key velocityvariables P (c) start of bond graph 1Qr

(d) geometric constraints represented

Figure 4.1i: Tutorial example

I ]

4.2.

SYSTEM MODELS WITH IDEAL MACHINES

223

Step 5 now directs that the constraints found under Step 3 be implemented, as shown in part (d), in the form of three transformers. This structure implies relations between the pressures, forces and torques that have not been directly considered. In place of these relations, however, the conservation of energy has been assumed implicitly. As a result, these relations must be correct. This is an automatic benefit of using the bond graph. Steps 6 and 7 do not apply to the present situation. Step 8 starts by asking about effort sources. The mass at the right is pulled downwardby a constant gravity force, which is represented in part (e) of the figure by an effort source. Acceleration of the mass and the piston requires or imparts an "acceleration force," which is represented by an inertance element bonded to the 1-junction for }. The fiywhe.el itself represents a substantive inertia, which is added to the 1-junction for ¢ by a second inertance. The constitutive relations for these added 1-port elements, requested by Step 9, are Se -- mg, Ir~ = m (ignoring the mass of the piston and piston rod, etc.) and I¢ -- J, where J is the mass momentof inertia of the flywheel. The bond graph is now complete, barring any refinements to represent friction, leakage, inertive or compliance phenomena. Step 10 suggests you consider streamlining or simplifying the graph. One possibility, shown in part (f) the figure, telescopes the transformers T2 and T3 into a s!ngle transformer with modulus T2T3, removes the unnecessary 1-junction for ¢, and straightens out the layout. Subsequent analysis is considerably simplified by combining the two inertances into a single inertance, as you will see later. 4.2.10

Common

Misconceptions

Should you experience difficulty in constructing bond-graph models, the following list of commmonmisconceptions and comments may help. Misconception element.

1: Each physical

component is represented

by a bond graph

Comment: Some physical components require several bond graph elements for their representation. On the other hand, one bond-graph element sometimes can represent more than one physical component. The bond graph is an integration of a set of relationships rather than a set of components. Never invent ad hoc bond graph elements such as three-ported T or G elements or (before readingChapter 10, at least) two-ported R, I or C elements; they have not been defined. And, never use ad hoc symbols in a bond graph, such as L for an inductance or b for a dashpot. Misconception 2: A "LOAD" always can be represented by a resistance element. Comment: Elements indicated by a word in a word bond graph do not have their type of behavior uniquely specified, unlike the standard bond graph elements. A "LOAD,"for exa~nple, could include inertance or compliance effects or a source.

224

CHAPTER 4.

INTERMEDIATE

MODELING

Misconception 3: One can usually draw a bond graph before defining variables.

the key

Comment: Defining the key variables on a drawing of the physical system is the key step in the modeling process, regardless of the choice of the modeling language. The definitions implicitly represent manyof the assumptions being made. You also should determine the relations between these variables before drawing the bond graph. Misconception 4: Gravity implies potential energy, and therefore is always represented by a compliance elelnent. Uomment: Oftentimes an Se element, which also can store energy, is much more appropriate, particularly to represent the (constant) weight of a solid object. In general, first determine the relation between the generalized force and the generalized displacement. Also, rememberthat the inertia of a mass is treated separately from its weight. Misconception 5: Each unique kinetic energy, T, has a unique inertance, I. 1 Comment:The unique kinetic energy equals ~.Iq-, presuming linearity applies, but there maybe several different possible definitions of 0. Thus, I is referred to 0- The same is true of resistances. Misconception 6: Each unique potential energy, ~2, has a unique compliance, C. Comment: C depends on the q it is referred to, just as I depends on its 0Sometimes the choice is important; for example, the formula w,~ = 1/x/T~ requires that C and the I are based on the same q and its time derivative, 0, respectively. Misconception 7: The bonds around a junction need not have either a commoneffort or a commonflow.

Comment: You know better than this when you think about it, but are you always consistent? Following the 10 steps of the modeling guidelines will minimize this type of accident. 4.2.11

Summary

Couplers such as electric transformers, fixed-field DCmotors and positive displacement fluid/mechanical transducers often are modeled with two efforts but with a single flow at one or both of their ends. The corresponding bond graphs have a 1-junction and two boundary bonds at these ends, joined by a transformer or a gyrator. Frictional losses can be represented by resistances bonded to lojunctions, while slip or leakage losses can be represented by conductances (or resistances) bonded to 0-junctions. A hard mechanical constraint between two velocities within a machine can be represented by a transformer. Such kinematical constraints often are between the sums or differences of the velocities used as the variables of the model. Such sums and differences can be represented by a structure of 0 and 1-junc~tions. If the structure is based on knownvelocity constraints, the conservation of energy implicit in the structure automatically ensures that the forces are properly related, apart from frictional losses, which can be added in the form of resistances,

4.2.

225

SYSTEM MODELS WITH IDEAL IVlACHINES

~

flywheel

pa~r [[ spring

Figure 4.12: Guided Problem 4.3 and kinetic energy storage, which can be added in the form of inertances. If the structure is based on known force relationships in the absense of friction and inertia, the proper velocity constraints also must result. Most students who have studied elementary mechanics tend to favor force-based approaches at first, but ultimately recognize that velocity-based approaches are more powerful. You are encouraged to develop skill in dealing with constraints in terms of velocities, and in practice to determine the relationships before attempting to draw a bond graph.

Guided

Problem

4.3

This is a relatively simple problem with geometric constraints. It is important that you attempt it before viewing the author’s solution. For the system pictured in Fig. 4.12, define variables for a dynamic model, draw the bond graph for this model and annotate the graph with the variables. Suggested

Steps:

1. Identify distinct components and the variables used in the definition their characteristics.

of

2. Represent the components by bond graph models. If you have not done so already, identify which variable is commonin the key junction of the model, and represent this by a 1-junction. 3. Assemble the elements of your model into a complete bond graph.

Guided

Problem

4.4

This straightforward problem involves finding a steady- state ~nodel for a circuit with hydraulic actuators. The hydraulic system shown in Fig. 4.13 drives two hydraulic cylinders and one rotary actuator. One of the cylinders is connected regeneratively, while the other has one side connected directly to a reservoir at a negligible pressure.

226

CHAPTER 4.

INTERMEDIATE MODELING

cylinder 2 area of cylinder1: area of cylinder 2: Ac2 area of shaft 1: area of shaft 2: As2 diameter of vane hub: dh diameterof vanetips: d, width of vanes (normal to plane of drawing): Figure 4.13: The hydraulic system of Guided Problem4.4 Drawa bondgraph for the system, identify its elements in terms of the geometric parameters listed in the figure, and relate the input pressure P and fiow Q to the forces F1 and F2, momentM, and velocities 21, 22 and ~. Neglect friction and leakage. Suggested Steps: Drawa bond graph for the system, representing each of the actuators as a simple transformer. (The details of the regenerative circuit are not addressed at this .point.) Pay special attention to the need for. any 0 or 1-junctions. Label the variables P, Q, F1, 21, F2, 22, Mand ¢. Find the modulusof the transformer for cylinder 1, which by convention equals the ratio of the output generalized velocity to the input generalized velocity as indicated by the powerconvention half-arrows that you have chosen. (Both half-arrows must be oriented the same way.) Find the modulusof the transformer for cylinder 2, whichis regenerative. This can be done either of two ways: (a) The brute-intelligence way employsa bond graph for the subsystem, as in Example4.7 (pp. 208-209), with a 0-junction and 1-junction. Note the negative transformer modulus(which could be madepositive by redirecting the two relevant power-conventionarrows). (b) The insightful genius way notes that for every net unit volume of oil that enters the cylinder, an equal volumeof steel shaft must emerge. 4. Find the modulus of the transformer for the rotary actuator. This is

4.2.

227

SYSTEM MODELS WITH IDEAL MACHINES

piston area:

Ap

shaft area:

A s,

cyhnde piston and rod mass: rn ~ tubing

dashpot

~

~__~

spring

pulley mass:

M

pulleyradius:

r

springrate:

k

dashpotcoefficient: b

mlley

Figure 4.14: Guided Problem 4.5 directly related to the volumetric displacement, D, defined as the volume of fluid displacedper radian of rotation. The three loads are attached to the core of the modelby 0 or 1-junctions. Write expressions for the efforts and flows on the load bondsconnected to these junctions in terms of the load variables and the transformer moduli. Perform the appropriate summationsfor these junctions in order to find expressions for the source pressure and flow in terms of the load efforts and velocities.

Guided Problem 4.5 This problemincludes a regenerative hydraulic cylinder as in the problemabove, but emphasizesmechanical constraints ~vith rotation and dynamics. Incompressiblefluid is ported to both sides of a piston, as shownin Fig. 4.14. A belt with a spring at one end and a dashpot at the other end is wrappedwithout sliding arounda pulley comprisinga disk of uniform thickness. Parameters are listed in the figure; phenomena not represented by these parameters, other than gravity, maybe neglected. Modelthe ~ystem, considering the pressure of the fluid, P, to be an independentexcitation and the motionsof the solid parts to be the responsesof interest.

228

CHAPTER 4.

INTERMEDIATE

MODELING

is connected in a regenerative circuit;

model it ac-

Suggested Steps: 1. The piston/cylinder cordingly.

The velocity of the piston is an obvious state variable. Something about the rotatiorl of the pulley also is needed; choose between its angular velocity and the velocity at its periphery relative to the.velocity of the piston. Place corresponding 1-junctions on your paper labeled with symbols for the chosen velocities. Write an equation to represent the kinematical constraint between the velocities you have chosen, and represent it by a structure of bonds, junctions and transformers. Take care with the power convention arrows. At this point you have completed the junction structure of the model, which is the difficult part of the modeling process. Complete your model by adding R, I and C elements, as suggested by the drawing and the list of parameters. Also, add something to represent gravity.

Guided

Problem

4.6

This is a typical problem regarding the interpretation of sales literature for a constant-field DCmotor. A permanent-magnet DC motor is advertised as behaving as follows for a rated armature voltage of 40 VDC:the no-load speed is 1153 rpm, for which the armature current is 0.27 amps; at a "rated" speed of 900 rpm, the torque is 120 oz-in and the current is 3.00 amps. Find the values of G, R and Mo. Also, estimate the stall torque, the maximumpower the motor can deliver and the associated speed and efficiency. Finally, estimate the maximumefficiency the motor can deliver and the associated speed and power. Assume that the model given in Fig. 4.7 (p. 211) applies. Suggested

Steps:

1. Extrapolate the current to the stall condition on a plot of current vs. speed. Compute the armature resistance, R. 2. ComputeG as the ratio of the change in torque to the change in current. 3. Compute M0from the no-load current

and the value of G.

4. Computethe stall torque by extrapolating the characteristic as defined by the two given conditions to zero speed, or use the plot in part (c) of the figure or equation (4.1) (p. 212).

SYSTEM MODELS WITH IDEAL MACHINES

229

The speed for maximuinpowerlies at the mid-point of the characteristic. The associated poweris the product of the speed and the torque. It can be found from equation (4.2), also. The associated efficiency can be found fromequation (4.3) or picked off fromthe plot in part (d) of Fig. The maximum efficiency and the associated speed and powercan be found from equations (4.4). The efficiency and poweralso can be picked off from the plots. PROBLEMS 4.8 Modelthe hydraulic system below with a bond graph. accumulator

DC motor whe ~~

el pump

~

,~,

hydra~lic/~ v.~e

motor

--~_

~o..~s~~

R

flY

I , coupling reservoir

reservoir

4.9 Represent the electric circuit below with a bond graph. C~

4.10 A motor drives a pumpto supply a hydraulic cylinder that lifts a 3000-1b weight. Thetorque-speedcharacteristic of the motoris plotted on the next page. The displacement of the pumpis 0.50 in3/rad, and the area of the cylinder is 5.0 in2. The flow passes through a valve with the characteristic also plotted. weight

cy ~ lind P~’

er val’~ve Q~

Pc

~

230

CHAPTER 4.

600 M, in.lb 4OO

INTERMEDIATE MODELING

1200 pressure, psi 800

200

0

0

400

40

80

120

~ 160 200 ~, rad/s

0

0

40

80 12 Q, in~/s

(a) Modelthe system with a bondgraph, neglecting inertia, friction and leakage. (b) Determinethe pressure Pc in the cylinder, and add it to the plotted pressure Pp-Pcto get an effective load characteristic as seen by the pump. (c) Transformthe given torque-speedmotorcharacteristic to plot an equivalent pressure-flow source characteristic at the outlet of the pumpon the sameaxes as the plot of part (b). (d) Determinethe equilibrium pressure Pp and flow Q, the speed at which the weight rises and the torque and angular velocity of the motor. 4.11 A motor, gear pump, mechanical load and two hydraulic loads are interconnected as shownbelow.

pair

~

(a) Modelthe system with a bond graph, identifying the loads as resistances. (b) Relate the moduliof any transformers in your bondgraph to physical parameters of the system. (c) Express the torque on the motor as a function of the shaft speed and the various parameters.

4.2.

SYSTEM MODELS WITH IDEAL

MACHINES

231

4.12 Modi~’ the solution of Guided Problem 4.4 (pp. 225, 239) to include friction in the three actuators and internal leakage in the rotary actuator. Write the corresponding equations, leaving the friction and leakage relations as unspecified functions. 4.13 A double-rod-end cylinder exhibits internal leakage across its piston proportional to the pressure difference across it. Model the device with a bond graph. 4.14 The three-port hydraulic system below can be used to increase or decrease pressure or flow. (It is used as the heart of a device knownas a hydraulic "intensifier.") Model the system for steady velocity with a bond graph, neglecting leakage, inertia and friction. Start by labeling the flows for all three ports. Relate all transformer moduli to the areas ALp, Asp and ASH. / / / / / / / //.

filled

with incompressible liquid

///////~ ~ shaft area: ASH large piston area: Ate ,,~ ~ small piston area: Asp 4.15 Water is pumpedfrom a well to an elevated tank without the use of electric power. The level of water in a well is significantly above the level of a nearby pond. So~ne of the well water (flow Qm)drives a hydraulic motor on the way to the pond. The motor and pump have displacements of D,. and Dp, respectively. Assmne steady-state conditions; the heights zw and zt may be assumed to be given. The water has density p.

Find a steady-state

bond-graph model. Identify

any parameters in terms of

232

CHAPTER 4.

INTERMEDIATE MODELING

the information given. Neglect friction and leakage. The seepage into the well (flow Qs) maybe represented by a general source. 4.16 A motor drives a fan through a belt drive. The torque-speed characteristics of the motor and the fan are plotted below. The pulley ratio is not specified, but the motorshould not be run for long with a torque higher than its continuous-dutyrating of 4.0 in..lb (to prevent overheating).

12 torque,in-lb

4 0

I

0

I

400

I

I

I

I

I

I

II

1600 800 1200 speed, rpm

(a) Drawa bond graph model of the system, neglecting energy storage elements. (b) Determinethe speed of the motor that maximizesits allowable power. (c) Determine the corresponding speed and torque of the fan, and the pulley ratio that achieves them. Neglect belt losses. (d) Re-drawthe bondgraph, including the effects of kinetic energy associated with the motorshaft and with the fan shaft. (e) Estimate the torque available to accelerate the system, reflected to the motorshaft, for speeds less than one-half the equilibrium. ~, (f) The motor and fan have rotational inertias of 0.2 and 1.2 lb.in..s respectively. Estimate (4-10%) howlong it takes the system to go from rest to one-half its steady speed. 4.17 Derive equations (4.4) (p. 212) that describe the operating point maximum efficiency of a DCmotor with constant field. 4.18 You are asked to purchase a permanent-magnet DCmotor which will be driven by a power supply that has a maximum voltage of 24VDC.The maximum load is to be 10 oz.in, at 2000rpm. A wide assortment of fair-quality motors with Q =_ RMo/Ge= 0.04 are available.

4.2.

SYSTEM MODELS WITH IDEAL MACHINES

233

(a) Determinethe values of G, R and Moand the associated stall torque and no-load speed that wouldrepresent the smallest motor possible (which would therefore have to operate at maximum power). (b) Determinethe valnes of G, R and M0and the associated stall torque and no-load speed that wouldrequire the. smallest powersupply possible (which wouldtherefore have to operate at maximum efficiency). 4.19 A commercial permanent-magnet DCmotor is advertised as having a speed of 2200rpm at 24 VDC and no load, a 4 oz.in, friction torque, a "torque sensitivity" of 14.8 oz-in./amp, a "back EMF"of 11.0 V/Krpmand an armature resistance of 1.9 ohms. (a) Find the value of G in volt-seconds from the "back EMF,"and compare to the value of G in N.m/ampfound from the "torque sensitivity." Do they agree? (b) Find the quality parameter Q =_RMo/Ge,assumingthe modelof Fig. 4.7 (p. 211). (c) Use the information from parts (a) and (b) to computethe no-load speed. Does this approximatelyagree with the advertised value? (d) Estimate the maximum power this motor can deliver, and the associated speedand efficiency. (e) Estimate the maximum efficiency this motor can provide, and the associated speed and power. 4.20 A shunt-controlled DCmotor is shown below. The gyrator modulus for the ideal machinepart of this systemis proportional to the magneticfield and thus to the current if passing through the field circuit; you mayassumeG = wherea is a constant. i~ R~

(a) Assumingconstant supply voltage, ea, find the torque Mand sketchplot as a function of the shaft speed¢. (b) Represent this system (with constant ca) by a bond graph with fixedparameter elements. (Hint: Twoversions are possible, one using a 1junction and the other using a 0-junction; you might try to find both. No gyrator or transformer is needed.)

234

CHAPTER 4.

INTERMEDIATE

MODELING

4.21 A Ward-Leonard drive comprises two DCmotors with their electrical leads interconnected. As pointed out in Example 2.6 (p. 74), the combination acts like a rotary mechanical transformer. The transformer modulus can be varied through adjustments in one of the circuits for the magnetic fields. Investigate the effects of the total of the resistance in the armature curcuits, R, and the brush frictions M0a and M02. The gyrator moduli can be written as G1 and G2, and assumed to be known. The following steps are suggested: (a) Draw a bond graph of the system, neglecting energy storage. Label the input and output speeds and torques, the friction torques and the electrical current. (b) Use the diagram to deduce expressions for the two moments as functions of the speeds and the parameters. (The use of causal strokes can provide helpful guidance in assembling the equations for the elements.) (c) Plot the output moment as a function of the output speed, assuming constant input speed. The use of nondimensional ratios is preferred. (d) For the particular "quality" values ql = RMo~/G~ = 0.01 q2 -- RMo2/G1G~.~ = 0.01, find and plot the efficiency of the drive function of the dimensionless speed r -- G~.~)2/GI~I. Also, find the efficiency, and compare the power delivered at this efficiency to the power that can be transmitted.

and as a peak peak

4.22 The pulley system shown in Fig. 4.10 (p. 217) has negligible moment, (implying negligible mass momentof inertia, since Mincludes inertial torque). Give a reduced bond graph model. (Does it make sense?) 4.23 The pulley system shown in Fig. 4.10 (p. 217) has negligible force, Fc (implying negligible mass, since Fc includes inertial force.) Give a reduced bond graph model. (Does it make sense?) 4.24 A cylinder rolls without sliding. One cable is wrapped around the periphery; another is attached to a small central shaft. cable fm ~cable

(a) Model the system with a bond graph, using the appropriate geometric constraint(s) directly. Omit all inertance phenomena. Give the modulus of any T or G element.

4.2.

SYSTEM MODELS I~TTH IDEAL MACHINES

235

(b) Find a. bond graph by a different route: specialize the result of Fig. 4.10 (p. 212) by noting that 2b = 0. Is the result compatible with that part (a)?

(c) Introduce translational

and rotational

kinetic energy into your model.

(d) Reduce the model of part (c) so it contains si ngle in ertance, an d gi ve its modulus. (Reference: Guided Problem 3.3, p. 107.)

4.25 Model the mechanical system below with a bond graph, and relate the elements in the graph to the given parameters. Assumethat the link rotates no more than a small angle.

4.26 Determine the speed ratio of the rotating shafts of the epicyclic gear train of Example 4.9 (p. 219) when the ring gear is clamped. Also, find the result when the spider is clamped, instead. 4.27 The lever pictured below can be assumed to be massless, and to rotate only through small angles from that shown. The physical dimensions of the mass m are small relative to the lengths rl and r2. Define variables, and model the system with a bond graph. Identify the moduli of the model elements in terms of the given parameters.

4.28 Repeat the above problem with the lever assumed to be slender and uniform with mass M. Let m = 0.

236

CHAPTER 4.

INTERMEDIATE MODELING

4.29 Oil passes through a valve to drive a hydraulic cylinder whichin turn raises a cable that lifts a heavy steel ingot, as pictured. Youare asked to define appropriate variables, using symbolsplaced directly on the drawing and explained by words. Also, relate the moduliof all elements to physical parameters of the systemwhich you should define. Includethe effect of the principal inertia.

to reservoir

valve

ingot

4.30 A DCmotor with constant field, armature resistance R and brush friction M0rotates a winchof diameter d that drives a vertical cable. The cable raises one end of a uniform heavy beamof mass m and length L that is pivoted about its other end, as shownbelow. Nevertheless, the beamremains not far from the horizontal position. Define variables and modelthe system with a bondgraph, including the principal inertial effect. Relate the moduliof bondgraph elements to the given parameters.

DCmotor

~

) cable

m

4.2.

237

SYSTEM MODELS WITH IDEAL MACHINES

4.31 A DCmotor with armature resistance rotates a pinion which drives a rack supported by a linear ball bearing. The rack in turn drives a second gear that is attached to both a flywheel and a spring. Note that the second gear can translate as well as rotate. Define appropriate generalized variables and label the sketch accordingly, state the geometricconstraints betweenthese variables, draw a bond graph model, and relate the moduli of your model to the given primitive physical parametersand any others whichyou define. Inertial effects apart from the mass m maybe neglected. (Hint: Makesure you identify four distinct mechanicalvelocities. DCmotor and pinion R

~/~

~wheel k

//II//////I/I/~I/I// 4.32 A motor drives a cable with a winchof radius rw, as depicted below. The cable is wrappedaround a massless pulley of radius rp, and is terminated in a dashpot of modulusb. The pulley pulls a weight W1which is supported by two rollers that each weigh

(a) Define key velocities and forces on a copy of the drawing. Note particularly that the centers of the rollers W2do not moveat the sameveloc!ty as the slab W1. (b) Drawa bond graph of the system, including the effects of both the weightsand the inertias of the slab and the rollers. (c) Evaluate the moduli of the elements in your bond graph in terms the given parameters. Suggestion: The kinetic energy method may be particularly helpful for the rollers.

238

CHAPTER4.

INTERMEDIATE MODELING

4.33 A proposed sound ~novie camera must have a steady flow of film past the sound head despite the discontinuous loads and motion at 24 Hz (plus multiples) of the gate, with its claw drive, which acts on the sprocket holes. One path for vibrations to reach the sound head from the claw drive is through the pulley system. To minimizethis transmission, flexible belts are proposedon the pulleys, as shown. Treat the claw mechanismsimply as an independent unsteady force on the top of the claw drive system. Find a bond graph model and identify each componentwith a couple words (or more). Neglect variations in the magnetic torque in the motor.

’top view

claw drive

I sounddrive i

!

sound

drive~ opticalaxis fixed pin flywheel

flywheel

side view pulleyswith"flexible belts

SOLUTIONS

TO

GUIDED

PROBLEMS

Guided Problem 4.3 1. Motorshaft: effort is torque, Mm; flOWis angularvelocity, ~m. Flywheel:effort is torque, Mr;flowis angularvelocity, Largergear and spring: effort is torque M~;flow is angularvelocity ~.

4.2.

SYSTEM

2.

MODELS

motor:

WITH

IDEAL

S e---L--~ 1 .---~-

239

MACHINES

G ---~-

Ro gear pair:

viscous coupling:

~ 0

flywheel:

T R,.

spring:

~ C

assembled bond graph:

Guided

Problem

4.4 I^.__._.,_T 2 F...

1.

",-Po

F~ l~ 2.

notes: Q1 is the flow to actuator 1, and Q2 is the common flow through actuator 3 to ~ctuator 2. The power to the reservoir shown is neglected, because its pressure is virtually zero

A" 2

M

T~-

3. (a)

~-.-~ ~

To-~I.

F.__~

r° -’-’- x: Q.,---7-° -E-~ 52 _ 1 Ta ~ Qa A’c2 Qe Q~ + Q~

Q~/~

(b) Q~ = A~; therefore, 4. Q2 =

D6; therefore,

Tb~

~2 Qb -

+ Q~/~z

1 Ac2 - As2

A~ - (A~

’

- A~z)

T Ts - Q2 - D - w(r~ - r~)’

where r~ is the radius

to

the tip of the vanes, rs is the radius of the shaft and w is the width (not the thickness) of the vanes. See Guided Problem 2.5 (p. 69) and its solution further explanation. P=TIFI = TzF: + T3M

T,F, + T,M

=~11 + ~ T,F, ~ =2,/T, (choice) TI

T.,,M T3

240

CHAPTER

Guided

Problem

At

INTERMEDIATE

4.5

1. detailedmodel:

A~-A~ T~ = A~

4.

Pep 0

alternativedetailedmodel:

To = 1/Ap: To =A~-As ~ T2

A~,= areaof the piston A, = area of the shaft

desirable simplemodel:

l/As =l/r =m+M =l/k R=b

MODELING

4.2.

SYSTEM MODELS WITH IDEAL

Guided

Problem

241

MACHINES

4.6

~---stall current = 12.71amps i, amps M~ --~ 1. e-Ri

0.27 0

R =

R

¯~

3.00

G ~ 1~

S~

9~ 1153 speed, rpm

e 40 -= 3.147 ohms istau 12.71

120 G = change of torque _ _ change of current (3 - 0.27) x 16 x 39.37 x 0.22481 = 0.310.4 N.m/amp The citing of the 120 oz.in, of torque is redundant. The value of G can be found without this datumas follows: e - Ri = G~; therefore, G

e-Ri= I

(e-.Ri

=40-3.147x0.27

0.3242 volt.s/rad

These units are the same as N-m/amp.The value is slightly different from that found using the given torque. The difference could be due to round-off by the advertiser, or errors in the assumedmodel. WhenM = 0, Mo = Gi = 0.3104 × 0.27 = 0.084 N.m M~t~,u = Gist,~u - Mo= 0.3104 x 12.71 - Mo= 3.86 N.m. The linear extrapolation gives the samevalue. 3.86 ~maw

=

T

1153 27r X 6~ = 117 watts

X ~

3.145 x 0.084 RMo - 0.0213 Ge 0.3104 x 40

= ~(~

-

~)~

= ~.~

242

4.3

CHAPTER

4.

INTERMEDIATE

MODELING

Model Equivalences

The bond graph models developed thus far often can be simplified readily, without any loss in accuracy despite someloss in detail. Such simplifications usually are highly desirable for two reasons: the essential character of the simpler model is easier to grasp, aiding design, and any subsequent analysis is likely simplified. The most important reduction step has been treated already: the combining ’ of multiple R, C and I elements bonded to a single 1-junction or 0-junction. It is essential to remember that each element can be combined only with other elements of the same type that happen to be arrayed on the same junction. Remember also that two junctions of the same type connected by a bond can themselves be reduced to a single junction. Further, a junction with only two bonds can be removed and the bonds connected, as long as the two power halfarrows are directed similarly. If they are not so directed, you may re-direct the half-arrow on one of the bonds and compensate by changing the sign of one of its variables and the sign of the modulus of any associated one-port element. Further key reductions are given below. First, however, an important elaboration is considered which allows you to dispense with all one-port elements with characteristics that do not pass through the origin, except for the special constant-effort and constant-flow sources. 4.3.1

Thevenin

and

Norton

Equivalent

Sources

and

Loads

This book has repeatedly represented sources with characteristics that are biased, that is do not pass through the origin. An electric motor, for example, has a torque at zero speed, and a speed at zero torque; an impeller pump unit has pressure at zero flow, and a flow at zero pressure. Resistive loads with biased characteristics also have been employed. The pressure of the water sprinkler system of Section 2.1, for example, is not zero at zero flow, and the drag of a vehicle does not vanish when the speed approaches zero. Rather than directly employ a source with these characteristics, it is usually better, for purposes of analysis or simulation, to construct the source by assembling more primitive elements: a constant-effort or a constant-flow source, an unbiased resistance and a junction. The same type of structure also can represent biased loads. The same approach also can replace compliances having characteristics that do not pass through the origin with those that do. These replacements are particularly useful for straight-line characteristics. The resistances which result are constants, instead of functions. The sources also are constants, or independent functions of time. The conversion from a bond graph to differential equations is thereby expedited, as you will experience in the next chapter. The creation of a biased sink from an unbiased resistance is shown in part (a) of Fig. 4.15. Linear examples are shown by solid lines, and nonlinear examples by dashed lines. In the first example of a sink, the effort eo for 0 = 0 is marked by a heavy dot. The Thevenin equivalent sink is based on the resistance characteristic that results whenthe whole curve, including this point,

4.3¯

243

MODEL EQUIVALENCES

I sinks/~/

e=eo+e~

~---[eoW---s~= R

q qo q=qR-qo10R qo ¯ . .-~.eO~Sf

Norton

R sources

S, -~--~-1~ eo e=eo-e~

Thevenin -~

Se~ eo

e~q

q

R

e[ I s~eo

qo

qo

0 ~/~" Norton .... [ilq~=qo-q~

e = S~

R Figure 4.15: Thevenin and Norton equivalent

sinks and sources

244

CHAPTER 4.

INTERMEDIATE

MODELING

e

eo

e=e,.+eo eo . lgt’---~S ~ ---

e0

C Figure 4.16: Replacing a biased compliance with an unbiased compliance

is shifted vertically until the point is at the origin. The biased characteristic is reconstituted by adding the effort e0 to this resistance curve with its effort This summation, suggested by vertical arrows in the figure, is achieved in the bond graph by bonding the unbiased resistance element R to an effort source, Se, using a common-flowor 1-junction. The Norton equivalent is an alternative that employs flow summation rather than effort summation. The second case shown in the figure has the same characteristic as the first. A heavy dot is drawn when the curve intersects the axis, defining the negative flow -O0- A passive resistance R with flow 0R results when this curve is shifted horizontally until the dot lies a.t the origin. The biased characteristic is reconstituted by summing ~ with -00, as suggested by the horizontal arrows. In the bond graph, this is accomplished by bonding the element R to the flow source S)~ with a constant-effort or 0-junction. Note that the resistance elements for the Thevenin and Norton equivalents are the same for linear cases, but are distinctly different for nonlinear cases. Most sinks have characteristics that pass through the origin, and therefore need no Thevenin or Norton equivalents. Sources, on the other hand, virtually always benefit from these equivalents. Examples are shown in the bottom half of Fig. 4.15. In the Thevenin version, a resistive effort eR is subtracted from a source effort eo to get the total effort. Effort summations are vertical, again, and accomplished by a 1-junction. In the Norton version, a resistive flow 0R is subtracted from a source flow 00 to get the total flow. Flow summations are horizontal, as always, and are accomplished by a 0-junction. Note again that the Thevenin and Norton resistances are the same if and only if they are linear. Compliance characteristics sometimes do not pass through the origin. A Thevenin decomposition can be applied here, also, as shown in Fig. 4.16. A Norton equivalent does not exist, since displacement sources are not defined. Nevertheless, it is possible to redefine the displacement variable of a compliance such that the characteristic passes through the origin, which is tantamount to

"4.3.

245

MODELEQUIVALENCES

eR ~R ~-R

(a) original characteristic

/ ¯

el eo~ ~ .N~

~

I - ’ el I e~

0

q~

e=e~-eo

(b) origin shifted to point

(c) source perspective

o qe~ R

R

Figure 4.17: Shifting the origin of a source or sink characteristic along its path horizontal shifting of the characteristic. 4.3.2

Passivity istic*

With Respect

to a Point

on a Character-

The Thevenin and Norton structures can be viewed as translating the origin of the characteristics in the effort-flow plane. Translating the origin of source and load characteristics to the equilibrium point at whichthey intersect can aid dynamicanalysis. Informationregarding stability can result, for example,as will be suggested in Section 4.4.7. Consequencesregarding general dynamicbehavior appear later. For the present, consider only howsuch a translation can be accomplished.If the origin is to be displaced along a characteristic, translation of both the effort and flow axes are necessary. This suggests combiningthe Thevenin and Norton structures. One way to combine them is illustrated in Fig. 4.17. The original characteristic, shownin part (a), employsthe variables eR and 0R. The shifted characteristic employsthe variables e and 0. The new axes for the point labelled "b" are shownby dashedlines. Note that the only difference betweenthe sink or load version, shownin part (b), and the source version, shownin part (c), lies in the signs of the flows. Aload or sink has been defined as passive if it lies entirely in the first and third quadrants, that is if it never generates powerbecause the product e0 is

246

CHAPTER

4.

INTERMEDIATE

MODELING

nowhere negative. This property can change when the origin is movedalong the characteristic; passivity becomesa property of the point chosen as the origin as well as being a function of the shape of the characteristic. Consider the given characteristic with point b as the origin, as shownin part (b) of Fig. 4.17. The characteristic resides partly in quadrants two and four, and therefore is said to be active or non-passive with respect to this origin. The same conclusion applies to any point between points a and c. If, on the other hand, a point to the left or below point a or to the right or above point c is chosen as the origin, the characteristic is said to be passive with respect to that origin. The characteristic in part (c) of the-figure is identical to that of part (b), except that the sign of the flow is inverted by employing a source rather than a sink power convention to the bond for e and 0- The resulting plot is a left-toright flip of that given in part (b). The property of passivity does not change as a result of this change in perspective. The test for passivity above, however, assumes the power convention of a sink; the power convention of a source must be inverted before the test is applied.

4.3.3

Truncation Bonded

of to

R,

Transformers C or

I

and

Gyrators

Elements

Frequently a one-port element is attached to a system model by a transformer or a gyrator. The combination of two elements can be replaced by a single equivalent element. This option simplifies subsequent analysis, and can enhance your grasp of the fundamental structure of a system. Table 4.1 gives the equivalences. A transformer does not alter the type of the combination; a resistance remains a resistance, a compliance remains a compliance and an inertance remains an inertance. On the other hand, a gyrator bonded to a compliance yields an equivalent inertance, and a gyrator bonded to an inertance yields an equivalent compliance. Although a gyrator bonded to 2a resistance remains a resistance, its modulus becomes inverted. You need not memorize these equivalences, since you can reconstruct them readily at least two different ways. In the first, or direct way, you first label either the effort or the flow on the left-hand bond with a symbol. Then, you employ the definitions of the transformer or gyrator and the one-port element to find the corresponding conjugate variable. Finally, you compare this result to the corresponding conjugate variable for the reduced case, and deduce the equivalent modulus. 2The gyrator is said to be a dualizing coupler. Thus, for example, if two inertances are bonded together by a gyrator (11 ~G~I.2), each inertance "sees" the other inertance as a compliance. As a result, this system would exhibit oscillatory behavior, since an inertance bonded to a compliance --- for example a mass attached to a spring -- is an oscillator. This is why two fluid vortices, neither of which has any "springiness," interact in an oscillatory fashion.

247

4.3. MODEL EQUIVALENCES Table 4.1 Equivalences for One-Port Elements Bonded to Transformers and Gyrators

Consider the following two examples, which are the first and the fourth entries in the table: e= T2R~ O

T RT~ R TO

e O CG2 = P~

G p/CG e/G

C

In the left-hand example,youstart by labeling the fiow variable on the left as 0- As a consequence,the flow variable on the right must be TO. The resistance element then requires the effort variable on the right to be RTO,and finally the transformer requires the effort variable on the left to be TRTO.Since by the definition of R~ this effort variable also equals R’(L it follows that R~ = T2R. Alternatively, you could start by labeling the effort variable on the left as e, and proceed around the bonds in inverse order. In the right-hand example, you start the procedureby labeling the effort on the left as e or the flow as 0. If you start with the effort, the gyrator requires the flow on the right to be e/G, the compliancethen requires the effort on the right to be p/CG(recall that p =- fedt), and finally the gyrator requires the flow on the left to be p/GCG. Since the inertance I’ is defined by the relation O = P/I~, its value is I ~ = G~C. The second methodfor deducing the equivalences is based on poweror energy, and is recommended as a superior discipline. For the first example, the powerdissipatedin the resistancesequalsboth~R1 ,.2q and~R(Tq)-,1 " ~ so that R~ = T2R. For the second example, the energy stored in Y, namely~lqlT~.2, 2, so that I ~ = G2C. equals the energy stored in C, namely ½C(G0) 4.3.4

Reduction

of Two-Pair

Meshes

Cascadedbonds interconnecting alternate 0 and 1-junctions to form a closed path is called a simple bond-graph mesh. Meshesoccur often. Meshes of the types shownin Fig. 4.18 can be reduced to equivalent tree-like bond graphs, simplifing subsequentanalysis.

248

CHAPTER 4.

INTERMEDIATE MODELING

Z~ ~.1 "Z~ w,¢

Z~ 0

=g>--,-o---~

""~-o~ z~

Z~,~,.. 0 1

.-o ~ "’~’o---~ Z2

(a) single meshes

(b) nested meshes Figure 4.18: Reduction of simple even bond-graph meshes

4.3.

MODEL EQUIVALENCES

249

A mesh is called even if the number of mesh bonds which have the same power direction convention in the clockwise sense is even, and odd if the number is odd. The same statement applies to the bonds with counter-clockwise power direction convention~ since the total number of mesh bonds is even. All the meshes in Fig. 4.18 are even. It can be shown that such evenness or oddness is a property of the mesh.3 This means that reassigning the power convention arrows and compensating by changing the signs of the efforts or flows, in a compatible fashion, always leaves the evenness or oddness unchanged. Two meshes that are identical except for the property of evenness vs. oddness therefore model markedly different systems. Four-bond even meshes, as shownin the left side of Fig. 4.18 part (a), can reduced to the tree-like forms shownin the right side of the figure. In the first case, the efforts on the left-most and right-most bonds are the same, because they both equal the sum of the efforts on the two 0-junctions. In the second case, the flows on the left-most and right-most bonds are the same, because they both equal the sum of the flows on the two 1-junctions. These reductions are extremely useful, because tree-like graphs are easier to analyze than graphs with meshes, as you will see later. Nested even meshes each having four mesh bonds, as shown in part (b) of the figure, can be reduced similarly, as shown. These equivalences apply to the junction structure regardless of whether the elements bonded to it are resistances, sources or something else. The figure employs the symbol Z to represent unspecified one-port elements, which could be resistances, compliances, inertances, boundary bonds or something more complex. Odd meshes are not reducible in this manner. As a result, their analysis is more involved. They tend to occur less often, fortunately. EXAMPLE 4.10 Apply the standard steps to modeling the circuit below with a bond graph, and show that an even mesh results. Then, reduce the graph to an equivalent tree-like structure.

aF.T. Brown,"Direct Application of the LoopRuleto BondGraphs,"J. Basic Engineering, ASME Trans., v 94 n 3 pp 253-261Sept. 1972.

250

CHAPTER 4.

INTERMEDIATE MODELING

Solution: The standard procedure gives the following results after steps 2 and 4 (p. 192), respectively:

after step 2

1

after

",,

step

4 ~ /6

1

/\

1"~-0"--~ 1 "-~’0

Z1

Z,

Z~

The mesh within this graph is even, since two of its mesh bonds have clockwise-directed power arrows and the other two are counterclockwise. The 0-junction is first split into two 0-junctions in order to removeelement Z1 from the mesh. The meshthen is in the form of the second meshgiven.in Fig. 4.18, and so the equivalencegiven there is used to get the final graph. Z7

1 -.,r- 0 -~- 1 ---~-Z3 Z~

0 Z 4

15 "--~-Z

z~

4.3.5

Transmission Matrix Reduction of Steady-State Models*

Often a two-ported modelor sub-modelcomprises a chain-like structure with no energy-storage elements. A commoncase is shownin Fig. 4.19. Such cases can be represented as a two-port element described by a single transmission matrix, whichfor somepurposes is a significant simplification. The transmission matrix concept is introduced in Section 2.5.6 (pp. 82:83). The transmission matrices for the shunt and series resistances are readily

4.3.

251

MODEL EQUIVALENCES ---4-~.’ 0 ~ 1~ ~> -~--~-. q~ l~1-~2 ~2es+e2 l q2

eI e~ TWO-PORT.-~-ql q2

R 2

Figure 4.19: Reductionof a chain of junctions with stub elements

deduced as ~1

e2

0

I~:] = I1}R 011 [~:];

e 1 ~1

e.~ l

q2

R The cascade of two such structures in Fig. 4.19 can be reduced by multiplying their respective transmission matrices:

l/R1 I+R.2/R1] " A cascade of such elements of any length can be telescoped in this manner. Moregeneral steady-state couplers are discussed in Section 9.3. EXAMPLE4.11 Find an equivalent transmission matrix for the static two-port model

252

CHAPTER

4.

INTERMEDIATE

MODELING

Solution: The combination of T.9 and/~3 is first reduced to an equivalent resistance R~. Also the left-hand 0-junction is expanded into two bonded 0-junctions, in order to isolate the even mesh clearly: R2

The next intermediate graph implements the standard reduction of this mesh. In the third intermediate graph, the resistances R.~ and R~ are combined to give R4.

The central portion of the bond graph, with its 0 and 1-junctions tached resistances, is now precisely in the form of Fig. 4.19:

and at-

e2 e3 e4 e~ ---~--~.- T~---~--~- 2-PORT ~ G .-~--~ ~i ~2 ~ q3 q4 -4e~EQUIVALENT OVERALL2-PORT e.--~ ~ q~ q4 The transmission matrix approach allows the transfomer and gyrator parts of the graph, as well as this central portion, to be telescoped in one step:

l/T1] [1/1~1 01] [~ ~141 [1/G = (I+R4/R1)/GT1 4.3.6

G/T1R1

(t4

Summary

Subsequent analysis is simplified if general sources and resistive loads are replaced by Thevenin or Norton equivalents, if compliance characteristics similarly are made to pass through the origin, if transformers bonded to one-port elements are replaced by equivalent one-port elements, if simple even meshes are replaced by three-like structures, and if steady-state portions of models

4.3.

253

MODEL EQUIVALENCES

F Figure 4.20: Guided Problem 4.7 comprising junction structures and resistances are reduced to simpler one or two-port models. The use of bond-graph equivalences and analytical reduction techniques have been presented toward this end. These equivalences are first used in the next section to help find the equilibium states of models. The use of graphical methods for nonlinear structures is developed, also. Guided

Problem

4.7

This problem typifies simple analytical reduction of a model comprised of steadystate elements. For the system of dashpots and a lever shown in Fig. 4.20, (a) draw a bond graph model, (b) draw an analogous electric circuit, and .(c) find an equivalent overall resistance. The lever may be assumed to rotate only through a very small angle. Suggested

Steps:

1. Follow the procedure suggested in Sections 4.2.6 and 4.2.7 for finding the junction structure, and note that the lever acts as a transformer. 2. Reduce any mesh, should your bond graph have one, to a tree-like

graph.

3. Invert the procedure of Section 4.1.1 (p. 192) to find the equivalent electric circuit. 4. Reduce the graph one junction-structure element at a time, starting presumably at the right end. At each step find the new terminal equivalent resistance. The final result should be the desired answer. 5. As a check you might reduce the electric

circuit

by some other means.

PROBLEMS 4.34 Represent the hydraulic power supply of Fig. 2.45 (p. 85) by (i) Thevenin and (ii) Norton equivalences. Sketch the characteristics of the resistances.

254

CttAPTER

4. INTERMEDIATE

MODELING

4.35 Continue Problem 4.15 (p. 231) with the well and pump. The flow Qs into the well decreases linearly as the level of the well is raised; represent this source with a Thevenin or Norton equivalent. Also, add to your model the compliances of the well and the tank, the resistance due to friction in the pump/motor subsystem, and an output flow Qt drawn from the tank. 4.36 Determine the criterion for a linear (constant) resistance to be active with respect to any point on the characteristic. 4.37 The flywheel with rotational inertia If depicted below is driven through pair of gears with ratio r > 1. The gears and bearings may be presumed to be frictionless and massless. Drawa bond graph for the system. Also, reduce the model to a simple equivalent inertia, with no transformer, and find its modulus in terms of If and r.

4.38 Show that the translating coil device of Guided Problem 2.6 (p. 70) comes a dashpot when the electric circuit is completed by an added resistor to give a total electric resistance R. Also, determine the dashpot coefficient, b, in terms of the given parameters. 4.39 A permanent magnet DC motor can be made into a rotary completing the armature circuit with a resistor. (a) If the motor satisfies the relation M = ai and the electrical of the armature curcuit is R, determine the dashpot coefficient,

dashpot

by

resistance b.

(b) Resistors come in different physical sizes depending on the heat they need to reject. What power rating is needed if b -- 0.1 N-m.s and the dashpot must tolerate a steady speed of 5 rev/sec? (c) State fu nctional advantage th is de vice ha s ov er a vi scous ro tary damper of the type pictured in Fig. 2.23 part (d) (p. 45). 4.40 Reduce the bond-graph mesh in Fig. 4.6 (p. 210) to a tree-like structure. Also, combine the two bond graphs to get a single model representing a positivedisplacement pump or motor with both friction and leakage. 4.41 Determine whether any of the meshes in the bond graphs of Figs. 4.10 (p. 217) and Example4.8 (p. 218) are reducible to tree-like structures.

4.3.

255

MODEL EQUIVALENCES

4.42 Reducethe mesh for the bond graph of Problem4.7 (p. 203). 4.43 Derive the secondand third entries in the right-hand columnof Table 4.1 (p. 247) using (i) direct interpretation of the elements, and (ii) considerations of poweror energy. 4.44 A rotary spring with rate k, massless and frictionless pair of gears with diameters dl and de, and flywheel with momentof inertia I are connected as shownbelow.

(a) Modelthe system with a bond graph. Relate all moduli to physical parameters. (b) Find an equivalent simpler bondgraph with only two elements. Relate the new modulusto physical parameters. (c) Find the natural frequencyat whichthis systemoscillates following non-zeroinitial condition. 4.45 Showthat the bond graph below represents an oscillator, natural frequency, w~. 1

C1

g

~

and find its

C2

4.46 A large charged capacitor drives an ideal DCmotor (neglect armature resistance and friction) after a switch is closed. The motoris connectedeither to a (i) flywheel, or a (ii) spring. Certain parametersare definedon the drawings. DC motor M=ai

(ii)

~~C

~ flywheel ~

"-’~ M=o, DCm°.t°r~ F S;~g

~

(a) Modeleach case with a bond graph, and state whether or not it oscillatory in nature.

256

CHAPTER 4.

INTERMEDIATE MODELLNG

(b) If either (or both) of the systemsare oscillatory, find its" natural quencyin terms of the parameters defined. 4.47 The electrical terminals of a DCmotor/generator for which e = a~ are connected to a large inductor with inductance L. In version (i), the shaft connected to a flywheel with rotational momentof inertia I. In version (ii), the shaft is connected to a rotational spring with constant k, as shownbelow. Electrical resistance and mechanicalfriction can be neglected for purposes of the following questions: flywheel (i)

~

._DC motor/generator ~’~-~

(a) Modeleach case with a bond graph, and state whether or not it oscillatory in nature. (b) If either (or both) of the systemsis oscillatory, determineits natural frequency in terms of the defined parameters. 4.48 A motor with a knowntorque drives a frictionless and leak-free positivedisplacement pumpthat in turn drives a system that includes a frictionless and leak-free single-rod-end cylinder and two valves (that might be adjustable). Assumethat all parameters are known,and that the load force F is known. Gravity maybe neglected.

~ t-.F ......... F....... IH...... valve

dl

~va~ve piston/cylinder ~]l~-L--II-- ]opentank (reservoir)

(a) Define parameters and variables and draw an appropriate bond graph model."

4.3.

257

MODEL EQUIVALENCES

(b) Write whatever relations for the elements that are not explicit in your annotated bond graph. (c) Solve for ~ in terms of the knowninformation. 4.49 The energy of an aircraft landing on an aircraft carrier is dissipated in an "arresting engine" located below decks. A hook on the aircraft grabs a cable which is connected to this machine by pulleys, which are not shown below. The cable wraps nine times around the two sets of sheaves (pulleys) on the machine itself. The nine sheaves at one end rotate about a fixed shaft; the nine pullleys at the other end rotate about a commonshaft which is attached to a single large piston via a moveablecrosshead. As a result, hyraulic fluid is forced out of the cylinder and flows through a flow restriction to an accumulator. The restriction is adjusted continusously throughout the process in order to maintain the desired pressure in the cylinder and therefore the desired tension in the cable. The accumulator contains a free piston to prevent mixing of the oil and the air (or nitrogen); the pressures of the oil and air are virtually identical. One end the cable is anchored. cross-section side view: accumulator/,free piston

;~I~

~diu~tahle

9 loops of cable -~

came

common fixed shaft

9 sheavest

attachedxto crosshead

~ension,’T "~ ancnorea ena or cable

9 sheaves

top view showing pulley system (reducedscale):

fixed shaft

.velocity, v movable shaft tension, T

(a) Find a model for the system, considering the tension T as the input variable and the velocity v as the output. Define parameters and variables as needed. The inertias of the crosshead assembly and the pulleys are important, but losses due to mechanical friction can be neglected. (b) Reduce your model to give a single equivalent resistance,

compliance

258

CHAPTER 4. and inertance.

INTERMEDIATE MODELING

Relate these parameters to your primitive parameters.

(c) Evaluate the compliance relationship assuming (i) adiabatic or isothermal conditions. The initial pressure of the gas, P0, can be taken as given. The resistance can be presumed to be a known function of the position of the cable.

SOLUTION TO GUIDED PROBLEM Guided Problem 1.

4.7

R.~ "~-5

1 "---"-~-

0 ~R

¯ -----------------~0 "~-1 "---’~-R x

2. The bond graph above is tree-like. However, should you draw the electric circuit on the right below, the formal procedure gives a meshwhich is best replaced by an equivalent tree-like graph. or R3 ~ R3"-=~"6

1 ~R

R7 ’

1 R6

T

R7 =R3 +R 6

R4

z~ 0 ~ 1 -----~R

~ 0 ~ I .----~-R~.

x

X

R~

~1 R9

"-~-R

=R8

~ F--~O-’-’~R~x

R8 RS=T~’R7 I t ~ 2 0 ~ 1 "--’-~-R

"

X

Thisis left to the student,butthe net resultshouldbethe same: R = RaR------2--9= Rt{R2 ÷T2[R3 +R,,Rs/(R,,+R~)]} z R~ 9 +R

4.4

Rt +R2 +T [R3 +R,,R 5/(R4

+Rs)]

Equilibrium

The equilibrium or equilibria of many models are stable. The state of such a model eventually settles down to one of these equilibria after time-varying excitation ceases. A model with one or more unstable equilibria might never settle downto a stable equilibrium, even if one exists, but oscillate in what is

4.4.

259

EQUILIBRIUM

knownas a limit-cycle oscillation. The location of an equilibrium can be affected by compliances,but not by inertances. The stability of the equilibrium states, on the other hand, is apt to be strongly affected by both inertances and compliances. This section first addresses the equilibria of modelscomprisingonly nondynamicelements. It then proceeds to consider the effect of compliances. How inertances affect the’stability of equilibria is deferred until the next chapter. 4.4.1

Reduction of Steady-State Source; Case Study

Models

With

a ,Single

A steady-state model (no energy-storage elements) with a single unspecified source or boundary bond can be reduced to a Thevenin or Norton equivalent load regardless of its original complexity. Sucha reduction allows the location and stability of its equilbriumstate(s) to be determinedfor any source characteristic. Analytical and graphical options for carrying out the reduction are illustrated below. The analytical approachimplies the simultaneoussolution of algebraic equations for the various elements of the model. It is the methodof choice for linear models;the constancyof the moduliof the various sources, resistances, transformers and gyrators lead to efficient matrix methods.Graphical methods, however,mayprovide greater insight, can deal directly with experimental data, and offer particular advantagesin addressing complicatednonlinear modelsfor which analytical models require fancy methodsor create confusion regarding multiple equilibria. Thus, even if you plan to employan analytical methodultimately, a preliminary rough graphical analysis can be well worth the effort. Either procedure employsstep-by-step coalescence of cascades, meshes and parallel combinationsof elements, until a single source-load synthesis remains. Section 4.3 abovedescribes most of the individual steps in a reduction. As a case study, consider the system of dashpots given in Fig. 4.21. As a first step, you maychoose to combinethe resistances R~and R3 to give R5, as shownin part (c) of the figure. Alternatively, you maydefer this coalescence, and proceed as shownin part (d) by adding a new 0-junction so as to isolate the meshin preparation for its reduction. The meshreductions are carried out with the aid of Fig. 4.18 (p 240). The resulting tree-like graphs are shown parts (e) and (f); the latter is reduced to the former by the delayed combining of R2 and R3. The two resulting 0-junctions are combinedin part (g). Next, R~ and R5 are combinedto give R6, as shownin part (h). Finally, R4 and are combinedto give the overall resistance, R, as shownin part (i). Note that the force is common to both boundarybonds. R is the resistance of an effective overall dashpot. Carrying out the indicated steps is especially simple if the resistances are constants. Each of the pairs R2, R3 and R4, R6 have a commoneffort, so 1

1

1

= +--; R--; R3

1

1 1 R- R4 + R6

260

CHAPTER4.

INTERMEDIATE MODELING

Rl

(a) system

F

F R~ F_~

1

~

0

~

1.~F

0

~

(b) bondgraph

R, F

R~ choicesg R~

~lter::te

[~0~1~0 ~

F

~ R~

~0~

~o ~ ~

[ ~4

R~ ~ 0 ~ R~

Figure 4.21: Determinationof the equilibrium for a steady-state model

4.4.

261

EQUILIBRIUM

(b) F

F

3~ R~ R

Figure 4.22: Graphical analysis of the steady-state

model

The pair R1, R5 has a commonflow, so R6 = R1 q-R5. Nonlinear characteristics can complicate the algebraic approach considerably. You may prefer to use graphical methods instead. Example characteristics for the five original dashpots are plotted in parts (a), (b) and (c) of 4.22. Horizontal addition is employed in part (a) to form R5 from R2 and R3. Vertical addition is employed in part (b) to form R6 from R1 and Rs. Finally, horizontal addition is employed in part (c) to form R from R4 and/~6. The characteristic for the particular R3 shown in the figure allows no motion until a threshold of force is exceeded; this is typical of dry friction. Then, the force decreases for a range of small velocities, before again increasing. This pattern is typical of some bearings and rubbing surfaces that become better lubricated as the sliding velocity increases. The resulting characteristic for the overall system also has a dip-then-rise. In part (d) of the figure, a constant force Fo is applied to the left end of the system, and the right end is grounded. The source-load synthesis, showngraphically, reveals three equilibria. All are valid, but are they stable? To answer this question it is necessary, as always, to consider non-equilibrium states. If the resistance force of the overall equivalent dashpot exceeds Fo, rapid leftward acceleration ensues, impededonly by the inertia of the parts. This case

262

CHAPTER 4.

INTERMEDIATE MODELING

is illustrated by the states associated with the arrows on the left and on the right. If, on the other hand, the resistance force is less than F0, rapid rightward acceleration ensues, illustrated by the states associated with the center arrow. This analysis reveals that the center equilibrium is unstable and the outer two equilibria are stable. Whichof the two stable states comesinto being depends on the history of the force and the initial velocity of the dashpot. 4.4.2

Alternative Steady-State

Approaches Models

to

Reducing

Not all very complex systems can be reduced completely through use of the above procedures. Recourse can be made to iterative numerical procedures applied to the set of corresponding equations. In the special case of linear systems, non-iterative matrix methodscan be used. Such approaches are the subject of textbooks and monographson algebra and numerical methods, and are not addressedhere, in favor of a simpler alternative. This alternative converts the static model to a dynamicor time-varying model. This is done by adding either real or fake compliancesto 0-junctions or inertances to 1-junctions. The revised modelleads to differential equations, with time as the independentvariable, in place of the original algebraic equations. Routine numerical simulation then applies. The solution usually settles down ¯ to a stable equilibrium state that is independentof the presence of the added compliances and inertances. If there are multiple equilibria, you may need to start with an initial condition somewhatclose to the desired equilibrium. Occasionally, ~the solution maycontinue to oscillate rather than settle down to an equilibrium state. Youshould then change the assumedcompliances or inertances until stability is achieved. The availability of this option sharply reduces your needto becomeproficient in the more exotic methodsavailable for reducing complexalgebraic or steadystate systems. It is a by-product of the modeling and analysis of dynamic systems, whichis the major subject of this book. 4.4.3

Removal

of Elements

for

Equilibrium

Equilibrium implies that each bond has a time-invariant effort and flow. An inertance with a constant flow has zero effort. Whenever the objective is merely the determination of equilibria, therefore, all inertances maybe removedfrom a bondgraph. Further, if an inertance is bondedto a 0-junction, the effort on the ¯ junction becomeszero, so all other bondsattached to it also maybe excised. The situation is morecomplicatedfor compliances. A complianceat equilibrium has a time-invariant effort and therefore a constant displacementand zero flow. Whenboth resistance and compliance elements are bonded to a common 0-junction, and therefore have a commoneffort, the flow of each compliance element is zero, whereasthe flows on the resistance elements are not. The compliance elements therefore can be removedfrom the graph; this action leaves the flow state of the systemunchanged.This situation is illustrated in Fig. 4.23

4.4.

263

EQUILIBRIUM C

(a)

,~ ~ ,x~

~?~

~ x:

i.xe" equilibrium~:> ,~._]~F~~._._~

R

x~ ~ x,

(b) ~ C x~

~ 2~

~,~’~-,~ R

~~-x~ R

C~I.~R

-

x~-x:

equilibrium C

Figure 4.23: Removalof compliancesand resistances for the equilibrium state part (a). The force compressingthe spring equals the total applied force. this F is constant, a constant displacementx3 - x2 results, so ~3 = ~72. ASa consequence, the power on the compliance bond is zero, and this bond maybe removedfrom the graph. In the situation illustrated in part (b) of the figure, on the other hand, both resistance and compliance elements are bonded to a common1-junction. Since both elements have a commonflow, this flow must be zero if the appled force is constant. Otherwise, the displacement of each compliance would increase without limit, requiring infinite force. Since the equilibrium flow is zero, the effort on each resistance is zero. Thus, removal of the resistances from the graph leaves the equilibrium state unchanged. Retention of the compliances then gives the proper displacements. The above erasures, when combinedwith the other steps for bond-graph reduction, eliminate either all resistance or all compliancefrom all or major portions of the bondgraph. If all complianceis gone, the resulting equilibrium involves a steady flow. If instead all resistance is gone, the resulting equilibrium is static. Problemsregarding the deflection of static structures, for example, are of this latter type. Youusually can tell whetherto delete compliancesor resistances by appealing to physical reasoning.. The examplesof Fig. 4.23 are no exception. 4.4.4

Case

Study

with

a Steady-Velocity

Equilibrium

The objective is to find the equilibrium state of the motor-driven mechanical system shownin Fig. 4.24, disregarding the dynamicsand assumingthe weight does not interfere with the pulley before the equilibrium becomesmeaningful. The model comprises some components with constant moduli and others with nonlinear characteristics. The winchhas a radius of 0.05 mand the pulley has a radius of 0.10 m. The weight weighs 150 N. The motor and the viscous coupling havenonlinear characteristics, as plotted. The bond graph given in part (b) of the figure includes a compliance and resistance Rc bonded to a common0-junction. WhenM= constant, the

CHAPTER 4.

264

INTERMEDL4TE MODELING

/~.~pulley M~

F

r°as’r’n ¯ spring| ; weight and ~ dashpoi

winc(h

(b) bond graph

~ motor

(a) mechanicalarrangement 15 moment,

Cr Ca.,~ 1.ta’~RSae

’, or N’mll

// Jviscous

(c) bondgraphfor equilibrium

(motor) moment N.ml( ~_.........~,,~"~Ulllbrlum )’- load 5

0 ~coupling 0 100 200 300 ~, rad/s (d) givencharacteristics

0

I

0

"

I

I

100

I

I

I

200 300 ~, rad/s (e) determinationof equilibrium

Figure 4.24: Case study for 9quilibrium with motion

4.4.

265

EQUILIBRIUM

compliance Cr has a fixed deflection (zero relative velocity), and therefore removed from the graph. The compliance Cd and the resistance/~d are bonded to a common 1-junction, so when F = constant the velocity ~d = 0; as a result, Rd is removed. The compliance Ca now becomes bonded directly to the 0-junction, since there is no significance to a two-ported 1-junction. The velocity ~2 equals ~1, so Ca also can be removed. Part (c) of the figure shows the resulting reduced bond graph for equilibrium. Given W-~ 150 N, T~, -- 0.05 m and Tp = 1/0.10 = 10 m,

’

= 2.0

Mm = T~,F = T~,(W + TpM) = 0.05(150

(4.10a)

+ 10M) = 7.5 + 0.5M (4.105)

Values of the viscous coupling mom.ent M= 0, 2.5 and 5.0 N m are now chosen, and the corresponding values of ¢c picked off from the plot for the coupling. Equation (4.10a) gives the corresponding values of ~,~, and equation (4.10b) gives the corresponding values of ~lm. The three resulting points are plotted as heavy dots on the axes for M:,, and ~),n. Fitting a smooth curve through these points gives a "load" characteristic that intersects with the "source" (motor) characteristic to give a close estimate of the equilibrium state. 4.4.5

Case Study Equilibria

with

Stable

and

Unstable

Static

Belleville springs are washer-like disks with a conical shape when they are unloaded, as drawn in Fig. 4.25. The force-deflection characteristic exhibits a rise-fall-rise shape as shown. Should a particular given weight Wbe placed on the spring, there are three equilibria. The stability of these equilibria can be determined, as usual, by considering off-equilibrium states. Examples of these, given in part (b) of the figure, indicate acceleration resisted only by inertia. The analysis reveals that equilibria A and C are stable, whereas equilibrium B is unstable. This is similar to the stability analyses carried out in Chapter 2 for constant-velocity equilibria. 4.4.6

Case

Study

with

Limit-Cycle

Behavior

The gear-head electric motor in Fig. 4.26 is presumed to have the plotted torquespeed characteristic. The torque-speed charactertistic plotted for the load that the motor drives is characteristic of a bearing heavily loaded in a transverse direction. The friction drops as the speed increases from zero and a lubricant begins to act. For higher speeds the torque again increases as viscous drag takes over. The shaft connecting the motor to the bearing is presumed to have significant torsional flexibility, modeledas a rotational spring. The model of the system comprises a source for the motor, a resistance for the load, a compliance for the spring and a 0-junction to recognize that all three componentsexperience the same torque.

266

CHAPTER 4.

i

~

INTERMEDIATE

fixed

MODELING

support

(a) cross-section through centerline

F

W X

(b) force-deflection characteristic

and .stability analysis for applied weight, W

Figure 4.25: Case study of Belleville springs with stable and unstable equilibria

The single equilibrium state is defined by the intersection of the source and load characteristics. This equilibrium is unstable! The spring allows the motor to run at a different speed from the load, but requires that the torques at its two ends be equal. The state of the system at a moment in time therefore can be described by a horizontal line segment, with the state of the motor at one end and the state of the load at the other. If the system is started from rest, it will initially jump to the pair of states labeled as 1 in part (e) of the figure. (All inertia of the rotor and gears is being neglecte.d, so. ~1 jumps to the given value virtually instantly.) The velocity difference ¢1 - ¢2 now causes the twist angle ¢1 - ¢2 to increase in time. Thus the spring winds up, and the torque Mincreases. When~1 reaches its equilibrium value, at state 2, the load still hasn’t budged; ¢2 is far from equilibrium. Thus the wind-up of the spring continues, at only a slightly smaller rate. Whenthe momentattempts to exceed that of state 3, however, the load finally is forced to break away. It accelerates virtually instantaneously (no inertia, again) to state 4. At this point ~2 >> (ill, so the spring rapidly unwinds and the torque decreases. It is still decreasing when it reaches state 5, at which point the load speed has no choice but to suddenly become zero, giving state 6. State 6 soon becomes state 2, and then state 3, and so on: the cycle is repeated. The endless cycling from states 6 to 3 to ~5 to 6 is knownas a limit cycle oscillation. This particular type of limit cycle often is called a stick-slip instability. It is also known as a commontype of "chatter." Should the system be started somehowat its equilibrium condition, any small disturbance will trigger the instability, and the same limit cycle results. In fact, any initial condition results almost immediately in the same limit cycle, as you can verify.

4.4.

267

EQUILIBRIUM

C frictional load (a) schematic

M

~

or), Mvs

(b) bondgraph

~1

(c) torque-speedcharacteristics of source and load

(d) shaft spring characteristics

M

II II II

II II II

II II II

t3.4ts.6 t3.4ts.6 time (e) state diagram

(f) timehistory

Figure 4.26: Anunstable S-R-Csystem that develops a stick-slip limit-cycle oscillation

268

CHAPTER 4.

INTERMEDIATE MODELING

This problem is continued in Section 5.1.4, with the inevitable load inertia added. Computer simulation and analysis will show how this inertia modifies and, if large enough, eliminates the limit cycle. EXAMPLE 4.12 An induction motor and load have the characteristics plotted below (which are the same as in Problem 2.4, p. 28.) The shaft and the load are connected by a spring coupling (rather than the rigid shaft of Problem 2.4). Determine the stability of the three equilibria, neglecting the effect of all inertias. Compare the result to the case in which the springiness is neglected and inertia assumed.

M

[

]~~~load

Solution: Model tile system with a bond graph, taking care to get the right junction type:

S .~. ~M

MO -~-~.

R

[

C The momentis common; moment is plotted on the vertical axis, so the vertical coordinates of the two states are the same, which means that a line segM ment drawn between the two states is horizontal. The arrows, which indicate the direction of change, are pointed /~ M upward if CdM/dt = ~M -- ~)L > O; ~( otherwise, they are downward. Equilibrium 1 is seen to be unstable, while both equilibria 2 and 3 are stable. If the initial value of ~Mis between that of equilibrium 1 and the peak of the curve, equilibrium 2 is approached. If it is to the right of the peak, equilibrium 3 is approached. If it is less than that of equilibrium 1, the system stalls altogether. For inertia but no spring, the 0junction is replaced by a 1-junction and the C element by an I element. The horizontal line segments are replaced by vertical line segments. These indicate that equilibria 1 and 3 are stable, whereas 2 is unstable.

4.4. F, QUILIBRIUM 4.4.7

269

Necessary Condition for Instability or Limit-Cycle Oscillation*

Thestability of an equilibriumpoint betweena static source characteristic and a static load characteristic, and the absenceof a limit-cycle oscillation, are assured if both characteristics are passive with respect to the equilibrium point. This passivity is defined in Section 4.3.2 (p. 245). Whetheror not an equilibrimn point that fails this test is actually unstable depends on the magnitudes of any compliances and inertances that maybe present. Further, an equilibrium point could be stable in the presence of small impulsive disturbances, and yet respond to a large impulsive disturbance with a limit-cycle oscillation. The tools necessaryto ~nakethis determination analytically, without simulation, are developedin later chapters. All points on the source characteristic in Fig. 4.26 are passive. Onthe other hand, the non-passivity of any point on the negatively sloped portion of the load characteristic can be observed readily. The graphical analysis presented with that simple exampleis sufficient to predict the limit cycle instability associated with any equilibrium point in this range. Points in the positively sloped portion of the load characteristic with momentsbelow the break-away momentare nonpassive, also, but the analysis reveals neither an instability nor a limit-cycle oscillation for an equilibrium in that range.

4.4.8

Summary

The equilibrium states for bondgraphs that modelconstant velocity systems can be found analytically or graphically. Whenthese methodsbecomeexcessively awkward,the simultaneoussolution of a set of nonlinear algebraic equations by iterative numericalmeansis a major alternative, but is beyondthe scope of this text. A more straightforward approach adds real or fake compliances and/or inertias to the model,leading to a solvable set of differential equationswith time as the independentvariable. The equilibrium state is approachedasymptotically in time, assumingthe modelis stable, using numericalsimulation (as introduced in Section3.7). The equilibria of modelswith both compliancesand resistances can be found by eliminating resistance elements that have zero generalized force, and eliminating co~nplianceelements with a non-zero generalized velocity. In somecases equilibria with steady velocity result. In others, static equilibria result. Anequilibrium state betweensource and load characteristics is knownto be stable if both characteristics are passive at that state. Otherwise,determination of the stability requires a detailed examinationof compatible non-equilibrium states. The presence of an unstable equilibrium maysimply favor one or more stable equilibria. In other cases a limit cycle results.

270

CHAPTER 4.

INTERMEDIATE MODELING

P~

~

~ P2 Q2 valve ~1 ~ --~.Q~ 7--~_ ~/ Q~ adjustablel : .~ relief ( "--~,’il~/i:,"~,’~i~i ~ valve

~~k

co on

.......

reservoir

cm3/s

/

0

cm3/s

5

10

15 20 P~, MPa

|

//~ 0

5

10

15 Pl-P~,

20 MPa

Figure 4.27: The system of Guided Problem4.8. Guided Problem 4.8 This problem typifies graphical reduction of a model comprising steady-state elements. Hydraulic oil is pumpedto high pressure by a motor-driven pump. The fluid in turn drives a hydraulic motor with an unspecified attached load, as shownin Fig. 4.27. A relief valve limits the pressure, partly for safety, and an adjustable "meter-in" valve can be set or modulatedto control the torquespeed characteristic seen by the load. (The drawinguses standard fluid-power symbols.) Characteristics of the motor/pump,relief valve and adjustable valve including the attached fluid line are plotted. The motor has a displacement D = 60 cm/rad, and suffers an internal leakage of 0.2 cm/s for each MPaof pressure drop across it. It also has a static or break-awayfriction of 60 Nmand an addedviscous friction of 5.0 Nm/(rad/s). Youare asked to find and plot the torque-speedcharacteristics of the system as seen by the load, for various angular positions, ¢, of the adjustable valve. Suggested Steps: 1. Construct a bond graph model, treating the motor/pumpas a one-port source. Presumablyyou will include several junctions, several resistances and possibly a fixed-effort source. This is probably the most critical step in your solution. Use this graph to guide the steps below. 2. Combinethe given characteristics of the motor/pumpand the relief valve to plot a source characteristic pertaining to a location betweenthe relief

4.4.

EQUILIBRIUM

271

valve and the adjustable valve. Combinethe results of step 2 with the characteristics of the adjustable valve to plot a family of source characteristics, each for a different angle ¢, pertaining to a location between the adjustable valve and the hydraulic motor. 4. Modify the results of step 3 to refer to the flow which actually displaces the motor but does not leak past it. Convert the characteristics plots of step 4 to get torque vs. angular velocity. Due to the simple transformer type of coupling, this can be done by merely replacing the scales of the ordinate and abcissas. Correct the characteristics of step 5 to account for the static and viscous friction on the shaft. The result should be the desired family of source characteristics as seen by the load. Subsequent specification of a load characteristic and an angle ¢ would give a specific operating condition. PROBLEMS 4.50 Slightly compressible fluid enters a cylinder, shownbelow, pushing a piston against a linear dashpot with modulus R. //(/////////////////// piston

--~p,_.~ fluid, pressure P ~/~ areaA ~’//I/////////////////

"-~i

"-’-"~ C

T "--~r-

dashpot, modulus R

]

R

~ --~’P

T=A

i "~" R ’

C R’= TeR = A~-R

(a) Model the system with a bond graph. Determine the moduli of the elements in terms of expressed and implied parameters. (b) Simplify the model as much as possible; elements.

find the moduli of any new

(c) Give the simplest bond graph valid for equilibrium, and the associated algebraic relationship.

272

CHAPTER 4.

INTERMEDL4TE MODELING

4.51 A motor drives a winch of radius 1.0 ft. which both lifts a 50 lb weight and pulls a massless pulley of radius 3.0 ft with spring and damperattached, as shownbelow. Characteristics of the componentsare plotted below. motor

~ 200g.Z~ .......... momenq~ ft-lblo01

force,~-

/I force,

~ 51!

06’’’~’’’~’’’1’ 2 angularvelocity, rad/s

’’~~’’’’~ velocity, ft/s

’’

winch

I-

12

~ -[

vO 4 8 12 displacement,ft

(a) Define appropriate variables, and identify and label the plots correspondingly. (b) Drawa bond graph of the system, and label its variables consistent with part (a). (c) Estimate the moduliof any elements not described by the plots. (d) Find the equilibrium state of the system(i.e. find the values of all constant, non-zerovariables at equilibrium).

4.52 A gear motor (relatively slow speed), shownat the top of the next page, rotates a winch which rolls a cylinder of weight W= 10 lbs. up a 30° incline at steady speed, without slipping, against a restraining dashpot with resistance coefficient b = 0.25 lb.s/in. Theindicated radii are rw = 6 in., rl -- 10 in. and r2 ---- 5 in. Thetorque-speedcharacteristic of the motoris plotted. (a) Define variables, drawa bondgraphfor the system, and find the values of the parameters (moduli) therein. (b) Estimate the equilibrium angular velocity of the winch and linear velocity of the center of the roller.

4.4.

273

EQUILIBRIUM

50 40 in-lb 30

motor~ "" "~I

rider, W

O0 I11 I 21 I 13 I 41 ~ 5~ ~ 6~1 ~, rad/s 4.53 Design problem. A four-passenger vehicle is to be given an engine scaled to an engine with a known displacement, plus a manual transmission. Most parameters can be taken as given below. You are asked to size the engine and propose the values of between three and five transmission ratios. Factors such as vehicle loading, hill-climbing capability, acceleration and fuel economyare left to your discretion. The work can be divided into two phases: (a) Model the system, presumably with a bond graph. Get explicit constants or functions which detail the meanings of the various one-port elements in the model. Someinterpolation is necessary. Include hill-climbing and the possibility of acceleration. The result should give the vehicle speed and engine torque which results from any specified gear ratio, inclination of the road and vehicle acceleration. (b) The design phase of the problem includes your compromises regarding acceleration, hill-climbing, fuel consumptionand relative cost. In addition to. stating the engine displacement and the various gear ratios, describe the performance of the vehicle in a form that an informed consumer would want and can understand. Vehicle weight: 2200 lb., exclusive of the engine and accessories. 4. Air drag coefficient: 0.35. Density of the air: 0.0024 lb.s2/ft 2Frontal area: 15.4 ft Tires: 1032 rev/mile; constant rolling resistance force equals 0.65% of the total road weight. Overall power train losses: 25%at 10 ft.lb engine torque; 15%at 20 ft.lb; 6%at 90 ft.lb. Axle gear ratio: reduction of 2.6:1. Engine: The weight, available torque, fuel consumption and needed power for the accessories are proportional to the engine displacement. An engine with a displacement of 2.0 liters weighs 240 lbs including accessories, and has

274

CHAPTER 4.

INTERMEDIATE MODELING

the characteristics plotted below. The solid lines represent constant values of the brake-specific fuel consumption(BSFC),defined as poundsof fuel per brake horsepower hour. ("Brake" refers to the meansof measurement.)The powerto run the accessories (alternator, cooling fan, power steering and brakes, etc.) varies linearly from 2 HPat zero speed to 6 HPat 80 mph.

lO0 shaft torque, ft’lb 75

~...- ..... ~...maximumtorque ~-’~solidlines: indicated values of BSFC // / . "

locus of lowest f~el consumption 0

100

200

300 400 engine speed, rad/s

500

4.54 For the Belleville spring shownin Fig. 4.23 (p. 265), sketch the stored energy as a function of the displacement. Describe howthe force is related to this plot. 4.55 Identify the portion of the load curve for the case study in Fig. 4.24 (p. 264) that is non-passive. Also, identify which portion of the load curve wouldbe unstable or producea limit-cycle oscillation should the source (motor) characteristic be modifiedto give an equilibrium there. 4.56 A submergedturbulent jet of water discharges into a region of still water, passing below a stand-pipe with a sharp edge or lip as shownon the next page. The jet entrains a constant rate of fluid, Qe, fromthe bottomof the stand pipe. Also, if the constant pressure of the still water, Ps, is higher than the pressure, P, at the bottomof the stand pipe (just above the jet), with a correction for the weight of the water in-between, the jet bends upward; if Ps < P it bends downward.Considerable upwardbending causes a return flow Q~ to be peeled off by the edge. If on the other hand the jet is not bent upwardvery much, a pressure difference P~ - P causes a flow Qp to enter or leave the stand-pipe region through the gap betweenthe jet and the edge. These flows are plotted. (a) Drawa bondgraph for the system, recognizing that the stand-pipe acts like a compliance,the sumof the flows Q~÷ Qpacts like like a multivalued - source of fluid to this compliance,and the flow Qe acts like a (negative)

4.4.

275

EQUILIBRIUM

(b) Replace the individual characteristics for Qr and Qp with their sum, according to the bond graph model. (c) Determinethe equilibrium states. (Hint: there are three of them.) (d) Determinethe stability of each equilibrium state. (Postulate compatible non-equilibriumstate pairs in the usual fashion, and identify which direction the pressure is inducedto changeas a result.)

psi //~cross-sectional[// ~/~area A=20ft2 ~/standg~//’~ (1density p ~-//pipe "94 lb’s2/ft4)

-4 ..O.~~OQ,

~ ~

"

ft3/s

-8

_Q~

still

-12

water

~

/i/i/i/i///i/ 4.57 A type of reverse check valve, constructed as shownbelow left, has the pressure-flow characteristic as plotted. The fluid is incompressible. This valve can be used as the heart of a "hydraulic ram" (see Problem6.16, p. 407). ~pu,rnp flow Qo

accumulator valve

x

"1

valve characteristic ~

Pm

~

reservoir testing system

am

(a) It is desired to test this valve over as muchof its characteristic as possible, without inducing an instability. A pumpwith constant flow Qo and

276

CHAPTER 4.

INTERMEDL4TE MODELING

the traditional accumulatoris used with an adjustable linear restriction placed as shownabove right. Choose a value of Q0 > Qmand plot this point on the Q axis of the plot above. Then, construct two or three useful sourcecharacteristics, correspondingto different restrictions, on the plot. Deducethe region of the (load) valve characteristic that can be tested stably with your flow source. (b) Removethe accumulator and repeat (a). (A tiny fluid inertance mains.) (c) The orifice in tile valve has area 2~rrx/x/~, wherer is a constant effective radius. The valve opening x equals xo whenthe spring (stiffness k) has no force. The fluid has density p. Find the characteristic of the valve in terms of the parameters given, and computeP,~ and Q,,~. 4.58 A motor.drives a load through a spring coupling. Assumingthe characteristics given belowand neglecting all inertia, sketch-plot the angularvelocities of the motorand the load as funtions of time. Hint: limit cycle behavior exists for a wide range of starting conditions.

~

flexible shaft

-~

4 momem, N.m 2 I

I

20

I

I

40

I

I

I ~I

I

60 \ 100 speed, rad/s \

4.59 Consider the two systems (i) and (ii) below, in which the charactistics the source and the resistance are a plotted. In each case, and for equilibrium points. 1 and 2, indicate whether the equilibrium is stable or unstable. Show your reasoning.

system(i)

system(ii)

characteristics

4.4.

277

EQUILIBRIUM

4.60 Axial-flow fans and compressors typically demonstrate an S-shaped characteristic. A fan with the characteristic plotted belowsupplies air to a chamber of volumeV = 4 ft 3. Anorifice in the chamberexhausts the air to the environment;two different possible characteristics are plotted. Assumeatmospheric conditions. Determinethe stability of the respective equilibria, and detail any limit-cycle operationthat results. 15Pgage

lOpsi 5-

orifice fan t_____._.a

0

10 flow, ft3/s 20

4.61 A motorwith a given characteristic and negligible inertia drives a load with a given characteristic and insignificant inertia through a gear pair with negligible friction and inertia. Thecouplingis (i) rigid or (ii) highly flexible. Estimate the gear ratio in each case that produces the largest possible stable speed of the load, and permits the load to reach this speedfrom start-up. Also, estimate the equilibrium speeds. 3O torque, fi-lbs 20 10 0

I

0

I

I

I

I

I

100

~1

I

speed, rad/s

I

I

I

I

200

4.62 A cylindrical workpieceis machinedon a lathe by a cutting tool fed axially by a screw drive at a constant velocity }, as shownon the next page. The tool is somewhat flexible, however,so its feed rate ~ at the cutting edgecan differ from ~. The reaction force on the tool in the samedirection, plotted as a function of 1), has a region of negative slope (due to heating and softening of the material). The conCernis that strong vibration might result (tool "chatter").

278

CHAPTER 4,

INTERMEDIATE

force vs. k

40

~ l/

force vs. 5~

axial 30 force, lbs 20

to01 k__~j

~ rotating

MODELING

workpiece

10

........ ..........o 0

case i I

I

0.1

case ii I

I

0.2

case iii I

I

I

I

0.3 0.4 velocity, in/s

(a) Model the process with a bond graph, including only phenomena mentioned above. (For example, exclude inertias and any variation in the rotation rate of the workpiece.) Hint: Your model should include but not be limited to a compliance, a resistance and a junction. (b) Find the equilibrium drive forces for the three drive characteristics plotted by the dashed lines above: case (i), 2 = 0.09 in/s; case(ii), ¯ = in/s; case (iii), 2 = 0.32 in/s. (c) Determine whether each of the three equilibria is stable. Show your reasoning. Also, should any limit cycle exist, show how the force and speed of the workpiece change. (d) If you were to add one more phenomenon or parameter to the model, what would it be? (There are several good answers.) 4.63 Review question: Make a table of the dimensions of the moduli of bond graph elements. Label five rows as follows: R, C, I, T, G. Label four columns as follows: mechanical translation, mechanical rotation, incompressible fluid flow, electric circuits. Assumethe transformers and gyrators involve no transduction. Employthe symbolsL for length, F for force, t for time and Q for electric charge, as given in Chapter 1. 4.64 Answer the previous force.

question using M for mass and eliminating

F for

4.65 Answer the previous question substituting SI units for the dimensions. Use Newtons, meters, secouds and coulombs only. 4.66 Answer the previous question using kilograms but not Newtons. 4.67 Answer the above question in terms of inches pounds force and seconds. Omit the electric circuit column.

4.4.

279

EQUILIBRIUM

SOLUTION Guided

TO GUIDED

Problem

PROBLEM

4.8

1

Rr Ro

RL

So Rj

R~: resistance to flow of the relief valve. R~,: resistance

to flow of the adjustable valve.

RL: resistance

of the hydraulic motor to leakage flow, 0.2 cm3/(s MPa)

R f: viscous frictional

torque resistance

Se: break-away frictional

on the shaft, 5.0 N m/(rad/s)

torque, 60 N m.

40O cm~/s 200 0

0

5

10

15 ~0 P~,MPa

P2 -- P - 1 - (P1 - P’2). Therefore, subtract the (horizontal) given in the plot for R, from those of P1 above, to get

Q3 6oo~ ’400 ~__150--’~..~

0

4. Subtract

5

10

15 20 P~-P~, MPa

QL = P/RL = P/0.2 from Q3 to get Qm:

400 ~.~ 0~"~ cm~/s I ~oo~6"~--.~~

0

5

10

15 20 P~-P_,, MPa

values of P~ -

280

CHAPTER 4.

INTERMEDIATE

MODELING

1 1 rad/s = T = ~ = 6-~ cma/’~-~; therefore ~ = Q.~/60 and ideal torque: Mi = 60P,~cma’MN-= 60 x 10-6P~_ x 106ma MNN = 60P.~ N.m ~ m m~ MN

rad/~n~ ’ 300

600 900 1200 M~, Nm

6. M = Mi - Se - Ry~ = Mi - 60 - 5.0q~ (Subtract torque loss from plot above.) 00° ~,10 ~.~

rad/~g

~ 1200 M~, Nm

Chapter 5

Mathematical

Formulation

The writing and simulation of differential equations is extended in this chapter from the simple cases given in Chapter 3 to the much broader class of models considered in Chapter 4. Models are categorized accordin~ to their causal status, and whether they are linear or nonlinear. Section 5.1 focuses on the writing and simulation of differential equations for the simplest causal classification, and Section 5.2 focuses on more difficult cases. Section 5.3 introduces certain special features of linear models that are identified by their behavior as well as their structure. An operator notation is introduced that allows linear differential equations to be treated like algebraic equations, and an efficient simulation program for linear models is accessed. Finally, in Section 5.4, methods for approximating nonlinear models by linear models are presented, along with reasons why such approximations may or may not be justified.

5.1

Causality

and Differential

Equations

The application of causal strokes to bond graphs leads to a categorization of models according to the types of equations that they produce. After these categories are identified, the discussion focuses on the basic category of "causal models," for whicha set of state-variable differential equations suitable for analysis or simulation is produced directly.

5.1.1

Applying

Causal

Strokes

A certain procedure must be followed in applying causal strokes to every bond of a bond graph to assure that a subsequent procedure gives a proper set of state differential equations. In most cases the following steps are sufficient: 1. Apply the mandatory causal strokes for each effort or flo~v source (but not general source) that is present. 281

282

CHAPTER

5.

MATHEMATICAL

FORMULATION

2. Apply any mandatory causal strokes that follow from the presence of 0 and 1-junctions, transformers and gyrators. (See below for the latter.) 3. Apply integral causality to the bond of one of the remaining unspecified compliance or inertance elements. 4. Repeat steps 2 and 3 as many times as possible. This procedure produces one of the following outcomes: Every bond. has assigned causality, and each compliance and inertance element is assigned integral cuasality. These models are called causal. They are the easiest to treat. Every bond has assigned causality, but one or more compliance or inertance elements is assigned differential causality. These models are called over-causal. They can be treated directly, or re-formulated to become causal models. o

5.1.2

Some bonds are not assigned any causality. These models are called under-causal. They are more difficult to treat than causal models. Differential

Equations

for

Causal

Models

Once causal strokes are placed on every bond, a four-step procedure produces a complete set of state-variable differential equations. The procedure is a modest generalization of that introduced in Section 3.4.4. Before you start you should remove any variable designations that may already be written on the bond graph or, since these designations may still be valuable to you, start with a fresh unannotated rendering of the graph. (I) Annotate the input effort on each effort source and the input flow on each ilow source with some appropriate symbol. Also, if the conjugate variable is of no interest, you mayplace a large cross (×) in the appropriate place to register your disinterest. (II) Annotate the fiow side of the bond of each element Ci with ~i, and place a circle around this designation. Annotate the effort side of the bond with qi/Ci or, if nonlinear, a functional designation ei(qi). Similarly, annotate the effort side of the bond of each element Ii with Pi, and place a circle around this designation. Annotate the flow side of the bond with pi/~i or, if nonlinear, a functional designation (ti(ei). It is better if you use more specific symbols than the generic qi and p~ to represent the respective generalized displacements and momenta, which become the state variables. (III) That part of the bond graph that remains without effort and flow annotations has, as causal inputs, only the input terms from step I and

5.1.

CAUSALITY

AND DIFFERENTL4L

EQUATIONS

(a) circuit

(b) bond graph with first causal stroke

(c) causal strokes continued

(d) step II completed

(~p/l

q/C~(~

I

Qlp/l 1

C

(e) junction causalities

employed

283

q/C~Q C

(f) final annotated bond graph

Figure 5.1: Electric circuit

example

the terms q~/C~ and pi/Ii (or their substitutes). Propagate these variable designations throughout the bond graph, following the dictates of the causal strokes and using the power-convention half-arrows to determine signs. This key procedure is highlighted in the examples below. Note that the circled terms are "bottled up," and do not contribute to any further labelling. (IV) Write one first-order differential equation for each circled Oi and ibi. The circled term appears alone as the left side of the differential equation. The right side of the equation comprises functions of the input and state variables as dictated by the annotations and power-convention half-arrows on the bonds in the immediate vicinity of the circled term. For example, if the circled term is bonded to a junction, the right side comprises the sum of the properly signed causal inputs of the proper type (effort or flow) the other bonds on that junction. 5.1.3

Case

Study:

A Linear

Circuit

Consider the electric circuit shown in Fig. 5.1. The bond graph has two junctions. It is driven by an effort (voltage) source, which demandsthe causal stroke as shownin part (b) of the figure. Next, integral causality is placed on the iner-

284

CHAPTER 5.

MATHEMATICAL FORMULATION

tance (inductance) bond. This also forces the causality on the bondto the right side of the 1-junction, as shownin part (c). Integral causality nowcan and must be placed on the compliance bond, forcing admittance causality on the resistance bond, as shownin part (d). This completes the assignmentof causal strokes, and the modelis seen to be causal, since both energy-storage elements haveintegral causality. Part (d) of the figure also showsthe application of steps I and II for the writing of differential equations; the input effort is designated as e, the associated flow is markedwith a ×, and the efforts and flows on the bonds for the C and I elemeuts are designated as directed. Step III starts by observing that the causal stokes around the 1-junction dictate the labeling of the flows onthe left and center horizontal bonds as p/I, and the causal strokes around the 0-junction dictate that the efforts on the center and right bondsbe labelled with q/C, as shownin part (e) of the figure. This leaves only the flow on the R bond unannotated. The presence of impedance causality for the R element forces this effort to be written as q/RC, as shownin part (f). The first-order state differential equations nowcan be written (step IV). The circled term dp/dt is written on the left side of one of the differential equations, and the causal inputs of the attached 1-junction provide the terms on the right side:

dp

1

d--~ = e - ~q. (5.1) The circled term dq/dt is written on the left side of the other differential equation, and the causal inputs of the attached 0-junction provide the terms on its right side: dq 1 1

~ = ~p- ~-~q,

(5.2)

The first term on the right side of this equation is positive because the power conventionhalf-arrow on the center horizontal bondis directed towardthe junction. The second term is negative because the powerconvention half-arrow on the R bond is oppositely directed. There are two dependent variables (p and q) and the same numberof firstorder differential equations. As a result, the equations are solvable as a set. This is always the outcomewhenthe procedures are applied to a causal model. Takingthe derivative of equation(5.2), substituting the right side of equation (5.1) for the resulting term dp/dt and rearranging, gives d2 q I dq IC-~_ + -~-[ + q = Ce.

(5.3)

This is a classical second-orderlinear systemof the form of equation (3.41) (p. 149), which is one of the few that you should have memorized.Comparingthe coefficients, the natural frequency is seen to be w, = 1/v/~-~ and the damping ratio is seen to be (R/2)V/~ (which happen to agree with the 1-junction case discussed in Section 3.5.5, p. 150). The steady-state solution, which emerges whenthe two derivative terms in the differential equation are set equal to zero, is q = Ce (whichis different from the 1-junction case).

5.1.

CAUSALITY AND DIFFERENTIAL EQUATIONS

285

EXAMPLE 5.1 A machine of mass m is isolated from its foundation by a spring and a dashpot, as shown below (and in Problems 3.28 (p. 131), 3.39 (p. and 3.47 (p. 158)). In the present case, the foundation itSelf has a known displacement zi = zi(t), and the resulting displacement of the machine, Zo(t), is to be found. The bond graph below has been drawn as the first step in the analysis. As the second step, define state variables, find a set of state differential equations, and relate zo to the state variables. Zo

~ foundation

I= m, C = 1/k, R = b

I

Solution: The causal strokes are drawn following the four steps given in Section 5.1.1, starting with the mandatory strokes on the bonds for SI and Se. Integral causalities on the bonds for C and I produce no causal conflicts at the junctions, and therefore also are required. Since there is no choice in any of the causal stokes, the model is causal. Steps I and II of the procedure for writing the differential equations, as given in Section 5.1.2, also are shown below as annotations on the bonds for Sf, Se, C and I:

The output variable is related to the state variables p and q by zi = f ~i dt = (1/I) fpdt. As shown at the top of the next page, the lower 1-junction and its causal strokes force the writing ofp/I for the flow on the bond to the left of the junction. The causal stokes on the bonds for the 0-junction mandate that the flow on the vertical bond be written as izi -p/I, which also becomes the flow on the bond for R. The effort on the bond for R becomes R(~.i-p/I), and finally the ettbrt on the vertical bond and the bond to the right of the 0-junction becomes q/C + R(i~i - p/I).

286

CHAPTER

5.

MATHEMATICAL

FORMULATION

C q/C 1 R(~-p/I) q/C+SR ~(~i"~ -p/I)l ii "p/I q/C+R(zl 7 zi

p/I

-p/I)

I mg

~ 1 t-~--S~ I

The two differential equations result from the respective causal strokes and annotations on the bonds for the respective 1-junctions: dq dt

= ~(t) - -~p dp 1 R dt - -~q- Tp + Rib(t) - mg.

5.1.4 Case Study: Nonlinear Stick-Slip The stick-slip system analyzed in Section 4.4.6 (p. 266) is nowrevisited, introducing the effect of the rotational inertance of the load that was neglected in Fig. 4.26 (p. 267). The system is re-drawn in part (a) of Fig. 5.2, and the given linear torque-speed characteristics of the motor and the resistance (frictional) load are reproduced in part(b). The bond-graph model shown in part (c) the figure is the same as given before, except for the added inertia, which also requires the use of a 1-junction to register the fact that the load resistance and load inertance refer to the same angular velocity. The model is now too complex for its behavior to be deduced directly; differential equations are needed. The source S is neither an effort source nor a flow source, so step 1 of the procedure for drawing causal strokes (p. 281) does not apply. In step 2, integral causality is applied to the compliance, as shown in part (d) of the figure. Step 3 recognizes the causal mandate of the zero junction, as shown in part (e). No such mandate exists for the 1-junction, however, one proceeds to step 4 and applies integral causality to the inertance, as shown in part (f). Finally, the causal mandate of the 1-junction dictates the causal stroke on the final bond, leaving the bond graph of part (g). Step I of the procedure for writing differential equations (p. 282) does not apply. Step II annotates both the efforts and the flows of the energy-storage elements; it is included in part (g) of the figure. In part III, the causal stroke on the compliance bond now dictates the annotation of the efforts of the lefthand and center bonds, and the causal stroke on the inertance bond dictates the annotation of the flows of the center and right-hand bonds. For convenience, the effort on the source is also labelled as Ms, and the flow on the resistance is also labeled as ~R, to give the situation shown in part (h) of the figure. Finally, the admittance causality of the nonlinear source gives a function in the

5.1.

287

CAUSALITY AND DIFFERENTIAL EQUATIONS

force[ 0.inertive ~ ’,’,load turn -II ¯ [ ~,o,or I nex’ble shaf~ ~ 8~motor, t°rque I [ ....... I ............. ~"

(a)sy--

frictional loadu

1~1

Cs

S~ s

0

1~1

~

1

1~1 g

~R

CR /

g ~ C 1 (c) bondgraph

.....

¯

MsvsCs

lb

0

~0 X

80 speed, rad/s

(b) torque-speedCharacteristics

S"-~-~ 0 -’----’~-

S I’-’-~’~

0 "~’t

120

1 ~R

1 "----’~R

C I (d) first causalstroke

C I (e) causal strokes fromO-junction

(f) causalstroke for the inertance ¢c/CI~ (~p/l _ C l (g) final stroke, andstep SI.Ms=. ¢c/C r ¢c/

M = (/)c/C ¢c/C ~ 0 ----~-~--~1 ~ .rR _ S~ s T,,~,p/I,..:.,,i p/I=CR ¢c/C1~.¢c)(~pj~p/l ¢c/C ~---~. 0~11 MR =MR(~e,) ¯ rR C I (h) applicationof junctioncausalities /I

C I (i) fully annotatedgraph Figure 5.2: Stick-slip systemwith load inertia included

288

CHAPTER

5.

MATHEMATICAL

FORMULATION

form q~s(Ms) for its flow, and the impedance causality of the resistance gives function in the form ]I.’/R(q~R ) for its effort. All the efforts and bonds are now annotated, as shown in part (i) of the figure. The writi.ng of the differential equations (step IV) now can proceed. The circled term ¢c is set equal to the sumof the efforts on the 0-junction: dec = (hs(Ms) dt

- ~p; ~s = Ss(Ms);

Ms = ¢c’/C.

(5.4)

Similarly, the circled term ~ is set equal to the sumof the flows on the l-junction:

~ = ,~/c - M~; M~= M~($~); ~ = dt

I"

(~.~)

These equations are solvable, presuming the functions ~s(M8) and Mn(~n) are known, since the only variables are the state variables ¢c and p. They are not readily solvable analytically, however, due to the nonlinearities of these functions. The source ~nd load torque-speed characteristics, as plotted, are given by

Ms= M~o- M~- M~$~, (5.6~) M~= sig~(p)M~oM~ Ip/Zl + M~ (p/I)~; p/ Z = ~,(5. 6b) Mso = 10.0 ft.lb Ms2 = 0.0056 ft.lb.s~;

M~0 = 5.9 ft-lb

Ms1 = M~ = 0.10 ft-lb.s

M~= 0.00086 ft-lb.s~;

C = 0.1 rad/(ft.lb).

(5.~)

Noting that Ms = M = ¢c/C, equation (5.6a) can be solved for ~s:

+s-

2Ms~ ~k~]

+ Ms~ CMs~

(5.7)

Substitution of equation (5.7) into equation (5.4) and equation (5.6b) equation (5.5) gives an integrable state-space formulation. A corresponding MATLAB function file follows. The particular value of I given (I = 0.00001 lb-ft s ~) corresponds to virtuMly negligable inertin. ~c~£on ~=stkslp(t ,x) ~S0=10; ~R0=5.9; ~SI=0.1; ~RI=0.1; ~$2=0.0056; ~R2=0.00086; ~=0.1; I=0.0000~; f (1) =-~S1/(2,~S2) +sqrt (SS 1/(2.~S2) ) "2+~0/~$2-x (1) -x(2)/~; f (2) =x (1)/C-~ ign (x (2)) *gR0+~Rl*abs(x (2) /I) -~R2* (x The model can be simulated and the output shaft speed ~c = p/I plotted for 0.1 second, from a resting start, with the following commands: [t,x]=ode23(’stks~p’ ,0,0.1, p~ot(t,x(: ,2/.00001))

[0 0])

5.1.

289

CA USALITY AND DIFFERENTIAL EQ UATIONS

Plots of load speedvs time: 120

...........

~ I = 0.00001lb.ft.s rad/~o 0i

i

0

]!

II

0.02

I|

0.04

II

II

II

II

II

0.06 0.08 t, seconds

/

0.10

6O 2I = 0.001lb.ft.s

40 rad/s 20 0 5O

2I = 0.0021lb.ft.s

I = 0.0022 and 20.0100lb.ft.s

40 rad/s 2O - 0.9100 lb.ft.s ~ 0.0022 lb.ft.s 2 [ rad/?OI/~/A A A A /~//~ /~ ~ 0.2

0.4

0.6 0.8 t, seconds

Figure 5.3: Simulationresults for stick-slip model

1.0

290

CHAPTER 5. MATHEMATICAL FORMULATION

Theresult, shownat the top of plot of Fig. 5.3, reveals a limit-cycle behavior almost the same as deduced graphically in Fig. 4.37, but with a very small effect of the inertia. The corresponding result for I = .0010 ft-lb s2, shown next, reveals a lowerlimit-cycle frequencyand a reduced fraction of time spent sticking. Whenthe inertia is increasedto I = 0.0021ft-lb s’~, the sticking portion is seen to have virtually vanished. Whenthe inertia is increased marginally to 0.0022 ft-lb s’~, however,the simulation reveals that the shaft speed converges toward the equilibrium speed, which is ¯ - +MRo _ 25.19 rad/s. ¢~q~i~ =~/Mso V Mm. Ms~

(5.8)

Thusthe instability has abruptly given awayto stability. A further increase in inertia increases the decayrate for Oscillations without any effect on ~equib. The exampleof I = 0.01 ft-lb s2 showslittle oscillation. It is instructive to plot one of the state variables of a second-ordermodel versus the other. This is called a phase-plane plot. In Fig. 5.4, phase-plane plots of the torque Ms = ¢c/C versus the angular velocity p/I are given for the various cited cases. Timeis supressed in these plots, but the directions of increasing time of the trajectories is indicated by arrows. In the present case, the underlying source and load torque-speed curves are shownby dashed lines, for comparison. EXAMPLE 5.2 RepeatExample5.1 (p. 285), substituting the nonlinear constitutive equations ec = av/-~ for the complianceand eR = b~ for the resistance. Solution: The only differences from the solution given for Example5.1 are in the notations for the efforts on the bondsfor C and R: C aC~q,. ,b(z~ -p/I)~. aC"q+b(~.i-p/I)2[~.i- p/I ~ X ~agrq +b(~i-p/[)

2 mg

I The differential equations therefore become dq_ dt

1 i p- ~ (equation unchanged)

(

d--i =,~vq - ~ ~(t) -

5.1.

291

CAUSALITY AND DIFFERENTIAL EQUATIONS

10 \\motor

8 torque 6 lb ft

\\

I = 0.00001lb ft s’-

2I = 0.010lb ft s

2

motor \40 80 ~ speed, rad/s

80 \40 ~ speed, rad/s

120

120

10 8torque 6 lbft 4

\\motor = 0.0021 lb ft _\ I

2 S

-

\\motor 8 \

2 0

~20

0 0 ~ speed, rad/s

,

,

i 80 ~ speed, ra~s

Figure 5.4: Phase-planetrajectories for the stick-slip system

120

292

CHAPTER 5. el ~T e.~ q:

MATHEMATICAL FORMULATION

e~ ~G e.~ t)~ q:

(a) transformer

or ~!--.~ q~

G~ q2

(b) gyrator

Figure 5.5: Causal constraints for bonds on transformers and gyrators

/ ¯

~

~off~’~)gear

reduction

compliance C

DCmotorwith fixed field note: bearings not sho~ [ flywheel,ine~ia I (a) system

/ R~ R~p /1 ~/I~’

:

~ ~ e--~e-R~Tq/GC~Tq/C~ - q/C~ ~ /~Tq/GC~e/G-R~Tq/G C-~Te/~RIT q/G C]

~ Rfl

C

I

(b) bond graph wi~ a~omtedvariables Figure 5.6: DCmotor driving a flywheel through a flexible shaft 5.1.5

Case

Study

with

Transformers

and Gyrators

The causal strokes on the two bonds of a transformer must be aligned the same, that is be asymmetricalwith respect to the element, as shownin Fig. 5.5. This follows directly from the definition of the transformer; specifying the flow on one side also specifies the flow on the other side, and similarly for the efforts. The causal strokes on the two bonds of a gyrator must be. aligned oppositely, that is be symmetrical with respect to the element, also as shown.Either both flows must be specified and the two efforts follow directly, or vice-versa. These causal constraints direct the propagationof causal strokes through transformers and gyrators in bond graphs. The electromechanicalsystemshownin Fig. 5.6 illustrates the use of causal strokes with transformers and gyrators. A DCmotor with a resistance in the armaturecircuit and a fixed field drives a flexible shaft througha gear reduction. The shaft in turn drives a flywheel. The non-idealites of the motorare ignored,

5.1.

CAUSALITY AND DIFFERENTIAL EQUATIONS

293

along with the inertia of all parts except the relatively massive flywheel. The springiness of the shaft is assumed to be important, however, and is represented by a compliance. The resulting bond graph model is shown in part (b) of the figure. Placing integral causality on the compliance determines the causalities of all except the two right-most bonds. You should carefully see that this happens, step-by-step, starting with the two horizontal bonds off the 0-junction. Placing integral causality on the inertance bond determines the final two causalities. Step I of the procedure for writing differential equations (Section 5.1.2, p. 282) places the effort e on the left-hand bond, and is completed by placing an × on the effort side of the bond for the flow source. Step II places the standard notations(~and q/C on the compliance bond, andGand p/I on the inertance bond. Step III, annotating the efforts and flows on all the bonds according to the dictates of the causal strokes, demandsparticular attention. First, the effort on all the bonds of the 0-junction are labeled as q/C, since this is the causal input to that junction. Similarly, the flow on all the bonds of the rightmost 1-junction are labeled as p/I, since this is the causal input to that junction. The resistance R2 has one of these flows as its causal input; its causal output must be labeled as R2p/I, consistent with its impedance causality. Dashed lines with arrows have been placed on the bond graph to underscore the meanings of the causal strokes and the consequent sequence of the determinations of the variables written next to each bond. You may find this practice helps you handle confusing situations. You now proceed to label the remaining efforts, and flows, following a sequence dictated by the causal strokes. The effort (~) is a causal output from the transformer; it equals T times the causal input effort on the right side of the transformer. The flow (~) then follows from the causal strokes on the gyrator. This flow is the causal input to the left-hand 1-junction, and therefore is written also on the bond for R1. The resistance R1 has impedance causality, like its cau~al output is the effort (~). The efibrt (~) is the causal output effort of 1-junction; it equals the difference between the effort e and the effort(~). Note that the power convention half-arrows determine the signs of the terms, while the causal strokes direct their content. Next, the causal strokes on the gyrator bonds dictate the flow (~). Finally, the transformer converts this to the flow (~). The differential equations for the system now can be written upon inspection of the annotated bond graph (step IV). The circled 0 is seen, from the causal strokes and power-convention half-arrows around the 0-junction, to equal the flow entering the 0-junction from the left minus the flow exiting to the right, or dq T ( R~T ~ (5.9) d--[ = -~,e - --~ q ) - -] p. The circled ~b is similarly seen, from the causal strokes and power-convention half-arrows around the right-most 1-junction, to equal the effort of the zero junction minus the effort of the resistance R2, or dp 1 -p" dt - -~q - --i

(5.10)

294

CHAPTER

5.

MATHEMATICAL

These equations are solved for particular lem 5.7 given in Section 5.3.

FORMULATION

parameter values in a Guided Prob-

EXAMPLE 5.3 The bond graph model below comprises elements with constant moduli. Apply causal strokes, define state variables and write a set of state differential equations. R

r 1

S s (~.(~)

~ T

~I

C Solution: The causal stroke on the left-most bond is mandated by the flow source, and the gyrator forces the causality of the next bond, as represented at the top the next page. The 1-junction allows integral causalities for the compliance and the inertance, which therefore must be chosen. Note the constraint on causalites imposed by the transformer. The state variables therefore are the displacement q on the compliance and the momentum p on the inertance; these are placed on the respective bonds, with dots to indicate differentiation and surrounding circles as flags for their significance. The conjugate variables for the I and C bonds also are shown. R

S~I

- G

,~,,,(t)

~1 1

-IT

p/I

r~I

q/CI @ C The causal strokes nowdictate the writing of the remaining efforts and flows. As always, all uncircled annotations must include no variables other than the state variables and input variables. (Parameters such as I, R, G etc. are constants in this problem, not variables.) The result is

5.1.

295

CA USALITY AND DIFFERENTIAL EQ UATIONS

R

s~i OplTS ,~,,,¢t)- G

plTl

-i

olTl ,Git,.(tJ-RplTI-qlC . - p/TI ~1 T -~----~I

q/C C from whichthe differential equations can be written directly: dq

1

dp 1 R G + dt - ~-~q- ~-~P ~qo(t). 5.1.6

Models

Reducible

to

Causal

Form;

Order

of a Model

The causal constraints for effort and flow sources, junctions, transformers and gyrators sometimesprevent one or more of the compliances and inertances of a particular system from being assigned integral causality. These over-causal modelscar/ be reduced to causal form, simplifying the subsequent determination of their differential equations, although sometimesthis reduction is not practicable. The energy stored in any elements found to have differential causality is not independentof the energy stored in the elements given integral causality. Specifically, the displacements on the complianceswith integral causality and the momentaon the inertances with integral causality not only determine the energystored in their respective members.Theyalso are sufficient, collectively, to determine the energy stored in the inertances and/or complianceswith differential causality. Thenumberof independentfirst-order differential equations, that is the order of the system, equals the numberof energy storage elements with integral causality only. This assumes that the standard procedure which favors integral causality has been followed. Consider two compliances bonded to a commmon 0-junction, as shown in Fig. 5.7. One of them must be assigned differential causality because of the causal constraint of the junction. The two displacements are not independent, because they both are proportional to the same effort. As a result, the two energy storages are dependent. This situation is easily simplified by combining the two compliancesinto a single compliance,as shownin part (b) of the figure: C =C1 +C.~. In contrast, consider two compliances bonded to a common1-junction, as shownin part (c) of the figure. Both can be assigned integral causality; two state variable displacementscan be defined, and two differential equations can be

296

CHAPTER 5.

MATHEMATICAL FORMULATION

(a) first-order withdifferential causality

~C2

e=el+e2

C=CI+Q ~ e

~

(b) first-order, nodifferential causality

1/C=I/CI÷I/C2

(c) second-order,no differential causality ~eo q

~

T

~

C~

~

.~’T~qC 2 ~,q 2C=C~+C2/T T 1/C=I/C~+T~/C~ C~ CI (d) first-order withtransformer (e) second-orderwith transformer 1

Figure 5.7: Combiningdependent compliances written. The contribution to the order of the overall systemby these compliances therefore is two. Nevertheless, it is possible to combinethe two compliancesinto a single compliance:1/C = 1/C1+ 1/C.2. If this is done, the order of the system is legitimately reduced by one. You can say that the minimumorder of the system(with the combinedcompliance)is less that the nominalor actual order of the system whenthe compliancesare treated as distinct. Parts (d) and (e) of the figure showpairs of compliancesthat are less obviously combinablebecause of the presence of transformers. The causal strokes reveal the same pattern as before, however: with a 0-junction, only one independent differential equation can be written, whereaswith a 1-junction, two differential equations maybe written, although the numbermaybe reduced to one by combiningthe two compliancesand the transformer into a single equivalent compliance. In either case the combinedcompliancecan be found either of two ways. In the first method, the compliance C2 is combinedwith the transformer to give an equivalent compliancewith modulusC2/T’2 (as noted in Table 4.1 in Section 4.3.3), which then is combinedwith C1. In the second method, the efforts or flows of the two compliancesare related to a common effort or flow. Their energies, expressedas a function of this commmon effort or flow, are summedto give the energy of the combined compliance. The second method tends to be more powerfuland general, and therefore is preferred. Similar considerations apply to two inertances bonded to a commonjunction, as shownin Fig. 5.8. As you have seen before, the combinationrules for inertances are the dual of those for compliances;those for a 0-junction in one case act like those for a 1-junction in the other case. It is not necessaryor even

5.1.

CA USALITY

AND DIFFERENTIAL

EQUATIONS

297

(a) first-order with differential causality e=e, +e~ I=4 +I2 ~ ~

~

(b) first-order, noI differential causality

e

1/I=1/11+1/I2

(c) second-order, no differential causality e ~ 1.1-----.~.

T~----"I2

~

0

e

q

~T r,12 ~ I/I=I/II+I/T~I~

~q l=Ii+T2l2’ TdI~

11

(d) first-order

with transformer (e) second-order with transformer

Figure 5.8: Combining dependent inertances

desirable to memorize these rules. You should work out each case by focusing on its variables, noting particularly which variable is commonand which variable sums, and utilizing the constitutive relations of the elements. In some cases the reduction of an over-causal model to a causal model is relatively difficult to accomplish. You have the option of foregoing this reduction, and proceeding directly to the writing of differential equations. This option is discussed in Section 5.2. EXAMPLE 5.4 Determine the order of the following model that comprises elements with constant moduli:

Sy-- O -----’--~-

Il

T-------..~

I .-------~-R

C

Solution: Causal strokes are placed, starting with the mandated flowsource bond. Any two of the three energy storage elements can have integral causality assigned, but the third must have differential causality. One of the three possibilities is

298

CHAPTER 5.

MATHEMATICAL FORMULATION

l

~’differential causality

Syl--~

O I~--.-~-~.-

T I~---~-

I ~---~-R

C As a result, there are only two state variables and two first-order equations. The model is said to be of second order.

5.1.7

differential

Summary

A four-step procedure for assigning causal strokes has been given which is sufficient for causal and over-causal models. Every compliance and inertance element in a causal model is given integral causality by this procedure; the over-causal models are those for which one or more differential causalities result. The model is called under-causal whenever the procedure fails to assign causalities for all the bonds. Another four-step procedure is given for subsequently writing a set of statevariable differential equations for a causal bond graph. This procedure annotates the efforts and flows on all the bonds, before the equations themselves are written, following the special causal rules. Previously written annotations may violate these rules, and should be removed or set aside whenever this happens. Even if an annotated effort or flow is functionally correct, if it violates the causal strokes it will frustrate the purpose of the causal method. Definitions of variables using different symbols can be reintroduced after the differential equations are written. Onedifferential equation of first order results from equating the derivative dpi/dt or dq~/dt for each energy storage element to the causal input of its bond. The signs are determined by the power convention half-arrows. The energy storages with differential causality in an over-causal model can be expressed as functions of the same state variables in terms of which the energy storages with integral causality are described. They are therefore not independent, and the number of differential equations and the order of the model equals the number of energy storage elements with integral causality, assuming the standard procedure which maximizes this number has beeen followed. It is advisable to reduce over-causal models to causal form whenever the reduction is not very difficult to carry out; simple examples have been given. Otherwise it is still possible to proceed, as described in the following section, which also addresses under-causal models.

5.1.

CAUSALITY

AND DIFFERENTIAL

EQUATIONS

299

Figure 5.9: Two-tank system for Guided Problem 5.1

Guided

Problem

5.1

Neededpractice in finding differential equations for linear models is provided by this problem. Fig. 5.9 shows two fluid tanks with an independent supply of liquid to the left-hand tank, an interconnecting tube with both fluid resistance and inertance, and a resistive drain from the right-hand tank. Draw a bond graph for this system. The resistances depend on viscosity, and may be identified by unspecified R’s. Other moduli should be expressed in terms of physical parameters such as areas, lengths and fluid density. Identify the order of the system, and write a corresponding number of state differential equations. Suggested

Steps:

Identify five flows: the input flow Qi~at the left, the net flow that fills the left-hand tank, the flow between the two tanks, the net flow that fills the right-hand tank and the flow which leaves the drain. Use these flows to establish a junction structure for the bond graph, and complete the graph. Apply causal strokes to the graph, using the proper priorities. Is integral causality possible for all the energy storage elements? Identify the order of the system. Label each compliance bond with t~i and qi/Ci, and each inertance with ibj and pj/I~, where the subscripts are chosen to distinguish different elements of the same type. Circle the terms Oi and ~bj. Place × for the effort variable associated with the input flow Qin. Propagate the causal inputs Q~n, qi/Ci and pi/Ii through the bond graph until all efforts and flows are properly annotated. Write the differential equations, using the causal stokes and power convention arrows to direct your work. You are not asked to combine these equations into a single equation of higher order, but note that this could be done.

300 Guided

CHAPTER Problem

5.

MATHEMATICAL

FORMULATION

5.2

Dynamics are added to a steady-state model in this problem, and the corresponding state differential equations are found. An important way to eliminate differential causality is demonstrated, which allows the differential equations to be found. Augment the model of the hydraulic system of Guided Problem 4.4 (p. 225) by adding masses or rotational inertia and resistances to the rotary actuator and the two pistons. Define state variables and write the associated set of state differential equations. Suggested

Steps:

1. Add 1-junctions, inertances and resistances the three inertances and three resistances. 2. Attempt to add integral causality violation of causal constraints.

to the bond graph to represent

to these inertances,

and observe the

3. Computethe kinetic energy of the offending inertance, and express it as a function of the generalized velocity of one or both of the remaining inertances. Use this calculation to transfer the removed inertance to one or both of these inertances, augmenting the value(s) of the inertance(s) appropriately. 4. Computethe energy dissipation of the corresponding resistance, and similarly express it as a function of the generalized velocity of one or both of the same inertances. Use this calculation to transfer the removed resistance to one or both of these locations, augmenting the resistance there appropriately. 5. Reapply integral causality (it works now), and prepare the energy storage bonds with the usual notations. 6. Write the differential manner.

equations, employing the causal strokes in the usual

7. Write auxialiary equations for the outputs Q and ~ as functions state variables, in case these are of interest.

Guided

Problem

of the

5.3

Even if your instructor suggests that you forgo carrying out this more sophisticated problem, you should at least study it and its given solution. It involves the determination of nonlinear differential equation models for the mechanical snubber shown in part (a) of Fig. 5.10, which is available commercially. The piston is made with a low-friction coating, and fits closely with the glass cylinder to give little leakage. An air valve acts like a small orifice for out-flow (which

5.1.

301

CAUSALITY AND DIFFERENTIAL EQUATIONS F drive rod (b) model neglecting Rs (air compressibility . piston

eval

spring in chamber I ~ 1 ~ Cspr (c) model with air compressibility Re( l

one-way

adjustable orifice

(a) system

0 -----~ C~

Rva!

Figure 5.10: Guided problem 5.3 in practice can be adjusted), but a one-wayvalve (check valve) presents a much larger effective orifice for in-flow. Asoft spring fits inside the cylinderto resist compression.Youare asked to find a differential equation modelwhich can give the position x(t) in response to an applied force F(t). The mechanicalrubbing can be characterized by a dry friction force F0, the fluid leakage across the piston by a constant resistance Rp (based on laminar flow), the spring by its rate and free length x0, the mass of the piston assemblyby m, the nominaldensity of the air by P0, the effective area of the orifice for out-flowby Ao, the effective area of the orifice for in-flow by A~and the area of the piston by Ap. (a) The bondgraph in part (b) of the figure neglects the compressibility of the air. Annotate the graph with symbols, and find an approximate differential equation model. Note: the symbolsign(y) maybe useful; equals +1 when y > 0 and -1 wheny < 0. (b) The bond graph in part (c) of the figure recognizes compression the air. Find an approximate differential equation model. The air compliance is more difficult to implementthan most nonlinear compliances, so significant details are proposedbelow. It is suggested that you retain the approximatior/Q .~ v/(2/p)lP- Pal for valve flow, since the interest is in pressures P which do not differ radically from the atmospheric pressure Pa. (A more accurate modeling requires considerations of compressible fluid flow, which are not addressed until Chapter 12.) Further, the compressionmaybe considered to be isentropic.

302

Suggested

CHAPTER

5.

MATHEMATICAL

FORMULATION

Steps:

1. Apply causal strokes to the bond graph of part (b), define state variables in the conventional way and annotate the conjugate forces or velocities on the I and C bonds with the standard computationally realizable functions. 2. The resistance Rfric acts something like a force source, but with the sign reversing with the direction of the motion. Write a computationally realizable function for this force adjacent to the bond. 3. Use an orifice equation to approximate the flow through the valve, add the flow for the leakage across the piston, and invert to find the pressure as a function of the total flow. Then relate this flow to a state variable. Also, relate the force Fair to the pressure and thereby to a state variable. 4. Write the two first-order causal strokes. 5. To start strokes applies. inverse need to

differential

equations,

with guidance from the

the solution of the model with compression of the air, apply causal to part (c) of the figure and adapt the prior model to the extent Note that the causality on the leakage resistance element is the of what you used for the incompressible model, so that you do not invert the equation for leakage flow.

6. Note that the assumption of an isentropic process means that the pressure in the chamber can be written as P = Pa(q/x) k, where q is the portion of the piston displacement associated with compression of the air and k = 1.4 (not to be confused with the spring constant). Use this relation to express the effort on the 0-junction as a function of state variables. Then, get expressions for the leakage flow in terms of this effo}t and fixed parameters, depending on the sign of the velocity. 7. Write all three state differential equations. Makesure the only variables that appear on the right sides of the equations are, directly or by substitution, the excitation variable F and the state variables x, q and p. PROBLEMS 5.1 Systems are modeled with the constant-parameter bond graphs shown below. Apply causal strokes, using integral causality. Define state variables, and write the corresponding set of state differential equations.

5.1.

303

CAUSALITY AND DIFFERENTIAL EQUATIONS

5.2 Answerthe question above for the bond graph shownbelow left. S~ el-~ i" 1 ~ 0 ----~R

S]- ~-~. 0 ~ 1 "-~-R I

C

C

I

5.3 Answerthe question above for the bond graph shownabove right. 5.4 Find electric circuits that correspond to the bond graphs of (a) Problem 5.2 and (b) Problem5.:3. 5.5 Find mass-spring-dashpot systems with x-motion only that correspond to the bond graphs of (a) Problem5.1 and (b) Problem 5.6 Write the state differential equations for the modeldeveloped in Guided Problem4.:3. Theelectric current is specified as a function of time. 5.’/ Answerthe questions of Problem5.1 for the bond graph given below: I ein

S~ O "~"7"

C

G"--"-~

R,

I

R~

5.8 For the bond graph below, in which the causal input is rC

0

rR

(a) Applycausal strokes, choosestate variables and label all the efforts and flows accordingly. (b) Write the correspondingset of state differential equations.

304

CHAPTER

5.

MATHEMATICAL

FORMULATION

(c) The output variable of interest is ~0. Relate this to the state variables and the input variable, e0. (d) Express, in terms of the given parameters, any time constants, natural frequencies or damping ratios that apply.

5.9 Find systems with (i) springs and dashpots, and (ii) electric capacitors resistors, that are analogous to the two-tank system of Fig. 5.9 (p. 299). Assume constant parameters. Find a commondifferential equation model, with primary interest on the response of the second compSance. 5.10 A simple I-C (or L-C) filter connects a voltage source to a resistive load, as shown below. Find a differential equation relating the load voltage eL to the source voltage es.

filter

5.11 Continue Guided Problem 4.5 (p. 227) by defining a minimumset of state variables, considering the fluid pressure, P, to be an independent excitation and the motions of the solid parts to be responses. Find the corresponding state differential equations in a proper form for integration. You may employ parameters of your own devising if they are defined explicitly. 5.12 Consider the mechanical system described in Fig. 4.24 (p. 264) and the accompanyingtext. Define state variables and write a set of first-order differential equations in a form suitable for solution. Continue to ignore the inertias of the moving parts. The nonlinear functions may be defined as unspecified functions, e.g. ]~/m:/~//ra(~m) or ~)m:~)m(Mrn). 5.13 A hydraulic shock absorber of the type described in Fig. 2.24 (p. 47) employed as the dashpot in the system of Problem 4.2 part (c) (p. 201). Write the state differential equations for the resulting nonlinear system. 5.14 Write a set of state differential engine" of Problem 4.49 (p. 257).

equations for the aircraft carrier "arresting

5.1.

CAUSALITY AND DIFFERENTIAL EQUATIONS

305

5.15 Design Problem. Carpenters often use pneumatically driven nailers for framing, roofing and finish nails, poweredby a large air compressorthrough a rather large hose. Battery-powerednailers have been considered, but the power a reasonable-sized battery with solenoid or electric motorcan deliver is vastly less than the instantaneous powerneeded to drive a large nail in one motion. The nail must be driven homein a few milliseconds, for otherwise the recoil momentum is too great. Some~neans is need to store energy for quick release. Springs weigh too much.Kinetic energy is one possibility. The energy storage of a compressedgas is another, whichis the subject of this problem. I~ is proposed to use a small motor-driven hyraulic pumpto force 0il into a small hydraulic accumulator. The oil also pushes on a piston which, whena trigger release is actuated, drives the nail. Someof the details are suggestedin the schematic drawingbelow. The plunger is returned to its initial position by a small spring, not shown.

~

accumulator

.... ~ three-way two-position valve restrlcuom~

~ mOt ~~

pis!onOr i

trigger

[

[ i

[ k~

) pump

~ ~

reservoir 1[

~

nail

The principal problemis that although sufficiently small commercialbladder accumulators exist, they have a vMvemechanismto prevent the gas pressure from extruding the bladder through the neck of the accumulator. (See Fig. 3.37, p. 167.) This presents an excessiveresistance to the flow of oil that surges out of the accumulator to drive the nail. Youare asked to investigate the problem and makedesign recommendations.The questions regard the size of the accumulator, the gas pre-charge pressure, the minimu~n allowable effective area of the flow restriction in question, and the diameterof the hydraulicactuator. It is suggested that a computersimulation wouldbe an effective tool in addressing these questions. Details about the electric motor, pump,valving and mechanism details are not to be addressed. The nails are the large 16-d size, 3.25 inches in length and 0.022 lbs in

306

CHAPTER

5.

MATHEMATICAL

FORMULATION

weight. They are resisted by a force that grows linearly with the penetration of the nail into the wood, for a total work of 100 ft-lbs. It has been decided that the accumulator can hold a pressure of up to 3000 psi, the charging pressure of the nitrogen (the pressure when no oil is in the accumulator) should be at least one-third of the maximumpressure to be used, to prevent excessive flexure of the bladder. It is suggested that the inertia of the nail and the other moving parts is important enough to be included in any dynamic model. Energy lost because of spent momentumand orifice flow must be kept small so that the battery and electric motor not become too heavy.

SOLUTIONS TO GUIDED PROBLEMS Guided

1-3.

Problem

5.1

Q i~ A.-2.~

"-Vl-"~,dt

~

A2

:~ ,~ -- V::./’Q:dt~

~porousplugdrain

Integral causality can be applied to the inertance and both compliances without producinga causal conflict. Theorder of the system, therefore, is three. Thestate variables are Vx, }~ and p.

5.1.

CAUSALITY

Guided

Problem

AND DIFFERENTIAL

EQUATIONS

5.2

1-2.

T~ F~.~, 1~-’-- 1

T3 M~ 1 I~-l~ ~ differential

RI

causality forced

R~

1 "2 2 1., .2 3. T = -~I~¢3 = l~I3(T3Q2)2 = ~I3(T3 ~ 2]T 2) z ~13x2 Therefore, I’3 = (T3/T2)213. Combine this inertance

with I2 to give the overall inertance

x~_=I2 +(T~/T2)2X~. Combine R~ with R2 to get R~ = -~2 + ¯(T3/T2)2R3

6.

dp~ __ dt = 1p_ R, -~l pl T~ t

dp__~_dt= I PT2- ~ P2 7.

Compare the two bond graphs above to get 1 Q= T~---~,2 +

1

307

308

CHAPTER

Guided

Problem

5.

I~

FORMULATION

5.3

1. FI 1

MATHEMATICAL

2. Let ~ > 0 for upwardmotion. Fftic = -sign(~,)Fo = sign(p)Fo C=l/k; I=m 3. volumeflow:

~C

~IF,~II=QIA~, R)~,. R,~,.

Q = Ao~(P-

P,)

+ ~(P-

P~),

~ = -A, (P. - P) - ~( a - P), Therefore,

F~=A~(P-P~)=A~-~+V

+

2p

P ~ ~ + IReAe ~ ~

(-RvA{,[

~,,.=-d~

’ p >0

p , p<0

~ +V~

4. Differenti~ equations (note that p is downwardmomentum): dx 1 p dt m _1 @ F

d~ =

~(~ - ~o) - F~

=F+k(x-xo)-Fo-Ap

~+ V 2p

’ >0

A~I] 2

a; h =F+k(x-x°)+F°+Av~+

V

~

+ A~I]

p< 0

k6. P = P~(q/x) Therefore, F~ir = ApPa[(q/x)~ - 1] and Q = Ao

+ RpAp’

P>0

Q =-di V --~--~ ~ F~.~ + R~,A’~-~, p
The differential equations are as given below, with the expressions given in t[ above step substituted for Fair and Q. dx 1 pdt m dPdt = F - +(x - xo) - Ffric - F~ir = F + k(x - xo) sgp(p)Fo - F~ir dq 1 Q dt - -]P Ap

5.2.

5.2

OVER-CAUSAL

AND UNDER-CAUSAL

Over-Causal

309

MODELS

and Under-Causal

Models

Over-causal models are described by fewer differential equations than there are energy storage elements, as noted above. As a result, direct use of the bond graph produces equations in which derivatives appear on both sides of one or more of the first-order differential equations. Under-causal models produce a typically more difficult problem: some of the equations produced are algebraic rather than differential. In many cases of both types, the equations can be reduced to a simple set of differential equations. The problem of a mixed set of generMly nonlinear differential-algebraic equations (commonly known as DAE’s) sometimes must be faced, however. Special software is available. Also, a method for treating these situations is described which approximates an under-causal model by a causal model, permitting use of standard methods for ordinary differential equations. 5.2.1

Treatment

of

Over-Causal

Models;

Case

Study

Differential equations can be found directly from most over-causal bond graphs. The same four steps given for causal models in Section 5.1.2 (p. 282) are used, with careful interpretation, and a fifth step is added. The care is in properly implementingthe differential causalities in the third and fourth steps. The fifth step is the rearrangement of the resulting differential equations to insure that, the derivatives of the state variables lie only on the left sides. An example is shown in Fig. 5.11. After the mandatory causal stroke is placed on the bond for the flow source, integral causality is placed on the como pliance. This placement forces the causalities of all the other bonds, as shown in part (a) of the figure. The inertance I~ receives integral causality, but the inertance I., is forced to have differential causality. As a result, the model is only second order; the energy in I.~ is expressible in terms of the state variables for the other energy storage elements. The modeler must decide at this point whether to find an equivalent bond graph with only two energy storage elements, or whether to press ahead with the existing graph. The following presses ahead. Step I starts with placing the notations O0 and x on the bond for the flow source. Then, step II proceeds with the placement of notations adjacent to the bonds describing integral causality: 0 and q/C on the compliance bond and ~5 and p/I1 on the inertance bond. This completes the step; the inertance I2 does not participate, since it does not have the required integral causality. The situation at this stage is shownin part (b) of the figure. The propagation of the terms 0o, q/C and p/I onto the remainder of the efforts and flows, step III, is shown in part (c) of the figure. Dashedlines with arrows describe the sequence followed. First, the remaining effort off of the 0junction is annotated as q/C. Second, the gyrator causalities are implemented by writing the flow on the 1-junction as q/GC. Third, this flow is replicated as the causal inputs to both the resistance R and the inertance I2. Fourth, the causal output of the resistance is written as Rq/GC, as mandated by its impedance causality. The causal output of the inertance 1.2 is written as 12

310

CHAPTER 5.

MATHEMATICAL FORMULATION

I

Sf~--.~- O .------~t

G b---.--~

l b---~-l: I (a) bond graph with causal strokes

I C

1

R

Sf~.-.~O.~..IGI-.------~I]~=.-I 2 q°q/CIq(q @ C

I ~P/~q/C-. ~o

t C

~ (b) a~ersteplI R e~-

T ~

I~/GC.~

~q/u~ /’T~ kx~// (c) ~lly annotatedgraph (step III)

Figure 5.11: Exampletreatment of an over-causal model

5.2.

OVER-CAUSAL

AND UNDER-CAUSAL

MODELS

311

times the time derivative of the flow, or I2(t/GC, as shown. This appearance of the time derivative of a state variable is peculiar to the differential causality, although it does not depart from the procedure. Fifth, the effort noted as e (to save space) now is written as shown to the right of the graph. This term is dictated by the causal strokes; the signs of the sub-terms are determined by the power-convention half-arrows. Finally, the flow to the left of the gyrator is written as e/G, completing step III. The two state differential IV). The equation for ib is

equations for the system now can be written (step

dp 1 dt - ~q"

(5.11)~

The equation for q differs in an important respect:

dq 1 1 1 I: dq R q" d-~ = (~o - ~P- -~e = ~o - ~P G’~C dt G:~

(5.12)

The difference is that the derivative term appears on both sides of the equation. This problem is rectified by collecting the two terms together on the left side:

I~ ) dq 1+~

1

R

(5.13)

This final step can be viewed as a usually simple step V that is necessary whenever derivative causality is implemented. Equations (5.11) and (5.13) comprise the state differential equation model of the system. This model is somewhat more complex than the corresponding model which results from subsuming the dependent inertance element into the independent compliance element. This is because different state variables are employed. The propagation of the state variables can be carried out even when the elements are nonlinear. You need to be especially careful in these cases to follow the dictates of the causal strokes meticuously. There is one potentially serious complication, however: if the causal output of an element with differential causality is a nonlinear function of the derivative of a state variable, it may be difficult or impossible to carry out step V. This rare situation sometimes can be avoided by proper choice of the energy storage element given differential causality. In the case considered above, for example, the element I2 could have been given integral causality, rather than the element C.

312

CHAPTER 5.

MATHEMATICAL FORMULATION

EXAMPLE5.5 Find a differential equation relating the displacementof the compliancein the modelbelowto the excitation ei(t). All elements have constant moduli. I

el(t) Se’------~

C

r G

. 0 -~-’--~R

Solution: Either the inertance or the compliance element may be given integral causality, but not both; the modelis over-causal. Since the question asks for the displacementon the compliance,written here as q, it is simplest to makethis the sole state variable by assigning integral causality to the compliance.The result at this point is I

S~

e,(t)

C

rlGI

~ 0 --~--~R

Propagating the variables according to the dictates of the causal strokes gives the effort on the right side of the gyrator, the flow on the left side of the gyrator, and the flow on the inertance, as shownbelow. The differential causality on the inertance tells you that its effort is I times the time derivative of the flow, or I(t/GC:

The completed annotations then become I

C

I?I/GC Iq/GC q/CIQ q/C q/C ei(t) ei(t)-l[l/GC S~ 1 q/GCr I G lei(t)/G_l~l/G2 ~ 0 -~-.~-~IRq/RC Thesole differential equation becomes,at first, dq 1 dt -- -~ei(t)

I dq q’ G~C dt

1 R~

5.2.

OVER-CAUSAL AND UNDER-CAUSAL MODELS

313

in which ~ has been written with the more formal notation dq/dt in order to emphasizethat this term appears on both sides of the equation. Collecting these terms,

The literal answerto the question is this equation with both sides divided by the constant within the parentheses. 5.2.2

Equations

for

Under-Causal

Models

Under-causal models result when the four-step procedure for placing causal strokes (Section 5.1.1, pp. 281-282) leaves somebonds without strokes. The author advocates a special procedureto handle these cases. As Step 5, a virtual inertance is bondedto a 1-junction that has somestrokeless bonds, or a virtual compliance is bonded to a 0-junction that has some strokeless bonds. These added elements normally will be considered to have zero moduli, so as not to change the meaningof the model. Step 5 is completedby applying integral causality to the addedvirtual element, and propagating the causal strokes as far as possible, consistent with the usual rules. The step is tentative, contingent on it not producinga causal conflict. In somecases involving a bond-graphmeshsuch a conflict appears at a junction; in this eventuality the step is aborted in favor of a Step 6, described below. Step 5 is repeated, using other junctions with strokeless bonds, until all bondshave causal strokes or causal conflict prevents further application. The numberof virtual energy storages added by step 5 plus the numberof implementationsof step 6 (below) can be called the degree of under-causality. For each degree, one algebraic equation results. Takentogether with the differential equations in the model,ghey forma set of differential-algebraic equations (DAE’s)that are said to be in semi-explicit form: dx d-~= f(x, y, t), 0 = g(x, y, t). 5.2.3

Algebraic

Reduction

Method;

(5.14a) (5.14b) Case Study

The algebraic equations (5.14b) potentially can be used to eliminate the variables y from equations (5.14a), that is to reduce the DAEto an ODE(ordinary differential equation). DAEmodelsfor whichthis reduction is possible are called index zero models, which is the simplest type of DAEto solve. An exampleis shownin Fig. 5.12. Steps 1-4 of the procedure for placing causal strokes (pp. 281-282)leave four bondswithout causal strokes, as shown in part (a) of the figure. The newStep 5 adds either a virtual inertance to the 1-junction or a virtual complianceto the 0-junction. The first option (it makes

314

CHAPTER

5.

MATHEMATICAL

FORMULATION

C~

(a) steps 1-4 for placing causal strokes

(b) step 5 with virtual inertance

T

T

qo /.:.,,[ ~p/l I (c) steps I-II of procedureto write the differential equations

(d) after step

Figure 5.12: Linear under-causal exa~nple

4o-.f,./ -pZ " Rz

5.2.

315

OVER-CAUSAL AND UNDER-CAUSAL MODELS

little difference) is implemented in part (b) of the figure. All four bonds given causal strokes as a result; the degree of under-causality is one. The four-step procedure for writing differential equations for causal bond graphs (pp. 282-283) now is implemented, with a critical difference that recognizes the zero value of the added virtual inertance, Iv. Specifically, in place of writing a circled/),, on the effort side of the virtual inertance bond and pv/Iv on the flow side, a circled 0 is written on the effort side to represent its zero effort, and a symbol fv is written on the flow side to represent its non-zero flow. The result after completion of step II is shown in part (c) of the figure, and after part III of the procedure is shown in part(d). The equations for the two real and one virtual now written (Step IV)

energy storage elements are

dq d-~ = fv, (t°dt dP=R’2(

(5.15a) fvT

0=~- 00 fvT R2(

~)’ ~) q_ -~ R1 f~,.

(5.15b) (5.15c)

The virtual inertia has given algebraic equation (5.15c) which, since it is linear, can be solved for the flow fv in terms of the input variable 00 and the state variables q and p. This fv is then substituted where it appears in the differential equations. The final result is the ODE dq

T T q] dt - I + R1T2/R.2[1 ~lo- ~p- ~C ’ dp _ T RIT 1 ] RIToo -[-p + -~q ¯ dt 1 + R~ T2 / R2 [

(5.16a) (5.165)

EXAMPLE 5.6 Write a differential equation for the sole state variable in the following model. All moduli are constants.

316

CHAPTER

5.

MATHEMATICAL

FORMULATION

Solution: Only the left-most and the right-most bonds have mandated causal strokes, so the model is under-causal. You have the choice of adding a virtual inertance to the 1-junction or a virtual compliance to the 0-junction. Doing the latter, the flow on the virtual compliance is written as a zero and is encircled, and the effort, which is the non-zero ratio of a zero displacement to a zero compliance, is written as ev:

The annotations

on all the bonds now can be completed to give the result

e,, _ e~(t) ei-Rle~,/G _ Se~ l ~,~ G 2~ e~]~ ,T e~tr lei/G_Rle~./G i p’-~l

Rae~./G~e~,/G

R~ Setting the encircled zero flow equal to ~he sum of the input flows to ~he O-junction gives ~he algebraic equation 1

Rx

1

1

which can be solved for ev. Setting the encircled ~ equal to ~he effort of ~he O-junction gives the answer

-- = = dt

5.2.4

Differentiation

e~

R~/G ~ ~ 1/~ Method;

Case

~(t)-

]P "

Study*

Equation (5.15c) is solvable for fv because it is linear. Multiple linear equations also are solvable if they are non-singular. In other cases, however, different methods usually are needed. A nonlinear example is the ground-effect machine or GEM(sometimes called a Hovercraft) shown in Fig. 5.13. This vehicle hovers over land or especially water, without direct contact, due to air pressure underneath caused by a large fan. A rigid skirt around the periphery constricts the outflow of air from the interior plenum, or chamber. If the craft is displaced downward~the restriction

5.2.

317

OVER-CAUSALAND UNDER-CAUSALMODELS

2OO top view

A

fan

A

160 volume flow 120 m3/s

lO.Om

80 40 0

0 0.5

1.0 1.5 2.0 pressure, kPa

section A-A

2.5 3.0

(a) systemand fan characteristic

(b) system model

(c)causal strokes after step 4 S (d) placementof virtual complianceand R annotation of bonds

P P ~l-,¢-a--~,~ ~0~. Qslr, y)

Se P/T T~I~I

Cv

Figure 5.13: Primitive GEM for vertical stabilization

318

CHAPTER

5.

MATHEMATICAL

FORMULATION

for this airflow increases, increasing the pressure and causing the craft to rise. The GEMpictured, with its single plenum, does not correct for the tilting motions of roll and pitch; embellishments to the design are needed for these, as described later. For now, attention is directed toward vertical translational motion only. The vehicle shown measures 4 x 10 meters, with a plenum depth of 0.8 meters. The pressure-flow characteristic of the fan, as plotted, can be }epresented by the equation Q f = Qo - a~ P + a~P"2 P4, - a4 (5.17a) Qo =180; m3/s;

a1=0.08m

5/Ns;

-SmT/N2s; a~=2.0×10

a4 = 1.482 × 10-12 m9/N3 s, (5.17b) 2 where the pressure P is in N/m (Pa) and the flow Q is in m3/s. The pressureflow characteristic for the skirt flow can be approximated by applying Bernoulli’s equation: Qs = LycdI~P.

(5.18)

The periphery of the GEMhas length L = 28 m, and y is the eleveition of the bottom of the skirt above the surface. The coefficient Cd = 0.65 represents a discharge flow coefficient, and p = 1.23 kg/m3 is the density of the air. The fan is represented in the bond graph of part (b) by a general source, and the resistance to skirt flow by a resistance. The effect of the compressibility of the air in the plenum is slight, and is neglected, so that the excess of the fan flow over the skirt flow, QI - Qs, produces the vertical velocity of the vehicle, ~. The relation is represented in the bond graph by the transformer with modulus T = l/A, where A = 40 m~ is the area of the vehicle. The weight Of the craft is represented by the element Se, and its mass by the element I = 4000 kg. Step 4 of the procedure for placing causal strokes (p. 282) ends without designations for the source and resistance bonds, as shown in part (c) of the figure. Therefore, step 5 (p. 313) is invoked, as shown in part (d), by adding a virtualcompliance C,, to the 0-junction. Steps I-III for writing the equations (p. 282) also are included in part (d). The resulting equations become dp = PIT - rag, dt dy 1 ~ = ~P,

(5.19a) (5.19b)

1 0 = Qf - Q8 - ~--~p = Qo - alp + a2P2 - a4P4 - Lycd

- ~-~p.

(5.19c)

The momentum of the vehicle, p, is a state variable. The fact that the resistance R is a function of y makesy a state variable, and requires the inclusion of equation (5.19b). (In general, state variables are contributed both by energy-storage

5.2.

OVER-CAUSAL

MODELS

319

five times actual compliance actual compliance

0.3 elevation, meters 0.2

~~

0.1

0

AND UNDER-CAUSAL

0

mpliance)

1

2 time, seconds

Figure 5.14: Three simulations of the lift-off

of the single-plenum GEM

elements and by resistances and transformers that are functions of displacements; other cases of the latter are illustrated in Chapter 9.) Equation (5.19c) theoretically can be solved for the pressure in the plenum, P, in terms of the net fiow Qf - Qs = p/TI, and the result substituted into the differential equations. As a result, the DAEmodel is officially of index zero, and P is not a true state variable. An analytic solution of such a high-order polynomial equation is not known, however. Algebraic equations become differential equations upon differentiation. The index of a DAEis defined as the number of differentiations theroetically needed to solve for the time derivatives of the variables represented by the algebraic equations. The higher the index, the more difficult a DAEis to solve, regardless of the method used. The derivative of equation (5.19c) cma be solved for readily for dP/dt as a function of p, x and P. This differential equation can be used as the third in a solvable set of ODE’s. Therefore P is treated as though it is a third state variable, despite the minimumorder of the model being two. The derivative of equation (5.19c) can be solved for dP/dt as follows: dR LCdx/~--~(dy/dt) + (1/TI)(dp/dt) d---~ = al - 2a2P + 4a4P3 + Lycd/ 2~-~ "

(5.20)

A solvable third-order ODEresults when this equation is combined with equations (5.19a) and (5.19b). The three dependent variables are p, y and P. elevation y is plotted in Fig. 5.14 for a simulation that starts at lift-off ("differentiation model"). The initial conditions are p(0) = 0, y(0) = 0.0001 prevent division by zero) and P(0) = 2000 Pa. Note that the number that represents the fan flow at P = 0 (180 m3/s) was lost when equation (5.19c) differentiated; its effect is recovered through the use of the proper initial condition P(0) = 2000 Pa. This value is mandatory. No other value has a legitimate meaning, given the values of p(0) and y(0), since the model has minimumorder two, permitting only two independent initial conditions.

320

CHAPTER 5.

MATHEMATICAL FORMULATION

5.2.5 Methodof Non-ZeroVirtual Energy-Storages; Case Study Continued* Regardless of the index of a DAE,it is possible to generate an approximate ODEsubstitute by imbuing the virtual complances Cv and inertances Iv (in the step 5 above) by small but non-zero values. Eachalgebraic equation is then converted to an approximate differential equation, so any ODEsolver can be used. For the current example, the annotations on the bond for the virtual compliance C~are changedto ~. on the flow side and q,/Cv on the effort side, like any real compliance. Therefore, P = qv/C~, and equations (5.19) are replaced by

dp 1 -~ - CvTq~, dy 1

(5.21a)

(5.215) ~-~ = ~p, dqv d---~ = Qo-al--~q, +a2(D C2v -~ - -C-~q~ - LyCd.. qv - 1~-~P. (5.21c) 3 2 a3 2~/-~_~ The principal questions noware what value should Cv be given, and howsignificant an error does this complianceintroduce into the modeland its solution? A more accurate model of the system, it so happens, would recognize the compressibility of the air in the plenum, resulting in an actual compliancein place of the virtual compliance. Treated as a constant, this compliancewould have the approximate value Vo/kPo = 0.265 × 10-3 mS/N, where Vo is the nominalvolumeof the plenum(a little greater than its minimum value, due to the meanelevation y of the craft), Po is the meanabsolute pressure (virtually 101 kPa) and k is the ratio of specific heats (1.4). The result of a simulation with this modelalso is given in Fig. 5.14. It is barely different from the solution with Cv = 0, demonstrating that compressibility is of virtually negligible effect. Executionof the simulationis relatively slow, however, due to the smallness of C.. (In absolute terms, however, the simulation is not excessively slow for this simple model.) Pressure excursions from the nominal movingequilibrium have very rapid dynamicsrelative to the basic motion of the vehicle. Simulations must proceed at ~ rate dictated by the need to have manytime steps per cycle of the fastest phenomenon in the model, even if this phenomenon is not of interest. The execution can be speeded up by using a larger value of C., but at a cost of inaccuracy. A faster simulation with five times the proper complianceis included in Fig. 5.14. The error is significant, although it might be acceptable for somepurposes. A trade-off between accuracy and computingtime is characteristic of the methodof non-zero virtual complianceor inertance. Modelscombiningrelatively very r~pid and very slow phenomenaare called "stiff." Multi-step integrators typically are superior to single-step Runge-Kutta integrators for handlingstiff models.

5.2.

OI/’ER-CAUSAL

5.2.6

Commercial

AND UNDER-CAUSALMODELS Software

for

321

DAE’s

The solving of DAE’sin general is a major enterprise. Should you deal with themoften, you probably should avail yourself of a software package designed 1 and LSODI. explicitly for them. Twopresently popular packages are DASSL 2 These are based on multistep backward-differentiation formulas (BDF). Extensive discussions of these formulas and the programsare available. 3 These programswork well for most index zero and index one systems, and someindex two systems. They are beyondthe scope of this book, however,whichfocuses on modeling rather than numerical methods. The MATLAB integrator odel5s is a variable-order integrator for "stiff" systemsthat you can readily use to handle 4a class of DAE’s,however. 5.2.7

Case

Study

With Meshes*

Somebond-graph meshes, or closed loops of bonds, produce special complications. Sophisticated or basic methodscan be used to overcomethese complications. For simple ever~ meshesa sophisticated methodsubstitutes a tree-like equivalent bond graph, as given in Fig. 4.18 (p. 248). (Care must attend treatment of the power-conventionhalf-arrows should they not be aligned initially in the mannerillustrated.) Moregenerally, an even or odd meshcan be re~novedby reformulating the modelwith different state variables, based upon examinationof relations for the stored energy, as described in Chapter 10. For the present, no such cleverness is attempted for a complexbond graph with a mesh:the basic methodfor finding differential equations is applied directly to the given graph. Step 5 Of the procedurefor designating causal strokes (the introduction of a virtual complianceor virtual inertance whena modelis under-causal) sometimes produces an uncorrectable causal conflict when a bond-graph mesh is present. In these cases, the offending virtual element must be removed. Some different junction should be chosen, the proper virtual energy-storage element appended,and the resulting graph checkedfor causal compatability. Occasionally, an apparent causal conflict can be corrected by switching the causality of an energy-storageele~nent from integral to differential. Youmight prefer, however, to retain the original state variable by ~selectinga different virtual element, if this is possible. The classical modelfor the small-motiondyna~nicsof an assumedrigid vehicle shownin part (a) of Fig. 5.15 illustrates the problemand its solution. Rolling and side-ways motions are omitted, so there are two "degrees of freedom": that 1L.R. Petzold," A description of DASSL: A differential/algebraic system solver", Scientific Computing, eds. R.S. Stepleman et al., North-Holland, Amsterdam, 1983, pp. 65-68. 2A.C. Hindmarsh, "LSODE and LSODI,two new initial value ordinary differential equation solvers" ACM-SIGNUM Newsletters, v 15, 1980, pp. 10-11. 3K.E.Brenan, S.L. Campbell and L.R. Petzold, Numerical solution of Initial-Value Problems in Differential-Algebraic Equations, North-Holland, Elsevier, New-York,1989. 4L.F. Shampine, M.W. Reichelt and J.A. Kierzenka, "Solving Index-1 DAE’sin MATLAB and Simulink," SlAMReview, v. 41, 1999, pp. 538-552.

322

CHAPTER

(a)

5.

MATHEMATICAL

FORMULATION

sc~

L2--~1 0 Tl /

,/.r

L2 T3 = L1 +L2

ly2 0"

(b) junction structure with mesh

r~

o

q~ /Cl~O 1-.r~--0~

R~

1

IT, ’’’’~

S~

0"

(c) bond graph before choice of unspecified causality

C~=l/k~

, "x"~" T2 ,~,,.."

S~

l p*/T3I~’ Tz (~IP~/1:,

; C2=1/k2 ; Ira=m;

q2/C2~ 1-~.-0-~.-

1

R2

(Can add gravity force or weight,but this is just a constant and doesn’t affect the dynamics.)

l¢=J ; R~=b~; R2=b2

Figure 5.15: Model of vehicle dynamics with bond graph representation

5.2.

OVER-CAUSAL AND UNDER-CAUSAL MODELS

323

is, two generalized displacements and their derivatives are required to specify the state of the system. The two displacements could be yl and y.,, which are the vertical displacements of the two axles. However,the vertical displacement of the center of mass, ycm,and the pitch angle, ¢, are moreconvenient, since the kinetic and potential energies are morereadily expressed in terms of these and their time derivatives. The ground under the front and rear axles has vertical displacementsyg~(t) and yg2(t), respectively, relative to somereference level; the time derivatives of these displacementswill be co~{sideredto be the excitations of the system. The suspension at each axle (springs, tires and shock absorbers) is modeledsimply (and rather crudely) by a parallel spring-dashpot combination, with parameters k~, b~, k2 and b2, as shown. The vehicle has a knownmass, m, and knownmomentof inertia, J, ~bout a known center of The geometric constraints betweenthe velocities of the ~les and the center of m~s are L2 L~ Vc~= ~~L~+ (5.22a) +~~’L~ +L~ 1 ~ - L~ + Le(92 - 9~). (5.22b) These lead directly to the junction structure of the bond graph shownin part (b) of the figure. Expressions for the kinetic and potential energies of the modelin terms of the defined displacements and velocities are written ~ the next step in the modelingprocess. These are T=1

~ " ~mv~ + ~ J¢~,

(5.23)

1 ~ 1 k ’~ (5.24) ~ = ~k~q~ + ~ ~q~, where q~ and q2 are the extensions of the springs from their nominallength: q~ = y~ - y~l,

(5.25a)

(5.25b) q2 = Y~- Y~. These allow the bondgraph to be completed ~ shownin p~rt (c) of the figure. Integral causality is applied successfully to all four energystorage elements, as Msoshownin p~rt (c), revealing that the modelis indeed fourth order. The causalities of manybonds, including the mesh bonds, are not mandated by this causality, however,indicating an under-causal model. Therefore, a virtual energy-storage element is to be bondedto one of the causally incompletedjunctions. If you try to appenda virtual inertance to either of the mesh1-junctions, however,a causal conflict results, ~ illustrated in part (a) of Fig. 5.16. This conflict is correctabl~ by switchingthe causality of the I¢ elementto differential form. Althoughacceptable, this approach is more complicated than necessary; a simpler methodexists.

CHAPTER

324

G

~-[[

1~0.-¢-~1 RI

MATHEMATICAL

10 __

IC~

1 ~-=,- 0 ~-~---~ 1

R2

1 SI

I \ ~,/T31¢, T3 ~’~[p -~_ ,/1~.

T1

q,/C~(~ ~Y 0 -.,~--

FORMULATION

q2

.’T"

(a) additionof virtual inertia with attempedretention of integral causalities

R|

5.

q~I

~

1

1 ----~-

S:

0 -~- 1 R~

0

(b) addition of virtual T3 compliance to a mesh bond ~[p~/l~

.C,

.

~e,TpmI1~...,,. C~l Cv~

v" 11-~---L2 0 ql -.~-I 1 qil e2_ I.

~0 ~ note:

’~,

~~Tle

C~

T,+T,=I 1,--~

0 ~ 1

s, (c) addition of second vi~al complianceto complete causality

--~

+il~ Note:for~and~ R~R~C~ seeequations(5.26)

Figure 5.16: Continuation of analysis from Fig. 5.15

5.2.

OVER-CAUSAL AND UNDER-CAUSAL MODELS

325

As a second attempt, a virtual compliance is added to the upper mesh 0junction in part (b) of the figure. This produces no causal conflict, but still leaves the bond graph under-causal. Thus, a second virtual compliance is added to one of the remaining 0-junctions with incomplete causality, as illustrated in part (c)¯ Now,the causal strokes are completed without conflict, and the efforts and flows on all the bonds can be annotated in the usual manner, as shown. To reduce the clutter on the graph, it is helpful to define the flows fl and f2 as functions of the state variables and the efforts el e2 of the virtual compliances: 1 fl : f2 - T3I---~’ ¯ f2=-yg’,.

(5.26a)

el + e~.

R~

q2

(5.26b)

R~.C.~_"

The algebraic equations associated with the circled zeros on the bond graph are q~ eu 0 ---fl --~ii+ /~IC-~--~ - #g~,

(5.27a)

Pm + T2fa im

¯0 =Tifi

(5.27b)

Substituting equation (5.26a) into equations (5.27), and using the fact T1 + T2 = 1

(5.28)

reduces these equations to pc T~Ie~

0=~.)

q~ e2 ~RI R~ C~

~gi,

0 = A Tlpo

(5.29a) (5 ¯295)

TaIo I,~ "

Solving equation (5.27a) for e2, substituting equation (5.26b) for f.) and noting equation (5.28) gives the effort e2 in terms of state variables, as desired: ~2 _

T~R~p¢ T3I¢

R~pm

q~

+ ~ + ~-

~l~gl-

(5.30)

To find the corresponding equation for e~, equation (5.26b) can be substituted into equation (5.29b) to eliminate f2, giving el

= R2~g2

--

e2

q2 C2

TIR2P¢ T31¢

R2Pm

(5¯31)

Substituting equation (5.30)therein givestheeffortel in termsof the state v~riab]es, as desired: el = R2~g2 + Rl~91 + (TzR1 - T~R2)p¢ T3I¢ Finally, the four state differential right sides, are

(RI + R2)Pm Ln

ql q2 (5.32) C1 C2"

equations, which employe~ and e~ on their

CHAPTER 5.

326

MATHEMATICAL FORMULATION

(5.33a)

(5.3ab) (5.33c) (5.33d) The choice of variables above and the resulting differential equations are not the only ones possible. A recast model presented in Chapter 10 for this problem has no mesh, saving effort at the cost of increased required competence. EXAMPLE 5.7 Determine whether the three mesh-type bond graphs below are reducible to tree-like equivalences. Then, determine whether each of the graphs is causal, over-causal or under-causal. Finally, define state variables and write the corresponding set of state differential equations for case (ii).

i

~

(iii) C

1

C~ 1

1

1 ~ (ii)

12

I R

Solution: Three of the mesh bonds have clockwise power arrows and one has a counterclockwise power arrow in all three cases. Therefore, the meshes are odd and not reducible to tree-like structures. The mandated causal strokes are as follows: (i)

./l differential .[~= causality

(iii) C

’~qi’(t)

0"-~

1 ~,,-"

/

1

S~

0 ~ (ii)

~ 1

)SY~ 0 ~I

} I~

C

} unspecified caus~ strokes R

1 ~strokes } u~quely specified I

5.2.

OVER-CAUSAL AND UNDER-CAUSAL MODELS

327

In case (i) the causalities of the two inertances could be inverted, but regardless of the choice only one of them can have integral causality, so the model is over-causal (and of first order). In case (ii), several bonds have mandated causalities, so the model is under-causal. In case (iii), both energy storage elements have integral causality and no bonds are left unspecified, so the model is causal. The under-causal case (ii) is the most difficult to address. An attempt place a virtual inertance on the lower 1-junction produces a causal conflict, as shown below left. A second attempt, this time placing a virtual compliance on the left-hand 0-junction, produces a satisfactory result, as shown below right. Therefore, the efforts and flows on this bond graph are annotated in the standard fashion, as shown. C q/C~(~

C I" 1 ~conflict S~

0 V""~" -)’~~ "~1

0 "~1I

0 ~2e~’/R-q/RC+p/I’~O~’-~I ~’- e,,-q/C~ S~ T-_~,,~-e~, p/l ~ e~) "~ l ~f C,, 2e,,_q/ci2e,,/R-q/RC

I,,~R R The algebraic equation associated with the circled zero is 4e~, 2q p 0 = q~ (t) - -h-- + n--~ - 7’ which gives q Rqi(t) e~ = -~ + 4 The state differential dq

2e,~

1

7 dp d-[ 5.2.8

41"

equations become

-~O~(t)

q_ 1 R = ev C ~q + -~?li(t)

R - -~p.

Summary

Over-causal models are those for which it is impossible to apply integral causality compatibly to all the inertance and compliance elements. The differential causality on an energy-storage element implies that its energy is not independent of the state variables for the other energy-storage elements. The basic 4-step methodfor writing differential equations still applies, but requires special attention to insure faithful application. Time derivative terms appear on both sides

328

CHAPTER 5.

MATHEMATICAL FORMULATION

of the differential equation that involves an element with derivative causality. In most cases these terms can be collected on one side, and the equation solved for the derivative. The result is a set of differential equations equal in number to the number of energy storage elements with integral causality plus the number of resistances or inertanccs that depend on a displacement that is not otherwise a state variable. Under-causal models are those for which the basic four-step procedure for placing the causal strokes (pp. 281-282) fails to define all of them. A fifth step appends a virtual compliance to an affected 0-junction, or a virtual inertance to an affected 1-junction. Whena bond-graph ~nesh is present, this step sometimes produces a causal conflict. Should this happen, a different junction must be chosen for appending a virtual inertance or compliance. One algebraic equation is produced for each virtual energy storage element in a bond graph, because of the zero value of its modulus. The algebraic equations emerging from step 5 may or may not be solvable for the defined effort or flow in terms of the input and state variables. If they are so solved, the writing of differential equations can proceed normally. Otherwise, the model comprises a set of one or more algebraic equations as well as differential equations, called a DAE(differential-algebraic equation). DAE’sare more difficult to solve numerically than ODE’s(ordinary differential equations). Special software is widely available, including the MATLAB integrator odel5s. Alternatively, it is possible to approximate a DAEby an ODEby conferring small non-zero values to whatever virtual compliances and inertances have been introduced by the procedure for placing causal strokes. The smaller the moduli chosen, the more accurate the approximation but the more difficult the solution. Simulation based on multistep BDF algorithms (backward-differentiation formulas) usually work better than single-step RungeKutta algorithms. In a more accurate procedure, algebraic equations within a DAEset are differentiated to give differential equations, and the resulting expanded set of differential, equations is simulated with a simple ODEintegrator. Care must be taken to employthe proper initial conditions fcir the added differential equations. Guided

Problem

5.4

The system of Fig. 5.17 is used here to illustrate the two basic methods for handling dependent energy storages. Start by drawing a detailed bond graph for the system, and defining the moduli of all elements. Then apply causal strokes, and note that differential causality appears for one of the inert’ances. Proceed to find the corresponding differential equations, using the standard method. Then, as an alternative approach, find a bond graph which has only one inertance; also find its differential equations, and comparewith the first set. Suggested

Steps:

1. Carry out the first

part of the solution; it is suggested that the angular

5.2. OVER-CAUSAL

AND UNDER-CAUSAL

MODELS

329

Figure 5.17: System of Guided Problem 5.4 velocity ~ describe one of the 1-junctions, and that the two radii become transformer moduli. To get the correct sign for the force source, note whether the input work is positive or negative for your defined velocity. Note that the inertance bond with differential causality has no state variable; its effort should be labeled as the product of the inertance and the time derivative of the flow. Annotate both sides of all bonds, and then write the differential equations. Re-do the bond graph with only one of the inertances. To find what this equivalent inertance should be, write an expression for its stored energy in terms of its velocity, and compare to an expression for the sum of the original two stored energies also reduced as a function of the same velocity. Find the differential equations. Are the differential equations the same? Are they equivalent? Which set more clearly reveals the behavior?

Guided

Problem

5.5

This relatively complex problem compares the the use of virtual energy-storage elements with the differentiation method in the treatment of under-causality. It extends the modeling and analysis for the ground-effect machine addressed in Sections 5.2.4 and 5.2.5. Roll stability is imparted to the GEMby introducing two internal skirts, as shown in Fig. 5.18. The lower edges of these skirts are in the same plane as the external skirts. Whenthe GEMrolls to the right, the resulting increase in the resistance to flow for the right-side skirt and the decreased resistance to flow for the left-side skirt produces a higher pressure in the right-side plenum than the left-side plenum. These pressures impart a correcting moment, Or roll stiffness, to the vehicle. You are asked to model the system with a bond graph

330

CHAPTER lO.Om i

5.

MATHEMATICAL

FORMULATION

~

ght plenum

]-~-~ ~ enter

:

plenum

i

_.~left skirts~

top view

section A-A

~-j A

Figure 5.18: GEMwith internal

skirts

plenum

for Guided Problem 5.5.

that ignores the small compliances of the plenums, and proceed to develop the corresponding describing equations. "tbu will see that the algebraic equations are too complex to permit direct reduction of the model to a minimal set of state differential equations. The simplest approach confers non-zero values to the virtual compliances, increasing the order of the model by three. The differentiation approach is more difficult and also increases the order of the model by three, but does not modify the model. The effort required for you to carry out the details may not be justified, but it is instructive for you to establish a plan. The steps below and the solution at the end of the section outline the procedure and describe the solutions, which illustrate important conclusions. The two models are simulated first with no roll angle, partly to determine the equilibrium elevation. The physical compliances and five times these values are employedseparately. Next, the behavior following an initial roll velocity of 0.05 rad/s at the equilibrium elevation is simulated. The radius of gyration of the GEMin roll is taken as 1.5 meters; other parameters are as before or as shown in the figure. The various results are compared with respect to the behavior of the physical design, the accuracies of the simulations and the computational time required. Suggested

Steps:

and parameters. Draw a bond graph for the system, 1. Definevariables omitting compliances. Each of the three plenums merits a 0-junction, and there are 1-junctions and inertias for both elevation and roll. 2. Apply the first four step; for placing causal strokes. Note that the model is under-causal. 3. Carry out step 5 of the procedure for causal strokes. 4. Annotate the bond graph in the recommended manner.

5,2.

OVER-CAUSAL AND UNDER-CAUSAL MODELS

331

Writethe state differential equations and associated algebraic equations. The algebraic equations cannot be solved analytically for the non-state variables in terms of the state variables. The simplest approach employs non-zero values of the virtual energy-storage elements. Write the correspondingset of state differential equations. Find the differential equations for the derivative method.This is done by taking the derivatives of the algebraic equations. A3 x 3 matrix equation can be written, with the time derivatives of the three pressures as the dependent variable. This equation is readily solved by MATLAB. Establish the initial conditions for lift-off (with no roll) for the three models. Also find the initial conditions whenthe physical complianceis increased by a factor of five. Notethat the pressures in all three plenumsis non-zero, and that the center pressure is higher than the others. A very small initial elevation is required to preventthe differentiation modelfrom trying to invert a singular matrix (which wouldmakeit an "index two" problem). Carry out three-second simulations for lift-off, and comparewith each other and with the response in Fig. 5.14 (p. 319). Note howmanyseconds your computertakes to carry out the simulation. 10. Establish the initial conditions for equilibirum height and roll angle but the specified roll angularvelocity. 11. Carry out the respective simulations, and again compare. PROBLEMS 5.16 Continue Guided Problem4.5 (p. 227) by defining a minimumset of state variables, considering the fluid pressure, P, to be an independent excitation and the motions of the solid parts to be responses. Find the corresponding state differential equations in a proper form for integration. Youmayemploy parametersof your owndevising if they are defined explicitly. 5.17 Augment the bondgraph belowwith causal strokes, define state variables, and write the correspondingset of state variable differential equations.

332

CHAPTER

5.

MATHEMATICAL

FORMULATION

5.18 For the bond graph given below left: (a) Apply causal strokes to find the order of the system, assuming constant parameters. (b) Find the set of state differential strokes.

equations corresponding to your causal

(c) Reduce the number of energy storage elements to a minimum and find the corresponding state variables and differential equations. Check to make sure they agree with the results of part (b),

5.19 Answer the questions of the preceding problem for the bond graph above right. 5.20 The bond graph below left

has constant parameters.

(a) Assign causality, identify state variables, and write state differential equation(s). (b) Find the output fiow 0o as a function of the input effort eo.

O~G~

R 1

e.o

0

C 2

S~

eo 1 ~. Sf

1 -..-~-G--..-----

R~

12

5.21 Answer the questions of the preceding problem for the bond graph above right, except note that the input is 00 and the output is eo. 5.22 A hydraulic shock absorber of the type described in Fig. 2.24 (p. 47) employed as the dashpot in the system of Problem 4.24 (p. 264). Write the state differential equations for the resulting nonlinear model. (This is a modification of Problem 5.12, p. 305.)

5.2.

OVER-CAUSAL

AND UNDER-CAUSAL

333

MODELS

5.23 Write the state differential equation(s) for the model shown in part (d) Fig. 5.7 (p. 296), using the differential causality given. Comparethe result the single equivalent differential equation. 5.24 Repeat the above problem for the model shown in part (d) of Fig. 5.8 (p. 297). 5.25 The motor of Problem 4.52 (p. 272) has a rotational in..lb.s 2 and the cylinder has a radius of gyration rg = 8 in.

inertia

J = 0.3

(a) Repeat part (a) of that problem to include dynamic effects. (b) Assign causal strokes to the bonds. Note the presence of differential causality. State the order of the model. (c) Carry out the procedure illustrated in Guided Problem 5.4 to combine the energy of the element with differential causality with that of a different element, so as to eliminate the differential causality. (d) Write state variable differential equation(s) for the model. The electromagnetic torque on the motor may be left in the form M= M(~). (e) Simulate the response of the system, starting from an initial motor speed of 1.5 rad/s. Plot the angular velocity for the motor and the displacement of the center of the roller as functions of time until equilibrium is virtually reached. The function M(~.) can be approximated

5.26 Augmentthe bond graph below with causal strokes, define state variables, and write the corresponding state variable differential equation(s).

5.27 The bond graph model below comprises elements with constant

moduli.

334

CHAPTER 5.

MATHEMATICAL FORMULATION

(a) Definestate variables, and write a set of state differential equations. (b) Sketch an engineering system for which the bond graph could be reasonable model, and relate the parameters of the bond graph to those of the engineeering system. Suggestion: Let the source be electrical, and the right-hand portion of the modelrepresent a mechanical subsystem. 5.28 The bond graph below has constant parameters I, C, R, T. The source is characterized by a knownfunction es = f(0s) or ?Is = g(es).

(a) Determinethe values of e~ and Os a.t equilibrium in terms of the given parameters and/or functions. (b) Definestate variables, and find a set of state differential equations. 5.29 Answerthe preceding question for the bond graph given below.

5.30 Consider the bondgraph given in Problem4.7 (p 202) for the effect of drain tube on humanhearing, except omit the transformer. (a) Definestate variables and write differential equations without changing the given bondgraph with its mesh. (Note that integral causality specifies the causalities of all the bonds, makingthis a relatively simple problem.) (b) Applya bondgraph equivalence from Fig. 4.18 (p 248) to eliminate mesh. Find the differentia1 equations in terms of the samestate variables as in part (a), and compareto makesure they are the same.

5.2.

OVER-CAUSAL AND UNDER-CAUSAL MODELS

335

5.31 Repeat the problem above for the bond graph below. Note that integral causality does not specify the causalities of all the bonds, makingthis a more difficult problem. C

5.32 Consider that the ground-effect machinediscussed in Sections 5.2.4 and 5.2.5 hovers over water that rises and falls uniformly according to the formula z = zo sin wt. Carry out simulations for frequencies wt = 2 rad/s, w2 = 8 rad/s and w3= 32 rad/s, all with the amplitude zo = 0.05 meters. The initial conditions can be for equilibrium conditions. 5.33 The ground-effect machinediscussed in Sections 5.2.4 and 5.2.5 lacks roll stability and pitch stability, and the improvementinGuidedProblem5.5 lacks pitch stability. The design pictured below employsfive plenumswith internal and external skirts all having their lower edges in the same plane. Modelthis systemwith a bond graph, as a first step towardan analysis. 10.0m A~,-~

top view A

section A-A

336

CHAPTER 5.

IVlATHEMATICAL

FORMULATION

SOLUTIONS TO GUIDED PROBLEMS Guided

Problem

5.4 T~

~-

r2

C = 1/k

dx T, dt - I¢ p= -~P (a) dp

Tx

( I~T~ dp

d-7= c x - T~ \ v, d*

]

(b)

Collecting terms in equation (b), I~T~2 "~ dp T~ 1+ I~ / ~-~ = -~-x+Turng

or

-- "~ ----

(c)

kr~x + rimg

Equations (a) and (c) are a completeset of state differential equations. Note that ~ = pig.

1

i , 242 V=~r~ =~-~¢

2

therefore, C’ -- 1/kr]

T= J~b ~ + ~rny = (J + mr~’~)(b ~ = ~ ~ ~ ,’therefore,

I=J+rnrl

1 de 1 , ’2p dt- ~P - J÷ rnr~ dp’ = r~mg- 1 -~7¢ = r~mg - kr/ ¢ d---~ 1 , 1 These are the same equations, since (b - j + ~nr~p = -~p orp’= l + ---~-)

p an(~ ¢ = x/r~.

Although the first method shows the structure in more detail, methodis easier and shows the behavior more clearly.

the second

5.2. Guided

OVER-CAUSAL Problem

AND UNDER-CAUSAL

337

MODELS

5.5

iCangularmotion(roll)

left

plenum~

plenum center

t~

~right

plenum

I

vertical motion(heave)

The area of the deck is .4 = 40 ms, and T~ = Tt~ = )-~ = 0.100 m-~; Tc~ = ~]~ = 0.050 The width of the deck is w = 4m, and -Tic = T,V --

1 (A/4)(aw/S)

1 m_a.

~. From the given information, I~ = 4000 kg; I~ = mr~ = 9000 kg-m Repeating equation (5.17), Qf(Pc) = Qo - al P¢ + a~Pf - a4P~; Q0 = 180 m3/s; at = 0.08 m~/N.s; a~ = 2.0 x 10-amT/N2.s; aa = 1,482 x The center plenumhas a periphery of outer skirt of length L~ = 4m, with flow Q~ = L~yCd~Pc. Outer skirt are of the left-hand plenum: At = ~,(L + ~) - ~w (~- + ~-~) Outer skirt are of the right-hand plenum: A, = y ( L + -~ ) + ¢w (-~ + ~-66 ) The respective

flows are Ql= Alcd~/~Pl and Q~= Arcd~P~.

The flows across the internal skirts are Q¢~= L y -

c~

(P~ - P~)

338

CHAPTER 5. MATHEMATICALFORMULATION

/

.

s

\

s~~

Thefoursta~ev~iablesp~, p~,g and~ giveNurdifferentialequationsin terms of the functionsdefinedin step1: d~ ~ = ~/I~

5.2.

OVER-CAUSAL

AND UNDER-CAUSAL

MODELS

339

--= --P~+ + dt T~u "1’~ dp ¢ 1 ~ p~ d~’~" = ~¢Pt + 77¢ There are also three algebraic equations: 0 = Qcl(Pc - Pt,y,¢) - Qt(Pt,y,¢) - ~b/Tl¢ - ~/Tt~ 0 = Qcr(Pc - P~;y,¢) - Q~(P~,y,¢) - ~/T~ 0 = QI(P¢) - Qc(Pc,y) - Q¢,(Pc P~, y, ¢) - Qc~(P~ - P ~,y,¢) - ~ /T Thethree state variables V~, V~and V~ are added to represent the three volumetric compressionsin m3. In the differential equations of step 5, the pressure Pt is replaced by V~/Ct, P~ is replaced by V~/C~and P~ is replaced by V~/C~. The following three state differential equations replace the derivatives of the pressures:

dV~ = Q.d’-~

1 Q~- T~--~V~ -

1

1 d--~- = QI - Q~- Q¢~- Q~- --T~uI~ p~ Note that C~ represents half of the total compliance, and C~ and C~ each represent one-quoter. The total compliance is about 0.265 x 10-am~/N, ~ computedin Section 5.2.5. IdP~/dt] =F-lf, [ dP~/dtJ f~

wheref

= OQ~ O(P~- P~)’

f¢ andF f~o f2

f~ f7

0 fS

f9

= OQ¢l + OQt O(P~ - Pt) OPt

+ 1 d(b + 1~ ~ I.~ f =i)Q~\ OQcl’~ o~ o~~ +) ( \OQi0¢OQct-~)~ ~ ~-~

340

CHAPTER

5.

MATHEMATICAL

FORMULATION

Duringlift-off, the pressures in the side plenumsbriefly drop belowzero because of the rapid rise of the craft caused by the high pressure in the center plenum. Consequently, it is necessary to employsign and absolute value functions in computing the flows Qt and Qr and their derivatives. This problem doesn’t quite arise in the simulation with the compliances~ since these compliances reduce the pressure decrease enoughto keep the pressures positive.

At equilibrium, the flow leaving one of the side plenumsequals the flow entering. Therefore, for ¢ = 0 and any y,

wherePs is the pressure in the side plenumand Pc is the pressure in the center plenum. Therefore,

(

Pc=l+ 1+ =2.44. P8 Wheny = 0 the fan flow is zero, so the pressure in the center is 3000 Pa. The reasoning above gives the pressures in the sides as 1042 Pa. The initial compressionof the center plenumis therefore Vc(0) = ~ x 0.265 x -3 x 3000 = 0.399 ma, and in the side plenumsis ~ (0) = V~(0) = ~ x 0.265 -ax 1229. 6 = a. 0.0818 m

9-10.

0.3 elevation, meters 0.2 0.1 0

0

~five times nominal compliances

--no compliances (differentiation model)

~

1

2 time, seconds

3

The lift-offs with and without the compliancesdiffer little, although increasing the compliancesby five times gives a significant error. The responsesare similar to those found in the absense of internal skirts, although the loweredpressures in the outer plenums reduces the equilibrium height. The final elevation is 0.1703 m, which is used as an initial condition for the simulation with roll. The final pressures in the center and side plenums, respectivbly, are 1392 Pa and 570 Pa. The roll rate of 0.05 rad/s corresponds to an initial condition for the angular’momentumof 9000 × 0.05 = 450 kg.m/s. For this roll, the proper initial pressure in the left plenumis 13 Pa greater than the 570 Pa, and the initial pressure in the right plenumis 13 Pa less. (These corrections can be found by trial-and-error; any other values gives an equilibrium roll angle different from zero. Note that the minimumorder of the differential model is four, so that only four initial conditions can be selected independently.)

5.3.

LINEAR

341

MODELS AND SIMULATION

11. 0.01 roll angle, radians 0 -0.01 0

1

2

4 3 time, seconds

5

The simulations for roll reveal significantly less dampingthan those for lift-off, a weaknessin the performance. Thethree simulations are nearly indistinguishable, unlike for lift-off. The reason is that the three pressures do not vary nearly as muchas in the case of lift-off. Thesix-second simulation for roll with the nominal compliancesrequired 32 seconds of computer time on a particular PC computer, using an error factor of 0.0001 with MATLAB’s ode23 in version 4. Increasing the compliances by a factor of five reduced the computationtime to 5 seconds. The derivative model took only about 2 seconds, despite the fact that each iteration required roughly twice the computation. Hadthe efficiencies of the computations been increased through, the use of script files and global variables, the relative ~dvantageof the derivative model wouldhave been even greater. Nevertheless, it is hard to justify the effort required to carry out its programming.Whenevercomputer time becomesexcessive for the virtual energy-storage methodto give adequate accuracy, the use of special software such as DASSL can be justified.

5.3

Linear Models and Simulation

The property of superpositlon defines the class of models called linear. Bond graphs with elements having constant moduli are linear models. Linear models also can be represented by linear differential equations. Operational notation allows these equations to be manipulated like algebraic equations. Unlike nonlinear models they are generally amenable to analytic solution. This fact allows numerical simulation to be carried out considerably more efficiently than through the use of a general purpose algorithm such as the Runge Kutta integration presented in Section 3.7. A MATLAB command that directs such a linear simulation is presented below, after the statements above are clarified.

5.3.1

Superposition

and Linearity

Linear systems can be identified merely by their response to disturbances; it is not necessary to have any analytic representation. Consider a "black box" system, as in Fig. 5.19 part (a), with a single disturbance or input, u(t), and single response or output, x(t). Suppose that you can choose u(t) and observe x(t) for a variety of separate "runs." First, you choose two arbitrary functions, ul(t) and u2(t), and observe the respective responses xl (t) and x2(t), as trated in part (b) of the figure. Then you choose a function u3(t ) which equals

342

CHAPTER 5.

MATHEMATICAL FORMULATION

(a) "black box" systemwith excitation and response excitation u l (t ).,_~ "~~esP°nsexflt)__.________ ~

rune

exci~t~ion k~~~g--.-

response

x2(t)

excitationu3(t) =u~(t) +u2(t) .

~

time =

~(t)

~e (b) signals Figure 5.19: Property of superposition

5.3.

LINEAR

343

MODELS AND SIMULATION

the sum of ul(t) and u2(t): (5.34)

u3(t) = ul (t) + u2(t).

The property of superpositlon is said to pertain to the system if the observed response x3(t) equals the sumof xl (t) and x~.(t), that is if X3(t) : Xl (t)

(5.35)

÷ X2(t).

More precisely, the property applies if this relation is true for arbitrary ul (t) and u2(t). Systems often have a domain of u,x within which superposition is satisfied and the system acts linearly. Outside of this domain, that is for sufficiently large signals, the system is nonlinear. A convenient special test employs (5.36)

u2(t) ul (t), which implies u~(t) = ul(t) + us(t) = 2ul(t),

(5.37a)

x2 (t) = xl (t).

(5.37b)

The property of superposition, if it pertains, requires x3(t) = xl (t) + x2(t) = 2x~ (t).

(5.38)

In words, superposition requires that the doubling of an excitation merely doubles the response at each and every time, t. Further, this happens regardless of the size of u~ (t), leading to the conclusion that the response to the excitation cub(t), where c is any constant, is cx~(t). This is what is meant by linearity. The simplest case of all eliminates time as an independent variable, leaving an algebraic relation between the excitation and the response. Now, if u~ produces x~, 2u~ must produce 2x~, so in general x = ku where k --" x~/u~. A plot of x versus u is a straight line through the origin; any other relation violates superposition and mandates the name nonlinear: 5.3.2

Linearity

and Differential

Equations

You may be most familiar with linearity in the context of algebraic and differential equations. The nth order differential equation d’~x an-~

d~-~x

+ an-l

=f(t)

~ + an-s

d~-~x d-~i"~_2

+ "’"

dm u d"~ - ~ u ,=b~,-~-~÷bm_~ d-~-~-ff_~

dx + al-~ + aox

du +.’.÷b~-~-[+bou

(5.39)

in which the coefficients ai and bj are constants, represents a linear relation between the independent function u(t) and the dependent function x(t~) because each term satisfies superposition and therefore is linear. For example, doubling

344

CHAPTER

5.

MATHEMATICAL

FORMULATION

x(t) doubles each term, and therefore corresponds to doubling u(t). Were any term raised to any power other than unity, for example, the linearity and its property of superposition would be lost. More explicitly, equation (5.39) expresses a stationary linear model. The model becomes nonstationary if one or more of the coefficients ai become explicit functions of the independent variable of time, t. Superposition still applies, however; the modelis still linear. The fact that the shape or form of the response of a linear system depends only on the shape or form of the disturbance and not on its magnitude allows the system to be characterized simply in terms of its operational behavior. This property of linear systems is developed in later chapters to give powerful analytic results. By comparison, characterization of most nonlinear systems in terms of their operational behavior is very difficult and usually impracticable; the response to a given disturbance may for example vary wildly depending on the initial condition. (The modern theory which goes by the name "chaos" actually is based on deterministic nonlinear dynamics.) Despite this complexity, nonlinear models may be deduced from direct considerations o] the physics almost as easily as linear models. Identifying a dynamic nonlinear model from observation of dynamic behavior rather than knowledge of the internal structure of the system, on the other hand, is apt to be rnuch more difficult than for a linear model of comparable complexity. The use of causality..on a particular bond graph results directly in a set of first-order differential equations. Whensuch a set of equations is linear and stationary, it can be placed in the canonical or standard matrix form dx

~ = nx +

Bu,

(5.40)

in which A and B are matrices of constants. The general excitation is the vector u(t) and the response is the vector x(t). Note again that the term "stationary" does not i~nply "static," but rather means that the dynamic characteristics are invariant in time. Thu_s, characteristics such as natural frequencies and time constants do not change over time. In matrix notation, nonstationary linear models are represented by the same equation with A = A(t) and B = B(t). This book considers analytical solutions for stationary models only. As you probably have noticed, bond graphs in which the moduli of all the elements are constants directly produce linear algebraic and/or linear stationary differential equations, with all their conceptual and computational advantages. A model described by a set of first-order differential equations such as equation (5.40) can be called complete, since the behavior of all the state variables is represented. Equatio_n (5.39), by _coptra~s_t~ descri_be~ only the relation between single input variable and a single output variable of the system. This incomplete model may nevertheless be all the modeler needs, and is the focus of the solutions developed in Chapter 6. (Matrix methods are largely deferred to Chapter 7.) The relation between input and output variables, for example, establishes the transfer function between these variables, which is sufficient for vibration analysis or classical control calculations.

5.3.

LINEAR MODELS AND SIMULATION

345

In general, the modeler maybe interested in one or more weighted sums of the state variables and the input vector. Representingthese sumsby the vector Y(Q, [y(t)

= Cx + Du.

(5.41)

This classical representation includes as special cases y = x (C equals the unit diagonal matrix, I, and D = 0) and a single output (C and D become row matrices). Usually, one wants D = 0. 5.3.3

Operator

Notation

The manipulationand solution of linear differentia t equations is simplified by ’~ the use of operator notation. The derivative operator, S, is defined by S =- dr" Withthis notation, equations (5.39) and (5.40) become,respectively, (arts n + a,-1S~-1 + ... + alS + ao)x = (bins m + bin_iS m-~ + ".. + b~S + b0)u,

(5.42)

(5.43)

(5.44) (SI - A)x = Bu, where I is the unit diagonal matrix. Theseequations also are written in terms of the transfer function operators, G(S), G(S) and H(S): X

= a(S)u;

bmSra + bm-~sm-1+ "’" + b~S + bo a(s) = a~s~ + an_~S~_~ +.. ¥’a-~-[oo

x= G(S)u;

(~’~)

G(S) = (SI - A)-~B,

y= H(S)u; H(S) = C(SI- A)-~B

(5.46)

wherein the last case equation (5.45) has been used to give the result in terms of the output variable y(t). Transfer functions commonly are placed in boxes, ~ shownin Fig. 5.20. The functions upon which they operate are indicated by an arrow pointed into the box, and the functions whichresult are indicated by an arrow pointed out from the box. The result is called a block diagram. Equation (5.45) is represented by part (a) bf the figure. Equation (5.46) is represented by part (b) figure; the double lines indicated that the associated variables maybe vectors rather than scalars. The circle with the summationsign inside indicates that the output variable is the sumof the two input variables. Block diagrams are most widely used in the specialized domain of automatic control, although somepractitioners use them more broadly to represent dynamicmodels. Mostuse in this text is restricted to control applications. 5Someauthors prefer to use the symbol D for the time derivative operator, which historically is known~ the Heaviside operator.

CHAPTER 5.

346

MATHEMATICALFORMULATION

(a) scalar u(t)

(b) vector

u(t)

,x(~

_~, y(t)

_

Figure 5.20: Block diagrams of equations (5.45) and (5.46)

5.3.4

Transformation

from State-Space

to

Scalar

Form

Oneuse of the operator notation facilitates the reduction of the state-space models of equations(5.40), (5.44) or (5.46) to the single-variable formsof equations (5.39), (5.43) or (5.45).

EXAMPLE 5.8 Find a transfer function relating the input u(t) to the output q(t) for the following differential equation, and give the correspondingsecond-orderdifferential equation:

-- = -2p + 2q + 24u(t), dt d-~ = 4p - 9q + 4

+ 2u(t).

Solution: In operator notation, these equations become (S + 2)p = 2q + 24u, (S + 9)q = 4p + (4S + 2)u. The variable p can be solved for in terms of q, algebraically: P--

2q + 24u S+2

This result then can be substituted into the secondoperator equation, giving the following equat, ion from whichp has been eliminated: (S + 9)q 4(2q + 24u) S + 2 + (4S + 2)u.

5.3.

LINEAR

347

MODELS AND SIMULATION

This equation can be multiplied by the denominator factor (S + 2) to give [(S + 9)(S + 2) - S]q(t) = [96 + (4S + 2)(S + from which G(S)=

2 + 10S + 100 96 + 4S2 + 10S + 4 4S = S ~ +11S+10 2+11S+18_8

This transfer function can be interpreted to give the differential du d2u d’2q + 11-~ + 10q = 4-~., dt---~.

equation

10 + dt + 100u.

Notice that algebraic operations have replaced derivative operations, which is a great convenience. The procedure 6 becomes even more streamlined if you choose to use matrix notation in addition to the operator notation. Pre-multiplying both sides of the equation x = (SI - A)-IBu by det(SI - A) (from equations (5.46)) gives general result det(SI

- A)x = det(SI

- A)(SI - A)-IBu.

(5.47)

This is a vector set of scalar differential equations: the first in terms of x~, the second in terms of x2 and last in terms of x~. EXAMPLE

5.9

Solve the problem of Example 5.8 using equation (5.47) Solution: The state-space

Note that the derivative the derivative operator,

equations in matrix form are

term 4 du/dt has been accomodated by the use of S. The square matrix (SI - A) becomes

which when substituted

S+9

’

into equation (5.47) gives

The second of these two scalar equations identically ferential equation of Example5.8.

gives the resulting dif-

6TheHeavisideproceduredescribed here is formalizedand extendedin the definitions and applications of the Laplacetransforml whichis introducedin Section 7.2.

348

CHAPTER 5.

5.3.5

Transformation

MATHEMATICAL FORMULATION

from Scalar

to

State-Space

Form*

It is possible also to transform a scalar differential equation into state-space form. Theresult is not unique; it dependsuponwhat definitions of are chosen. The simplest schemeassumes x~ = x and an = 1, and employs

A--

--an-i-an-2 1 0 0

0

0

.... ...

al 0

-ao o| ] (5.48a)

...

1

(5.48b)

C = [bn-1 bn-2 "" b0] D=0

(5.48c) (5.4Sd)

whichapplies as long as m_< n - 1. EXAMPLE5.10 Find a state-space differential equationconsistent with the scalar differential equation du d3 x 3 d~ x~ dx 2 dt ~ + dt~_ + 4-~ + 2x = dt + u" Equation (5.48) gives

-d

Xl = dt x2 x3 5.3.6

-3

-4 0 1

Transformations

Xl x2 x3 Using

~,(t); ~=[o21]x_~ .

x3

MATLAB

MATLAB provides translation from the state-space format to the scalar format. This is accomplished with the command [hum, den]=ss2~f (h,B,C,D,i The ss2tf can be read "state-space to transfer function" (which really means. scalar transfer function). The state-space matrices A, B, C, D must be defined numerically before this command is issued. The argument± is an index designating which of possibly multiple inputs is to be considered; if you have only one input, as in Examples5.8 and 5.9 above, insert 1. The output argumentnum (whichcould be designated with any symbol)is a row vector of the coefficients of the numerator polynomial of G(S). The output argumentden is a row vector

5.3.

349

LINEAR MODELS AND SIMULATION

of the denominator polynomial. If the matrix C has more than one row, more than one transfer function is being requested. Each has the same denominator, which therefore is reported only once. The respective numerator polynomials are reported as the rows in num. EXAMPLE 5.11 Use MATLAB to find the transfer function from the input u(t) to the output x.2 (t) for the third-order model dXl 1 dt - 4x~ + ~x.~ + u(t), dx.~ -- = 2Xl -- 2X’2 + 2X3, dt dx3 d--~ = x2 - 4x3, Solution:

(5.61a) (5.61b) (5.61c)

The MATLABentry

A=[-4 1/2 0;2 -2 2;0 1 -4]; B=[1;0;0] ; C--[O~ 0]; D=[0] ; [hum,den] = ss2tf(A,B,C,D,1) gives the response IlUm =

0

0

1.0000

t0.0000

2.0000

8.0000

den =

29.0000

20.0000

which indicates the transfer function 2S+8 G(s) = $3 10S2 + 29S + 20" The inverse transformation

is given by the command

[A ,B,C, D]=tf2ss(num,den) in which tf2ss can be read "transfer function to state-space."

5.3.7

Simulation

of Linear

Models Using

MATLAB*

Numericalsolutions to linear scalar differential equations are plotted, with little effort on your part, by the MATLAB command Isim(num,den,u,t) Related options given below include the treatment of matrix differential equations, lsim stands for "linear simulation." The argument t is a row vector

350

CHAPTER

5.

MATHEMATICAL

FORMULATION

of the times at which the response is to be computed. The argument u is a vector of the input values at precisely those same times; it must have the same length as t. Normally, the. program assumes that the actual input for intermediate times is given by a linear interpolation of the immediately surrounding given values. (In some cases it concludes that a step-like zero-order-hold function is intended, and excercises that assumption.) Exact analytical solutions of the differential equations are evaluated numerically, a process which is potentially both more accurate and more efficient than the numerical simulation scheme presented in Section 3.7.

EXAMPLE 5.12 Secure a plot of the solution of the differential 5.10, repeated here:

equation used in Example

dUx d2X 4 dx du~ 2 4- 3~. + dt 4-2x= dt + u’ dr--~ with the particular

excitation

(5sin(2~rt/10),

u= 0,

0 < t < 5, t_>5.

Consider the range 0 < t < 10 seconds. Superimpose a plot of u(t). Solution: You start by defining a vector of discrete times at which the solution will be found. The commands tl = [0:.i:5];t2 = [5.l:.l:lO];.t=[tl establish vectors for the two segments of time and the combined time duration of 0 to 10 seconds. Next, you specify a corresponding vector for the excitation signal, u(t): ul = 5*sin(2*pi*tl/lO); u2 = zeros(size(~2));u=[ul The coefficients on the right side of the differential equation, which are given in the numerator of the transfer function format for the model, are recognized by the statement hum = [2 1] ;

¯

Similarly, the coefficients on the left side of the differential equation or the d4nominator of the transfer function are recognized by the statement den = [1 3 4 2]; A plot of the response results isim(num,den,u,t)

from the single added command"

5.3.

LINEAR

351

MODELS AND SIMULATION

The plot below results.

The excitation

u(t)/2

is added by the commands

hold plot(t,u/2,’--’) 3.5

2.5

0.5 0 -0.5 -1 0

1

2

3

4 5 Time(secs)

6

9

10

The commandl sim handles matrix differential equations through use of the standard A,B,C,D notation. The options below include creating an accessible file of the results by adding a left side to the statement: [y,x] = lsim(A,B,C,D,u,t); [Y,X] = lsim(A,B,C,D,u,t,x0); [y,x] = lsim(num,den,u,t); The second case allows you to introduce non-zero initial conditions by first defining the vector x0. In the third case, the output vector y and the state vector x actually are identical. EXAMPLE 5.13 Repeat Example 5.12 using the state-space formulation (as given in Example 5.10). Solution: The matrices A, B, C, D are defined with the commands A = [-3 -4 -2;1 0 0;0 1 0]; B = [1;0;03; c = [o 2 ~]; D=O The plot results from the command lsira(A,B,C,D,u,t) or you could use the commandwith the left plot (t,y)

side [y,x] and then type

352

CHAPTER

5.

MATHEMATICAL

FORMULATION

lsim converts the differential equation into a difference equation which for initial-value problems is accurate regardless of the size of the time interval. (It does this using a matrix exponential, which is described in Sections 7.3.1 and 7.3.2.) Because of the assumed linear interpolation of u(t), however, the time interval might have to be rather small to give acceptable accuracy. 5.3.8

Summary

Systems that exhibit a ~lomain of input and response variables satisfying the property of superposition are, by definition, linear within that domain. They may be represented by linear algebraic or differential equations. The domain cannot be infinite for a real physical system, of course; if the variables get large enough there must be a catastrophic failure, which is a nonlinearity of the most severe type. Operator notation allows a linear differential equation to be treated like an algebraic equation. This expedites the transformation of differential equations from one format to another, including the elimination of unwanted variables. Transfer function and state-space formats are of particular interest. MATLAB carries out these transformations with little effort on the part of the analyst. Operator notation also leads to analytic solutions using transform methods, as is developed later. MATLAB capitalizes on some of this with the linear simulation Commandls±m, which you can use even before you understand its mathematical basis. This simulation is more efficient than a corresponding one using a nonlinear integrator such as a Runge Kutta algorithm, but it applies only to linear models. Guided

Problem

5.6

You will use operator methods in this problem to combine two coupled first-order linear differential equations with the objective of characterizing the dynamics of the model. A model is described by the two state-space differential equations below. Determine its natural frequency and damping ratio.

dp --

= -2p-q dt dq = 7p - q dt

Suggested

Steps:

1. Write the two equations with operator notation. 2. Solve one the the equations for p as a function of S and q, or q as a funciton of S and p.

LINEAR MODELS AND SIMULATION

353

Substitute the result of step 2 into the other equation so as to eliminate either p or q. Cast the result of part 3 into the form f(S)p = orf(S )q = OThe function f(S) should be a polynomial. Interpret the result of step 4 as a differential equation with dependent variable p or q. Use the result of equation (3.41) (p. 149) to determine the natural quency and the dampingratio.

Guided

Problem

5.7

This guided problemgives neededpractice in the reduction of state-space differential equations into differentiM equations with a single dependentvariable, using both direct and operator analytical methods, and using MATLAB. It also offers practice in linear simulation with Considerthe electromechanicalsystem of Fig. 5.6 (p. 292) and equations 5.9 and 5.10 (p. 294). Combinethese equations directly to get two single secondorder differential equations with dependentvariables p and q, respectively. Then use operator methods (but not MATLAB) to accomplish the same objective. Next, find the matrices A, B, C, D, assuming that the input variable is the voltage e, and that the output variable of interest is the angular velocity of the shaft. Use MATLAB to find the two second-order differential equations a third way. Finally, use ls±m to plot the shaft speed as a function of time for 1.0 second, assuminge = eosinwt with e0 = 5 volts and 03 = 8~r rad/sec. The -1, 2, parameters are T = 0.2, G = 1.0 volt sec, C = 0.05 N-~m I = 0.2 kg m Rl=4ohmsandR2=l.0Nms. Suggested Steps: 1. Place equations (5.9) and (5.10) in the forms p = p(q, dq/dt) and q = q(p, dp/dt), respectively. Take the devivatives of the equations found in step 1. Then, substitute the results of step 1 into these equations to get two resulting differential equations of secondorder, each having only Onedependentvariable, p or q, respectively. Rewrite the original differential equations using operator notation. Combine these algebraic equations to get one equation with dependentvariable q only, and one equation with dependentvariable p only. 4. Interpret the results of step 3 in terms of differential equations, and check to see that they agree with the results of step 2.

354

CHAPTER

5.

MATHEMATICAL

FORMULATION

Place the original equations in the state-space matrix form, extract the coefficients A and B and evaluate these matrices numerically. Relate the desired output variables to the state variables p and q, so as to define the coefficients C and D. Evaluate these matrices numerically. Use the MATLAB function ss2tf with the results of steps 5 and 6; check to see that the differential equation that emerges agrees with those of steps 2 and 4. Define a new vector of times from 0 to 1.000 seconds with an interval of perhaps 0.005 seconds, using MATLAB. Then, express the excitation voltages at these times as a row vector. Employ the isimcommandto secure the resulting plot. You may annotate this plot by hand if you do not care to learn MATLAB labeling procedures. Are the results reasonable? PROBLEMS 5.34 A system responds as shown below to the given excitation. Examine the data carefully and report whether there are any signs of nonlinearity.

excitatlon, t

o

’i’’ time, seconds

7 ~ time, seconds 5.35 Consider the IRC model of Fig. 3.26 (p. 137), as represented by equations (3.32) (p. 139):. (a) Combinethese equations to get single second-order differential tions with dependent variables p and q, respectively. (b) Repeat part (a),

using operator

equa-

methods (but not MATLAB).

(c) Find the state-space matrices A, B, C and D, assuming that (~ = p/I and q are the output variables of interest.

5.3.

355

LINEAR MODELS AND SIMULATION

(d) Use MATLAB to convert your answer to part (c) to single differential equations in ~1 and qc, using R = 1, C = 2 and I -= 3. Verify whether your answers are consistent with those of part (a). (e) Characterize the dynamics of the model with the parameter values given in part (d) by finding its natural frequency and dampingratio. Hint: Use equation (3:41) (p. 149).

5.36 Answer the questions of the preceding problem for the IRC model of Examples 3.14 and 3.15 (pp. 138, 140). 5.37 Answerthe questions of the preceding two problems for the electric of Fig. 5.1 and equations (5.1) and (5.2) (pp. 283-284).

circuit

5.38 Answer the questions of the preceding problems for the example of Fig. 5.11 and equations (5.11) and (5.13) (pp. 312, 313). In part (d), let I2

5.39 Consider the linear model of the two-tank system of Guided Problem 5.1, (p. 299) with A1 = 2.0 2, A2= 4 .0 ft 2, Ap = ~ra : = ~r/ 4 in 2 and L = 20 ft. Assuming quasi-equilibrium laminar flow of water at 70°F, this gives (from equation (2.14)) RI~. = 8pL/~ra 4 = 7056 lb.s/ft ~ and I = 4pL/3Ap = 9480 lb-s2/ft 5. The resistance of the outlet is Rd -= 2RI:. (a) Evaluate the matrices A, B, C, D, assuming the output variables interest are the depths of the water in the two tanks. (b) Find the equilibrium depths of the water in the two tanks for steady flow Qi : Qo = 3/s. 0.025 ft (c) Using the MATLAB command ls±m, find and plot the depths of the water for 8000 seconds following an abrupt dumpingof 2.0 ft 3 of water into tank 1 (giving an initial depth 1.0 ft above the equilibrium) with Qi = as above. Suggestion: First make sure the unexcited system produces no response. (d) Using lsi~, find and plot the depths of the water for 300 seconds when, at t = 0 seconds, the input flow Q~ switches from Q0 to Qo[1 + sin(2~rt/100)]. 5.40 Consider the vehicle of Figs. 5.15 and 5.16 and equation (5.33) with equations (5.30) and (5.32) (pp. 322, 324, 325). The values of the parameters m = 1250 kg, L1 = 1.4 m, L2 : 1.6 m, J = mr~" with r 2 = 1.6 m2, kl = 30 kN/m, ks = 32 kN/m and R1 = R2 = 4000 Ns2/m.

356

CHAPTER

5.

MATHEMATICAL

FORMULATION

(a) Evaluate the matrices A, B, C, D assuming that the input variables are ~gl and ~g2 and the output variables are ~cm and ~. (b) Using the MATLAB commandls±ra, plot the res~)onses of the vehicle to a bumpwhich passes under the first axle and then under the second. Let y~l = yo sin(2rt/T), valid for 0 < t < T/2; the road is otherwise smooth. The rear wheels pass over the same bump T seconds later; y0 = 0.05 m and T = 0.2 seconds. (c) Using MATLAB, find the scalar transfer function between the excitation ~gl and the response 9c,~. Give the corresponding differential equations. (This part can be done independently of part (b).)

5.3.9

SOLUTIONS

Guided

Problem

TO GUIDED

PROBLEMS

5.6

(S + 2)p -- -q, (S + 1)q = 7p. 2. Fromthe first equation, q = -(S + 2)p. 3. Substituting the result above into the second equation, -(S + 1)(S + 2) 7p. 4. Rearranging, (S2 + 3S + 9)p = 0. d2p

5.

+ 3 ~ + 9p = O.

6. Comparingthis differential .equation to equation (3.41) (p 149), w~= vf~ rad/s, and 2(w~= 3 so that ~ = 0.5. Guided

Problem

5.7

RiTeI p= -O~ q - 7i idq + -~IT e

(a)

CR ~ q -- --~--p ÷ C (b) ~tt dp _ R~T2I dq d’-)q IT de (c) dt G~C dt I-~ + --~--~ dq _ CR~ dp d~-p (d) dt I dt + C-~ Substitution of equations (b) and (d) into (a) R~T:I

dep~

[CR~ + cdP~

IT

v =- a~----5~,--i-~ "~) - I(CR: \-i- dp -~ + c T~) + "5"e or, collecting terms,

~ c -~ + C R ~ + --~-- ] -~ + 1+~ ] ~ = -~ ~ Substitution of equations (a) and (c) into (b) CR~ f R,T~I

q=-i-t,--drO-q-~+~

IT)

+c ( R~T~Idq a~c ~ ~-~+-~-~] d~q ITde’~

5.3.

LINEAR

MODELS

or, collecting IC-~ 3.

357

AND SIMULATION

terms,

+ \’--’GT--

S + -~-,]

2

+ CR’~

q + ~p = ~e

G dt

and

~q = 0

Solving the second equation for p and substututing

into the first,

eS+ -~U-~-,] R1T2"~ q+ 1I Sq/C + R~/ I - TG Multiplying

both sides by (S + R~/I) gives

Solving the second equation for q, on the other hand, and substituting first,

into the

( o,

s +tN+

s+t

+N

The differential equations resulting from step 3 ~e the same as those resulting from step 2 (apart from the multiplicative factor IC in p~t 2).

4.

5.

~

=[

l/C

2Substituting

-R~IIJ

+

~G

e

values into the two matrices, A_ = [-’23d

The output variable

is ~ = p/I,

so C = [0 5] and D = 0.

The MATLAB programming and its .a. = [-3.2

-5;20 -5];

response

B = [.2;0];

is as follows:

C = [0 5]; D=O;

[z,p] = ss2tf(A,B,C,D) Z =

0

0

20. 0000

1.0000

8.2000

116.0000

p =

2O This means that ~+8 .2S+116e ~= S or (S 2 + 8.2S + l16)p = 20Ie(t) above.

= 4e(t),

which agrees with the result

358

CHAPTER

5.

MATHEMATICAL

FORMULATION

7.-8. The MATLAB programmingand response is ms follows: t=[O: .005:1] ; e=5*sin(8*pi*t) isim(A,B,C,D,e,t)

0.4 ~, rad/s 0.2

r~

0 -0.2 0

5.4

0.2

0.4

0.6 0.8 time, seconds

1.0

Linearization

Nature is nonlinear. Nevertheless, for small perturbations about some nominal state, as is typically the case for vibrations, most nonlinear systems act as though they were linear. The simplicity of the characterization and analysis of linear models further motivates the modeler to overlook nonlinearities. Methods for approximating a nonlinear system by a linear model are considered in this section. A model may be linearized before or after it is converted to differential equations. In most cases the latter procedure is preferred. The meaning of linearization usually is clearer when it is carried out directly on the individual elements of the model, however, so this approach is presented first. It also allows a graphical alternative approach, and therefore sometimes is preferred. 5.4.1

Case

Study

With

Linearization

of

a Resistance

Consider a fluid tank with uniform area At and a Bernoulli square-law orifice flow, as shown in Fig. 5.21. The pressure at the bottom of the tank is related to the orifice flow by ¯ P = 21-0

~00 ’

(5.49)

where Q is the volume flow through the orifice and A0 is its effective area (including the vena contracta). This characteristic is plotted in part (b) of the figure. A bond graph model of the system is given in part (c). It shows that the flow is computed with the causality Q = Q(P), inverting equation (5.49). An analytic solution to the nonlinear differential equation -

Ao v ~

(5.50)

happens to exist if the input flow Qin is a constant, or changes in discrete steps. This is not usually the case, however; simulation would be the natural recourse. The solution of the corresponding problem with a linear or constant resistance is muchsimpler. Further, the property of superposition would apply if the

5.4.

359

LINEARIZATION

P volume V

pressureP tl~orifice

a (a) system

(b) resistancecharacteristic

C P= WC 1R P

ss

*_ P*I~

~---~--~0

LQ(p R (c) bondgraph

(d) linearizedcharacteristic 4 3

s;

°

linearized: ~--

20%increase in flow

(e) linearized bondgraph t/r (f) responsesto step changesin Qi,, Figure 5.21: Linearization of a fluid resistance

360

CHAPTER 5.

MATHEMATICAL FORMULATION

model were linear, enabling the response to a complicated excitation O,~n(t) to be assembled from a sum of the responses to simple components of that excitation. Can an approximate linear model be substituted for the nonlinear model in order to gain these advantages? The answer to this critical question depends on how greatly the depth of the water varies, and how accurate the solution needs to be. A linearization now will be carried out, and the behavior of the linearized model compared with that of the nonlinear model. A linearized model describes the behavior of perturbations, or changes, in the state variables relative to some nominal condition. The water tank has a single state variable, the volume of water in the tank, V. This volume is represented as the sum of a nominal volume, V, and a perturbation volume, V*:

y = v + y*.

(5.51)

The nominal volume is the equilibrium state for a particular flow rate, which also is designated by an overbar: Q. Note that at equilibrium, the input flow equals the output flow. Further, the pressure at the bottom of the tank then has its nominal value, designated as P. The non-equilibrium values of these variables are designated as Qin = ~ + Qi*~, Q = ~ + Q*,

(5.52a)

P = ~ + P*.

(5.52c)

(5.52b)

where Qi*~, Q* and P* are the perturbations. The three pressures and three output flows are indicated in part (d) of Fig. 5.21. The linearization corresponds to replacing the curved characteristic for the resistance by a straight line, as shown in the figure. The usual choice is a tangent to the nonlinear characteristic passing through the equilibrium point. The linearized characteristic is described by P* ~- R*Q*;

(5.53)

R* = ( d-~Q ) ¢2= ~.

The slope R* is called the linearized or perturbation or tangential resistance, which is very different from the nonlinear or chordal resistance as defined in Fig. 2.22 part (b) (p. 44) and equation (2.15) (p. 46). The value of be estimated by drawing the tangent with a pencil, and measuring its slope. It also can be computed mathematically by evaluating the derivative, as given in equation (5.53), from equation (5.49). The linearized or perturbation variables and parameters can be placed on a special linearized bond graph, as shown in part (e) of Fig. 5.21. This graph gives a linear differential equation in terms of the perturbation state variable, V*: dV* , ~ dt - Q~n -R*c

dV* V* or

~---~-

V* * + = ~-Q~n,

T = R*C.

(5.54)

5.4.

361

LINEARIZATION

This equation is in the standard form for a first-order model, as given in Section 3.5 (p. 145). Solutions to the cases in which the input flow is increased by 20%and by 100%from initial equilibrium values are plotted in part (f) of Fig. 5.21. The nonlinear solutions are found by simulation. The linear solutions are plots of the analytic solution of equation (5.54) for Qi*,~ = constant, namely V* = (1 - e-t/r)~-Q~ n.

(5.55)

The solution of the linearized model for the 20%increase in flow is seen to be close to that of the nonlinear model; agreement would be better for smaller perturbations. The solutions for the 100%increase, on the other hand, are significantly different; the results of the linear model are only of qualitative usefulness. Nonlinear models with small disturbances about some equilibrium act virtually the same as their simpler linearized models predict, in most cases. Thus, for example, linearized models are used almost exclusively in the field of acoustics, which deals largely with very small pressure and velocity perturbations. The same situation commonly applies regarding mechanical vibrations. The engineering analyst must excercise vigilence, however, for nonlinear effects that are large enough to require abandonmentof the simplicities of linear analysis. 5.4.2

Linearization

of

a Function

of One Variable

The function y = f(x),

(5.56)

if continuous in x, can be expanded in a Taylor’s series expansion as follows: Y = f(~) + -~xdf x-~- (x - ~) + ~d2f x=~ (x -2~)2 ~- higher-order terms. (5.57) It is convenient to designate some nominal values of x and y as ~ and ~, such that ~ = f(~).

(5.5S)

The actual values of x and y are designated as x = g + x*,

(5.59a)

y = y + y*.

(5.59b)

The differences x* and y* between the actual and the nominal values are the perturbations. With this notation, equation (5.57) becomes y. = df -~-+ 2 ~=5 x*2 higher-order -~x ~=~x* + dx d2f

terms.

(5.60)

362

CHAPTER 5.

MATHEMATICAL FORMULATION

*1 _p*

C

P

V Figure 5.22: Linearization

of a compliance

This relation can be approximated, at least for small values of x* and y*, by truncating the series after the first, or linear, term. The resulting linearization can be written

y* ~_ax * ;

df a : -~x ~=.~"

(5.61)

The linearization of the resistance relation applies this more general result; in that case, x = Q and y = P. Another application is to linearization of a compliance relation between a generalized displacement x = q and a generalized force y = e. In this case, the derivative (df/dq)q=- 4 equals the reciprocal of the linearized compliance, or l/C*. This linearization is pictured in Fig. 5.22 for the special case of a gravity fluid compliance for a tank of nonuniform area, in which x equals the volume of liquid in the tank, V, and y equals the pressure at the bottom, P. The derivative can be found either analytically or graphically (by drawing the tangent to the curve at the nominal point, and measuring its slope). The latter procedure does not require an analytical expression for the nonlinear function, but can proceed based on a plot created from experimental data.

EXAMPLE 5.14 Linearize the compliance and the resistance of the conical tank of Example 3.22 (p. 165) about some nominal volmne, ~7. The relation between the volume of the liquid in the tank and the pressure at its bottom is reproduced here for convenience: V = 7~tan2~ (~9)3

5.4.

363

LINEARIZATION Solution: The slope of the characteristic

at a nominal volume ~ is

pg 1 _ dP ~ 3 1/3 ~-~7/" v=V- 3 (~7-~tan’~(~) C* The corresponding linearized (5.53),

resistance

(cdA) 2-

becomes, from equations (5.49) and

cdA -- CdA \ntan2a)

EXAMPLE 5.15 Write the linearized differential equation corresponding to the results of Example 5.14 above, and identify the time constant that characterizes the response to changes in the input flow. Solution: The compliance and the resistance of the combined system are at equilibrium if the flow into the tank, Qin, equals the flow out at is bottom, Q. A linearized bond graph for perturbations about this equilibrium is as follows:

\ u IflCR*"

IR*

Although this has the same structure as the bond graph for the actual variables, the products of its conjugate efforts and flows are not the true powers, and are not even proportional to the true powers. Such a graph sometimes is called a pseudo bond graph to underscore the distinction. The graph gives dV* 1 Q~n C’R* V* . dt This is a linear first-order differential equation of the form of equations (3.36) and (3.37) (pp. 145-146) with the time constant

364

CHAPTER 5.

(a) value

MATHEMATICAL

(b) slope

FORMULATION

(c) curvature

Figure 5.23: Types of discontinuities

5.4.3

Essential

Nonlinearities

There is one case in which the model for the resistance to flow of an orifice may not be considered accurate within acceptable bounds for any departure of the arguments from their nominal values. This occurs when linearization is performed for perturbations about the state of zero pressure drop and zero flow, that is about the origin of the characteristic plot. The calculation in Example 5.14 gives R* = 0 or l/R* = c~, which implies no resistance whatever. This is correct; for precisely zero pressure drop and flow the linearized pressure-vs.-flow characteristic is horizontal. Thus, if one is truly interested only in extremely small flows one could choose to neglect the orifice restriction altogether. (This assumption is indeed commonly made in acoustics.) On the other hand, the resistance departs from zero very rapidly as the flow increases. You could well decide that this characteristic comprises an essential nonlinearity, that is a condition for which linearization is not justified. A characteristic whichsuffers a significant discontinuity in value at a singular point clearly demonstrates an essential nonlinearity at that point. This case is illustrated in part (a) of Fig. 5.23. If, instead, it suffers discontinuity in a f ir st derivative at the point, as shownin part (b) of the figure, the linearization still cannot be carried out formally by taking the derivative. If a discontinuity of value or of slope is small enough, however, one might choose to overlook it in favor of some linear compromise. The issue faced at zero pressure drop for the orifice and in part (c) of Fig. 5.23 is even more subtle: di scontinuity of a second derivative, which is a kind of borderline essential nonlinearity. Discontinuities in the third or higher order derivatives are not considered essential nonlinearities. A possible resolution of the question for the orifice results from use of a more accurate model of the orifice flow. The discontinuity in the second derivative disappears, and linearization about the origin becomes fully justified for srnall disturbances. This happens because when the flow is very small the effect of viscosity, which is neglected in equation (5.49), is no longer negligible compared with the effect of inertia. The actual characteristic for a fixed orifice would resemble the plot shown.in Fig. 5.24. The resistance to flow (slope of the characteristic) never goes below some value associated with "creeping viscous flow." Details can be found in texts on fluid mechanics or hydraulics.

5.4.

365

LINEARIZATION

Bernoulli model

O

Figure 5.24: Correction of Bernoulli modelfor viscous effects Generalizing, someapparent essential nonlinearities are consequencesof our modelingabstractions rather than the physics, although the more commondanger is the overlooking of significant phenomenathrough excessively crude abstractions. Modeling,as noted before, is an art.

5.4.4 Linearization of a Function of TwoVariables The function y = f(x, u) (5.62) can be linearized about the nominal values x = 5, u = ~ and y = ~ = f(5, ~) by retaining only the first or linear terms in a Taylor’s series expansion: Y~-~ +O-~xx=5(x - ~)+

-~uOf ~=~_ (u -

(5.63)

Defining perturbation variables as before, this result can be written ~* ~ c~z* + c~u*;

c~ =

;

cu =

(g.64)

It is applied belowboth to single resistances and entire differential equations. Resistances not infrequently vary in response to Changesin somedisplacement. Such cases are perhaps best understood in graphical as opposedto algebrNc terms. The adjustable hydraulic vane, as re~resented in ~ig. g.2g, is an example. Note that flow is plotted on the vertical ~is and pressure on the horizontal axis, contrary to the normalusage in this bookbut consistent with the valve industry. Such a valve is essentiMly an adjustable orifice. The area of the ven~ contr~ct~ is considered a function of somevalve displacement, labeled here a~ x:

~ = ~(~.

(s.~)

366

CHAPTER

5.

MATHEMATICAL

FORMULATION

X

P~

O_ -= "--0

pa=p~-p2 Figure 5.25: Characteristics

of variable orifice (valve) and linearization

The pressure-flow characteristics, assuming the model A = Ao(x/xo) 1"~, are plotted in the figure for three equally spaced values of x. This plot could be refined by adding any number of added characteristics for interpolated or extrapolated values of x. Mathematically, it is of the form Q = Q(P~, x),

(5.66)

which is a function of two variables. The family of characteristics may be approximated in a region about some nominal pressure difference Pd, valve displacement 5 and associated flow Q by a set of parallel, equally spaced straight-line characteristics, as shown in the figure. These represent a linearized model. Mathematically, equation (5.63) can be appled to give ~, + kx*, Q, = P~ X* ----

X --

~,

1 _ OQ R*

k =~

OPaP~=-f ~ ’

(5.67a) (5.67b) (5.67c)

x-~’~

p~=_~ x=’~

For most cases this linearization is accurate within any desired bound for sufficiently small departures of x, Pd and Q from the nominal 5, Pd and Q, respectively. Where to set these bounds, that is how much inaccuracy to accept in order to reap the benefits of linearity, ultimately is a matter of engineering judgement. Often an engineer will elect to start an analysis with a quick, rough and simple linearized model, and when satisfied with the general nature of the result, refine the analysis with a relatively expensive-to-use nonlinear model.

5.4.

LINEARIZATION

367

EXAMPLE 5.16 Linearize the drag and thrust characteristics of the boat described in Guided Problem2.3 (p. 52) about the equilibrium condition at 45 feet per second. Incorporate the results into a linear state-space modelof the boat, assuming a virtual inertia 50%higher than the mass of the boat and its contents to approximate the effect of the water which is accelerated along with the boat. The boat and its contents weigh 2500 pounds. Finally, for howlarge a changein thrust wouldyou judge that the linearization gives a reasonable representation? Solution: The first step is to drawtangents to the drag and thrust curves at the given equilibrium point directly on the. given plot. Then, a set of equally spaced parallel lines can be drawnto approximatethe thrust for other values: 800 100 ~c~~9 force, F ~ 600 [

0’0 10 20 30 40 ~ 5~0 ~ 60 This plot helps yourepresent the linearized characteristics with linear algebraic equations. Let the thrust be Fa and the drag be Fa, both in pounds force, and ~he percent throttle be P: Fa = 324 + 5.59(P - 70) - 11.6(~ - 46.2) = 5.59P - ll.6k + Fa = 17.3(~ - 27.5) = 17.3 k - 476 The differential equation for the speedof the boat simplyrepresents that ~heacceleration equals the ra~io of the net thrus~ (F1 - Fa) ~o the effective inertia: d~ _ F~ -Fa - z-0.248k + 0.0480 P + 8.11 ft/s dt m The linearization applies exclusively in the region wherethe boat planes, workingfairly well for speeds between42 and 52 ft/s. It totally fails to represent the decrease in drag associated with the onset of planing.

368

CHAPTER 5.

MATHEMATICAL FORMULATION

5.4.5 Linearization of a First-OrderDifferential Equation The differential equation for a first-order and excitation u(t) can be written

dynamic model with state variable x

dx

d--i=f(x,u(t)).

(5.68)

The term on the left side equals a function of two variables, so this equation is in the form of equation (5.82), and can be linearized in the same way. Thus, the variables x and u are decomposed into two parts: x(t) = ~(t) x*(t),

(5.69a)

u(t) = ~(t) + u *(t).

(5.695)

The novel feature is that x and u are functions of time, but this does not upset the linearization. The terms x* (t) and u*(t) are generally small variations or perturbations of x(t) and u(t) about their nominal values Z(t) and ~(t). These nominal values, as always, must satisfy the original nonlinear equation. In general, they represent some nominal "trajectory" in time about which the linearization is taken. Usually, the engineer is interested in perturbations about an equilibrium or steady state, in which ~(t) = constant and the derivative d-2/dt = O. This is the only case considered in detail below, although the more general case is hardly more involved. All time-dependent phenomena are therefore left to x* (t) and u* (t). The first order of business is to solve for the which corresponds to the given ~. For the case of a constant equilibrium, this implies solving the algebraic equation

f(~,~)=

(5.70)

The second step applies equation (5.64) to give the linearization dx* -d~ ~- Ax* + Bu *,

Of ; B = ~u Of ~=~ " A = ~x ~:~

(5.71)

One approach to linearization linearizes the individual nonlinear ele~nents before constructing the differential equation. This has been done above for the example of the conical tank and orifice, with its nonlinear resistance and nonlinear compliance. A simpler and more reliable approach, in most cases, is to find the nonlinear differential equation and then linearize. This approach now is illustrated.

5.4.

LINEARIZATION

369

EXAMPLE 5.17 Return to the conical tank and orifice of Examples 3.22 (p. 165), 5.14 (p. 362) and 5.15. Linearize directly the nonlinear differential equation, which is dV A0x/~ (3V)1/6, d--i = Qi, (~ tan~ O01/6 and compare the result to that found in Example 5.15 by linearizing resistance and compliance elements individually. Solution: The equilibrium solution corresponding to Qin = -0 is

the

_

0 = Q - A0

~t~

a

Using equation (5.71), the linearized differential dV* ~ ~ Qi~

6 Ao~(

equation becomes

~5 ~ 3 tan2 )~/~V*.

This is a first-order linear differential equation of the form of equations (3.36) (p 145) and (3.37) with the time constant

r

= Ao~

-

It agrees with the results of Example5.15, and is reasonably valid for perturbations in the volume, V*(t), that are considerably smaller than the mean volume, V. 5.4.6

Linearization

of State-Variable

Differential

Equations

Linearization of a higher-order model usually is best carried out on the individual first-order differential equations in state-variable form, or dx d~- = f (x, u(t),

(5.72)

The discussion that follows drops the final argument t, restricting consideration to stationary models. In this case the equilibrium and perturbation variables are X:~+X * ~ (5.73a) u = ~+ u*,

(5.73b)

The equilibrium solution to equation (5.72) 0 = f (~,~).

(5.74)

370

CHAPTER 5.

MATHEMATICAL FORMULATION

The Taylor’s series expansionof equation (5.73) dx dx* _ _ Of -- = O+ = f(x,u)+

x*+

Of

u* +higher order terms. (5.75)

This result can be cast in traditional form as follows:

(5.76) The matrices of derivatives that comprise A and B are called Jacobian matrices. EXAMPLE 5.18 A mass rests on a spring which in turn rests on a rigid foundation. The spring satisfies the nonlinearstiffening relation Fs = kl X + k3x3;

kl = 2000 rmN/m;

k3 = 340,000 N/m

(in order to support a range of massesand applied forces without bottoming out). The particular mass of interest compresses the spring 0.100 meters whenthe applied force F is zero. Modelthe system, determinethe mass, find a single linearized differential equation relating F = F* to x*, and determine the linearized natural frequency. Solution: The bond graph

gives the state-space model d-~

= F+mg-klX-kaz

The equilibrium condition for F = F* = 0 is 0 : 0 + mg- kl5

-

~3, k3

a "

5.4.

371

LINEARIZATION from which !(k1~ + k3~ 3) = 2000(.1) + 40,000(.1) 3 = 24.5 kg 9.81 g Thelinearized differential equation is m=

2-kl

~ p*

p.

-3k~

+

F*,

so that

d~2x*_ 1 dp* _ kl4:3k~-~ 2x, + dt 2 m dt m The linearized natural frequencytherefore is ’~ o)~ ~/kl + 3k3~~" -~/2000 + 3 × 40000(.1) m 24.5 5.4.7

Case Study ria

With Three

Different

IF*. m

= 11.44 rad/s = 1.820 Hz.

Types of Equilib-

The second-order system -- = -q + sinp dt d-~ = p- ~rq 1-

(5.77)

has three equilibria, each of a different type. The behavior can be studied globally, using simulation, to achieve a picture of the overall behavior. With muchless effort, the behaviorsfor states close to equilibrium can be determined using linearization. The two approaches are developed and comparedbelow. Equations(5.77) represent an autonomoussystem, that is with no excitation. The entire behavior depends on whatever initial state is imposed. A phaseplane plot of the system is shownin Fig. 5.26. Trajectories in state-space are given by solid lines with arrows that designate direction with increasing time. These trajectories were found by simulation of the equations using the standard MATLAB routine ode23, using various initial states on the boundaries of the plot. All trajectories starting above the line designated separatrix A sweep across the state space, remaining above the separatrix, and continue toward infinite values. This region is unstable. A secondregion in state space is virtually surroundedby the line designated as sepm’atrix B. Anyinitial state in this region producesa trajectory that approaches and then encircles point Pl, ql in the clockwisesense. This point and this region are called meta-stable, since no trajectory ever reaches the point itself, but rather encircles it indefinitely. Lookedat as functions of time, both p and q oscillate virtually sinusoidally. All trajectories in the third and remaining

372

CHAPTER 5.

MATHEMATICAL FORMULATION

Figure 5.26: Phase-planeplot of systemwith three different types of equilibria

5.4.

373

LINEARIZATION

region, which lies between separatrix A and separatrix B, end at point P3, q3This point and region are stable. Equilibrium points satisfy the conditions dp/dt = 0 and dq/dt = O. Labeling such points as ~ and ~, equations (5.77) give ~ -- sin~,

(5.78a)

~ = n~(1 - ~/2).

(5.78b)

These two equations are plotted in Fig. 5.26 by dashed lines. Their intersections represent the equilibrium points. In addition to points Pl,ql = ~r/2, 1 and P3, q:~ = -2.4886,-0.60757, noted above, there is point p.), ~ = 0, 0. This is an unstable equilibrium point, because any initial condition represented by a point near but not at it produces a trajectory that heads away from it. The behaviors in the local neighborhoods of the three equilibrium points can be determined by linearization, without any need for extensive numerical computations such as those that produced the trajectories and separatrixes shown in the figure. Equations (5.77) linearize about any nominal state state trajectory ~,~ to become

dp*

- q* + (cos~)p*, dt dq_~_*= p. + ~r(~ - 1)q*. dt

(5.79a) (5.795)

(You are urged to derive these equations yourself.) This result now will particularized to give linearizations about the three equilibrium points. Substituting the values ~ = ~r/2, ~ = 1 for the first equilibrium into equations (5.79), there results

dp*--

q*

dt dq* _= p, " dt Combining these two equations to form a second-order differential

(5.80) equation in

p*,

*d~-p dt ~ +p* = O.

(5.81)

This equation describes an oscillation with natural frequency wn = 1 and damping ratio ( = 0. The absense of damping renders this equilbirum meta-stable. Substituting the values ~ --= 0, ~ = 0 for the second equilibrium into equations (5.79), there results

d~_Z* =~.-q*, dt dq* p. d--i- = - ~*’

(5.82)

374

CHAPTER 5.

MATHEMATICAL FORMULATION

In operator notation these become (S - 1)p* = -q*, (S + ~)q* = p*.

(5.83)

Combiningthese two equations to give an equation in p* *d2p * = 0 or dt ~-(~r-1) 2

[S2+(~r-1)S+(1-~r)]p The solution is

p* = p~e-2"ss41t

+(1-~r)p* = 0. (5.84) d_~_~_~

+ p~e0"7425t,

(5.85)

where p~ and p~ are arbitrary constants dependent on the initial conditions. The secondterm, with its positive exponent,reveals the instability. Substituting the values p = -2.4886, ~ = -0.60757 for the third equilibrium into equation (5.79), there results

dp* -- =

-0.7943p*- q*, dt d* q . -~- = p - 5.0503q*.

(5.86)

Combiningthese equations, (the use of operators methods as above is suggested), gives *d: P (5.87) dt 2 + 5.1297d_~t + 1.4011p* = 0. This has a natural frequency of 1.184 rad/s, but no oscillations appear because it is overdamped with ~ = 2.167. It is rock stable. 5.4.8

Case Study:

Stick-Slip

System

With Inertia*

The stick-slip systemaddressed in Section 5.1.4 (p. 286) is linearized below order to demonstrate the methodand someof its advantages and limitations. Recall that the system, as shownin Figs. 5.2 and 5.3, exhibits rather exotic limit-cycle behavior whenthe load inertance is small. Whenthe inertance is large, however,the state convergeson the equilibrium point, whichin this case is stable. Could this stability have been determinedwithout carrying out the nonlinear simulation? The nonlinear modelis represented mathematicallyby the differential equation (5.4) with the algebraic equation (5.7) substituted, and differential equation (5.5) with the algrabraic equation (5.6b) substituted. Thelinearization of these equations is d¢5 1 -(1/CMs~)¢~: dt -- 2 ! ~/(Ms1/2Ms2)°" + Mso/Ms2 -

alp*1~M~.___~ p .__[_-~ -- ~¢c’ +

2. MR~eq~b.p,.

1 , - ~p , (5.88a) (5.SSb)

5.4.

375

LINEARIZATION

Substituting

all the given parameter values except for the inertance I gives d¢~ _ 26.17¢~- (1/I)p*, dt dp_~*= lobby: + (0.0567/I)p*. dt

The state-space

(5.89a) (5.Sgb)

formulation therefore gives

If. A scalar second-order differential equation with either of the state variables as the unknowncan be found several ways, as outlined in Section 5.3.4 (p. 346). There is no excitation (i.e. u(t) = 0). Equation (5.47) (p. 347) is a one-step method that gives -~ dt det(Si_A)¢~,_d2¢~+(26.lT_O.O__~67)d¢~,

-~-+

-~b c=0. (9.9~17)

(5.91)

This is a standard linear second-order differential equation with constant coefficients. As discussed in Section 3.5.3 and given in equation (3.41) (p. 149), can be characterized in terms of a natural frequency, wn, and a damping ratio, if, as follows: d ¢c , dec (5.92) dt--~-- + 2;w,~--~- + n~,6, = O. Thus the natural frequency is ~/I, which for I = 0.0021 ft-lb s z is 63.7 rad/s = 10.14 Hz = 1/0.0987 s -1 and for I = 0.0022 ft-lb s e is 62.2 rad/s = 9.90 Hz = 1/0.1010 s -1 . These values can be seen to agree with the corresponding plots in Fig. 5.3. The damping ratio can be either positive or negative, depending on the magnitude of I. Rememberthat the solution of the linear equation is a sinusoid oscillating within an envelope with the factor e-;~"t, as discussed in Section 3.5.4 and given in equation (3.44) (p. 151). Therefore, a positive value 6f ~ indicates stability, and a negative value indicates instability, since the envelope decays to zero in the first case and explodes to infinity in the second. Thus the model indicates instability for small values of I (note the instability of the original model with I = 0) and stability for large values of I. The borderline case, for which ff = 0 is given by 0.0567 ~, I - 26.17 - 0.00217 in.lb-s

(5.93)

which is between the value of 0.0021 ft lb s 2 that gave an unstable simulation and the value of 0.0022 ft lb s "~ that gave a stable simulation. What happens if the linear simulator :~sira is used for this system? The results that correspond to the three largest inertances considered before are given in Fig. 5.27. The linear model is inaccurate for large perturbations from

376

CHAPTER 5.

MATHEMATICAL FORMULATION

21 = 0.002t lb.ft-s

40

.........

/

|

:FAAAAAAAA

2I = 0.0022lb.ft.s

.........

4o / 21 = 0.010lb.ft.s -20 K 0

0.2

0.4

0.6 0.8 t, seconds

Figure 5.27: Linear simulation of stick-slip system

1.0

5.4.

LINEARIZATION

377

equilibrium, which is essentially all there is in the first case: the result merely certifies that the system is unstable. The results for the second and third cases also also inaccurate at first, when the perturbations from equilibrium are large, but becomeincreasingly accurate as the oscillations decay. The only substantive advantage of the linear simulation is its computational speed and efficiency, which can be quite important for very complex models. Coding of the MATLAB ls±m command for the case with I = 0.0022 ft-lb s 2 can be as folows: I=O. 0022; A=[-26.17 -l/I; lO 0.0567/I] ; B=[0;0];C=[0 1/I];D=0; t=[O:.001:1] ; u=zeros (i,1001) x0=[0;-25.19.I] ; [y,x]=lsim(A,B, C,D,u,t,xO) plot(t,y)

5.4.9

Summary

Most nonlinear systems behave as though they were linear for sufficiently small disturbances about some equilibrium, motivating the linearization of nonlinear models. Linearization usually is applied best upon the set of nonlinear state differential equations. Alternatively, nonlinear resistances and compliances can be linearized directly, before differential equations are formulated, by replacing their nonlinear effort-flow or effort-displacement characteristics, respectively, with linear or straight-line characteristics. The linearized characteristic passes through the origin defined by these incremental or perturbation variables, which is located at the point of tangency of the two characteristics and may be different from the origin of the full variables. Graphs that act much like bond graphs may be constructed using the linearized variables, but such pseudo bond graphs do not represent the actual power flows and energy storages, or even their incremental variations. Analytical linearization of a term in an algebraic equation or of a differential equation is carried out formally by truncating a Taylor’s series expansion after the linear or first-order terms. Graphical linearization is carried out by replacing curved characteristics by straight lines, and families of curved characteristics by familes of equally spaced straight lines. The graphical option is especially useful when you have experimental data rather than an analytical model. Even if you perform a linearization analytically, a comparison of the graphical representations of the nonlinear and approximate linear characteristics conveys a sense of how the error imposed by the linearization varies with the span of the changes in the state variables. A third way to linearize a system is to expand the energy storage functions 12 and 7- and the power dissipation function :P in Taylor’s series, and drop all

378

CHAPTER

5.

MATHEMATICAL

terms of third and higher order. This method is illustrated 5.10 on p. 379. Guided

Problem

FORMULATION in Guided Problem

5.8

This first linearization problem employs the straightforward analytical procedure on a set of differential equations. Linearize the following state differential equations about the equilibrium state. The variables are p and q; the coefficients a, . .., ] are constants. [1 + (a + bq)~’]~t "~ = el - cq + d(a + bq)p dq

d-~=ep- f v~

Suggested

Steps:

1. Find the equilibrium state or states by setting the two time derivatives equal to zero. Label these as p and ~. 2. Expand each term in a Taylor’s series expansion. Note that the equilibrium value of dp/dt is zero. (Derivatives do not always have a zero equilibirum value, however.) 3. Truncate the expansions to the linear,

terms in p* and q*.

4. Subtract the terms of the original differential equation, with p and ~ substituted for p and q, from the truncated expansion, to leave.only terms proportional to the incremental variables p* and q*.

Guided

Problem

5.9

This second linearization problem employs graphical linearization with a subsequent mathematical analysis, a very commonprocedure. An induction motor drives a load that has the resistive characteristic plotted in Fig. 5.28 plus a rotational inertia of 0.05 kg.m2. The drive may be considered to be inflexible. Perturbations are imposed on the load torque equal to M = 2.0 sin(40 t) N.m. You are asked to make a close estimate of the resulting perturbations of the angular velocity. Suggested

Steps:

1. Linearize the steady-velocity characteristics of the motor and the load, determining numerical values for the resistances. 2. Drawa linearized bond graph for the system, including the inertance the disturbance. Apply causality, and define the state variable.

and

5.4.

379

LINEARIZATION

0

10

20

30 ~, rad/s

40

Figure 5.28: Guided Problem 5.9

Write the differential

equation.

This text has not yet addressed the solution to differential equations with sinusoidMexcitations. The differential equation is only of first order, however, and the part of the solution you want is so simple that the most primitive method taught in your course on differential equations can be applied. Substitute an assumedsolution of interest p~ sin 40t + p~ cos 40t into the differential equation, and determine the coefficients pl*. and p~. The answer you want is the magnitue of the simusoidal motion ¢*, which

equals

+

Checkthat the resulting perturbations of the angular velocity are not large enough to impart significant error to the linearization.

Guided

Problem

5.10

This third linearization problem is intended-to show how linearization can be carried out in terms of energy relations. A simple pendulum has a momentof inertia I about its center of mass and Io -- I ÷ md~ about its center of rotation, where m is its mass and d is the distance between the centers of rotation and mass. (a) Neglecting dissipation, find a nonlinear characteristic based compliance, and sketch a plot of this characteristic. (b) Linearize this characteristic, characteristic for comparison.

for the gravity-

and plot on the same axes as the nonlinear

(c) Write the nonlinear and linear state-based differential

equations.

380

CHAPTER

Suggested

5.

MATHEMATICAL

FORMULATION

Steps:

1. Sketch the pendulum, and find its gravity energy V relative to its minimum for the vertical equilibrium, using the angle of departure from vertical, ¢, as the displacement. 2. Compute the torque M = dl)/d¢

and plot M versus ¢.

3. Carry out the linearization two different ways. First, directly linearize the relation between Mand ¢, and plot. 4. For the second approach, do a ~aylor’s series expansion of the function V = V(¢). Drop terms higher than the quadratic. The result gives the linear relation between M = d)2/d¢ and ¢. 5. Write the state differential equations, using as state variables either the standard pair (p, q = 4) or an alternative (¢, q~).

Guided

Problem

5.11

The last guided problem involves linearization in two variables. The resistance to the flow of oil through a hydraulic spool valve is associated with an orifice in the shape of a very short segment of a cylinder of diameter d, as shown in Fig. 5.29. The fluid can be considered to be incompressible with density p, and viscous effects can be neglected. (a) Find the volume flow rate Q through the valve as a function of the pressure drop P and the spool position, x. Sketch the characteristics using appropriate dimensionless coordinates. (b) Linearize the characteristics about some non-zero operating pressure drop and flow. Addthe resulting linearized characteristics to your plot. Suggested

Steps:

Assumethe Bernoulli equation for the flow, representing the area of the orifice as the area of the segment of the cylinder of length x. A flow coefficient Cd can be applied, which has a theoretical value of 0.611. Group the variables and parameters such that a dimensionless group proportional to the flow Q is a function of a dimensionless group proportional to the pressure drop P and a dimensionless group proportional to x. An arbitrary maximumor reference pressure Ps can be defined to make this possible.

5.4.

381

LINF~ARIZATION .*-x

etry "~+Ap

.

Figure 5.29: Guided Problem 5.11 3. Sketch-plot the results with the first group above as the ordinate, the second group as the abcissas and the third group as the parameter which distinguishes three of four different curves. 4. Carry out the linearization about someoperating point Q, P, ~. Equation (5.67) (p. 366) can be adapted to the dimensionless groups. 5. Drawa set of equally spacedparallel lines that correspondto the linearized characteristics. Notethe limited-area of applicability.

PROBLEMS 5.41 A dashpot obeys the relation F = a~3/2. Find its linearized resistance R* for perturbations about a nominalvelocity ~. 5.42 A "stiffening" spring satisfies the force-displacementrelation ’~ F .= kx + ax find its linearized complianceC* for perturbations about the displacmentz = 5.

382

CHAPTER 5.

MATHEMATICAL FOR_MULATION

5.43 A vehicle weighing3000lbs. is driven at a constant speed of 50 ft/s, The thrust force then is increased suddenly by 10%,and maintained at that level. Youare asked to find howthe speed of the vehicle increases with time. The drag characteristic is plotted below. (a) Estimate the initial and terminal thrust forces. (b) Linearize the drag characteristic, and estimate the terminal speed. (c) Drawa linearized pseudo-bondgraph and write a differential equation describing the transient behavior. (d) Solvefor the velocity as a function of time; sketch-plot the result. 6O drag force, lbs. 40 2O

0

I

0

I

20

I

I

40

I

I

60 80 speed,fl/s

5.44 A series-wour/d DCmotor has the torque-speed characteristic plotted below. It drives a load with a total massmomentof inertia of 0.25 in.-lb-s 2 and steady-state torque of 5.0 in..lb at its equilibriumspeed. 15 torque, in.-lb 10

1000

(a) Determinethe equilibrium speed, ~0-

2000 speed, rpm 3000

383

5.4. LINEARIZATION

(b) At time t = 0 a brake applies an additional torque of 1.0 in..lb. Determine its final speed. (c) Drawa linearized bondgraph and write a linear differential equation that whensolved approximates the transient change in speed, ~*(t), for ti~ne t > 0. State the values of all constants. (d) Sketch the speed vs. time, noting the values of any time constants natural frequenciesthat exist. 5.45 Briefly describe under what circumstances the following situations give rise to essential nonlinearities: (a) coulomb(dry) friction (b) fluid checkvalve (c) electrical diode (d) finite-sized fluid gravity tank of uniformarea (e) a simply mountedcompression spring 5.46 Linearize the following differential equation about the equilibrium corresponding to ~ = 4. dx -- = -6v~ + 3u dt 5.47 Linearizethe followingset of state differential equations about the smallest positive equilibrium correspondingto ~ = 2. dq 2 x/~ p2 4pq u dt 7r dp 10 cos q dt 5.48 A system has been modeled using a bond graph and characteristics shownbelow. :

C

Q, m3/s

3V, m

as

384

CHAPTER

5.

MATHEMATICAL

FORMULATION

(a) Annotate the bond graph as needed and find the state equation(s).

differential

(b) Determine the equilibrium state(s)

of the system if P0 = 50 2.

(c) Linearize the resulting differential the equilibrium state.

equation(s) for disturbances about

5.49 The model described

by

+ 2x = u(t)

dt~-~ + 2 has a nominal input u(t) = ~(t) = 2 +

(a) Find the nominal response, ~(t). Hint: ~ = a + bt . (b) Linearize the differential equation for the responses x*(t) = x(t) -~(t) to the disturbances u*(t) = u(t) -~(t). 5.50 The system with an axial-flow fan and tank given in Problem 4.60 (p. 277) is. modified by placing a duct of length L = and cross-sectional area 3 × 3 inches, as pictured below. Consider the case of orifice #1. fa~n

L

~ m[~ orifice

tube, areaA

(a) Model the system with a bond graph. (b) Write a set of differential equations for thee model. At this point you may leave the source and load charateristics as unspecified functions of specific variables. (c) Linearize the model of part (b). Take particular

care with the signs.

(d) Determine the natural frequency and damping ratio of the linearized model. (e) Show that a small value of L produces instability (and by inference limit-cyle oscillation), and a large value of L produces stability. (f) Determine the borderline value of 5.51 Consider the ground effect machine (GEM)described in Section 5.2.4 (pp. 316-319) and Fig. 5.13 and modeled by equations (5.17) - (5.19).

5.4.

385

LINEARIZATION (a) Determine the equilibrium values of the state variables y and (b) Linearize the differential librium.

equations for perturbations about the equi-

(c) Determine the natural frequency and damping ratio of the vertical (heave) behavior. Comparethese results to the corresponding nonlinear simulation in Fig. 5.14. 5.52 A memberwhich moves at a controlled speed ~ pushes one end of a linear spring, the other end of which pushes a weight at velocity 3. The weight slides over a horizontal surface with the friction force as plotted below. 2friction force, lbs. I

-2

W = 5 lbs

-1

1

2

3

k=21b/in.

I

I

4

I

I

I

I

5 6 ~, in./s

~

(a) Draw a bond graph for the system. (b) For the case & = 1 in/sec, determine whether the corresponding equilibrium value ~ = 1 in/sec is stable, neglecting the effect of the inertia. (c) Linearize the friction characteristic for the equilibrium state of part (b), and estimate the corresponding incremental or linearized resistance, R*.

(d) Determine whether the inertance affects the stability state, and describe the character of the behavior.

of the equilibrium

5.53 Consider the standpipe and fluid jet of Problem 4.56 (p. 274). The net upward flow in the standpipe has been plotted as a function of the deviation of the pressure just above the jet from its nominal value. Three equilibrium states have been found, the outer two of which are stable. (a) Linearize the characteristic P- Ps < O psi.

for the upward flow for the region -2.5

(b) Represent the dynamics of the system by a bond graph, allowing the linearized characteristic to substitute for the actual characteristic. Inertial effects in the standpipe may be neglected.

386

CHAPTER

5.

MATHEMATICAL

FORMULATION

(c) The jet is suddenly turned on, following a quiescent state. Compute the transient and steady-state response of the level of the free surface, using the model of part (b).

SOLUTIONS TO GUIDED PROBLEMS Guided

Problem

5.8

1. If~-~=0and

~-=0,

then

and

~=

0 = el - cqo + d(a + b~)-~, or

+f

e2

--

~, C2

or

q +

dbf2

Therefore,

~+~=0.

~ = -~

db]~

¯

~

db]~ ]

dbf~.

+d(a + b~)2~p* + *.db~:q dq*a d~

Since ~ ~ = 0,

4. After the equilibrium terms are substracted, this gives [~ + (~ + ~)~] *, dp" ~ ~ -ca* + 2d(a + b~)~p* + db~2q dq* ~ , ~ _ ep -~ q*. Guided

Problem

5.9 ’

’

I ~,~

I

o[or

40

’

N.m 20 01

0

I

~

10

~

I

20

30 ~, rad/s

Fromthe linearized characteristics

40

superimposedon the plots above,

motor: M~, = -a~*, a = 3.94 N.m.s load: M* = -b~*, b ~- 1.86 N.m.s

5.4.

387

LINEARIZATION

MOTOR

LOAD

! dp* a- b, -~- = M" - a(~" - (-~’) = M* - ---i--p

3.

alp*

Therefore, -~- + 41.6p* = 2sin40t. 4. The suggested substitution gives 40(p~cos 40t - p~ sin 40t) + 41.6(p~sin 40t + p~ cos 40t) = 2 sin from which 40p~+ 41.6p~---- 0, 41.6p~ - 40p~p2= 2, giving p~ = ,0.025 andp.; = 0.024 so that the magnitude of the sinusoidal response of ~ equals X/(0.025)~ + (0.024)~/0.05 = 0.693 rad/s. 5. Fromthe plot it can be seen that the magnitude of the perturbations ~* are so small that the linearization introduces negligible error. Guided

Problem

5.10 7- = mgy = mgd(1 - cosO)

linearized M]characteristic/~_..l~

dT = mgsinO 2. M = -~

actual . .

mgd~ 0

z/2

t?

3, ~1 = (dM)_~_ o=o=mgdc°sO=mgd;theref°re’C°= lmg-~ O~ 04 06 4. T = m#d 1 - 1 + -~. - -~. + ~ ..... ( ) dT M = --~ = mgdO

p/ o =

-~ = Co dO

1

-~ Top

~~for small 8. -~madO

-mgdO

d-~ =- Io or

dO

-~ o

z

388

CHAPTER, 5.

Guided

Problem

MATHEMATICAL FORMULATION

5.11

1. The area of the orifice Bernoulli’s equation, is

is A = x~rd. The volume flow rate,

Q,

according to

= cdAV/~

where Ap is the pressure drop between the inlet and outlet channels, and Cd = 0.611. The dimensionless flow rate is CdX,~Trd

x,~ V Po

where xm and P0 are the nominal maximumvalve opening and the maximum applied pressure, respectively.

Q

~r

~~

=x/x,,

~0.8 0.6

~ .......

,

0.4 0.2

00

AP/Ps

)Q = qo + O(A-~/po

AP/ps ~ ( AP - Po

eo,~o\ "P-:

OQ

J + ~)Po,~o

x AP For the special case -- = 0.6 and = 0.5, Zm --~0 there results Q0= 0.6 0yz~--~.5= 0.424, 2x~ 0(A~Ps) Po,~o =1x~1 (oQ) O(~)

=

= ~(0.6)~

= 0.424;

= 0.707.

5. See the set of parallel lines superimposedin the plot above (step 3).

Chapter 6

Analysis Part 1

of Linear Models,

Linear modelshave alwaysenjoyed a special status in engineering, despite being a small subset of all possible models. There are two reasons. First, linear modelscan be analyzed mathematicallywith relatively little difficulty. Before the advent of the computerthis fact forced linear modelsto the forefront, but the present ubiquity of simulation routines weakensthe motivation. Second, the behavior of linear modelscan be characterized in vastly simpler ways than almost all nonlinear systems, regardless of whether mathematics is employed. The significance of this fact has not waned;well engineered systems ought to behavein simple,, predictable ways, so linear behaviorremains a frequent object of design. This chapter deals with scalar differential equations,that is differential equations in terms of a single dependentvariable. Sucha formulation alwayscan be deduced from a state-variable formulation by using the methodsintroduced in Section 5.3.4. Classical or direct methodsof solution are given. Later, operatorbased methodsusing an expansion of the transfer function are seen to be more efficient for high-order models. Anextensive study of the response of linear models to sinusoidal inputs is based on complexnumbermethods. Application of the property of superposition is used to find the response of a system to complicated inputs. Several of the calculations are expedited by MATLAB. Thedirect solution of matrix differential equations is deferred to Chapter 7, along with the formal use of transform methods.

6.1

Direct Solutions of Linear Differential Equations

This section reviews the solution to the general nth-order scalar linear ordinary equation with constant coefficients, which is given as equation (5.39) and 389

390

CHAPTER 6. ANALYSIS OF LINEAR MODELS, PART

repeated here: dn-lx

d~-2x

dx

an~Z+a~-ld~-~_~+a~-~d-~ + "’" + ~1~+ ao~ = f(t)

d~-lu =b ,~d~udtm + bm-~ ~ +...

du + bou. + b~-~

(6.1)

The consideration of impulses as disturbances or input excitations maybe new to the student. Someresults are given without derivation; you are referred to almost any mathematicaltextbook on linear differential equations for a rigorous development. The homogenousequation is the differential equation with f(t) = O. Its solution, designated xh(t), plus any particular solution of the full equation, designatedas xp(t), equals the general solution to equation (6.1):

x(t) = ~h(t)+ ~p(t). I

(6.2)

The existence and uniqueness theorem states that this solution is unique, provided that precisely n independentconditions are specified. Theseconditions most commonly comprise values at the initial time: x(0),2(0),..., d’~-lx/dt’~-~It=o. 6.1.1

The

Homogeneous

Solution

The homogeneous solution is the complete solution whenthere is no excitation f(t). This solution contains undeterminedcoefficients equal in numberto the order of the system. Their values are determined by an equal numberof conditions that must be specified to make the solution unique. Most commonly, these conditions are specified at time t = 0 and are called initial conditions. The most common initial conditions specified are the values of x and its first n - 1 derivatives. If all these conditions are zero, all the coefficients and the solution are zero. Whenthe excitation f(t) is non-zero, the homogeneoussolution forms a part of the complete solution, as indicated by Equation (6.2). In this case the undeterminedcoefficients dependpartly on the excitation, and usually are non-zeroevenif all the initial conditions are zero. The homogeneousdifferential equation can be written in an algebraic form through the use of the operator notation introduced in Section 5.3.3 (p. 345): (a~S~ + a~_~S~-~ + a,~_~S~-~ +... + a~S + aO)Xh

=

0.

(6.3)

Since the general Xh ~ O, the following characteristic equation results: ~n an

+ a~_lS~-~ + a~_~S~-2 + ... + a~S + ao = O. I

(6.4)

This is an nth order polynomialin S whichpossesses n solutions, or roots, that are labeled here as S1, S~,..., S~. It is necessaryto find these roots, regardless of whetheryou use the direct solution techniqueor a fancier transformtechnique.

6.1.

DIRECT SOLUTIONS OF LINEAR DIFFERENTIAL

EQS.

391

The homogeneous solution comprises a sum of terms associated with the individual roots. A real unrepeated root Si contributes the term cie s~t to this solution, where ci is an undetermined coefficient. The simplest example is the first-order differential equation, for which the only root is $1 = -ao/al =- -1/~-, giving IXh

= ce-t/’.J

(6.5)

This solution is consistent with equation (3.38) (p. 146). Real roots $1 and of a second-order equation give the solution Xh = cle -t/~l -t + ITs, c2e

T1 ~ -1/S~;

~ ~ -1/S~,

(6.6)

which is consistent with equation (3.48) (p. 153). Recall that the symbol called a time constant; while Si h~ the dimension of inverse time, its negative reciprocal Ti has the more palpable dimension of time. Second and higher-order charateristic equations often have complex roots, which come in complex-conjugate pairs. If the second-order characteristic equation is written with the familiar notation

(s

Is +

(6.7)

=0,[

its roots are, when the dampin~ratio ~ < 0, 8~,2 = -¢wn ~ j~ - ~2Wn = -a ¯ jwd. The relations

(6.8)

between the two forms of the coe~cients can be seen to be

The homogeneous solution

becomes

Xh = c~e(-a+j~")~ (+ a-j~")~. c~e

(6.10)

To be meaningful physically, this solution must be real. This happens ff and only if the coefficients c~ and c~ are also complex conjugate: 1

(6.~)

c~,:= ~(c~¯ ~c~). With the use of the identities ej" ~ cos~ + j sin~, e-i~ ~ cosa - j sin~,

(6.12a) (6.12b)

there results [Xh = e-at(C~COSW~t-- c, sinwat),

~ < 1 or w, > 0.]

The result agrees with equation (3.44) (p. 151). Alternative with conceptual and plotting adv~tages are Xh = ce-at COS(Wdt+ a) = -at si n(walt + ~)

(6.13)

equivalent forms

(6.14)

392

CHAPTER 6.

ANALYSIS

OF LINEAR MODELS, PART 1

A multiple real root of the characteristic equation, S~, repeated m times, contributes to the homogeneoussolution the terms (c~o + c~lt + cet 2 slt. +... + C~mt’~)e The example of a second-order equation with two equal roots $1,2 = -w~ (corresponding to a = -w~, for which Wd = 0 and ~ = 1) gives Xh = (c~ + -~"~, c~t)e

(6.15)

which corresponds to equation (3.47) (p. 151). A repeated pair of complex roots (which doesn’t happen very often) produces a combined contribution of the form [c~0 sin(wrist + ~o) + C~ltsin(wd~t + ~) +’’ -at, similar to the case with repeated real roots. EXAMPLE 6.1 Find the homogeneou.ssolution of the fifth-order

differential

equation

d2x d3x d5x 9 d4x dx 24 dt-- [ + -~ + 43-~-f + 71~-~ dt - 2. 100x = 2 + 3t Solution: The characteristic

equation is

S5 +9S 4 +43S3 + 71S ~- 24S- 100 = 0. No general analytic solution exists for a fifth-order polynomial; resort to a numerical solution is necessary. To solve by MATLAB, you enter and receive noindent>>

roots(J1

9 43 71 -24 -100])

-3.0000 +4.0000i -3.0000 -4.0000i -2. 0000 -2. 0000 i. 0000 TherootS = 1.0000contributes thetermc~e~ to the homogeneous solution. Since this root is positive, Xh --~ oe as t --~ oo, and the model is unstable. The pair of roots S = -2.0000 contributes the terms (c2 + c3t)e -2t, and the pair of roots S = -3.0000 =h 4.0000i contributes (c4 sin 4t + c5 cos -3t 4t)e which.also can be written as c~ sin(4t +/~)e -3~. Collecting these terms, the homogeneoussolution can be written in either of the equivalent forms Xh = c~et + (c2 + c3t)e -2t + c4e-3t sin(4t + ~), Xh = c~et + (c2 + c3t)e -~t + e-~t(c4 sin4t + c5 cos4t), which have five’undetermined coefficients.

6.1.

393

DIRECT SOLUTIONS OF LINEAR DIFFERENTIAL EQS.

0

time

Figure 6.1: Unit step waveform

area = 1 lIT

area = 1

area

0

0

t

0

t

Figure 6.2: Unit impulse waveform A methodfor determining the undeterminedcoefficients is given in Section 6.1.6, followingthe discussion regarding the particular solution. Shouldthere be no excitation or input disturbance, the proceduredescribed there can be applied directly. If there is an input, however,the determinationof the coefficients must be deferred until the particular solution is addedto the homogeneous solution. 6.1.2

Singularity

System

Inputs

A family of singularity functions that is zero for all time t < 0 is particularly useful. The unit step function us(t), pictured in Fig. 6.1, is defined as u~ (t)

for t < 0 {01 for t > 0

(6.16)

Step changes of other amplitudes or dimensions can be represented by multiplying the unit step by a constant. The unit impulse function 5(0 approximates a sharp pulse disturbance. As pictured in Fig. 6.2, it is defined as the limit of the square pulse of width T and integral (area) 1 as T -~

394

CHAPTER 6. ANALYSIS OF LINEAR MODELS, PART 1

0

time

Figure 6.3: Unit ramp waveform

~(t)= limiT(t); T-,o

~r(t)=

0 for t < 0 lIT O 0.

(6.17)

The time integral of the unit impulse is the unit step: (6.18)

’_ ¢J(t) dt =us(t).

Conversely,the time derivative of the unit step can be considered to be the unit impulse, although discontinuous functions are not differentiable in the formal sense: 5(t) = dus dt

(6.19)

The unit rampfunction uT(t), pictured in Fig. 6.3, is defined as increasing linearly with derivative (slope) 1 from the origin: t <0 ur = t0 for fort>0

(6.20)

The rampis the integral of the unit step, and the unit step is the derivative of the ramp: ur = us(t)

dr; u,(t)

= --~.

(6.21)

OO

A singularity can be located at a time t = t* different from t = 0 by writing t - t* in place of t (or t - 0) as the argumentof the singularity function. exampleis shownin Fig. 6.4. This allows functions composedof a series of steps, rampsand impulses to be represented as a sum of singularity functions. The property of superposition then can be invokedto represent the response of a particular systemto such a function as the sumof the responses of the individual singularity functions. This possibility considerably enhancesthe usefulness of singularity functions.

6.1.

DIRECT SOLUTIONS OF LINEAR DIFFERENTIAL EQS.

395

0.5 ur(t) 1.5 u,(t-l) - 2 6(t-2) - 0.5 ur(t-2)

0

1

2"".,. 3 4 ~",, -0.5 ur(t-2)

Figure 6.4: Time-shifted singularites and waveformsynthesized therefrom 2 r plots u/c

l 1

0

~

~-

of .....

e ~’

~

~~.~,.~’~=0.05 s=O

]~/--

s=-O.05

0

2

s 4

=-0.4 6

8

l

10

~

12

Figure 6.5: Real exponential waveforms 6.1.3

Exponential

and Sinusoidal Inputs The exponential input u(t) = st occurs naturally as theoutput of a fir st order process and as componentsof the outputs of higher-order processes. The coefficient s can be real, imaginaryor complex. Positive real values of s produce functions which grow "exponentially" in time, whereasnegative real values of s producefunctions which decay exponentially, as indicated in Fig. 6.5. Complexvalues of s and c must appear in complexconjugate pairs to produce a real function u(t), just as the complexconjugate pairs S and c are required to contribute real componentsto the homogeneous solution of equation (6.13). Therelation c1,~ = ½(cr =l: jci) will be assumed,as before. The special case of pure imaginary values of s, written as s = =l=jw,gives the sinusoidal waveform u(t) = cr cos~t - ci sin wt,

(6.22)

which is plotted in Fig. 6.6. The more general case of complexconjugate pairs

396

CHAPTER6. ANALYSIS OF LINEAR MODELS, PART 1 cr cosag - cl sin ag

Figure 6.6: Sinusoidal waveform "--.. envelopeae """-’---.._ e~ (cr cos a~t - qsin 6)0

~

Figure 6.7i Waveformfrom complex conjugate pair s = a =l: jw gives 1 1 u(t) = -~(cr + jci)e (~+~)t + ~(c~ - jci)e =eat (cr cos wt - c~sin wt).

(6.23)

This waveformcan be viewedas a sinusoid oscillating at frequency w within an envelope of amplitude ~eat, as shown in Fig. 6.7. The excitations or inputs to engineering systems frequently are assumedto be sinusoidal, particularly whenthe concern is vibrations. Virtual sinusoids with amplitudes that grow or decay exponentially in time also are important, since they often emerge from commonsystems with singularity inputs to becomethe inputs to other systems. 6.1.4

Power-Law

Inputs

Occasionally a memberof the power-lawfamily u = ct n, where n is a positive integer, is used as an input to a system. Power-lawfactors occur naturally in the outputs of modelswith repeated roots, as noted above.

6.1. 6.1.5

DIRECT SOLUTIONS OF LINEAR DIFFERENTIAL EQS. The Method

of Undetermined

397

Coefficients

This is the classical methodfor finding the particular solutions to linear ordinary differential equations with the various types of inputs f(t) detailed above. The particular solution can be any solution to the completedifferential equation; it might as well be the simplest possible solution. The forms of the terms in the particular solution that correspondto the various input terms are given in Table 6.1. All that needsto be done is to affix initially undeterminedcoefficients to each of the applicable terms, substutite themfor Xp in the differential equation, and evaluate the coefficients by satisfying the resulting equationfor all times, t.

Table 6.1 term in f(t) 1 st e nt

Method of Undetermined Coefficients corresponding terms in Xp c st ve

Cntn q- Cn-1tn-1 ~- ¯ " ¯ -~ Clt -~" CO

sin wt or cos wt or cos(~t + #) ca sin wt + c2 cos wt or c cos(~vt + # +

The particular solution for the step input can be seen to be xr = constant for t > 0. The exponential input u(t) = uoest gives the form f(t) = foest. This correspondsto the secondentry in Table 6.1; the particular solution is of the form xp(t) = cu0est. Carrying out the methodof undeterminedcoefficients by substituting these into the differential equation (equation (6.1)) and performing indicated differentiations, (an8 n q- an-18 n-1 -~’"

~t = (bins m + bm-lsra-1 +.’. + bo)uoest. (6.24) +ao)cuoe

Solving for c and thus zp(t), c =

binsmq- bin-1sin-1 + "’" +bo an8 n q- an-18 n-1 + " ¯ " ¯ ao

"t. [ [xp(t) = G(s)uoe

= G(s),

(6.25a)

(6.25b)

Thusc equals the sametransfer function G(s) as given by the operator notatiort of equation (5.45) in Section5.3.3 (p. 345) to represent the general linear differential equation with constant coefficients, except that the original argumentand derivative operator S is replaced by the exponential coefficient s. This fact is very useful; it is employedbelowextensively, particularly for sinusoidal inputs.

398

CHAPTER 6. EXAMPLE 6.2 Find the particular 6.1.

ANALYSIS

OF LINEAR MODELS, PART 1

solution to the differential

equation given in Example

Solution: The terms on the right side of the example, namely f (t) = 2+ ~, are of the power-law form t ’~, with n = 0 and n = 2. Table 6.1 gives the form of the particular solution as follows: Xp -- c2t2 + Clt + Co. Substitution of this solution into the differential

equation gives

71(2c2) - 24(2c2t + cl) - 100(c2t 2 + 2, clt + co) = 2 + 3t from which -100c2 = 3;

-48c2 - 100ct = 0; 142c2 - 24c~ - 100co = 2.

The result is x~ = -O.03t 2 + 0.0144t - 0.046056.

6.1.6

Application

of Initial

Conditions

The undetermined coefficients in the homogeneoussolution must be found after the complete solution is formed by adding the particular solution, if it is nonzero. The particular solution itself should at this point have no undetermined coefficients. The undetermined coefficients in the homogeneouspart of the solution are resolved through the introduction of the same number of independent conditions. In the most commoncase these conditions are given at t -- 0 and therefore are called initial conditions. This case is illustrated below. Whenone or more conditions are specified at a different time, such as a final time, the procedure is essentially the same. EXAMPLE 6.3 Solve the following second-order differential conditions: d~ x 2 dx dt ~+ dt +Sx=3(1-e-2t); Solution: The characteristic

equation for the given initial

x(0)=0;

5(0)=0.

equation and its roots are

S~ + 2S+ 5 = (S + 1) 2 + (2) 2 = 0; $1,2 = -1 =t:j2, so that the homogeneoussolution is Xh = e-t(c~ cos2t + c2 sin2t).

6.1.

DIRECT SOLUTIONS OF LINEAR DIFFERENTIAL EQS.

399

Treating the input as u(t) = ul (t)+ u2(t) with ul (t) u2(t ) = -e -2~, the particular solution can be found from equation (6.25b) as follows: xp = 3G(0) - 3G(-2)e -2~ - 3 2 3 e_2~ = 0.6(1 - e-2~). 5 s + 2s + 5 8=-2 Note that ul is in the proper exponential form with s = 0, giving simply G(0) = 0.6 for its response. Thus the complete solution and its time derivative are -2~. x =Xh + Xp = e-~(Cl cos2t -+ (2cl c2 sin2t) + 0.6(1 e-2~), d__~_x = e- Cl)cos2~ - c2)sin 2t] + -~ [(2c2 dt Substitution of the initial conditions into these equations specifies the coefficients Cl and c2: 0=c~+0 0=2c2-c~+1.2

or c1=0 or c2=-0.6.

Substitution of these values gives the unique result x = 0.6(1 - e-2t - e-t sin 2t). This is plotted below, along with its homogeneousand particular components. Notethat the given initial conditions are satisfied, and that virtually any other initial conditions also could be satisfied by different weightingsof the two components. 0.8 0.4

-0.4

6.1.7

0

Solutions

1 to Impulse

2

3

4

t

5

6

Inputs

The impulse function f(t) = 6(t) is not included in Table 6.1, since it acts in a special way.The entire excitation lasts but an instant; it is useful to say that its duration is betweentime t = 0- and time ~ = 0+. The excitation is zero for t > 0+, which meansthat the total solution has identically the same form as

400

CHAPTER6. ANALYSIS OF LINEAR MODELS, PART 1

the unforced or homogeneous solution. The effect of the impulse is to impart an initial condition for this particular solution at time t = 0+ that is different from the condition at time t = 0-. The jump x(0+)-x(0-) can be determinedby integrating the full differential equation betweenthe times t = 0- and t = 0+. For the first-order case, dx/dt+ (1/T)X = C~(t), this integration is

_ dt +rJo-

xdt= c~I(t)dt,

(6.26)

= c.

(6.27)

which gives + T

Note the use of the fact that the integral across a unit impulseis 1. Theintegral term in equation (6.27) must be zero unless x were infinite over the vanishingly small interval betweenthe limits of integration, whichit is not. Since x(0-) is by definition zero, the result is x(0+) = c, whichgives -t x /r = ,ce

t > +. 0

(6.28)

For the nth-order case,

° Onlythe highest order derivative contributes to the left side of this equation, with the result that the impulse imparts a non-zero value to only this term:

dt,~_~

= c.

(6.30)

EXAMPLE 6.4 Find the response of the general underdampedsecond-order model to an impulseof amplitudec, for both general initial conditions and for zero initial conditions. Solution: The complete solution (for t >_ +) i s t he s ame as t he homogeneous solution, as given by equation (6.13) and repeated here: x =e-at (cr cos Wdt-- ci sin Wdt), This gives

x(0+) =or ~(0+) = -acr - ~dC~

+ t . >_0

6.1.

401

DIRECT SOLUTIONS OF LINEAR DIFFERENTIAL EQS.

Since only the highest-order derivative changes across the impulse, x(0+) -- x(0-). Fromequation (6.30), +) - ~(0-) -- c. Therefore,

c~ = x(0-), -~[ax(0-) + ~(0-) Should the initial conditions x(0-) and ~(0-) be zero, the solution plifies to x = (1/w4)e-a~ sin w4t.

6.1.8 Differentiation

and Integration Properties

If an excitation u2(t) of a systemis the time derivative of an excitation ul(t) to the samesystem, then the response x2 (t) is the time derivative of the response x~(t). That is, dul if u2=--~- then x2----’dr (6.31) Thus, for example, the response of a given model to a unit impulse input, traditionally written 9(t), equals the time derivative of the reponse to a unit step, traditionally written h(t), since the impulse is the time derivative of the step; 9(t) = dh($)/d~. Inversely, if the excitation u2(~) of a systemis the time integral of the excitation u~(t) of the samesystem, the response x2(t) is the time integral of the responsex~ ($), if u2=/Ul(t)dt

then x2(t)=Ixl(t)dt.

(6.32)

Thus, similarly, the response of a given modelto a unit step input equals the time integral of the responseto a unit impulse; h(t) = fg(t) Theseproperties allow you to extend a knownsolution to certain other cases. Rather than go through the analysis above to find the response to an impulse, for example, you have the option of finding the response to a step, and then computingits derivative, taking care not to overlookany discontinuities.

6.1.9

Step and Impulse

Responses

Using MATLAB

MATLAB automatically computes the step and impulse responses of linear models. Youenter the coefficients b,n,’ "’, bo of the numeratorpolynomialand the coefficients an," ¯, a0 of the denominatorpolynomial, and then simply call the routine of interest. The programchooses its own(default) scales for the axes (which you can override). EXAMPLE 6.5 Using MATLAB, plot the unit impulse and step responses of the model d3x

d2x dx dt a + 3~- + 102~ + 100x =50du dt + 200.

402

CHAPTER6. ANALYSIS OF LINEAR MODELS, PART 1 Solution: The transfer function is 50S + 200 G(S) = $3 3S2 + 102S + 100" To get the plots below, you enter >>nam= [50 200]; >>den = [1 3 102 100]; >>step(num, den) >>±mpulse(num,den) ~r~

2

1.5

1

0

0

0.5

1

1.5

2

O

O.S

1

1.5

2

2.5

3

3.5

4

2.5

3

3.5

4

;;me (ram)

4.5

4.5

6.1.

DIRECT SOLUTIONS OF LINEAR DIFFERENTIAL EQS.

403

Pressing the enter key will erase the present plot, unless you enter hold on, whichwill superimposesubsequent plots. Further options to control the details of the plots and secure data files for further use are available. If you substitute [y, x, t] =step (num, den) or [y, x,t] =impulse(num, den), the times used stored in a vector t, the output values are stored in a vector y and the state variables chosenby the routine are stored in a matrix x. Valuesof t specified in advancewill be used if t is used as the third argumentof the command, i.e. step (num, den, t). Tospecify 201values of time evenly distributed over ten seconds, for example, you enter either t=linspace (0,10,201) or t= [0 :. 05 : i0]. 6.1.10

Summary

Theclassical representation for the solution of a linear differential equationwith constant coefficients comprisesthe general solution to the homogeneous equation (with zero right side) plus any particular solution to the completeequation. The homogeneous solution is found using the roots of the characteristic polynomial formedfrom the coefficients of the various terms. The particular solution can be found, for the cases of primary interest, using the methodof undetermined coefficients. The final step in finding the unique solution to a particular input is the determination of the coefficients of the homogeneous part of the solution throughthe use of the initial conditions. Singularity waveformsincluding impulses, steps and ramps, and exponential and power-lawwaveformshave been featured as simple system inputs with readily calculable particular solutions. Thesewaveformsare particularly useful, because they can be combinedto represent muchmore complicated waveforms. Carrying out this promiseis the subject of most of the remainderof this chapter and Chapter 8. The first case addressed in detail, in the following section, is that of sinusoidal excitations. Guided Problem 6.1 This straightforward problemreviews the classical methodfor solving linear differential equations. It also illustrates an important phenomenon. Find and plot the solution x(t) to the followingdifferential equation, subject to the initial conditions x(0) = 0, ~(0) i d2x ( 0 w-~dt--~ + x =~ sin(0.95cant)

"t <0 t > O.

Suggested Steps: 1. Note that the equation can be simplified by employingt ~ = wnt rather than t as the independentvariable. 2. Write the characteristic equation and find its roots. Use these to write the homogeneoussolution with its two unknownconstants.

404

CHAPTER 6. ANALYSIS OF LINEAR MODELS, PART 1

3. Use the methodof undeterminedcoefficients as summarizedin Table 6.1 (p. 397) to find the particular solution. 4. Substitute the initial conditions into the total solution to evaluate the unknownconstants. 5. Plot the results using MATLAB or someother software package~ The result demonstrates a "beating" phenomenonbetween motion at the natural frequencyand motionat the slightly different forcing frequency, which is perpetuated because of the absence of damping. PROBLEMS {}.1 Solve the followingdifferential equation with the initial condition x(0) = and the forcing function u(t) 10. Sketch-plot yo ur answer.

d’-~ +5x = u(t) 6.2 Answerthe precedin~question substituting u(t) 10sin2t. 6.3 Answerthe preceding question substituting u(t) 10(1 6.4 Solve the followingdifferential equation with the initial conditions x(0) = and 5(0) = 0. Sketch-plot your answer. d2x 4dx dt ~ + dt -b 8x = lO 6.5 Solve the preceding problemwith the coefficient 4 replaced by 6. 6.6 Solve the preceding problem retaining the coefficient 6 but changing the coefficent8 to 9. 6.7 For the bond-graph model below, Se

e---£-~

I

1 ~ 0 ~R

C

(a) Definestate variables and find a set of state differential equations. (b) Find the step response of the displacement on the compliance for e = eous(t), wheree0 is a constant.

6.1.

DIRECT SOLUTIONS OF LINEAR DIFFERENTIAL EQS.

405

(c) Find the responsefor e = e0 sin wt valid for t 6.8 A particular system responds to a step excitation with a ramp response. Characterize the dynamicsof the system generally. {1.9 The excitation u(t) and impulseresponse g(t) of a linear systemare plotted below.

(a) Carefully sketch and label the step response, h(t), of the system. (b) Representthe input u(t) as sum ofsteps. (c) Sketchthe response of the systemas the superposition of the responses to the individual steps in part (b). 6.10 An input to a system is as plotted below: u(t)21

/ I

0

1

2

I

t

3

(a) Representthis waveformas a sumof singularity functions. (b) Find the response to the first-order system dx/dt + x = u(t). 6.11 Carry out the above problem for the input plotted below:

I I

0

I

2

4

t

6

406

CHAPTER6. ANALYSIS OF LINEAR MODELS, PART 1

{1.12 A spring pushes a piston whichforces a viscous liquid through a tube, as shown.

(a) Find the effective resistance of the systemwith respect to the motion of the piston, in terms of the symbolsgiven. Assumefully developedflow. (b) Find the time constant for the motion. Evaluate for k = 50 lb/in., dl = 1 in., d2 = 0.05 in., L = 40 in. and # = 2.9 x 10-6 lb.s/in 2 (water at 70°F). 6.13 The combination of two springs and a dashpot shownbelow are hit with an impulse (perhaps from a hammer)with f F dt = Z lb.s. Find the resulting motion. k k/ F(t)

~~x

6.14 Find F(t) for the system shownbelow with x = xous(t).

6.15 One way to measure viscosity is to observe the velocity v at which small spheres settle in still fluid. The terminal velocity sometimesis not reached quickly, however,due to the inertia of the sphere and someof the surrounding fluid. 3Consider a sphere of radius r = 0.13 cm and density Ps = 7840 kg/m (steel) settling in a fluid of density p = 1200 kg/m3 and viscosity # = 0.20 N.s/m2. AssumeStokes law, which gives a force equal to 67r#rv and applies with considerable accuracy for Reynold’s numberNR= 2pvr/# < 2 and can be adequate for 2 < NR< 10.

6.1.

407

DIRECT SOLUTIONS OF LINEAR DIFFERENTIAL EQS. (a) Find the resistance with respect ot the velocity

(b) Find the effective inertance, which includes the mass of the sphere plus the "virtual mass" of the fluid that also is accelerated. The latter approximately equals the mass of fluid that occupies a volumeequal to one-half that of the sphere. (c) Find the time constant, the velocity v(t) starting from rest, and the time at whichv reaches 95%of the terminal velocity. Checkthe Reynold’s numberto makesure Stokes relation applies. (d) Find the correspondingdistance.

{1.16 A penstock such as in Problem3.52 (p. 160), but smaller, is terminated in a reverse check valve such as given in Problem4.57 (p. 275) to form what knownas a hydraulic ram. A drawingis shownbelow. Water accelerates through the long tube until its velocity is so great that the valve slams shut, whereupon the pressure rises and opens a simple check valve to a load tank with a constant high pressure PL, pumpingwater to that tank. After the flow has stopped, the check valve closes to prevent downward flow, the reverse check valve opens automatically, and the cycle repeats. Water passing through the reverse check valve is dumpedinto a stream at that level. The device is attractive whenno electrical poweris available and a cheap, simple systemis preferred.

PL= pghL check valve

Find the quantity of water pumpedto the load tank in one cycle, neglecting losses, in terms of the parameters given aboveand in Problem3.52. Also, find the cycle time. 6.17 Guided Problem5.5 (pp. 329-331) finds nonlinear models for a groundeffect machine(GEM)with roll stiffness. Linearize the modelwith the actual compliancesfor roll motion. Determinethe characteristic values numerically, using MATLAB. The equilibrium values given in step 9 of the solution maybe used.

408

CHAPTER

SOLUTION Guided

OF

Problem

1. With t’

6.

ANALYSIS

GUIDED

PART 1

6.1 d2x dt,~ -~- x = 0 t’ < 0

= w~t,

The resulting

t’ >

equation is s 2 ÷ 1 = 0, which has roots sl,2 -- =l=j.

The characteristic

3. Since f(t)

MODELS,

PROBLEM

= sin(0.95t’) 2.

OF LINEAR

homogeneous solution

= sin(0.95t’),

is xh = csin(t’

xp Xpl si n(0.95t’) +

+ ¢).

xp2 cos(0.95t’),

d2x~ = -(0.95)2[x~1 sin(0.95t’)

÷ x~ cos(0.95t’)].

Substituting into the differential equation, [-(0.95) 2 ÷ 1][sin(0.95t’) ÷ x~2 cos(0.95t’)] -- sin(0.95t’). 1 2 = 10.256, Therefore, xp2 = 0, xpl = 1 - (0.95) giving x = csin(t’ + c~) 10.256 si n(0.95t’). d2 x 4. dt,~ -- csin(t’

÷ ¢) - 9.256sin(0.95t’).

At t’ -- 0, x -- 0 = csin¢; therefore, ¢ = 0. dx At t’ = 0+, ~- = 0 -- c cos t’ ÷ 9.744 cos(0.95t’) therefore,

= c ÷ 9.744,

c -- -9.744, giving x -- -9.744 sin t ~ ÷ 10.256 sin(0.95t’).

X/Xol©

-2O

6.2

Sinusoidal

Frequency Response

Periodic excitations of natural and engineered systems abound. Aerodynamic and water wave forces vibrate bodies, acoustic signals comprise music and speech, and rotational imbalance shakes machines. The simplest representation of excitations such as these is sinusoidal. It will be shown in Chapter 7

6.2.

409

SINUSOIDAL FREQUENCYRESPONSE

that any periodic waveformcan be decomposedinto a sum of discrete sinusoidal oscillations at multiples of the base frequency, and any waveformcan be decomposedinto an infinite sumof sinusoidal oscillations. "Frequencydomain" methodsform the basis of the Fourier and Laplace transform methodsdeveloped in that chapter. Further, systems often are characterized by howthey respond to sinusoidal inputs. Physical experimentsare carried out using electrical function generators and mechanical shaker tables that sweepan excitation over a range of frequencies. This section is concernedwith the response of linear modelsto the sinusoidal excitation [u(t)

= u0 cos(wt + ~). J

(6.33)

The focus is on the particular solution, whichoften is called the steady-state response since it persits as long as the input persists. The homogeneous part of a complete solution, on the other hand, usually decays quickly. All that is neededis stability and a pinch of damping.Withoutstability, there is no point in studying any particular solution, which wouldquickly be overwhelmedby the exponential growth of the homogeneous solution. The tacit assumption is made below, therefore, that the systemis stable, whichmeansthat the real parts of all the roots of the characteristic equation are non-positive. 6.2.1

The

Phasor

Method

The sinusoidal excitation of equation (6.33) can be recast as the sum of two complexconjugate terms: u(t) = 1 [ej(~+~ ) ÷ e_jC~t+~)] ~uo

=

+

.

The equivalence of equations (6.33) and (6.34) follows from trigonometric identities (equations (6.12), p. 391). Eachof the terms is exponential, ~d therefore contributes a term of the form of Equation (6.25b) (p. 397) to the particular solution, with s = ~jw: xp(t)

= 1 [eJ~G(jw)e ~t -j + ~t] e-~ZG(-jw)e

(6.35)

G(jw) is called the frequency transfer function. The terms G(jw) and G(-jw) are complexconjugate, so that they can be written -~¢, G(j~) = ~G(j~)~e~¢; G(-jw) = ~G(j~)le

(6.36)

where ¢ ~ gG(jw) is called the phase angle of G(jw), which has as its tangent tan¢ = Im[G(jw)] Re[G(jw)]

(6.37)

410

CHAPTER 6. ANALYSIS OF LINEAR MODELS, PART 1 a =Im[G(]w)] b =Re[G(]w)]

~a>Oa>O[N~

b<~

>0 aR<eo

Re FORe ~Ol ~e a
fourth quadrant

Figure 6.8: Determination of the quadrant of G(jw)

Equation (6.35) becomes xp(t) = ~]G]uo[ej(~°t+~+~)+ e-J(~t+~+O]

(6.38)

or

xp(t) IG(jco)lUo cos(wt + ~3+ ¢), ¢(w) =/G(jw)

= tan -1

[Im[(G(jw)] [ Re[(G(jw)]j

(6.39)

Thus, x~(t) is sinusoidal with an amplitude IG(j~)I times the amplitude of the excitation; IG(jw)[ is called the amplitude ratio or magnitude ratio gain. The phase angle of x(t) is advancedrelative to that of u(t) by the angle ¢(w), knownas the phase shift. Most commmonly this angle is negative; is knownas the phase lag. Examples of the complex number G(jw) are drawn in Fig. 6.8 as vectors or phasors in the complexplane. It is essential that ¢ be recognized in its proper quadrant. For example, if both ~3[G(jw)] and ~[G(jw)] are negative, ¢ lies in the third quadrant, or 180° < ¢ < 270°. Youare urged to find the quadrant of phasors by first determining the signs of the both terms, and then sketching the phasor. Be warned that manysimple calculator and computer routines automatically assumethe first or fourth quadrant. EXAMPLE 6.6 Find the steady-state response of the modelwith the transfer function 50S + 200 G(S) = $3 + 3S2 + 102S+ 100 to the excitation u(t) = 2 sin(10t).

6.2.

411

SINUSOIDAL FREQUENCYRESPONSE Solution: At the specified frequency, 50(j10) 200 2++ IG(jl0)I = (jl0) 3 + 3(j10) 102(j10") + 100 I I

200+500j = ](100 - 300)+2(1020-

1000)I/(200)2+(500)2 = V ~-~-~-U¢~ = 2.679

¢ = tan-1 \200] - tan-1 = 68.20° - 174.29° °= -106.09 The particular solution, therefore, is xp(t) = 2 x 2.679sin(lOt- 106.09°) = 5.358 sin(10t- 106.09°). Note that the numeratorof G(jl0) is a phasor in the first quadrant, whereas the denominator is a phasor in the second quadrant. The phase angle of G(10j) equals the angle for the numerator minus the angle for the denominator. Plots of u(t) and x(t) are given below. °106.09

6 4 2 0 -2

0

6.2.2

0.2

0.4

0.6

t

0.8

1.0

Bode Plots

Magnituderatios and phase angles expressed as functions of frequency serve to characterize a linear transfer function completely. Plots of the magnituderatio or gain and the phase angle as functions of frequencyconveyto the analyst not only the meaningof a frequencytransfer function, but also the very character of the dynamicmodel, in a clear, visual way. The information can be deducedfrom direct experimenton the physical system, with no use of modelsor mathematics; the response to sinusoidal excitation at manydifferent frequencies is observed and plotted. A shaker table ofter is used to producethe sinusoidal disturbances on mechanical systems. Both linear and logarithmic axes have been used widely for the plots. An exampleusing logarithmic axes for the frequency and the magnituderatio and

412

CHAPTER 6.

0.1

ANALYSIS

OF LINEAR MODELS, PART 1

1.0 10 o), rad/s

100

Figure 6.9: Example Bode plot

a linear axis for the phase angle is given in part (a) of Fig. 6.9. Such a representation is known as a Bode plot. The transfer function chosen is the same as in Example 6.6 above. You should check to see that the magnitude ratio and the phase angle as plotted at 10.rad/s agree with the calculated values. The particular Bode coordinates have become the standard for system analysts interested in dynamics and control. This is largely because they allow a model and its transfer function to be decomposed into components and reassembled graphically, as you will see later. Further, the use of the logarithmic axes allows small but important amplitudes to be seen clearly despite the presence of vastly larger amplitudes at other frequencies. Finally, widely available software, including MATLAB,employs Bode coordinates. The decibel scale has become a conventional representation of the logarithmic magnitude ratio. If the magnitude ratio is m, the decibel scale, abbreviated "db," is ~ " [ db -- 20 log lo m]

(6.40)

Both the m and the db scales are shown in part (a)of the figure. 1Some older references use 10 log10 m. The use of the multiplier 20 rather than the more natural 10 appears to be a carryover from the prior and continuing use of decibels to represent the power of an acoustic wave or an acoustic field. Power is proportional to the square of the sound pressure (or the product of pressure and velocity, as we have seen), and the logarithm of a square gives a multiplier of 2.

6.2.

413

SINUSOIDAL FREQUENCYRESPONSE EXAMPLE 6.7 Use MATLAB to produce a Bode plot for the model of Example 6.6. Solution: >> hum = [50 200]; >> den = [1 3 102 100]; >> bode(num,den) Compare theresulting plotwiththatofFig.6.9:

0 10 Frequency (rad/sec)

"1 10

0 10 Frequency (rad/sec)

101

= 10

The alternate command[mag,phase,w] = bode (hum, den) directs that the calculated values of frequency, magnitudeand phase be stored in the vectors w, magand phase, respectively. It also lets you specify the frequencies. For example,to specify 200 values of frequency with even logarithmic spacing over the band from 1 to 40 rad/s, you enter w=logspace(1,40,200). 6.2.3

Models

Without

Damping

Modelswithout dampinghave only even-ordered derivatives in equaiion (6.1) (p 390), that is a~ = 0 and b~ = 0 for i = 1,3,5,.... As a result, G(jw) is real, no phase angle is needed, and equation (6.39) is simplified xp(t) = G(jw) u(t). (6.41) Note that G(jw) can be negative, °. which corresponds to a phase angle of +180 At any particular frequency, the response is said to be "in-phase" or "out-ofphase" with the excitation. The second-order IC modelis a simple example. The equation 1 d2x 1 (6.42) w--~ dt --~ + x = Uo cos(wt + 8); wn = --~,

414

CHAPTER6. ANALYSIS OF LINEAR MODELS, PART 1

gives

a(S) 1 +

(6.43)

Equatioa (6.41) yields the solution Uo ~ zP = 1 - (u/un)

cos(wt+ fl).

(6.44)

Note that xp(t) is in-phase with u(t) for (w/wn) < 1 and (180°) out-of-phase for w/~,~) > 1. EXAMPLE 6.8 A machinecontains an unbalanced rotor whichproduces a vertical component of excitation force Fe = -ml (d2/dt2)[r cos wt]. The machineis constrained to movevertically, and is supported by a spring with rate k in order to reduce the vertical force F transmitted to the floor; such a force promotes undesired noise and vibration. Give a Bodeplot for F normalized to (divided by) the fixed force mlrw~n, where wnis the natural frequency. Also, give a second Bodeplot for F/F~.

Solution: A bond graph for this system,

gives the state differential equations ~ dPdt mlrw coswt - ~x, dx 1 These equations can be combinedas follows: ICd2x 2dt = cd. p.

= Cmlrw 2 coswt

-

x.

6.2.

SINUSOIDAL

415

FREQUENCY RESPONSE

With uo -- 1, this gives 2 rw ~ xp- 1 Cml - (~/~n) cos wt.

The force on the floor is F -~ Xp -- ~--

2 mlrw

cos

The solid line plotted below represents the nondimensional ratio of the amplitude of this force to mlrw~. It shows how the amplitude of the force varies with the applied frequency, w. As the frequency of the excitation is increased well above the resonant frequency, this ratio approaches unity.

normalized force on floor 0.1

o.(

0.I

1.0 :~/~o~10

The dashed line represents the ratio of the amplitude of the force on the floor to the excitation force mlrw2, which grows rapidly with frequency. As the excitation frequency is increased well above the natural frequency, or the natural frequency is reduced well below the excitation frequency~ this ratio rapidly decreases toward zero. At high frequency, the center of gravity of the masses ml and ms remains virtually fixed; the amplitude of the force on the floor therefore is proportional to the resulting fixed amplitude of x times k. The design objective therefore is to make k so small that the natural frequency is considerably below the actual excitation frequency. Whena shaft-driven machine that is operated aboye its natural or resonant frequency is started from rest, or is shut down, the resonant frequency is traversed. An unacceptable resonance can result. You may have noticed a brief vibration when you looked at an auto engine being started or stopped. The vibration can be minimized by passing the engine speed through the resonant frequency quickly.

416

CHAPTER 6.

ANALYSIS

OF LINEAR MODELS, PART 1

A second-order mechanical model, such as described above, is said to have one degree of freedom. This is because there is one geometrically independent position variable. A model with two degrees of freedom has two geometrically independent positions. In general, such a model would be fourth order, since there would be two momentumstate variables in addition to the two position state variables. A third-order mechanical model, then, still has two degrees of freedom, but is degenerate in the sense of having only one momentumvariable, because of a neglected inertance but a considered resistance or damping. Similarly, a first-order mechanical model still has a degree of freedom but also is missing an inertance. Models without damping, however, are always of even order, as the following example with two degrees of freedom demonstrates. EXAMPLE

6.9

A vibration absorber comprises an added mass and spring on the system modeled in Example 6.8. Find a transfer function between the input cos wt and the output force, made nondimensional by normalizing by rmlw~ as in Example 6.7. Show that this force vanishes at one frequency of excitation. Also, produce the associated magnitude Bode plot to show what happens for other excitation frequencies. To get a particular result, consider only the special case in whichthe natural frequencyc~n-- 1/x//x//x//x//x//x//x~ equals1/x/7"-2-~2, and the mass ratio 12/11 is 0.2.

z cos cot F~ = m~rco

me (~gca

Solution: From the bond graph of the idealized model with no damping, the state differential equations corresponding to the four energy storage elements are dx

1

6.2.

SINUSOIDAL

417

FREQUENCY RESPONSE

The dependent variable of primary interest is x; all others should be eliminated, to give a single fourth-order differential equation. This can be done various ways, with or without the use of operator notation, retaining the unspecified values for the parameters. None of these are nearly as easy as using the MATLAB function ss2tf, which requires specific values of the parameters. Because of the restrictions given, the following specific numbers can be assigned: I1 = 1, C1 = 1, I2 = 0.2, and C~ = 5. These give wn = 1 tad/s, so that regardless of the units of the individual parameters, the results can be interpreted with time considered as the nondimensional w,t and frequency considered as the nondimensional w/wn. With this interpretation, the results apply to any values of the parameters that satisfy the given restrictions. The matrices needed for ss2tf are

A=

0 -1 0

0 1 -0.2 0 0.2 0

c=[1000]; The MATLAB coding to get the transfer

¯ ’

B=

¯

’

D=0. function

from Fe to F is

A=[0 0 1 0;0 0 1 -5;-1 -.2 0 0;0 .2 0 0]; B=[0;0;1;0]; C=[1 0 0 03; D:0; [num, den] =ss2tf (A,B,C,D) The response is num= 0 0 1.0000 0 1.0000 den = 1.0000 0 ~.~000 0 1.0000 which means that $2+1 G(S) = S4 2. 2S~ + 1 -w 2 + 1 G(jw) = w4 _ 2.2w2 + 1 The force therefore vanishes when w = 1 (or in the general interpretation, when w/w~ = 1). This transfer function has as its input the force Fe = -rml w2 coswt, which increases in magnitude with frequency, whereas the requested Bode plot is to have the constant-amplitude -rmlw~ coswt as its input. Therefore, the G(S) needs to be multiplied by the ratio (w/w~) 2, which can be accomplished by multiplying the numerator by S2, changing it to S4 2. +S The MATLAB coding continues: num=[1 0 1 0 0]; bode(hum,den)

418

CHAPTER6. ANALYSIS OF LINEAR MODELS, PART 1 The resulting magnitudeplot, with someannotations, is given below. The existence of two resonant freqencies is revealed, one at 80%of Wnand the other at 125%.Therefore, the idea of the vibration absorber has merit only whenthe excitation frequency is fixed or nearly fixed. The two resonances can be spread further apaxt in frequency by increasing the mass I~ (and decreasing the complianceC2 correspondingly). Similarly, if the mass were reduced, the separation woulddecrease, and in the limit of zero mass the behavior (and the system) wouldbe identical to that of Example6.8. 4O

I

I

i

I I

I

I

i

I

I

I

I

I

special case for ~o~= ~ F 20

note:

decibels

ding segments

-20 ire °ut °f Phase I

-40 O.1

1.0

co/a4

I

I

I

I

I

10

The vibration absorber is analyzed in the alternative style of classical vibrations in AppendixB. This style somewhatsimplifies the analysis for the restricted class of modelsto whichit applies, makingit easier to retain unspecified parameters. 6.2.4

Models

Comprising

a Single

Pole

or Zero

The transfer function G(S) = is sai d to have a z er o at theorigin, sinc e G(0) = 0. The model G(S) = 1IS is said to have a pole at the origin, since it becomesinfinity whenS = 0. Bodeplots for these cases are given in Fig. 6.10. For the square magnitudeformat used, the slope of the magnitudecurve is +1 in the first case and -1 in the second case. (With the decibel format, it wouldbe ±20 db/decade.) The phase angle is a constant 90° in the first case and a constant -90° in the second case. Youshould verify that the substitution S ~ jw gives these results. Multiplyingthe transfer functions by a constant factor k, to give G(S) = and G(S) = k/S, would merely shift the magnitude plots up or down, leaving their slopes unchanged. The phase plots would be unchanged. The transfer function G(S) k(1 ± S/wz) is described as a zero at S =~:wz, since G(~Wz)=O. Bodeplots for this modelare given in Fig. 6.11 for the special case of k = 1. As before and for all transfer functions, a value of k different from zero simply shifts the magnitude curve up or downand leaves the phase

6.2.

419

SINUSOIDAL FREQUENCYRESPONSE 20"

1£

db 0 -2O

0.01 °90

0.1

1 10 ¢o, rad/s

100

G=S

OO -90°

G=I/S

Figure 6.10: Bodeplots for a pole or a zero at the origin curve unchanged. Theseplots follow from the substutution S = jw: logxo [G(jw)[ = log~o [1 + jw/wz[ = log~o V/1 + ~, (w/wz)

(6.45a) (6.45b)

,/G(jw) = +tan-l (~z)

Note that the magnitudeplot is the sameregardless of whether the plus or the minussign is used. Onthe other hand, the phase angle increases whenthe sign is plus, and decreaseswhenit is minus.This is a critical difference. The transfer function G(S) = k/(l+S/wp) is described as a pole at S = since G(-wp)-~ ~. Its Bodeplot, again for k = 1, is given in Fig. 6.12. Again, these plots follow from the substutution S = jw: I 1 loglo IG(jw)l = lOglo 1 + j-w/wp[ 1 _ = l°gx° ~/1 + 2(w/w~) / G,i(jw) = - tan-~ (-~ .

l°gl° V/1 + (w/w~)2’

(6.46a) (6.46b)

Note that only a plus sign is used in the transfer function, rather than the + sign used in the transfer function for the zero. The reason is that a minussign wouldrepresent an unstable system. As noted before, the particular solutions of unstable systemsare of no interest.

420

CHAPTER6. ANALYSIS OF LINEAR MODELS, PART 1

20 - 10 magnituderatio

db " 10-

0 - I.C O. ° 90 ° 50

,

asymptotes 1.0

10

phaseshift

-" left-half plane ° 30 - (+ slgn).~ linear approximation,, (- sign) ° .50 _900 Figure 6.11: Bodeplots for single real non-zero zeros

6.2.

SINUSOIDAL FREQUENCYRESPONSE

421

asymptotes 0 - 1.0 db "

magn~

:\/

........

-10 -

-20 - O. 1 O.

1.0

~- ..........

10

"" l~:ar approximatio~

Figure 6.12: Bodeplot for single real non-zero pole

The magnitudeplots for the modelswith the single pole and the single zero are top-to-bottom flips of one another. So are the phase angles, for the common plus-sign case. The magnitudeplots for the real non-zero poles and zeros can be approximatedby asymptotes, as shownin Figs. 6.11 and 6.12 by dashed lines. The low-frequency asymptote corresponds to the neglect of the term w/wz or W/Wprelative to 1; it produces a horizontal straight line. The high-frequency asymptotecorresponds to the exact opposite: neglecting the term 1 relative to the w/wz or w/wp. The high-frequency asymptote therefore is the same as a pole or zero at the origin, scaled so that it equals 1 at the break frequency w = w~ or w = Wp. The two asymptotes intersect at the break frequency. This fact enables you to closely approximatea break frequency if you are given the plot (e.g. experimental data) rather than the equation, a common situation for an engineer. The greatest departure of the actual magnitudefrom the closest asymptoteoccurs at the break frequency, where it equals the ratio v~ or 1/~/~. Asymptotesalso exist for the phase angles, corresponding to the neglect of the corresponding terms. The asymptote for low frequency is 0°, and for high frequency is +90°. The greatest departure between the actual value and the value of the nearest asymptoteis 45°, again at the break frequency. The phase asymptotes represent the actual phase more poorly than the magnitude asymptotes represent the actual magnitude, however. A muchbetter approximation for the frequency range 0.1wi < w < 10wi is a straight line (in the semi-log

422

CHAPTER 6. ANALYSIS OF LINEAR MODELS, PART 1

coordinates of this plot) betweenthe two asymptotesat the two respective frequencies a decade below and a decade above the break frequency. These linear approximationsare represented in Figs. 6.11 and 6.12 by dashedlines. 6.2.5

Models ros

Comprising

a Pair

of Complex

Poles

or Ze-

The modelwith the transfer function k 2G = 1 + (2~p/Wp)S + (S/wp)

(6.47)

has a pair of complex conjugate poles wheneverthe dampingratio ¢ < 1. The Bodeplot variables are

I

1

log10 IG(jw)[ lo g10 1 - (w/w,~) 2 + = -l°g~°

zc(j,,.,)=- tan-’

\w,~]

\ wn

]

Plots for various values of ~ are given in Fig. 6.13. Althoughcases with ~ _> 1 are included, these cases are better computedby factoring the transfer function into a product of two real poles, as described later. Lowand high-frequency asymptotes are drawn as dashed lines. As with the first-order poles, the low frequency asymptote corresponds to the neglect of all terms in any sum except those proportional to the lowest powerof w. This leaves simply a horizontal line with value 1, regardless of ¢. The highfrequency asymptote neglects all terms in any sum except those proportional to the highest powerof w. This leaves a straight line with a slope of -2 which intersects the low-frequencyasymptoteat the break frequency, which equals the natural frequencywnregardless of ~. The dampingratio ~ strongly affects the behavior only for frequencies near the break or natural frequency. For low dampingratios, the magnitudepeaks or resonates at approximatelythe break frequency. At precisely the break or natural frequency, the magnitude ratio equals 1/2~, which can be deduced simply from equation (6.48a). Use of this fact plus the asymptotesallows you to sketch the magnitudecarve for a particular dampingratio with adequate accuracy for most purposes. The phase curves are harder to sketch with accuracy (you will likely refer back to this plot for this purpose), but observe that the drop in phase totals 180° and is relatively abrupt for small dampingratios and gradual for large dampingratios. Since the case of ~ = 1 corresponds to the product of two equal real poles, the straight-line approximationused for real poles applies, with the total phase changebeing double the 90° for a single pole.

6.2.

423

SINUSOIDAL FREQ UENCY RESPONSE

-

~-0.1 mag

-2O

~ - 0.25

0.1 0.05 ,/ ,/ ~

-

o0

° -90

-4O

~ 1 +2 IS~ton+(S/a~n)

o 10

Figure 6.13: Bodeplots for second-order poles

1011800

424

CHAPTER 6.

ANALYSIS

OF LINEAR MODELS, PART 1

A system with a pair of complex-conjugate zeros rather than poles has a transfer function that is the reciprocal of the transfer function above. The logarithm of the magnitude of this transfer function therefore equals precisely minus the logarithm of the transfer function above; this relationship is the same as you saw before with first-order zeros. Consequently, the magnitude plot for the complex pair of zeros is exactly the top-to-bottom flip of the magnitude plot for the complex pair of poles. For this reason, and because such second-order zeros occur relatively infrequently, a separate plot is not given here. The phase angle plots for the zeros are similarly top-to-bottom flips of the phase angle plots for the corresponding poles; the phase increases by a total of 180° rather than decreases. On the other hand, the case of the zeros permits the damping ratio to be negative without causing an instability. In this case, the phase plot is identical to that of the pole with positive damping. 6.2.6

Factorization

of

Higher-Order

Models

Transfer functions of second and higher-order models are profitably follows: k(S - zl)(S z2)... (S -- Zm) G(S) -- SN(S _pl)(S_ P2)"" (S --pn_N)"

factored as (6.49)

The denominator of G(S) is the characterisitic polynomial, and the values Pl,’",Pn-N, known as the poles of the transfer function, are precisely the roots $1,. ¯., Sn-N of the characteristic equation. The integer N represents the number of additional poles with zero value (or at the origin), if any. The numerator is factored similarly; the values z~,..., Zm are the roots of the numerator polynomial, and are known as the zeros of the transfer function. Should there be one or more zeros at the origin, equation (6.49) still applies, but N becomes a negative integer. Individual poles and zeros can be real, imaginary or complex numbers. They often are represented in an S-plane plot, in which the abcissas is the real part and the ordinate is the imaginary part. Poles are indicated b~ crosses (×), and zeros by (0), as illustrated in Fig. 6.14. The pole labeled 1 in t~is exampleis real and negative. It contributes a term to the homogeneous solution proportional to e-2t; the minus sign in the exponent is associated with the presence of the pole in the left-half of the S-plane. Right-half-plane poles produce instability, but left-half-plane poles do not. The pole 2 is at the origin. It allows the homogeneoussolution to contain a constant (or c e°t). Imaginary and complex poles and zeros come in complex-conjugate pairs. The pair of poles labeled 3a and 3b in the example S-plane lie in the left-half plane; they contribute a term porportional to e-t sin(3t + f~) to the homogeneous solution. Again, their presence in the left-half plane is associated with the minus sign in the exponent and the stability of their contribution to the unforced behavior. The imaginary parts +3 give the damped natural frequency, Wd, and the real part -1 gives the exponent a = -~wn. The length of the chord from ¯the origin to either pole is the natural frequency wn = V~d

6.2.

SINUSOIDAL FREQUENCYRESPONSE

425

ImS

0 5b

-I

× 3b Figure 6.14: ExampleS-plane plot The zero labeled 4 is real and positive; those labeled 5a and 5b are complex with conjugatewith negative real parts. Zeros, unlike poles, can exist in either half plane without affecting stability. For most purposes it is convenient to replace the factors for a complexconjugate pair by its real product, as was done above for the underdamped second-order model. They are then described in terms of its dampednatural frequency and dampingratio. Also, the non-zero poles and zeros are interpreted in terms of postive frequencies w~i = m k l-I(1 ± s/~:~) II [1 + ~=~+1 G(S) = ~=~k n-N

(6.50)

s l-I(1 + Slo: ) H i--~l

i=k+l

The notation l-I~=l(1 + S/wzi) implies the product (1 ± S/w~i)(1 ± S/wz2) ¯ .. (1 ± S/wzj), with plus signs being used for left-half plane zeros and minus signs for right-half-plane zeros. The parameters z~,..., zj and p~,- ¯ .,p~ are real zeros and poles. The dampingratios ~zd+~,’",~zm and Cp,k+~,’’’, ~n-Nall represent subcritical behavior; they are less than 1. MATLAB will factor a transfer functipn with the numerator and denominator specified in terms of the polynomials numand den with the following command: [z,p,k]

= tf2zp(num,den)

The "tf2zp" can be read as the "transfer function to zero-pole" transformation. Should you start with the state-space format A, B, C, D, the command [z,p,k] --- ss2zp(A,B,C,D,~.)

426

CHAPTER 6.

ANALYSIS

OF LINEAR MODELS, PART 1

produces the same result. The index ± is an integer that indicates possibly several scalar transfer functions is being requested. EXAMPLE 6.10: scribed by

Find a factored

~ =

transfer

-~2

function

which of

for the model de-

x+

with the output y = x2 (t). Solution:

The MATLABcoding

A=[-4 1/2 0;2 -2 2;0 1 -4]; B=EII0;03; C=£0 1 0]; D=[0]; [z,p,k] = ss2zpA,B,C,D,1 gives the response -4 -5.0000 -4.0000 -1.0000 2 which indicates Notice that the should be canceled, zeros rarely cancell plane pole-zero pair equal right-half-plane

the factored result

2(s+ 4)

G(S) = (S 1) (S + 4)(S +

example has a zero and a pole for which S = -4. These but MATLAB is not that smart (yet). In practice, poles and exactly. Cancellation of an approximately equal left-halfproduces little error, but cancellation of an approximately pole-zero pair covers up an instability.

6.2.7 Bode Plots for Higher-OrderModels* Bode plots for models with any number of poles and zeros can be assembled readily from the Bode plots of the individual poles and zeros. The ordinate of the magnitude Bode plot equals the logarithm of the magnitude of G(jw), or log10 IG(jw)l. From equation (6.46), this becomes J loglo IG(j~)I = log~o k + ~ log~o I1 ¯ Jw/wzil i=1

+ ~ log10 tl - (w/wzi) 2 + j2~iw/w~il i=j+l

- Y log Ij~l

6.2.

427

SINUSOIDAL FREQUENCYRESPONSE

The ordinate of the phase Bodeplot is the angle of the phasor G(j~), with mathematical symbol /G(jw):

L[I1- (~/~:i)~ + 5j2¢:~1~]

LG(jw) = ~ L(1 :t: jmlw:i) i=1

i=lq-j

k

n-N

i=1

i=k+l

- N 2

- ~ Z(1+ jwl~p,) - ~ L[1- (~/~,)~ + j2~,,~l~,,]. (6.52) Theseequations expedite the determination and interpretation of Bodeplots. Equation (6.51) shows that the ordinates of both the.magnitude and phase plots equals the suin of the ordinates of the componentpoles and zeros. The magnitudeplot also is shifted vertically by loglo k. Decibelssumdirectly. These summationsapply to the asymptotes and linear approximations as well as to the exact curves. The two examples below regard the third-order model d~ x

a3 -~ q-

d~ x . dz a2 -~ + al ~ -t-

aox = fo cos(wt

q-/~).

(6.53)

EXAMPLE 6.11: Find analytic expressions for the Bode plot variables in the case of three real roots with break frequencies ¢ul, w2and wa. Then, for the special case ofw~= 10 rad/s, ~vz= 50 rad/s and w3= 200tad/s, plot the asymptotic approximation for the magnitude curve and the approximation for the phase curve, and sketch the complete Bodecurves. Solution: The transfer function becomes

Y0

G(S) = a3S3 q- a2S2 + alS

-t-

ao = (1 q- S/Wl)(1

-t-

kSIw2)(1

The Bodeplot variables are, in light of the equations abovefor individual poles, logIG(jx)l=lOglok-lOglo

1+ ~11 -l°gl°

z, -log,oTl+(~) ZG(jw)=-tan:’(~l)-tan-’

(-~2)-tan-l(-~3).

1+

428

CHAPTER 6.

ANALYSIS

-20 ~- magnitude

OF LINEAR MODELS, PART 1

O° \...._.~.N..~ linear iappr°ximati°ns :

db -40

~3

zGO’~o)

°-180 100,000 ~ ~ = (S+ 10)(S+~0)(S+200) o(~)

°lO

....................... ~ lO

~ 10

~0~70°

The Bode plots above are for the given values of ~, ~ and ~ and ~ = fo/ao = 1. The sum of the asymptotes for the magnitudes of the individual poles is shown by a dashed line. Below the first break frequency all the asymptotes equal 1. Between the first and the second break frequency only the asymptote for the first pole is sloped, and therefore the sumof the three has the slope of -1. For frequencies between the second and third breaks both the asymptotes for the first two breaks have slope -1; their sum has slope -2. Above the third break frequency all three poles have asymptotes with slope -1, so their sum has slope -3. The linear approximations to the three individual phase curves also are shown. Recall that these approximations equal 0° for frequencies less than one-tenth their break frequencies, -90° for frequencies greater than ten times their break frequencies, and a straight-line interpolation in between. The sum of these three approximations can be seen to approximate the actual phase shift rather nicely.

A generalization can be made regarding the magnitude asymptotes: in the absence of zeros, the slope of the asymptotic approximation to the magnitude curve at any frequency equals -1 times the number of poles that have lower break frequencies, including poles at the origin.

6.2.

SINUSOIDAL

429

FREQ UENCY RESPONSE

EXAMPLE 6.12 Find analytic expressions for the Bode plot variables for the case of one real root at Wl and a pair of complex roots with damping ratio ( at w2. Then, set wl = 10 tad/s, 0;2 = 100 rad/s and ( = 0.065, plot the asymoptotic approximation for the magnitude curve and the approximation for the phase curve, and sketch the complete Bode curves. Solution: The transfer

function can be factored as k

G(S) = (I+ S/~I)[I (~ .(/~2)S (S /~2)~]" + The Bode variablesbecome

log~o Ig(jw)l = loglo k - log~o 1 + ~-~

log~0 ZG(jw)

1- (~-~

+ \ w-’-~/

=-tan -1 ~ -tan -~ 1-

(w/wz) ~] "

The case with the given parameters and k = I is plotted below:

"~agnitude

~\

-20 db -4O

° -180

-60

-8O

°10

101

102

~’270° 10

430

CHAPTER6. ANALYSIS OF LINEAR MODELS, PART 1

The magnitude asymptote for the model has the slope -1 for frequencies betweenwl and w2, and the slope -3 for frequencies above w2; the drop in slope of -2 results from the double break, or two poles, located at w = w2. The model resonates about this natural frequency w2. The value of the magnitudeat the break frequency is almost exactly a factor of 1/2~ = 7.69 greater than the intersection of the asymptotes at that frequency; about one-half percent error results from the effect of the break frequencywl. The linear approximation to the phase curve can be seen to apply reasonably well for w < w2. The phase drops precipitously 180° in the vicinity of w = w2, because the dampingis fairly small. The total phase shift at infinite frequency is the total numberof poles times -90°, or -270°. Note that the same result applies to the other third-order models in Examples 6.11 and 6.13. EXAMPLE 6.13 Repeat Example6.12 for the case in which wl and w2are switched. Solution The procedure is the same as in the previous example. The resulting plots are given below. The resonance nowappears at 10 rad/s, and the first-order pole appears at 100 rad/s. In general, the lowest-order singularies tend to be the most important. In the present case, the presence of the first-order pole in this case might be neglected for practical purposes (as also noted below).

o0

° .90

o .180

6.2.

SINUSOIDAL

431

FREQUENCY RESPONSE

A time constant in a homogeneous solution or an impulse or step response is the reciprocal of the break frequency for a real pole, and a natural frequency is the break frequency for a complex conjugate pair of poles. Step and impulse responses for the cases of Examples6.11, 6.12 and 613 are given in Fig. 6.15, with the respective designations (a), (b) and (c). These plots can be secured readily through the MATLABcommands step and impulse. In case (a) (Example 6.11), the dominant time constant is rl = l/w1 = 0.1 second, which can be seen in both the impulse and step responses. In case (b), (Example 6.12) oscillations occur at the damped natural frequency w2v/i -- (2 = 98.2 rad/s 15.63 Hz, and decay according to the damping ratio ( = 0.065. In addition, the effect of the time constant r = 0.1 second is similar to that of case (a). case (c) (Example 6.13), the oscillations occur at the lower resonant resonant frequency of virtually 10 red/s, and decay with the damping ratio ~ = 0.1. The time constant in this case is so relatively short, only r = l/w2 = 0.01 seconds, that it effects the responses imperceptibly. For manypurposes, therefore, this model could be simplified by omitting the real pole (as also noted above).

EXAMPLE 6.14 Sketch-plot the Bode curves for the fifth-order

model

200,000

a(s) = (s + 1)(s + lo)2(s~2000) Also, use MATLAB to plot the impulse response of the model. Solution: This case includes a pair of identical first-order poles, which is the same as a second-order pole with critical damping. There is also a firstorder pole at low frequency and a second-order pole at high frequency. The impulse response shows the effects of all these poles, but the pole at the lowest frequency produces the dominant relatively slow exponential delay. The double break at w = 10 controls the rise in the first one-half second. The resonance at about w = 45 produces the small rapid oscillation superimposed on an otherwise smooth response.

432

CHAPTER6. ANALYSIS OF LINEAR MODELS, PART 1 15

0

1. 0’1 ~] ~.2~ ~J 0.3"~ ~.~.0.4 0 for E: ~-nple6.11 for Example6.12

I

I

I

I

0.1

0.2

0.3

0.4

t 0.5

I

t 0.5

1C

-5 2

for Example6.13

I

00

1

I

I

2

t

3

Figure 6.15: Impulse (g(t)) and step (h(t)) responses for the modelsof Examples 6.11, 6.12, and 6.13.

6.2.

SINUSOIDAL

433

FREQUENCY RESPONSE,

-90o

’1

phase

~

-6o

~)= G:S’

ZG(jeo) °-270

200,000 (s+ 1)(s+ ~o)~(s%ss+ 10°

lift

lOt

co 10~

MATLAB coding that gives the complete Bode plot decibels) ~s well as the impulse response is as follows:

(although

hum = [200000] ; den = [1 26 2225 42700 240500 200000]; bode(hum,den) impulse(num,den) 0.8 0.6 0.4 0.2 0

0

1

2

t

3

4

with

434

CHAPTER6. ANALYSIS OF LINEAR MODELS, PART 1

Zeros always contribute an increasing value to a magnitudeplot as frequency is increased. A transfer function with a zero produces a magnitudeBode plot that is unchangedif the sign of the real part of the zero is changed,that is if the zero is movedan equal distance into the opposite half-plane. On the other hand, the right-half-plane zero contributes an increasing phase as the frequency is increased, whereasthe left-half-plane zero contributes a decreasingphase (or increasing phase lag) as the frequency is increased. Therefore, you can use the phase plot to determinethe signs of the real parts of zeros. Modelswith no zeros in the right=half plane are called minimumphase-lag models. It is possible to deduce the transfer function of a knownminimum-phase-lagmodel from its magnitudeplot alone. A modelwith one or more right-half-plane zeros is called non-minimum phase-lag.

EXAMPLE 6.15 The Bodeplots below correspondrespectively to the following transfer functions with zeros: 0.01(S2 4- 2S 4- 100) 1 -4- 0.1S Ga(S) - $2 4- 0.2S ÷ 1’ Gb(S) S24- 0.2 S ÷ 1

20[ 10: G(/~,

I

dbOI 10 o0

-20 I

-401 10" 10-1

° 1~

101 ~

10~ -I 10

° 10

~180 101 ° (~

~0

Resolve whether the 4- signs should be plus or minus. Solution: In both cases the behavior in the vicinity of w = 1 is caused by the identical poles (in the denominator). Both zeros have break frequencies of w -- 10. In both cases the breaks cause the slopes of the asymptotes to increase, from -2 to -1 in the first-order case and from -2 to 0 in the second-order case. Therefore, both d= signs should be +. The magnitude asymptotes and phase approximations also given in the figure themselvesare sufficient to determinethe transfer functions.

6.2.

SINUSOIDAL

435

FREQUENCY RESPONSE

Models often are classified according to their numberof poles or zeros at the origin. The generic designation is type N, where N is the net number of poles at the origin. The models presented previously have been type zero. Type-one models (with the factor S in the denominator) are also fairly common;typetwo and type-minus-one models (with the factor S in the numerator) are not uncommon, but others are rare. The presence and number of poles or zeros at the origin is readily identifiable by the slope of the magnitude curve as the frequency approaches zero, which is -N or -20N db/decade. Note that the °. phase angle at zero frequency is directly affected also; it is -N x 90 EXAMPLE 6.16 Verify that the model with the first Bode plot below has a single pole at the origin, and that the model with the second Bode plot has a single zero at the origin.

lO’t I

~

~~’~,a~.mptotes I -90’ appr°ximati°n’~

G~(S)~,I

10"~ .. ~ti~~ ~ magNm~

~

~]~ -180_ q

10-~ ~

~

-~

~’" "~

phase

-270° 10

~ 10

10 ~ ~ ~ 10

104 , ,,,,,,, ,,,,,,,,~ , I0° I0~ I0~ ~

N~lu~m~he slope of ~he magnitude curve ~ ~ ~ 0 is -1 in the first c~e and +1 in ghe second, directly leading to the indicaged conclusions. ~rther, the phase angle for ~ ~ 0 is -90~ in the first case and 90~ in the second, corroborating the conclusions. The magnitude ~ymptotes and the ph~e approximations are also shown above. These are su~cient to give the complete transfer functions. A summ~y of the Bode ~ymptotes and line~ ph~e approximations for first-order poles and zeros is given in ~ig. 6.16. Recall that an approximation any higher-order model results from appropriately summing these ~ymptoges ~d approximations, and shining the magnitude result vergicNly to account for the coe~cient ~. The ph~e of a second-order pole or ~ero that h~ low d~ping

436

CHAPTER 6. ANALYSIS OF LINEAR MODELS, PART 1 10 key: a, o

magnitudeasymptote:

1

phase approximation: ’~g

oi _90 o.1 O.lcoo olOco ° lOO ~ _~’1 I 90 1+s/69 o lO l

1~ I’S/~v ~

1

] [

10

O.lcoo co o ! i/~

1

,J ~ 90° ~ °0

1 10 I+S/~o 1

° -90

o.1

]1 ~ ~

~ ’

i

I

lOco o ~ 90° ~

~ 0

j

°90

~ °0

°_90

Figure 6.16: Summaryof Bodeasymptotes and linear phase approximations

is a partial exception, however.Rather than spread its 180° phase change over two decades of frequency, it is better to concentrate the change in an abrupt jump at the break frequency. You would also wish at least to approximate the resonance or anti-resonance in the magnitudeof such a case, which is not included in the asymptotes, by noting the factor of 1/2~ at the break frequency. It is possible to estimate a transfer function from a Bodeplot drawnusing experimental data, as suggested in Examples6.15 and 6.16. An understanding of the asymptotes and the discrepancies mayhave its greatest Utility in the interpretation of experimental data, since knowntransfer functions can be plotted readily by software programsincluding MATLAB. The slope of the magnitude asymptoteat infinite frequencyfor any transfer function equals -1 times the numberof poles minus the numberof zeros, called the pole-zero excess. Youguess the type and locations of the break frequencies, drawthe associated asymptotes and the phase approximations, and note the discrepancies between the approximation and the actual curves using Figs. 6.11, 6.12 and 6.13. You then makeadjustmentsuntil a satisfactory fit results. This process is an example of what is called the experimental identification of a dynamic model.

6.2.

437

SINUSOIDAL FREQUENCYRESPONSE

EXAMPLE 6.17 The Bode plot below might represent experimental data, Approximatethe associated transfer function, and thereby write the associated differential equation. 10 ................................. magnitude ratio 1.0

0"10.01

0.1

1.0 10 ~o, rad/s

100

°90 phase

0o

°_90 °-180 Solution: First, examine the magnitude ratio and observe that the lowfrequency asymptote is horizontal. This meansyou have a type-zero model; N = 0. Next, observe that you also have a horizontal asymptote at high frequency; the pole-zero excess is zero, or the numberof zeros equals the numberof poles. The most prominent feature of the magnitude plot is an apparent resonance at about 2 rad/s. Youconfirm the presence of a double pole by examinationof the phase plot, whichshowsthe telltale rapid drop of well over 90° in the vicinity of 2 rad/s. The magnitudestarts to rise noticeably for frequencies lower than one-quarter of this, however, which doesn’t happen for sharp resonances, as can be verified by examination of Fig. 6.13 (p. 423). Youalso knowthat there has to be at least two zeros in the model(because of the pole-zero excess). Finally, observe that the phase rises at low frequencies, which only can happen because of a zero there. Puting this information together, you annotate the magnitude plot as follows:

438

CHAPTER 6.

ANALYSIS

OF LINEAR MODELS, PART 1

magnitude ratio 1.0

~ second-order polefirst-prd.e,r.ze.r,?, 0’ 10.01 ....

(~.1

1:0 10 100 to, rad/s The break frequency for the zero is merely guessed; in any case it is followed by an asymptote with slope +1, as shown. The break for the second zero must occur at some frequency above the resonance, for otherwise the magnitude curve would not bend from a sharply downward slope toward zero slope. Therefore, you try an asymptote with slope -1 before this pole, as also shown by a dashed line above. You note with pleasure that the two sloped asymptotes have a difference in slopes of -2, which is precisely what the second-order pole demands. Further, the intersection of these asymptotes is close to the peak of the resonance at about 2 rad/s. You now make adjustments to the asymptotes to make their intersection more precisely align with the resonant frequency, and to take advantage of the fact that at their respective breaks the deviation of the actual curves from the intersection of the first-order zeros is v~, or about 3 db, plus something more for being near the second-order pole. The result is:

magnitude ratio 1.0

O. 0.01 °90 phase

0o -90 °

°-180

10 1.0 a), rad/s

100

zero’in’e’ft’-l~i ~ ’plan~~:::~’;~ ........... [......~........zeroin right-halfplane second-order po~le.................

6.2.

SINUSOIDAL FREQUENCYRESPONSE,

439

Your value of k = 0.5 is determined so that G(jO) 0. 5, as plotted. Not e Msothat your transfer function gives G(joo) 0.2, a confirmation. The damping ratio can be estimated by comparing the value of the magnitudeat the resonanceto the intersection of the two asymptotesthere. The ratio of the values is 5:1, whichequals 1/2¢, so that ~ = 0.1. There is one ambiguity left to be resolved: the sign of the zeros. The phaseincreases in the vicinity of the lowerzero, so it mustbe in the left-half plane. Onthe other hand, the phase change due to the higher-frequencyzero is negative, whichdictates the right-half plane. Thepresence of one zero in the right-half plane is confirmedby the fact that the overall phase change is -180°. Were this a miniumum-phase-lagsystem, the final phase would be zero, because the pole-zero-excess is zero. The phase approximations associated with the zeros are shownabove by dotted lines. The sum of these, plus the 180° phase drop approximatingthe pole, is shownby dashed lines. The differertces betweenthe actual curves and this approximationare consistent with the plots of Figs. 6.11 and 6.13 (pp. 420, 423). The resulting transfer function can be interpreted to give the following differential equation, whichin turn could be used to solve for the response of the systemto any other input: du d2u 1 8 d2x + 0.2~-~ + 4x -~ dt = -0"2~ 5 + " dt + 2u" The procedure used in Example6.17 is one of manythat would arrive at the same answer. The information is redundant. You should not attempt to discover one procedurefor all cases. If you understand the individual effects, you should be able to identify the wholepicture, as with a picture puzzle having several pieces. 6.2.8

The Pure

Delay

Operator*

Imagine a pure-delay model, for which the step response is a step delayed by T seconds, and the impulse response is a impulse delayed by T seconds, etc. Transport and wave-like phenomenaare among those often approximated by this model. To find its transfer function, imagine a sine waveof frequency w which also is delayed by T seconds. If the frequency is one cycle in T seconds, or ~ = 2~r/T rad/s or Tw= 2~r tad, the phase shift is -360° or -2~r radians. If for examplethe frequency is halved, the phase shift is halved. Thusthe phase shift, in general, equals Twradians, and the frequencytransfer function is G(jw) = -jTw.

(6.54)

This is representable by a phasor of unit length and angle Tw, since the pure delay neither amplifies nor distorts the signal. The more general pure-delay operator becomes G(s) = e -Ts, (6.55)

440

CHAPTER 6.

ANALYSIS

OF LINEAR MODELS, PART 1

so that x(t)

= e-Tsu(t)

= u(t -- T).

(6.56)

Since the magnitude of the pure delay operator is 1 while the phase lag is non-zero, it must be a non-minimum-phase-lag operator. That is, it must have zeros in the right-half plane. There are an infinity of such zeros, in fact, each matched by a pole of the same magnitude in the left-half plane. A crude approximation for a pure delay phasor is offered by the combination of a single right-half-plane zero in combination with a single left-half-plane pole of equal magnitude: 2IT Gl(jw) _jwjw +- 2IT’ [Gl(jw)[ = /Gl(jw)

=-2 tan-X (-~-~).

(6.57a) (6.575) (6.57c)

This is the crudest of a series of approximations to the pure delay in the form of ratios of polynomials known as the Pade approximants (pronounced Pahday). The Pade approximant with third-order numerator and denominator is G3(jw) -( Tjw)3 + 12(Tjw)2 - 60Tjw + 120 (Tjw) 3 + 12(Tjw) 2 + 60Tjw ÷ 120 ’

(6.58)

which also has unity magnitude for all frequencies, identical to the exact phasor e-jTw. The phase angles are reasonably valid only for a restricted band of frequencies which increases with the order of the approximant2, as indicated in Fig. 6.17. 6.2.9

Summary

Sinusoidal excitations abound in the physical and engineering world, so the response of engineering systems to these inputs is a major concern. The transfer function G(S), when evaluated with S = jw, gives a complex number G(jw) called a frequency transfer function. Its magnitude equals the ratio of the amplitude of the sinusoidal response x(t) to the sinusoidal input u(t) at the frequency w. Its phase angle equals the phase shift between the sinusoidal response and the input. Plots of the logarithm of the magnitude ratio IG(j~)I and of the phase shift versus the logarithm of the frequency are called Bode plots. The quantity 20 loglo IG(jw)l, called decibels (db), often is chosen for the magnitude, including by the bode routine of MATLAB. Transfer functions for systems described by ordinary linear differential equations are ratios of polynomials which can be factored to reveal poles and zeros. Models without damping have only even powers of S, so their frequency transfer functions are real and their poles and zeros come in plus-and-minus imaginary 2The Control SystemToolboxof the professional version of MATLAB includes a command l~ade whichgenerates Pade approximantsof any order.

6.2.

441

SINUSOIDAL FREQ UENCY RESPONSE o0

phase °-180

° -360

°-540

0

I

I

I

I

2

4

6

8

~,

T~o

I

10

12

Figure 6.17: Phase lags of a pure delay and Pade approximants pairs. The numberof resonant frequencies that the poles represent equals the numberof degrees-of-freedomand one-half the order. Models with damping include real poles and/or complex poles. Complex poles and zeros comein complexconjugate pairs; it is convenientto leave their polynomials in unfactored quadratic form. Each pair is describable in terms of its natural frequency and dampingratio, and is associated with a degree-offreedom.Real poles occur in all first, third and other odd-orderedmodels, and contribute exponential behavior that is either dampedor explosively unstable. Fourth-order models, then, have two degrees of freedom, two natural frequencies and two dampingratios, although if a dampingratio exceeds unity it is preferable to factor the associated second-orderpolynomialinto two first-order polynomials. Bodeplots can be sketched quickly by first drawingtheir straight-line asymptotes and phase approximations. Conversely, the transfer function corresponding to a given Bodeplot can be estimated by approximating these asymptotes and phase approximations. As the frequency rises past a pole the slope of the magnitude asymptote breaks downwardby an added -1; for a zero it breaks upward by ÷1. This applies even to poles and zeros at the origin. Complex poles cause resonancesin the vicinity of their second-order break frequencies, while complexzeros cause notches. Each left-half-plane pole reduces the phase asymptote by 90°, as does right-half-plane zeros. The opposite changes occur for right-half-plane poles and left-half-plane zeros, although only zeros can be in the right-half-plane without causing instability. As a consequence, models

442

CHAPTER6. ANALYSIS OF LINEAR MODELS, PART 1

q=ql sin ~ Figure 6.18: System for Guided Problem6.2 with only left-half-plane singularities are knownas minimum-phase; the phase angles follow directly from the magnitudes,and vice-versa. Modelswith right-half-plane zeros are non-minimum phase. The pure delay function includes an infinity of right-half-plane zeros. Its approximationshave a finite numberof such zeros. Guided

Problem

6.2

This is a classical elementary probleminvolving the undampedvibration of the single-degree-of-freedomsystem shownin Fig. 6.18. The objective is to comparethe amplitude of the steady-state displacement of the block, ql, with the amplitude of the displacement of the bottom end of the springs, q = q0 sin Suggested Steps: 1. Model the system With a bond graph. Note that its input is the time derivativeof q(t). 2. Write the state-variable differential equations for your model. 3. Combinethe differential equations to give a second-orderequation in terms of a momentum. 4. Express the output variable of interest, ql, as a function of the momentum. 5. Determinea transfer function betweenthe input variable, q, and q~.:~Convert this to a frequencytransfer function. 6. Either sketch the Bodeplot correspondingto the transfer function, or use MATLAB. 7o

Presumingthe mass is given, observe the approximate range of spring constants, k, that would(i) keep the amplitudeof ql to within ten percent of that of q, (ii) wouldkeep the amplitude of q~ to less than ten percent of that of q.

6.2.

SINUSOIDAL FREQUENCYRESPONSE

Guided

Problem

443

6.3

This problemregards the use of the phasor methodto find a frequencyresponse. Combining the four first-order differential equations into a single fourth-order differential equation is the mostdifficult part, unless matrix software is used; both the use of MATLAB and detailed manual steps are indicated below. Plotting is particularly simple if MATLAB is used. The mechanical system shownin part (a) of Fig. 4.1 (p. 198) is excited with the force F = F0 sin wt. Find the frequency transfer function for the displacementof the massml relative to the exciting force, and plot its magnitude and phase in Bodecoordinates. The parameters are as follows: rnl = 1 kg, ms = 2 kg, kl = 25 N/m, k2 = 100 N/m, R = 0.25 N s/m. Suggested Steps: 1. Find a bondgraph modelfor the system. (This is done in the solution to GuidedProblem4.1, p. 203.) 2. Applycausal strokes, and write the four first-order differential equations in the standard fashion. If you wish to use MATLAB for finding the scalar transfer function of interest, find the elements A, B, C, D of the state-variable formulation. Then, use the MATLAB function ss2tf to find the answer. If you wish to practice manualdetermination of the scalar transfer function, start by convertingthe differential equations to algebraic equations, using the operator S =- d/dt. Steps 5-8 below complete this part of the problem. At this point it is possible to apply equation(5.47) (p. 347) to get a single fourth-order equation in terms of one of the variables; your interest is in the momentum of the first mass, which equals mlle. Since the matrix operations required are awkwardwithout the use of appropriate software, a different approachis suggested. Multiply the two most complexalgebraic equations by S, and substitute the other two equations to eliminate two of the four variables. Youshould be left with two equations in terms of the two momenta. Solve one of the equations to get the second momentum as a function of the first (and the operator S), and substitute this into the other equation to get the first momentum as a function of the input excitation. 7. Write pl/F as a ratio of polynomialsin S. Convert this to the ratio xl/F by noting that Sx~ = p~/m~. 8. Substitute jw for S. The result is the frequencytransfer function G(jw).

444

CHAPTER6. ANALYSIS OF LINEAR MODELS, PART 1

0.1

0.001

]0

~o, rad/s

100

400

Figure 6.19: Bode plot for Guided Problem6.4 9. To compute and plot the frequency response, you can apply equations (6.39) (p. 410), or use the bode command of MATLAB as illustrated above. You have nowcompleted what, if done analytically in an introductory course in vibrations, is usually considereda very difficult problem. Guided Problem

6.4

This is a basic problemin the identification of a linear modelfromits frequency response. A Bodemagnitudeplo~ for a modelis given in Fig. 6.19. Assumingthe model is minimumphase, (a) sketch-plot an approximation for the corresponding Bode phase plot, and (b) estimate the correspondingfrequency transfer function. Suggested Steps: 1. Estimate the asymptotes of the magnitudeplot, noting that their slopes must be whole integers. The deviations of the curves from the asymptotes can be appreciated with the help of Figs. 6.11, 6.12 and 6.13 (pp. 420-423). 2. Identify the break frequencies from your magnitudeasymptotes, and start the phase diagramby sketching its corresponding horizontal asymptotes. 3. Note the sloped straight-line approximationfor the phaselag of first-order poles and zeros given in Figs. 6.11, 6.12 and 6.16 (p. 436). Incorporate this improvementinto your sketch whereverit applies.

6.2.

445

SINUSOIDAL FREQUENCYRESPONSE,

° 90

° _90

0.1

° -180 ° -270 0.01

10

~ rad/s

100

r’-q~6360°

Figure 6.20: Bode plot for GuidedProblem6.5 4. Completeyour phase estimate with a smooth curve. 5. Usethe break frequenciesyou have already identified to estimate the transfer function. Guided

Problem

6.5

This problemillustrates the effect of right-half-plane zeros on the frequency response of linear models. Three stable models have the same Bodemagnitudeplot given in Fig. 6.20, but different phase plots as shown. Estimate the corresponding transfer functions. Suggested Steps: 1. Estimate the asymptotesof the given magnitudeplot. Note that a secondorder zero has precisely the same form in Bodecoordinates as the corresponding second-order pole except for an opposite sign. Identify the dampingcoefficient from the degree of inverse peaking. Write the transfer function or functions, assumingall poles and zeros affect the plot. Recognize any ambiguitywhich nevertheless results in a stable system, which the existance of the plot assumes. 2. Predict the phase angles using the results of step 1. Can a non-minimum phase lag exist? Comparewith the given plots for the three cases, and drawconclusions.

446

CHAPTER 6. ANALYSIS OF LINEAR MODELS, PART 1

3. Additional phase lag with no effect on the magnitude can result from symmetricplacementof one or more pairs comprisinga zero in the righthalf-plane and a pole in the left-half-plane, as in the Pade approximants for a pure delay. A pure delay itself is included as a possiblility. Estimate the actual case by focusing on the differences betweenthe phaselag of the remaining unknownsystem and the phase lags of the two knownsystems. PI~OBLEMS 6.18 Evaluate, directly from the given transfer function, the magnituderatio and phase angle of the steady-state response of the modelgiven in Fig. 6.9 (p. 412) for sinusoidal excitations at 1 and 100 rad/s. Checkto see that they agree with the plots. 6.19 Showfor what ratios of parameters the torsional system shown below corresponds to the tuned vibration absorber shownof Example6.9 (p. 416).

6.20 A shaft with angular velocity ~ and applied momentMdrives the central memberof a torsional vibration absorber that has a momentof inertia Id = O. 1 ft.lb.s. This memberin turn drives, through four springs as pictured below (bearings not shown),a flywheel of momentof inertia I = 1.0 ft.lb. Eachspring has stiffness k -- 500lb/ft, and the length a is 0.25 ft.

(a) Modelthe system with a bond graph. (b) Evaluate the parameters of your model. (c) Find a single differential equation relating $ to M. (d) Find the steady-state amplitude of ~ in response to.an Mthat~ oscillatory at frequencyw. Sketch-plot this amplitudeas a function of w. Does this system act as an ideal vibration absorber at somefrequency?

6.2.

SINUSOIDAL FREQUENCYRESPONSE

447

6.21 Return to Problem4.7 (p. 202) and Problem5.30 (p. 334) in which effect of a drain tube placed through an eardrumis modeled.Find the transfer function from the input Pin to the output Pe/Ie. (Your may use MATLAB’s ss2tf.) Find and sketch the Bode magnitude asymptotes for the model, which ignores dissipation, and compareto the corresponding modelwithout the tube. Briefly discuss the anticipated consequencesthe tube has on hearing. Also, speculate on the waythe inner ear compensatesfor the natural response, using whatever knowledgeyou may have about hearing. 6.22 Find the steady-state force F(t) for the system shown below with x = x0sin

6.23 An instrument transducer produces a signal that contains an unwanted periodic noise signal at 60 Hz. The part of the signal of interest lies below5 Hz. The signal is received by a recording and display instrument that requires negligible signal current, i.e. its input impedance is very high. A simple potential solution employsan RCfilter, as shownbelow. Standard electrolytic capacitors of 2, 5, 10 and 20 mf are available, as are the standard resistors of 10, 12, 15, 22, 27, 30, 33, 39, 43, 47, 51, 56, 62, 68, 75, 82, 91 kf~ and all factors of 10 lower and higher. R

Find a design, if it exists, that reduces the unwantedsignal by a factor of at least 10 without reducing any of the wantedsignal by more than 40%,while ¯ keeping the current small. The instrument is inaccurate if more than 0.1 ma peak-to-peak is drawnfrom the transducer for a signal of 4 volts peak-to-peak at the 5 Hz. 6.24 Consider the classical case of sinusoidal position excitation of a massspring-dashpot system shown on the next page. The objective is to choose springs and the dashpot so that the motion of q produces little motion of ql. (Note that this problemis similar to GuidedProblem6.2 (p. 442), except that a dashpot has been introduced. The use of similar steps is suggested.)

448

CHAPTER 6. ANALYSIS OF LINEAR MODELS, PART 1

q =qosln ~ (a) Model the system with a bond graph. (Hint: Two junctions needed.) (b) Determinea transfer function betweenthe input, q(t), and the output, ql (t). Write the correspondingfrequencytransfer function. (c) Determinewhether b should be large or small if the frequency is (i) muchsmaller than the natural frequency of the system, wn, (ii) about the sameas wn(iii) muchhigher than wn. Also, describe a design strategy k as well as b can be chosen. Suggestion: Consider the nature of the Bode plots for large and small amountsof damping,b. 6.25 A duct leads from a acoustic source to a resistive load; an exhaust pipe from an engine is an example. In order to reduce the propagation of acoustic pulsations in the flow, a Helmholtzresonator is attached, which comprises a chamberwith fluid compliance, C, attached to the duct by a narrow entrance with fluid inertance, I, as shownbelow.

duct Q~

~

-~ Helrnholtz resonator az

~-~

(a) Modelthe system with a bond graph. (b) Writedifferential equations. (c) Combinethe differential equations to get a second-order differential equation with a momentum as the dependent variable. (d) Relate this momentum to the input flow, Q1. Also, relate the output flow Q2 to this momentum and Q~, in order to get a transfer function between Q1 and Q~. (e) Sketch, or plot using MATLAB, the Bodeplot for the particular case which I = R2C/4 (The frequency Wn= 1/x/7-~ can be used as normalizing frequencyto give a dimensionlessfrequencyratio.)

6.2.

SINUSOIDAL FREQUENCYRESPONSE

449

(f) Use the aboveresults to suggest the design of a muffler that would effective over a range of frequencies. 6.26 A system with the transfer function I) G(S)- $2lO0(S +2S+10 0 is excited by the disturbance

u(t) -- 4sin |-’" t|

\5]

=0

0 < t < 2.5sec t _> 2.5sec

(a) Find the homogeneous solution xh(t). (b) Find the particular solution xp(t) valid for 0 < t < 2.5 seconds. (Use of the phasor methodis suggested.) (c) SumXh(t) and xp(t) to get the completesolution valid for 0 < t < 2.5 seconds. Evaluate the undeterminedcoefficients using the knowninitial conditions x(0) -- 0, 5~(0) (d) Find x(2.5) and :~(2.5), and use these as initial conditions for solution for t >_ 2.5 seconds. 6.27 Find an approximate numerical solution to the previous problem using the :[s±m commandof MATLAB. 6.28 The systemrepresented in Fig. 6.9 (p. 412) is excited by the samedisturbance as the two previous problems. (a) Find the response analytically, following the steps given for the first of these problems. (b) Find the response numerically, following the steps given for the second of these problems. 6.29 The impedanceof a spring is 1/CS in terms of the operator S, and 1/Cjw in terms of the frequency w. The factor 1/j = -j describes the fact that the force lags the velocity by 90°, or is in phase with the displacement. Sometimes a servo system is designed to imitate a spring, but produces a force that lags the displacementsomewhat,or lags the velocity by a little morethan 90°; it can be described as a "spring with a lag." Yourtask is to determinethe stability of a system comprising a mass connected to ground through such a "spring," and what effect is producedby adding a dashpot. The following steps are intended to help, but are not mandatory.

450

CHAPTER6. ANALYSIS OF LINEAR MODELS, PART 1 (a) Represent the system (without the dashpot) with a bond graph. can invent a symbol,such as Z, to represent the "spring." (b) Characterize the "spring" by its impedanceas a function of frequency; choose some reasonable assumption to represent the small added phase lag. (c) Transformthe result of part (b) into a function of (d) Write the differential equation for the system in operational form, using the result of part (c). (e) Use the result of part (d) to find the characteristic values that determinethe stability. (f) Modifyyour modelto represent the addition of a dashpot betweenthe mass and ground. Repeat step (d) and (e). (g) Repeat part (f), instead placing the dashpot between the mass the spring.

6.30 You are asked to determine the transmission of sound through a wall comprisinga single sheet of plaster 3/8 inch thick, whichis very thin compared with the acoustic wavelengthsof interest. Thespecific gravity of plaster is 1.8, and its material dampingis virtually nil. The acoustic impedancefor plane wavesentering a large space such as a roomis a resistance of magnitude2.67 lb.s/ft 3, whichrepresents a ratio of pressure changeto velocity change. (a) Modelthe wall so as to permit estimation of the ratio of the transmitted to incident sound pressure, Pt/Pi. Neglect the effect of motion of the plaster on the incident soundpressure, and neglect the effect of any mechanicalsupports for the plaster. (b) Evaluate the ratio for (i) 100 Hz (deep bass) (ii) 2000Hz (soprano). Express the answersin decibels (db), whichare defined as 20 loglo(Pt/Pi ). (Note that humanscan detect loud sounds despite attenuations of 50 db and occasionally even 100 db.) Whatwouldrock music sound like on the quiet side? (c) A double wall is constructed with the same plaster sheets and a inch air space in between. Repeat pa~ts (a) and (b) above, assuming mechanicalconnection betweenthe two layers. Do your results suggest the acoustic significance of the mechanicalconnections conventionally used to strengthen the wall? (d) Design problem. Designa practical non-load-bearinginterior wall for sound isolation, using plaster sheets and steel supports. Defendyour design in terms of strength and other practical considerations as well as its acoustic properties.

6.2.

SINUSOIDAL FREQ UE‘NCY RESPONSE,

451

6.31 Sketch the Bodediagramfor the modeldescribed by the following transfer function: S2 + 20S G(s) = $3 4S2 + 104S + 200 6.32 Sketch Bode plots for the models described by the following minimumphase and nonminimum-phase transfer functions. Find and plot their responses to a unit step input. 1 G~(S)= ~(I+ TS);

G2(S) = ~(1-TS)

6.33 Find the transfer functions for the two systems characterized with Bode plots in Example6.16 (p. 435). Note that magnitudeasymptotes and straightline phase approximations have been drawn already. 6.34 Estimate the transfer function for the system that produces the Bodeplot below. 100 magnitude ratio 10

0.1

1.0 10 frequency, rad/s

100

452

CHAPTER6. ANALYSIS OF LINEAR MODELS, PART 1

6.35 Answerthe previous question for the Bodeplot below.

1.0 ° I00

........

I0

I000

~, rad/s I00

_

.

°: _-100 -2000

:

~"~ .............

.:

~ I

.

~

6.36 Answerthe previous question for the Bodeplot below.

0ilO0

o-150

01 01

I I I IIIll 1 0

I I I Illlll\ I I I lllll 100 ea rad/s 10~, o-250

6.2.

453

SINUSOIDAL FREQUENCYRESPONSE

6.37 A simple untuned viscous damper replaces the spring of the tuned vibration absorbers above with a dashpot, as shownbelow.

(a) Showthat the ratio of the force on the support to the force F Fosinwt applied to the mass Mhas the amplitude 2~kxo[ = fl(1

~ + 4~2/r - ~)2 + ~’4~’- (1 +r)f~]

]1/2 and =r m/M.

where ~ = (w/wn)2; n2 = k/M; ;= b/2v~(kM); (b) The best design maybe the one that minimizesthe peak value of the force ratio found in part (a). For this, the derivatives of the force ratio with respect to fl and with respect to ¢ should vanish. Showthat the latter requirement gives the following frequency at the minimumpeak: tip = 2/(2 + r). Also, showthat the minimum value of the peak is 1 + 2/r, showingthe need for a heavy mass m. (c) Employthe first derivative condition noted aboveto establish the followingrelation for the associated dampingratio: ¢ = v/r/2(1 + r)(2 + 6.38 A tall building sways excessively in the wind. It is proposed to place a heavy mass on the roof that can moveback and forth on rollers, and which is constrained by a hydraulic damperas shown: ~per

A soft spring (not shown)tends to center the mass, without materially affecting the dynamics, so the dampernever gets over or under-extended. The building swaysin its first modeof vibration, whichhas a period of ten seconds; the wind force produces a synchronouscomponent.The stiffness of the building referencedto the lateral motionat the roof is 4.0 x 10s lb/ft. (a) Model the system with no internal dampingand no added mass and damper, and evaluate the effective mass. (b) Addthe inherent internal dampingwhich gives a dampingratio 0.01, and find the ratio of the maximum displacementat the roof level to the amplitudeof an effective sinusoidal force applied at that level.

454

CHAPTER 6.

ANALYSIS

OF LINEAR MODELS, PART 1

(c) Introduce the added mass and damper, and write the corresponding differential equations. Solve for the ratio of the displacement amplitude of the top of the building to the amplitude of the applied force.

6.39 Design problem. Specify the parameters of a damping system for the building of the previous problem. Tuned and untuned damper designs are possible. An equivalent damping ratio (for a building with no damping system) 0.02 is desired (doubling the inherent damping), and 0.03 is excellent. There is concern that the added mass not be too heavy for the structure to support, and that the motion of the mass relative to the building not be too great. Use the information given and whatever judgements you can defend. Describe the behavior of your solution quantitatively, and report the method by which it was chosen.

SOLUTIONS TO GUIDED PROBLEMS Guided

Problem

6.2

I

1-2.

Don’t confuse q with qc = q- ql.

0 --~

C

dqc 1 d--~ = O(t) - ~p.

I~=qoea cos

s~ d2p ldqc 1 [ ~ = dt C dt - C (l(t) In operator 4.

ql

1 ] -

form, S 2 + ~ p(t)

(

1)

= -~q

=q-qc=q-CSp

q~=q-CSsz:Cl/1Cq=

IG(Jw)l

[ ’ ] = 1 - (w/u,):

1 8~¢T/1C q; ~ = --~

q= S~-~-~-IC,]q

6.2.

SINUSOIDAL

FREQ UENCY RESPONSE

455

0"1.~

0 o.o~

(i) For the amplitude ratio to be within ten percent of 1, the frequency square of the ratio, (w/wn)2, should be less than about 0.1. (The small range of alternative possible values at slightly above 1.0 is likely of no practical use.) 2. Therefore, k > lOmw (ii) For the amplitude ratio to be less than 0.1, (w/w,~)2 must be greater than about 3.3. Therefore, k < 2. 0.3row Guided

Problem

6.3

1. l~rom the solution to part (a) of GuidedProblem4.1 (p. 203):

rx~

i-~x2

F

1~ =

tn I =

1 kg CI

=

1/k~ = 1/25 m/N

12 = rth = 2 kg C2= 1/k2 = 1/100 m/N

~

R = 0.25 N s/m ez) = R(p~/ll-p2/1

~

F

1.~

~ .o

I, 1

d~

1

dp~

1

~

~ [ql] ~ p~

~ I2

~ R

dq~

d

~ 1.~

q~/C~ 2/C2 +e~/It-pz/12x 0 ,--’~~. 1 ~-~-~C2 c~ ~ x(x~ P2/~.I meshreducti°~-"/~[L~°l/I’ I, P~p2 /I2

1

1

=

~ [

-~/c~

L 1/C~

1

1 0 0

o -1/C2

0

R

R

1/I~

-RH~ R/h 1/11-1/I~] R/I~ -R/h

~ +

[ql]

456

CHAPTER 6.

Therefore,

ANALYSIS

0 0 -100

A= 0 -25 25

Lety=pl--[0010]x,

0 -.25 .25

sothat

OF LINEAR MODELS, PART 1

. 0.125 | and B --.125J

C--[0010]andD=0.

With these values, the MATLAB command[z,p] =ss2tf (,~,B,C,D) gives the response 0 1.0000 0.1250 62.5000 0 p = 1.0e+003 * 0.0010 0.0004 0.0875 0.0125 1.2500 The accuracy of the second coefficient in the last line above can be increased b: typing p(2), whichgives the response 0. 375. 1

1

Sx2 = 1 1

R

R

1

1 (1

R

R

1 )

1

RS

RS

or, combining terms, R ( $2+

12

+~+

S+ = (5

6. Since it is desired to keep pl, solve the second equation above for p2 = p2(pl): (R/11)5"-}- 1/I~CI 0.255’ + 25 p2 -- S2 + (R/I2)S + 1/I~C1 + 1/I2C~pl : S2 p~ + 0.125S + 62.5 Substituting this into the first equation above gives I

(0.1255’÷ 12.5)(0.255’25).J Pl __-- 5"f S2 ÷.25S ÷ 25 S~ ÷0.125S + 62.5

S2 +0.125S+62.5 7. x_~: p__L~: F SF S’~ ÷ 0.375S a ÷ 12.5S ÷ 1250 (jw)2 ÷ 0.125jw-t- 62.5 8. G(jw) = (jw),~ 0. 375(jw)3 ÷ 87.5(jw) 2 ÷ 12.5jw ÷ 1250

6.2.

SINUSOIDAL The

FREQUENCY

457

RESPONSE

MATLAB commands

hum=J1 .125 62.5]; den=It .375 87.46875 12.5 1250]; bode(num,den produce the screen display below, which shows two modestly damped resonant frequencies:

Guided 1.

Problem

IIII ~ 10 Frequency (rod/sec)

0 10

10-~

I

I I I I IIII~ 10

6.4

1

80\

20",\ 0.1 :

"\~

50

0.001

2 straight-line

° °90

approximations

10

~, rad/s

100

°180 400

458

CHAPTER 6.

ANALYSIS

OF LINEAR MODELS, PART 1

The magnitude plot clearly indicates the presence of a first-order pole at roughly 5 rad/s and an underdampedsecond-order pole pair at approximately 80 rad/s. Attempting this combinationby itself produces an asymptote for the high freqencies whichis far belowthe data, however.A closer look also reveals that the asymptote for the upper frequency range has a slope of -2, not the -3 associated with three poles and no zero. These facts identify the presence of a first-order zero. A goodapproachto locating all the poles and zeros starts with approximating the asymptotes immediately after the first pole and after the complexpole pair. Then, a horizontal asymptote is drawn to locate the resonance at about 80 rad/s. The intersection of this asymptote with the first asymptote locates the zero at about 20 rad/s. Finally, the peak of the resonace is at a factor of roughly 4 above the intersection of the corresponding asymptotes. This gives a dampingratio of ~ = 1/(2 x 4) = 1/8. 2.-4. The horizontal asymptotes for the phase drop down90° at the first pole (5 rad/s.), rise back to ° at t he zero ( 20 tad/s), a nd drop t o -180° at t he complex pole pair (80 tad/s). The straight-line approximationsfor the first-order pole and zero, shownin dashed lines, spreads each rise or fall over two decades. The sumof these approximations is given by solid straight-line segments. The actual phase curve nowcan be estimated, as shownby a curved straight line. The behavior near the second-order pole pair is estimated with the help of Fig. 6.19, recognizing the value of ¢. 1600(S + 20) 5. The transfer function is very close to G(S) = (S -I- 5)(S2 + 20S ÷ 6400) Note that the number6400 is w~, the number1600 is chosen to give G(jO) = and the coei~icient 20 for the damping equals 2¢w,. The substitution S jw can be used to check the magnitude plot and possibly refine the transfer function, and to refine the phase plot. Guided

Problem

6.5

1. Five features of the given magnituderatio are evident. First, the asymptote for low frequencies has a slope of -1, which meansghat G(S) has a factor of 1IS. Second, there is a second-order zero at 10 rad/s, which meansthat the numerator of G(S) has a factor of S~ + 2~10S + (10) 2. The corresponding asymptote in the region immediately above 10 rad/s has a slope of 1. Third, the value of 1/2¢ is 5, since this is the ratio of the asymptotesat w -- 10 to the actual magnitudeat that frequency. Therefore, ¢ -- 0.1. Fourth, the asymptote for high frequency is horizontal at the magnitudeof 0.5. The intersection of this asymptote with the asymptote in the region immediately above the zero gives a pole with break frequency 50 rad/s. Fifth, the value of [G(S)[ at w = is 1.0. Putting these five conclusions together gives 82 ± 2S + 100 G(S) = 2S(S+ 50) so that

100 - w2 + j2w G(jw) = 2jw(jw 50)

The ± in the numerator recognizes the possibility that the zero could lie in the right-half plane. The pole, on the other hand, must lie in the left-half plane; otherwise, the system would be unstable and the frequency transfer function never could have been measured. This transfer function overlooks the possibility of pole-zero combinations for which the magnitudescancel, but the phase angles do not.

6.3.

TRANSFER FUNCTION EXPANSION

459

( ±2w / G(ju~) = tan-l \ ~-~--~ ] - tan-l ( 5_--Ow Note that a negative denominatorfor the argumentof an inverse tangent places the angle in the second or third quadrant, whereasa negative numerator places the angle in the third or fourth quadrant. A check of a few values shows that the plus sign in the numerator gives the proper phase for case (a), and the minussign gives the proper phase for case (b). The phase for case (c) is virtually identical to that for case (a) for low quencies, but a gradual discrepancy apprears shortly above the notch or antiresonance which appears to accelerate for higher frequencies. Apparently some magnitude-cancellingpole-zero pairs exist. If the phase difference betweenthe two cases is examinedas a function of linear frequency, on the other hand, it can be seen to increase at a steady rate of 0.9 degrees per radian/second, reaching 90° at 100 rad/s and 360° at 400 rad/s. This reveals a pure delay of T = 2~/400 seconds, so the transfer function becomes G(S)

6.3

Transfer

S2 + 2S + 100 -(~/2oo)s 2s(s + 50)

Function

Expansion

The direct method for finding the response of a high-order model to an impulse or step requires tedious application of the initial conditions. This awkwardness can be circumvented by decomposing the model into a sum of first and secondorder models, each of which has a relatively simple response. The decomposition is called a partial fraction expansion. A complex excitation signal usually can be considered to be the response of a special system to an impulse excitation. The response of a model to this excitation then can be found by multiplying the transfer functions of the special system and the model, applying the partial fraction expansion to the product, finding the impulse response of the simple component models, and summingthe results. Almost all time responses to transient inputs of linear models can be found this way. Details follow. The partial fraction expansion also is a tool in the application of Laplace transforms, a powerful formalism closely related to the present techniques. Laplace transforms are treated in Section 7.2. 6.3.1

Impulse

Responses

You have seen that the MATLAB commands ±repulse and step produce plots and/or data files for the impulse and step responses of linear models. The commandls±m extends this capability to more general disturbances. In this subsection, analytic functions of time are found for the impulse response, and in the following subsection analytic responses are found for the step and other excitations. The algebraic details that establish the coefficients can be found with the aid of MATLAB,reducing drudgery.

460

CHAPTER 6. The partial

ANALYSIS

OF LINEAR MODELS, PART 1

fraction expansion of a transfer function G(S) with distinct poles

is

G(S)- rl

S

A pole pj of multiplicity

Is rj

tpl

-

~

r2

÷’"+~+rnkiS).l S - p,~

I

(6.59)

m expands as follows:

÷ rj+l

÷ ...

(s- p )2

÷ rj+m-1

(6.60)

~

Analytical methods for carrying out this expansion are given in Section 6.3.3. You alternatively can use the MATLAB command residue, which requires that you have previously defined the numerator and denominator polynomials of G(S), hum and den, in the standard way. EXAMPLE 6.18 Use MATLAB to find the partial-fraction expansion of the fifth-order fer function of Example 6.14 (pp. 431,433), for which 200,000 G(S) = $5 26sa + 2225S3 + 4900S: + 240,500S + 200,000 Solution:

The coding

hum = 200000; den = [1 26 2225 42700 240500 200000] ; [r,p,k] = residue(num,den) produces the response r =

0.0234 0.0234 -1.2838 -10.8401 1.2370

+ 0.00721 - 0.0072i

-2.5000 + 44.6514i -2.5000 - 44.6514i -10.0000 -i0.0000 -1.0000 = k

[]

’

trans-

6.3.

461

TRANSFER FUNCTION EXPANSION The term k is blank; there is no remainder. Thus, 0.0234 + 0.0072j 0.0234 - 0.0072j G(S) -S ÷ 2.5 - 44.6514j + S + 2.5 + 44.6514j 1.2838

10.8401

1.2370 2 S ÷ I0 (S ÷ 10) S ÷ 1 Eachtermin a partial-fraction expansion contributes separately to theimpulseresponse. Thepossible termsarelistedin the left-hand columnof Table 6.2,andthe corresponding impulse responses arelistedin the right-hand column.Entry#I represents an identity. The mostusefulentryis #4,r/(S-pl). As notedin Section 6.1.7(p. 400),thisgivesthe impulse response rep’t;the polePl equalsminusthereciprocal of thetimeconstant, T, asgivenin equation (6.28).Entry#2 is merelythe special caseof Pl = 0. Entry#7 represents complex-conjugate pairof polesin theformthatresults physically, thatis the formwhichhas onlyre~ltermsin the impulse response. Itsimpulseresponse canbe found,again,as a special caseof entry#4. It is helpful to notethe identity S: - 2prS + (P~r + P~) =- (S - pr) 2 + p~. (6.61) A pure imaginary pair of poles is represented by the special case with Pr = 0. Entries #3, #5 and #6 represent multiple real poles. They can be verified by direct substitution into the differential equations that they represent. Table

6.2 Transfer

Functions

G(S)

and Associated

Impulse Responses

ImpulseResponses,

r

g(t),

t >_ 0

rS(t)

r

rtm-1

r

(m - 1)! r

S trteP~

~ (S - pl) r

m (S - pl)

(m - 1)!

trn_lePl t

rr + rij rr S - Pr - Pij +S - Pr + p~j _ 2rr(S - Pr) - 2ripi

(s - p~)2+

2ep’t [r~ cos(pit) - ri sin(pit)]

462

CHAPTER 6.

ANALYSIS

OF LINEAR MODELS, PART 1

EXAMPLE 6.19 Find the impulse response of the fifth-order model of Examples 6.14 (pp. 431, 433) and 6.18 using the results of the latter and Table 6.2. Check to see if it agrees with the plot in Example6.14. Solution: The respective terms in the impulse response are given by entries #7, #4, #5 and #4 in the table: g(t) =2e-2’st [0.0234 cos(44.6514 t) - 0.0072 sin(44.6514 - 1.2838e-l°t - 10.8401 te -l°~ + -~. 1.2370e Nearly all the terms in expansions of any order are addressed in the example above, so very few situations require additional information or calculation. 6.3.2

Step

and

Other

Transient

The unit impulse is the time derivative 6.1.2. In operator notation, ~(t)

= S u,(t)

Responses of the Unit step, as noted in Section

or us(t)

1 = -~ 5(t).

(6.62)

The step response, designated h(t), is therefore the time integral of the impulse response, designated g(t), as noted in Section 6.1.8 (p 401). As a result, h(t)

= G(S)us(t)

= -~ 5(t).

(6.63)

If you want to find h(t), therefore, you can first find g(t) and then integrate over time. More simply, you can replace G(S) with G(S)/S and carry out the process for finding the impulse response as detailed above. The replacement is accomplished by multiplying the denominator polynomial in G(S) by S. This increases the order of each term in the polynomial by one, and adds a zero as the new zeroth-order coefficient. EXAMPLE 6.20 Find the step response for the model of Examples 6.14, 6.18 and 6.19. Solution: First, a zero is added to the end of the list of polynomial coefficients in the denominator, as follows: den = [i 26 2225 42700 42500 200000 0]; The resulting poles are unchanged, except for the addition of p -- 0. There is again no remainder. The numerator coefficients change, however: 0.0001 - 0.0005± 0.0001 + 0.0005± O. 2368 i. 0840 -1. 2370 1.0000

6.3.

TRANSFER FUNCTION EXPANSION

463

The complete step response nowcan be asssembled as h(t) =2e-2’5~ [0.0001cos(4.6514t) - 0.0005 sin(4.6514t)] ÷ 0.2368e-l°t ÷ 1.0840re-mr - 1.2370e-~ ÷ 1. The sine and cosine terms happento be essentially negligible. Excitations other than steps also can be related to an impulse by transfer functions. The excitation u(t) is related to the impulse ~(t) in Fig. 6.21 by the transfer function G~(S),and in turn is related to the output variable x(t) by the transfer function G(S). The two transfer functions are cascaded. Their product then becomesan overall transfer function G~(S) between(~(t) x(t) EXAMPLE 6.21 Find the transfer function G~(S)that gives the same response whenexcited by a unit impulse ~(t) as a modelwith transfer function G(S) gives when excited by u(t) = Uoe-~/~, ~ > O. Solution: Accordingto entry #4 in Table 6.2, the given input is the unit impulse response of a systemwith transfer function

=

u0

The relation between the impulse and the response becomes z(~) = ~(S)~(~) = C~(S)C(S)~(~)

S ÷ I/T

Tocompute theresponse, therefore, youintroduce a factor (S+ 1/~)tothe denominator ofthetransfer function andproceed asifsolving foranimpulse response. This procedureis developedfurther through the use of the Laplace transform in Section 7.2. It has broad utility in the consideration of control systems, as presented in Chapter 8. In particular, a table of "Laplace transform pairs," given in AppendixC, relates transfer functions G~(S) or G~(S) to a considerably greater variety of signals u(t) or x(t), respectively, than those given in Table

or

Figure 6.21: Cascading two transfer functions to relate .the response of the second to an input impulse rather than its actual input

464

CHAPTER 6. ANALYSIS OF LINEAR MODELS, PART 1

6.2. (The operator S is written as a lower case s.) Youmayuse this table now, without formal study of the transforms. 6.3.3

Determination

of Partial

Fraction

Expansions

Partial fraction expansions can be found through the use of the MATLAB function residue, as described above. They also can be found analytically, as described presently. A transfer function G(S) that is a ratio of polynomials in S, i.e. G(S) N(S)/D(S), with the order of N(S) being no higher than that of D(S), can be expanded with no remainder. In the unusual cases where the order of N(S) exceeds that of D(S), D(S) can be divided into N(S) to give a new N(S) with the same order as D(S) plus a remainder polynomial in S. This first step, if necessary, is presumedbelow to have been executed already. ’ In the simplest cases the polynomial D(S) has distinct real roots, as in equations (6.59) (p 460), and the remainderk(S) is zero. Thecoefficients ri are called residues. They can be determinedby multiplying both sides of equation (6.59) by S - Si and then setting S equal to S~. EXAMPLE 6.22 Evaluate the residues rl, r2 and r3 of the transfer function r2 G(S)= S(S+21)(S 5)= rl~+ ~ + S +~’ r3

which has the impulse response g(t) = rl + r2e-t -5t. + r3e Solution: To find r~, for whichS~ = 0, multiply both sides of the equation by S and then set S -- 0: 2 (r r2S r3S "~ lim SG(s) = lim s-~o s-~o(S+ 1)(S+-5)-s-~olim\1 -~~--~ ~-S -[- 5,] This gives rl-

2 (0+1)(0+5)

’

- 0.4.

To find r~, for which $2 = -1, multiply G(S) by S + 1 and set S = -1: s]i_,m_~(S + 1)G(S)= sli_,m_~ S(S+ 5) This gives 2 r2 - (-1)(-1+ The same procedure gives ra = 0.1.

-" --+-~+

r~. r~(s+ 1)] s+5 J"

6.3.

465

TRANSFER FUNCTION EXPANSION

Whena pair of distinct roots are complex conjugate, the same procedure can be applied to get the real and imaginary terms rr and ri of entry #7 in Table 6.2 (p. 463). It is simpler to get these coefficients, however, by leaving the sum of the complex conjugate terms of the transfer function in the alternative real form. A compromise strategy starts by evaluating the residues for any distinct real poles using the method above. Then, multiply both sides of the equation by the denominator of the transfer function (so the left side of the equation becomes the original numerator) and proceed to evaluate the coefficients. This procedure will provide one redundant check on the results for each real root previously evaluated.

EXAMPLE 6.23 Find the partial fraction expansion and the unit impulse response for the model with the transfer function 5S2 + 22S + 39 G(S) = 2 + 2S+ 5)( S + 2

Solution:

Expand G(S) as follows: 5S2 + 22S+ 39 aS + b c + 2 (S + 2S+ 5)(S + 2) (S 2 + 4 S + 2

Next, find c by multiplying both sides by S + 2 and then setting S -- -2: c-

2O - 44 + 39 -3. 4-4+5

Then, multiply both sides of G(S) by the denominator: 5S2 + 22S+ 39 = (aS + b)(S 2) + 3 (S2 + 2 s + 5 ) = (3 + a)S2 + (2a + b + 6)S + (25 + 15). A comparison of the coefficients of the polynomials in S gives a = 2 and b -- 12, plus a check. (Had c not been found first, this comparison would have given three equations with three unknowns, permitting c to be found anyway, without the check.) The part of the impulse response corresponding to the second-order pole is found using entry #7 in Table 6.2 (p. 461), which gives 2rr--a--2;

2ri-

-b - 2rrPr Pi

The complete impulse response becomes -t. + 5si n2t)e g(t) = -2t + (2cos2t

5.

466

CHAPTER 6. ANALYSIS OF LINEAR MODELS, PART 1

The same methodcan be used in the case of multiple roots, except table entries #5 and #6 are used instead of #7. EXAMPLE 6.24 Find the partial fraction expansion and the unit impulse response for a modelwith the transfer function 3G(S)= $2+2S+3 (S+1) Solution: First, G(S) is expandedacdording to equation (6.60) (p. 460): S2 + 2S + 3 a3 a2 (S + 1)3 = (S + 1)----~ + (S + 1)----~ + (~ + 1----~; Multiplying by the denominator, (S + 1)3G(S) 2 + 2S + 3= a3+ (S + 1)a2 + (S ÷ 1)2 al = aiS2 + (2a~ + a2)S + (a~ + a2 a3). Therefore, a~ = 1;

2a~ + a2 = 2;

al + a2 + a3 = 3,

from which a~ = 1, a~ = 0, a3 = 2. The impulse response (entries #4 and #6 in Table 6.2, p. 461) is g(t) = t2e -t + O + e-t -t = ,(l+t~)e 6.3.4

t >_O.

Summary

The transfer function of most high-order models can be represented as a sum of transfer functions of first-order and second-order modelsthrough a partialfraction expansion. (The only exceptions are models with more than one repeated real root or a repeated complexpair.) Thus the step and impulse responses of most high-order models can be found without having to determine the coefficients of the respective terms by the relatively laborious procedureof substituting the initial conditions into the general solution and its several derivatives. The work is expedited through the use of Table 6.2, or a corresponding table of Laplace transforms such as given in AppendixC. The formalismof these transforms, along with a broader class of applications, is given in Section 7.2. Systeminputs other than impulses often can be treated as the output of a second system with an impulse input. The product of the associated transfer function with the transfer function of the systemitself transforms the problem into one with an impulse input. The cascading of transfer functions can be generalized to address a variety

6.3.

TRANSFER FUNCTION EXPANSION

467

of problems regarding the control of engineering and physical systems, as is introduced in Chapter 8. Guided Problem 6.6 This is a basic problemin which the temporal response of a linear modelto a disturbance is found using operators and a partial fraction expansion. A modeldefined by the differential equation d2 x 2 dx dt "-~ + dt + 8x = u(t) is excited by the disturbance u(t) = 3(1 - e-t/2). Find an analytic expression for x(t). Suggested Steps: 1. Write the differential equation in operator form, and deduceG(S). 2. Notethat the first term in u(t) is a constant, so its contributionto x(t) is that of a step response. Find the correspondingoperator Gul (S) by which G(S) is multiplied in treating the problemlike an impulse response. 3. Note that the other term in u(t) is a decaying exponential. Find the corresponding operator Gu2(S) by which G(s) is multiplied to give an equivalent transfer function for an impulseinput, using Table6.2 (p. 461). 4. Sumthe operators from steps 2 and 3 to get G~(S). 5. Multiply G~(S) by G(s) to get the equivalent transfer function G~(S) which treats the problemlike an impulse response. Determineits numerator and denominator polynomials. 6. Use the MATLAB commandresidue to implement a partial pansion of the modifiedtransfer function, or

fraction ex-

7. Determinethe partial fraction expansionanalytically. 8. Assemblex(t) from the information given by steps 6 or 7, again using Table6.2. PROBLEMS 6.40 Verify the equality of the complexand real expressions for G(S) given in entry #7 of Table 6.2 (p. 461). 6.41 Find partial fraction expansions for the following:

468

CHAPTER6. ANALYSIS OF LINEAR MODELS, PART 1 1 (a) G(S) = (S 1)(S + 2)(S +~ S+ 1 (b) G(S) = S(S 2)(S+3)

(c) ¢(s)=

(S+1)2(S+2)

(d) G(S)=

2S2 + 6S + 7 (S + 2)(S2 + 2S + 2)

6.42 For the model, input, and initial conditions d3x d2x "3 dx dt 3 + 3-~y + -~ + 2x = 6(t), x(O) = O, ~c(O) = O, Y;(O)

(a) Identify G(S). (b) Factor G(S). (c) Performa partial fraction expansion of G(S). (d) Find x(t). 6.43 Find the transfer function G~(S) that relates an impulse input 6(t) the function u(t) c(1 - e-t/T). Hints: This u(t) is proportional to thetime integral of e-t/T , whichis treated in Example6.21. Alternatively, the procedure used in Guided Problem6.6 can be used. 6.44 Find the solution to the second-order problem

d2z 2dz

dt "-T + dt + 5x = 3(1- e-2t);

x(0)= 0; ~(0)

using the techniques Of the present section. Compareyour solution to the classical methodapplied to this problem in Example6.3 (pp. 398-399). Hint: See the preceding problemfor help in finding Gu(S). 6.45 Revisit Guided Problem 6.1 (p. 403), substituting present section. The following steps are suggested:

the methods of the

(a) Theinput signal, u(t) = sin(0.95wnt)for t > 0 but u(t) = 0 for t < 0, can be considered to result from an impulse exciting an undampedsecondorder model. Determinethe transfer function Gu (S) for that model.

6.3.

TRANSFER FUNCTION EXPANSION

469

(b) Find the transfer function G(S) for the given system, and multiply it by Gu(S) to get the equivalent transfer function G~(S). (c) Employa partial-fraction

expansion on G~(S).

(d) Use the result of part (c) to find the response, x(t). 6.46 Determine the response of the system with the transfer the given input excitation. You may use MATLAB. 12 G(S) = (S 1)(S 2 + S + 4)

SOLUTION Guided

Problem

TO GUIDED

function below to

u(t) te -3t.

PROBLEM

6.6

1 ~ + 2S + 8" S 2. The step response is the time integral of the impulse response, 1. (S~ + 2S + 8)x = u(t). Therefore, G(S)

so G~I(S) = 3/S. 3. u~(t) -3e-t/~, soent ry #4 in Table 6.2 (p. 461)appli es with r = - 3 an p = -0.5. Therefore, G~(S) = -3/(S 0. 5). 3 3 1.5 4. au(s)=GuI(S)-I-Gu2(S)= S+0.5 S( S+0.5) 1 1.5 G~= ~+2S+8 S S~+0.5S 6. The MATLABprogramming 5.

n = 1.5; d=[1 2.5 9 4 0]; [r,p,k]=residue (n,d) produces theresponse 0.0194+ 0.0318i 0. 0194- 0.0318i -0.4138 0.3750 p= -I.0000+ 2.6458i -I.0000- 2.6458i -0.5000 0 k= []

1.5 4+ = 2.5S3+9S2+4S+0" S

470

CHAPTER 6. 1.5 ~ 7. (S 2q_2S_bs)S(S.b0.5)

ANALYSIS

OF LINEAR MODELS, PART 1

1.5 = S2q-2Sq-8 + (0q-0-bS)(0q-0.5) aS+b[

]1

+ [(-.05)~ + 2(-0.5)+ 8](-.05) o.5 aS ÷ b 3/8 12/29 S~+2S+8 S S+0.5 Multiplying both sides by (S2 + 2S q- 8)S(S ÷ 0.5) gives 12 3 ~+0.5s)+ ~+28+8)(s+0.5)-~(s -

1.5= (aS +5)(S

~_(S

~+88) +28

= (a - 0.0388)Sa q- (0.5a -b b q- 0.1099)S~ ÷ (0.55 q- 0.0647)S-b 1.5 Setting the coefficient of the S3 term equal to zero gives a = 0.0388, and setting the coefficient of the S term equal to zero gives b -- -0.1293. These values satisfy setting the coefficient of the S~ term equal to zero, whichserves as a check. The values of p~ = 1 and pl = 2.6458 come from the factoring S2 + 2S + 8 = (S 1)~ + (2.6458)2. The values of rr and ri are rr = a/2 = 0.0194 and ri = (-b/2 rrp~)/pi = 0.0318. These values agree with the results of MATLAB found in step 7. 8. From entries #2, #4 and #7 in Table 6.2, x(t) 2e-~[0.0194 cos(2.6458t) - 0.0318 sin(2.6458t)] - 0.4138e-°’St ÷ 0.375

6.4

Convolution*

The property of superposition has been emphasized repeatedly, most notably in Sections 3.5.2 (pp. 148-149), 5.3.1 (pp. 341-343) and 6.1, as a meansto determine the response of a linear system to an arbitrary input. If you understand this concept, the present section becomes an optional development that leads to both formal methods of analytical solution and numerical approximations that are readily implemented using MATLAB.If, rather, you would merely like a more thorough grounding in superposition, without the detailed analytical and numerical methods that usually can be replaced by other means, you can restrict study to Sections 6.4.1 and 6.4.2. Convolution is a process of applying superposition, in which the input signal to a linear system is decomposed into a sum of steps or a sum of square pulses or impulses equally spaced over time. If a limited number of steps or pulses are used, the decomposition may be approximate; if an infinite number of steps or pulses is used, the decomposition becomes exact. The response to a unit step or pulse is then determined, from which the responses to each individual input step or pulse is found. Finally, these responses are summedto give the response to the original input.

6.4.1

Decomposing Signals into a Sum of Steps

An arbitrary function of time f(t) can be approximated by either a sum of pulses or a sum of steps. Steps may be preferred, since fewer are required to form a decent approximation, as suggested in Fig. 6.22. If the successive steps

6.4.

CONVOLUTION*

471

~_~(t) ~ ~ df~ ~

~,

,,

Af~ (negative) ~

Figure 6.22: Approximation of a continuousfunction by a series of step functions are distinguished by a subscript, k, their various amplitudesare designated by Ark and the times at which the jumps occur are designated by Tk, the entire step-function approximationfor t > to can be written f(t) ~_ f(to) ~-~A fkus(t -- Tk)

(6.64)

k

The steps in the figure are placed at varying intervals of time, as appropriate considering the varying slope of f(t). Alternatively, the steps could be uniformly spaced with a constant interval AT. In this case, multiplying and dividing the summationterm by AT(so as not to change the equation) gives Ark . Tk)AT. f(t) ~- f(to) + ~ -~--~us~t

(6.65)

k

Nowas AT -+ dT (i.e. becomesinfinitesimally small), the ratio A flAT becomes the slope df/dT and the summationbecomesan exact integration:

ft?

rt df ~ u,(t

](t) = ](to)

T)dT.

(6.66)

The interval t < T < ~x~ contributes nothing to the integral (since #l(t - T) = 0 for T > t), so the upper limit of integration can be reduced from ~ to t if desired. 6.4.2

Discrete

Convolution

Anarbitrary input or disturbance u(t) is to be imposedonto a general stationary linear system as represented by its operator G(S). Either a crude step-wise approximationor an exact respresentation of u(t) can be used. In both cases, it is assumedthat u(t) is zero for t < 0. In the case of the approximation, u(t) ~ Au(Tk) us(t -- Tk) k

(6.67)

472

CHAPTER 6. Table 6.3

unit step at origin t, sec h(t) 0.02 0.04 0.06 0.16 0.10 0.40 0.14 0.60 0.18 0.74 0,22 0.82 0.26 0.88 0.30 0.94 0.34 0.96 0.38 0.98 0.42 1.00 0.46 1.00 0.50 1.00 0.54 1.00 0.58 1.00

ANALYSIS

Equation (6.68) time ~ sea 0.04 0.08 0.12 0.16 0.20 0.24 0.28 0.32 0.36 0.40 0.44 0.48 0.52 0.56 0.60

OF LINEAR MODELS, PART 1

Applied to Example 6.25

Au(Tk)h(t -- Zk) for TI~IAu(Tk) .06/.80 .10/.16 .14/-.14 .18/-.02

.02/.20 0.008 0.032 0.080 0.120 0.148 0.164 0.176 0.118 0.192 0.196 0.200 0.200 0.200 0.200 0.200

0.032 0.128 0.320 0.480 0.592 0.656 0.704 0.752 0.768 0.784 0.800 0.800 0.800 0.800

0.006 0.026 0.064 0.098 0.118 0.131 0.141 0.150 0.154 0.157 0.160 0.160 0.160

-0.006 -0.022 -0.056 -0.084 -0.104 -0.115 -0.123 -0.132 -0.134 -0.137 -0.140 -0.140

-0.001 -0.003 -0.008 -0.012 -0.015 -0.016 -0.018 -0.019 -0.019 -0.020 -0.020

8tim 0.008 0.064 0.214 0.460 0.669 0.793 0.858 0.907 0.955 0.975 0.988 1.004 1.004 1.000 1.000

The key idea is that the linear operator G(S) satisfies superposition. The response to the sum of steps is simply the sum of the responses to the individual steps: x(t) ~_ Ek Au(Tk) h(t I - Tk).

(6.6s)

EXAMPLE 6.25 Consider the system with the response to a unit step, h(t), as measured experimentally (perhaps by throwing a switch) and plotted in part (a)of Fig. 6.23. The system is excited by the signal u(t) as also plotted. Use step decompostion of the input with At = 0.04 seconds and discrete convolution to approximate the response. Solution: The first step is to approximate u(t) by a sum of steps. This is done graphically in part (a) of the figure. The times T~ and amplitudes the steps Au. Tk) also are given in the first numerical row of Table 6.3. The use of only five steps may seem rather crude, but in view of the sluggishness of the step response its approximateness is not bad for most purposes. The next step is to find the responses to the five componentsteps. These are shown by dashed lines in part (b) of the figure, and are given in the five associated columns in Table 6.3. The sum of these responses is the approximate system response, x(t), plotted as a solid line and tabulated in the right-most column of the table. An alternative discrete convolution is based on decomposing the excitation signal into a sum or sequence of square pulses of amplitude u(Tk)AT, which is the product of the height and the width of the kth pulse. Therefore, u(t) ~- u(Tk)AT JT(t -- Tk) k

(6.69)

6.4.

473

CONVOLUTION*

1.2

~

1.0 0.8

citation and stepwiseapproximation,u(t)

0.6 0.4 0.2 0

0

0.1

0.2

0.3

0.4 time, seconds

0.5

0.6

(a) informationgiven 1.2 systemresponse, x(t) shed curves)

1.0

~

0.8 0.6

¢ //The dashedcurves represent the responses

0.4

/~~/// of the five componentsteps.

0.2 0

0

0.1

-OT:2"-S ]] 0?3 074 time, seconds (b) response Figure 6.23: Example6.25

075

0.6 --

474

CHAPTER6. ANALYSIS OF LINEAR MODELS, PART 1 Table 6.4 Equation (6.70) Applied to Example 6.26 for unit pulse centered at t = 0

t,sec g(t)

.02 2.5

.06 .10 .14 .18 .22 .26 .30 .34 4.5 5.5 4.25 2.75 1.75 1.5 1.0 0.5

.38 0.5

.42 .25

.46 0

.50 0

.54 0

.58 0

center of pulse T~, sec (ls_t line); pulse area (2nd line) .04 .08 .12 .16 .20 .24 .28 .32 .36 .40 .44 .48 .52 .56 .60 .008 .04 .0464 .0408 .04 .04 .04 .04 .04 .04 .04 .04 .04 .04 .04

Au(T~)g(t - T~) .06 .10 .14 .18 .22 .26 .30 .34 .38 .42 .46 .50 .54 .58 .62

.O20 .036 .044 .034 .022 .014 .012 .008 .004 .004 .002 0 0 0 0

.10 .18 .22 .17 .11 .07 .06 .04 .02 .02 .01 0 0 0

.1160 .2088 .2552 .1972 .1276 .0812 .0696 .0464 .0232 .0232 .0116 0 0

.1020 .1836 .2244 .1734 .1122 .0714 .0612 .0408 .0204 .0204 .0102 0

values sum (horizontally) .10 .18 .22 .17 .11 .07 .06 .04 .02 .02 .01

.10 .18 .22 .17 .11 .07 .06 .04 .02 .02

.10 .18 .22 .17 .11 .07 .06 .04 .02

.10 .18 .22 .17 .11 .07 .06 .04

.10 .18 .22 .17 .11 .07 .{)6

.10 .18 .22 .17 .11 .07

.10 .18 .22 .17 .11

to give

.10 .18 .10 .22 .18 .10 .17 .22 .18 .10

0.0200 0.1360 0.3400 0.5648 0.7308 0.8256 0.8830 0.9314 0.9650 0.9816 0.9960 1.0036 1.0020 1.0002 1.0000

The response to a single pulse of unit amplitude, centered at time Tk, is defined as gT(t -- Tk). Therefore, the overall responseis x(t) ~_ Z gT(t - Tk)U(Tk)AT.

(6.70)

k

EXAMPLE 6.26 Applypulse convolution to the signals of Example6.25, using the unit impulse response g(t) plotted in part (a) of Fig. 6.24 and the sameexcitation u(t) as repeated in the figure. Use the sametime intervals. Solution: Fifteen square pulses as represented in the figure serve to approximate u(t). Their times and areas are listed in the first two numericalrows of the maintable given in Table6.4. Valuesof g(t) are listed just underthe figure title. The responses to the individual pulses are plotted with dashed lines in part (b) of the figure, and listed in columnsin the table. Thesums the responses at a given time t represent the completeresponse x(t) at that time; these are shownin the plot by dots and in the table by the right-most column. 6.4.3

Discrete

Convolution

by MATLAB

The MATLAB commandcony(p, q) carries out discrete convolution of the signals p and q, which must be vectors of equal length. For time convolutionbased on steps, p represents the vector of values of Au, and q represents the vector of values of h. For time convolutionbased on pulses, p represents the vector of

6.4.

CONVOLUTION*

475

0.4 ~ ~ I "~lx°ne-tenth the response to a unit impulse, g(t)/lO ~ 0.2 0

~ 0

0.1

0.2

0.3 0.4 time, seconds

0.5

0.6

(a) informationgiven 1.2 systemresponse, x(t)

1.0

ed from dashedcurves;solid line is a smoothedinterpolation)

0.8 0.6

~

0.4 0.2 0

0

0.1 (b) response

~_,_

0.2

0.3 0.4 time, seconds

Figure 6.24: Example6.26

0.5

0.6

476

CHAPTER 6.

ANALYSIS

OF LINEAR MODELS, PART 1

values of the pulse areas uAt, and q represents order of p and q can be reversed.

the vector of values of g. The

The computation produces a vector for the output that is twice the length of the procribed vectors. For purposes here, the second half of this vector is spurious, and should be discarded.

EXAMPLE 6.27 Carry out the step-wise Solution:

convolution of Example 6.25 using MATLAB.

The MATLABcoding

du = [.2 .8 .16 -.14 -.02 0 0 0 0 0 0 0 0 0 0]; h = [.04 .16 .4 .6 .74 .82 .88 .94 .96 .98 1 1 1 1 1]; x = conv(du,h) in which du is really Au, produces the response = X

Columns 1 through 7 0.0080 0.0640 0.2144 Columns 8 through 14 0.9076 0.9552 0.9748 Columns 15 through21 1.0000 0.8000 0 Columns 22 through28 0 0 0 Columnn 29 0

0.4600

0.6688

0.7928

0.8584

0.9884

1.0036

1.0036

1.0004

-0.1600

-0.0200

0

0

0

0

0

0

The first 15 numbers are correct; all the remaining numbers should equal 1.0000, but do not. This is because the vector h is truncated at 15 numbers, whereas the real function continues at unity. The vector can be truncated to exclude the spurious tail by entering x = x(l:lS), and plotted by entering plot (x).

EXAMPLE 6.28 Carry out the pulse-wise Solution:

convolution of Example 6.26 using MATLAB.

The coding

udt=[.2 1 1.16 1.02 1 1 1 1 1 1 1 1 1 1 1]*.04; g=[2.5 4.5 5.5 4.25 2.75 1.75 1.5 1.0 0.5 0.5 0.25 0 0 0 0]; x=conv (udt, g) gives virtually

the same response as the previous step-wise case.

6.4. 6.4.4

477

CONVOLUTION* Convolution

Integrals

For an exact analysis you can let Au -~ du and let T becomecontinuous, so equation (6.68) (p. 472) gives the pair of equivalent equations

x(t) = h(t - T) i~(T) dT =

(6.71)

h(v) i~(t

The dots over the u’s in these expressions imply differentiation with respect to T and r -- t - T, respectively. These key equations are knownas the superposition or convolution integrals of the time functions h(t) and u(t), often written h(t) fi (t), wh ich means th at the output equals th e convolution ofthe step responsewith the time rate of changeof the input. Equations (6.71) can be integrated by parts to give an alternative pair convolution integrals that are morewidely used. Recall the formula

f

(6.72)

u dv = uv -Iv du,

and let du = -h(t - T) dT = -g(t - T)

(6.73a) (6.73b)

dv = u(T) dT. The first term on the right side of equation (6.72) becomes

(6.75)

h(t - T) u(T)Itto =h(O)u(t) - h(t - to) = 0,

whichvanishes becauseu(to) was given as zero, and, for a causal system, gs(O) 0. Thusequation (6.72) gives x(t)

~o

g(t T) u(T) dT-f_t-to

g(~)u(t~)

dT.

(6.76)

These express the convolution g(t) u(t), which re presents th at th e output equals the time convolutionof the impulse responsewith the excitation signal. The discrete convolutionbased on square pulses, as given by equation (6.76), can be seen to be the approximationof equation (6.70) for uniformly separated discrete times.

478

CHAPTER6. ANALYSIS OF LINEAR MODELS, PART 1 EXAMPLE 6.29 A simple system is described by its impulse response: g(t) =-at,

t >O.

Use a convolution integral to computeits response to the input signal u(t) = 1 - -bt,

t > 0.

Thesesignals are plotted in parts (a) and (b) of Fig. 6.25. Solution: Fromequation (6.76), the response

= -ot

e(~_b)r~ t= 1

X--7]

a(a-b)

The meaningof convolutionis presented visually in part (.c) of the figure. The functions g(t-T) and u(T) are plotted vs. the commondummytime T; their product also is shown. The shaded area of the product is the value of x(t). Thefirst function g(t-T) can be movedright or left to give any desired time t. It can be perceived as a function whichweights the contributions to the present state of the various parts of the excitation signal as a function of the age of those parts, that is howlong before the present time they occured. In the example,the older the segmentof excitation u(T) considered, the less it contributes to the response x(t) (regardless of the function u(T)), since g(t - T) decays monatonically as T is reduced backwardin time from the current time t. (The other three convolution integrals can be interpreted similarly.)

6.4.5

Summary

The superposition property dictates that the response to a signal comprising a sum of componentsignals equals the sum of the responses to the individual componentsignals. Approximateimplementationof this property can be carried out by decomposinga disturbance into a sum of discrete steps, and summing the responses to each of these steps, in a process knownas discrete convolution. An alternative discrete convolution employs the impulse response instead of the step response, and the input signal directly in place of the changesin the input signal. In both cases the result approachesthe exact solution as the steps are made smaller and smaller; the summationsbecomeintegrations knownas convolutionintegrals.

6.4.

479

CONVOLUTION*

u(t) 1/b t (a) input signal

(c) computation

1/a t (b) impulse response

1~ .... u(T)

-)~

I/b T product of abovefunctions: 0 area = x(t) g(t-T)u(T)l~=fg(t-T)u(T)dT T Figure 6.25: Exampleof a convolution integral

480

CHAPTER6. ANALYSIS OF LINEAR MODELS, PART 1

I

I

t, sec l.O

0

0.5

I

t, sec

Figure 6.26: Excitation signal for GuidedProblem6.7 1.0~ t~2.0

_~.F~ i ~ul t, sec.

Figure 6.27: Excitation signal for Guided Problem6.8 Guided

Problem

6.7

This is a mandatoryproblem involving superposition which can be done graphically, without mathematics. A system G(s) responds to a unit impulse with a square pulse of amplitude 2.0 and duration 1.0 seconds. The system is excited with a double-step signal as shownin Fig. 6.26. Find and plot the response of the system. Suggested Steps: 1. Since the excitation comprises steps, it is most convenient to work with the response of the system to steps. Integrate the unit impulse response to give the unit step response. 2. Find the responseof the systemto the individual steps of the input signal. 3. Superimposethe responses of step 2 to give the complete response.

Guided

Problem

6.8

The system in the previous guided problem is excited by the signal shownin Fig. 6.27. Find its response. Analytical convolutionis intended. Suggested Steps: 1. Since the excitation is not a finite sumof steps, the superposition must be carried out either by representing the excitation approximately by a

6.4.

481

CONVOLUTION*

sum of small steps or or square pulses, or by carrying out a convolution integration. Choosethe integration so as to get an exact answer to the problem as posed. You have the choice of computing the convolution g(t) * u(t) or h(t) * i~(t). If youopt for the latter, find ~(t) and h(t). Set up one of the fourconvolution integrals as given in equations (6.71) and (6.76) (p. 477). Sketchthe factors in the integrand versus the variable of integration, Takespecial care of the limits of integration. Completethe integration and sketch the result. Doesit makesense? PROBLEMS 6.47 The response of a particular linear systemto a unit impulse is 2 sin 3t.

00

T t

(a) Find the response of the system to unit st ep, assuming zero in itial conditions. (b) The system is excited by the square pulse shown below. Find the smallest value of T for whichthe response of the systembecomeszero for t>T. 6.48 Carry out suggested step 9 of GuidedProblem5.7 (pp. 353-354) using the MATLAB commandcony rather than isim. 6.49 Find and plot the solution to Problem6.46 (p. 469) numerically, using the MATLAB commandcony. Let 0 < t < 10. 8.50 Do part (b) of Problem 5.40 (p. 355) using the MATLAB command rather than lsim. 6.51 Use MATLAB to plot an approximate response x(t), t _< 8 seconds~ the modelwhich has the step response h(t) and the time derivative/~(t) of the input disturbance u(t), as given below.

it(t)l 0

0

2

I

I

4 t, sec

482

CHAPTER 6.

ANALYSIS

OF LINEAR MODELS, PART 1

{}.52 Find the exact analytic solution to the preceeding problem.

SOLUTIONS TO GUIDED PROBLEMS Guided

Problem

1. h(t)= f

6.7

g(t) at=

2t, / 2, 0,

O
2. Response to step att--0:

x~=0.5h(t)=

0, t_<0 t, O
t<0.5 Responseto step att = 0.5 s:x2 = 0.5 h(t - 0.5) 0, t, 2t-0.5, t ÷ 0.5, 2,

3. x(t)=xl(t)+x2(t)=

x2I O00.5 Guided

Problem

t - 0.5, 0.5 l.5

{°,

t_<0 0 < t < 0.5 0.51.5

y 1

1L’~ "~ ’ )5t,s ~

6.8

1. As in the previous problem, g(t) 2[us(t) - ur(t - 1)] an d h(t) = 2Jut(t) ur(t-1)] where ur(t) is a ramp with slope 1 whichstarts at t = 0. The function l (t - i sin~rt) t < 2 is given, from which ;1 ’ t_>2 ,~(t) = -

t

~2(t)

J"½(1- cos~t) t <2 0

t>2

2. For the choice x(t) = g(t - T) ~(T)

u(T)

case shown: l
¯

t-1 1 t 2

~

r ~

6.4.

CONVOLUTION*

fort _<1,

483 t 1 (T-

x(t) = 2. 1 2 ~(_~ =~-t

forl
+

x(t)=

~(t)= l 5

2.~

= -~ +3tfor ¢ ~ a,

~(t) =

~ +

2. ldT = 2 1

Plot of ~he resulg:

T--sin~ t2

=

2 1co sa-t]

+

+cos~t)

2.~ T--sin~t ~ ~ 1 2 5 + ~cosrt

=tfor2
~sin~rT)dT

dT=

dT+ [1+cos

~T +

2.1dT

cos~t

Chapter

Analysis Part 2

7

of Linear Models,

Excitations other than sinusoids are treated in Chapter 6 as discrete or continuous sums of impulses or steps. This is awkward for many signals. Periodic signals are more conveniently represented in Section 7.1 by a discrete sum of sinusoids, called a Fourier series. Then, pulse-like signals are treated as a continuous sum of sinusoids, called a Fourier transform. A twist on this latter transform produces the Laplace transform, which is developed in Section 7.2. The Laplace transform is ideal for finding analytic expressions for the responses of linear models to transient excitations, disturbances that are zero for t < 0. Matrix methods for addressing linear inodels are extended in Section 7.3 by the decomposition of complex total behavior into modal components, each of which behaves very simply, with a single time constant or combination of natural frequency and damping ratio. Certain matrix-based computational methods also are developed, including the basis of the MATLAB commandls±m. A method to convert bond graphs directly to transfer functions is given in Section 7.4. Th.is method is a considerable convenience for those who use bond graphs extensively, but is logically an optional topic. The section starts by presenting signal-flow graphs, which are closely related to block diagrams; both are widely used by control engineers.

7.1

Fourier Analysis

An arbitrary signal is decomposedinto a sum of steps or impulses in the timedomain analysis emphasized in Sections 6.3 and 6.4. The response of a linear model to such a signal is reconstructed by summingthe individual responses to the individual steps or impulses. A different approach is now introduced, which often is called frequency-domain analysis. Instead of steps or impulses, the signal is decomposed into a sum of sine and cosine waves. The superposition property is invoked, as before: the response of the linear model equals the sum 485

486

CHAPTER 7. ANALYSIS OF LINEAR MODELS, PART 2

-T

-T/2

0

T/2

T

3T/2

2T

time, t Figure 7.1: Periodic signal with period T of the responses to the individual sine and cosine waves. A signal which is periodic in time comprises a sum of sinusoidal wavesat discrete frequencies, called a Fourier series. A non-periodic signal comprises a sumof sinusoidal wavesat all frequencies, called a Fourier transform. 7.1.1

Fourier

Series

Consider a periodic function of time u(t) with period T, such as that shownin Fig. 7.1. The frequencyof the periodic function is

(7.1)

=

Anysuch function can be represented by a sumof sine and cosine waves, or by an equivalent sumof phase-shifted cosine waves: u(t)= ao + ~ [2a~ cos(nwot) + 2b,~ sin(nw0t)], u(t)= ao + ~ 2c~ cos(nwot + On), n----1

-1 + 0n =- tan

¯

(7.2)

These sums are knownas Fourier series. The exampleof a particular square waveis plotted in Fig. 7.2. This is an even function of time, that is symmetricabout time t = 0. As a result, only the Fourier cosine series coefficients are non-zero. (All cosine wavesare even, and all sine wavesare odd, or anti-symmetricfunctions of time.) Thesecoefficients are plotted in part (b) of the figure, and the first three harmonicsare plotted in part (c). Their sum, plus the time-average term, is comparedto the actual square wavein part (d). The sumof the first seven harmonicsalso is shown. The more harmonics that are added, the closer the sum approaches the square-wave function. Youneed to evaluate the Fourier coefficients an and b,~ in order to use a Fourier series. The general formula for computingthe coefficients an results

7.1.

FOURIERANALYSIS

4~ time

487

average I

time, t

3T/2

(a) time function a, 1 2 3 ,~ : t ’ ~ 910 1~ "1~1 " /’1

(b) Fourier coefficients (for r T/4, ti me av erage =1) 2 ~..¢ ¢’~/~time...., av~rage~,x

-T/2

4

0

~

2

-T/2

~’/~"~ (- first h~rmonic

T/2 T 3T/2 2T time, t (c) the first three harmonics -sumof ~lrst , three harmonics,~.,sum ~seven harmonics, of the first plus time plus the time.., average j average exact signal

0

T/2 T (d) sumsof harmonics

3T/2 2T time, t

Figure 7.2: A particular square-waveperiodic signal

488

CHAPTER7.

ANALYSIS OF LINEAR MODELS, PART 2

from changing the index n in equation (7.2a) to k, multiplying both sides cos(nwot), and integrating over one period. To computethe coefficients bn, you replace the cosine with a sine. The respective integrations are

The period chosen here is the symmetricinterval -T/2 < t < T/2. All terms on the right-hand side of these equations equal zero, except for those that integrate the square of a cosine or sine function, which exist only whenk = n. The terms that vanish are all sinusoids with zero meanvalue integrated over an integral numberof cycles. The multiplied functions are said to be orthogonal over the time interval. The functions 1 cos2(nw0t) = 311 + cos(2nw0t)],

(7.4a)

sin2 (nwot) " "~ [1 - cos(2nwot)

(7.4b)

on the other hand, have an average value of 1/2; their integrals over the time interval T equal T/2. As a result, equations (7.3) give

(7.~)

Equations (7.2) and (7.5) comprise the Fourier series pair.

7.1.

489

FOURIER ANALYSIS

EXAMPLE 7.1 Find the Fourier series expression for the square-wavesignal addressed in Fig. 7.2. Solution: Equations (7.5) give 4cos a,~= T--!J-r~2

dr- 4T 2~rn T sln~-~--) 4~" sin(~rnT/T) =-¯ T ~rnv/T ’

n rel="nofollow">l -

bn =0. Substituting T = T/4 gives the Fourier series u(t) : 1 +

~

sin(Trn/4) cos(27cnt/T). ~n/4

An alternative expression for the Fourier series pair employscomplexnumbers:

jn~°t, u(t)= ~ Vne 1 fT/2 Un=~ J-T~2 u(t) e-J’~°tdt.

(7.6)

Since u(t) is a real function, it is necessarythat U_, be the complexconjugate of Un. In particular, Un=an -jb,~,

n >_1;

U-n= an + jbn,

n >_1.

(7.7)

whichgives the explicit relationship betweenthe coefficients Unand an, bn. This morecompactrepresentation of the Fourier series pair is used belowto generate the Fourier transform, and in Section 7.2 to generate the Laplace transform. 7.1.2

Response tion

of a Linear

System to a Periodic

Excita-

Recall that the response of a linear systemwith the transfer function G(S) to an input signal u(t) = cos(wt + 8) is x(t) = IG(jw)l cos(wt + ~), where~ /G(jw) is the angle of the phasor G(jw). Eachterm in a Fourier serier is of this form, with w = nwo. The property of superposition, then, requires the response to the Fourier series of equation (7.2b) to

490

CHAPTER7.

ANALYSIS OF LINEAR MODELS,PART 2

¯ (t) = Xo(O) + ~ ~ IX,dcos(nwot + a~), nmO

IX~] = IG(jnwo)lc~;

(7.8)

am = O~ +/G(jnwo).

Note that the constant time term is treated as a cosine of zero frequency. For the morecompactnotation of equation (7.6), the equivalent result

~ XneJn~’°t; x(t) =n=-~z

Xn = a(jnwo)Un.

(7.9)

EXAMPLE 7.2 Find the steady-state response of a linear systemwith the transfer function G(S)

2 - $2+6S+5

to the square-waveexcitation of Fig. 7.2 and Example7.1. Solution: The response, from equation (7.8), x(t) = G(0) x 1 + 2 si n(~rn/4) n=l ~rn/4 IG(2~rjn/T)lcos(2~rnt/T

an);

am = /G(2~rjnt/T). The magnitudesand phase angles of the transfer function contained in these equations are, from the given G(S), IG(2~rjn/T) l

2 V/[5 - (27cn/T)2]2 ~’ + [12:rn/T]

127cn/T ] /G21(2~rjn/T) = - tan -~ 5- (27rn/T) ~ " The response x(t) is plotted below for the case T = 5 minutes. Its mean value is G(0) × 1 = 0.4. Eachof the first three harmonicsalso are shown, and it can be seen that their sum plus the meanvalue comprises the bulk of the completesolution.

7.1.

491

FOURIERANALYSIS

response sumof first three harmonicsplus mean

meanvalue second harmonic o third harmonic -0.2 -0.4 0

4 5 time; min Care must be taken to insure that the phase angles of the individual harmonicsare identified in their proper quadrants. Use of a standard inverse tangent routine to evaluate the phase angle might place the angle for n = 1 in the fourth quadrant and all the other angles in the first quadrant. The angles should be in the third quadrantfor n >_ 2, however,so it is necessary to add (or subtract) rr to the calculated angles in these cases. 7.1.3

Fourier

1

2

3

Transform

Anon-periodicsignal can be viewedas part or all of a single cycle of a periodic signal with an extremely long period, T. If you let T ~ ~, any signal can be so represented. In this case, 27r (7.10a) ~o = -~- = A~ --~ 0, nwo ~ ~. (7.10b) The discrete frequencies nwo becomethe continuous frequency w. Equation (7.2) becomes u(t)

= lim ~ ~ U~e~’~

492

CHAPTER 7.

ANALYSIS OF LINEAR MODELS, PART 2

or

u(t) =

2~ J_~

U(jw)=

]

f U, ~ . ~olim t, A~/2~ l = ~li~o(U.T)

(7.12a)

(7.12b)

Here U,, is known as the frequency or spectral density I , which is the content per Hz (per cycle per second). The formula for U(jw) as a function of u(t) is found from equation (7.6b): U(j~)

(t) dr.

0".~3)

This computation of U(fi¢) is known as the Fourier transform of u(t), and the computation of u(t) using equation (7.12a) is known as the inverse Fourier transform of U(jw). Indicating the Fourier transform with the symbol ~’,

u(j~)= ~:[u(~)]; u(t) = Y--~[u(j~)].

(7.~)

The response x(t) of the linear system G to the input u(t) can be found by ~ modifying equation (7.9) after the fashion of equation (8.12): x(t) = ~’-~[X(j~)]; X(j~)

= G(jw)U(j~).

Although the Fourier transform is very powerful in addressing manysignals and the responses of linear systems to these signals, it has a serious limitation because it is not defined (becomes infinite) if f_~]u(t)ldt = oc. Thus, for example, the method fails for step functions and the responses of most systems to steps, and for oscillatory functions that do not decay. The transform applies only to "pulse-like" signalv. This limitation will be remedied, in Section 7.2, by employing a modification of the Fourier transform known as the Laplace transform. This text does not ask you to solve problems analytically using the Fourier transform. Should you ever wish to do so, however, you should know that many commontime functions have transforms that are given in widely available tables, relieving the analyst of analytical drudgery. 1Energy or power is usually proportional to a signal squared, and thus ]U(jw)] 2 is known as the energy density spectrum. 2This result also can be proven by applying the convolution integral, noting that G(jw) ~[g(t)], where g(t) is the impulse response. A similar proof is given in Section 7.2.2 for the Laplace transform.

7.1. 7.1.4

493

FOURIER ANALYSIS Digital

Spectral

Analysis*

The availability of an extremely powerful experimental tool known as the dynamic analyzer or spectrum analyzer is sufficient to justify learning the concepts of the Fourier transform, and partly justifies the development above. In its basic form, one voltage represents an excitation signal u(t), and a second voltage represents a system response x(t), as measured by instrument transducers. These voltages are connected to the input terminals of an analog-to-digital converter ("A/D converter"), which generates N digital values to represent each of these voltages at equally spaced times over an interval of time, T. In a basic analyzer, N = 2l° = 1024 or N -- 211 = 2048. The Fourier series for each signal is then computed, using an algorithm called a fast Fourier transform (FFT), which is much more efficient than a direct approximation of equation (7.3). By using a dedicated computer chip these calculations are completed a small fraction of a second. A sampling theorem limits the number of possible frequencies fn = ~n/2~ = niT for which Un and Xn can be estimated to N/2, or 512 for N = 1024 or 1024 for N = 2048. In practice only perhaps the first 400 for N -- 1024 and 800 for N -- 2048 are found, since the remaining harmonics are increasingly susceptable to error. The work of a dynamic analyzer can be performed on a general purpose computer cot~pled to an A/D converter, also, particularly when high speed is not critical. Software is widely available. MATLAB, as described below, can compute the fast Fourier transform and its inverse for data provided to it. The computation is more generally called a discrete Fourier transform (DFT). The signal need not have any periodicity. On the other hand, the duration of the measurementsis finite, so this duration also can be viewed as the period, T, of a periodic signal; the transform then becomes an approximate Fourier series, with the fundamental harmonic frequency lIT. This fact is exploited below to allow the FFT statements in MATLAB to closely estimate the results plotted in Figs. 7.2 part (b) and 7.3. In practice, most dynamic analyzers offer a choice of special automatic "windowing" modifications to minimize errors that result when a non-periodic signal overlaps the time boundaries of accepted data, appearing to give it an abrupt start and stop. The details are beyond the scope of current interest. The analyzer can display the real and imaginary parts of U(2~rjfn) and X(2~rjf,~), or alternatively their magnitudes and phase angles, as a function of the frequency ],~ = nit Hz. With its typically 400 or 800 points or steps per signal, the displays resemble continuous functions. Numerical values also can be displayed and digitally read out. As an example, assume that the square-wave signal of Fig. 7:2 and Examples 7.1 and 7.2 has a period of 5 minutes and is sampled over a time of T = 200 minutes. 3 The true Fourier transform would show peaks at the fundamental frequency of 1/5 cpm and multiples thereof, as given by equation (8.7). The 3Commercially available dynamicanalyzers do not usually take data at such a slow rate. At the upper end, a rate of 256,000data points per secondand a maximum display frequency of 100,000 Hz is common.

494

CHAPTER 7.

ANALYSIS OF LINEAR MODELS, PART 2

0.8 ReU 0.6 0.4 0.2

-0.2

1:2,:4 frequency, cycles/minute

Figure 7.3: Display on the dynamic analyzer for the square-wave signal

values between the peaks would be zero~ The actual screen display for N = 1024 and lIT = 0.005 cpm, as shown in Fig. 7.3, is different from the ideal shown in part (b) of Fig. 7.2, however. The first ten harmonics appear properly, to the maximumfrequency of 400/T = 2 cpm; twenty would appear (up to cpm) if N = 2048. The principal difference is the appearance of much smaller components at many other frequencies. These result from the discretization of the signals. Note that in the case shown, the signal is an even function of time so that the imaginary part of the transform is zero. On the other hand, a non-periodic signal with a single peak, such as shown in Fig. 7.4 part (a), gives a transform that is essentially continuous, such shown in part (b) of the figure. A very important special case is the impulse (of zero width).

7.1.

FOURIERANAL~rSIS

495

0.4/ 0.2 0.1 0.3~ 0 0

1

2

3

4 5 time, seconds

6

(a) pulse in time domain

0.8 0.6

~

0.4 0.2

°o ’ ~ ’ ,;, 0

-45

’ ~, ’

10

,

~

-90 -135

-18o ° , ~ ’ ~ ’ ; ’ frequency, rad/s (b) pulse in frequencydomain Figure 7.4: Exampleof a pulse and its Fourier transform

496

CHAPTER 7. ANALYSIS OF LINEAR MODELS, PART 2 EXAMPLE 7.3 Find and plot the Fourier transform of the unit impulse ~impulse~(t)

(a) time domain

t

Solution: Equation (7.13) gives U(jw) =

e-~tS(t)

Since the integrand is non-zeroonly for the instant at whicht = 0, the factor ej~t -~ e° = 1. Further, the integral of a unit impulseis identically 1, so the result is V(jw) = (b) frequency domain-716(t~] note:( _7is positivePhase anglereal)is zero

% Thus, an impulse contains all frequencies equally, and a hammerblow, which approximatesan impulse, excites all frequencies belowsomelimit associated with its actual non-zero duration. This makesthe combinationof a dynamic analyzer and a hammer,instrumented with an accelerometer to give an electrical pulse whenit strikes an object, a powerfulmeansfor testing mechanical vibration. Equation (7.15b) gives the frequency transfer function G(jw)-

X(jw) V(jw)"

(7.16)

If the analyzer is used to measure x(t) and u(t) synchronously, it also can computethis ratio. In practice, for each of the 400 or 800 frequencies, the computations [X(27rjfn)l" IG(27rjfn)l =[U(27rj f,~)[ (7.17a) / G(27rj fn) = (~z (2rj fn) - (2r jf~).

(7.175)

are madeand displayed. The results are meaningfulfor each frequency fn that has an experimentally significant content U(2rcjf~). For a hammerblow, for

7.1.

497

FOURIERANALYSIS

example, this means all frequencies, showing its special virtue (aside from ease

ofuse). Digital spectral analysis, then, is a powerful tool for examining the frequency content of time-based signals, and determining experimentally the transfer function of a linear system or process by monitoring its input and output. It can predict the responses of systems. The discussion above serves merely as an introduction to the possibilities, techniques and limitations.

7.1.5

Fourier Analysis

Using MATLAB*

The commandsU-fft (u) and U=f~t (u,ra) return the discrete Fourier transform (DFT) of the vector u, with a complex conjugate pair of numbers for each frequency. It is theoretically impossible to compute more than half as many frequency components as there are data points, so if the integer m exceeds the length of the vector u, U is padded up to the length m with zeros. If m is less than the length of the vector u, U is simply truncated at m elements. If ra is a power of two, a radixol fast Fourier transform (FFT) algorithm is employed. This choice is recommended, for otherwise a much slower mixed-radix algorithm is used. The commands±fft(U) and lfft(U,m) return the inverse transform. The discrete Fourier transform is the same as an approximate Fourier series for a periodic signal of period equal to the duration of the data vector u. One cycle of the square wave of Fig. 7.2 and Examples 7.1 and 7.2, for example, could be entered as 1024 data points evenly distributed over the T = 5 minutes as follows: ul=4*ones (i, 128) u2=zeros (I ,768) u=[ul u2 ul]; The commandU=fft (u) returns the Fourier series, with each element ~nultiplied by 1024. The first element in U is for zero frequency and therefore is real. The second element in U is for the frequency lIT cyles per second or 2~r/T radians per second, the third is for the frequency 2IT Hz, and so on up to the 513th which is for the frequency 512/T Hz. The remainder of the numbers up to the 1024th are redundant: the complex conjugates of the 2nd through the 512th numbers. They are in inverse order; the 514th is the complex conjugate of the 512th, and the 1024th is the complex conjugate of the 2nd, which means the fundamental frequency lIT Hz. The instruction U(I: 16)/1024 displays the first sixteen elements as follows: 1.0000 0.9003 0 -0.1800 0 0.1000 0 -0.0692

+ + -

0.00281 0.0028± 0.00281 0.00281

0.6366 -0.2122 0.1273 -0.0909

+ + -

0.00391 0.0039i 0.00391 0.00391

0.3001 -0.1286 0.0818 -0.0600

+ -

0.0028i 0.0028i 0.00281 0.00281

The small imaginary components correspond to a time shift of one-half a sampling interval, and are of little significance. Nevertheless, they can be eliminated

498

CHAPTER 7.

ANALYSIS

OF LINEAR

MODELS, PART 2

with the instruction U=real (U). The result agrees with that of Example 7.1 and part (b) of Fig. 7.2, up to at least four significant digits for the first several Fourier harmonics. The Fourier series can be multiplied by the frequency transfer function G(jw) to get the Fourier series of the ouput signal, x(t). The function G(jw) must have 1024 elements corresponding to the same sequence of frequencies as U, however. Therefore, you can construct a frequency vector w: w0=2*pi/5 ; for i=1:513 w(i)=(i-l)*w0; end The 512 values of G(jw) for the case addressed in Example 7.2 are found as follows: hum=[2]; den=If 6 5]; Gl=freqs (num,den,w) Thecomplex conjugates of the secondthrough the lastvalues, in inverse order, areneededto complete thevectorG: for i=1:511 G2 (i) =conj (G1 (513-i)) end G=[GIG23 ; The theoretical limit of 1024 + 2 -- 512 frequencies means that there can be 511 complex values of U (the second through the512th, and the 514th through the 1024th) plus two real values (the first and the 513th). To be completely consistent, therefore, you should eliminate the imaginary part of the 513th value of G: G(513)=real

G(513)

Nowyou are ready to compute the Fourier series for the output signal: X=U. *G; Fina/ly, a plot consistent with the complete response as plotted in Example7.2 results from the inverse Fourier transform command Pl0t(real(ifft(V)) The inverse Fourierroutineproduces verysmallspurious imaginary partsdue to round-off errors. Theplotting routine requires realnumbers, explaining the use of the sub-command real.Shouldyou ever find a non-trivial imaginary partof an inverse transform, lookfora mistake.

7.1.

499

FOURIER ANALYSIS

7.1.6

Summary

A periodic signal can be represented by a Fourier series comprising a sum of sine and cosine waves. The frequency of each component is an integral multiple of the frequency of periodicity. The response of a linear system to a periodic signal is the sum of the responses to the individual sine and cosine waves. The different frequencies are apt to be filtered quite differently, so that the ouput signal might look quite different from the input signal. High-frequency noise often is purposely stripped from a measured signal, for example, by passing it through a "low-pass filter"; what emerges is the lower-frequency signal of interest. A broad class of aperiodic signals, such as pulses, can be represented by a Fourier transform. The criterion for its existence is that the time integral of the absolute value of the signal over its entire duration be finite, ruling out step-like functions and oscillations that do not decay. The Fourier transform is the same as the Fourier series except it includes a continuous distribution of all frequencies rather than the multiples of a base frequency. The response of a linear system to such a signal is therefore the sum (or actually the integral) the responses to the individual frequency components. Digital spectral analysis may be carried out by a special instrument called a frequency or spectral analyzer, or on a general-purpose computer. MATLAB has a capability in this regard. A discrete Fourier transform is computed for a train of numbers that represents a signal measured over a period of time. The result approximates the Fourier transform for a non-periodic signal, and a Fourier series for a periodic signal with period equal to the duration of the data train. The analyzer can process both excitation and response signals simultaneously. It then can compute and display the ratio of the two, expressed as a function of frequency. This is the frequency transfer function, G(jw). Guided

Problem

7.1

This problem provides a basic experience in computing a Fourier series for a periodic signal, and then finding the periodic response whenthis signal is applied as the excitation to a linear first-order system. The saw-tooth disturbance plotted in Fig. 7.5 acts on a system modeled as dq T-~ +q = Ce (which corresponds to the RC system of Guided Problem 3.9, p 142). Find the responseq( t ). Suggested

Steps:

1. Find the average value of the disturb~ce e(t) and the consequent constant term in the response of qc(t).

500

CHA’PTER 7.

0

ANALYSIS

T

OF LINEAR MODELS, PART 2

2T

time t

3T

Figure 7.5: Disturbance for Guided Problem 7.1 2. Find an analytic expression for e(t) valid for at least one complete period of the disturbance. 3. Evaluate the Fourier coefficients an and bn for the disturbance using the function e(t). Write the Fourier series using the cosine form with its coefficients cn and phase angles fl,~. 4. Find the response of the system to a cosine wave of arbitrary

phase angle.

5. Assemble the response of the system to the complete disturbance by combining the results of steps 3 and 4. Guided

Problem

7.2

This problem both illustrates basic concepts regarding the use of Fourier series and transforms and explores a very practical appliction of the dynamic analyzer. Should you have access to a dynamic analyzer with an instrumented hammer, you would likely benefit from carrying out a similar experiment. The spectrum of the response of a linear system to a perfect impulse is proportional to its transfer function G(jw), as described in the text. Experi: mentally one attempts to generate the impulse with an instrument6d hammer. The acceleration or force signal from this hammer and the signal from an accelerometer or other motion transducer attached to the object under test are sent to a dynamic analyzer. The analyzer displays the Fourier transforms of these two signals as shown in Fig. 7.6. Describe qualitatively in what way the actual excitation pulse differs from a perfect impulse. Also, sketch-plot the transfer function (which the analyzer would do for you if asked). Discuss briefly where the analyzer with this excitation is prone to give large errors in G(jw). Suggested

Steps:

1. A perfect impulse would have a constant magnitude spectrum. The actual signal decays with frequency and goes negative at a particular frequency. Show that a square pulse of duration T seconds could produce a hammer signal similar to that given; evaluate T. (The actual pulse likely would be rounded, but the difference is not significant for the low frequencies shown.)

7.1.

501

FOURIERANALYSIS accelerometersignal o0 phase °180 I instrumented

unOer/ test

k,,

log amplitude

~

~ hammers~gnal: acce1 rom~eter eome analyzer dynamic phase0°1 °180

,og

amplitude

0

200 400 frequency, Hz

Figure 7.6: Guided Problem7.2

2. There are no scales given for the magnitudeplots, so the best you can do is sketch-plot G(S) to the same scale with no absolute amplitude specified. The interest presumablyis in the dynamiccharacter of the response, that is the natural frequencies and relative magnitudesof the various resonances. Notethat the log of a ratio of twoquantities equals the difference betweenthe logs of the quantities, and sketch-plot the magnituderatio.

3. The phase of G(j~) equals the difference between the phases of the numerator and the denominator. Sketch.

4. Note that large errors in a computedratio occur whenboth the measured numerator and denominator are small.

502

CHAPTER 7.

ANALYSIS

OF LINEAR MODELS, PART 2

PROBLEMS 7.1 A linear system is excited by the signal u(t) and responds with the output variable x(t). The system is described by a~ x dx dt ~ + -~ + 81x = u(t), and the excitation, plotted below, is the periodic signal (valid for all t) u(t) = 2 cos(3t) + cos(9t).

0

1

2

3

t

4

5

(a) Find G(jw) for the system. (b) Evaluate

IG(Jw)lfor the frequency(ies) appropriate to this problem.

(c) State which frequency component predominates in the response x(t). (d) Evaluate /G(jw) for the frequencies appropriate

to this problem.

(e) Assemble the response x(t). 7.2 A linear system responds to a unit step excitation with the output 2(1 e-t/T). Find the response of the system to the symmetric square wave shown below. Let T = 2T.

(a) Evaluate the first

three non-zero Fourier series components of u(t).

(b) Find the transfer function of the system, or the corresponding differential equation.

7.1.

503

FOURIER ANALYSIS (c) Find the response of the system to a sinusoidal excitation.

(d) Find the first three Fourier series components of x(t). Plot these emd their sum; either sketch-plots or the use of MATLAB are acceptable. (e) Since the signal comprises repeated upward and downward steps equal amplitude, the exact solution to this problem can be solved by assuming an undetermined initial state, using the given step response to find the consequent state at the end of one cycle, and setting the two states equal to one another in order to evaluate the actual state at the beginning and end of each cycle. Carry out this procedure, and compare the result to that of part (d). 7.3 Find the response of the problem above using the fast capability of MATLAB. 7.4 Find the steady-state

Fourier transform

response of the second-order system 1 d2x 2~ dx w~dt ~- +~,~-- -~ + x = u(t)

to the square-wave disturbance u(t) shown in Fig. 7.2 (p. 487) and described with the Fourier coefficients found in Example 7.1 (p. 489). AssumeT = and ~ = 0.2. (a) Note the time-average value of the disturbance, and find the response to this zero-frequency component. (b) Write expressions for the magnitude and phase angle of the nth harmonic component of the response. (c) Plot the first five harmonic components for one cycle. Which component is the largest, and why? (The use of MATLAB is suggested.) (d) Plot the sum of a least the first overall response.

five harmonics to approximate the

7.5 Find the response of the problem above using the fast Fourier transform capability of MATLAB. 7.6 A mass-dashpot system is excited by a triangular periodic force, as shown below. Find the consequent velocity as a function of time, neglecting any transient behavior associated with the initiation of the force.

,/

%, A /’

VtV"V,

504

CHAPTER 7.

ANALYSIS

OF LINEAR MODELS, PART 2

(a) Write the differential equation with 2(t) as the state variable; determine A and B, which in this case are scalars.

(b) Deduce G(S). (c) Find the Fourier series representation for the applied force F(t). The calcuation can be simplified by noting that F(t) is an even function, so the interval of integration can be limited to positive values of time, and only the cosine terms are non-zero. You may wish to use a table of integrals; otherwise, integration by parts can be used. Note finally that the even numbered harmonics vanish.

(d) Find the response of the system to the nth harmonic. Represent this terms of a phase-shifted cosine or sine wave rather than separate sine and cosine waves, since this reduces the number of summations if you intend to make a sketch-plot. Sum these responses to get an expression for the overall response. (e) For T = 1 s, m = 1 kg, a = 1 N and b = 10 N s/m, evaluate the magnitude and phase of the first harmonic of 2 and all others that are as large or larger than one percent of the first harmonic. Sketch these terms and their sum. Note that very few terms contribute significantly.

7.7 Plot the velocity ~ of the problem above, for the values given in part (e), using the fast Fourier transform capability of MATLAB. 7.8 An instrumented hammer strikes two specimens which are freely suspended and instrumented with accelerometers. With one of the specimens the phase angle between the excitation and the response is 0° or +180° at virtually all frequencies, with abrupt jumps in between. With the other the corresponding phase angle varies continuously with frequency.

(a) Characterize the difference between the two transfer functions, assuming both can be represented by ratios of polynomials. (b) What causes the difference? (c) What categorical nitude Bode plots?

difference would you anticipate

between the two mag-

7.1.

505

FOURIER ANALYSIS

7.9 An accelerometer placed on the rim of an axisymmetric bration absorber, shown below left in cross-section, produces below right on a spectrum analyzer when shaken with a purely at frequency w0. Both vertical and horizontal axes have linear

disk-and-rim vithe display shown sinusoidal force scales.

accelerometer rim

hub ~sk

~F=Fos~

~

(a) State if and why the data indicates linear or nonlinear behavior. (b) If damping is neglected there are two time functions which the data could be describing. Find and sketch-plot these accelerations, and state why the ambiguity exists. (c) (extra credit) Explain, qualitatively, the cause of the non-fundamental harmonic revealed in the display. Also, suggest any practical implications this might have for the absorber. [Hint: Consider the nature of the spring characteristic between the hub and the rim, and particularly the effect of radial as well as bending stresses. Does the natural frequency of the absorber vary with the amplitude of its motion?]

SOLUTIONS

TO GUIDED

Guided

7.1

Problem

PROBLEMS

0 T 2T 3T time~ Theaverage value of the disturbance is seen from the plot to be eo/2. 2. e(t)=

eo

eo

~-t+eo, -T
’°

~ ~ tcos T2 ~-T/2

= -~t,

dt+~

O< t <

cos

~ dt

506

CHAPTER 7.

= T~

cos

~

ANALYSIS

--

OF LINEAR MODELS, PART 2

sin

+

-T/2 "0

q_.~ [2~_~sin (2_~_~t)] -T/2

=eo(~[cosQrn)--cos(--Trn)]+4~n[Si~(Trn)--sin(--Trn)]) + 2~n[Sin(0)-- sin(--Trn)]---This answer can be anticipated since e(t) is an odd function of time and a~ represents only the cosine or even parts of the signal.

~oi ~’~(~_~)+~oo (~) dt

b. = ~ ¢-~/e tsin

.=T~

~

-~cos

~ ~T/esin

--

--

-T/~

dt

T ~cos

=eo(~[sin(~n)-sin(-wn)]-~[cos(O)-cos(-~n)]) = {-eo/wn nodd 0 n even Therefore,

~

eo e(t) = ~

(~)

2e~osin 2 t 7rn --

dqc v-~- + qc = Ce.

e0 = -~ +

The steady-state

~

2eo --zrn

(~)

2 t cos --+90 ° .

Ceo is qc = ~.

If e = e~ cos(wt + ~ - tan-r(T~), then

c qc=

~+1

qc = ~ +~=~

cos[~+ ~ - ta~-~(~)]. ~1 + (2wriT/T) e cos

+ 90° + ¢~ ,

Result for r/T = 1:

1

2

tiT

3

7.2.

THE LAPLACE

Guided Problem

507

TRANSFORM

7.2

1. Crude model of pulse:

U(jw)

e-J’l dt = 2 ]

cos wt dt = - sin ~t)

d --T/2

This resembles the given spectrumfor the hammerif wt/2 = ~r for w = 2~r x 300, or T = 1/300see.

0

7.2

100

The Laplace

200 300 400 frequency, Hz

Transform

The Laplace transform is a formal extension of the operator methods that have been used, starting in Section 5.3 and continuing most particularly in Section 6.3. It is used to aid the solution of ordinary linear differential equations, which it converts to algebraic equations. Problems with non-zero initial conditions are treated, as well as the class of excited systems with zero initial conditions addressed in Section 6.3. The Laplace transform is used in Chapter 11 to aid the solution of partial differential equations, which it converts to ordinary differential equations. The Laplace transform can be addressed at different levels. You will satisfy the mini~numrequirement for practical use by starting with Section 7.2,3. If rather you start with Sections 7.2.1 and 7.2.2, you should gain a deeper appreciation of the transform and its relation to other methods; it is shown to be adaptations of the Fourier transform and the convolution integral. 7.2.1

Development

from

the

Fourier

Transform*

The integral f_~lu(t)ldt often doesn’t exist, as you saw in the last section, precluding the use of the Fourier transform of u(t). If, however, the function u(t) is multiplied by e -~t, the Fourier transform of the product more likely does exist, at least for some range of a; that is, ~_~lu(t)e-~tldt < ~. This is virtually always true for signals that are zero for t < 0, which is the traditional

508

CHAPTER7.

ANALYSIS OF LINEAR MODELS, PART 2

transient signal; it is merely necessary that a be greater than somefinite positive number.Therefore, the function U(a + jw) = :~ [u(t)e -~t] =

(t)e-~te-J~tdt,

(7.18)

or

U(s)

t)e-Stdt =_£[u(t)];

s = e + j~,

(Z.19)

is defined. This is knownas the two-sided Laplace transform of u(t). The inverse formula is u(t)e -~t = ~-~[C(a + jw)], (7.20) or, from equation (7.12a), u(t)

= ~_~(e + j~)e~t~

=

U(~ + j~)e(~+~td(j~), 2~j ~

(7.21)

or

~(t)

2~ J~_~

This is integration in the comple~plane, usually involving residue theory, ~hich is included in the study of f~nctio~s of a complex wri~ble. Although some readers mayknow about this subject, it is beyond the scope of your present need. There is a great deal that can be done without this knowledge. This book employsthe simple expedient of a library of commonly occuring pairs of functions of time and their Laplacetransforms, in table form. Partial fraction expansions also help. Two-sidedLaplace transforms are used infrequently because of convergence problemsthat often arise whenu(t)¢ 0 for t < 0. ~or the cases of usual interest in which ~(t) = 0 for ~ < 0, however, they reduce to the commonone-slded Laplace transform, which is the only Laplace transform referred to in most textbooks:

The inverse formula, equation (7.22), unfortunately is not simplified. The response of the linear process G(~) to the input u(t) nowCanbe generalized from equation (7.1g):

This is a key result that you will employto computethe response of linear systems to explicit excitations. The Laplace G(~), like the Heavisideoperator G(S) used repeatedly before, and the special G(S) used for exponential functions st e (including sinusoidal signals whenS = j~), is knownas the transfer function of the system. They are related intimately.

7.2.

509

THE LAPLACE TRANSFORM

~region

of

d( J $ integration

.L--Z.. J .....

\

t Figure 7.7: Changingthe order of integration in equation (7.28) 7.2.2

Development

from the

Convolution

Integral*

Equation(7.24) (and its counterpart for Fourier transforms) also can be derived from a convolution integral. Presuminga causal process, that is excluding outputs that occur before their causal inputs, the convolutionintegral for the response of a system with impulse response g(t) to the excitation u(t) can be written as x(t) = u(t - 4) g(~) (7.25) Note that ~ is a dummyvariable; any symbol could be used. The one-sided Laplace transforms of the excitation u(t) and the response to a unit impulse g(t) are, respectively,

The objective is to derive the Secondof equations (7.24), with the addedobservation that G(s) is the Laplace transform of the impulse response. Therefore, the product

is computed,taking care to use dummytimes, ~ and ~. The definition t ~ ~ + ~ gives

The key idea nowis to changethe order of integration, as suggested in Fig. 7.7. This gives

510

CHAPTER 7. ANALYSIS OF LINEAR MODELS, PART 2

G(s)U(s) = fo~ ( fote-Stu(t - ()g(()d~)

The integral in the brackets { } above equals x(t), as indicated by equation (7.25). Therefore, G(s)U(s) = fo~e-St x(t)dt =

(7.30)

establishing the objective. 7.2.3

Definition

and Inverse

The Laplace transform F(s) of a function of time f(t) usually is defined as F(s) =- £[f(t)] = f(t)e-Stdt,

s = a + jw.

(7.31)

This transformexists (does not becomeinfinite) for all physically possible (finite) signals if a has a real value larger than some minimumnumber. Note that the transform is unaffected by the history of f(t) for t < 0. Either you restrict its use to functions that are zero for all t < 0, or you represent that history solely by its effect on f(0) and the time derivatives f’(0), f"(0), The inverse Laplace transform is written as f(t) = ~1I ~+~F(s)e+~tds. 27r3 J~_j~

(7.32)

Its direct evaluation is a majortopic in the study of the functions of a complex variable, whichis not addressed here since you need to knowonly a restricted set of easily computedtransforms. A set of transform pairs sufficient to ham die linear differential equations with constant coefficients is given in Table7.1; several of these are derived below. A moreextensive set of Laplace trangform 4pairs, given in AppendixC, reduces the need for partial fraction expansions. 7.2.4

The Derivative

Relations

The Laplace transform of the time derivative of a function is related to the Laplace transform of the function itself in a special way that can be found by an integration by parts:

4The pairs given on the last page of Appendix C apply to distributed-parameter such as in Chapter 11.

models

7.2.

511

THE LAPLACE TRANSFORM

Table 7.1 A Short Table of Laplace

Transform Pairs

Laplace transform

time function, t _> 0 af(t)

aF(s)

f(t) +g(t) ~ d dt,~f(t)

F(s) s’~F(s)

s" -lf(0) _

s’ -2f(O) ~F(s)

f_t~f(t)dt

¯. ¯

e-~t f(t)

+1 0

f(t - T)

F(s + a) e-T~g(s)

unit impulse (f(t) unit step us(t)

1/s

1

1 n+l 1/8

(n = 1,2,3,..) " -at e n

1/(s + a)

at

1/(s + "+~ a)

~.~t e- (n = 1, 2, 3,...) ne -~ cos(~t - ¢) = Ae-at sin(wt + ¢) A = ~/b 2 + [(c

J~

~ d (-1)n d-~F(s)

tnf(t)

ltn n!

~,

- 2ab)/w]

(bs + c)/[(s + ~ + ¢ --- tan-][(c

- ab)/bw]

~b ---- tan -1 [bw/(c -- ab)]

Jt=O

512

CHAPTER 7.

ANALYSIS

OF LINEAR MODELS, PART 2

=~ Sfo~e-stf(t)dt

+ e-Stf(t) 0

= sF(s) - f(O).

(7.33)

As a result, s often is called the time derivative operator, a description which is most apt in the most commonapplication for which f(0) = 0. It can be viewed as a ibr~nalization of the operator S used in Chapters 5, 6 and 8. Applying this property again to find the Laplace transform of the second derivative, d2 f(t) L:[--d~]

(7.34)

= s£[~]-f(O)=sg"F(s)-sf(O)-](O).

Thus, for the nth derivative, ~[~]

= snF(s)

- sn-l

f(O)

- sn-2

f(O)

- ...

- (dn-l

f

The inverse of s can be considered to be the time integral -

8

~

)dr.

/dtn-1)t=

o.

(7.35)

operator: (7.36)

The lower limit of integration in equations (8.31) and the transform pair (7.36) is given as t = 0; there also is f(0),f’(0) ,fn-l (O) in th e g eneral pairs for first and higher-order derivatives as given in the table. This time is ambiguous in cases in which the function jumps from one value to another at precisely t -- 0. The two values can be distinguished by referring to the one approached from negative time as occuring at time 0-, and the one approached from positive time as occuring at time 0 +. In practice one may choose to use either O- or 0+ in a given equation, but it is essential that all terms in the equation be interpreted consistently. Whenone deals with functions f(t) that equal zero for t < 0, it is simplest to employthe 0- option, since then the values f(0-), if(0-),..., fn-l(0-) are all 7.2.5

Singularity

Functions

and Discontinuities

It is also important to have at your disposal the Laplace transforms of a handful of key functions. The transform of the unit step us(t) is £[us (t)] = fore -st dt = -8 1e-St ~=8"1

(7.37)

0

The unit impulse 5(t) is the time derivative of the unit step, as you have seen. Thus, equation (7.33) gives £[5(t)]= £[dus,!t)] [ a~ j

= 1-us(O).

(7.38)

513

7.2. THE LAPLACE TRANSFORM

But what is u~(0)? At time t = 0 the function jumps from 0 to 1; u~(0-) and us(0+) -- 1. Whichdo you use? The answer, again, is that either is correct, as long as you choosethe sameO- or 0+ in the Laplace transform of all terms. In a particular case, one maybe more useful than the other, however. In the present case the use of 0+ gives £[(~(t)] = 0, whichisn’t very useful. Examination of the defining integral reveals that the impulsethen resides entirely outside of the integration interval. To include the impulse in that interval you use the 0option, whichgives/:[5(t)] = 1. This result maybe checkedby direct calculation: + 0

(7.39) Theuse of the 0+ option in effect transfers the role of an excitation to that of an initial condition, whichin somecases maybe helpful. 7.2.6

Other

Key Relations

The Laplace transforms of other key functions given in Table 7.1 are computed -a~, here. For the exponential decay e -at ~.. e

[

]=

e -ate - stdt=

e-(a+~)tdt

_ _e_(~+~)~ oo_

(7.40)

For a product of the exponential decay and a function f(t), Z; [e-~t:(t)]

e-(~+"~tI(t)dt = F(s +

(7.41)

For the product

~ d

=

Finally, for a function shifted in time by the delay T, £[f(t - T)] = foo°~e-~tf(t - T)dt

---- e-TSF(s)¢

where f(t) = if t < 0.

(7.43)

514

CHAPTER7.

ANALYSIS OF LINEAR MODELS,PART 2

The function e -Ts is knownas the time delay operator, since whenmultiplied by the Laplace transform of a function of time it gives the Laplace transform of the same function delayed in time by T. Note that the switch in the lower limit of integration from -T to 0 requires that the function be zero-valued for t < 0. The time delay operator maybe viewed as a generalization of the frequency pure-delay operator presented in Section 6.2.8 (pp. 439-441). The Pade approximantsintroduced there apply here, also, with S = s.

7.2.7 Finding Laplace Transformsof Output Variables Thefirst of twosteps in solving a differential equation using Laplacetransforms is to find the Laplace transforms of the desired output variables. The second step, evaluating the ouput variables themselvesfrom the transforms, is addressed subsequently. The Laplace transform of the response of an initially quiescent system to a transient excitation is especially simple to find, because all of the initialvalue terms x(0), ~(0),..., (dU-lx/dt~-l)t=o are zero. Therefore, the Laplace transformof the scalar differential equation nd d’-lx dx (7.44) an ~--~ + an-~ ~ +"" + a~-~ + aox : f(t) becomes (ans’~ + a,~_ls n-~ +... + als + ao)X(s) = F(s),

(7.45)

from which X(s)-

F(s)

p(s)

(7.46)

wherep(s) is the familiar characteristic polynomial. The other major situation is the initial-value problemin whichf(t) = for t > 0. In this case, the Laplace transform of equation (7.44) gives 1 [ +a~-~s~-2 - a X(s) =~~L(a’~sn-~ +""+a~)x(O)+(ansn-~- -~ +... + a~)~(0)(artsn-3+’" +a~)~(0) +. + (a~s

+ an - 1) \dt ]t=o

+ a,~ \ dt-~-~Y-~ ] t=0J

n-3 8

(7.47)

In almost all cases of interest, very few of the terms within the square brackets above are non-zero; usually the terms are best found by direct application of the third entry in Table 7.1. To address the linear state-variable formulation, recall the matrix notation: (7.48) [ dx(t)~ : Ax(t)+ Bu(t). ] Takingthe Laplace transform of both sides gives sX(s) - x(0) = AX(s) +

(7.49)

7.2.

515

THE LAPLACE TRANSFORM

Solving for X(s), X(s)= G(s)U(s) G(s)= (sI

+ (sI - A)-Ix(O)

(7.50)

- A)-~B.

Setting B = 0 or u(t) = 0 leaves an initial leaves a forced response problem. A more general case traditionally variable is defined as

value problem, and setting x(0) =

is cited in the literature.

y(t) = Cx(t) + Du(t).

The output

(7.5~)

In this case, Y(s)= H(s)U(s)

+ C(sI - A)-Xx(0),

H(s)_ = C(sI - A)-IB +

(7.52)

The matrix H(s) is simply a more generalized transfer function which allows output, y(t), to be different from the state vector, x(t). The elements of the matrices G(s) and H(s) scal ar tran sfer func tions or, in the language of most elementary textbooks in automatic control which do not employ matrices, just "transfer functions". Note that the expressions for G(s), G(s) and H(s) are identical to the operators first introduced in Section 5.3.3 (p 345) merely to represent differential equations, except for the use of in place of S. The scalar G(s) also is noted in Section 6.1.5 (p 397) to represent the ratio of the response x(t) to the excitation est. The subsequent development extended this result to frequency response, using s = jw. These facts are closely related. EXAMPLE 7.4 Find the response of a system with the scalar transfer function 2

a(~) (s + 1)(s+ (which is the factored version of the transfer function used in Example7.2, p. 490) to an input unit Step, using the Laplace transform table in Appendix C. Solution: The Laplace transform of the unit step is 1

u(~)= 8

516

CHAPTER 7.

ANALYSIS OF LINEAR MODELS, PART 2

The Laplace transform of the response x(t) is the product of G(s) and U(s), or 2 X(s) = s(s 1)(s + 5) This is proportional to transform #15 in the table of Laplace transforms in Appendix C, with the coefficient values a -- 1 and b = 5. The corresponding time function therefore is x(t) 0. 4 - 0. Se-t + -5t. 0. 1e EXAMPLE 7.5 Use Laplace transforms to find the response x(t) of the general underdamped second-order model to an impulse cS(t), in the presence of general initial conditions x(0) and 2(0) (same case as in Example 6.4, pp. 400-401). Solution: The Laplace transform of the differential

equation gives

[(s + a)"~+ ~]X(s)- ~x(0)- 2(0)- 2a~,(0)

a = ~,~.

Solving for X(s), 1

x(~)- (~ + ~)2~ [(~ + ~)x(0) +c + ~(0) +~( which from items #18 and #19 in Appendix C (p. 608) gives

:~(t) = x(O)cos~t

-at .

c+ ~(0)+ax(0)sin~dt]

[

EXAMPLE7.6

~

Find the behavior of the state vector of a third-order autonomous linear system described by its matrix A and initial conditions x(0):

A=

-2 2 1

;

.

~(o) 1

-4

Solution: From equation (7.50),

x(s) = (~ - A)-~x(0). Carrying out the indicated operations,

X(s)=~[

1 [ s2 +6s+6 2s+8 lg~a) 2

p(s) =-- [sI -

s/2+2 2 (s+4) s+4

2s+8 s:+6s+7

AI = s ~ + 10s~ + 29s + 20 = (s + 1)(s + 4)(s

; 1

7.2. THE LAPLACE

517

TRANSFORM

Carrying out the indicated matrix product,

X(~) =

1

+6s+5 0 ~+6~+5) -(s

1 (s+l)(s+4)(s+5)

-

s+4

~(0)

(s+l)o(S+5) _(s+l)(s+5)l

z(o).

The inverse transform, from either the tenth item in Table 7.1 or item #5 in the table of Appendix C, is

x(t)=x(0) From the present perspective the cancellation of so many nu~nerator and denominator factors to give so simple an answer seems rather an incredible coincidence. Later, in Section 7.3.4, this problem will be shown to correspond to an excitation of a "mode of motion" of a particular system. One rarely chooses initial conditions corresponding to a mode of motion accidentally; they are very special. The MATLAB co~nmand [num,den]=ss2tf(~.,B,C,D,±) gives the numerator and denominator polynomials of the matrix H(s) = C(xI-A)-IB+ D given in equation (7.52). Thus, the coefficients of the nine polynomials in p(s) above are given (as well as the denominator coefficients) by setting C equal to the unit diagonal matrix and D equal to the null matrix. The coefficients of X(s) above result from setting B equal to the vector x(0) and D equal to the null vector. 7.2.8

Partial

l~raction

Expansions

Laplace transforms comprising ratios of polynomials so as to represent ordinary linear differential equations with constant coefficients can be expandedin partial fraction expansions to reduce the complexity of the individual terms. The usual result comprises terms of first and second order only, a great simplification. Partial fraction expansions are presented in Section 6.3 (pp. 459-470). Although that discussion preceeds the formal introduction of Laplace transforms, it applies directly to them; the Heaviside S merely is replaced by the Laplace s. You are urged to review the material at this time. Table 6.1 (p. 461) of transfer functions and their associated impulse responses should be recognized also as a table of functions of time and their Laplace transforms. All but the final entry in that table can be found in Table 7.1 and on the first page of the table in

518

CHAPTER7.

ANALYSIS OF LINEAR MODELS,PART 2

Appendix C. The final entry in Table 6.1 (entry #7) relates the residues and poles of general second-order transfer function or Laplace transform to a simple analytic expression of the correspondingimpulse response or function of time. It relates nicely to the output of the residue operation of MATLAB. It suffers from one practical complication, however:there are two sinusoidal functions of time where only one is needed. A slightly different approach to the same problem is taken in the final entry in Table7.1, resulting in a single equivalent phaseshifted sinusoidal function. To see the contrast, the following examplehas the same transfer function or Laplace transform as Example6.23 (p. 465). EXAMPLE 7.7 Find the inverse Laplace transform of 5s2 + 22s + 39 2s + 12 G(S) -~ F(s) ~ + 2s+ 5)( s + 2) (s -~ + 4+ --’s+ 2

3

Thefinal entry in Table 7.1 gives f(t) = 3e-2~ + v~-~e-t cos(2t - 68.2°); f(t) = -2t + x/-~e-t si n(2t + 21.8°). The approachin Example6.23 (Section 6.3, p. 465) finds the values 2rr = and 2ri = -5, from which f(t) = 3e-2t + (2cos2t + -t. 5sin2t)e Althoughall three answers are equal, the versions with a single sinusoidal term reveal their meaningmoreconcisely. The same problemof a second-order Laplace transform, or subsets of it, are treated in yet slightly different waysin items #8, 9, 18, 19, 20 and 21 of Appendix C, which are the most commonlypublished forms. You have choices.

7.2.9

Initial

and Final

Value

Theorems

Sometimesyou maybe satisfied with a quick calculation for an initial value, y(0), and a final value, y(o~), rather than the moredifficult determination the entire function, y(t). Thesevalues also can be used as a partial check on the complete response. The initial value theorem is lim y(t) = ~li~m~[sY (s)], t~o

(7.53)

in whicht = 0- or 0+ according to the choice employedin the original Laplace

7.2.

519

THE LAPLACE TRANSFORM

transform. The final value theorem, which assumes stable outside the left-half plane), is

behavior (no poles

EXAMPLE 7.8 Apply the initial

and final value theorems to the X(s) of Example 7.4: 2 X(s) = s(s 1) (s + 5)

Solution: 2 lim[( s-~ s+l)(s+5 2 x(~x~)=lim s-~0 (.s+l)(s+5) x(0):

)]

:0,

] =0"4’

which agree with the x(t) found in the example. 7.2.10

Summary

Laplace transforms reduce ordinary differential equations to algebraic equations. In particular, the Laplace transform of the response of a linear system with zero initial conditions equals the product of the transfer function and the Laplace transform of the excitation. The inverse Laplace transform of the transfer function itself is the impulse response of the model. Both problems with excitations and with non-zero initial values are treated by finding the Laplace transform of the desired output and then computing its inverse. The Laplace transfer functions for lumped models are ratios of polynomials in s. The same is true for the Laplace transforms of most of the simple excitations signals considered herein, such as impulses, steps, ramps and sinusoidai signals that commenceat time t = 0. (Signals involving pure delays are exceptions, as noted in Section 7.2.6. Another class of exceptions, illustrated in Guided Problem 7.4, is developed for distributed-parameter models in Chapters 11.) In these cases the defining integrals for the transform and its inverse can be avoided by the expedient of table look-up, using a partial fraction expansion if necessary. Multiplying a Laplace transform by s is equivalent to taking the time derivative of the corresponding time function. This fact permits simple determination of the Laplace transformof a differential equation. A transfer function G(s) is algebraically equal to the function G(s) found earlier for the ratio of the response of the system to the exponential excitation est . Initial and final value theorems permit rapid evaluation of a time function at t = 0 and t = ~c, respectively, from examination of its Laplace transform. Software packages such as MATLAB dispatch most analytical drudgery.

520

CHAPTER7.

ANALYSIS OF LINEAR MODELS,PART 2

The Laplace transform can be viewed as a Fourier transform modified so that it can treat nearly any transient signal. The fact that G(s) is the Laplace transform of the impulse response of the system also follows directly from the superposition property. Guided Problem

7.3

Find the Laplace transform of the function a(1 - e-bt), whichis one of the most com~nonlyencountered in engineering and science. Suggested Step: 1. Substitute this function for u(t) in equation (7.31) (p 510) and integrate. Guided

Problem

7.4

The purpose for this optional problem is to suggest that Laplace transforms apply to a muchbroader class of signals than is considered in this chapter. Find the Laplace transform of the function av~. Suggested Steps: 1. Lookup the integral that defines the transform in a table of definite integrals. 2. Notethat the result is not a ratio of polynomials,unlike the other Laplace transforms considered in this volumeof the text (except for the pure-delay operator). Guided

Problem

7.5

This problemillustrates the application of the transform of a pure delay. Find the Laplace transform of the truncated ramp: f(t)l

Suggested Steps: 1. Represent the function as the sum of an upward sloping ramp and a delayed downwardsloping ramp. 2. Computethe Laplace transform of a single ramp. 3. Use the delay operator sum of the two ramps.

e -Ts

to assemble the Laplace transform for the

7.2.

THE LAPLACE TRANSFORM

Guided

Problem

521

7.6

You are urged to start your experience solving differential equations by the Laplace transform method with a straight-forward scalar initial-value problem. Consider the example d3 x d’) x 8 dx dt 3 + 5-~ + -~ + 4x = F(t). Find the Laplace transform of x(t) given the values x(0) = 1, x(0) = 2 x(0) = -3. Then, for F(t) being a unit impulse, find x(t). Do not refer to the Laplace transform table in Appendix C. Suggested

Steps:

1. Take the Laplace transform of the differential for dnx(t)/dt n given in Table 7.1.

equation using the formula

2. Substitute the function for £[F(t)] and the initial

values as given above.

3. Solve for ~:[x(t)] 4. Use a partial fraction expansion to find terms in the forms given in Table 7.1. Note the existence of a repeated root. 5. Use the table to evaluate x(t).

Guided

Problem

7.7

This problem uses the matrix formulation, and compares the procedure for a forced response with that for an equivalent initial-value problem. Solve the third-order system example of equations (7.89) and (7.90) (in following section, p. 536) for the responses of the three state variables to an impulse of force F(t) of magnitude I. Set this up in two ways: as a forced response, and as an initial value problem. Suggested

Steps:

1. Define the matrices B, C and D, and note A from equation (7.90). 2. For the forced response, find U(s) and compute Y(s) from equations (7.52)

(p.515). 3. For the initial value problem, find x(0) by integrating the differential equation over the infinitesimal time interval 0- to 0+. Showthat the first of equations (7.52) gives the same result as it does for the forced response. 4. Find the inverse transform of Y(s) using one of the tables. fraction expansion allows the use of simpler tables.

A partial

522 Guided

CHAPTER 7. Problem

ANALYSIS

OF LINEAR MODELS, PART 2

7.8

This problem offers a comparison between the response of a simple system to a transient sinusoidal excitation and the response of the system to a steady-state sinusoidal excitation (or frequency response). The system is identified by its response to a step excitation. A linear system gives the response 3(1 - e-2t) when excited by a step signal at t -- 0 of amplitude 2.0. Find the response of the syste~n to the sine wave 4sin(3t) which starts abruptly at t = 0. Compare the result for t -+ ~ that found from a steady frequency excitation using the frequency response techniques. Suggested

Steps:

1. Find the Laplace transforms of the step and its response using the table in Appendix C. Find the transfer function G(s) which is their ratio. 2. Find the Laplace transform of the sine wave. Multiply this by G(s) to get the Laplace transform of the desired response. 3. Find the desired response from its transform. A partial may help.

fraction expansion

4. The response cannot suffer a discontinuity at t = 0, and therefore must be zero there. Verify this with the initial value theorem. Note that the final value theorem fails, however, since the response has no asymptotic value as t --~ ~x~. 5. Find the result for t -+ ~x~, and interpret its magnitude and phase angle relative to the excitation. Compare these to the IG(jw)l and /G(jw) the frequency response technique. PROBLEMS 7.10 Find x(t) given ~ + 4k + 3x = h + 2u and u(t) = ~(t). 7.11 Repeat the above problem given u(t) = us(t). 7.12 The system ~ + 2k + 5x = 0 has initial Find x(t), t >_

7.13 Find the inverse

transform

conditions

2s"~ + 8s ~- 10 of X(s) - (s 2) 3

7.14 Find the Laplace transform of F(t) defined by

x(0) = 2, k(0)

7.2.

523

THE LAPLACE TRANSFORM

~(t) = o, t < = tsinwt,

t>_O

7.15 Find the Laplace transform of F(t) for a single sinusoidal pulse defined by

f(t)

= O, = =

sinwt, 0

7.16 Find the solution of the differential k+ax=bsinwt,

t

< 0 O<wt<~r wt, > ~ equation x(0)

7.17 A secoffd-order system with damped natural frequency w and time constant 7 is excited by a transient signal f(t) = -at. Find t he r esponse. 7.18 Estimate the impulse response in analytical form for the system characterized in Bode form in Guided Problem 6.4 (p. 444). 7.19 Estimate the step responses in analytical form for the three systems characterized by Bode plots in Guided Problem 6.5 (p. 445). 7,20 Find the Laplace transform of the square pulse plotted below.

0 7.21 The square pulse of the above problem excites a system with the transfer function G(s) b/ (s + a) . Fi nd and pl ot th e re sponse.

SOLUTIONS

TO

Guided Problem

7.3

GUIDED

PROBLEMS

U(s) = ~o °°a(1 - e-~t)e-~tdt(~s 1= a l_~_~_~t~e_st[ + s +b ]

Io =~a (1 1 s s~-b

524

CHAPTER 7.

Guided Problem

ANALYSIS

MODELS, PART 2

7.4 a

U(s)

OF LINEAR

avffe-~tdt = ~s

~r

(This result is given in AppendixC as transform pair #36.) Guided

Problem

7.5

1. f(t) = (a/T)[u,(t)

- ur(t T) a/T 2. For the first ramp, Fl(s) = 3. F(s) = -~-(1 - -T~) e Guided

Problem

7.6

d3 x 5 d2 x 8 dx 1. ~ + dt 2 + -~ -= 5(t)

[dx(s)- 82x(0)- s~(0)- ~,(0)]+ 5 [dx(8)- ~x(0) +s[sx(~)- ~(0)]+~ [.¥(8)][~(~)] 2-3. (x 3 + 582 + 8s + 4)X(s) (s 2 + 5s+ 8)x(0) + ( s + 5)~(0) + ~ 2 + 78 + 16 "X(s)(s~ +58+ 8)1 + (s + 5)2+ (-3) X3 "~- 582 "~- 88 ~- 4 (s+ l)(s -+-2 a b c s+l (s+2) 2 8+2 4. Can use the MATLAB’command [c ,p, r] =residue (aura,den), as follows: L [

Is2 +7s+ 16] ~;2"~ .j

or solve by hand

=a=10

s2 + 78 + 16= 10(s + 2)2 + b(s + 1) + c(s + 1)(s + = (10 + c)s 2 + (40 +b + 3c)s + (40 +b + 2c) Therefore, c = -9, b = -6 and check: 40 - 6 - 18 = 16 x(~) s+l10 (s+2)6 ~ s+29 5. x(t) = lOe-t - (6t + -2t 9)e Guided

Problem

7.7

1. Equation (7.90) gives A

-2

1

. Fromequation (7.89),

Presumingall three state variables are of interest, C=

1 0

,D=0

7.2. THE LAPLACE

TRANSFORM

525

u(s)= ,[u(t)] -Y(s) = H(s)U(s) -~ s

0 s+2 6 s 3 p(s) = ~ +5s~ + 20s + 16

-~ 0 =

H(s)=

~(0)

;

1 P-~

s~ +3s+8 8

;

Y(~) = e(sI - ~)-~x(0) = same ~

¯ he denominatorof the Laplace ~ransforms for p~ and pb can be facgored into ghe product of a first-order ~ermand a second-order term. Transformpair in the ~able of AppendixC ~hen gives p~(~) directly. ~inding p~(t) the same wayis more awkward,because of ~he numerator term s~; a summagionof terms from pair ~a2, pair ~aa ~d the time derivative of pair ~g ~e necess~y. Alter~mtively, ~he use of the MATLAB commandIt,p, r] ~re~id~ (~, c~ries out a p~gial fr~tion exp~sion, after which the ~sociated time funcgion can be found more readily. ~he developmentbelow c~ries ou~ ~he p~tial fraction expansions by hand (which is relatively uneconomicalin engineering practice): sa + Ss~ + 20s +16 ~-~ s ~ + 4~ + 16 6= ~(s~

Therefore,

la

+4s+16)+(s+

6 b=-~;

Thus:p~(t)=~_~

16. s~+as+18= ~(s

l)(bs+c)

24 c=-~+ [

~+

s~l

~ +4s+160=

6 18 13 13 check: 6s+18 ] (s+2)2 +12

(s+l)(bs+c)

: Therefore,

3 22 b = - ~; c : - ~

16 _~

9~ -2~ sin [2v~t - tan-l(3v~/16)]

96 18 13 13

78 13 =6

526

CHAPTER 7.

ANALYSIS OF LINEAR MODELS, PART 2 e-t -

Finally, q(t) = -~p, (t) Guided

Problem

e-st sin(2vf~t + tan-1 2x/~)

7.8

4x3 2. U(s) = s~ + 3~ ;

X(s) = G(s)U(s) 3~)(s +

36

3. ~ansform pair ~32 can be used directly, with a = 0, c = 2 and w = 3. Alternatively, the following p~tiM ~action exp~sion (which could be found using MATLAB) can be used: k bs+c X(s) = ~ + s~ +

Therefore,

36 b=-~;

72 c= ~;

~om Table 7.1 (p 511), 36 -2t 12 -) x(t)=~-~e. ÷~sin(3t÷¢),

324 2 x 72 468 (check) 36= ~+ 1~= 1~

¢=tan-l(

~)=-tan-l(~

x(O) = Ls-~ [sX(s)] = 0 Note that the time function above agrees. Ls~o [sX(s)] -- 0 the origin.

This is not x(oc), however, since there are two poles on

As t -~ ~, x(t) -~ 12 -~ sin " [3t-tan-l(~-)] The frequency response method gives ~0

uo]G(j3)~

= 4 ~

12 = ~+22

= ~; 12¢= AG(j3)

=-tan-l(

~

x(~) = xo sin(3t + ¢), whicha~ees with the result above.

7.3

Matrix ior*

Representation

of Dynamic Behav-

The previous chapters and sections have made a restricted use of matrix concepts, despite frequent use of matrix notation and MATLAB commands. In this section, matrix exponential solutions to linear models and the modal behavior of these models will be developed. The result is an enhanced understanding of the behavior of linear models as well as the development of certain practical computational methods. The linear state differential equations are, as before,

7.3.

MATRIX

REPRESENTATION

OF DYNAMIC BEHAIqOR*

527

dx

-~= Ax(t) + Bu(t), y(t)= Cx(t) + Du(t).

.. (7.55)

The special case of the stationary model, for which A, B, C and D are constants, is assumed. The autonomouscase, for which u(t) = 0, is treated first.

7.3.1

The Matrix Exponential

The general solution of equation (7.55) for the homogeneouscase u = 0 can written as

x(t) = eAtx(0), in which eAt is known as a matrix exponential. It has the same series sion as a scalar exponential:

(7.56) expan-

e At = I + At +9~A2t 2 + IA3t3 + .... (7.57) 2. 3! This equation can be used to evaluate the matrix exponential, but there are much better approaches that do not require evaluating many terms and do not suffer a truncation error. Mathematics texts give various ways for evaluating functions of matrices in general and the matrix exponential in particular. A particular method for finding the matrix exponential is developed below that also elucidates the behavior of the system. First, however, it is important to recognize two uses of the matrix exponential. The obvious use is to give analytic functions of time for the various states in x(t). The second use employs a fixed time interval, At, which can be chosen arbitrarily:

x(t + At) = eA~x(t).

(7.5S)

In this application, the matrix exponential eAAt is evaluated numerically. Then, x(At) is found from equation (7.58). After this, x(2At) is found ~he same that is by premultiplying x(At) by the same numerical matrix exponential. Again, x(3At) equals x(2At) premultiplied by the matrix exponential, which can be called a state transition matrix. This process may be repeated indefinitely, to evaluate x(nAt) for n = 1, 2,3,... It is important to note that regardless of the magnitude of At, this process entails no inherent error (beyond round-off error), assuming the matrix exponential is evaluated properly. By contrast, general simulation routines such as the Runge-Kutta method presented in Section 3.7 suffer an error that grows rapidly as the time increment is increased. The general simulation schemes, on the other hand, have the advantages of handling arbitrary forcing functions as well as nonlinearities.

7.3.2 Response to a Linearly Varying Excitation Equation (7.55) with a linearly varying excitation U -~ 110 -}- Ult,

(7.59)

528

CHAPTER 7.

ANALYSIS

OF LINEAR

MODELS, PART 2

freqnently is employed to represent linear interpotation of an excitation between known data points u(tk). This is the basis of the MATLAB function ls±m, the use of which is described in Section 5.3.7 (pp. 349-352). Note that the assumption of a zero-order hold corresponds to ul = 0. The Laplace transform of equation (7.59)

U(s) uo + = --s

(7.60)

~T"

The Laplace transform of the equation (7.55a) (same as equation (7.48), p. is given by equation (7.49), and its solution for X(s) is given by equation (7.50). Substitution of equation (7.60) into this result gives the transform that has its inverse transform x(t)

/o /0’//

e At Buo dt ’ +

eAt’’ Bul dt" + eatx(0),

in which t’ and t" are dummyvariables of integration. from carrying out the integrations: x(t)

(7.61)

The desired result follows

=A-l(e At-I)Buo+A-2(eAt-I)Bul-A-xBult+eAtx(0).

(7.62)

The accuracy of :~s±m can be seen to depend on the fidelity of the linear interpolation of the excitation between the beginning and end of each time step. Whenthe excitation is constant or varies precisely linearly, there is virtually no error. The command:~s±m also has a feature that is not well documented: it attempts to identify inputs that involve zero-order holds (step-wise excitations), and to represent them accordingly. This can be useful in treating systems that have sample-and-hold control signals, for example, but also is a potential source of error. For more information, see the Help instructions that accompany MATLAB.

7.3.3

Eigenvalues, Eigenvectors and Modes

Rather than giving the solution to the autonomous or homogeneous problem in the form of equation (7.56), the inverse transform of equation (7.50) (p. gives in general

[x(t)----~ixi(t)=~ieS~txi(O), (7.63) assuming that the roots si of the characteristic polynomial are distinct. Each term in this summation, called a mode of motion, satisfies the equation (7.64)

[(siI - A)xi = 0, ] in which si is an eigenvalue that satisifies

Ip(s) = det(sI -

the characteristic

A) = 0, /

equation (7.65)

7.3.

MATRIX REPRESENTATION

OF DYNAMIC BEHAVIOR*

529

and has its corresponding eigenvector, x~. Equation (7.63) reveals that the individual modes of motion can be superimposed in any proportion, depending on the initial conditions, without interaction with one another. It can be very instructive to grasp the significance of the individual eigenvalues and eigenvectors. The eigenvalues customarily are assembled in a square diagonal eigenvalue matrix, S:

(7.66)

If one or more of the roots of the polynomial are repeated, equation (7.63) produces fewer than the requisite number of terms. For a double root s~ the associated terms are eSt[x~(O) v/ t], (7.67) where v~(0) is a vector of constants. For a root repeated m times, this generalizes to est [x~(0) + V~lt + v~.~t2 +... + Vimt’~]. (7.68) The other terms in equation (7.63) are unchanged. 7.3.4

Case

Study:

Three

Fluid

Tanks

As a case study, consider the system shown in Fig. 7.8, which has three independent energy storage elements and thus is of third order. Applying integral causality in the usual manner, as shown, gives the state equations d V2 dt V3

An = --~

+

=A

V2

+

;

,

(7.69a)

LQuJ

V3

A~

= -R~C~

A~3 = O,

" Aa~ = O; Aa~ = R~C~

Rb¢ C~ -~

.

(7.69b)

To make equation (7.69) correspond to equation (7.~) (p. ~27), the Q = [Q~, Q~, Q~]r somehowmust be removed. This is accomplished by defining constant values ~o = [Qm, Q~o, Qao]r and Vo = [I%, ~o, ~o] r such that

~hus all changes in time reside in the & terms. The volumes ~o, ~o, Vao can be chosen such that they represent equal levels of water in the three tanks,

530

CHAPTER 7.

ANALYSIS

OF LINEAR

MODELS, PART 2

a3-~

Rc

I

water: pg = 62.4 ~ lb/fi 3Ab = 1.248 ft 3Ac = 0.624 ft A~ = 0.312 ft 3 A~ Co = ~-ff-= 0.0050 frS/lb Cb = 0.0200 ftS/lb Cc = 0.0100 ftS/lb Rc = R~c ~ = 50.0 lb min/ft

~ R~ = R~ = Ra~ = 100.0 lb min/ft

(a) system and parameters l

Vl/C 0 Co

R~

S~2

Ra,

R~

R~

(b) bond graph with causal strokes Figure 7.8: Three-tank system

Rc

7.3.

MATRIX REPRESENTATION OF DYNAMIC BEHAVIOR*

1

[~.~V.

op-.2-..-.,--_,-.,-....... _~l ,v/vl’ , , 0

5

0.4

I

I

I

I

I

0.8

1.2 1.6 2.0 time, minutes (c) response,first experiment andmode

~%

4 3 2 .,... V~/V,

O0

0.4

0.8

1.2 1.6 time, minutes

2.0

(d) response, secondexperimentand mode 2

~ V.2/Vii

1 0 -1 -2

~"

0

I

I

0.4

I

I

0.8

I

I

I

I

I

1.2 1.6 time, minutes

(e) response, third experimentand mode

2.0

531

532

CHAPTER 7. ANALYSIS OF LINEAR MODELS, PART 2

as indicated in part (a) of the figure. The corresponding depth of the water is V,o/Aa = V~o/Ab = V~o/Ac, which is proportional to ~’~o/Ca = V2o/Cb = }~o/Cc. Thus, Q,oRa = Q~oRb= Q3oRc, or

1 Q10---- Q2o= ~Q3o.

(7.71)

0 = AVo + Q0

(7.72)

The equilibrium equation

5nowcan be subtracted from equation (7.69) to give the desired result dAV - h AV + AQ. dt

(7.73)

The immediateinterest is the special case of AQ= 0. In particular, howdo the three levels approachtheir equilibrium with its equal depths if they start with different depths? The numerical parameter~sin the figure give the matrix A and characteristic polynomialalready used in Example7.5 (p. 516). The roots of the characteristic polynomial therefore are 81 :

-4;

s2 = -1;

s3 = -5 min-’.

(7.74)

The order of these roots is arbitrary. The solution can be found from equation (7.63):

I-1,",,1 + LtS()J LV LSJ.

.

Theessential characterof the dyn~icscanbe seento be representedpartly by three different exponentialdecayrates. Decayrates of linear models,as noted before, traditionally are expressed by e-~/~, wherev hasthe dimensions of time and is called a time constant. In the present, systemthe three time const~ts ~e 0.25, 1.0 and0.20 minutes,respectively. Thesolution to equation(7.75) also is characterizedby three eigenvectors given in the squarebracketson the right side. Eacheigenvectorcan be factored into a product of a scalar and a normalizedeigenvector.For simplicity it is assumed herethat the first elementin the normalizedeigenvectoris unity. (Later it will be seenthat MATLAB makesa different choiceto get aroundthe possibility that the first elementis zero.) Thenormalizedsecondeigenvectorof SThe approach is consistent with linearization, ~ discussed in Section 5.4. Equatio~ (7.69) is "linear bi~ed," however, and ~ a result the model of equation (7.73) introduces no additional error.

7.3.

MATRIX REPRESENTATION OF DYNAMIC BEHAVIOR*

533

the three-tank problem, for example, can be found by substituting the second eigenvalue, s2 = -1, into equation (7.64) (p. 528), as follows: = 0.

(7.76)

The first of the three equations imbeddedhere requires V22(O)/V,2(O) The third then gives V32(O)/Vl~(O)= 2. The second is redundant, but affords an opportunity to check the results. Repeating the sameprocedure for the two other normalizedeigenvectors and assemblingthe results into a modalmatrix, P, gives P-- V21/VI~ V22/VI2 V231V~a = L v31/v~l va2/vl~. Va3/VIa

6 -2 . 1 2 2

(7.77)

Withthis result equation (7.75) can be written V~(t) = V3(t)

V,~e-4t +

V,~e -~ + --2 V~3e-SL

(7.78)

1

Imaginethe following experiment. Initially, the three tanks are at equilibrium, all at the samelevel with the rate of flow out of each equal to the rate of flow in. Suddenly, you removea bucket of water of volumeVll from the third tank and dumpit into the first tank. This establishes the initial condition

yl~,

y2(0)/ = v3(0)

1

(7.79)

which, whencomparedwith equation (7.78), showsthat V12and V13equal zero. The subsequent behavior must be

vdt)/ = Va(t)

e-4t ,Vll 1

(7.80)

whichis plotted in Fig. 7.8 part (c). Note that the entire responseinvolves only the first time constant, and the ratios of the three excess volumesalwaysretain the proportions [1, 0, -1], respectively. This is the behavior of the first mode; the vector [V1(t.), V2(t), V3(t)]T rema.insparallel to its initial direction in state space, and decays at the rate e-4~ dictated by the first eigenvalue. This same initial condition was given in Example7.5 (p. 516). The cancellation of the numerator and denominatorfactors of X(s) is nowexplained, leading to the simple result for the x(/) of that problem.

534

CHAPTER 7.

ANALYSIS

OF LINEAR

MODELS, PART 2

If as a second experiment you suddenly dump one unit of water of arbitrary volume V~2into tank 1, six units into tank 2, and two units into tank 3, the resulting behavior would be

-~, v.2(t)=

v,12e

(7.81)

which is plotted in part (d) of the figure. This is behavior of the second mode; the vector [t~ (t), V2(t), l/~(t)] T remains parallel to its initial direction in state space, and behaves as directed by the second eigenvalue, 82, that is the decay rate e-t. The f~test decay would result from a third experiment thut estnblished conditions for the third modeof behavior, which is plotted in part (e) of the figure. Equation (7.78) can be generalized

(7.82) z(t) = Es,z(0); s2t e

...

¯

...

0

...

Est =

(7.83) ~

.: e t

(7.84)

¯

Equation (7.82) represents tr ans]ormation of variables bet ween z(t ) and x(t) The variables z(t) are very special, as revealed by equations (7.83) and (7.84): each component zi(t), i 1, ..-, n, act s independently of the other components, employing only its unique time constant, ~-i -- -1/si. Thus z(t) is the set modal variables. Inverting equation (7.82) and setting t = (7.85)

z(0) = P-ix(0).

Substituting this result into equation (7.83) and the resulting z(t) into equation (7.82), gives the final general result x(t)

(7.86)

= PEstP-~x(0).

A comparison of this result to equation (7.56) (p. 527) reveals that the matrix exponential can be computed by evaluating (7.87)

l e*’ = PEstP-’. ] For the three-tank problem, equation (7.85) gives

..

[ V~ ~

. ]v~:~ = y(0) [ ~]~j

’1

16

0

a -3

~:(0)/ t~(0),j

7.3.

MATRIX REPRESENTATION OF DYNAMIC BEHAVIOR* Ra

Se

535

C= 1

~"=1/8

(L~

Ia~’:~-’~ 1 p~/l~ 0 "--’--~ y~ T pb/l~ q/C~ ~ C R~

I~ = 1/6 Ra=Rb=I/3

(a) bond graph 2.0

(b) responseto impulseor initial condition Figure 7.9: Third-order system example

[ 16V1(0) +0- SV3(0)

= 1/2v1(0 ) + 3v2(0)2V3(0) , 24 L 6v1(0) - 3v2(0) + 6v3

(7.88)

whichcan be substituted into equation (7.78) to give the general result for that problem.

7.3.5

Case

Study

With

Complex

Roots

The three-tank problem is rather special in that its roots are all real. The methodabove is applied next to a problem with complexroots. Consider the system shownin Fig. 7.9, which has three independent energy storage elements and is thus of third order. Applyingintegral causality in the

536

CHAPTER 7.

ANALYSIS OF LINEAR MODELS, PART 2

usual manner, as shown, gives the state equations

-dt

=

yRb/I~ --1lib

-1/Ia

1/C -1/RaC

(7.89)

The numerical parameters in the figure give

A= so the characteristic

0 -2 -8 -6

(7.90)

[oo

equation (7.65) becomes 0 -1 s + 2 -1 6 s+3

i

p(s)

= 0,

(7.91a)

or

p(s) =_ 3 +5s~ + 20s + 16= (s + 1)( -~ + 4s+ 16)= 0. The roots of the characteristic

(7.91b)

polynomial are s2,3 = -2 ¯ 2jv~,

s~ = -1;

(7.92)

where j is the unit imaginary number, Even if you don’t continue the analysis further, these eigenvalues tell you the essential character of the dynamics of the system, as represented by the decay rates e-t and e--~t with their time constants of 1.0 and 0.5 seconds, respectively, and the damped natural frequency, Wd = 2v~ rad/sec. Youfind the first normalized eigenvector by substituting the first eigenvalue, -1, into equation (7.64), and setting xl~ equal to unity:

0 8

1 6

-1 x12 2 LX~3

= 0.

(7.93)

The first of the three equations imbedded here requires x~3 = -1. The second then gives x~2 = -1. The third equation is redundant, but affords an opportunity to check the results. Repeating the same procedure for the two other normalized eigenvectors and assembling the results into the modal matrix gives P= -1

l + j/v~ -1 -2 + 2jv~

The determinant IPI is -2jx/-~, p_~ _ Adj(P) det(P)

1-2-

j/v~ 2jx/~

.

(7.94)

and the inverse of P is

_ 1 -jv/-~ 26 +jx/~

6- jv/-~ 6+jx/~

-1- 2jv~ -l+2jv~J

(7.95)

7.3.

MATRIX REPRESENTATION OF DYNAMIC BEHAVIOR*

537

As an example, consider as the initial condition the result of an impulse of the force F(t) whichoccurs at the time t = 0. The force is designated as ZS(t), where Z is the magnitudeof the impulse or time integral Z = fF dr. This time integral is a step of amplitudeZ and can be designated as Zus(t). The immediate effect of the impulse excitation appears only on the state variable p2, as can be seen by integrating the differential equation (7.73) over the infinitesimal time interval from t -- 0 to t = 0+, whichby definition includes the entire impulse. The result is p~(0+) = 0;

pb(0+) = Z;

q(0+) = 0.

(7.96)

Equation (7.85) nowgives z(0) = p-1

= ~o

- jx/’~ ¯ + jx/SJ

(7.97)

(Note that if you knew in advance that only pb(O+) would be non-zero, you could forgo computingthe first and third columnsof p-l, saving significant effort if hand calculation is used. For most practical situations above the order two, however, the calling of the MATLAB inversion command,±nv(P), recommended.) Next, you find the matrix PEst, the separate columnsof which represent the separate modesat time t: eSlt

*’t PEst = -e

s’t e

eS2t (1 + j/v~)e ~ (1 (-2 + 2jv~)e *~t (-2 -

es3t j/v~)e ~3t . s~t 2jvr~)e

(7.98)

Multiplying the vector of modalamplitudes, y(0), given by equation (7.97), this matrix gives x(t) = PEstP-~x(O) (-2+~jv%t + (6 + jv~)e(-2-2jv~)t (-2-~jv%t ] [ -12e-t -t + +(7 (6 +- jv/-~)e jx/-~)e (-2+eiv%t + (7 - jx/~)e [ = Z__ ] 12e 26 L 12e-~ - (6 - 14jx/~)e(-~+’~jv~)t - (6 + 14jx/~)e(-2-~jv%t J

z ~ -~e-t + e-*t[~ cos(~t/+

= i~ 6e-t-t + e-~t[Tc°s(2~/~t) v~sin(2~/gt)] . 6e - e-2t[6 cos(2v~t) - 14~/~sin(2v~)]

(7.99)

These results are plotted in part (b) of Fig. 7.9. As a partial check, you can verify that they satisfy the initial conditions given by equations (7.96). Whennumerical as opposed to analytical results are satisfactory, MATLAB can relieve the analyst of most of the drudgery above. Whenanalytical results such as eqtiation (7.99) are desired, it is possible to avoid having to deal with the complication of complexnumbers. An optional methodis presented next.

538 7.3.6

CHAPTER 7. Modified

Method

ANALYSIS for

OF LINEAR

Complex

MODELS~ PART 2

Eigenvalues*

The complex conjugate nature of the terms in the modal matrix permits them to be coded with real numbers, saving effort. For example, the modified modal matrix for the third-order system above is

Pm

=

-1 1 1/ -1 -2 2~/~

.

(7.100)

J

The first eigenvector or column is real, so it is reproduced without change from equation (7.100). The second and third eigenvalues are complex conjugates one another; their commonreal parts are placed in the second column and the imaginary parts, with the signs of the second eigenvector, are placed in the third column. In general, complex eigenvectors are paired off, the real parts being placed in one column and the imaginary parts being placed in the adjacent column to the right. This modified modal matrix is readily inverted to give

p~l

13118 1 3vcg =-5 -66 ~ -11 .

The key result of equation (7.86) (p matrices as follows: x(t)

(7.101)

~a4) can be modified to apply to these

= P,~EsmtP~lx(0),

(7.102)

where, for the present problem, the matrix Esmt is

Es,~t =

e -2t cos(2~/~t) -e -2t sin(2x/-~t)

e -2t sin(2v~t) e-2t cOs(2V~t)

(7.103)

In general, the exponential terms for the real roots appear on the diagonal, and the terms for each complex conjugate pair appear as a 2 × 2 array, centered on the diagonal, and given in the form e-st coswt e -st sinwt -e -at sinwt e-c~t coswt where a is the real part and w is the imaginary part of the associated eigenvalue. Application of equation (7.102) to the third-order system now gives the modal decomposition and the function x(t) relatively simply. Somesoftware automatically reports eigenvectors in this modified form, and computes only the inverses of real matrices, further motivating its use. MATLAB, on the other hand, handles complex eigenvectors and modal and other matrices directly. The modified form has the advantage of giving analytic solutions directly in terms of real functions.

7.3.

MATRIX REPRESENTATION OF DYNAMIC BEHAVIOR*

539

(a)system

(b) pitch mode

¯

¯

4.843 m

(c) heave mode Figure 7.10: Undampedvehicle modes 7.3.7

Case

Study:

Vehicle

Dynamics

As a final problemregarding autonomousmotion, consider the vehicle dynamics of Figs. 5.15 and 5.16 (pp. 322, 324), which has a fourth-order modelwith two degrees of freedom. The excitation (the toad bumps) is removedto make autonomous,and the dampingis removedto clearly reveal the modalbehavior. (The dampingis reinstated in Section 7.3.11 below.) The reduced model shownin Fig. 7.10. The equation of motion becomes d

y2 pc

=

~ Llk~

0 1/m -L2k~

L2/J 0

I

y2 pc

(7.1o4)

540

CHAPTER 7. ANALYSIS OF LINEAR MODELS, PART 2

The parameters are given as m= 1250 kg, L1 = 1.4 tn, L.2 = 1.6 m, J = 2mr with r 2 = 1.6 m2, kt = 30 kN/mand k2 = 32 kN/m. Therefore,

s

o

0 30000 -42000

sI-A=

-o.ooo80.0007]

s -0.0008 -0.0008[ 32000 s 51200 0 s0J’

(7.1o5)

and p(s) = I ]sI- A = s4 + --1 m

1+k2+

L~kl --2+L~_k2~s + 2 r

m2r2 (L~ + L2)2 = 0, (7.106)

or

p(s) = s4 + 119.96s2 + 3456= (s 2 - 71.88)(s2 - 48.08) = 0, si,2 = ±6.934j; s3,4 = ±8.478j. Theith normalizedeigenvector[1, P2i, P3i, P4~]T satisfies 0 30000 -42000

-0.0008 si 32000 s~ 51200 0

-0. 008 si

P2i

/p3,/ = O.

(7.107) (7.108)

(7.109)

J LP4iJ

The fourth scalar equatio.n embeddedabove gives P2~= (42 000 -

s~m4i)/51

200,

(7.110)

which, whensubstituted into the third equation, yields P4i = -(1250s~ + 56 250)/(0.25s{).

(7.111)

The first equation gives

(7.112) 1250si + 7~m4i. Equation (7.111) can be used to find the fourth element in all four eigenvalues, whereuponequation (7.110) gives the second element and (7.112) gives third. It is simpler to computeonly the first and third eigenvectors this way, however, and note that the second and fourth eigenvectors must be the complex conjugatesof the first and third, respectively. Werethis not true the state variables themselves wouldbecomenon-real; whichis impossible. The resulting modalmatrix is P3i

I

=

1 1 0.5195 0.5195 P = | 6724j -6724j L-2221j 2221j

1 -1.8047 -3274j -]5853j

1 " -1.8047 .3274j ’ 15853j

(7.113)

7.3.

MATRIX

REPRESENTATION

OF DYNAMIC BEHAVIOR*

541

and the corresponding modified modal matrix is pm = O. 95

0 6724 -2221

-1.8047 0 0

-3274 " -15853J

(7.114)

These matrices reveal that the first complex conjugate pair of roots combine to give a mode of motion in which the amplitude of y2(t) is 1.8047 times the amplitude of y~ (t), with a 180° phase difference. This requires that a node (point with zero velocity)be located between the two points, as shown in Fig. 7.10 part (b), suggesting the designation pitch mode. Similarly, the second complex conjugate pair of roots combine to give a mode of motion in which the amplitude of y~_(t) is 0.5195 times the amplitude of y~(t), with a 0° phase difference. This requires that a node be located at some distance from the vehicle, as shown in part (c) of the figure, suggesting the designation heave The eigenvalue matrix is jw~ 0 S = 0 0

[

0 -jw~ 0 0

6.934 tad/s, from which

ejwlt

Est -:

[i

0 0 jw2 0

0 ] 0 0 ’

w2=8.478 rad/s, 0 -j~lt e 0 0

0 0 j~2t e 0

0 ] 0 0 e-j°~2t

(7.115a)

(7.115b

(7.116a

0 -sinw~t cosw~t 0 O0 Esmt = (7.116b 0 0 cosw2t sinw_~t " ] [ coswlt sinw~t 0 0 - sinw~t cosw~tJ The inverses of the modal matrices are -5 ~0.11176 -0.21513 0.9753j x 10 2.9526j x 10-5 ] -5 -5 0.11176 -0.21513 -0.9753j x 10 -2.9526jx 10 -5 ; (7.117a) -5 0.3882 0.21513 -6.961j × 10 1.4376j × 10 -~ -5 0.3882 0.21513 6.961j × 10 -1.4376j × 10

0

-0.22352 -0.43026 0 0 0 -0.1951×10 -4 -4 -0.5905x10 P~I = 0.77648 0.43026 0 ’ 0 . (7.117b) -6 0 0 1.3922× 10 -0.2875 x 10-~J The matrix exponential caa be found ~om either equation (7.87) (p. 534) (7.102) (p. 538): e ~t = PEstP -~ = P,~E~tP~ ~. (7.118)

542

CHAPTER 7.

0

0.5

1

1.5

ANALYSIS

OF LINEAR

2 2.5 3 time,seconds

3.5

MODELS, PART 2

4

4.5

5

Figure 7.11: Motion of undamped vehicle This square matrix can be used either to find analytic functions of time in responses to given analyti c excitations, or it can be used as a state transition matrix to compute the responses at times nat numerically. As an example of the latter, consider At ---- 0.1 sec. This gives 0.74506

eo.~A

=

I

0.04333 -2788.3 L 1104.9

0.04623 0.68565 -2905.4 605.66

-4 0.7436x 10 0.7263 x -4 10 0.7622 -4 -0.0332x10

-4 -0.6265x 10 -0.7037 x -4/ 10 -0.0208 0.6685 J

/"

(7.119)

Further, consider as an initial condition yl(0) = 1, y2(O) = O, pro(O) pc (0) = 0. This gives the analytic, solution + 0.40338cosw2t[ y2, =/-°.4°338c°swlt (7.120) 731.8sinwlt5221sinw2t Pm pc i)

/

J L 3543.5sinwlt

+ 1724.6sinw2t

/’

J

which is plotted in Fig. 7.11. Successive .pre-multiplication .of the state vector by the constant matrix exponential above gives the highlighted points. The. coarseness of the time interval itself produces no error in the calculation.

7.3.

7.3.8

MATRIX REPRESENTATION

Application

OF DYNAMIC BEHAVIOR*

543

of MATLAB

Eigenvalue problems are simply treated with MATLAB.The procedures are illustrated here by application to the three problems above. Once the matrix A is entered, the commande±g(A) then gives the eigenvalues. To get the modal matrix as well as the eigenvalues, you enter [P,S] = e£g(A). The response is the modal matrix

0.7071 -0.0000 -0.7071

-0.3333 0.6667 -0.6667

-0.1562 -0.9370 -0.3123

and the diagonal eigenvalue matrix

-4.0000 0 0

0 -5.0000 0

0 0 -1.0000

The three eigenvectors are normalized to unity length; that is, the sum of the squares of their elements is 1. This is different from the modal matrix of equation (7.78) (p. 533), in which the elements in the top row were arbitrarily chosen 1. It accomodates the use of a zero value in this row, amongother advantages, but would have entailed extra calculation if done manually. The diagonal matrix es~ can be found for the specified value of t = 0.1 by entering ES = ex!0ra(. 1.S). The corresponding matrix exponential of equation (7.87) can be found by entering P*ES*£nv(P). It is simpler, however, merely to enter expm(, l*~). Note that expm(X) returns the matrix exponential x, whereas exp (X) returns with = f or e ach of t he e lements i n X. Equation (7.86),(p. 534) allows the determination of the analytic response as a function of t. Finding this result is expedited by determining the vector z(0) from equation (7.85) by invoking z0=±nv(~i)*x0. The initial state x(0) = x0 must have been defined first. MATLAB treats systems with imaginary or complex eigenvalues and eigenvectors the same way; standard notation is used. For example, the modal matrix corresponding to equation (7.105) (p. 540) but normalized to give real unit tors is given as

0.5774 -0.5774 -0.5774

-0.1666 -0.2611 -0.2339

+ 0.1637£ + 0.0675± - 0.9044±

-0.1666 -0.2611 -0.2339

- 0.1637i - 0.0675i + 0.9044i

The normalized vector z(0) of equation (7.108) is found by typing z0=£nv(P)* [0;1 ;0]. Construction of the analytic solution of equation (7.99) (p 537) would be directly in the first or complex form of the equation.

544

CHAPTER 7.

ANALYSIS

OF LINEAR

MODELS, PART 2

The use of the modified modal matrices with real elements is not necessary, but might be preferred as a meansof finding analytic results directly in the form of real functions. This is an advantage for finding the motion of the vehicle as expressed in equation (7.120), for example. The highlighted points in the plot of Fig. 7.11 can be computed through the repeated use of the numeric matrix exponential of equation (7.119). The points and plot can be programmed in MATLAB as follows: A = [0 0 .0008 -.0007;0 42000 -51200 0 0];

0 .0008 .0008;-30000

-32000 0 O;

ES = expm(O.l* A); x(:,l) = [1;0;0;0]; t(1) = for k = 2:51 x(:,k) = ES * x(:,k-l); t(k) = 0.1 * k; end xlabel (’time, seconds’) ylabel (’yl (blue) and y2 (green)’) plo’6(t,x(l:2, :), ’*’) Notethatonlythefirsttwoof thefourelements in x arechosenforplotting. The studentversionof MATLABpermitsindividual matrices to have up to 8192 elements. Thispermitsthe timeinterval for the five-second run aboveto be reduced to 0.005seconds, morethanenoughto givevirtually continuous curves. In thiscase,ofcourse, theasterisk should be leftoutof theplotstatement.

7.3.9

Response to Exponential

and Frequency Excitations

The response of scalar differential equations to exponential and to frequency excitations was addressed in Sections 6.1 and 6.2. For exponential excitations of the form uoes~, the response, taken from equation (6.25) (p. 397), st. x(t) ~ G(s)uoe For frequency excitations equation (6.39) (p. 410),

(7.121)

of the form uo cos(wt +/~), the response, taken from

x(t) = uoIG(jw)l cos(~t +/~ + ¢)

(7.122a)

¢(w) = tan-l ~[G(jw)]}~(G[jw]"

(7.122b)

These results can be generalized to the A, B~ C, D matrix formulation of equation (7.55) (p. 527), with the only changes being that the scalar transfer fuction

7.3.

545

MATRIX REPRESENTATION OF DYNAMIC BEHAVIOR*

or Laplace transform G(s) is replaced by the array of transfer functions H(s) of equation (7.52) (p. 515), and s becomeseither the coefficient of the exponentor the frequencyjw of the excitation.

Example 7.9 The undampedVehicle discussed in Section 7.3.7, with parameters given by equations (7.104) and (7.105) (pp. 539, 540), is excited by the force the forward axle and the force F2 on the rear axle, where

F2= F2ocos(~t+ ~).

F1= Flocos(~t),

Find the matrix transfer function H(s) that relates the state variables yl, y2, Pmand pc to these forces. Further, plot the magnitude of the transfer functions betweenthese forces and the two displacements. Solution: The equation of motion is the same as equation (7.104) except for the addedforcing terms: d

y2

=A Y2

PC

PC

whereA is given by equations (7.104) and (7.105),

o1 = o . -/51 /52

-1.4

(You might wish to verify the added terms by placing bonds Se F1

and

Se F2

on the respective lojunctions for ~ and ~2, and carrying out the standard procedure.) Since all four state variables are considered to be the output variables of interest, C -- I, D = 0 and H(s) = G(s). The matrix transferfunction H(s) becomes, using equation (7.52) (p. H(s) = G(s)

1 [Adj(sI-

A)]B

-0.0008 0.0007 -0.0008 -0.0008 s 0 0 s

1 -1.4

(s I - A)-IB = p~

0 s = ~1 Adj p(s) [ 30,i 00 32,000 ~. L-42,000 51,200

" 1

546

CHAPTER7.

ANALYSIS OF LINEAR MODELS, PART 2

The first two columnsof (sI - A)-1 or Adj(sI - A) do not affect the result, because of the zeros in the matrix B, and need not be found. The result is the 2 × 4 matrix F0.0017882+ 0.1152 -0.000328 ’~ ] -0.0003282 1 | 0.0020882+ 0.108| H(s) = p-~ s3 +76.88 + 638 / ’ L -1.483 - 76.88 1.6s3 + 728 J ’ p(s) = 4 +119.968~ + 3456. The resulting displacements are

y2(t) = IH~IF~ocos(wt + Oe~)IH~.IFzo cos(wt + ~ + 01e), Hij = Hij (jw),

Oij =/ Hij (jw).

Only even powers of s appear in the numerator and the denominator polynomials of each Hij, resulting in a real number.If this numberis positive, the phaseangle ¢~j is zero; if negative, ¢~j is :1:180°. (Thereis no difference between +180° and -180°.) The magnitudesof IHI~I and IHell are plotted in Bodecoordinates below:

-100

.0

10 frequency, rad/s 100

The two natural frequencies, w = 8.478 and 6.934rad/s, which were found in equations (7.108) (p. 540) as the imaginary eigenvalues, appear as resonant frequencies with infinite amplituderesponses. The eigenvalues or poles are s = +6.934j and s = =t:8.478j. The zeros of GI~ are at w = 8.045 rad/s (s = +8.045j) and of G2_~are at co = 7.206 rad/s (s = =t:7.206j). A phase angle jumps by 180° wheneverthe frequency passes through either a pole or a zero, since there is no damping.

7.3.

MATRIX Example

REPRESENTATION

OF DYNAMIC BEHAVIOR*

547

7.10

Introduce the damping resistances R1 and R.o into the model of the vehicle above, to make it more respectible. Determine the modified matrix of transfer functions for the special case R1 = R2 = 4000 N s2/m, and make Bodeplots of the positional responses to the force on the first axle. Solution:

The matrix A is modified to become 0

A =

0 -kl Llkl

0 0 -k2 -L.2k2

1/m -L1/mr’2 2L2/mr 1/m | -(L2R~ - L1R~)/mr 2 | " -(R1 + R.2)/m -(L2R2 - L1R1)/m - (L~.R2 + L~R1)/mr 2 J

With the given values of the parameters, this becomes ’ s -0.0008 sI - A = 30 000 32 000 s + 6.4 [_-42

000 51200

0.64

-0.0008 0.4

’

s ÷ 9.04

from which the values of HI~, H~_~, Hmand H22 are 1 [ 0.00178s~ + 0.0144s + 0.1152 - 0.00032s~ ] H(s) = ~ L -0.00032s2 o.o020ss2 + 0.0144s + 0.1082 J ’ p(s) = 4 +15.44s 3 + 177.56s ~ + 892.8s + 3456.0. The Bode plots comprising the new magnitude ratios ¢ij for i = 1, 2 and j = 1 are as follows:

IHij[ and phase angles

-5O gain, db l

-150 1.0

10

frequency, rad/s

100

phase

1.0

10

frequency,rad/s

100

548

CHAPTER 7. ANALYSIS OF LINEAR MODELS, PART 2 Theseresponses are clearly preferable from the viewpointsof comfort and safety. The resonances are sufficiently dampedto be almost unnoticeable. The transfer functions still have poles, whichcorrespondto the eigenvalues sl,2 = -3.161=t: 6.179j,

s3,4 = -4.559 -1- 7.139j.

Thevarious scalar transfer functions still have zeros, also, whichare either real or complex. For example, the numeratorof Hll vanishes for s = zl,2 = -4.045=t: 6.954j.

7.3.10

Representation

in the

s-Plane

It is customaryto plot the poles and zeros of a transfer function in the s-plane, whichhas the real part of s as its abcissas and the imaginary part of s as its ordinate. Such a plot is given in Fig. 7.12 for the H~ of the exampleabove. Poles are indicated by x’s and zeros by O’s, which is the convention for such plots. Oneapplication for an srplane plot is to visualize the response to an exponential disturbance u = uoe"~. This is illustrated in the plot for the special case of frequency reponse, i.e. s -- jw. Solid arrows are drawnfrom the poles to the point on the imaginary axis, whichrepresents the particular sinusoidal disturbance of interest. Dashedlines are drawn, also, fromthe zeros to the same point. Now,the polynomials in the numerator and in the denominator of the transfer function can be factored into products of first-order polynomials:

Im,

-jl0

IRes, s

--jlO Figure 7.12: Pole zero plot for H~(s) for the dampedvehicle

7.3.

MATRIX

REPRESENTATION

OF DYNAMIC BEHAVIOR*

(8 - zl)(s z2) Hll (8) = (8 - 81)(8 - 82)(8 - 83)(8

549

(7.122)

The solid arrows, then, represent the denominator factors, and the dashed lines represent the numerator factors, both with s = jw. Considering the arrows as vectors 1.~, I/~,...,lt as shown, the magnitude and phase angle of Gu (jw) becomes

ZHu(jw) = Z~ + Z~ - AV~ - Z5 - AV3 - AS.

(7.123b)

It is possible, then, to determine the magnitude ratio by measuring the lengths of the vectors and multiplying and dividing as indicated, and to determine the phase angle by measuring the angles of the vectors and adding ~d subtracting as indicated. The frequency w c~ be changed and the process repeated. It is more efficient to compute the transfer function algebraically, and MATLABdoes it automatically. It can be very instructive, nevertheless, to be able to look at an s-plane display of poles and zeros and visualize the form of the frequency transfer function. 7.3.11

Summary

An unexcited linear system with nonequilibrium initial conditions exhibits behavior that can be decomposed into a sum of simple modal behaviors. The dynamics of each mode is represented by its eigenvalue. The shape of each mode, that is the proportions of its component variables as represented by its eigenvector, remains fixed in time. Thus the contributions of each mode to the total behavior depends on the initial conditions. Standard matrix methods for treating the details have been presented and illustrated. Whencomplex roots are present, you can choose to use either real or complex matrices. MATLAB accomodates complex matrices, whereas certain other software packages use real matrices only. Initial value problems can be addressed in a more routine if somewhat less insightful way using Laplace transforms, as presented in the previous section. Cases with repeated roots are best handled this way, so details of their modal behavior have been omitted here. The simplest approach for finding the autonomous or forced behavior of a linear system employs the numeric matrix exponential. These solutions obscure modal behavior, but particularly in heavily damped systems the modal decompostions may be of minor interest. MATLAB allows the matrix exponential to be requested directly, or indirectly through the use of the commandlsim which treats excitations that vary linearly in time over each time interval. Exponential and frequency responses for a model represented by a matrix transfer function are found in essentially the sa~ne way as a model represented by a scalar transfer function. The matrix transfer function is simply treated as an array of scalar transfer functions.

550

CHAPTER 7. ANALYSIS OF LINEAR MODELS, PART 2 R= I0~ I = 200 ~h C = 50/~f E= 5.0V Figure 7.13: Guided Problem7.9

Guided

Problem

7.9

The first guided problemdeals with a second-order system, and can be treated analytically without unduedifficulty so as to reinforce the key concepts. The two-wayswitch in the circuit in Fig. 7.13 replaces the battery at t = 0 with a conductor. Find the subsequent voltage e(t). Suggested Steps: 1. Represent the after-switch circuit with a bond graph. causality and define state variables in the usual way.

Applyintegral

2. Write the state differential equation; find the matrix A. 3. Solve IsI- AI = 0 to get the characteristic polynomial.Solvefor its roots. 4. Writethe general solution for the charge, q(t), of the capacitor. 5. This problem is simple enoughthat you can avoid having to evaluate the eigenvectors and modalmatrix, etc. Instead, specialize the result of step 4 by applying the knownboundary conditions q(0) = q(~) 6. Find the remainingConstant by noting that the initial current through the inductor is E/R and p(O)/I, and p(0) is related to q(t). 7. Sketch-plot the resulting e(t) = q(t)/C to see if it makessense. Guided Problem 7.10 This probleminvolves a fourth-order, ~wo-degree-of-freedomvibration model. It is treated analytically, and therefore maybe too ambitious for your needs. The beating phenomenon that occurs is highly instructive, however,So at least you ought to examinethe given solution. As an alternative, you could apply somereasonable set of parameters, makingsure the spring is weak, and find the behavior using MATLAB. Twootherwise simple oscillators are not infrequently weaklyinterconnected by a compliance. One exampleof this is two identical pendulumsconnected by a weakspring, as shownin Fig. 7.14. Considering small motions only, examine the modalbehavior of this systemfollowing arbitrary initial conditions.

7.3. MATRIX REPRESENTATION OF DYNAMIC BEHAVIOR*

551

Figure 7.14: Guided Problem7.10

Suggested Steps: 1. Drawa bondgraph, label variables, and find the consequentdifferential equations. Evaluate any inertances and compliances. If you write a differential equation for the energy storage associated with the spring, note that it is not independentof the other differential equations and does not introducea fifth state variable. 2. Convert the result of step 1 to give the matrix A. 3. Find the characteristic polynomial,using IsI - A1 = 0. 4. Solve for the four characteristic values, whichcan be written as j~l, -jwl, Jw2, -jw2. 5. Find the four eigenvectors, letting mli -- 1. Assembleinto either the standard complexmodal matrix or the modified real modal matrix. 6. Find the eigenvalue matrix Est or Es,nt. 7. Compute the matrix product PEst or PmEsmt. Each column of this matrix represents the behavior of its respective mode.Therefore, the general solution can be written as the sumof an unspecified coefficient times the content of each column. If you used the complexmatrices to get the general solution, the sum of the first two columns and the sum of the other two columnsmust be real; this meansthat the unspecified coefficients must be complexconjugate pairs. Combinethese pairs of columns to get the functions in real terms (sines and cosines). This awkwardstep is unnecessaryif youuse the real matrices from the start. 8. Interpret the results. Describea motionfor whichonly the wl terms exist. Repeatfor the w2terms. Now,note that if the spring coupling is weak, wl and w2are almost the same. The sumof two sine or cosine wavesof nearly equal frequency can be represented by a sine or cosine waveof frequency equal to the average of the two frequencies modulated in amplitude by an envelope that is a sine or cosine wavewith a frequency equal to the difference betweenthe two frequencies. This is sometimesdescribed as a beating phenomenon.

552

CHAPTER 7. ANALYSIS OF LINEAR MODELS, PART 2

m=l kg k=400 N/m b=40 Ns/m F0=10 N

Figure 7.15: Guided Problem 7.11 Guided Problem

7.11

This frequency response problememploysa familiar simple model, but asks you to treat it by starting with the matrix format. The prototypical vibration problemwith a single degree of freedomcomprises a mass, spring and dashpot excited with a force F(t) = Fo sin wt, as shownin Fig. 7.15. Find the responseof this systemin the form x(t) = Xo(W)sin[wt+¢(w)] and sketch the correspondingBodeplot for the parameterslisted in the figure. Suggested Steps: Drawa bond graph, write the state differential equations and place in standard matrix form. Find the matrices A and B, using symbols rather than numbers. Extract the scalar transfer function G(s) that relates the force of amplitude F0 to the displacementof amplitude xo from the moregeneral matrix G (s) given by equation (7.50) (p. 515). 3. Substitute jw for s in G(s), and find the magnitude IG(jw)l and phase angle /G(jw) as functions of the frequency w and the parameters. 4. Substitute the values of the parametersand employthe value of Fo to find the functions xo(w) and ¢(jw). 5. Evaluate these functions for several values of w in the range of 1.0 to 100 rad/s. Plot in Bode coordinates. You may prefer to use MATLAB. Guided Problem 7.12 The modelneeded for this second freqency response problemis of higher order; as a result, the use of the state-space matrix methodsis more efficient than scalar methods. Consider again the two weakly interconnected pendulumsof Guided Problem 7.10. A small horizontal excitation force F(t) = Fo sinwt is applied to the left mass. Find the response of this mass (¢1) and represent it by a Bodeplot for the parameters L = 0.5m, a = 0.15m, rn = 1.0kg, k = 8.0N/m and Fo = 1.0N.

7.3.

MATRIX REPRESENTATION

Suggested

OF DYNAMIC BEHAVIOR*

553

Steps:

You already have the matrix [sI-A] and its determinant, and B is a vector with a single non-zero element. Thus only one element in the inverse of the matrix [sI - A] needs to be found to determine the transfer function G(s). Determine which element this is. 2. One possibility is to solve this problem using MATLAB. Otherwise, follow the steps below. 3. Find G(s) using the determinant p(s) = IsI - AI and the adjoint matrix Adj(sI - A). 4. Substitute the values of the parameters and substitute the magnitude and the phase angle of G(jw).

jw for s; evaluate

Plot FoG(j~) in Bode coordinates, including a frequency band starting at least one decade below the natural frequencies and ending at least one decade above. PROBLEMS 7.22 The mass-spring system shown on the left in Problem 7.23, known as a Wilberforce spring, exhibits both translational and rotational motion. The parameters include the radius of gyration, rg. The static equations for force and torque include a weakcoupling coefficient, k~¢, which is typical of coil springs: M = k¢¢ + kxcx ;

F = kx¢¢ + k~x

(a) Model the system in standard matrix form. (b) Find a modal matrix and describe the individual

modal motions.

(c) Find analytical expressions for the response given initial deviations from equilibrium of x = 0.1 m, ¢ = 0, ~ = 0, ~ = 0, and the values of the parameters assigned in Problem 7.23 below.

7.23 Address the problem above using MATLAB with the initial conditions given in part (c) and the parameter values given at the top of_the next page. (a) Do parts (a) and (b) numerically. (b) Find the numerical state transition matrix for a time step of 0.1 second. (c) Plot x(t) and ¢(t) for several cycles.

554

CHAPTER 7.

ANALYSIS

OF LINEAR MODELS, PART 2

kx = 400 N/m k¢ = 0.400 N m/rad kx¢ = 1.00 N/rad rn -- 1.00 kg rg = 0.031 m

7.24 The double pendulum shown above right can be modeled as two pointmasses of weight 1 lb at the ends of two massless wires each of length 10 inches. The angles ¢1 and ¢2 should be assumed to be small, permitting a linear analysis. Answerthe same questions as stated in the preceding problem, substituting ~1 and ~b2 for the displacements x and ~b, respectively. 7.25 A five-story building is assumed to rest on a rigid foundation, but each floor can movelaterally with a relative deflection per floor proportional to the shear force. Model the system with a bond graph, define state variables, find the matrix A, compute the natural frequencies of vibration, and describe the positional modal shapes. Use of MATLAB is suggested. The stiffness per floor is 1 × 10s N/mfor all five floors. The mass is concentrated in the floors, and equals 5 × 105 kg per floor. Damping may be neglected.

7.26 The building of the above problem is subject to a sinusoidal lateral force of amplitude 2 × 106 N at various frequencies, uniformly distributed over the height. Youare to investigate the deflection of the top floor.

(a) Evaluate the matrices A, B, C and (b) Give a magnitude Bode plot of the response. scales for this plot may require special attention.)

(Getting reasonable

7.27 Repeat the above problem if, instead of a wind force, an earthquake shakes the ground laterally with a sinusoidal amplitude of 0.2 meters.

7.3.

MATRIX

REPRESENTATION

OF DYNAMIC BEHAVIOR*

555

7.28 The building of the above problems has a damping coefficient per floor of 1 x 106 N.s.m. Find the decay rates and damped natural frequencies of the individual modes. 7.29 Repeat Problem 7.26 assuming the damping added in Problem 7.28. 7.30 Repeat Problem 7.27 assuming the damping added in Problem 7.28.

SOLUTIONS TO GUIDED PROBLEMS Guided

Problem

1.

q/C ",I-.--:-r~-~ O-,~-~--~.I aq/~c T.~.P q/C~) C 1 1

R

dq 2.

dp

7.9

1

therefore, ~

A=[ llC

= A

,

[_ooo _:oo1

= 20000

3. s_20000 +2000 5000s =0, ors 2+2000s+1 s=0 x 10 Therefore, s = -1000 5= V/(10O0)e- s = -1000 4-j9950 s-1 4. q = qoe-l°°°t sin(9950t 5. q(0) = 0 = qo sin c~; therefore, (~ ° 6. p~ -- --~---~q-1 ’~dq -1°°°~ _- [-(2000 ÷ 1000)sin 9950t - 9950cos 9950t] qoe

p(0)= ~ 7.

I e(t)=

~

= -9950qo; therefore, qo = -50.25 x 10-6 amps 50.25 q(t)50 e-lOOOt sin 9950t

556

CHAPTER 7.

Guided

Problem

ANALYSIS

OF LINEAR MODELS, PART 2

7.10

= 1

spring: ~ k(deflection)’ 1 therefore, C~e = ka ~ gravity: therefore, inergia:

1

C~ = C~ -

~mv- = 1

dp~

1

dp~

1

~ + (~

at -

~)

2. The state v~iables p~ and p~ ~e replaced below with ~ = p~/I~ ~d ~ = p~/I~ since this helps cl~ify the results. d

4~

0

=

Therefore, ~ =

3. lsI-

g

therefore, w

5.

0 al --a2

jwi --a2 al

0 jcai 0

-al

0 a2

0 0

a2

--a~

0

0 s -a: a~

~ Al = a~ -ae

0

-1 0 s 0

~ 2ka

--1 0 jwi

g

] m3i k m4i

a~

=

a2

~ 2

a 2 mL~

’

0 ~+( -1 a~-a~)=O =sa+2a~s 0 s

m2i

0

= 0

mn

7.3.

MATRIX

REPRESENTATION

OF DYNAMIC BEHAVIOR*

Fromthe first equation represented the third equation, P2i = (al/a~.) P4i = (jwi/a~)(al + ~). The fourth of ~ and ~w2simplifies the result

P= j~ -jw~ k-J~

Est =

jw~

jwtt e 0 0 0

[ [

Esmt -=

0 -y~t e 0 0

cosiest - sinwlt 0 0 ej~l t

j~ -j~ jw~

557

in this matrix relation, P3~ = jwi. From (wi2/a.~). Thesecond equation then gives equation serves ~ a check. Substitution to

~

P~ =

-j~

~1

0

-w~

0

~2

~

0 0 0 0 j’‘et e 0 0 e-~t ] sinw~t 0 0 cosw~t 0 0 0 cos w2t sin ¢v2t 0 - sincv2t cos¢a2t

’t PESt = ]I --e3~ j~e3~t

te-JWl

eJw2t

--e-J~o’t

eJ~°~t

e-Jwo.t

_jw~e-J~t jw~eJ~t _jw2e_j~t [-jw~C~ ~ jw~e-~ ~ jw~e~ t t-jw~e-J~

- cosw~t -sinw~t coswet sinw~t -wl sinwlt Wl COSWlt --w2 sinw2t w2 cosw2t wi sinwlt --Wl coswlt --w2 sinw2t w2cosw2t The result in either c~e can be written PmEsmt =

¢~ ~ ~

=

-b~ cosw~t- b~ sinw~t + ba cosw~t + b4 sinw~t -b~w~sinw~t + b~w~cos w~t - baw~sinwet + b4w2cosw2t b~w~sin w~t - b2w~cos w~t - haw2sin w2t+ b4w2cos w2t

in whichbt,.. ¯, b4 are coefficients determinedby the initial conditions. Specig ically, b~ = P;tx(O)= ba

b~

g

/¢~(0) / o ~/~ ~/~ J L¢~(o)J 0 1

~/~ 0

-~

~

Note that the third and fourth rows above equal the derivatives of the first two rows, respectively, which we anticipate anyway since ~ = de/dr and ~b~. = d¢.~/dt. Whenb3 = b4 = 0, only the first modewith wl exists. In this case ~b~ = -4~, indicating motion that is symmetric about the vertical centerline, as shownbelowleft. This implies that the center of the spring does not move, so each pendulumsees an effective spring rate of 2k, whichexplains the natural frequency ¢v~ = g/L +.2ka’Z/mL~. Whenb~ = b~ = 0, on the other hand, only the second modewith w~ exists. In this case ¢~ = ¢%indicating

558

CHAPTER 7.

ANALYSIS

OF LINEAR MODELS, PART 2

no deflection or force whateverexists in the spring; as shownbelowright. The natural frequency of the motion therefore is the same as a single free pendulum, or w.~ = g/L. The general solution is of course the sumof both modes. The beating phenomenonreferred to in the suggested steps is particularly interesting.

motion of firstmode

I

Guided

Problem

motionof ~’] second mode_ / . _/

/-~]

I . .

7.11

/

dp dt

= F(t) -

d~

<,[:] [:]

d~ .= A

+ BF(t);

2. G(s)= (sI-A)-’B

therefore,

rLs

+ r/I

A

1

r-.,-, l 1/1

,

B=

[1o]

1/sC]-~ [10] =p-~ [1;i]

p(s)= ~ +(R/Z)s + x = GF; G(S) = 1/.~_I p(s) 1/I 3. G(jw) = 1/IC - w2 + (R/I)jw

Ia(jw)l

1/I ~/(1/IC - w2)’)- + (Rw/I)~ ’

za(j~) = - tan-~ \ 1/IC - w2]

4. x = xo sin[cot + ¢(w)] 10 ’ ~o = FolG(jw)l=~/(--’400 - we)~ "~’ + ( )’40w

( 4~

¢(w) = - tan -1 \400

7.3.

MATRIX REPRESENTATION

OF DYNAMIC BEHAVIOR*

559

5. The following plots are given by MATLAB:

Guided

Problem

"8010o

~ 10 Frequency (rad/sec)

10o

~ 10 Frequency (rad/sec)

~ 10

7.12

1. The third differential equation has a forcing term added: dt - C1¢1 - (¢1 - ¢~) LF(t)

Therefore,

B = L~I = 1/--L

first state variable, so you need to find

2. After the matrices A~ B, C, D are entered into MATLAB, the command bode(h,B,C,D,1) produces plots like those below. Alternatively, the actual plots were produced using the results of steps 3 and 4 below with the commands u=[1:.05:4 4.01:.01:4.42 4.429 4.4297 4.43:.01:4.58J; ~=[~ 4.588 4.5889 4.59:.01:5 5.1:.1:10]; num=[’2/3 0 40.68/3] ; den =[1 0 40.68 0 413.2J; bode(hum, den, w)

Frequency (rad/sec)

560

CHAPTER 7.

ANALYSIS

OF LINEAR MODELS, PART 2

3. Fromstep 3 of the solution to Guided Problem 8.5, 2-a~2 p(s) )= s4+2als.)+(al G(s) = (sI - A)-IB = p--~s) Adj(sI Youneed only the element of Adj(sI - A) in the first row and the third column, whichis the cofactor of the element in the third row and first column. Since 0 -1 this is s 0 (S 2 +al)/mL Therefore, G~3(s) = s4 + 2als. ~ + (a~ - a~) (sI-

A)

0 al -a2

s -a2 al

-1 0 0

0 -1 = (al + 2) s

.) ka2 8.0 x (.15) s_ 4. a2- mL~ -- lx (0.5) 2 --0.720 2 g 1 1 2 _~ -2 al--a2+~=0.720+--,.o =20’34s mL- 1 ×1.5- 3 0.6667(-w2 + 20.34) G~3(jw) = w4 40.68w2 + 413.2 This is a real numberfor all w; its phase angle is 0 whenthe numberis postive and 2=180° whenit is negative. 5. See plots given above. The poles and zero were pinpointed for better plotting by the MATLAB commandsr=roots (den) and r=roots (num). The zero ka2 + occurs at the frequency w = ~ = ~--~-. It means that the forced pendulumis immobile at that frequency; the spring and second pendulumresonate, and therefore appear to the forced pendulumas an infinite impedance.

~

7.4

The Loop Rule*

You have employed the causal stokes on bond graphs tO direct the writing of state differential equations, and e~nployed matrix methods to extract differential equations with a single dependent variable. This procedure is sufficient to address any linear lumped-parameter model. An optional graph method for deducing transfer functions is available, however, which can facilitate the deduction of transfer functions without the use of matrices, and lend greater insight into the structures of particular and general system models. It also can be used to expedite the basic method of finding the ~tate differential equations. This method is based on the loop rule. As originally presented by Shannon and elucidated by Lorens, ~ this method applies directly to signal flow graphs. Bond graphs can be converted to signal flow graphs readily, although no general inverse procedure is known. The author has extended the loop rule to apply directly to bond graphs, ~ wiih con~sequent simplifications. The description below 6c. Lorens, Flowgraphsfor the Modelingand Analysis of Linear Systems, McGraw Hill, 1964. ¯ 7F.T. Brown,"Application of the LoopRule to BondGraphs," ASMETransactions, J. of DynamicSystems, Measurement and Control, v. 94 n. 3 (Sept. 1972) pp. 253-261.This work is used as a basis for the LorenzSimulationsoftwarepackageMS1(seefootnote p. 136).

7.4.

THE LOOP RULE*

561

starts with the definition of the signal flow graphand proceedsto the conversion of a bond graph to this form. The loop rule is then presented. Application of the loop rule to find transfer fuctions directly from bondgraphs follows. The procedure is particularly siinple to apply for the determination of the state differential equations, whichis the final subject.

7.4.1

Signal

Flow Graphs

A signal flow graph represents a set of linear differential and algebraic equations, s None of the power and energy constraints assumed with bond graphs are required. The graphs comprise variables, represented by nodes drawn as conspicuousdots which are labeled alongside by the algebraic symbolsfor tile variables, and relations betweentile variables, represented by directed lines between the nodes. The relation each such line represents is labeled next to an arrowheaddrawnnear the center of the line. One or more directed lines emerging from a node carry the value of the variable represented by that node:

The linear algebraic relation x = ul + u2 + ... + un is represented by a node with incident lines that represent each of the componentterms: 21~ U

U ~~’~

U3

The linear algebraic relation z = au is represented by the arrow itself on the associated line; the symbola, or whateversymbolrepresents the coefficient, is labeled next to the arrow. Minussigns can be used where appropriate. If the line conveysa variable directly, without any modifyingcoefficient a, the symbol 1 is substituted: ~t

a :

~c=au

Finally, the differential operation z = a du/dt is represented by placing the notation as (or aD) next to the associated arrow, and the integration operation z = afudt is represented by placing the notation a/s (or a/D) next to the arrow: as ~ =- du ~=a

SNonlinear algebraic

functions

dt

a/s ~ " J=afu

also can be incorporated,

dt

but are avoided in this text.

562

CHAPTER 7.

ANALYSIS

OF LINEAR

MODELS, PART 2

Example 7.11 Drawa signal flow graph for the following set of linear algebraic, differential and integro-differential equations, in which the coefficients a... k are constants: dxa xl = ax3 dt + b -~- + cu2; dx6 ; x’2 =-~d --~ dx6

x,~ = ex5- f ~-; x4 = gx5 + hUl; x5 = ixo. +jue; X6 = kXl

+ X2.

Solution: These equations are assembled in the signal flow diagram below. Each variable is represented by a node, and the constant coefficients appear next to the arrows, along with the operator s to represent differentiation and 1Is to represent integration. You ought to identify the individual equations embedded in the graph.

The signal flow graph is a very close relative to the block diagram, which may be knownto the reader. The block diagram uses boxes in which the coefficients and operators are placed, instead of writing them next to an arrowhead. It uses circles instead of nodes or dots to indicate summation, and places plus or minus signs next to these circles rather than incorporating negative signs into the coefficients. Finally, it uses straight lines, most of which are horizontal or vertical; the tradition in signal flow graphs is to use curved lines, so as to identify loops visually more easily.

7.4. 7.4.2

THE LOOP RULE* The

Loop

Rule

563 for

Signal

Flow

Graphs

The loop rule gives the transfer function between any input and any output in one sometimes simple and sometimes giant step. If the input in question is labeled as uj, and the output of interest as xi, the loop rule can be expressed

(7.124)

The symbol Lijk represents the path gain for the kth directed path from the input to the output. This is the product of all the coefficients and operators on the path. Paths are distinct if any one segment is distinct; separate paths may traverse some commonsegments. One entire path does not qualify as part of another path, however. Identifying all the paths and writing a list of their path gains is a major part of the procedure for implementing the loop rule. The symbol A in equation (7.124) is called the graph determinant. The graph determinant is in general a complicated function of the gains of loops inside the signal flow graph:

= - + Ly - +...).

(7.125)

A loop is a closed directed path from one node back to itself; the gain of the lth loop is L~. Therefore, the term ~L~ represents the sum of the gains of all the loops in the graph, where separate loops are identified the same way as are separate paths. The superscript 2 means the products of loop gains for pairs of non-touching loops. Loops are said to touch if they share one or more common nodes. Loops so often touch one another that, even if there are several loops in a graph, there may be few pairs of non-touching loops. The superscript 3 means the it products of loop gains for sets-of-three non-touching loops. Different setsof-three loops may share two loops but have distinct third loops; nevertheless, these combinations are usually few, if any. The subscripts continue, representing set-of-four and more nontouching loops, although in practice such combinations are rare. Finding the graph determinant is apt to be the hardest and therefore most critical part of the application of the loop rule, and must be done with care.

The final symbol in equation (7.124), Aijk, is called a path dete77ninant. This represents the determinant of the graph in which the kth path from the input node uj to the output node xi is expunged. Erasing this path usually eliminates most of the terms in the overall graph determinant, particularly those for products of loop gains. If there are no loops left at all, the unity first term in the series survives, nevertheless.

564

CHAPTER 7. ANAL}SIS OF LINEAR MODELS, PART 2 Example 7.12 Find the transfer function between the input u2 and the output x3 of the graph of Example7.11 (p. 562). Solution: First, you identify the three loops and their gains: Loop1: Loop2: Loop 3:

(-ds)ig = -dig k(-fs)(a/s) = -kfas; ~. k(-fs)(bs) = -kfbs

Loops 2 and 3 touch (share one or more commonnodes); loop pairs 1-2 and 1-3 are not touching, and no set-of-three nontouchingloops exists. The system determinant becomes A = 1 + dig s + kfa + kfb s~ 3. + digkfa s + digkfb s The first path from the input u2 to the output x3 has the gain -ckfs. This path touches loops 2 and 3, so whenit is erased it leaves the path determinant A1= 1 + dig s. The only other path has the gain je. It touches all of the loops, so A~= 1. Theresulting transfer function therefore is je- ckf s(1 + digs) (1 + kfa) + (dig + digkfa)s + kfbs 2 3" + digkfbs The value of x3 responds to input Ul also, of course. Its transfer function G31shares the same graph determinant, however, and both of its paths (with gains -hdies and hdfs 2) touch all three loops. Thus both path determinants are unity, and 631 :

2-hdie s + hdf s (1 + kfa) + (dig + digkfa)s + kfbs ~ 3" + digkfbs

The overall behavior of the system becomes X3 : G31Ul -b G32u2.

7.4.3

Converting Bond Graphs to Signal Flow Graphs

Each non-activated bond in a bond graph represents a bilateral pair of signal flows: e

Anactivated bondcuts one of the signal flows: eI

lq

e2

q~

e~=

1

= e2

7.4.

THE LOOP RULE*

565

A transformer Ti simply represents a coefficient T~:

el

q~

_

e~

el:

q2

t~:

e2 q2

e~.,,.~e2

A gyrator is very similar:

e~ q*

The 0-junction has a common effort, whichis represented in the signal flow graph by a node with a single input and two or more outputs, as dictated by the causal strokes on the bonds. There are, on the other hand, two or more input flows and one output flow; the latter is representedby a nodewith a single output. The result is

el

lel!

The 1-junction is the dual of the 0-junction. Its signal flow graph is the same as for a 0-junction, therefore, except the roles of effort and flow are inverted:

e~

566

CHAPTER 7. ANALYSIS OF LINEAR MODELS, PART 2

Example 7.13 Consider a two-tank system such as in Guided Problem5.1 (p. 299), but with the inertance of the tube that interconnects the two tanks included in the model,as well as the resistance. Find the transfer function that relates the input flow, Qi, to the volumeof the second tank, ~.

Qi.~

A2

A ~ density,o ",----L

I~ porouspiugdrain

-~ "

Solution: A linear bond-graph model is drawn below left, and integral causality is applied. The correspondingsignal-flow graph is drawnbeside it S

Oi

1

P PIo

1/C,

~ C~ -1 p

I@I [ ~--’--~

R~

9, ~/s ’ 1/C

P 1/I

P21 v~/G

-1/~2 R2 This graph contains four loops, but only two non-touching pairs and no set-of-three nontouchingloops. The gains of the loops and the system determinant are -1 -R1 -1 gains of loops: -1 ~ ; ~ : "-~: Csls ¯ non-touchingpairs graph determinant:

+R~+Is A = 1 1+R--~

1~

CI1S

2+ ~

1RzCIs + 1~+ R1 ~ -~ R2CICIs

There is only one path from the single input, Q~, to the output 12.2; this touches all .loops, so the corresponding path determinant, A2, is unity. There also is only one path from Qi to the output ~], but this path touches only one of the four loops; as a consequence,its path determinant, A1, is the

7.4.

567

THE LOOP RULE* same as the determinant A except for the deletion involve the gain of that loop (-1/ICls2): path determinants:

dI = 1 +

~s

RI ÷7~- ÷

~

of the two terms that

1 . ÷ R2---~’

,42= 1

The resulting transfer functions are

V1(s) fq (s) Qi(s)

sOi(s)

= s’~ + (-fi +tt.,c’)s +(y~+R~Ic) v:2(s) O~(~) --1

7.4.4

Direct Application without Meshes

of the Loop Rule to Bond Graphs

Before the loop rule can be applied, all multiport fields must be purged. Mutual elements can be handled by the full method, but to simplify the presentation they also will be assumed to be replaced by bond graph equivalences having only one-port I, C and R elements. Next, causal strokes are applied to the graph. Impedance or admittance causalities may be chosen in any compatible combination. If you prefer to find polynomials in s directly, rather than polynomials in 1Is as were first found in the example above, employ differential causality for the C and I elements. You wish to forego the drawing of the signal flow graph. The principal task is the determination of the system determinant. Five ~teps can be distinguished: (i) locating each loop; (ii) determining which loops touch; (iii) ’determining magnitudes of the gains of the loops; (iv) determining the signs of the gains of the loops; and (v) accounting for the effect of any bond activation that may present. In practice these steps can be carried out virtually simultaneously. Bond graphs without meshes, which are tree-like, produce only what are called flat loops. A flat loop includes both signal-flow paths for each involved bond. Thus, these bonds form a simple cascade with one-port non-source elements at each end, as illustrated in Fig. 7.16. The signal flow paths in this figure and the next two figures are shown as dashed lines superimposed on the bond graph. This practice is not customary, but occasionally is instructive. The causality of bonds in the chain must be directed unilaterally through each 0 and 1-junction. Reversals in causality, in fact, occur only at gyrators. Each R, C and I element can be singled out as a potential termination for a cascade of

568

CHAPTER 7. ANALYSIS OF LINEAR MODELS, PART 2 bondgraph: loop gain:-R/Is signal flowgraph:

bondgraph:

ZI~{<

/ 0 ~----T~----~-0 signal flow graph: Z~TT~=i ~t~Z~i~ t~-~ lOOPl ,gain:.

T~Z~

Figure 7.16: Examplesof flat loops bonds corresponding to a flat loop. Proper causal paths to other R, C and I elements are sought. Anybondactivation in a cascade voids the loop. If carried out to completion,this process will identify each flat loop twice. Flat loops touch if and only if they involve a common0 or 1-junction. The loop gains include the impedance (R, 1/Cs or Is) or the admittance (l/R, Cs or l/Is) of the terminal R, C and I members,according to the indicated causalities. Theyalso include either T/2 or 1/Ti2 for each transformer Ti in the cascade, and G~ or 1/G~ for each gyrator Gi in the cascade, again according to their causalities. Finally, the gains of all fiat loops are negative, presuming only that the powerconvention directions on each of the two one-port elements is directed into the element. Inspection of the bondgraph in Example7.12 reveals directly the existence of fiat loops between CI and I, I and R1, I. and C2, and C2 and R2. Only two pairs of these four loops do not use a common junction, and therefore do not touch. I and R2 have admittance causality; the others have impedancecausality. There are no transformers or gyrators. The system determinant follows directly fromthese facts. Paths from an input to an output variable can be seen directly from the causal consistency of their cascaded bonds. In Example7.12, the path from Qi to ’(’1 simply passes through the upper zero junction; the powerconvention arrows on the two bonds are aligne.d similarly, so there is no sign changeand the gain is 1. Thepath fromQi to V2is shownin part (c) of the figure. It starts the same way, but passes through C1 (impedance causality) and back through the upper zero junction. Since this second pass is on the effort sides of the

7.4.

THE LOOP RULE*

569 _R

-’~, ,~" meshstub

c t ~ CCWopen (gain loop= RT/G) ",//¢ --i’~._~_~’O._~S_~.~_~j___~I 2 ]1

Figure 7.17: Anopen loop with a mesh stub bonds, there is again no sign change. The path then proceeds downthrough the 1-junction to the I element; the causal strokes aroundthe 1-junction allow no choice. The path continues through the I element, which has admittance causality. It returns through the 1-junction on the flow side; again there is no change in sign. The path then passes through the lower 0-junction, remaining on the flow side of the bonds, and emergesas the desired V2. There is no sign change again, this time because the powerconvention arrows of both bonds have the same orientation. The path gain becomes (1/Cls)(1/Is). This path and gain can also be seen in the signal flow graph. To generalize, there is no sign changefor a path traversing a 0-junction on the effort side or a 1-junction on the flow side. Similarly, there is no sign change if the two powerconvention arrows are aligned the same way. Since these two conditions are independent,on average one in four traverses of junctions produce a sign change. Note that the path to ~’1 clearly touches only one loop, whereasthe path to ~r2 clearly touches all four loops. Thus the path determinants A1 and A~2can be found and the transfer functions completed. 7.4.5

Bond Graphs

with

Meshes

The presence of one or more meshesin the bond graph considerably complicates the application of the loop rule. As a result, youusually are well advisedto apply any conversion that would eliminate such meshes. Otherwise, the following procedure can be used. Each mesh produces a pair of counter-directed open loops. Someopen loops involve a mesh stub, such as shown in Fig. 7.17. Other open loops do not; a stub is involved if and only if the meshbonds on either side of the associated junction have opposite causalities in the clockwise-counterclockwise sense. The magnitudeof an open loop equals the product of the moduli of each meshtransformer or gyrator, or its inverse as determinedby the causalities, and the impedance or admittance of each meshstub. Both counter-directed loops o] a mesh without gyrators have the samegain, whichis positive if the meshis even and negative i] it is odd. (Recall that an even meshhas an even number

570

CHAPTER 7. ANALYSIS OF LINEAR MODELS, PART 2

activated bond

(a) bondgraph showingnon-touchingopen loops (dashed lines) 1/R

e -1

~11

-1 el

1

1/C2s 1/Cls (b) signal flow graph Figure 7.18: Mesheswith two non-touching open loops

of branches betweensuccessive junctions with a clockwise or counterclockwise causality; evenness and oddness are immutible properties of a mesh.) Loop gains must be dimensionless, a fact whichhelps identify mistakes. A gyrator inserted into a branch of an otherwise even or odd meshforces a mixed causality for that branch. The result is called a neutral mesh, for which the two gains o] the open loops have the same magnitude but opposite sign. The loop with the positive gain is directed similarly as are the power convention arrows of those non-gyrator branches which are odd in number. A second gyrator inserted into the meshconverts it back to even or odd status, and a third gyrator reconverts it to neutral status, and so on. Neutrality is als0 a property of a mesh; it cannot be changed without changing the meaning of the model. The two open loops of a meshtouch each other if and only if one or more mesh stubs are present. Twoopen loops of different meshes might not touch one another even if they share a commonbond, however. An example is shown in Fig. 7.18. This examplealso illustrates the fact that activation of a meshbondcuts out one of its-two open loops. (Activation of bondsis not considered until Section 9.2, but reference is madehere for efficiency and because Chapter 9 can be addressed before the present chapter.)

7.4.

571

THE LOOP RULE*

Example 7.14 Reconsiderthe heaving and pitching vehicle presented in Fig. 5.15 (p. 322), Substituting an exciting force F(t) on the left axle for the original positional disturbances. Determinethe transfer functions from this force to the elevation of the center of mass, Ycm, and to the pitch angle, ¢. To get you started, a bondgraph of the situation in given at the top of the next page, with integral causality applied. The associated signal flow graph is drawn next to it, for reference. It maybe tempting to employthis graph for the analysis, but at this point you are urged to work directly from the bond graph instead. Onceyou are used to identifying paths and loops directly from the bond graph augmentedwith causal strokes, your process becomes quicker and less proneto error.

l),, 2.0__

C -’*--~/

F(t)

"1 ~--~-C

Solution: All four mesh branches counterclockwise causalities, so the mesh stubs. The rules above tell mesh loops are -T~/T1, and that one another. The bond graph also

21/1.# llT, A 1~,. ~/1/T, ""....~...k~-l/C~s

1/Ics~,..J have the same clockwise or mesh is even and involves no you that the gains of both these loops do not touch has eight flat loops, with

~, -RI/T~~m~,-T~/ C~T~T~ gains- t / C~T~~m, ~,~,-n~T~/T~~,

-1/C2T~I~s2, -R2/T~I~s, -1/C~T~2Ims~ and -R~/T~Ims; only the firs~ ~nd second do no~ ~ouch~he fifth and sixth, and all eight touch both open loops. As a result, the systemdeterminan~is

~ =~ + ~ ~ ~/c~ + n~ (~/c~ + n~)~ T~ + ~ + T;~Z,~, + T~T~, ~/~ + ~ ~/C~s + ~ (~/C~ + n~)(~/C~+ + T~, + Ti~, + ~T~Z~,~

572

CHAPTER 7.

ANALYSIS OF LINEAR MODELS, PART 2

The path from the input force; F, to the velocity of the center of mass, Sin, passes through the transformer T1 and the inertance L,. Its gain is therefore 1/TlI,~s. This path touches the clockwise mesh loop, but not the counterclockwise loop. It also touches all of the flat loops except those that pass between C_~ and I¢ and between. R2 and I¢. Its path determinant is therefore T~. I/C~.s+ R~ Aura = 1 + ~ + T~2Ics , and the transfer

function becomes ~_ 1 A~m F A T~I~s

The path from F to the angular velocity, ~, passes through T~, T2, T3 and finally the inertance I,. The gain changes sign at the right-most 1junction, since the path is on the effort side of a common-flowjunction and the power convention arrows reverse direction relative to the path. The p~th gain, therefore, is -T2/T~T3I~s. This p~th also does not touch the counterclockwise mesh loop, but does touch all eight of the flat loops. Thus its path determinant is A¢=1+~, and the transfer

function becomes

F A TITaI~s

7.4.6 Determination of State Differential

Equations

Finding state differential equations usually is considerably easier than finding transfer functions. The reason is that most if not all of the loops in the full signal flow graph are excised. The procedure presumes integral causality has been applied to the bond graph, unlike above where integral causality is ~nerely one option. The application of standard state-variable notation also is presumed. Next, all I and C elements are removed. This can be indicated by crossing out these elements with X’s. The flow pi on the ith inertance and the effort qi/Ci on the ith compliance are treated as inputs to the system. The corresponding dpi/dt for the inertances and dqi/dt for the compliances are treated as outputs. The expression for each output in terms of all the inputs becomes the associated state differential equation.

7.4.

THE LOOP RULE*

573

Example 7.15 Write the state differential 7.13(p. 566). Solution:

equations for the two-tank system of Example

The causal bond graph with the I’s and C’s removed is S

Removal of the elements C1, C2 and I eliminates all three loops, so the system determinant as well as all path determinants equals unity. The path gains from the five causal inputs to the three outputs are assembled in a table given below left. These directly give the differential equations given to the right.

output

input Q, eo V/C, V2/C2 1

o

0

0

0

0

0

-1/R:

0

1

1

-1

dV~ = Qi - ~ P

p/I -1

dV2

1

dp

1

1

V: ’87 = Tp- ~c%

1

1

1

~[ = C~ v,- ~V2- .T p- Po

-R,

Example 7.16 Write the state differential Example 7.14 (p. 571).

equations for the heaving and pitching vehicle of

Solution: The causal bond graph with the energy-storage elements crossed out is given at the top of the next page, with identification of the loops. Removalof the four energy-storage elements eliminates all eight fiat loops, but leaves the pair of mesh loops intact. These loops do not touch one another, so the system determinant becomes A = 1+2T~ T~.~-’ T, + T?

"

(2 TxJ

= 1+

574

CHAPTER 7.

ANALYSIS

OF LINEAR

MODELS, PART 2

flat loops: none gains of meshloops: -T2/Tt

F(t)~ .i.T~ ..~.y,/C,~ ~./

"T~\

(non-touching pair; negative gain since mesh is odd)

R,

R z

~

T~

The path gains and path determinants are given in the table below. Some paths touch both loops ~d therefore have a path determinant equal to 1. Others touch only one path, ~d therefore have a path determinant equal to 1 + T~/T~. input output 9,

y~/Cl y~/C~ p.,/l., 0

0

1/T~ dI

92

0

-l/T, d=At

p ~/I ¢,

0

=d

1/T~

F(t

-Tz/TI A=A l

0

1/T 3 A=Ai

0

At-= 1 + T2/T~

d=d t -1/T~ -R,/Tt ~ -R2/T3T~ -1/T 1 d=d~ A=I A=I A=dt

A=At A=A~ d=l The resulting differential

d=l

d=At

equations axe

~ A = (1 + T:/TO

dpm 1t [_ T~_.~..~ly~ + A R~ Rz dt - A -~F(t] ~lC~ y2 - ~I~ p’’- T~p* + 2 TaCI -~"~ = -~1 [[ T,ArT

A R~T~ l ~3Y2 + TI"~3 P,.-

R z _ A~T2] T3Zl’~ pC TtT3J

7.4.

THE LOOP RULE*

7.4.7

575

Summary

The loop rule facilitates the writing of state differential equations for linear systems. It also is efficient in finding one or two transfer functions for a complex system, precluding the need to deal with the matrix~equations. Should one wish to find several transfer functions for a particular system, however, the basic matrix approach may be preferred, particularly if a program such as MATLAB is available. The rules for applying the loop rule to causal bond graphs are based on the special energy constraints implied by the graphs. They aid in finding paths and loops, determining which loops touch, and finding the loop gains without laborious examination of the corresponding signal flow paths.

Guided

Problem

7.13

Consider the drain tube placed through an ear drum as described in Problem 4.7 (pp. 202-203), including the given simple model that neglects damping and other effects. Find the transfer function relating the input pressure to the displacement of the eardrum. Find and plot the magnitude ratio for frequencies between 20 Hz and 10,000 Hz, and compare this to the case with no drain The following paramete rtube. Does the tube impair hearing significantly? values may be used: Ce = 0.5 x 10-6 ft5/lb, Ie = 1.5 lb.s2/ft 5, Cm= 1.2 × 10-s ft~/lb, and It = ~. 5.8 lb.s2/ft Suggested

Steps:

Since the mesh is even, it is appropriate to find the equivalent tree-like bond graph. Apply causal strokes in any consistent fashion. Differential causality actually simplifies the intermediate expressions somewhat. Identify and label the input and output variables. Find the loops (which are all fiat), write their loop gains and identify pairs of non-touching loops, etc. Find the graph and path determinants. Write the transfer functions with and without the tube in place. Substitute s = jw or use MATLAB to plot the frequency response. Plot the magnitude Bode asymptotes to the two cases, and draw conclusions.

576 Guided

CHAPTER 7. Problem

ANALYSIS

OF LINEAR MODELS, PART 2

7.14

This problem offers a fairly general and complex bond graph; it is appropriate only if you have already studied Section 9.2 on activated bonds. Repeat Guided Problem 9.5 using the loop rule method as applied to bond graphs. Suggested

Steps:

1. Copy the causal bond graph from part (d) of Fig. 9.11, or spread it out shown in the solution to Guided Problem 9.5. Annotate the appropriate bonds with the time derivatives of the state variables and the conjugate variables, as also shown in the solution. Cross out or erase the energy storage elements. 2. Identify the flat loops, if any. Note that any potential flat loop must involve two of the three resistances, since the graph has no other honsource stubs. 3..Identify the open loops, if any. Note that a mesh stub with a deleted energy storage element will preclude any open loop that would use that stub. Also, pay attention to the activated bonds, which eliminate one of an otherwise pair of open loops. 4. Find the graph determinant using the results of steps 2 and 3. 5. Makea table of the various path gains from the six graph inputs to the five graph outputs. This is the most difficult part of the solution, and this is a complexcircuit. First, locate the various paths starting at each input and following those bonds which have compatible causal strokes. Hint: there are a total of twenty paths, including more than one from certain inputs to certain outputs. Second, find the magnitudes of the path gains, and enter these into the table. Only the resistances and the transformers cause the magnitudes to be different from unity. Third, find the signs of each path, and enter minus signs where appropriate in the table. In this endeavor it is helpful to note that signs change only when a path goes through a junction, and then only if an effort path passes through a 1-junction or a flow path passes through an effort junction and the power sign arrows of the bonds on either side of the junction are oppositely directed. 6. Find the path determinant for each path, and enter these into the table above. Path determinants are the graph determinants of the graph with the path deleted; since the determinant of the full graph is quite simple, these path determinants are simple also, a majority equaling unity. 7. Write the desired differential equations directly by assembling the infor’ mation found in steps 4, 5 and 6. This should be the simplest part of the job. It is suggested that you check your answers.

7.4.

THE LOOP RULE*

577 PROBLEMS

7.31 A permanent-magnet DC motor drives a gear pump through a reduction gear, forcing a virtually incompressible fluid from a vented sump, past a hydraulic accululator, through a long rigid tube, into a piston-cylinder load connected in parallel to an effective mass, spring, and dashpot, and from the other side of the piston back to the sump. The elements can be assumed to act linearly.

(a) Identify the meanings of the various elements in the bond graph model above. Also, identify on the graph the flow that emerges from the gear pump as Q1, the flow through the rigid tube as Q.,, and the velocity of the load as 2. (b) Simplify the bond graph by changing the definitions of some of the one-port elements so as to eliminate the need for transformers. (c) Draw a signal-flow graph for the bond graph of part (b), considering the voltage e as the input and the velocity 2 as the output. (d) Find the transfer

function between the input and the output.

(e) Repeat (d) using the bond graph directly. (f) Is oscillatory Why?

behavior still

possible if all compliances are removed?

7.32 The device shown near the top of the next page is designed to insert highfrequency sinusoidal pulses into a very long pipe which is shownin cross-section. These pulses generate waves which propagate downthe pipe. If the pipe is long enough and the frequency high enough, the apparent impedance of the pipe, as seen locally, is a pure resistance Rs known the the surge or characteristic impedance. (This phenomenon is addressed in Chapter 11.) The mean pressure in the pipe for a real device was as high as 800 psi, so that a single bellows would extendaxially and burst unless it were overly stiff and massive. An adequate back-up spring would be too massive. Thus, a highly compliant air-filled metal bellows with an equalized steady-state pressure was used. The compliance of the air kept the pressure in the bellows from fluctuating widely, reducing the surges in the small equalizing tube and allowing the bulk of the input power to be transmitted downthe pipe as desired.

578

CHAPTER 7. ANALYSIS OF LINEAR MODELS, PART 2

cross-sectionof verylong pipe separationplate~ driven by electromagnetic shaker

metal bellows Q2

mean pressure equalizing tube

~ air

_ __ _ __ m water

Twobond graph models are given below. The first is a direct model, and contains an even bond graph mesh. The second is equivalent to the fir’st, but employsthe bond-graphequivalence of Fig. 4.18 (p. 248) to eliminate the mesh. Intermediate steps are shownwhichdeal particularly with the problemof aligning the powerconvention half-arrows so that the standard meshreduction can be applied.

7.4.

579

THE LOOP RULE* (a) Find the transfer should be the same.

function P(s)/F(s) directly

from both graphs. They

(b) Draw the signal-flow graphs which correspond to the application differential causality to both bond graphs. Check your answers to (a) from these new graphs. (c) Find a set of state variable equations. (Work directly from one of your bond graphs, but use integral causality in the standard fashion.)

7.33 The sound movie camera of Problem 4.33 (p. 238) could be modeled shown below. pulley ratio

pulley ratio

t,--,-~-~ rs--~--- o -.--flywheel inertia, etc.

1 -~----

0 -~----

I,, C Cs c belt motor belt compliance inertia compliance

_ force, F Tc~ 1.ea~----

Ic claw-shaft & flywheel inertia

(a) Applydifferential causality, in so far as possible, and find the transfer function between the input force, F, and the output fluctuations in the angular velocity of the sound shaft, (b) The force perturbations quencies. Does the analysis tion? SOLUTIONS

TO

Guided

7.13

Problem

GUIDED

---~ 1 --’--~C. C,

presumably are well above both natural fresuggest good performance under this assump-

PROBLEMS

notedifferentialcausality

0 .~--.-.~ 1 i~--~. T xl~.-~

l, 3. loop gains: ~ -ItC~s~ ; -ItC,~s2; -IeCes The first loop above touches each of the others, but the other two do not touch one another.

580

CHAPTER 7.

4.

ANALYSIS

OF LINEAR MODELS, PART 2

A = 1 + I~C¢s2 + ItC,~s ~ + IeCes2 4+ ItC,~IeC~s 3 path from P to ~: (+C,,,s)(+I~s)(+C~s)T = C,~ItC~Ts

The path touches all three loops, so the path determinant is 1. ~ if(s) C,,,I,C~Ts 5. ~(s) = 1 + (I,C, + ItC,,~ + I, Ce)s~ 4+ (I,C,~I,C,)s 33.48 x 10-14Ts = 5.22 x 10-14s4 + 3.6568 x 10-682 + 1 Absenceof the tube gives I, -~ c%resulting in ~C(s) C,~C~Ts 0.6 x lO-l~Ts P(s) = (C, + C,~) + C,,I,C,s 2 = 8.64 x 10-1~s2 -B + 0.501172 x 10 6. No tube present: double-pole at ~ =

5.801 x 107 = 1212.2 Hz

tube present: double poles at ~= 3.6568x10-B " 7( 3.6568x10-~ ’~2 w 2 × 5.22 × 10-14 4- 2~ 5.--~ ~( 1--~14/]

1 5.22

×

-14 10

or ~ = 83.4 Hz, 1329.5 Hz 1212Hz . ,,. .... ~withoutmbe[ IoglP(jw) ~x(~°~ I "w~

asymptotes

log frequency The response is greatest in the mid-range; the inner ear must compensatefor this. The tube has no effect at high frequencies (> 1330Hz), causes a small loss at lower frequencies (a factor of 1.2 whichimplies 1.6 db) and a severe loss at extremely low frequencies (< 80 Hz). Guided

Problem

7.14

7.4.

THE LOOP RULE*

581

2. There are no sequences of bonds with unilateral causality interconnecting any of the three pairs of resistances, and therefore no fiat loops exist. 3. Loops for the left-hand mesh are blocked causally at both the upper 0 and 1-junctions. The right-hand meshhas a clockwise loop with gain -1/T1T2; the counterclockwiseloop is blocked by the bond activation. 4. The graph determinant is Ag = 1 + 1]T1T2 5-6. In each entry below the path determinant is given below the path gain outputs 1 R1

151

~2

-1 R,

1

1; -1

1

A~; 1

A~; Ag

-

-

1 A 9

1

-

-R3;

-R2

-R2

T~

T~

1; 1

1

R3 -Re TI T~ ’ T~ 1; 1

-R~ T~ 1

Chapter 8

Introduction Control

to Automatic

The purposeful managementof the state of an engineering system over time is called system control or, to emphasize that the details are carried out by machine rather than by a person, automatic control. In one scenario, the state is varied in some desired manner: a. vehicle is accelerated or.turned automatically, a machine tool is directed to follow a prescribed path, or a metal casting is cooled in such a manner that its properties achieve a desired standard. In the other most commonscenario, the state is held constant despite buffeting from the environment: a boat is prevented from rolling in the waves, a rolling mill for sheet steel produces uniform thickness product despite thermal expansion and contraction of the rollers, or the temperature in your house is held at 72°F despite outside temperature fluctuations. These two scenarios often coexist. Although ancient examples of automatic control exist, the modern discipline of control system science can be dated to the WorldWar II with its special impe-. tus for gunnery and vehicle control. The science comprises many mathematical and computer techniques. Its application to engineering systems requires the development of mathematical or computer models of these systems. This is an important application of the modeling that is the primary subject of this book. Often, the control science that needs to be added to the model!ng is relatively modest. This single chapter, which focuses on continuous as opposed to discrete control, can be adequate for manysituations, despite its presentation of only a fraction of the discipline of system control. Should you wish to learn more, you can elect a course or series of courses devoted exclusively to it, or study any of a score of textbooks.

8.1

Open and Closed-Loop

Control

The major components of an automatic control system are the the system to be controlled, the controller, which is the brains, the actuator, which is the muscle, 583

584

CHAPTER 8.

INTRODUCTION

TO AUTOMATIC CONTROL

power command input ~ error

disturbances

1I

I

~----I _1 system being[I

~

~~)-’~ c°ntr°ller

~1]1. a~tuat~lant~’]

c°ntr°lled

response or output

II I

Figure 8.1: Basic feedback control system and signal transducers, which are the senses. The actuator together with the system being controlled is known as the plant. A block digram showing the basic interconnections is given in Fig. 8.1. This is classic feedback control: the behavior of the system as observed by a signal transducer or sensor is compared with the desired or commandvalue of that variable, and the difference or error is sent to a controller, which in turn directs the actuator such that it reduces the error. In complex systems the actuators themselves are imbedded control subsystems, with their own brains, sensors and muscle. 8.1.1

Example

Plant

An example plant is used repeatedly in this chapter to help develop the basic concepts of control systems. In this example, shown in Fig. 8.2, the position of a rotational load with inertia I is to be controlled. The muscle or actuator of the system is a DC motor with gyrator modulus G = a and armature resistance R coupled to a gear drive with reduction ratio T = r. The mechanical friction in the motor and the gear drive will be neglected in order to give a linear model. The plant is the combination of the actuator and the inertial load. The transfer function from the input to this plant, which is the armature voltage v, to the angular velocity ~, is shown to be 1 r / a

GI(S)- (1 +

r2 RI

~=-~-" "a’~

(8.1)

This transfer function is represented in the block diagram in part (c) of the figur e. The angular velocity ~(t) is then integrated in time, as indicated by the block with transfer function 1/S, to give the angle ¢(t). The overall transfer function of the plant relates the input v(t) to the output ¢(t): ¢(t) 8.1.2

v( _~t) = Gg.(S)v(t), G~

Open-Loop

and

Optimal

(S) -

~/a S( 1 + rS

(8.2)

Control

To movethe load from rest in one position to rest in another position in minimumtime, maximumallowable voltage of the appropriate sign is applied for a 1Thereader maysubstitute for the operator S the Laplace s, throughoutthis chapter.

8.1.

OPEN AND CLOSED-LOOP

585

CONTROL

(a) schem~~/

(b)

bond

graph

v ~ !i~=.

Gp/TI G(v-Gp/TI)/R __ ~ p~l ~ ~ T1-’.¢v_Go/TI)/R G

|

v-Gp/TI~ R dP-T~(V:-~p); dt

G=a

Sincep=/~,

T=r a 2 ~; a (S+ ~-~]~p=~-~v

thisgives

1/r (c)

block

diagram

--~G 1(S)

I

r/ra

= r/va

rrv/a

(d) optimal control (particular example)

-1 ¢, degrees 20 I

0 (e) hypothetical removal (cancelation) of pole origin

(f) behavior hypothetical system

0.5

1.0

~ Im S~L -~r z~’r Re S cancelation /

t/r

1.5

~

30~degrees 0

(g) nominal implementation of hypothetical system

0.5

controller

1.0 motor

t/r

1.5

Figure 8.2: Open-Loop position control system

I

I

2.0

586

CHAPTER

8.

INTRODUCTION

TO AUTOMATIC

CONTROL

certain period of time, followed by maximumallowable voltage of the opposite sign for a second period of time. The result for a particular maximumvoltage and angle shift is plotted in part (d) of the figure. The minimization of some performance index, in this case the total time for the movefrom the first state to the second, is called optimal control. It is relatively difficult to compute the periods of maximumeffort, even in the present simple case; suboptimal control strategies are more commonlyadopted because they typically require much less on-line computation and are otherwise easier to implement. Nevertheless, the optimal control serves as a benchmarkagainst which simpler control schemes can be judged. Most practical control schemes involve one or more feedback loops. The more primitive scheme without any feedback loop is called open loop control. This schemeis addressed first. The steady-state response of the motor-load combination to a constant voltage input is an output with constant angular velocity. This fact is associated directly with the pole at the origin in the transfer function Go.(S). If this pole somehowwere eliminated, to give the transfer function kG1 (S) (with k being a constant), as shownin part (e) of Fig. 8.2, a single real pole with would be left, and the desired constant angular position would result instead. The remaining pole would impose an exponential approach to the final position, with time constant T, as shown in part.(f) of the figure. (The rate of change of the commandsignal ¢c(t) is limited in the plot, producing the initial ramp, to prevent the armature voltage from exceeding the limit imposed in the earlier optimal control. This allows a meaningful comparison of the results. Without the limit, the initial voltage would be infinite.) Could this desirable situation be achieved? The open-loop approach achieves the goal, at least theoretically, by superimposing a zero directly on top of the undesired pole, canceling it. The controller, therefore, comprises a differentiator; its transfer function can be written kS. This controller operates on the commandsignal to produce the armature voltage, as suggested in the block diagram of part (g) of the figure. The resulting overall transfer function between the commandsignal and the output signal therefore is the desired kG~ (S). If the commandsignal is taken to be the desired dimensionless angular position of the load (¢c), the ratio kr/a equals 1, so that 1 (8.3) ¢(t) = 1 + ~¢~(t). As a practical matter, however, this control scheme is virtually unworkable. The controller in reality would not be a perfect differentiator, and the pole at the origin would not be perfectly cancelled. It is likely that the load either would not come to a perfect stop following an extended zero value of the command signaL(that is would "drift" unacceptably), or that friction would cause it stick prematurely at the wrong angle. Errors would accumulate. The idea of open-loop control itself allows no correction for the inevitable errors that exist because no controller is matched perfectly to its system. Most systems also are susceptable to disturbances from the environment,

8.1.

OPEN AND CLOSED-LOOP

587

CONTROL

Controller . . Motor and load signal

T measured output signal (a) system arrangement

(b) example with proportional control

r(t)

G(S) r(t) c~t) =1"~-~

" ,~

(c) generalized unity-feedback loop Figure 8.3: Position control system with feedback

against which an open-loop control system is impotent. For example, the rotary output member in the example might interact with its environment such that a small torque is placed on it. Some means to compare the actual output of the system (the angular position of the load) with the commandinput is needed to address both this problem of disturbances and the problem of imperfect controllers. This means is called feedback control. 8.1.3

Feedback

Control

Measurementof the actual position of the load, by a rotary potentiometer, a digital resolver or some other means, makes knownthe difference e(t) =- Cr(t)-¢(t) between the input or desired position and the actual position: Since minimization of this error is the objective, it makessense to use e(t) as the input signal to the controller. This scheme is represented by the block diagram given in part (a) of Fig. 8.3. The choice of the controller is left to the design engineer. The class of linear controllers, the simplest and most common,is assumed here. The simplest linear controller is a constant coefficient, here given as k. With this proportional controller, shown in part (b) of the figure, the transfer function between the error e = ¢~ - ¢ and the output angle ¢ is kG~(S). This block diagram is generalized in part (c) of the figure as the universal single-loop system with an input or reference variable r(t) = Cr(t), output or controlled variable c(t) = ¢, an error signal e(t) = r(t) - c(t) and unity

588 feedback.

CHAPTER

8.

INTRODUCTION

TO AUTOMATIC

CONTROL

Since

c(t) =a(s)e(t) a(S)[r(t) - c(t)],

(8.4)

the universal transfer function relation between the input and output signals is c(t) = G(~!_.. r(t) 1 + G(S) For the particular

case at hand, substitution

(8.5) G(S) = kG9.(S) gives

1 ¢(t) = 1 (a/rk)S + (Ta/rk)S "~ Cr(t)

(8.6)

You should recognize this as a classical linear second-order inodel with two poles and no zeros. Most important, there is no steady-state error; when Cr is constant, the output angle ¢ is equal to it. The transient behavior is represented by the two poles. For small values of k these poles are real, so the response to a step change contains only exponential time-constant terms, and is smooth with no overshoot of the final value. For large values of k these poles are complex conjugate, and can be represented by a natural frequency and a damping ratio. The response to a step oscillates and exhibits overshoot. In general, one can say that too small a value of k renders the response too sluggish, and too high a value of k renders it too oscillatory. What is the best value of k? Creating a root locus plot helps answer this question in general for scalar unity~feedback loops. The locations of the poles of the closed-loop system G/(I+ G) in the complexS-plane are plotted as functions of the gain k. The fixed poles of the open-loop system G also are plotted. The root-locus plot for the present case is shown in Fig. 8.4. The two closed-loop poles emerge from the openloop locations as k grows from zero, at first approaching one another along the real axis. After they meet, further increases in k drive these poles apart into complex territory, symmetrically, along a vertical line that will be shown is the perpendicular bisector of the line segment connecting the two open-loop poles. Thus the envelope decay coefficient ~Wnremains constant, while the damping ratio decreases and the natural frequency increases. As k ~ ~x~, ff -+ 0 and Responses generally are dominated by the pole nearest the origin, which contributes the slowest transient term. Usually this means that you seek to keep the closed-loop poles as far from the origin as possible, particularly with regard to their real parts. For the example, therefore, the value of k ought to be chosen such that the real parts of the two closed-loop poles lie on the vertical part of the locus, which has the real part --1/2T. It is less clear how large the imaginary parts of the poles should be; different practitioners could disagree. The high natural frequency that follows from large imaginary parts is attractive, but the envelope function e -;~"t that bounds the decaying oscillation and largely establishes the settling time is a function only of the fixed real parts. Many engineers would choose a damping ratio of about 0.7 as a compromise between fast initial response, modest overshoot and rapid settling time. The settling

8.1.

OPEN AND CLOSED-LOOP CONTROL

589

Im S arrows show increasingk

l

j~od

" Re S

-l/z"

-l/z-

-l~2r- ~k

Re S

-j/2r t

(a) open-looppoles and loci for the closed-looppoles

10°I/.,

(b) suggestedclosed-looppoles (for k = a/2 rr)

optimal control (from Fig. 8.2)

~" 3°°Ioo ’ 0

2 4 6 8 t/r 16 (c) responseto particular step command

12

Figure 8.4: Root loci and behavior of position control system with feedback

590

CHAPTER 8.

INTRODUCTION

TO A UTOMATIC CONTROL

time is defined as the last moment at which the error of the step response is larger than some percentage, perhaps 2% or 5%, of the step size. The cosine of the angle between the real axis and a radial line from the origin of the S-plane equals the damping ratio of the pole pair. Therefore, the value ~ = 1/v~ = 0.707 makes the angle 8, shown in part (b) of Fig. 8.4, equal °. With this value, 02d = 1/27 and k = a/2~-r. The associated step response of this suboptimal strategy is plotted in part (c) of the figure. The response for the sample step is quite inferior to that of the optimal response. The most important contributing reason for the inferiority is that never more than one-quarter of the armature voltage allowed in the optimal control is in fact used. Since the system is linear, the control effort (and the armature voltage) is proportional to the size of the commandsignal. For example, were the commandsignal four times larger (116.9 ° instead of 29.2°), the initial armature voltage would equal the maximumallowed for the optimal control, and the response curve would simply be four times larger but no slower. The settling time for the optimal control, on the other hand, would increase markedly relative to its given plot, although the response would remain somewhat superior to that for the proportional control. Since the optimal control exerts maximum effort regardless of the size of the commandstep, it completes small excursions muchfaster than large excursions. The linear controller will exceed the allowable voltage for yet larger commandsignals; some kind of nonlinear limit must be imposed either naturally or by design to prevent damage. The summation employed in a feedback control system sometimes is carried out mechanically or with a pneumatic/mechanical system, but more often is carried out electrically. The same is true of the proportional controller action discussed above, and the other types of controller actions discussed below. The various actions can be carried out with continuous electrical signals using circuits based on operational amplifiers; see for example Ogata. 2 Some can be carried out with passive electrical circuits. Alternatively, the use of analog-to-digit~], and digital-to-analog converters allows the use of digital controllers. ~. EXAMPLE 8.1 A plant with the transfer function G(S) = 1/(S ~ + 2S - 3) is unstable by itself, since its homogeneoussolution or unforced response to non-zero initial conditions includes a term aet that grows exponentially. Place this plant in a unity-feedback loop with an adjustable proportional gain k, and find the transfer function of the overall system. Also, find the value of k above which this sytem is stable. Then, plot the root locus for the system as k varies from 0 to ~, using the example given in Fig. 8.4 as a guide. Finally, choose a value of k that gives responses that are relatively fast yet not excessively oscillatory, and sketch-plot the response of the system with this value to a unit step excitation. 2K. Ogata, ModernControlEngineering,Third Edition, Prentice Halt, 1997, especially p. 270.

8.1.

OPEN AND CLOSED-LOOPCONTROL

591

Solution." The closed-loop transfer function Gc(S) is, from equation (8.6), G(S) k Go(S) - 1 + G(S) - ~"- + 2S + (k Oneof the roots of this equation is negative for k < 3, but both are positive fbr all k > 3. The open-loop ploes are at the zeros of S2+2S-3 = (S + 3)(S - 1), namelyS = 5:1,-3: /mS X -3

’ -2

’ -1

’~Re S

0

This is the samepattern as the open-looppoles in Pig. 8.4, except they are shifted to the right, movingone of theminto the right-half plane. The root locus therefore has the samepattern: locations of poles proposed closed-loop I

-3

-2

Im S] 1

0

"~Re S

Whenk = 4, both roots equal -1, which is close to the best response but slower than can be achieved with a larger value. As k becomesvery large, its behavior becomescncreasingly oscillatory. A good compromisewouldbe k = 5, for whichthe roots are S + -1 + j. This gives Gc(S) = S" + 2S + whichdictates the sameresponse to a unit step as the examplein Fig. 8.4 if -1/2~- is interpreted as -1 s-1, that is "r = 0.5 seconds, and if the steadystate is replaced by 5/2:

0

0

2

,

’~ t, seconds

592

CHAPTER 8.

"~-~]i~=(v-Gp/TI)/R~ v-Gp/Tl~ R dp_G

INTRODUCTION TO AUTOMATIC CONTROL

I

p/TI

T=r

p--~-~M~ i Se (torque disturbance) 2+ a "_ a 1

G

(a) motor-load system

r/ra

11r

G(S) c(t) =1 +G(S)H(S)

(c) generalized non-unity feedback loop

Figure 8.5: Position Control System With Disturbances

8.1.4

Response to Disturbances

Unpredictable disturbances in the form of torques on the load may produce undesired errors. These disturbances are labeled as Md in the bond graph and the block diagram of Fig. 8.5. If these disturbances are considered to be the input to the system, rather than Cr, the system can be viewed as having a forward path transfer function or gain (from the summation point to the output ¢) G(S) G2(S), an d a fe edback ga in (f rom th e ou tput to the summation with the input) H(S) = k. Such non-unity feedback loops occur frequently, and are generalized in part (c) of the figure. Equation (8.5) for the simpler unity-feedback system is modified to become c(t)

= G(S)[r(t)

- H(S)c(t)],

(8.7)

which whensolved for r(t) gives the general result forward path gain = G(S) r(t) 1 + loop gain 1 + G(S)H(S) For the particular

]

(8.8)

system under consideration this gives

rR/ak ¢(t) = 1 ~" (a/rk)S + (Ta/rk)S

MD(t).

(8.9)

8.1.

OPEN AND CLOSED-LOOP

CONTROL

593

The poles of this transfer function, and therefore the dynamics of the behavior, are identical to those fbr the response to ¢~. On the other hand, the output ¢(t) of equation (8.9) can be considered to be an error, and increasing k decreases this error. If the error is critical to the success of the system, you could justify using a larger value of k than that chosen previously, despite the.reduction in damping.

8.1.5

Root

Locus

Basics

Control engineers tend to becomeexperts at sketching root locus plots, following several detailed rules that are given in virtually any text on automatic control. The discussion below, in contrast, gives only the basics. Instead, emphasis is placed on the use of the MATLAB command rlocus, which does the detailed work for you. The root locus is a plot of the values of S which satisfy knum(S) 1 + den(S~ =

(8.10)

for values of k ranging from zero to infinity. For the general single-input singleoutput feedback system with a proportional controller with gain k, comparison to the denominator of equation (8.8) gives k num(S) = G(S)H(S), den(S)

num(S)

= (S - Zl)(S

z2 )... (S

(8.11a)

- Z m)

den(S) = (S - p~)(S - P2) " " (S -

(S.llb)

(8.11c)

Other controllers are considered later. The locus as a whole therefore is defined by the phase condition /[num(S)/den(S)]

= (1 5= 2n)180 °. n = 1, 2,...

Points on the locus corresponding to a particular magnitude restriction nmn(S) I 1 den(S)I

(8.12)

value of k have the additional

(8.13)

594

CHAPTER 8.

INTRODUCTION TO AUTOMATIC CONTROL

EXAMPLE 8.2 Showthat both the horizontal and vertical segments of the root locus of the postion control exampleof Figs. 8.3 and 8.4 satisfy the defining relation of equation (8.12). Then, use equation (8.13) to find the values corresponding to the points $1 - 1/47 and $2 = (-1 + j)/27- on the locus. Solution The segment of the real axis between the two open-loop poles is part of the root locus, since num(S)= 1 and, as shownon the left below for any point thereon, den(S1) equals the product of the vector $1 - p.~ positive real number)and the vector $1 -pl (a negative real number), therefore satisfies equation (8.13). The perpendicular bisector of this line segmentalso is on the locus, since for any of those values of S., the sumof the angle of the vector $2 - p~ and the angle of the vector S~ - p~ equals either 180° °or. -180 Im S

i S2-~

lmS -1/2r

, "137o -1/r

$1

Re S

pole on real segmentof locus

-l/z-

-1/2r

Re S

pole on perpendicularbisector of real segmentof locus

The value of k in equations (8.10)-(8.13) equals the product of the magnitudes of the twovectors. For the particular point $1 = -!/47-, the resulting value of k is 3/167’~; for S.9 = (-1 + j)/2T it is 1/27 -2. This k incorporates the factor r/Ta included in G~ (S) and Gz(S); the respective values of the of equations (7.10) and beforeare 3a/16rT and a/2rT. Certain general rules are especially helpful in anticipating the migrations of the closed-loop roots as k progresses from zero to infinity. The locus for a closed-loop pole emerges from each open-loop pole as k grows from zero. One of the loci is terminated, as k ~ 0%at each finite open-loop zero. Let the numberm represent the number of opemlooppoles minus the number of openloop zeros; mis normally assumedto be zero or positive. The mloci which do not terminate at a finite zero terminate at infinity. Specifically, they become asymptotic to radial lines focused at a particular point on the real axis, and they are arrayed uniformly about the full 360° of arc and sy~nmetrically with respect to the real axis. If mis odd, one asymptotelies on the negative real axis, and if it is even, none lie on the real axis. For cases with ra > 2, the centroid of all closed-loop poles remains fixed, independentof k. The focus of the asymptotesis at the centroid of all open-loop poles and zeros, with zeros treated as negative poles.

8.1.

OPEN AND CLOSED-LOOP

595

CONTROL

-4

-2

RealAxis

m=O

o

2

RealAxis

dashed lines represent the asymptotes

/

/

’,, 0

RealAxi:

rn =3

m =4

2

4

RealAxis

Figure 8.6: Root loci for systems with various numbers of the pole-zero excess The position-control system has two poles and no finite zeros, so rn = 2. Examples of root loci with m = 0, 1, 3 and 4 are offered in Fig. 8.6. 8.1.6

Use

of

MATLAB

The command rlocus is included in the Student Version of MATLAB,and resides in the Controls Toolbox of the professional version. It has the following forms: rlocus(num,den) rlocus(num,den,k) rlocus(A,B,C,D) rlocus(A,B,C,D,k) [r,k] =rlocus (num,den) It,k] =rlocus (num,den,k) Jr,k]=rlocus (A,B,C,D) [r,k]=rlocus(A,B,C,D,k)

596

CHAPTER

8.

INTRODUCTION

TO AUTOMATIC

CONTROL

Wheninvoked, the vector k is user-defined; otherwise, a default vector is used. Whenthe left-hand arguments are not specified, the root loci are plotted directly; otherwise, the values of the complex closed-loop poles and the associated values of gain are stored in the vcctors r and k, and are displayed in response to typing r or k. The command plot(r,’’i plots the root loci. The alternative plot(r,’o’)

forms

or plot(r,’x’)

marks each pole with a small circle or cross. In regions where the pole locations change slowly with increasing k, these markings will be densely spaced; where the pole locations change rapidly, they will be sparsely spaced. EXAMPLE 8.3 Plot the root loci for the transfer function G(S) = (S 2)/(S 2 + 4) , us ing MATLAB. Make sure that the entire region of interest is included. Then~ mark the locations of the closed loop poles for several values of the gain, k. Solution:

The MATLABcommands >> num=[1 2]; >> den=[l 0 4]; >> rlocus(num,den)

produce the plot

-4

-3

-2

-1

0 Real Axis

1

2

3

8.1.

597

OPEN AND CLOSED-LOOP CONTROL

The axes happen not to include the entire region of interest. This problem can be remedied by the axis command, which allows the extent of the axes to be specified. Then, specific closed-loop poles for specific values of the gain k can be added to the plot through the use of the command r=rlocus (hum,den,k). The coding >> v=[-8 2 -4 4]; >> axis(v) >> hold Currentplot held >> grid >> rlocus(num,den) >> k=[2:2:12]; >> rlocus(num,den,k); >> plot(r,’o’) >> title(’Root-locus plot of k(S+2)/(S\^{}~+4)’)p

produces the result Root-locusplot of k(S+2)/S^2+4)

2 RealAxis

Perhaps the best behavior is that associated with the poles for k = 8, namel# r = -4 =i= 2j.

598

CHAPTER

8.

INTRODUCTION

TO AUTOMATIC

CONTROL

EXAMPLE 8.4 Repeat the above example for the case G(S) = S/(S 3 + 3S2 + S + 3). Solution: The codings on the left produce the plots on the right below.

>> num=[1 0]; >> den=J1 3 1 3]; >>rlocus (hum,den)

RealAxis

>> v=[-4 1 -4 4]; >> axis(v) >> hold

~...............

= :

:

>> rlocus(hum,den) -2 >> grid Current >> k=[l:6]; pl°t heldi s >>r=rlocus (hum,den, >>plot (r,’ o ’ 44 -3.5 -3 -2.5 >> r(4,:)

,

.

, ::

: ................

-

-2Re~,1

0.5

a!qs =

-i.0000

-1.0000+ 1.4142i

-i.0000~ 1.4142i

The pole-zero excessis rn. = 2, producing two lociapproaching inanity vertically withthe asymptote passing throughthecentroida] location S -].5.(Thezerois treated as a negative entityin the calculation of this centroid.) Theexplicit closed-loop locations pinpointed in thesecondplot suggest thatk = 4~ r = -i, -I 4- 1.414jwouldbe a goodchoice.

8.1.

OPEN AND CLOSED-LOOP CONTROL

599

EXAMPLE 8.5 Repeat the above example for the case G(S) = (S 0. 4)/(S 4 + 6S3 + 8S2). Solution." The codings on the left below produce the plots on the right.

>> num=[l.4] ; >> den=[i 6 8 0 0]; >>rlocus(num,den)

4

RealAxis

>> v=[-5 1 -3 3]; >> axis(v) ~ >> hold ~ Current plot held >> grid >> rlocus(num,den) -2 >> k=[2:2:12]; >> r=rlocus(num,den,k) >> plot(r,’o’) -3 -5 -4 >> r=rlocus(num,den,7) r = -4.5502 -0.3746 + 0.8590i -0.3746 - 0.8590i

-3

-2 RealAxis

-1

0

~ %

1

-0.7006

This example has a double pole at the origin, from which two root loci emerge vertically. The pole-zero excess is m = 3, so three loci extend to infinity at uniformly separated angles. The zero near the origin acts as an attractor to the complex-conjugate loci. The larger-scale drawing with explicit pole locations shownfor k = 2, 4, 6, 8, 10 and 12 suggests that the optimal value is about k = 7, giving r = -5.55, -0.701, -.375 + 0.859i.

600

CHAPTER

8.1.7

Criteria

for

8.

INTRODUCTION

TO AUTOMATIC

CONTROL

Stability

.~

Feedback control can make an unstable plant stable, and can make a stable plant unstable. A root locus clearly reveals whether or not all the poles of the closed-loop transfer function are in the left-half plane, as they must be for stability. Alternatively, with less effort, you can determine qualitative stability by examining the characteristic equation directly. Use of the root commandof MATLAB will determine the poles readily, as noted before, if you have numerical values for the coefficients of the characteristic equation. If on the other hand one or more of the coefficients is to be chosen, or if you don’t have a computer handy, use of the RouthoHurwitz algorithm will determine the number of poles in each half-plane. This algorithm is given in virtually any textbook that goes beyond an introduction to automatic control. Rather than reproduce the full algorithm here, certain special results are given. The characteristic equation for linear lumped models can be written in the form anS ’~ + an-~S n-1 + "" + alS + ao = O. (8.14) A necessary condition for stability is that all coefficients an,an-1,’",ao have the same sign. This condition suffices for the first order system, in which it is tantamout to requiring the time constant to be positive, and for the second order system, in which it either requires the two time constants be positive or the natural frequency be real and the damping ratio be positive. For higher-order systems, however, the condition is not sufficient for stability. Stability is assured for a third:order system if and only if an additional condition is satisfied: {a2a~ - a3ao >_ O. I

(8.15)

Stability is assured for a fourth-order system if and only if two additional conditions are satisfied: [ a3a2 - a4a~ >_ 0; a3a~.a~ - a4a~l - a]ao _> 0. ] EXAMPLE 8.6 A fourth-order system satisfies

the characteristic

(8.16)

equation

S4 + 5S3 + 10S2 + (k + 5)8 + k = in which k is to be chosen. Find the range of k for which the system is stable. Solution: The requirement that the fifth coefficient have the same sign as the others demands k ~ 0. This automatically satisfies the requirement that the fourth coefficient also be non-negative. The first of equations (8.16) gives the additional requirement 5 x 10 - 1 x (k + 5) ~ 0; combining these requirements, 0 < k < 45. The second of equations (8.16) gives the further requirement 5 x 10(k + 5) - l(k + 2 -52k ~ 0, which giv es k ~ 20.9 6. The complete result therefore is 0 < k < 29.63.

CLOSED-LOOP CONTROL

601

amary the key concept of automatic control. In a simple system, it is ~t someoutput variable match a reference signal, which either is a point or a dynamicallyvaried input. The difference, or error, between ~ed and the measured and "fed back" output variable is used as the j Jr actuation. This error becomesthe input to a controller. The output /~,~tl~e controller commandsthe actuator to move,generally in the direction of reduced error. Outside disturbances can buffet, a system, causing undesired changes in its output. The presence of a feedback loop acts to minimizethese changes. Control systems without feedback can be designed, but cannot effect these corrections (unless they directly sense the disturbances and act on this information.) The controller of such an open-loopcontrol system also wouldbe ineffective unless it is based on a precise modelof the system being controlled. By contrast, a feedbacksystemis comparativelyforgiving, since the error itself is independent of the modeluponwhichthe controller is designed. Thelocation of the poles of the transfer function of a completeor closed-loop linear systemis the most critical indicator of the quality of its behavior. The zeros influence the relative contributions of the various poles, but not their essential time constants, natural frequencies and dampingratios whichdependon the poles. The pole locations dependon the constants chosenfor the controller, as well as the parametersof the system being controlled. In deciding howlarge to makea particular controller constant, it is helpful to plot the loci of these poles as the constant is varied from zero to infinity. A few of the morebasic rules for constructing these root loci havebeen given, but principal reliance has been placed on use of the MATLAB commandrlocus. A few constraints on the coefficients of the characteristic equation also havebeen given. The assumptionof linearity leads to manyrelatively simple analytical methods. including those discussed above. The control engineer must keep his eye out for gross errors that this assumption can produce. A common error is due to the controller commanding the actuator to act faster than its capability. The example of a DCmotor acting through a pair of gears on an inertive load (such as a flywheel) has been most prominentlyused to illustrate the ideas above. Guided Problem 8.1 This first guided problemgives basic practice with proportional feedback control, using the MATLAB rlocu.~ as the major tool. Three fluid gravity tanks, each of area A = 0.1 ms, are interconnected by two flow restrictions, each with linear resistance R = 9.81 × 106 N s/m5, as pictured in Fig. 8.11. The object is to control the level in the third tank by controlling the flow Qin into or out from the first tank. The relation between this flow and the level of the third tank is readily deducedto be a~/AQin;)~. y =S(S+a,,S+3a, )

a= Pg = 0.01 s. R--~

602

CHAPTER 8.

/

ai,,

INTRODUCTION

tank

areas

TO AUTOMATIC

6

A ~

~k ’fl tube resistances R

fluid density p

Figure 8.7: Guided Problem 6.8 You are asked to investigate the possibility of simple proportional feedback control, based on measurement of the level in the third tank. The value of a reasonable gain, k, is desired. Suggested

Steps:

Draw a block diagram of the system, including the feedback. Label all variables, and pay particular attention to their signs. Determine the units of the gain k (which is not a pure number). Sketch-plot the three poles of the open-loop system on S-plane coordinates. Since there are three poles and no finite zeros, the pole-zero excess m = 3. Note what this says about the asymptotes of the closed-loop root loci as

Use MATLAB to plot the loci of the closed-loop poles. Note that too large a value of k produces instability, whereas too small a value produces an unnecessarily sluggish response. The response, y, to changes in the desired level, Yr, is faster when two of the poles are complex-conjugate, but the damping ratio should not be so small that excessive oscillations result. A good compromiseis ff = 0.5, which gives an angle between the real axis and a radial line drawn from the origin to one of the poles equal to cos-1(0.5) = °, as suming th e vertical and horizontal axes have the same scale. Enter an appropriate vector of gains k into the MATLAB commm~dr:~ocus, and estimate the desired value of k and the corresponding closed-loop transfer function. 5. Use the MATLAB command step to demonstate feedback system.

the step response

of the

PROBLEMS 8.1 A closed-loop system has the characteristic equation $3+ 3S2+ kS + 6 = O. Determine by the simplest method available the range of k for which the system is stable.

8.1.

OPEN AND CLOSED-LOOP

603

CONTROL

8.2 The angular velocity of the motor-load system considered in this section is to be controlled, rather than the angular position. Thus, the role previously taken by G~ (S) is replaced by G1 (S). The parameters are a = 0.006 V s, ~- = s and r = 0.2, and the maximummotor voltage is vm = 3.0 V. (a) Determine the optimal open-loop control signal which produces angular velocity of 2000 degrees per second, starting from rest, in as short a time as possible. (b) Draw a block diagram for linear proportional feedback control. the variables. (c) Draw the root locus for the closed-loop pole(s). say about the choice of the gain k?

Label

What does this

(d) Find the closed-loop transfer function for the motor voltage, v(t). Determine the maximumvalue of k for which this voltage does not exceed the stated limit for the commandsignal of part (a). (e) Find the closed-loop transfer function for the angular velocity. the step response for the value of k determined in part (e).

Plot

(f) Find the clbsed-loop error between the commandand response angular velocities for the case considered in parts (d) and (e). Do you expect error is acceptable, or should design changes be sought? If the largest commandinputs were smaller, could the value of k be acceptably changed to give a reduced error? 8.3 A proportional controller and unity feedback is applied to the system with transfer function G(S) = S/(S ~ + 1). (a) Determine whether the open-loop system is stable. (b) Plot the root loci for the controller infinity.

gain k increasing from zero

(c) Determine the range of k for which the closed-loop system is stable. (d) Choose a value of k which gives a good combination of speed of response and relative stability. Determine the corresponding closed-loop transfer function. (e) Determine from the transfer function whether the response of the system to a step input has a non-zero final value. If it does, report the steady-state error between that value and the unity input value, if any. (f) Plot the step response of the closed-loop system, using your value k.

604

CHAPTER 8.

INTRODUCTION

TO AUTOMATIC

CONTROL

8.4 Answer the questions of Problem 8.2 for G(S) = 1/S(S 1) ~. 8.5 Answer the questions of Problem 8.2 for G(S) = (S + 1)/S’2(S 4) 8.6 Answer the questions of Problem 8.2 for G(S) = 2 + 1) /(S + 4)(S 8.7 Answerthe questions of Problem 8.2 for G(S) = (S + 0.2)/S(S- 1)(S + "~.

SOLUTION TO GUIDED PROBLEM Guided

Problem

8.1

~] " I =-1S3 +0.04S2+0.0003S[ I

unitsin’~/s

~W.~

X -0.03

X lmS+ ~ Re S, s -0.01

Since m = 3, one asymptote is the negative real axisi and the other two have angles of +60° relative to the positive real axis. (Whenthe scales of the real and imaginary axes are not the same, as below, the slopes of the asymptotes remain at V~). The intersection of these three asymptotes is at the centroid of the three open loop poles, or -4a/3 = -0.0133 rad/s.

The MATLAB coding is hum = 0.001; den = [1 0.04 0.0003 0]; rlocus(hUm,den)

The fbllowingMATLAB commandsgives results k = [0:.25:3]*1e-3; [r,k] = rlocus(num,den,k); plot(r,’x’) r

for 13 values of k:

8.2.

605

DYNAMIC COMPENSATION

-3 Lines drawnat the slope :t:~/~ throughthe origin showthat the value k --- 1.5 x 10 m2/s gives virtually the desired ~ --- 0.5. The associated values of the closed-loop roots are rl = -0.0325 s -1, r~.3 = -0.00377 + 0.00642j s -1. Therefore, the closed-loop transfer function is

-6 1.8 x 10 G(S) 1 + G(S) S3 -6. + 0.04S: + 0.003S + 1.8 x 10 0.01 0.008

line

with slope equal to ~" --

0.006

4closed-loop pole for k= 1.8 x 10

0.004 0.002

xxx

x

x

x

-0.004 °0.006 !5 -0.008

.o%o

.o.o;

-0.025

-0.0~

-0.01~5

-0.01 -0.00~5

The MATLABcommands n = 1.8.1e-6; d = [I .04 .003 1.8.Ie-6]; step (n, d)

o.~

give the step response shown here:

8.2

Dynamic Compensation

The proportional controller employed above is but the simplest in a pantheon of possibilities. The most commonfamily of controllers in use today directly sums integral or derivative actions (or both) to the proportional control. Phase lead and phase lag compensators also are simple and popular. These examples of compensators are introduced in this section and applied to the motor-driven position control system of the previous section, along with other examples, to illustrate the richness of the possibilities for improving over proportional control.

606

CHAPTER 8.

INTRODUCTION TO AUTOMATIC CONTROL

8.2.1 Proportional-Plus-Integral

Control

The steady-state error in the output angle of the motor-drivenposition control system caused by a constant disturbance torque (such as producedby friction) has been investigated in Section 8.1.4. This error can be eliminated by integrating it over time, which can be accomplishedusing an operational amplifier. Should the error persist indefinitely, its integral wouldgrow towardinfinity. This integral is then multiplied by a constant and summedwith the proportional action of the controller to produce the control signal which drives the system toward zero error. The transfer function relation of the PI controller becomes u(t) = k(1 + 1/T~S)e(t), (8.17) in which T~ is knownas the integral time. A block diagram for the system is given in part (a) of Fig. 8.8. The overall transfer function from the moment disturbance to the angle of the output becomes (rR/a)G2(S) ~tld(t) ¢(t) = 1 + k(1 1/ TiS)G2(S) (rRT~/ka)S (8.18) = 1 + T~S + (aT~/rk)S~ + (TaT~/rk)S3 Md(t). The derivative operator S in the numerator operates directly on the disturbance momentMd(t). If the momentis constant, this produces a zero value ofthe steady-state response of ¢(t). Otherwise, the response is proportional to the numerator coefficient rRT~/ka; increasing k and decreasing Ti reduces its magnitu~le. Unfortunately, increasing k and decreasing T~ also reduces the relative stability of the system, and can even produceinstability. A compromise is necessary. Let the particular value of k chosen earlier (k = a/2r-r) be assumed.It is helpful to plot the locus of the poles of the closed loop systemas the inverse time 1/T~is increasedfrom zero towardinfinity. Sucha root locus can be plotted if the characteristic equation (the denominatorof the closed-loop transfer function) is placed in the form of equation (8.10), as always; the k of that equation is nowr/Ti. The numerator coefficient 7 makes this ratio dimensionless; the product ~-S also becomesa dimensionless derivative operator. The characteristic equation can indeed be placed in this form by dividing the denominatorof the transfer function in equation (8.18) by the operator S[1 (a/rk)S + (Ta/rk)S’~]: 1+

~ /T~

~ /T~

=1 + = 0. (8.19). S[1 + 2] (a/rk)S + (Ta/rk)S 2(TS)3 + 2(TS)-~ + ~-S The "open-loop"poles of this equation include one pole at the origin plus precisely the same ,two complex-conjugate poles of the closed-loop system with proportional control only. They are plotted in part (b) of Fig. 8.8. The resulting root locus showsthe pole at the origin migrating to the left, which is desirable. It also shows the complexpair of poles movingtoward the right, which reduces the damping. The sum of ,the three closed-loop poles remains

8.2.

607

DYNAMIC COMPENSATION

(a) block diagram with PI controller

~

k(1

+

1

)

[~

v(t)

~_+r/a

0.8 0.6 0.4

(b) root locus plot

0.2

-0.4 -0.6 -0.8 -1 -1

-0.8

-0.6

-0.4

-0.2

.

0.4

0.6

0.8

RealAxis

0.8 0.6

(c) closed-loop poles for r/T~ = 0.1, 0.2, 0.4, 0.7, 1.0

0.4 0.2

best solution: r/T~=0.2’ r = -0.3114+0.4100 i = -0.3773

E -0.2 -0.4 -0.6 -0,8

RealAxis

Figure 8.8: Motor-driven position system with proportional-plus-integral

control

608

CHAPTER 8.

INTRODUCTION TO AUTOMATIC CONTROL

(a) responseto step change of momentMa

1

2Rrer Ma / a 0 0

5’

00

5

10 ’

1 ’5 t/r

20

(b) responseto step change of command angle ¢c

10

15 20 t/r Figure 8.9: Responsesof position system with PI control constant (rememberthe rule for cases in whichthe pole-zero excess is 2 or more), so the complexpoles cross the imaginary axis and the system becomesunstable whenthe real pole hasmovedto the location 7-S = -1. The closed-loop poles are replotted in part (c) of the figure for five specific values of the gain TITs: 0.1, 0.2, 0.4, 0.7 and 1.0. The MATLAB coding which produces this plot and reports the values of the respective roots is as follows: num= 1; den = [2 2 1 0]; k = [.1 .2 .4 .7 1]; [r,k] = r:tocu.~(n,d,k) rlocus (n,d,k) Thus, r/Ti = 1.0 (or Ti = 7-) is the case of incipient instability. Thebest overall behaviorresults with 7-/Ti ~- 0.2 (or Ti -~ 57-). The associated responsesof the system to a step disturbance of momentand to a step commandare plotted in Fig. 8.9. Theseresponses are probably quite satisfactory, althoughthey are distinctly inferior to optimal responses, largely because they do not utilize muchof the available torque of the motor. Choosing a larger value of k would increase the speed of response, but at a probably unacceptablecost of decreased relative stability and potential for insensitivity to momentdisturbances. A muchsmaller value of k wouldmakethe response more sluggish. If somemeansof increasing the dampingof the system were introduced, on the other-hand, the values of k and 1/Ti could be increased, giving significantly faster responses without sacrificingstability.

8.2.

DYNAMICCOMPENSATION

609

EXAMPLE 8.7 A first-order thermal plant satisfies the differential equation dO 7"-~

1¯

+ o = -ffQi,,+ oe;

~-= mc --h-’

where 0 is the temperature of the plant, 0e is the temperature of the environment, ~i,, is the heat input used to control 0, H is the heat-transfer coefficient between the plant and the environment, mis the mass of the plant and c is its specific heat. The plant is placed in a feedbackloop that includes a measurementsystem with a simple lag and a proportional-plusintegral controller, as shownbelow:

controller

plant

mea..surementsystem

l

Considerthe case in whichthe measurementlag time "Fmequals v. Find the transfer functions betweenthe reference input 0r and 0, and between0e and 0. For the case of simple proportional control, i.e. T -~ ~, find the smallest possible steady-state errors without the dampingratio of the system being less than 0.5, and the associated value of k. Thenre-introduce the integral control, showthat it eliminates the steady-state errors, and find acceptable values of k and T. Solution-" The transfer functions can be found using the basic equations (8.5) and (8.8): 0 Or

k(TS + 1)/(7S 1) TS (1 + k(TS + 1)/(rS 1)TS(TmS +1)+ k(TS + 1) k(TS + 1)(rS + (TS + 1)TS(T,~S 1)+ k(TS + 1)’ 0 1/(’rS + 1) O~ 1 + k(TS + 1)/(TS + 1)TS(TmS 1) TS(rS + 1) (TS + 1)TS(~-mS 1)+ k(TS + 1)

WhenT ~ ~ and S = 0, the steady-state responses for simple proportional control result:

610

CHAPTER

8.

steady-state

=

INTRODUCTION

TO AUTOMATIC

1 + k =1 1 +k ;

steady-state

CONTROL

1 +k"

(These result by dividing the numerators and denominators by T before substituting T = 0.) Both error ratios are the same, and are reduced by making k as large as possible. With Trn : T, the characteristic equation becomes T2~q2q-27S+(l+k) = 0, which describes a second-order system with a natural~ frequency w, = v~ + k/7 and damping ratio of ~ = 1/x/~ + k. For ~.>_ 0.5 as required, k <_ 3. For k = 3, the steady-state error ratios are 0.25, which likely isn’t very satisfactory, and wn = 2/~-. With the integral control reintroduced, both steady-state errors become zero regardless of the values of k and T. The characteristic equation is third order; it becomes (TS) 3 ~- 2(7S) 2 -~ (1 + k)(rS) + kr/T = This can be factored as (TS + Tp)[TS):2 q- 2~TWn(TS)q- (Veda)2] ---- 0. A large value of Wnrequires a small value of p; to maximize the distance of all poles from the origin, set p = wn. Keeping ~ = 0.5, this gives (~-S) 3 + 27con(7S)2 + 2(TCOn)2(TS) (T~ dn) 3 = 0. Com paring ter m by ter m, TO. ) n : 1, k =1 and T = T. Thus, the elimination of the steady-state error without reducing the effective damping comes at a cost of reducing the natural frequency by a factor of two. 8.2.2

Proportional-Plus-Derivative

Control

The addition of derivative action to the controller is one of the oldest, simplest and most effective means of increasing the damping of a feedback control system. This is considered first in the absence of integral action. The transfer function relation of the PD controller is u(t) k(1 + T~S) e( t),

(S.20)

in which k is the coefficient for the proportional action of the controller, as before, and Td is the derivative time. The term kToS e(t) =- kTd~(t) is proportional to the instantaneous rate of change of e(t). (When implemented by an analog integrator network, the transfer function of the derivative term is actually TdS/(TdS+1), where Vd is set so small that its effect can be neglected.) The transfer function equals zero for S = -1/Ta; this zero is represented in the S-plane of the overall forward-path transfer function G(S) by a symbol o drawn at this location on the negative real axis. Application of the PD controller to the motor-driven position control system gives the forward-path transfer function kr(1 + TdS)/a G(S) S( 1 + ~- S) ’

(8.21)

and the closed-loop transfer function relation ¢(t)

G(S) 1 + TdS - 1 G(S) 1 + (Td + a/ rk)S + (a r/rk)S 2 ¢r (t).

(8.22)

8.2.

DYNAMIC COMPENSATION

611

The introduction of the derivative control action has left the order of the system unchanged at two. The natural frequency remains at V/-ff/a~ -. A positive value of Td is seen to increase the damping ratio by increasing the middle coefficient of the denominator quadratic. This equation shows that since the designer can freely choose the values of both k and Td, he has carte blanche to achieve any natural frequency and damping ratio he or she wishes. Design by pole placement is frequently practiced by control engineers. So why not achieve an almost instantaneous response by setting k at an enormous value? Think for a moment. Whyshould this behavior be better so-called optimal control presented in Section 8.1.2?

than that of the

The answer is that a very large value of k would produce an impractically large value of the armature voltage for any but the smallest commandstep. The optimal system recognized a practical limit to this voltage, presumably imposed by the regulated voltage of the electrical power supply. 3 The optimal system in fact responds exceedingly fast for very small commandsteps. What if the output voltage of the linear PD controller were similarly limited? Could an extremely large value of k then indeed be chosen, along with an appropriate time Td? This scheme, shown in Fig. 8.10, features a nonlinear saturation element. Responses are plotted for the same step as before (29.2°), a smaller step (10°) and a much larger step (180°). These were found by nonlinear simulation. The value of k used (1000) is so large that either the positive or the negative saturation limit is imposed almost all the time, giving almost the so-called bang-bang or switching control characteristic of the optimal scheme. The derivative time Td is chosen at Td = 0.226T, which gives barely zero overshoot for the 29.2° step. The response to this step is virtually indistinguishable from that of the optimal control as plotted in part (d) of Fig. 8.2. Larger steps produhe a modest transient overshoot of the final position, whereas smaller steps produce a tail in which the response creeps up to its final value more slowly than an optimal control. Raising Td to T log(2) = 0.693~" (or higher) prevents overshoot for all responses, but also increases the overdampedtails for small steps. The responses are not critically dependent on the value Of Td, however. Note that time required to approach the final angle increases markedly as the step size increases. This nonlinear behavior has the distinct advantage over linear behavior of taking advantage of the full motor torque, no matter how small the commandstep.

3Damage to the motor alternatively could be prevented by use of a powersupply and amplifier that limits the armaturecurrent rather than its voltage. Theresult wouldbe somewhat better performance,at a higher cost.

612

CHAPTER 8.

INTRODUCTION TO AUTOMATIC CONTROL

D controller saturation element (a) block diagram

180 150 ¢, 120 degrees 90 60 3O 0

0

J ~ 2,9.: (same as optimal control, Fig. 8.1) 1

2

3

4 t/r

5

6

(b) responsesto step commands of varioussizes Figure 8.10: Position control system with a PDcontroller and saturation

8.2.

613

DYNAMIC COMPENSATION EXAMPLE 8.8

A controllable force, F, acts on a mass m = 1 kg of a simple mass-spring system with spring rate K = 4 N/mso as to control its position, x: (mS2 + K)x = F. The postion x is monitored by a transducer, the ouput signal of which is compared to a commandinput, x~, to produce an error signal. This signal acts on a controller that produces the ibrce. Proportional control gives imaginary roots, a virtually uncontrollable condition. A PD controller is proposed: S) k(1 + rd

~

Select values of T and k, and discuss the resulting behavior. Solution’~ The transfer function of the forward path is x_..= k(TS + 1) _ kT(S + l/T) ’ F $2+4 $2+4 and the closed-loop transfer function is x kT(S + l/T) xr S "2 + kTS + (k + 4)’ which has a steady-state value of k/(k + 4). Thus, a large value of k is desired. A root lous for the case of T = 0.5 seconds is given in Example 8.3 (pp 596-597). The value of kT suggested there is 8, or k = 16, giving a steadystate error of 20%.The dynamic behavior is attractive: the transfer function is 8(S + 2)/($2 + 8S + 20) which has the fast and well-dampedroots -4 :t: s -1. This solution would be satisfactory if compensation can be provided for the steady-state error. Otherwise, the steady-state error can be reduced and the response made faster by increasing k. The value of T is reduced to keep the response from becoming over-damped. For example, k -- 96 and T = 1/8 second produces a 4% steady-state error with a natural frequency of 10 rad/s. There is no limit to the apparent improvement available in the response. There must be a catch, and it is this: the larger k, the more powerful and expensive the controller must be. Further, as with the motordriven position control system, a rapid change in xr will saturate any real controller; the linear model then gives a false prediction of the response. The solution with T = 0.5 and k = 16 may be acceptable, however.

614

CHAPTER

8.2.3

8.

INTRODUCTION

Proportional-Plus-Integral-Plus-Derivative

TO AUTOMATIC

CONTROL Control

The elimination of the steady-state error produced by integral action is unaffected by the presence of derivative action. The combination of proportional, integral and derivative control actions is called PID control. Adjustable PID controllers are available commercially, and represent by far the single largest type of dynamic compensation in use today. They are particularly prevalent in the field of process control, which largely refers to chemical processes, where they are called process controllers or three-mode controllers. The presence of the derivative action directly imparts an impulse to the output of the controller whenever the commandsignal is given a step change. This impulse inevitably saturates the plant being controlled, resulting in nonlinear action. This saturation was utilized in the previous section on the PD controller to give nearly optimal responses to input steps. Nevertheless, most users of PID control do not consider nonlinearities. Applications usually are restricted to systems for which the simple superposable behavior of linearity is favored over the quickness of response of the near-optimal nonlinearity. In practice, steps of enough amplitude to cause marked saturation ~ are rarely encountered anyway; the consideration of step inputs, becomes more of a conceptual tool than a physical reality. Certain rules-of-thumb are given in textbooks on control for achieving satisfactory PID control of processess, based on approximate observations of the behavior of the plant being controlled. Part of the popularity of these controllers stems from the simplicity of their application; it is not even necessary to have a detailed model of the plant. Designers of most control systems prefer linear behavior to the fastest behavior possible, which typically is nonlinear. One of the reasons is that the life of hardware would be reduced if it always exerted maximumeffort, no matter how small the commandsor the disturbances. Saturation is normally reserved only for the relatively infrequent occurances~ of the largest disturbances anticipated; the power of the system is sized to accomodate these disturbances. Other reasons for favoring linearity include the simplicity of analysis, and the consistency of the behavior in terms of the classical measures of natural frequencies and damping ratios. Non-PIDcontrollers are often designed to avoid saturation and thereby validate linear behavior. Thus, simple derivative actions are avoided. The motor control system under consideration is in fact not a good candidate for PID control, unless saturation and the resulting nonlinearity is directly recognized and utilized. A different kind of controller is introduced below, instead. EXAMPLE 8.9 The steady-state error of the system addressed in Example 8.8 is eliminated by adding an intergral action to the controller, which already has a proportional gain k = 16 and derivative time Td = 0.5 second. Find the integral time Ti that produces marginal stability, and show that a good response results from a value three times larger.

8.2.

DYNAMIC COMPENSATION

615

Solution: The controller has the transfer junction k(1 + TdS + 1/TiS), and the plant has a transfer function 1/(S 2 ÷ 1), giving for the values indicated above, x 8T~S2 + T~S + 16 3 xc S + 8TiS 2 + (4 + 16Ti)S + 16" Equation (8.15) gives the condition for marginal stability: 8Ti(4 + 16Ti) 16, or Ti = 0.25 second. For three times that, or Ti = 0.75 second, the roots of the characteristic equation are S = -2,-2 4- 2j, which is reasonably fast without significant oscillation (~ = 0.707). Increasing Ti to 1.0 second gives S = -4,-2,-2 (double pole), which is slower and somewhat inferior. Increasing Ti further continues to slow the response. The proposed solution may be quite acceptable despite the saturation caused by the derivative action when the commandedposition suffers an abrupt change.

8.2.4 Phase Lead Controllers A phase lead controller possesses a single zero and a single pole, for a total of three parameters. Its transfer function is

plus a gain,

S+z a > 1. (8.23) = k--; S+az The pole and zero are plotted in part (a) of Fig. 8.11. The requirement a > means that the phase angle of the transfer function of the controller for sinusoidal excitation is positive, that is a phase lead, giving the controller its name. The response of a phase lead controller to a step input is rather like that of a highly muted PD controller. Rather than an impulse followed by a constant value, a constant value is preceded by a non-infinite but larger value of non-zero duration, blended by an exponential decay. It is no surprise, then, that the consequences of the two controllers are similar. The phase lead control can be achieved by the simple passive RC circuit shown in Fig. 8.12, explaining part of its historical popularity. The gain k must be added through use of an amplifier. A phase lead controller can improve the performance of the motor-driven position control system. The transfer function for the DCmotor plant, taken from equation (8.1), r/aT (8.24) G2(S) - S(S l/ r) Gc

Its pair of poles is plotted in part (b) of Fig. 8.11. Whena phase lead controller acts directly on this plant, the overall open-loop transfer function becomes

a(s)(~,./,~-)(s + (s + ~z)(S1/1-)

(8.25)

The simplest and most desirable application sets a = 1/1-, so that the zero of the compensator cancels the non-zero pole of the motor. The result is a pair of poles, one at the origin and the other at S = -az. The example with a = 2 is

616

CHAPTER 8.

INTRODUCTION TO AUTOMATIC CONTROL

(a) pole-zeroplot of phaselead controller

(b) pole-zeroplot for DCmotor (G2(S)) and its productwiththe proportional controller (kG2(S)) (c) pole-zeroplot for product of phaselead controller and DCmotor, with X z= 1/rand tr=2 -2/r

(d) root loci and choicesfor closed-looppoles

T Re S

-l/z-

pole and zero cancel hn SI Re S

-1/r

j/r}

ImS j/2v ~

proportionalcontrol (dashedlines): k=a/2rr; tad=l/2r; (’=0.707 phaselead control(solid lines): k=2a/rr; tad= l/r, (=0.707

!

-1/r

-1/2r

Re S

:j/2r ¢ !

-J/r I (e) response particular step command

30°I - --20° l ~/¢~~se lead control 10o~ ~ --proportional control " (from Fig. 8.4) ~]/ Xoptimal control (from Fig. 8.2) 0 2 4 68 t/r 10 12

Figure 8.11: Phase lead compensation applied to the DCmotor system

8.2.

617

DYNAMIC COMPENSATION

ein

R2

e°ut

Figure 8.12: RCcircuit

S+ a7

q-R a’=-R~ 2 R2

for phase lead compensator

shown in part (c) of the figure. Note that it is acceptable to ignore any slight error in this cancellation, since the pole and zero lie in the left-half plane; should they have been in the right-half plane such an error would directly represent an instability. The root loci for the closed-loop poles are shown in part (d) of the figure. The loci for the original system with proportional control only are shown by dashed lines, and the loci for the system with the lead compensator is shown by solid lines, again for the special case of z = lit and a = 2. In both cases the closed-loop poles are chosen to give damping ratios of 1/v~. The net effect of the phase lead compensator can be seen to double the speed of the system, as represented by the distance of the poles from the origin (the natural frequency). The resulting responses for the 29.2 ° step considered earlier are compared in part (e) of the figure. The fact that the response is somewhatmore sluggish than the optimal indicates that saturation is not a problem. Were the value of aa made much larger, however, or were the step made much bigger, saturation would in fact occur, and the linear prediction would be in error. The larger the value of a, the more closely the phase lead controller k(S z)/(S + az) can resemble the PD controller kd(1 + Td$). The resemblance is otherwise could be said to be closest when the responses of the two controllers to a unit ramp disturbance approach each other as t -+ ~. This happens when k = kd~ and z = (a - 1)/aTd. One result is equal steady-state errors. EXAMPLE 8.10 Replace the PD controller employed in Example 8.8 (kd = 16, Td = 0.5 s) with the phase lead controller giving the "closest" equivalence, as described above, and compare the two step responses. Let a = 10. Solution: The "closest" equivalence gives k = kda = 160 and z = (c~ 1)/C~Td = 1.8 s -1. The two transfer functions become PDcontroller :

X

-

Xr

phase- lead controller :

X

--

Xr

=

8(S+2) S~ + 8S + 20’

k(S+z) mS3 + azmS2 + (K + k)S + z(aK + 160(S + 1.S) S3 + 18S~ + 164S + 360"

618

CHAPTER 8.

INTRODUCTION TO AUTOMATIC CONTROL

The two step responses are 1.2 ¢/¢r

I ~2~,/PD

control

0.4 t//lead control ~] 0

I

0

I

1

I

t, seconds

Unless you already have designed a PDcontroller, however,it is better to design a phase lead controller directly, for exampleby using a root locus. The rough equivalence above deteriorates when a is small, mandating a separate analysis.

8.2.5 Phase Lag Controllers A phase lag controller resemblesa phase lead controller, but the pole is closer to the origin rather than farther away: +/3p. Gc = kSS+p ’

/3 > 1.

(8.26)

Phaselag controllers are used to reduce steady-state errors; they are to integral controllers what phase lead controllers are to PDcontrollers. They.canalso be fabricated with a simple RCcircuit, apart from the amplifier neededto achieve the gain k, as shownin Fig. 8.13. A phase lag controller can be used with the motor-driven position control system, substituting for the PDcontroller used in Section 8.2.1 (Fig. 8.8, p. 607). This is carried out in Fig. 8.14. The overall transfer function from the momentdisturbance kid to the angle of output ¢ becomes ~ = Rr2z/ra2 . Md S3 + (p + 1/r)S 2 + (pit + rk/Ta)S + rkpfl/Ta

(8.27)

Substitution of S ~ 0 gives the steady-state response, or error due to the disturbance, ofRr/ak~. This is the same as with proportional control alone, as originally determined,reducedby the factor of ft. Phase lag controllers generally do not eliminate steady-state error, unlike PI controllers.

l

ei"

R1 R C.-T:e°"GC=~ S+p

1 S+ZpP=-~c fl= RI+R,

Figure 8.13: RCcircuit for lag compensator

8.2.

619

DYNAMIC COMPENSATION

(a) block diagram with phase lag controller Cr(t~

k

S+~p

v(t)

~_+

rr/a ¢(t)

-I S+p l I

(b) root locus plot 0.8 values of rS for varying values of rp 0.8

/

0.4

0.2 --0.2

rp=O01.

~ 003. 0.02

rp=O.03

0.02

0.01

-0., -0.8 . poles for rp = 0.02..~~ -0.8 rS~ =-0.3504 rS23 =-0.3348 +0.4163j -1

-0.5

~

RealAxis

0.5

(c) responseto unit step change of moment M d

10

(d) response to unit step change of commandangle

~PI

t/r

15

20

controller -."N lag controller

[~~oogn:~ornCt r o 11 e r lb

t/r

1’5

20

Figure 8.14: Motor-driven position control system with phase lag controller

620

CHAPTER

8.

INTRODUCTION

TO AUTOMATIC

CONTROL

To be consistent with the analysis for the PI control, the value k = a/2r~- is assumed. To make the lag control effective without an unreasonably large value of ~, the value ~ = 10 will also be assumed, reducing the error to one-tenth of its original value. The value of p remains to be determined. In order to plot a root-locus for increasing values of the dimensionless ratio ~-p, the characteristic equation is solved for Tp and then placed in the usual form of equation (8.10): 1 + Tpnum(S)/den(S)) = 0. This gives num(S) (TS) ~" + (~’S) + ~/2 3 den(S) (TS) + (rS) 2 +(1/2)(TS)

(8.2s)

The critical part of the resulting root locus plot near the origin is shown in part (b) of Fig. 8.14. One pole migrates to the left along the real axis for increasing values of Tp, while two other complex poles approach and eventually enter the right-half plane. The value ~-p = 0.02 is close to the best compromise. Use of this value gives the response of ¢ to a step change of the disturbance 2FZd plotted in part (c) of the figure. The corresponding responses of the system with the proportional-plus-integral controller (from Fig. 8.9) and of the system with the original proportional controller are plotted for comparison. The PI controller has the advantage of reducing the steady-state response to zero, while the lag controller has the advantage of a smaller maximumexcursion. The system without dynamic compensation behaves relatively poorly. The final set of plots in part (d) of the figure shows the comparative responses of the three systems to a step change in the command angle, Cr. Neither system with dynamic compensation behaves quite as well as the original system without dynamic compensation. This is because the dynamic compensators were designed explicitly to address the problem of sensitivity to disturbances in Md. This is unlike most design situations, where dynamic compensation is employed to improve the response to the commandsignal. 8.2.6

Phase

Lead-Lag

Controllers

Phase lead andlag actions often are used together, under the name of lead-lag compensation; the combination is akin to the PID control: kl+a7~S 1 + ~-2S

(8.29)

The simple RCcircuit shown in Fig. 8.15, plus an amplifier, serves as a simple implementation for the case when fl = a. Note that the impedance of any one element can be set independently, and the values of the other elements are determined by the desired values of the time constants ~-~ and T2 and a. The impedances cover an excessive range if a gets too large; the value 10 sometimes is cited as a practical limit if passive analog components are used. Digital and operational-amplifier implementations avoid this limitation. In practice, the phase lag portion of the controller is used to reduce steady-state error; the value of T2 normally is large, so the frequency range at which it acts is lower

8.2.

621

DYNAMIC COMPENSATION Gc= l+~zrtS l+r2S l+rlS 1 + ~r~-2S

¯ out

G-T"

R2C2 =

~’2

R~G= (u-1)(r2-

Figure 8.15: RCcircuit for lag-lead controller

reference, digital L_.J digital-to-analog ~ Imicroprocessorl q converter V-I v’~’" I

I [ anal°g-t°-digital ~_~-~-n~’~’-L~ I converter I I I

l

signal

~

outotit

input~

~

F sample-and-hold output

time Figure 8.16: Digital control system with sampleddata signals than the principal frequencies associated with the plant dynamics. The value of rl, on the other hand, is chosento be considerably larger, so the phase lead action provides dampingat frequencies whereit is needed, permitting a larger gain and increased bandwidth. 8.2.7

Digital

Control

Systems

A majority of control systems today employdigital processors that are coupled to the actuators and systems being controlled by analog-to-digital and digitalto-analog comverters,as suggested in the block diagramof Fig. 8.16. This allows greater flexibility in the choice of control algorithms, and avoids the need for analog circuits with their impedancelimitations, amongstother advantages. Ananalog-to-digital converter producesa train of numbersat discrete intervals of time. These sampled data signals normally are treated as though the signal were step-wise in time, as illustrated by the plot given in the figure. The signal sent by the digital-to-analog converter to the plant is also in the sample

622

CHAPTER 8.

INTRODUCTION TO AUTOMATIC CONTROL

and hold format. If the sample time is considerably shorter than any of the significant time constants of the plant, the effective time delays producedin the samplingprocess are insignificant, and the analysis results of this chapter can be applied directly. This is often the case today, since processor speeds have been increasing for manyyears. Whenthe sampling interval is not greatly shorter than the smallest significant time constant of the plant, the time delays, imposephase lags (nonminimum phase-lag processes) that normally reduce the relative stability of systemor cause outright instability. Thus, at the least you should allow an extra marginof stability if you are employingthe analog methodsof this chapter. Better yet, you can roughlypredict the effects of sample-and-holdusing the frequency response methodsof Section 8.3 below. Best of all, you can employan analysis based on the z-plane, wherez = est with s being the Laplace operator. The formal study of this methodologyincluding the design of digital compensators involves the z-transform, a modified form of the Laplace transform. This subject appears in manytextbooks on control. 8.2.8

Summary

Manytypes of dynamiccompensators or controllers are used in feedback systems. Introducing an additional integral action to a proportional controller usually eliminates any steady-state error that results from an unwanteddisturbance. Large values of the gain and the coefficient of the integral action desirably reduce the steady-state error. They also generally decrease the relative stability of the system, with the consequenceof undesirable resonancesand overshoot and, for large enoughvalues, outright instability. The introduction of an additional derivative action to the controller is one wayto increase the dampingof the system. Whether a control system is linear or nonlinear is the most fundamental distinction. Since the powerof a systemis limited, the responses to small commandsor disturbances potentially can be faster than the responses to larger commands or disturbances. A system that implementsthis potential is nonlinear. Theuse of derivative control also invokesnonlinearities in the responseto a step command. Nevertheless, linearity is usually preferred by control engineers, to save on wear and to give simple, consistent behavior. A phaselead controller acts rather like the proportional-plus-derivative(PD) controller, but has less tendency to saturate. The phase lead controller has a gain, a zero, and a pole of larger value than the zero. The controller typically is used to replace a real pole with a real pole of a larger value, increasing the speed of response of the system without decreasing relative stability. A phase lag controller serves to reduce steady-state error, acting partly like a proportional-plus-integral (PI) controller, whichusually eliminates steady-state error. A phase lead-lag controller combinesboth actions, somewhat-like a PID controller. All these controllers can be implemented.withdigital, operational amplifier : or mechanical/pneumaticcircuits. The phase lead, lag and lead-lag functions

8.2.

DYNAMIC COMPENSATION

Figure 8.17: Inverted

623

pendulum for Guided Problem 8.2

also can be implemented by passive RC circuits. Guided

Problem

8.2

The first part of the final guided problem relates to the control system that keeps a rocket vertical at launch despite the applied force being applied at the lower end. Without control the rocket would fall over. The second part of the problem relates to the position control of the rocket. In both cases proportionalplus-derivative control is central to success. It is not necessary to carry out the second part to benefit from the first. Both parts give practice with root loci as implemented by MATLAB. The inverted pendulum system shown in Fig. 8.17 comprises two equal masses separated by a rigid massless rod, a constraint such that the lower mass moves horizontally only, and a means of applying a control force to that mass. The parameters are mg = 1 lb, g = 32.2 ft/s 2 and L = 0.644 ft. (a) The relation

between the force F and the angle ¢

assuming linearity (small angles). Proportional feedback control cannot stabilize this system, but PD control can. You are asked to find appropriate values of the gain k and derivative time Td, and the associated poles of the closed-loop system. (b) The relation

between ¢ and x, again assuming small angles,

Unless an appropriate addional control is introduced, the position x can wander widely in response to disturbances. Place the system resulting

624

CHAPTER

8.

INTRODUCTION

TO AUTOMATIC

CONTROL

from part (a) inside a second feedback loop. Again, apply PD compensation (proportional control again won’t work) so that x(t) stably follows a commandinput, xT(t). Part (b) of this problem is revisited chapter. Suggested

in Guided Problem 8.3 at the end of the

Steps:

Find the poles of the transfer function between -F and ¢, and sketchplot them in the S-plane. Sketch the simple root locus that would result from the use of proportional feedback, showing that it fails to produce the desired stability for any value of the gain k. Draw a block diagram for a system with unity feedback. Include a reference signal Cr, even though your interest may be only for a zero value. Also, include a block for the compensator, and label all variables. Does the compensator relate the error to F or to -F? A PD controller has the transfer function k(1 + TdS), which gives a single zero in the left-half plane. It is tempting to superimpose this zero on one of the poles of the system, canceling it to leave a very simple system. This is acceptable, but somewhat misleading since perfect cancelation never really happens; further, better results follow from the choice of a zero somewhat to the left of this pole. Place the zero 50% further out on the real axis, and plot the resulting root loci as the gain increases from zero to infinity. Use MATLAB. Choose points on the root loci Which are close to being the best. Use MATLAB also to specify these points and the associated gain precisely. You now have completed part (a). Extend your block diagram to give the position x, and place the result inside a second loop with unity feedback, a reference signal xr(t) and second controller. 6°

Use MATLAB to give the root loci for this complete system, using proportional feedback. You should discover that no value of the second feedback gain gives stable behavior. Replace the proportional controller with a PD controller. Its zero will have to be rather close to the origin (a large derivative time constant on the order of one-quarter to one-half second) to be effective. Repeat the root locus procedure, and note that a range of satifactory gain values exists. Choose points on the loci and find the associated gain. Determine the roots of the characteristic equation for the overall system.

8.2.

DYNAMIC COMPENSATION

625

8. Write the resulting tranfer function between xr(t) and x(t), and use the step commmandof MATLAB to plot the response to a unit step change. Note that x starts by moving away from its final value, rather than toward it, unlike most of the systems you have investigated. Does this make physical sense? PROBLEMS 8.8 Show that PI or PID control acting on any plant G(S) with unity feedback produces zero steady-state error in response to an input and to a disturbance acting directly on the plant, as long as G(0) ~ 8.9 A motor-load system of the type featured in this chapter is reconfigured to control angular velocity, rather than angular position. A feedback loop with proportional-plus-integral control is proposed in order to reduce or eliminate the steady-state error that proportional control alone gives (as found in Problem 8.2, p. 603). Let a = 0.006 V.s, r = 1.0 s and r = 0.2~ and restrict the motor voltage to 3.0 V for commandedstep changes from zero to .2000 degrees/s (as in Problem 8.2). The proportional gain k and the integral time Ti are to be chosen to giv.e good performance. (a) Draw a block diagram of the system. Include the motor voltage v(t) as an internal variable. (b) Find transfer functions from the commandinput @(t) to the output velocity q~(t) and from the commmand input to the voltage v(t). (c) Is there a steady-state error in the output velocity for a step change the commandvelocity? Hint: There is no such error if the transfer function approaches 1.00 when S ~ 0, which corresponds to the frequency response for zero frequency. (d) Determine the natural frequency and the damping ratio in terms of and (e) Set the damping ratio equal to a reasonable value, such as 0.4. Determine the resulting relation between k and k/Ti. (f) Note that the largest allowable value of the ratio k/Ti gives the fastest response. Program in MATLAB the numerator and denominator polynomials in the transfer function with v(t) as its output, substituting the relation from part(e) for every k that appears alone, so the ratio is the only unknown. Multiply the numerator by the command input corresponding to 2000 degrees per second, and use the step commandto secure a trace of the voltage as a function of time for various values of the ratio. Determine the value that gives the largest allowed voltage. Then determine the values of k and Ti individually.

626

CHAPTER 8.

INTRODUCTION TO AUTOMATIC CONTROL

(g) Use MATLAB to plot the output ~(t). If you have done part (a) Problem8.2, comparethis response with the optimal. 8.10 Determine which of the closed-loop systems in Problems 8.3-8.7 have steady-state errors in response to a step input. For each of these, specify the gain and the integral time of an integral controller that eliminates the error while giving acceptable dynamics. 8.11 Repeat part (a) of GuidedProblem8.2 (p. 622), substituting a lead troller for the PDcontroller. 8.12 The system of Problem 8.7 (p. 604) exhibits either small damping negative damping(instability) for all values of the gain k. Determinethe gain and derivative time of a substitute PDcontroller that gives better responses. 8.13 Solve the previous problemusing a phase lead controller in place of the PD controller. The ratio of the magnitudesof the pole to the zero of the controller should not exceed3:1. 8.14 The steady-state error in the thermal system of Example8.7 (pp. 609610) was eliminated by the introduction of integral action, but at a cost of reducing the natural frequency. Expand the controller by adding derivative action (makinga PIDcontroller) so as to regain the original w,~ = 2/’r while retaining ¢ = 0.5. Find the "best" values of the derivative time Td, integral time Ti and the gain k. 8.15 A correspondance between a phase lag controller and a PI controller can be defined in the sense that their responsesto a unit step agree for small values of time. Find the correspondanceassuminga value of ~3 is chosen already: Apply the result to the PI controller of the motor-drivenposition control system, as carried out in Section 8.2.1 (pp. 606-608), finding the associated value of Tp for the corresponding lag controller. Comparethe result to the lag controller with ~3 = 10 for the sameplant but chosen by other meansin Section 8.25 (pp. 618-620), namelyTp = 0.020. 8.16 Derive the relations for the values of the following RCcircuits as given in the respective figures. (Note: The simplest methodstarts with a bond graph, and then finds the transfer function using the loop rule as developedin Section 7.4.2-7.4.4.) (a) the circuit of Fig. 8.12 (p. 617) for the phase lead compensator. (b) the circuit of Fig. 8.13 (p. 618) for the phase lag compensator. (c) the circuit of Fig. 8.15 (p. 621) for the phase lag-lead compensator.

8.2.

SOLUTION Guided

627

DYNAMIC COMPENSATION

Problem

TO GUIDED

PROBLEM

8.2

The open-loop poles are at S = 4-V/~/L = 4-10 rad/s. Thegain of the plant is 50/(S"~ 100). For a small feedback gain, the closedloop poles lie on the real axis, with one in the right-half plane indicating instability. For a large gain, both closed-loop poles lie on the imaginaryaxis, indicating oscillation without necessary dampingor decay.

~

)

The output of the compensator is defined above as -F.

Go(S)= k(1 + TdS)

3O

Td = 1/1.5V/’~/L = 1/15 s The open-loop gain becomes k50(S/15 + 1) S~ - 100 The coding for the root locus plot is nmn= 50*[1/1B 1]; den = [1 0 -100] rlocus (r,k)

The following additional coding gives the plot shownopposite, showing the closed-looppoles for the values k = 3, 6, 9, 12, 15, 18, and 21. The value k = 15 is chosen, for which r = -25 4- 5j rad/s. k = [3:3:21]: [k,r] = rlocus(n,d,k); plot(r, ’x’)

-~o

selectedclosed-looppoles

628

CHAPTER

8.

INTRODUCTION

TO AUTOMATIC

CONTROL

The block diagram below gives the open-loop gain -32.2k(S a + 15S~ - 50S - 750) S4 + 50S3 2+ 650S

@-F,

50(S+15)I ~-0.644(S-’-50)

1s-’+5os+65o1 -I, s:

(’ li

The root locus showstwo closed-loop poles in the right-half plane, producing an instability, for all values of the gain, k. 2O ~S, rad/s

-20-40’

Re S, rad/s

The zero of the PD controller is placed at S --- -a. This multiplies the numerator of the open-loop transfer function by the factor (S ÷ a) to give -32.2k[S4 ÷ (15 +a)S3 + (15a - 50)S~ - (50a ÷ 750)S - 750a]. The root-10cus plot below is for a = 3 rad/s. Oneroot approaches -oo on the real axis before the critical pole pairs near the origin emergevery far along their paths in complexterritory. The root reappears at +oo on the real axis, and ends at the zero in the right-half plane. This pole must be confined to the left-half plane to give stability. A reasonable solution is k -- 0.02, for whichr -- -87 tad/s, -20.1 tad/s, and -0.39 ± 1.50j rad/s. 10 ImS, rad/s

f

-3’0 -io -;0

0

;0 20

Re S, rad/s

0.644(-S 4 - 18S3 + 5S2 + 900S + 2250) x~ 0.356S4 ÷ 38.41S3 ÷ 653.2S~ ÷ 579.6S ÷ 1449" X

The step response is plotted on page 656.

8.3.

629

FREQUENCY RESPONSE METHODS

8.3

Frequency

Response

Methods

The design of feedback controllers has been based thus far on the placement of the zeros and especially the poles of the open and closed-loop transfer functions, with emphasis on the root locus plot. The use of frequency response methods is a valuable complement or alternative. You have been exposed already, through the extensive material in Section 6.2, to the concept of frequency response and its primary graphical tool, the Bode plot. In this section the Bode plot, and its siblings the polar (Nyquist) and magnitude-vs.-phase (Nichols) plots, will used to predict closed-loop performance from open:loop performance, and to aid in the design of controllers. Should you wish a deeper exposition or illustration of the methods, many standard textbooks are available.

8.3.1 Polar or Nyquist FrequencyResponse Plots The Bode plots comprise separate representations for the magnitude and phase angle as functions of frequency. Often it is desirable to combine magnitude and phase angle into a single curve. Oneway to do this is the polar plot, also called the Nyquist diagram. For a plot of G(jw), the x-axisrepresents Re[G(jw)] and the y-axis represents Im[G(jw)]. Therefore, the magnitude IG(jw)l equals the distance or radius from the origin, and the phase angle is the angle from the real axis. Since frequency is not used as an axis, it is helpful to place tick marks along the curve, labeled withthe corresponding values of w. EXAMPLE 8.10 Draw polar or Nyquist plots for the functions Ga(S) = S + 1, G2(S) 1/(S + 1). Solution: The function Ga(jw) can be plotted directly. The function G2(jw) = 1/(jw 1) is its reci procal, that is ha s r ecip rocal magni tudes and oppositely signed phase angles for corresponding values of w. The latter curve can be shown to be a semi-circle. 1 Im

magnitude/ of ~ ~.

G~(S)=S+ .0=0.5

0.5

/phase angle~ / of ~,(]1) 6)=0

-02

~ro=2 G,.(j ~)~/0"5 - "~=0.5

1 G2(S)= 1

Re

MATLAB draws polar plots in response to the commandNyqu±st (nura, den) or Nyquis~c(k,,B,C,D), where hum, den or /~,B, C,D are defined as before. MATLAB includes the plot for negative values of frequency, which has the

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same real part and minus the imaginary part of the corresponding values of frequency; ~t is the same curve flipped about the real axis. The reason for including negative frequencies will appear shortly. EXAMPLE 8.11 Use MATLAB to plot the Nyquist diagram for the unstable open-loop transfer function G(S)H(S) = k(S 2)/(S ~ + S + 1) (S - 0. 5) wi th k = 0. 2, 0. and 0.6. Solution: The following code produces the plots as given (except for the annotations). See the "help" window for optional arguments of Nyqu±st, including a frequency vector. den=If .5 .5 -.5]; for i=1:3 k=.2*i ; hUm=k*[i 2] ; Nyquist(num,den) hold on end

I

0.6

.-"’<’, ,/

0.6

0.2

dashed lines:

w
~

:

~02

-0.4 -0.6 -0.8

RealAxis

8.3.2

The Nyquist

Stability

Criterion

The objective is to determine the stability of a closed-loop system from knowledge of its open-loop frequency response, without having to carry out computations. The open-loop frequency reponse could be from experimental data, with no analytic model. In particular, it is desired to knowwhether any of the roots or zeros of the closed-loop characteristic equation F(S) = 1 + P(S) = 1 + G(S)H(S) = 0

(8.30)

8.3.

FREQUENCY RESPONSE

631

METHODS

segment first segmen

~hird segment

Figure 8.18: Contour in the S-plane that encircles all potential unstable ,roots (zeros) of the characteristic equation

lie in the right-half plane, and therefore represent an instability. A famous theorem by Cauchy refers to a closed path or contour in the S-plane for a function such as F(S], and the plot of the complex function F(S) for the same path values of S. It states: If a clockwise contour in the S-plane encircles Z zeros and P poles of the complex function F(S) without directly passing through any of them, the mapping of the contour in a plot ofF(S) encircles its origin Z-P times in the clockwise direction. An encirclement of the origin in a plot of F(S) becomes an encirclement of the point S = -1 in a plot of the open-loop transfer function P(S) = F(S) Since the stability test regards the entire right-half of the S-plane, the contour of choice encircles the entire right-half of that plane, as shownin Fig. 8.18. Part of the contour traverses the imaginary axis in two segments, as labeled. The first segment is S = jw, with w ascending from zero to infinity. The third segment is its complex conjugate, S -- -jw, with w proceeding from infinity to zero. The mapping of this third segment in the P(S) plane is the complex conjugate of the mapping of the first segment, which means it is the mirror image about the real axis. This is why MATLAB includes both segments in its Nyquist plots. A further simplification results whenever the second segment of the contour gives zero values of P(S) for all such infinite values of S, which happens whenever the denominator polynomial of P(S) is of higher order in S than is its numerator polynomial. EXAMPLE 8.12 Interpret the Nyquist plots for the three cases in Example8.11 to give the stability of the closed-loop system. Also, plot a root locus that shows the locations of the closed-loop poles for the three cases. Finally, use the RouthHurwitz criteria for stability (p. 600) to determine the range of k for which the system is stable.

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Solution: There is one open-loop pole in the right-half plane of the openloop transfer function, so P = 1. For the case with k = 0.2, the -1 point is not encirled by the Nyquist contour, so Z-P = 0. Thus, the numberof zeros in the right-half plane for the characteristic equation is Z = 0+ 1 = 1, which meansthat the closed-loop system has one pole in the right-half plane and therefore is unstable. For the case with k = 0.4, the -1 point is encircled once counterclockwise, so Z-P = -1 and Z = -1+1 = 0. There are therefore no closed-loop ploes in the right-half plane, and the system is stable. For k = 0.6, on the other hand, the -1 point is encircled once clockwise, so Z-P= 1 and Z = 1+1 = 2. There are two poles in the right-half plane of the closed-loop system, therefore, and the systemis again unstable. The root locus plot below shows howthe poles migrate in the S-plane as k is increased. It showsthat all poles are in the left-half plane only for a single band of values of k.

Z0. O.5

k=0.2~’ff \ 0.5 (cros~

poin:)

0.25 0.4 0.3 \~’k- 0.2 0.25 (cross,

point)

-0.5

"1’~2.5

-2

-1.5

-1 -0.5 Real Axis

0

0.5

1

The closed-loop transfer function is G(S)/(1 + G(S)H(S)), which gives the characteristic equation (S2 + S + 1)(S- 0.5) k(S + 2)= O Comparingthis "to the standard form a3S3 A-

a2,-.q’2 -4- ai S +ao=0

gives a3 = 1;

a2 = 0.5;

al = 0.5 +.k;

a0 = 2k - 0.5.

The necessarycondition for stability that all coefficients havethe samesign requires ao > 0 or k > 0.25. Equation(8.15) (p. 600) additionally requires a2al - a3ao= 0.25 + 0.5k - (2k - 0.5) > which is satisfied only for k < 0.75/1.5 = 0.50. Therefore, the system is stable only between the crossover points at which k = 0.25 and k = 0.5.

8.3.

FREQUENCY RESPONSE ImS

633

METHODS ite radius ent

~m

}infinitesimal "~

imaginar) axis ~’rsa~Re segments

S

Figure 8.19: Modified contour in the S-plane to avoid poles on the imaginary axis

The Cauchy theorem requires that the contour in the S-plane not pass through any of the poles of the open-loop transfer ruction. The contour must be modified, therefore, if one or more poles lie on the imaginary axis. The usual strategy employs a 180-degree arc of vanishingly small radius as a detour around such poles, excluding them from the encompassed region and the number P, as illustrated in Fig. 8.19. The mappings of these counter-clockwise arcs in the S-plane become clockwise arcs of infinite radius in the P(S) plane, as the next three examples demonstrate. MATLAB does not display any segments having infinite radius, however, so it is up to you to determine their effect on the number of encirclements of the -1 point.

EXAMPLE 8.13 Plot the Nyquist diagram for the type 1 system having G(S) = k/S(S 2 + S ÷ 1) and H(S) = 1 when k = 1. Determine from the plot what values of k give stability. Solution: The Nyquist plot at the top of the next page assembles a directlydrawn pseudo-infinite arc with a MATLAB-createdplot. Since there is one pole at the origin in the S-plane of a type-1 system, the infinite arc traverses 180° . It swings clockwise, as can be verified by checking the mid-point on the real axis S = ~ -~ 0 and G(~) The contour passes directly directly through the -1 point; for k > 1 it would encircle the point clockwise once, indicating one unstable pole. The criterion of stability, therefore, is k < 1.

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"’-,..infinite |

-2

I-1

CONTROL

radius

/mG

--12

1 ReG

EXAMPLE 8.14 Repeat the above example for the type 2 system G(S) -- k/S2(S 1) and H(S) -- 1 with k = 1. Solution: The presence of two poles at the origin produces an infinite arc of 360° in the Nyquist plot, as shown below. It is critical that you attach it correctly to the finite portion of the contour; placing clockwise arrows on the infinite segment helps. There are two encirclements of the -1 point, regardless of the value of k. This system is unstable regardless of the value of k; it has two right-half-plane poles.

""~’~’~..infinite "¯ from infinitesimal radius segment

"’"’""

¯

¯

¯ "4

ImG -,,

¯ arc on S-plane ¯ ¯ ¯

8.3.

FREQUENCY RESPONSE

635

METHODS

EXAMPLE 8.15 Plot the Nyquist diagram for the system with kS G(S) = (S’; I)(S + 1)(S + H(S) =I; k = I. Determine therangeof k forstability of theclosed-loop system, anddeterminetheanswerindependently usingtheresults of Section 8.1.7(p.600). Solution:The MATLAB coding is num=[1 0]; den=[1 4 4 4 3]; nyquist (hum, den) axis(I-.3 .3 -.3 .3]) The resulting contour approaches infinity at each of the two open-loop poles on the the imaginary axis. The arcs of infinitesimal radius around these poles in the S-plane become180° clockwise arcs of infinite radius in the plane, and are added below to the finite segments of the. locus produced by MATLAB to complete the locus. The -1 point is not encircled for the value k -- 1 of the locus, but would be encircled, twice, if the plot were expanded by the factor k = 8 or more. The criterion for stability, therefore, is k < 8. -"-"-’’-/ ,"

lm G

circular arcs at ",,~mfimteradii .,

,/’~.2 ..... plot for. k= 1 t,

-0.125

The characteristic

equation 1 + G(S)H(S) = 0 can be written S4 + 4S3 + 4S~ + (4 + k)S + 3 =

so that a4 = 1, a3 = 4, a.., = 4, al = 4 + k and a0 = 3. The first of equations (8.16) gives 16 - 4 - k > 0 or k < 12. The second of equations (8.16) gives 64 + 16k - (16 + 8k 2) - 48 > 0, fr om whichk < 8. The lat ter is controlling, and agrees with the Nyquist conclusion.

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CHAPTER 8.

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1/gain margin -1

~

10°[

~0"+

I

I

~

~

I

I

IIII

’ Fmquency(mdlsec)10+

~

I

I

I

IIII

~ i phase

Frequency (md/sec)

(a) Nyquist plot

(b) Bodeplot

Figure 8.20: Example system showing gain and phase margins

8.3.3 Measures of Relative Stability A control system must not only be stable, it must also be reasonably robust, which means insensitive to at least modest unplanned variations in its parameters. Further, the dynamics should not be excessively oscillatory: These requirements can be addressed effectively from the perspective of the frequency response. Most type 0 and type 1 open-loop systems produce stable closed-loop systems for small values of the proportional gain k, and becomeunstable for large values. The gain margin for a particular design is defined as the factor that k could be increased before the system becomes unstable. Most designers would like to see a gain margin of 2 or larger, although they are willing to shave this in special circumstances. The gain margin is readily apparent from the Nyquist plot; it is the reciprocal of the magnitude of the contour at the point it crosses °the negative real axis. The phase lag of the open-loop system traverses -180 at this point, so the gain margin also is readily determined from a Bode plot. Its value often is given in decibels. An example is given in Fig. 8.20. The phase margin is defined as the additional phase lag of G(jw), at the frequency w where IG(jw)l 1, tha t wou ld bar ely pro duce ins tability. Thi s also is shown in Fig. 8.20. A phase margin of 45° or more is desirable, although sometimes it is reduced to as little as 30° . Phase margin also can be read from the Bode plot, as shown. A third and particularly meaningful measure of relative stability is the maximumvalue, Mm, of the magnitude ratio, M, of the closed-loop frequency response. Assuming unity feedback (H(S) = 1) as before, the ideal value at all frequencies is M = 1. The deviation for M from 1 at zero frequency is the

8.3.

637

FREQUENCY RESPONSE METHODS

Figure 8.21: Mand N circles in the Nyquist plane, with an examplecontour steady-state error. A value Mm -~ 2 may be tolerated, depending on the application. The value of Mm can be determined upon inspection of the Nyquist plot, although this is not as obvious as the gain and phase margins. Let P(j~.,) = u + jv. Then, the square of the magnituderatio of the closedloop system is 2 M~ =I G(jw) (8,31) 1 + G(jw) (1 + u)’) ~’ +v which can be rearranged as u~ 2+ v

(u 1 - M:) + v~ =1 _- ~_~

"

(8.32)

For a fixed value of M, this is the equation for a circle in the Nyquist or. u,v 2) that is centered at v = 0 and u = M2/(1-M:). plane with radius M/(1-M resulting family of M-circles is plotted in Fig. 8.21. A sample Nyquist contour also is plotted, which has the value Mm= 2. A secondfamily of circles is plotted by dashedlines in Fig. 8.21. Theseare lines of constant closed-loop phase angle, ¢, called N-circles whereN -- tan~b. the relation ¢ = tan-~

(~)

- tan-1

(l~u)

(8.33)

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CHAPTER 8.

INTRODUCTION TO A UTOMATIC CONTROL

Open-Loop Phase(deg)

Figure 8.22: Nichols chart (MATLAB version closed7loop gain and phase added)

5.3, with annotations

for the

can be rearranged into the form (u+.05)

"°+ (1)

v-~

2

= a l+~g 1(1)

,

(8.34)

which shows that the loci for constant values of N are indeed circles, at u = 0.5, v = 1/2N and having radius 0.5V/1 + "1IN’2.

8.3.4

centered

Nichols Charts

A particular set of coordinates is rnore helpful for design than either the Bode or Nyquist coordinates. The MATLAB version of the Nichols chart is shown in Fig. 8.22. ’A more detailed paper version also is available. The log of the magnitude of G(S) is plotted vs. the linear phase of G(S). The M and N "circles" for the closed-loop transfer function G(S)/(1 + G(S) are mapped into this plane, ~vhere they no longer are circular. One advantage of the Nichols coordinates is that the value of k can be changed merely by translating the contour vertically. If done with a paper chart, the contour can be placed on an overlay that is slid vertically. In MATLAB, you. start by requesting the Nichols contour for G(jw), which appears without the M and N contours. You then

8.3.

FREQUENCY RESPONSE

METHODS

639

right-click on the figure and select "grid" to get these contours. Youlikely will wish to change the boundaries of the plot using the axis command. Thus, to achieve a particular value of Mmfor a particular system, you plot the Nichols contour for k = 1 and translate it vertically until it is tangent to the M-contour with the value of Mm. The distance translated, in db, is the gain sought. You then can observe values of M and N (or ¢) at several frequencies, thereby rapidly gaining an appreciation of the entire closed-loop frequency response. EXAMPLE 8.16 Use the Nichols chart to find the closed-loop gain k for the unity-feedback system with open-loop transfer function G(S) = k/S(S 2 + S + 1). Let the criterion be a maximumpeaking ratio M,~ = 1.5 (3 db). Solution:

The MATLABcoding

num= 1; den=[1 i i 1 0]; nichols(hum,den) produces the Nichols plot of the open-loop system, but does not directly show the Mand N contours. Right-click on the figure, and select " grid." Then, to expand the plot vertically, issue the commandaxis ( [-360 0 -45 40]), which gives the result below. (You may- need to refer to Fig. 8.22 identify the M and N contours, which MATLAB version 5.3 does not label.) From: U(1)

-350

-300 Open-Loop Phase(deg)

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CHAPTER 8.

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The vertical translation of G(jw) needed to makethe contour tangent to the locus for a maximum closed-loop magnitude of 3 db can be seen to be about -6.5 db, giving k = 0.47. The plot of G(jw) corresponding to this value is plotted as a dotted line; you can superimposethis plot by issuing the commandho:td on, followed by a repeat of the n±cho:~s command.You have nowmaximizedthe bandwidth of the closed-loop system, consistent with the specifications. 8.3.5

Dynamic

Compensation

Using

Nichols

Charts

The Nichols chart allows you to design a dynamiccompensatorvisually. Informationis given belowto help you select phase lead, phase lag, phase lead-lag, PD, PI and PIDcontrollers. The phase lead controller has the transfer function (from equation (8.23)) . S+z G~e~d= ~S---~-~az; ~ > 1,

(8.35)

and the phase lag controller has the transfer function (from equation (8.23)) +~3P, G~ag=kS --~--~-p"

~ > 1.

(8.36)

Recall that values of a or ~3 muchlarger than 10 likely wouldbe impractical, due to impedanceconsiderations, and values less than two wouldhave little effect. Nicholsplots for three sizes of each of the controllers are plotted in Fig. 8.23. Varyingk merelyshifts these plots vertically, so the value k = 1 is chosenin all cases. Use of the dimensionless frequencies f~ = w/z and f~ = w/p allows these curves to represent all values of z and p. The scale of this plot is the same as the scale of Fig. 8.22 The transfer function G(S) is the product of the transfer function of the controller and that of the plant. The magnitudeat a particular frequency is phase lag compensator phase lead compensator 20 magnitude, db 10

Note: scale agrees wi~ MATLAB Nichols cha~s

D= ~/ fl= ~ b~ ,

~ D= a~/z

~"~-A b 1(~--~

0

b

a = =

=2. Of

10b

~

-I0 ~5 -1~*

-50*

0~

~=

[a 10

° 50 ph~e ~gle

100"

Figure 8.23: Nichols plots for ph~elead and ph~elag controllers

8.3. FREQUENCY

RESPONSE

641

METHODS

[

without compensator~

~with compensator

20 ,,

.- ..... ~--;’-~’-" ~, ,’....... ~_,_ ,,.o~ ,.~,~,s

"-,-",--, ...... .... r .......

= , , ,, ;-- .... : ".,,,,, ,--~.* ~ o " " "’" ">:" "" ’"*~"-"--~ o-’° r’ ;" ~--, ---7’- .... ~-, .... ~___~ ......

’

’ ,

’

’, I "

,

I

’, , 1 , ,;, ;.--...:.: "’-’" "’"" "" ",- " ’, I~ ’,

, _.---’r--"

,--’-

.........

Open-Loop Phase(deg)

Figure 8.24: Nichols plots for a mass-sprxng system without and with a phase lead controller

the product of the two magnitudes, and the phase angle is the sum of the two phase angles. With the Nichols axes, therefore, both the vertical and horizontal components of G(jw) are the sums of the vertical and horizontal components of the controller and the plant. This fact is illustrated in Fig. 8.24 for the system of Example8.10 (p. 617) in which a lead compensator with a = 10 is applied to the mass-spring system of Example 8.8 (p. 613). The gain k = 160 in this example is taken as part of the original system transfer function. The branch of the frequency transfer function shown represents frequencies above the 2 rad/s resonant frequency of the system, for which the phase angle is a constant -180° and its Nichols plot is a vertical line through the -1 point. The compensated open-loop frequency function is pushed to the right, and can be seen to nicely wrap around the -1 point, keeping the closed-loop gain below 2 db for all frequencies. The frequency transfer function of the controller is shown in two postions: the upper position represents its effect at the "nose" frequency of 5.79 rad/s, and the lower position represents its effect at ten times that frequency. The phase lead controller exerts its maximumdesired rightward push for the frequency at the nose of its Nichols contour. The contoller also changes

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the magnitude of the contour, in a direction that usually is not helpful. As a result, the controller is less useful when the slope of the plant contour is small. Whensuccessful, the lead controller allows the gain k to be increased, making the system faster. EXAMPLE 8.17 A plant has the transfer

function 1 GB=

$3 Jr

2S 2 Jr

2S Jr

1

Whenunity feedback is employed with proportional feedback and gain k = 1, the resulting steady-state error is 50%. It is not possible to reduce this error muchby increasing k, without causing relative or absolute instability. Investigate what can be done using a phase lag controller, with a < 10. Solution:

The commands

num=l; den=[1 2 2 1]; nichols(num,den) axis([-360 0 -45 40]) produce the Nichols chart for the plant below, presuming you right-click the figure and select "grid" before issuing the axis command.

-2O

-3O

-350

-300

on

8.3.

FREQUENCY RESPONSE

643

METHODS

The largest allowed lag compensator should be chosen, which has ~3 = 10. The "nose" frequency should be large from the perspective of the speed of response of the system, and small from the perspective of stability. As a compromise, the point having the open-loop phase angle of -150° is chosen to have ten times the nose frequency.. This point is labeled on the chart as "~u = 1.112 rad/s." This frequency value is found by entering [m,p ,w]=nichols (num,den) [loglO (w), and looking downthe resulting table to fin d the frequency "w" listed opposite p = -150 °. (The use of the commonlogarithm precludes scaling problems in the table; you raise 10 to the value of the log to get the frequency.) At this frequency, w/p = 0.1b = 0.1x/~, from which p = 0.035 rad/s. The phase lag controller therefore has the t~ansfer function (equation (8.36)) S + 0.35 S + O.O35’ and the overall open-loop transfer function is S + .35 G(S) = kGcGp = $4 2. 035S3 + 2. 07S"- + 1. 07S + 0. 635" Entering "hold on" and then setting these new polynomials gives the second plot above labeled "phase lag compensator added." Leaving the present value k = 1 seems a reasonable compromise between reducing steady-state error and preserving relative stability. The closed-loop transfer function becomes G(S) S .3 5 1 + G(S) S4 + 2.035S 3 + 2.07S 2 + 2.07S + .385’ The step response, found using step in MATLAB,is plotted below. The steady-state error is 9.1%, which is rather large but may be acceptable. If a smaller error is needed, some other compensator is needed; a PID controller would do admirably. 1.2 1.0 0.8 0.6 0.4 0.2 0

10

20

30 time, seconds

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CHAPTER

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left half: PI compensator 3o gainin db 20

CONTROL

right half: PD compensator whole curve: PID compensator

°.45

45° 900 PI compensator: ~o=~, PD compensator: to=0 PID compensator: reo= 1

(a) Nichols chart

8o degreesPhase angle,40 r= "*a,*,

rT~,,~ 0.05~

0

(b) ’_ phase angles PID compensator

0.5

for ~_~_~__C_~ -80 ~ ~ ~ ~ ~ o.ol o.1

Figure 8.25: Frequency characteristics The transfer

function

PD : Gc(S) PI:

Gc(S)

~/~

1.o

~ ~ "’

1o

lOO

of PD, PI and PID compensators

for the PD, PI and PID compensators are = k(1

+ TdS),

= (1) k 1+ ~(~S+ = k \

PID : Go(S) = k 1 + TdS; q( T~-~)

r=~;

r = ~..

(8.37a) "~S = k q- r TS [1 (TS)2 + 1]

(8.37b)

(8.37c) (S.37d)

Nichols plots for these compensators, with k = 1, are plotted in part (a) Fig.8.25. Note that the contours for the PI and PD controllers are mirror images of each other, and the PID contour traverses precisely their combined contour. The mid-points in the phase angles of the PI and PD compensators, at -45 ° and 45°, respectively, occur when the dimensionless frequencies Tdw and Tiw equal 1. The gain and phase shifts for the PID controller are zero at its central frequency rw -- 1, with T = Tv/~-~. Howrapidly this dimensionless frequency changes with changes in the phase angle depends on the "strength" of the controller, r = v~/Ti, as plotted in part (b) of the figure.

8.3.

FREQUENCY RESPONSE

645

METHODS

EXAMPLE 8.18 The open-loop plant 50(-S + 100) Gp = $3 + 101S 2 + 150S + 5000 is particularly difficult to control because of its right-half-plane zero and the small damping of its second-order pole pair. It is desired nevertheless to eliminate steady-state error. Design a controller. Solution: Following the procedure of the above examples to plot the Nichols chart for the plant gives the result given by the solid line below: Nichols Charts From:U(l)

,

|

c, ompensat.e~ G(ja~), k=4.47 6)=7.036 rad/s

Integral action is needed to eliminate the steady-state error. Taken alone, however, this would require a very small value of k, on the order of 0.1, to give a reasonable relative stability. The system would becomevery sluggish. To improve the response significantly, it is proposed to include derivative action as well. The information needed to describe the resulting PID controller is given in Fig. 8.25. The point at which TW= 1 is chosen to lie at the maximumamplitude point, as labeled. The controller leaves the contour unchanged at this point, assuming k = 1; to the right of this point, the phase is shifted leftward, and to the left of this point it is shifted rightward. The magnitude is shifted upward in both cases, slightly for modest phase

646

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shifts and greatly for phase shifts in excess of 45°, as the plot in part (a) the figure shows. The choice of the controller shown was made in one pass, without iteration, by visual examination of the plant contour and the characteristics given in Fig. 8.25. The point at which the contour passes through the -1 point (labelled as w = 10) was chosen for a 350-40° rightward phase shift. This decision is implemented by proper selection of the "strength" parameter, r. It was first necessary to determine the frequencies of the -1 point and the maximumamplitude point. This was done with the commands [m,p,w]=nichols(num,den); [loglO(w),20*loglO(m),p] The resulting table shows that the peak occurs when lOglo(~ ) = 0.8477, or ~ = 10°s477 = 7.036 rad/s, and the -1 point occurs were the phase angle is -180 °, which happens to be 10.00 rad/s. Since ~-w = 1 at the maximum point, 7~ = 10.00/7.036 = 1.421 at the -1 point. The plot in part (b) the Fig. 8.25 indicates that r = 1 to achieve the 35°-40° phase shift. Using this value, then gives Td = T~ = 0.1421 seconds. The controller becomes, from equation (8.34c), Gc = k0"02019S2 + 0.1421S + 1 0.1421S ’ and its product with the plant transfer

function becomes

G(S) k (0 ’02019Se +4 0. 1421S3 + 1) 50(-S + 10 0.1421(S + 101S + 150S2 + 5000S) The new numerator and denominator polynomials, MATLAB as

for k=l, are given to

num2=50*[-.020191.877 13.215 i00]; den2=.01421*[l 101 150 5000 0]; with the result labeled as "compensated G(jw), k = 1" in the Nichols chart. This contour now can be translated upward to give a reasonable maximumvalue for the closed-loop frequency response. A 13-db shift, or k : 1013/20 -~ 4.47, gives the final contour. The polynomial hum2 was multiplied by 4.47 to get this curve. The step command of MATLAB was used to verify the desirability of the controller, with the result below. The overshoot is larger than might be desired, but might be acceptable considering the challenge in finding a reasonable solution. The steady-state error is zero, as specified. Note that the response starts out in the negative direction, which is a result of the zero in the right-half plane of the plant.

8.3.

FREQUENCY RESPONSE

647

METHODS

0.8 0.6 0.4 0.2 0 -0.2

8.3.6

0

0.1

Approximate

0.2 Correction

0.3

0.4 for

0.5 0.6 time, seconds Digital

0.7

Sampling

Digital control systems, as noted briefly in Section 8.2.7 (p. 621), normally employ samplers with holds. A sinusoidal signal of relatively low frequency thus treated becomes a series of steps that resembles a sine wave. The principal effect is a phase lag in the output of the sampler relative to its input. The phase lag equals 180° times the ratio of the sampling time to the period of the sine wave. If ten samples are taken per cycle, for example, the phase is 18°. If the open-loop frequency contour of a system closely passes the -1 point at this frequency, the sampler would have a significant deleterious effect on the relative or absolute stability of the closed-loop system. The amplitude of the signal emerging from the sampler is reduced from that of the input, but only slightly if there are ten or more samples per cycle, so for most purposes this effect may be neglected. If you have fewer than ten samples per cycle at a frequency critical to stability, you n~ed to look more deeply into the subject, including the role of digital filtering used to a~ldress problems of signal noise. EXAMPLE 8.19 The controller in the system of Example 8.16 (p. 639) gives a sample-and-’ hold output, with a rate of two samples per second. Modify the Nichols contour approximately to account for the effect of the sampler, and estimafe the revised value of the gain k that gives the same maximumpeaking ratio Mm= 1.5 (3db). Solution: The phase lag introduced by th~ sampler would be 180° at a frequency of w = 27r/T = 27r/0.5 = 47r rad/s, or Crag = 180~/47r degrees. The phase angles of the v£kious frequencies along the Nichols contour can be determined with the command[m,p,~] =nicho:~s (hum, den), as done for example in Examples 8.17 and 8.18. Ad~ting the phase lag from the sampldr for the critical region near -180° gives the contour indicated by thh leftward shifted curve in the plot given on the next page. This curve has to be translated downward by about 7.5 db to give the desired peaking, so k = 10-7.5/20 = 0.42.

648

CHAPTER 8.

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TO AUTOMATIC

CONTROL

From: U(1)

-350

-300

-200

-150

- 1O0

-50

0

Open-LoopPhase (deg)

8.3.7

Summary

This section has added frequency response methods to your arsenal of tools available for the design of feedback controllers. Which method or combination of methods works best for you depends on the particular situation. Frequency response methods are especially attractive when all you have to describe your plant is frequency-response data, for example found using an oscillator or a shaker table. It would be possible the use the identification methods of Section 6.2 to get an analytical model from such data, making the analytical pole-placeznent methods such as root locus available, but that is not necessary. Design methods based on frequency response, particularly using the Nichols chart but also using the more basic Bode diagrams, provide a powerful predictive insight that warrants their use even when analytical models are known. Polar or Nyquist plots of frequency response combine magnitude and phase information into a single curve, an advantage over the Bode plot. Although these plots can be used to predict quantitatively the behavior of closed-loop behavior, emphasis has been placed merely on the prediction of absolute stability via the Nyquist criterion. For more refined design and prediction, the alternative mapping called the Nichols. chart is usually more convenient, and has been emphasized. All these forms are readily available in the control toolbox

8.3.

FREQUENCY RESPONSE

METHODS

649

of MATLAB and in the student version of MATLAB.They all can be used to determine the gain aud phase margins of a system from its open-loop transfer function. These meassures, plus the inaximum peaking of the closed-10op frequency response, often are cited by control engineers in describing the behavior and the robustness of their designs. PI, PD, PID, phase lead, phase lag and phase lead-lag controllers are characterized in Figs. 8.23 and 8.25 (pp. 640, 644) in the form of Nichols contours. Their application is given in several examples, including the Guided Problem below. Small digital sampling delays introduce an effective phase delay equal °to. the ratio of the frequency to the sampling frequency, times 180 Guided

Problem

8.3

This problem involves the Nyquist stability criterion and the design use of the Nichols chart in an interesting case for which a phase lead controller outperforms a PD controller. The control of an inverted pendulum in Guided Problem 8.2 (p. 623) has two parts. The first is the quite successful use of a PD controller to keep the pendulum upright. The second is the use of a feedback loop with another PD controller to allow the position, x, of the pendulumto be controlled as well. The second control involved a trade-off between the speed of response and the relative stability, with a result that one would like to improve, if possible. Further, you were guided rather strongly in the key suggested step 7; now that you have further tools in your arsenal you can assume more initiative. Leave the result of the first part of the contol unchanged. Youalso may retain the block diagram of step 5 (p. 628), except replace the proportional controller (which doesn’t work, as step 6 revealed) with a lead controller. Choose the two parameters of this controller, find the resulting roots of the closed-loop performance, and compare to the results of the PD controller. Suggested

Steps:

The plant of this system has the force -F as its input, the position x as its output, and a transfer function equal to the product of the transfer functions given in the two right-most blocks in the block diagram of the original step 5. Verify the instability of proportional control on this plant through the use of a Nyquist plot. Pay particular attention to the infinite segments of the Nyquist contour, which are not given by MATLAB. Examine the Nichols chart contour for the same transfer function, and determine qualitatively how the controller must change this contour to achieve stability. Does the phase lead controller potentially satisfy the requireraent? (Note that sometimes the MATLAB program n±chols gives phase angles that are 360° larger than desired, with the result that the M and N contours are not available: You may therefore want to refer to these contours as given in Fig. 8.22 (p. 638). This difficulty will vanish when the contoller is inserted.)

650

CHAPTER 8.

INTRODUCTION

Choose the size of the controller value for the "nose" frequency of the plant contour and the controller You may wish to magnify the plot

TO A UTOMATIC CONTROL

(the parameter a) and an approximate the controller, through examination of contour as given in Fig. 8.23 (p. 640). through the use of the axis command.

Deduce the value of the frequency parameter z that follows from your choices, and then write the transfer function for the resulting controller. At this point the value of the gain k is not yet determined; set it at 1. Find the complete open loop transfer function, using you controller, and plot its contour using n±chols. Repeat this step and the last if you are not satisfied with the result. Examine the contour to see how many decibels it should be raised or lowered in order to give the smallest maximummagnitude to the closedloop frequency response. Adjust the gain, k, correspondingly, and repeat the contour to verify your choice. Find the characteristic equation for your closed-loop system, and use the command roots to determine its dynamic behavior. Compare with the results for the PD controller as given in step 7 on page 628. Deduce the closed-loop transfer function, and plot the step response. Is the result qualitatively consistent with your earlier findings? PROBLEMS 8.17 The following Bode plots represent with a proportional controller. A gain the associated phase margins, the gains errors. Work from the given Bode plots experimental data.

the plants in unity-feedback systems margin of 2.5 is specified. Estimate of the controller and the steady-state directly, as though they represented

(a) Bode plot of Example 6.11 (p. 428). (b) Bode plot of Example 6.14 (p. 433). (c) Bode plot for GI(S) in Example 6.16 (p. 435). (d) Bode plot of part(c) in Fig. 6.20 (p. 445).

8.18 The following Bode plots represent the plants in unity-feedback systems with a proportional controller. A phase margin of 30° is specified. Estimate the gain of the controller and the steady-state error. Workdirectly from the plots, as though they represent experimental data. (a) Bode plot of Example 6.11 (p. 428).

8.3.

FREQUENCY RESPONSE METHODS

651

(b) Bodeplot of Example6.14 (p. 433). (c) Bodeplot for GI(S) in Example6.16 (p. 435). (d) Bodeplot of Example6.17 (p. 437). 8.19 Hand sketch Nyquist plots for the systems represented by the following Bodeplots. Indicate whetherthey wouldbe stable if placed as the forward path in unity feedback systems. (a) Bodeplot of Example6.12 (p. 429). (b) Bodeplot of Example6.13 (p. 430). (c) Bodeplots (both) of Example6.15 (p. 434). (d) Bodeplot of Problem6.35 (p. 452). (e) Bodeplot of Problem6.36 (p. 452). (f) Bodeplot of part (c) of Fig. 6.20 (p. 445). 8.20 Consider the system represented by the block diagram

in which~(t) is a velocity and x(t) is its integral, a position. (a) For the case with k = 0 (zero velocity feedback), plot the Nyquist contour, including any infinite radius segments, and deduce from there the numberof any destabilizing poles in the right-half plane. (b) State the net numberof clockwiseor counterclockwiseencirclements the -1 point that the contour modified by velocity (derivative) feedback must exhibit in order to represent a stable system. (c) Plot the Nichols contour for the systemwith k = (d) Plot the Nichols contour for other values of k; determine what value gives stability with an approximatepeaking of 3 db. (e) Confirmthe stability with a Nyquist plot.

652

CHAPTER 8.

INTRODUCTION TO AUTOMATIC CONTROL

8.21 Use MATLAB to plot the Nichols contour for the system

G(S)

(I+ S/IO)(I + S/50)(I + S/200)

with k = 1. If G(S) is the forward path in a unity-feedback system, estimate the gain k for which the maximum peaking is 6 db. (Note: This is the G(S) of Example6.11, pp. 427-428.) 8.22 Use MATLAB to plot the Nichols contour for the plant

c(s)

1 (1 + S/10)(1 + 0.0013SS2/lO, 000)

(which is the system of Example6.12, p. 429). Specify a lag controller with ~ = 10 that would allow the output to better track the input than would a proportional controller. Reportthe resulting steady-state error. 8.23 Carry out the above problem, substituting a PI controller. 8.24 Carry out the above problem, substituting a PID controller. 8.25 A PIDcontroller is applied to a particular plant in Example8.18 (p. 645). Lag-leadcontrol is not as effective, but substitute such a controller and at least achieve a fairly small steady-state error and plausible dynamics.The following steps are suggested: (a) Securea Nichols contour of the plant. (b) Examinethe contour with an eye toward finding an acceptable lead controller. The nose of a lead controller works best where the contour is most steeply sloped, so pay most attention in your selection of the nose point to the region below and to the left of the -1 point. Youobviously will need the maximum value of a available, which is assumedto be 10. Note that the maximum phase shift of the controller is about 55°, so you can’t go too far downthe contour. (c) Determinethe frequency of your selected nose point, using MATLAB. (d) Deducethe value of z, and write the transfer function for the lead compensator, leaving k = 1. Multiply this by the transfer function of the plant to get the transfer function of the combination. Secure a new Nichols contour for the combination,and see if it looks promising.(Thelag portion to comewill tend to push the contour back to the left somewhat, so it is best to leave somemargin. Also, issue the "hold on" command so you do not have to repeat the original plant contour.)

8.3.

FREQUENCY RESPONSE

METHODS

653

(e) The phase lag compensator comes next. Choose a nose frequency that is at least ten times (and perhaps more) smaller than that for the phase lead part of the controller, so as not to undo what already has been accomplished. The largest value of ~ available (10) should be used. (f) Determine p in essentially the same way you found z, write the transfer function of the lag controller, and multiply it by the result of step (d) get the overall open-loop transfer function. Plot its Nichols contour. (g) The final design step is to increase the gain as high as relative stability permits, in order to reduce the steady-state error. This also makes the system as fast as practicable. Report the maximumvalue of the closedloop frequency response from examination of the final open-loop Nichols contour. (h) Write the closed-loop transfer function, and insert it into the step commandto get a better comparison with the result of the PID controller in Example 8.18.

8.26 In Fig. 8.24 (p. 641) (and Example 8.10, p. 617), a mass-spring system is controlled with unity feedback and a phase lead controller. If the control implementation is changed from continuous analog to digital sampling with a zero-order hold signal sent to the plant, estimate from the plo~ the minimum sampling rate necessary to keep the Nichols contour from causing a peaking greater than 3 db. Ignore any digital filtering that might in fact require a higher sampling rate.

SOLUTIONS Guided

Problem

TO

GUIDED

PROBLEM

8.3

1. The following commands define the plant and give the Nyquist plot at the top of the next page (except for the infinite-radius circles, the text and the dashed lines): num=32.2*[-1 -15 50 750];den=It50 650 0 0]; nyquist (num,den) axis(I-5 5 -5 5]) The -1 point is encircled twice, clockwise. Since there are no poles of the plant transfer function in the right-half plane, this meansthat a unity-feedback system with this plant has two right-half-plane poles, and is unstable. If the segment of the locus that is movingrightward toward the origin were pushed downward,passing below instead of above the -1 point, there would be no encirclements and the closed-loop system would be stable. This idea is suggested by the dashed lines. Specifically, the contour would have to have its phase angle reduced below -180° in the vicinity of the -1 point.

654

CHAPTER

[

[

8.

INTRODUCTION

TO AUTOMATIC

ha_s zero net encirclements

The Nichols plot below (apart ~he subsequent commands

]

from ~he text and ~he d~hed line)

nichols (n~, den) axis ( [90 270 -20 20] NiChols arts

~o s

contour of ~ given system]

CONTROL

need contour such as \this to achievestability

Phase (deg)

follows from

8.3.

FREQUENCY RESPONSE

METHODS

655

Whena Nichols contour resides exclusively in the first and/or second quadrants, MATLAB (version 5.3) assigns positive values to the phase angles, which unfortunately makes the Mand N contours unavailable. Youshould interpret the phase angles as being 360° less than the reported values, so the -1 point is at the center of the diagram. The type of contour needed to achieve stability is indicated by the dashed line; the phase lead needed is small enoughto be provided by a phase-lead controller, as long as the nose of the controller is placed appropriately. 3. Thelarge size lead controller with a --- 10 is chosen. Withthis controller, the nose point on the contour gets pushed downwardby 10 db as well to the right by about 55°. This suggests choosing a point at about 5 db and -185° °. = 175 The commands [mag,phase, w] =nichols (num,den) [log10 (~), 20*logl0 (mag), phase] produce a table of the plotted points, from which the value log~o w .... or w .... = 10°s = 6.3 rad/s can be inferred.

= 0.8

4. FromFig. 8.23, a = vr~ and ~ .... =_ w .... /z = a. Therefore, z = 6.3/~/~-~ = 2.0 rad/s. The transfer function of the controller becomes Gc(S)=k

8+2 -~~6 "

5. The open-loop transfer function including the controller becomes 32.2k(-S 4 - 17Sz + 20S2 + 850S + 1500) G(S) = $5 + 70S~ ~_ 1650S3 + 13,000S2 + 0S + 0" With this G(S), MATLAB gives the contour on the right below:

656

CHAPTER.

8.

INTRODUCTION

TO AUTOMATIC

CONTROL

The plant contour has been repeated from above, and the nose points on the two contours are indicated.. Raising the contour would worsen the high-fequency peak where the contour passes directly below the -1 point. Lowering the contour would worsen the ’low-frequency peaking where the contour also passes somewhat close to the 6 db closed-loop contour. The best compromise might be to lower the contour by about 1 db, setting k -- 0.9. For present purposes, however, k -- 1 is retained, so no new contour is found. The characteristic equation of the closed-loop system becomes S5 ÷ 37.8S 4 ÷ 1102.6S 3 ÷ 13,644S 2 + 27, 370S ÷ 48,300 -- 01 The roots

command of MATLAB gives

the characteristic

values

-9.84 :t: 24.75i -16.1057 -1.01 :t: 1.79i The PD controller, as reported on page 627, produced a somewhat slower and significantly less damped low-frequency oscillation. The advantage of the lead controller results from the limit it places on the magnitude of the controller gain for high frequencies. The higher phase lead that the PD controller has at these frequencies is wasted. The denominator of the closed-loop system comprises the characteristic polynomial above, and the numerator is the same as for the open-loop system. The use of step gives the plot (solid line) 1.5

0.5 0

, ~--~..~.~

~

,

phase lead controller ~

,

,

PDcontroller

-0.5 -1.0

~

-1.5

starts at -1.81 with PDcontroller ~ , , , , , 2 4 5 1 3 time, seconds

_2.0 0

6

which represents the position of the bottom of the pendulum given a commanded step displacement. Notice that the bottom must start by moving in the opposite direction from the commandedfinal change, which should be intuitive (try it with a ruler or other stick). The slower oscillation can be seen be likely more critical than the faster oscillation, even though it is somewhat more damped, because of its effect on the settling time. The dashed line represents the inferior control provided by the PD controller (from Guided Problem 8.3). The instantaneous jump to -1.81 at t = 0 would require an infinite force, and is not realistic.

Chapter 9

Models with Static Couplers Static couplers include the simple bonds and the transformers and gyrators with constant moduli used throughout earlier chapters. The range of systems that can be modeled expands dramatically when more complex couplers are allowed. This chapter treats more general couplers that are static, exhibiting no energy storage or dynamics. Dynamic couplers are treated in Chapter 10. An ideal static coupler conserves energy; its output power equals its input power instantaneously. Further, it must be reversible in the thermodynamic sense: it generates no entropy. This is the definition of the ideal machine, which you know as transformers and gyrators. The modulus of such an element can vary, becoming a function of the displacement at either end of the element or a function of some remote displacement. Another useful class of couplers conserve energy but are not reversible. Irreversible couplers expedite the modeling of systems with heat conduction, including the heat generated by friction and electrical resistance. The proper fiow or generalized velocity for heat conduction is identified, and thermal machines and thermal compliances are introduced. Models with mass transfer or convection, on the other hand, are deferred to Chapter 12. Modelers often wish to employ two-port elements that, like one-port resistances, do not conserve energy. The activated bond is introduced as an elemental non-conservative coupler. More general active or dissipative couplers als0 are allowed, which can be either linear or nonlinear. The effects of these various types of static couplers on the dynamics of systems also are considered.

9.1

Modulated

Transformers

The definition of the ideal machine in Chapter 2 permits its modulus to be nonconstant, a function of one or more variables. The two most commoncases are a 657

658

CHAPTER 9.

MODELS WITH STATIC

COUPLERS

fork

disk

cone hand wheel

sheave Figure 9.1: Adjustable mechanical drives transformer modulated by a remote displacement, and a transformer modulated by a local displacement. Local displacements are the displacement variables on the two sides of the transformer; any other displacement is called remote. The distinction is very significant. 9.1.1

Remotely

Modulated

Transformers

The mechanical friction drives shown in Fig. 9.1 can be modeled as ideal transformers with moduli that can be changed by adjusting the fork, the handwheel or the axial postion of the wheel. (The disk-and-wheel drive may be familiar because of its commonusage on self-propelled lawn mowers.) The transmissions in vehicles are in the same catagory, although the modulations usually are discrete rather than continuous. The modulus of electrical transformers (or "variacs’) can be changed by adjusting a slide-wire that varies the turns ratio. Simple purely fluid transformers do not exist, but the moduli of many mechanical-tofluid pumps and motors are adjustable; one example is the variable swash-plate pump/motor shown in Fig. 9.2. In all these cases the modulating variable is a remote displacement, as contrasted to the rotational angle of one of the drive shafts, the electrical charge on one side of the transformer, or the angle of the pump/motor shaft or the integral of the volume flow. These later displacements are local. Somedevices mimic transformers but are not; they behave like transformers only as long as the modulusis held constant, at any value, but require significant

9.1.

MODULATED TRANSFORMERS

659

Figure 9.2: Pump/motorwith adjustable displacement~ swash-plate type Reprinted with permission and courtesy o] Eaton Corporation

poweror energy to eIfect a change in modulus. An exampleis given in Chapter 10. For the present, you should simply recognize that a true transformer is strictly a two-port device in terms of powerand energy. Anymodulationmust be achieved without any powerand energy. There is no generalized force associated with the modulation. A third port would be needed to represent a modulatiing force and its associated powerand energy, which the transformer does not have. Youshould carefully consider the examplesof Figs. 9.1 and 9.2. Is there any essential force required to effect the modulation?A real device alwayspresents somefrictional force to overcome,but this is theoretically infinitely reducible, and can be modeledseparately (by an added junction with. appendedresistance element) if it is deemedimportant, leaving a transformer ~s the core element. The control of large powerflows from motors, IC engines, turbines and electrical and fluid powersupplies dependsupon the existence of physical devices that very closely approximate the ideal remotely modulatedtransformer. Such devices are modeled,therefore, by transformers, with resistances added as appropriate to account for the minorlosses. 9.1.2

Locally

Modulated

Transformers

The camdrive system shownin Fig. 9.3, whenidealized to neglect friction and elasticity, is an exampleof a locally modulatedtransformer. The transformer modulusis the ratio of ~ to ~, and this ratio is a function of the position (angle) of the cam, or T = T(¢); ¢ is one of the two local displacements. Alternatively, you could specify T = T(x); this is an unwise choice for most applications, however, since there are two values of ¢ for most values of x. The following exampleis similar.

660

CHAPTER 9.

MODELS W’ITH

STATIC

COUPLERS

F

Figure 9.3: Drive with cam and roller

follower

Example 9.1 Show that the slider-crank mechanism pictured below can be modeled as a locally-modulated transformer, and express its modulus as a function of the crank angle, ¢.

Solution: The output position x is a function of the input angle, ¢, and therefore the output velocity } can be expressed as the product of a function of ¢ and ~, qualifying the device as a locally modulated transformer. The calculation of T(¢) is as follows. The geometry requires sin ¢ = L2 sin ¢, z = L1 cos ¢ + L2 cos ¢. From the first

equation, cos2 ¢ -- 1 - sin 2 ¢ = 1 - (L1/L2)2 sin 2 ¢,

so that the second equation can be written in terms of ¢ as x = x(¢) = L1 cos¢ + L2V/1 (L ~/L2) ~ si n2 ¢.

9.1.

MODULATE, D TRANSFORMERS

661

The modulusof the transformer is the ratio of ~ to ~: ~ = T(¢)~. This moduluscan be found by differentiating the expression for x above, with the result r(0)=-

l+L2__l_

Thedrive with the circular camand roller follower pictured in Fig. 9.4 acts just like the slider-crank mechanism.Asshownby the dashedlines, the effective length of the crank, L1, is the eccentricity of the cam, and the effective length of the connecting rod, L2, is the sum of the radii of the camand the follower. This assumes the camfollower maintains contact with the cam.

9.1.3

Increase

in the

Order

of a Model Due to Modulation

The order of a model with no modulated transformers or gyrators equals the number of independent energy-storage elements. Whena transformer modulus is modulated locally, however, the order often increases by one. This means that an additional first-order differential equation is needed. Onecase of this type is examinedbelow, followed by a counter case with no such increase and a third case in whichthere is a choice. All employthe slider-crank mechanism.

Figure 9.4: Drive with circular camand roller follower

662

CHAPTER 9.

MODELS WITH STATIC

COUPLERS

Example 9.2 A slider-crank mechanismis driven from a flywheel with significant inertia, and drives a load comprising a series connection of a spring and a dashpot. Other inertias are to be neglected.

Model the system with a bond graph, and find the state differential equations. Note the order of the system (the number of independent differential equations). Solution: The bond graph, annotated with the usual integral M:,..t lq~l-~-~- TI~..Fo Q~p/I

is

-------~R ’=J

x/C~

I

causality,

C=l/k

R=b

C

The effort of the 0-junction is labeled F, for convenience. The energy storage state variables p and x produce in standard fashion the differential equations

dt dx

= T¢¯-

1

=T -

1

The modulus T in the right sides of these equations is given by the result of Example 9.1 as a function of the angle ¢. To make the equations solvable, therefore, ¢ must be added to the list of state variables, which becomes p, x and ¢. The associated state differentiM equation c~ be seen from the annotations on the bond graph to be de

~

1 P"

The model, then, is third order: one for each of the energy storage elements, and one for the modulated transformer.

In general, a modulated transformer will add one order, unless the argument on which it depends is already a state variable or a function of state variables. This exception is exemplified in the following:

9.1.

663

MODULATED TRANSFORMERS Example 9.3

The slider-crank drive from the previous two examples is retained, but the crank is attached to ground through a spring (and therefore cannot be rotated indefinitely) and a substantive mass is placed at the right end: L M coi~ Again, model the system with a bond graph, and find the state differential equations. Note the order of the system. Solution:

The bond graph with the usual annotations

I=rn

is

C=l/k

R=b

The compliance (spring) renders the angle ¢ a state variable. Therefore, the argument of the function T(~b) is calculable, and the second-order model

dt dt

- T(¢) -

T(¢)

- ~-~P

+ RT’~(¢)

M-

M-

is complete.

Sometimes there is a choice of the order of a model for a given system. The minimum order of a model is the smallest order that can be achieved by choosing amongalternative state variables. Consider the following variation on the application of the slider-crank mechanism: Example 9.4 Model the slider-crank system below considering ~ (rather than M) as the causal input. Also, negelect inertia. Write the state differential equation(s) and identify the order of your model.

664

CHAPTER 9.

MODELS WITH STATIC COUPLERS

Solution: The bondgraph reveals a single energy-basedstate variable: C Fc=x/C~ SyI M=T(.Fc+F4 T I Fc÷F~- -~=T~

¢

F,~=~-~ R

The differential equationfor this x is dx = T~, T = T(¢). dt The modulation T = T(¢) introduces ¢ as the second state variable, with the differential equation

de d-~=,~(t). This equation appears to be an identity, but since ~(t) is given, it must integrated to get ¢. The simulation modeltherefore is secondorder. The mimimum order of the model actually is one rather than two. This is because it is possible to express T as a function of x rather than ¢, eliminating the need to compute¢:

T =T(z). In order to carry this out, the expression for x = x(¢) (found in Example 9.1, p. 661) must be inverted to give an equation of the form ¢ = ¢(x). inversion is so awkwardalgebraically, however,that almost surely it is wiser to stick with the second-order model.

9.1.4

Dependent Inertance Transformer

with

a Locally

Modulated

Anover-causal modelwith its dependentenergy-storage element and its differential causality can be treated either of two ways, as described in Sections 5.1 and 5.2. First, you mayimplementthe differential causality directly. Second, you mayeliminate the dependencythrough the use of an equivalent bond graph with one fewer inertance. Theseapproches can be applied to modelswith locally modulatedtransformers, as the following pair of examplesillustrates.

9.1.

665

MODULATED TRANSFORMERS Example 9.5 Consider a slider-crank mechanismwith inertias at both ends:

Modelthe system with a bond graph, apply causal strokes in the usual fashion, and write the state differential equations. Solution: The bond graph reveals that one of the two inertances must be assigned differential causality. The solution belowis based on choosing integral causality for I1, forcingdifferential causality for I2: S~’~:~’~1 ",1~"-"~" T ~ 1 J----:~’.- 0 ,~’-v-~-~R "~x=l~ T q/~t~

Q;pll~ Il

C

12

Note particularly the annotations for ~ = T~ and h~, which are mandated by the causal strokes. (5 and ~ are temporaryvariables used for algebraic convenience.)The differential equations for the two energy storage elements with integral causality becomedirectly dq 1 T 1 = ic - -ff~q = T(p - -h-dq, d-~ dp q + I2~ M- T dt "

---

)

Youneed to find 5~ as a function of the state variables p and/or q in order to place the r.ight side of the secondequation in solvable form. Since 2 = (T/I1)p and ¢ = p/I~, you can write T dp dT d¢ 1 T dp dT 1 ~

~r~~t + ~¢-~ T~p = Y1-~+ ~ ~ " Substituting this result, and collecting the two terms in dp/dt, gives dp 1 T -~q T I~ I’~ dTp.2 dt - I + (I2/II)T ( ~" Mde ]" Addingthe followingthird state differential equation to the first and fourth equations abovegives a solvable set with the state variables q, p and ¢: de

1

CHAPTER 9.

666

MODELS WITH STATIC

COUPLERS

Example 9.6 Find a set of state differential equations for the example above, except this time reduce the number of inertances so as to preclude any differential causality. Solution: You have a choice of relating the inertance energies to either ~ or ~. Electing ~, the total kinetic energy becomes ~ I 1(i 1 .~ 1 ~ .~ = = + T I~)¢-, T = T(¢). Were T a constant, you could simply define an equivalent inertance equal to I1 + T219_. If you try this now, however, you get an inertance which is a function of a displacement. The scheme for writing differential equations does not permit this; all inertances must either be constants or (if nonlinear) functions of the local generalized velocity. To accomodate this restriction, you insert a new transformer indicated as TI:

Se-~--~

i~)~’-

T~q/

OCI~)-’~R

c

7", @~P/Ix

I1 You could choose to use any constant inertance, but assume you keep the ’~. Setting this equal to original 11 as shown. The energy stored is ½T~Ix~b the proper value as given by the equation above for 7-, there results T1 = v/1 + T216./I1,

T = T(¢).

Upon examination of the annotated bond graph you now can write the complete set of state differential equations: dq 1 T 1 dt - I~ T~p- -~q’ dP- 1 (M--~Tq) dt T1 de 1 p" dt - I1T1

’

The result of this method is somewhat simpler than that produced by the method used in Example 9.5, particularly due to the absence of the derivative dT/d¢. It also is easier to find, and provides a more concise perspective of the essential dynamics of the system. The first method has one advantage, however: it is less sophisticated.

9.1.

667

MODULATED TRANSFORMERS

The general method, applied above to generalized kinetic energies or inertances, also applies to generalized potential energies or complianceswith differential causality, usually with less difficulty. Thereare the sametwochoices: using differential causality on the original bondgraph where necessary, or restructuring the graph through an energy analysis to eliminate dependent compliances. 9.1.5

Inertance Dependent Case Study

on Local

Displacement;

The technique employedin Example9.6 above can be applied also to the broader class of modelsin which an inertance dependson a local displacement. Specifically, the introduction of a modulatedtransformer into a bondfor an inertance allows the generalized kinetic energy being represented to be a function of the local displacement,while retaining the requirementthat the inertance itself not be a function of any displacernent. The procedure is best understood through a case study. The rigid rocker shownin Fig. 9.5 rolls without slipping on a cylindrical surface of radius R. Since the instantaneous center of the rocker changes with the attitude ¢, the inertance of the rocker is not constant but dependson ¢. The integral p = f Mdt, where Mis the momentacting on the rocker, is not a function of the state of the system, so any attempt to define the inertance I so that the velocity ~ equals p/I will fail. Instead, the insertion of a locally modulatedtransformer allows the inertance to be kept constant. The approach starts by writing the kinetic energy 7- as a function of the generalized displacement(in the present case, the angle ~b) and the generalized velocity (the angular velocity ~):

7- =½[y +m(b2 +

(9.1)

J is the rotational inertia, of the rocker about its center of mass;the first term on the right represents the rotational kinetic energy. The mass of the rocker is m; the other term represents the translational kinetic energy. Note that the displacement ¢ appears in this term. Uponexamination of equation (9.1) might seem reasonable to define the inertance as I = J + m(b.2 + R2~2), but if this is done the momentis not I~ and the angular velocity ¢ is not p/I with /5 = M. Therefore, you should not makethis definition. Rather, you define a new flow Oc and transformer modulusT such that the inertance is defined as a constant:

This gives the bond graph of part (b) of the figure with the simple constant inertance I = J. (9.3)

668

CHAPTER 9.

MODELS WITH STATIC COUPLERS

~

note: equilibriumis at ¢=0

enlarged ~" view

(a) geometry

C (b) bondgraph 80 ¢, degrees 60 40 20 0 -20 -40 0

1

2

3

4

5

6 7 8 time, seconds

9

(c) particularresults

Figure 9.5: Rigid memberwith planar surface rocking on a cylinder

9.1.

669

MODULATED TRANSFORMERS

Since this inertance is a constant, the flow 0c equals p/I where p is the generalized momentum associated with 0c, and p is the generalized force acting thereon. Supposeequilibrium occurs whenthe planar surface of the rocker is horizontal, that is ¢ = 0. The gravitational potential energyequals the masstimes the elevation of the center of mass, or V = mg[(R+ b)(cos~b- 1) + R~bsin ~b],

(9.4)

so that the momentis M = ¢ - d~ _ mg(R¢ cos ¢ - b sin ¢).

(9.5)

c de

Causal strokes nowmaybe applied to this graph in standard fashion, giving as the state differential equations ¢, ~-~ = de 1 d~ = ~--~P’

(9.6a) (9.6b)

whereT is as given in equation (9.2). If you prefer to find a single differential equation in terms of the angle ¢, the second derivative of ¢ can be found from equation (9.6b) by substituting equation (9.6a) into the right side: (9.7)

dt -~ = Tl dt IT 2 d¢ dt - , ,~¢- ~ d¢ \ dt

[

= j + m(b~ + R2¢~) g(R¢cos¢- bsin¢) + R2¢

] .

Solutionsfor various initial conditions are plotted in part (c) of the figure. Therocker is unstable if it is given too large an in[tiM displacementor velocity.

9.1.6

Summary

The powerin machineryand vehicles, etc., is largely controlled through the use of remotely modulated transformers. By definition, these elements have only two power ports and two bonds; the modulation is effected without any energy transfer through the modulatingvariable. Locally modulatedtransformers model the way manyengineering devices, particularly mechanisms,transmit generalized force and motion in a variable way depending on the generalized position of the device. Also, the kinetic energy of manyengineering devices, again particularly mechanisms,dependson their positional state. The standard schemefor writing state variable differential equations requires that inertances have constant moduli, or at most a dependenceon a local velocity. Use of the modulatedtransformer therefore greatly extends the domainof devices you can

670

CHAPTER 9.

MODELS WITH STATIC

ou~u.t position, x, inches ~ 1b I 01

0

COUPLERS

i~ ~ ~ i I

~ ~ i I

~ ~ i I

~ ~ i I

90 150 240 360 shah angle, ¢, degrees

Figure 9.6: The cam system of Guided Problem 9.1

model. Sometimes you start with knowledge of the constraint, and sometimes you start with knowledge of the stored energy. The modulation of a transformer usually adds one order, that is one firstorder differential equation, to the equaivalent model with no modulation. There is no increase, however, if the variable that would be added already is a state variable, or could be expressed as a function of these variables. Sometimes you have a choice of which set of variables to use; a set with the minimumorder is not necessarily the most practical. Guided

Problem

9.1

This first problem is intended to emphasize the primitive idea that locally modulated transformers display the same type of reciprocal behavior as transformers with constant moduli. The drive shaft of a cam systern pictured in Fig. 9.6 rotates at a constant 180 rpm, while the linear ouput displacement moves as plotted. The load force is a constant 200 lb, and the forces of inertia and friction are to be neglected. Sketch a plot of the torque on the drive shaft as a function of the angle of the shaft, labeling the values ¢ = 45°, 120°, 180° °. and 270 Suggested

Steps:

1. Drawa bond graph, labeling the moment, angular velocity, force and linear velocity. 2. Sketch a plot of ~ vs. ¢. Note that this plot comprises well defined straight-line segments connected by short smooth curved segments that are relatively poorly defined. 3. Write an expression for the conservation of energy, and solve for the moment. Alternatively, use your knowledge about ideal transformers. 4. Combinethe results

of steps 2 and 3 to find the desired moment.

9.1.

MODULATED TRANSFORMERS

671

deck pulley

arresting

engine ~

(below deck; conceptually)

~

~

shown ~./ ""

Figure 9.7: Aircraft arresting Guided

Problem

~"

[ [ W

// /

system for Guided Problem 9.2

9.2

It is important that you experience the determination of the modulus of the transformer in this geometric.problem. An aircraft landing on the deck of a carrier extends a hook that catches a cable stretched across the deck between two pulleys, as shown in Fig. 9.7. The cables extend below deck, where they are attached to an "arresting engine" that for present purpuses can be modeled as a giant adjustable dashpot in parallel with a giant (air) spring. (More refined details of such an engine are given in Problem 4.49, p. 257, but are neglected here.) The motion of the cable forces oil through an orifice, the area of which is continuously varied to shape the deceleration profile of the aircraft appropriately. This gives the effective dashpot. The displacement of the oil also compresses the air in a tank, providing the effective spring. The aircraft lands under full throttle, just in Case the hook doesn’t catch. Since the deceleration takes only a couple of seconds, the aircraft engine isn’t shut downuntil afterward. You are asked to model the system, and write a set of state differential equations. The problem is continued in Problems 9.13 and 9.14 (pp. 679-680) with simulations and design of a position-dependent damping profile to give desirable performance. Suggested

Steps:

1. The geometry of the pulleys makesthe velocity of the cable below deck less than the velocity of the aircraft. Represent the relationship by a locally modulated transformer. To find its modulus, define the two displacements at each end, express the geometric interrelationship algebraically, and compute the derivative of this relation. 2. Draw a bond graph model for the system, including the inertia of the aircraft, its thrust and drag, the cable transformer, the resistance of the variable "dashpot" and the compliance of the "spring." Define whatever parameters are needed.

672

CHAPTER 9. MODELS WITH STATIC COUPLERS

3. Applycausal strokes, define state variables and apply the usual annotations to the graph. 4. Write the differential equation(s), makingsure that the right side(s) your equation(s) are in terms of the state variable(s).

Guided Problem 9.3 This problem shows the important fact that dependent compliances can be treated in essentially the samewayas dependentinertances. The familiar slider-crank mechanismwith series spring-dashpot load, this time with negligible inertias, is driven by a significantly flexible shaft characterizable by a complianceCs. This system therefore can be represented by the bond graph SS

¢1

0

Cs

¢2

T

~

0

R

C

Use of one of the knownrelationships x = x(¢2) or T = T(¢2) is preferred use of the moreawkwardinverse relationships. Define a set of state variabies and write the state differential equations. Do this two ways: employdifferential causality as it results with the given bond graph, and combinethe two compliancesinto a single compliance. Suggested Steps: Observe that both coInpliances cannot be given integral causality; one of them must be considered a dependent energy storage. (This problem wouldvanish, ironically, if the inertia of the load wasnot neglected.) First, use the methodwhich carries out the instructions of the causal strokes including the differential causality. Either compliancecan be given integral causality; presumeyou chooseCs. Therefore, differential causality is applied to C in the process of completingthe casual assignments. The state variables becomethe qs for the compliance.Cs and ¢2 for the modulatedtransformer; their time derivatives can be written and circled on the relevant bonds. Applythe usual notation for the effort on the bond for Cs. Label all the remainingefforts and flows on the graph consistent with the dictates of the causal strokes. Be especially careful with the flow on the bond for C, which should equal C times the time derivative of the effort. The terms T and dT/d¢2 maybe left as they are; in an actual simulation, of course, it wouldbe necessaryto substitute the detailed functions of ¢2-

MODULATED

TRANSFORMERS

673

Write the differential equations for the two state variables, using the causal strokes and the effort and flow notations in the usual way. The differential equation for qs is integrable, simplifying the procedure. Carry out the integration, and substitute the resulting expression for qs into the differential equation for ¢2. Collect terms with the commonfactor d¢2/dt. The result is a first-order differential, equation with dependent variable ¢2 and independent variables ¢1 and ¢1. Nowyou can proceed to the second method. Define the efforts of the two 0-junctions as es and e, respectively, and write an expression for the total stored energy as a function of these efforts. The total stored energy can be represented by a single compliance bonded through a transformer to either of the two zero junctions. The following steps assume the left-hand junction is chosen. Drawthis bond graph, label I. the new compliance as C’ and the new transformer as T Equate the energy stored in CI to the sums of the energies stored in the original Cs and C. (This is most easily accomplished through energy expressions using efforts.) Any value of C’ can be chosen; ~ - - C is acceptable. Evaluate the necessary consequent value of T~. Also, relate dT’ / d¢2 to dT / d¢2. Redraw the new bond graph, apply causal strokes and annotate the efforts and flows using integral causality, in the usual way. Write the differential equation in terms of the state variable q on the bond for C. A differential equation for ~b2 is. needed, since the moduli of both transformers are functions thereof. ¢2 equals none of the flows on the graph, complicating the matter. The effort on the left side of the transformer T equals the actual effort on the actual compliance Cs, however and the difference q~ - ~2 is proportional to the time derivative of this effort. Write the corresponding differential equation, and integrate it to get q as a function of ¢1, ¢2 and T’. 10. Substitute the result of step 9 into the differential equation of step 8. The result should be a first-order differential equation in ¢2. 11. The differential equation should contain derivative coefficients dT/d¢2 and dT’/d¢2. Use the results of step 7 to express the latter in terms of the former. 12. Rearrange the differential equation for comparison to the result of the first method, as found in step 4. They should agree.

674

CHAPTER 9.

MODELS WITH STATIC COUPLERS

PROBLEMS 9.1 Sketch-plot the transformer modulus(i.e. speed ratio) for the adjustable conefriction drive below,as a function of the position of the idler shaft or wheel. (Get the end points and the shape about right.) ~r- ~"~..~one) 8..! ..................

2 ~ i.n

........... ~--

12

~ lengths

in

mm

9.2 Identify which of the three devices below could be approximated as remotely modulated transformers, which have two powerports only. Justify your conclusions. (a) variable pulley drive.

(b) electrical transformer with variable coil size and fixed magneticcircuit.

9.1.

675

MODULATED TRANSFORMERS

(c) electrical

transformer with fixed coils and variable magnetic circuit.

~----~ ~

¯ "-

top view

variable shunt reluctance

]

] front view

9.3 The hydraulic power supply on the left below provides a nearly constant pressure source of hydraulicfluid to drive several loads. Oneof these, as shown, comprises a "modulated hydrostatic transformer" (MHT)through which a load is driven. The MHTcomprises two back-to-back hydraulic pump/motorswith displacements D1 and D2 which are adustable by means of the control lever labeled with the angle/9. In particular, D1 = Dosin 0 and D2 = Docos 0.

accumulatorf~r’~to other loads...t 0 (adjustment)

motor

-/pump

tank (same)

" "tank

Define key variables and parameters and give a model structure. The rotating memberwithin the MHT (not shown)has significant rotational inertia, and there is someexternal leakage (to the reservoir) at both ends of the MHT. Internal leakage maybe neglected.

676

CHAPTER 9.

MODELS WITH STATIC COUPLERS

9.4 The Scotch-yoke mechanismbelow has a crank arm 0.05 mlong, and drives a load compressionforce of 500 N for the rightward stroke and a load tension force of 100 N for the leftward or return stroke. Find the modulus of the transformer which can model the system, and plot the momenton the crank as a function of its angle. Neglectfriction.

~load

9.5 A motor with a torque-speed characteristic as plotted below drives a cam with inertia Jc, whichin turn drives a follower systemwith inertia Jr, spring constant K and dampingcoefficient B. The shape of the cam can be described by a knownfunction ¢I = f/~(¢c). Other effects maybe neglected.

(a) Forma bond graph modelof the system, and relate the moduli of its elements to the given information.

(b) Define state variables, and give the order of the system.

(c) Writea set of state differential equations in a formsuitable for simulation.

9.1.

MODULATED TRANSFORMERS

677

9.6 Consider the slider-crank mechanism with the series RC load as shown in Example 9.2 (p. 662), but with all inertances neglected. Computeand plot the driving momentand the displacements of both ends of the spring as functions of time, assuraing the effective crank starts in the fully extended position, runs ¯ with ¢ = constant = 10 rad/s for 90°, and then stops dead. The parameters are as follows: r -- 6 in, L = 10 in, k = 20 lb/in and b = 1.0 lb s/in. All masses are negligible. 9.7 The effort el of the system represented by the bond graph below is specified, the modulusof the transformer is a function of the displaceinent ql, and all other parameters are invariant. Identify the order of the system, define state variables, and write a complete set of first-order differential equations.

9.8 The bond graph of the preceding problem models a hydraulic positivedisplacement pump with shaft angular velocity ~1 and torque el, volumetric displacement T = To ÷ T1 sin(6ql) that includes sinusoidal pulsations due the individual pistons (To and T1 << To are constants), rotor inertance I1, fluid inertance I~, a fluid compliance C due to an accumulator and a load resistance R. Repeat the problem for the case in which the accumulator and its compliance are eliminated. 9.9 A Scotch yoke is driven from a flywheel and in turn drives a mass, linear spring and linear dashpot as shown.

cotch yok~

(a) Model the system with a bond graph, and express all moduli in terms of the given parameters. (b) Compare your model to that given for the system shown in Example 9.5 (p. 665). Commenton the applicability of the solutions deduced for that system.

678

CHAPTER 9.

MODELS WITH STATIC

COUPLERS

9.10 A cylindrical disk of mass md and radius r is mounted on a shaft with eccentricity e tO form a cam drive with a fiat-faced follower, as shown below. The follower system has total mass mf, and a linear spring with rate k and a linear dashpot with coefficient b are attached.

(a) Model the system with a bond graph, and state the moduli of the elements. Rotation of the drive shaft is resisted by a linear resistance, R, but friction between the cam and the follower may be neglected. (b) Apply causal strokes. Can. integral causality be employed on all of the energy storage elements? (c) Write a set of state differential equations. Warning: Special care must be taken to adcount for differential causality in the face of a non-constant transformer modulus. The results should be nonlinear. Also, make sure the number of first-order differential equations equals the number of state variables. (d) Report the order of your model, and the minimumorder of the system.

9.11 Continue the preceding problem using the parameters md = 1 kg, R = 0.1 N.m-s, r = 0.1 m, e. = 0.03 m, rnl = 3 kg, k = 100 N/m and b = 5 N.s/m. Carry out a simulation assuming that the applied moment is a constant M= 2.0 N.m. Find the minimumvalue of the spring precompression necessary for the cam to remain in continuous contact with the follower. (You are not asked to analyze the behavior for higher torques and speeds.) Plot the angular velocity ¢ and the normal contact force between the cam and the follower as functions of time. Hint: Use of the MATLAB d±ff function can expedite the plotting of the contact force. 9.12 A plastic disk of radius b, thickness w and density pp rolls back and forth without slipping on a fiat horizontal surface. A steel cylinder of radius a, length L and density Ps is pressed into a hole in the disk centered a distance c from its center. The equilibrium position is shown. To save you work, the mass of a solid plastic disk (without the hole) defined as rap, and the excess mass for the steel cylinder is defined as ms: mp= 7~wb’2pp;

ms = ~ra2(Lps - wpp).

9.1.

MODULATED TRANSFORMERS

679

The center of~mass of the system is defined to be at a distance r from the center of the disk, r = cms/(mp + ms), and the mass momentof inertia the parallel axis theorem,

of the system about the center of mass is, from

~ = mp (~b~" + r~’) + ms [~a2 + (c- r)2]

(a) Express the gravity potential energy of the system as a function of the angle of rotation from the equilibrium, ¢. (b) Express the kinetic energy of the system as a function of q~ and (c) Draw a bond graph for the system that employs a constant inertance, and define this inertance and other moduli therein. (d) Write state differential

equations.

(e) Simulate the motions following release from rest at the angles of °, 60° and 120° (relative to equilibrium) for the parameters b = 3 in, w a = 1 in, L = 2 in, c = 1.5 in, ppg = 0.0430 lb/in 3 and psg = 3. 0.283 lb/in (f) Qualitatively identify any limitations of your model for very large angles or large velocities. (g) If you have access to a physical model of this system, verify the predicted cycle times experimentally. 9.13 Carry out a simulation for the aircraft landing on the deck of a carrier, as described in Guided Problem 9.2. The aircraft weighs 35,000 lbs, the thrust also is 35,000 lbs. The initial velocity is 180 ft/s, at which the drag is 5000 lbs. The drag is proportional to the square of the speed. The two deck pulleys shown in Fig. 9.7 are 100 ft apart. The arresting engine essentially comprises a set of pulleys to reduce the velocity of a large piston to which the cables are attached, an orifice through which oil is forced due to the motion of the piston,

680

CHAPTER 9.

MODELS WITH STATIC

COUPLERS

and an air tank that gets compressed by the displacement of the oil. Therefore, the force of the dashpot on the cable can be approximated as a~2, where s is the displacement of the cable where it is attached to the dashpot. (a) Determine the coefficient a which brings the aircraft to a halt in distance of 300 ft. Let the spring constant be fixed at 300 lb/ft. Plot the velocity as a function of time. Does the result represent satisfactory performance? (b) Design problem. It is possible for the coefficient a to be varied as a function of the cable position s (by continuously varying the area of the orifice). Through trial-and-error or other means propose and demonstrate a desirable function a = a(s). Hint: A solution with a good terminal "flare" will require reducing the spring constant, k. 9.14 Repeat the above problem using the more refined model of the arresting engine given in Problem 4.49 (p. 257). The diameter of the hydraulic piston 1 ft. and the initial volumeof the air in the tank can be set at 13 ft 3. You will need to choose the charging (initial) pressure, Po, in the tank. The cross-head assembly weighs 2000 lbs not including the pulleys, each pulley weighs 75 lbs, and the hydraulic fluid weighs 50 lb/ft 3. Neglect the mass of the cable. For part (a), find an acceptable combination of Po and the fixed effective fixed orifice area for the fluid restriction, A0. For part (b), choose a reasonable, value for (of necessity smaller than in part (a)) and plot the orifice area as a function the cable displacment. A solution to Problem 4.49 should be provided to you.

SOLUTIONS Guided

Problem

1.

TO GUIDED

PROBLEMS

9.1

M. .-

F ~ CAM DRIVE

x

~ 2.-%in/sl

~ 360xlS0×3=5424060x60 360 90 150 \¢.de~ 360 x 180 x(-3) =_27 60 x 120 Given:~=360x 180/60= 1080deg/s; F =200lbs; ,~ as plotted above. 0~ 0

I .0

Ol

IIi I,~ 90150

240 d \¢,deg

360 ’,

9.1.

681

MODULATED TRANSFORMERS

Guided Problem

9.2

1. J/~~2

x~ ~= -- =T~ 8

Therefore, R

2-3.

S~ F~

RTp/I~

c

,%

-~ = Ft - Fd -- T(x) dx

T = T(x) =

X

+ L’~/4

Ft is the thrust of the engines Fd is the air drag on the aircraft I is the mass of the aircraft = m R is the (variable) resistance of the dashpot C is the complianceof the spring p + ~s

1

pd--~=i

Guided Problem 9.3

d~,

1 [

d--~-

q,

C ( l dq, 1 dT d¢a ~ +-~ = ~ ~ T dt T~ d¢~

1 C dTd¢~ C dqs - Te~IC, CsT3 d¢.~ dt) qs ÷ T’~Cs dt 4. qs = ¢1 - ¢~. Substituting this into the second equation, 1+ + ~---~-~-~¢~ =1---hil T:RC8,[ CsT----~d¢--~C dT(¢I-¢2)+T--~Cs CI d¢2-~z-nws1 1 2 1 2 5. V = ~Cse~ + 7Ce

+¯ TC_~c~

682

CHAPTER 9.

MODELS WITH STATIC

6-7,

v: 1c,

(e~2=

Therefore,

C’ C ~ = Cs + ~-

COUPLERS

~C~e~+ 1

1 C~ 1 lettingC r=C, Tr ~= C +T2’ Computingderivatives, this also -2 dT ~ -2 dT gives Tt3d¢~ ~d¢~ - T ~ dT T~3 dT or -dee Ta dee or,

e,=T’qTC e 21" ~R Sz~--,--0 --------,.~. T e,/RT 2 ~! T.2 e~/RT ¢1-es /RT

d.__~’ = T’ ~)I T’2 q’ dt T2 RC

~

T’ q ’/C ~@ C 9. ~

~-~2=C,~t~

=C~

Integrating, ¢~ - ¢~ = ---~---q, giving q = (¢~ - ¢~). ’eC dT’dt (¢~ - ¢~) = T’~ T~RC (¢~ - ¢e) I0. dq~dt ~C~T’(~ - ~) C~T ~ 11. Substitution of the relations between T’ and T and between dT’/d¢~ and dT/d¢~ found in step 7 into the above equation gives

+ + ~(e~

+

~ = ~6~ +

which, whenmultiplied by C/C~, is identical go ghe result from the firse me~hod(step 4).

9.2

Activated

Bonds

The bond graph models employed thus far have demanded the conservation of energy, except for the one-port sources and sinks (resistances) where energy is explicitly created or dissipated. The junction structures, including simple bonds, junctions, transformers and gyrators, have been strictly conservative. Occasionally one wishes to model a coupler that violates the conservation of energy, however. This need may arise because the details of the actual system are both too complex and too unimportant to justify faithful representation; an overall simplification suits the purpose better. The activated bond is the simplest non-conservative coupler.

9.2.

ACTIVATED

683

BONDS SYSTEM [~

signal

INSTRUMENT

-~

CONTROLLER

Figure 9.8: Instrument with activated coupling The use of such a coupler is illustrated in Fig. 9.8. An instrument measures a pressure or a velocity, etc, and thereby stimulates an electromechanical or hydraulic control system. The power drained from the system by the instrument likely is trivial, but including it in the model would introduce unnecessary complexity in the form of a much higher order model, etc. Yet, the measured signal plays a major role in the behavior of the system; it cannot be ignored) You want the system to power the instrument, but you do not want the instrument to drain power from the system model. Activating the bond between the system and the instrument accomplishes this objective. The activation and its directivity is indicated by the arrow in the middle of the bond. 9.2.1

Definition

and Application

The role of activated bonds can be grasped through their application to electronic amplifiers, which often are modeled as non-energy-conservative devices. The power supply is ignored, so what is actually a three-port device is treated as though it were a two-port device. The circuit symbol for an electrical amplifier is a triangle, as shown in part (a) of Fig. 9.9. A model in the form of equivalent circuit is given in part (b), showing a voltage gain, T, and input and output resistive impedances R1 and R.~, respectively. In the model, the power is produced by a dependent voltage source as if out of nothing. (The power supply is not directly described.) Further, any effect of the output current on the input is neglected. The bond graph for this voltage amplifier is shown in part (c) of the figure. The bonds on both sides of the transformer (either side alone will do) are activated, as indicated by the full arrows drawn near their centers. The meaning of the activation is suggested in part (d) of the figure. activated bond is a partially broken power bond. On the upstream (left) side of a bond emanating from a 0-junction the flow (current) is zero, and thus the power is zero. On the downstream (right) side of the bond the flow (current) is not zero, but equals whatever the downstream side of the system demands. The efforts, on the other hand, are identical, as for a regular bond. Thus the bond itself creates or annihilates power; it does not conserve energy. 1Theterm "signal" is used by systemsandcontrol engineersto focus on a single variable, excluding its conjugatevariable andthe associated powerfrom immediateand possibly ultimate consideration. Arrowsare employedin block diagrams(see for exampleany of the several examplesin Chapter 8) to indicate the directivity of signals. Thearrows in block diagramsand signal flow graphs are closely related to the arrowson activated bonds. These relationships are examinedin the Section7.4.

684

CHAPTER 9.

MODELS WITH STATIC COUPLERS

(a) electric circuit symbol(b) electric circuit equivalent el

(c) bond graph

el e2 /-~ 1 ~ T-~-- 0 ,---re-

1 (d) activated bonds i1~

(e) outputcurrentproportional to inputcurrent el t’~’~’2 ~ 0 "*’,-T’-’--- 11

e2

R2 (~ ou~mcurrent propogional to inpm voltage

(g) case (a)-(c) -oo

Figure 9.9: Activated bonds and electronic amplifiers

9.2.

ACTIVATED BONDS

685

For an activated bond emanatingfrom a 1-junction, the flow is identical at both ends; the effort is zero at the upstream end but not at the downstream end. Oneapplication is to a current amplifier, as shownin part (e) of the figure. Anamplifier in which the output current is proportional to the input voltage, using an activated bondand a gyrator, is shownin part (f) of the figure. If it is desired to neglect the effect of the input impedance,that is to assume that the amplifier draws no powerat all from its input circuit, the resistance R1 can be simply removed. Examplesare shownin part (g) of the figure. It important to retain the upstream 0 or 1-junction in the bond graph; otherwise there is ambiguity regarding whether the efforts or the flows at the two ends of the activated bonds are equal. Sometimesone needs to insert a 0- or 1junction into a bondgraph purely to avoid this kind of ambiguity. If the effect of the output impedanceof the amplifier is to be ignored, the resistance R~. and its bond also maybe removed, and the junction there also maybe removedif nothing but the load is attached. Fluid flow with an irreversible jet can be modeledwith the help of activated bonds, as the following exampledemonstrates.

Example 9.7 The water tank shownbelowis filled with water from a pipe whichdischarges below the waterline. The system shownon the right differs by the removal of the section of the pipe downstreamof the restriction, which becomesa nozzle. Modelthe two systems with bond graphs. p

1

nozzlerestriction ¯~. :.:;

.~~restriction

Solution: The first system can be modeledby the bond graph on the left at the top of the next page, in whichthe resistance to flow of the restriction is represented by an R element, the gravity head pgh is represented by an Se element, the gravity complianceof the tank is represented by a 6’ element and the interconnection between the pipe and the tank is represented by a 0-junction and its attached bonds. Note that the pressure P1 is affected by both the height z and the pressure P2 whichis proportional to the depth of the water in the tank.

686

CHAPTER 9.

MODELS WITH STATIC

COUPLERS

el

C

R

C

For the second system, the pressure P1 is no longer affected by either the height z or the pressure P’z. Rather, P1 is affected only by the resistance, R. On the other hand, the tank is still receiving the flow Q1. Energy is being dissipated in the free fall of the water between the level of the nozzle and the level of the water surface in the tank, and the tank is receiving power of magnitude P2Q~. This power is referenced to the level of the bottom of the tank. The new situation is represented in the bond graph by removal of the gravity head effort source and activation of the bond interconnecting the two junctions, as indicated by the full arrow. This activation cuts off all recognition by P1 of downstream effects; the effort on the upstream end of an activated bond driven by a flow (i.e. emanating from a 1-junction) is zero. It ~naintalns the flow interconnection of a regular bond, however, and allows the effort on the downstream side to be whatever it needs to be to satisfy conditions there. The power P2Q1is generated by the bond itself.

9.2.2

Causality

The causal stroke applied to an activated bond unambiguously shows whether the equality of effort or of flow is being enforced:

flow effort

equality: equality:

1 ~ 0 ~

~ ~

~ (zero effort on left side) ~ (zero flow on left side)

The application of causal strokes and the writing of state differential equations follows in a normal manner, as illustrated by the following examples. In general, the differential equations for a system with activated bonds are the same as they would be without the bond activation, except for the deletion of the effort or the flow at the upstream end of each activated bond.

9.2.

687

ACTIVATEDBONDS

Example 9.8 Causal strokes have been applied to the bond graph of the first system in Example9.7, assumingthe pressure P1 is the input variable and the pressure P’2 is the output variable. Applycausal strokes to the bond graph for the second system, assuming the same input and output variables, and write and comparethe state differential equations for the two systems. Solution: Note that integral causality is used for the complianceelement in the first system. If the sameis done for the secondsystem, the causality of the activated bond becomesas shownbelow: P1 -~ 1 ~----~P20

-~-~

C This is compatiblewith flow equality at both ends of the activated bond, as given above, and zero effort on the left side of the bond. This is precisely what is intended. Both systemsare first order and have a single differential equation with state variable V. In both cases this equation can be written initially as dV Pn d---~= Q~- Q~- R Q2, R=R(P~). The difference lies in the pressure P~. In the case with the submergedpipe, the bond graph reveals that P~ is the weighted sumof the efforts on three bonds, as follows: Pn = P~ + pgh - P2 = P~ + pgh - V/C. In the case with the cut-off pipe, however,only one bonddetermines PR: Pn = P~ The resulting differential equation is changedonly by the deletion of the additional terms in the equation for Pn resulting from the forced setting to zero of the effort at the left end of the activated bond.

The representation of gear friction is simplified throughthe use of activated bonds, as the next exampledemonstrates.

688

CHAPTER 9.

MODELS WITH STATIC COUPLERS

Example 9.9 Agear pair often is assumedto have a constant efficiency, 7. M~

Assumingthat the driving poweris from the left side,

M252 = Thus, the relations between the two speeds and the two torques can be expressed in terms of ~? and a transformer modulus,TV:

= M~ = TC M2 = TMM~;

TM= Tell.

Represent this model by a bond graph. Solution: There are in effect two constant transformer moduli: T¢ is the ratio of the speeds, and TM is the ratio of the moments.This leads to either of the two alternative modelsbelow, dependingon whether the input variables are ~ and M~or M~and ~2:

Note that should the direction of the powerbe reversed, the equations must be inverted and the moduli of the transformers changed. Certain other mechanical drives such as pulley and chain drives can be represented similarly. When~? = 1, the two transformer moduli becomeidentical, so the two activated transformers becomethe sameas a single ordinary or bilateral transformer.

9.2.

689

ACTIVATED BONDS

o roof ~tackpipe

Figure 9.10: Toilet for GuidedProblem9.4 9.2.3

Summary

Anactivated bondis like a simple bondexcept that the effort or the flow at one end is definedas zero. Thepowerat this end therefore is zero; an arrowis placed in the middle of the bonddirected awayfrom this end. Activated bonds permit simple representation of modelsthat purposefully violate the conservation of energy. Theyare particularly useful for modelingamplifiers, the attachment of low-powertransducers to high-powersystems, free-falling liquids and the friction in drive systemssuch as gear pairs. The activation of a bond changes the differential equations simply by the deletion of terms associated with the bondvariable that is forced to equal zero. Guided Problem

9.4

This is an excellent problemto cut your teeth on. It can use all of the basic elements including an activated bond, and deserves somenonlinearity; yet the result should be fairly simple. It is continuedin Problem9.20 (p. 694). Consider the traditional toilet system shownin Fig. 9.10. Modelthe basic system, neglecting any refill of the tank or the bowl. Define variables, and identify any elements which ought to be nonlinear, showingthe functional dependence. Write a set of state differential equations, defining parameters as needed. The valve can be considered to open widely, but there are flow restrictions around the rim of the bowl. Swirl maybe neglected. The stack pipe is so large that water never forms a plug in it. For simplicity it maybe assumed that wheneverthe channel betweenthe bowland the stack pipe is flowing, it is full of water.

690

CHAPTER 9.

Suggested

MODELS WITH STATIC COUPLERS

Steps:

1. Note that the state of the bowl has no effect on the state of the tank. Model the tank accordingly. Define needed parameters, and identify a key nonlinearity. 2. Note that the flow out of the bowl is impeded by the restricted area of the channel and possibly by the inertia of the flow. On the other hand, there is a distinct siphoning action. Complete the model of the rest of the system, defining parameters as needed, and note the key nonlinearities. equa3. Apply causal strokes to your bond graph and write the differential tions. All that should be missing from your result are numbers for the fixed parameters and specific functions for the nonlinear parameters, although you ought to be able to give reasonable forms for the resistances.

Guided

Problem

9.5

The electric circuit of Fig. 9.11 illustrates the use of a bond graph model of some complexity, including meshes and activated bonds. The modeling and the reduction is given; you are asked to write a set of state-variable differential equations. Assume that node a is grounded. Suggested

Steps:

Copy the bond graph of part (d) of the figure. Apply integral causality. Temporarily, overlook the presence of the bond activization. Note that the graph is undercausal. Apply the usual annotations on the bonds with integral notation, which includes the definitions of state variables. Annotate the input effort as ei, also. Add a bond with a virtual compliance to the 0-junction on the left side of the activated transformer, as befits an undercausal bond graph. Label its flow with a circled zero and its effort as e, in the usual manner. Applying integral causality to this compliance then dictates the causalities of all remaining bonds. Note that the resulting causalities on the bonds for the activated transformer are consistent with the bond activation, and in fact are required by it. The use of the virtual compliance therefore is not mandatory, but it resolves in routine fashion the usual difficulty associated with undercausal bond graphs, which the graph retains despite the activation. Annotate all of the bonds with the proper symbols as dictated in the usual manner by the causal strokes. The flow on the bond to the left of the activated transformer should be designated as 0, as the consequence

9.2.

691

ACTIVATED BONDS

(a) circuit

(b) bondgraph as first drawn -~- 0 -~- 1 --~ 0

at0 -

~

1 --’-~R1

1 "-’~-0-~-TI’~-I’----’-~T2-----~0---~ R,3

R 2

C~’~---1 "~"12

(c) eliminationof simplemeshes

0 -~"- 1 --~ 0 ~ 1 ------~ 0 -~"- 1

1 ----- 0-~--- 1 ----’- O-~-Tt-~-1-----R~ 0 (d) node (a) grounded

c, (e) node(b) grounded

Figure 9.11: Guided Problem 9.5

692

CHAPTER9.

MODELSWITH STATIC COUPLERS

of the activation. Someof the annotations maybe too long to fit in the confines of the graph, so you maywish to define one or more temporary symbols.In particular, it is suggested that you define the effort on the 0-junction toward the lower-right part of the graph as Writethe five first-order state differential equations in terms of variables labeled in steps 2 and 3. The causal strokes identify the proper terms, and the power convention arrows determine the proper signs, as always. Note that these ec)uations are not solvable until the effort e on the virtual complianceis defined in terms of the five state variables and the input voltage. Write the algebraic equation associated Withthe virtual compliance. This equation should contain the unknowne in two places. Collect these two terms, and solve for e as a function of the state and state variables. Substitution of this ~esult whereindicated in the state differential equations renders these equations solvable. PROBLEMS 9.15 Drawa bond graph for the electrical amplifier shownbelow, which has voltage gain g and resistive input and output impedancesR1 and R2, respectively.

9.16 A DCmotor with constant field, torque-to-current ratio G, constant applied voltage e0 and armature resistance R drives its load through a gear train with speed reduction ratio T and efficiency ~. Drawa bond graph of the system, and find a general relation between the output torque and speed of the gear reducer in terms of the given parameters. 9.17 Showthat a bond graph model for the system of Problem9.16 can avoid the use of activated bondsmerelyby changingthe effective value of R, presuming that the i of the graph need not represent the actual electric current. 9.18 A stream with given flow Qi feeds a dammedreservoir. Someof the water flows through a power-generatingturbine, and someoverflows a sluice gate at ¯ the top of the dam.Both of these flows exit to a bodyof water so large that its level is invariant. Drawa bondgraph of the system, employingelements with as simple characteristics as possible. (It is suggested that your resistance element fox the sluice gate be defined so that its characteristic goes throughthe origin.) The case with no flow over the sluice gate maybe omitted from consideration.

9.2.

ACTIVATED BONDS

693

~ _ _ ~_.__

~.~,uice

reservoir

~erflow dam~~

~..~.

gate with

9.19 Manytypes of transistors used in amplifiers can be modeled by the bond graph below. Find the associated transmission matrix.

9.20 A decorative water display comprises an upper pond fed by a water fountain, a waterfall overflow into a lower pond, and a hidden recirculating pump, as pictured below. You are asked to prepare for an analysis to determine how quickly the system approaches steady-state when started from rest.

area A~--

(a) Define appropriate variables and parameters, and model the system with a bond graph. Make appropriate simplifications, but do not ignore key nonlinearities. Relate the parameters of the bond graph to the direct physical parameters which are given or you have defined.

(b) Write a set of state differential

equations that describes the dynamics.

694

CHAPTER 9.

MODELS WITH STATIC ¢OUPLERS

9.21 Design problem. Carry out the simulation of the toilet given in Guided Problem9.4. Chooseparameters by trial-and-error or any other meansuntil you are satisfied with the response; report the physical dimensionsimplied by these parameters. Then add a refill feature to your design and simulation. (Note that conventional toilets have a separate refill flow.) Other features maybe added, if desired. Considerations should include but not necessarily be limited to the time required, the vigor of the flush and the water usage. Governmental regulations have reduced water consumptionin stages from 5 gallons per flush to 3.5 to 1.5 gallons per flush, requiring very careful design. 9.22 Design problem. A small passenger vehicle is to be poweredby an IC engine connected through a 2:1 reduction spur gear pair to a variable displacement pump.Hydraulic flow from a reservoir through the pumpgoes to hydraulic motor/pumps geared by other 2:1 reductions to the axles of the four drive wheels. A hydraulic accumulator also is attached to the pressure line from the pump in order to store excess energy from the pumpand energy from braking. This allows the engine to be smaller than it wouldhave to be otherwise. The throttle is adjusted automatically to keep the torque of the engine virtually constant, which improves fuel efficiency. The volumetric displacement of the pumpis varied automatically, within limits, to attempt to keep the pressure in the accumulator constant. The driver controls the volumetric displacement of the hydraulic motors, effectively determining the thrust force on the vehicle. When this displacement is madenegative, the motors becomepumps, and the resulting braking energy is stored in the accumulator rather than being dissipated. A relief valve prevents the pressure in the accumulatorfrom exceeding 3500psi for safety, although such a pressure should not be reached in normal operation except during braking. The speed of the engine is limited automatically to 500 rad/s, and the torque of the engine is allowed to decrease in order to prevent its speed from falling below100 rad/s. throttle control of torque [ pressure

gears~

control

,

, driver control accummator ..... o~ a~sp~acement

~

~ relief

~

~__~va.lve ,I, reservoir

wheels ~ load

Yourultimate task is the specification of the size of the accumulatorand the maximumdisplacements of the pumpand the motors. Other parameters are given. The performancespecifications are for you to choose, within the range of possibilities. Youwill defend your design based on its predicted performance. The vehicle complete with engine (but not the accumulator, the driver and the payload) weighs 2500 ibs. The engine characteristics and fuel consumption

9.2.

ACTIVATED BONDS

695

are as plotted in the statement for Problem4.53 (pp. 273-274). The air drag, tire properties and accessory power are as given below that figure. (Other parametersor features stated there, including the weightof the vehicle, the axle gear ratio, drive train losses and the possibility of changingthe size of the engine do not apply here.) The accumulatorwill weigh 20 poundsfor each gallon (231 in3) of total internal volume. The gear pairs have an efficiency of 99%. The hydrualic pumpand motors can be approximated as having a friction torque equal to a constant plus an added torque proportional to speed; the constant equals aD and the linear resistance equals bD, where D is the maximumvolumetric displacement and a = 20 lb/in 2, b = 0.50 lb s/in 2. Theyalso suffer leakage Correspondingto a constant leakage resistance of c/D, wherec = 400 lb s/in ~. Note that since one side of each machineis at the pressure of the reservoir, or zero psig, there is no operational difference betweeninternal and external leakage. The work can proceed in the following phases. You maybe asked to work in a small group. (a) Modelthe system with a bond graph, neglecting the gear losses and the auxiliary powerconsumption. Consider the accumulator to be a simple nonlinear compliance, and neglect minor inertances and compliances. Evaluate all parameters or functions in your model except for the key accumulator volumeand machine volumetric displacements. (b) Choose an operating torque for the engine and nominal operating pressure and charging pressure for the accumulator. Use this information, plus other information given above, to choose tentatively the control of the volumetric displacement of the pumpand its maximum value. (c) The engine speed ~ is controlled by the fuel throttle. Whenthe accumulator pressure is low, the engine should be .driven fast, and when it is high, the engine should idle. Choosean automatic control function that relates the engine speed to the pressure. (d) Choosea maximum desired thrust on the vehicle, and with information developed earlier choose a tentative maximum volumetric displacement of the motors (or the sum of their displacements). (e) Developa computersimulation of the vehicle, and demonstrate its behavior starting from rest on a level road with maximum thrust and a full accumulator. Youwill have to assume somevolumefor the accumulator, and mayprefer to keep it large enough so it doesn’t empty before the vehicle reaches highwayspeeds. (This avoids your having to change the model.) (f) Simulate, as you think appropriate, various schedules of acceleration, braking and hill climbing, in order to refine your tentative choices for the volume of the accumulator and the maximumdisplacements of the pump and motors.

696

CHAPTER 9.

MODELS WITH STATIC

COUPLERS

(g) Repeat the steps above, including the consideration of the power loss to the accessories and the friction in the gears. (h) Report your results according to whatever instructions are given you. Include a discussion on the merits and demerits of the basic scheme.

SOLUTIONS TO GUIDED PROBLEMS Guided

Problem

9.4

Vt is the volumeof the water in the tank. ~ !Q,----~Rt Qt is the volume flow rate out from the tank; Qt = -~t. G -v~ Ct = At/pg, where At is the tank area. R is the nonlinear resistance to the flow, which occurs largely in the small orificies aroundthe rim of the bowl. to bowl

F

I

Guided 1-2,

Problem

9.5

The flow Qt enters the bowl, which has a c0m’pliance such that Pb = Pb(Vb) is not linear. This pressure equals the pressure due to the resistance of the outlet channel, Rb, plus the pressure due to the inertance of the outlet channel, I, minus the constant siphon head, pgh. Rb is likely close to a nonlinear resistance of the Bernoulli type.

9.3.

NONCONSERVATIVE

dql 1 1 d’-~ = R1C1ql + -~-[~ei dq2 1 dqa

697

COUPLERS

dpl 1 dt - C~ q~ e dp~ 1 d-~- = e~ -

1

Setting the flow on the virtual complianceequal to zero, -~

11 12 R~ Solvingfor e, T2

[ql

dFq3

(

-ei

+ ~ + ~ -e

R2)pl

"2p21

e = T,-~ T. 2 -~-~l - ei Ca+ R3+~’~2~ "~"~+~"~’~’~j R~ p~ R~ T~ This givese~- T, + Tz -e~ + ql+ q~ ~ + R~ - ~ I~ T~Tz ~ These terms ~e substituted into the differential equations ~bove.

9.3

Nonconservative

~

Couplers

Engineers often wish to model highly complex two-port devices in an approximate summaryfashion, without reticulating them into a junction structure with one-port elements attached. Dynamic hydraulic machinery such as the pumps and turbines pictured in Fig. 9.12 are examples. One port is the mechanical shaft; the other is the net fluid power with its commoninput and output flow and its pressure rise or drop. This section considers general static rnodels of the form ~

el STATIC ql

2-PORT

e2. q2

in which Rf is called a two-port resistance used as a case study.

or field.

el

--O~

e2

Rs ~ ~2

The impeller pump will be

698

CHAPTER 9.

MODELS WITH STATIC

COUPLERS

Guidevanes

(a) radial-flow turbine

(c) radial-flow (centrifugal)

(b) axial-flow turbine

(d) bond graph

Fig~e 9.12: Examples of dynamic pumps and motors Adapted from Engineering Applications of Fluid Mechanics by J.C. Hunsaker and B.G. Ttightmire, copyright 1947, with permission from The McGraw-Hill Companies

9.3.

NONCONSERVATIVE

699

COUPLERS

i~ MI__M~_IMPELLER ~P PUMP

INDUCTION..~e~ MOTOR

LOAD iQ~,-

~,,,

~o

Figure 9.13: Bond graph for impeller pump with dynamic source and load

9.3.1 Causal Relations Four possible causal patterms are possible: impedance

e.b----~-Rf ql

e.---~

el =

e2 (9.8a)

el

admittance

~Rf~

e2 (9.8b)

immittance

: e.~---~-R ql

1 ~ el = el(ol,e~); q2

(9.8c) adpedance: e,-~-~---~Rs e,---~-~--I 01 = ql q~

e2 =e2(e1,02) (9.8d)

The environment of the coupler determines which causal pattern to use. As an example, consider an induction motor driving an impeller pump which in turn forces water through a load: IND.

MOTOR

M ~ -

P IMPELLER

PUMP

Q -

HYD. LOAD

If transient behavior is of interest, the inertia of the motor and the attached pump rotor, and the inertia of the fluid being pumpedis important. This gives the bond graph shown in Fig. 9.13. Application of integral causality shows that the pump should be represented by the impedance causality given in equations (9.8a). The manufacturer of the pump almost surely provides information graphical form only. 9.3.2

Equilibrium

The determination of equilibrium is considered, as a first step, using graphical characteristics. Each of the relationships within equations (9.8) can be generalized to the form R(Xl,X2,X3) = 0, where R can be read as "relation of." Represented graphicMly, this means a family of characteristics; one variable is plotted on the abcissas, another on the ordinate, while the third becomes a discrete parameter. Twosuch families of curves are needed to characterize

700

CHAPTER 9. MODELS WITH STATIC COUPLERS

the pump. There are manychoices, any of which is acceptable for purposes of finding equilibrium. Twochoices are particularly convenient, however,since they match with the ways the characteristics of the motor and the load are represented. Youare accustomedto representing a hydraulic load with a pressure-flow (P - Q) characteristic; presumethe load curve is as plotted in part (a) of 9.14. Similarly, you are accustomedto representing the motor with a torquespeed (M- ~) characteristic; presumethe motorcharacteristic is as plotted part (b) of the figure. The same coordinates can be used to represent the pump. The first of these therefore is either R1(P, Q, ~) = 0 or R1 (P, q, M)= 0; the formeris -convent!onial, and is shownin part (a) of Fig. 9.14. The secondof these is either R2(M,¢,Q) = 0 or R2(M,~b,P) = 0; the former is closer to most commonly available information, and is shownin part (b) of the figure. The resulting equilibrium state for the system can be found in two steps. First, the intersections of the load characteristic with the various P - Q curves are cross-plotted onto the M- ~ plot, as shownby the dashed line. Second,the equilibriumis noted as the intersection of this line and the motorcharacteristic, as noted by the cross. This result also is cross-plotted onto the P - Q plot, as shownby the cross there. This is a general procedure, whichapplies to finding the equilibriumof any two-port device that connectsa source to a resistive load. Pumpmanufacturers rarely report characteristics in a form as complete as those given in Fig. 9.14: unfortunately. Morelikely, they give a single P-Qcurve for a particular speed ¢, and a single M- ~ curve for a particular flow Q. They also maygive results on efficiency, althoughthis informationis redundant,as will be shown.Althoughthis situation is not particularly defensible, at least without further explanation, it results from the manufacturer’simplicit assumptionthat the application engineer can very closely extrapolate the given information. How can you rise to this challenge without using a more specialized approach than necessary? As a systems engineer you cannot be expected to becomeexpert on all types of engineeringdevices. 9.3.3

Dimensional

Analysis

It will be assumedthat the inlet pressure ("suction head") is high enough preclude cavitation, and that bearing friction is negligible. For a set of geometrically similar pumpsof different sizes, any characteristic length (typically the diameterof the impeller), the fluid density p and the viscosity # are dimensional parameters that may need to be combined with the variables P, Q and ~ to define the state. Thusthe impedancerelation of equation (9.8a) becomes M=M(~,Q,L,p,#),

(9.9a)

P = P(~, Q, L, p, #).

(9.9b)

9.3.

701

NONCONSERVATIVE COUPLERS

120 100 P, PS~o

60 40 20 0

0

I00

200

300 Q, in3/s

400

(a) pressure-flowcharacteristics

160 M, 120 80

k,,/foo 50

40

i

0

20z

40z

~, rad/s (b) torque-speedcharacteristics

I

I

60z

Figure 9.14: Steady-state matching of motor, impeller pumpand load

702

CHAPTER 9.

MODELS WITH STATIC COUPLERS

Various nondimensional groups can be formed from the dimensional variables and parameters. Perhaps the most useful is M ( P

(9.10a)

Q pQ)

pQ) (LQ--~

(9.10b)

where f and g represent as yet unknownfunctions of their two arguments. With this representation the results should apply regardless of the size of the pump or the density or viscosity of the fluid. The group pQ/L#is a Reynoldsnumber, which is proportional to .the ratio of forces due to inertia to the forces due to viscosity. Mostfluid machineryis operated at so large a Reynoldsnumber, that is so large a flow or so small a viscosity, etc., that forces due ~o viscosity are virtually negligible. Assuming this applies to the pump,equations (9.10) specialize to the case of Eulerian similitude: ’

pLS~ P

,

Q

Here, f~ and g~ represent Nnctions of a single variable or the group which often is called the flow eoe~cient of the pump. The flow coe~cient is proportional to the ratio of the velocity of the fluid at somelocation to the velocity of the tip (for example)of the impeller. The e~ciency of the pump,~, is the ratio of the output powerto the input power, or PQ-(Q/L~)g’(Q/L~)

:h( Q ).

(9.12)

Therefore, if a manufacturergives you just two curves or equations, each interpretable in terms of any two of the three equations (9.11a), (9.11b) or (9.12), you can deduce the third curve or equation directly, and go on to plot the two families of curves in Fig. 9.14 and the equations in any of the forms above. The relations used in the present case are given in Fig. 9.15. The characteristic ~. length used in Fig. 9.14 is L = 1 ft, and the fluid density is p = 1.94 lb s~/ft 9.3.4

Dynamic

Simulation

For the unsteady case of Fig. 9.13, the operating points on the motorcharacteristic and the pumpcharacteristic (in part (b) of Fig. 9.14) no longer coincide; one is directly abovethe other, as required by the left-hand 1-junction, and the difference is the momentthat accelerates or decelerates the motor. Similarly, the operating points on the load characteristic and the pumpcharacteristic (in part (a)) no longer coincide; one is directly above the other (because of

9.3.

703

NONCONSERVATIVE COUPLERS

1.6 1.2 0.~ 0.4

0

0

~

2lb s

~ ~,, I

I

I

I

2

4

6

8

I

I

10 12 14 Q/L3~b Figure 9.15: Dimensionlessbehavior of impeller pumpassumingEulerian similitude right-hand 1-junction), and the difference is the pressure causing the fluid to accelerate. The assumptionof Eulerian similitude allows ~he state differential equations to be written with the only nonlinear functions for the pumpbeing the f’ and g’ of equation (9.11). The causMstrokes dictate --= dt

M=

-

pLS~)~_f,( O

(9.13a)

’(

Q )-P~(Q).

(9.13b)

The prominenceof Q and ~ on the right sides of these equations suggests their use as the state v~riables in place of Pm= Im~ ~nd pQ = IQQ. Thus the equations become d~_dt I,,~I [Mm(~b)-pL5~b~f’(LQ-~o) ’ ] (9.14a) dQ- I [pL2~bZ g’ (L-~)-PR(Q)]dt IQ

.

(9.14b)

This result is implemented for a starting transient in Fig. 9.16. The following equations are used to represent the static characteristics of the motor, pumpand load, whichconformclosely to the plotted characteristics in Figs. 9.14 and 9.15: 4 - Ms~ ~ - ~, Mm~-~ -/~/o + ~¢1~- M4~ M~ 4 q ~ (Q/L3~) x 10

(9.15a) (9.15b)

704

CHAPTER 9.

MODELS WITH STATIC COUPLERS

250

2OO 150 100 50

0

~

0

I

0.2

I

0.4

I ~

0.6

I

0.8

I

1.0

|

I

1.2 1.4 t, seconds

1.6

Figure 9.16: Simulation of start-up of impeller pumpsystem .g’ = 0.174 - 0.000328q2 _ 0.0000281q3, y = 0.2 - 0.0236q + 0.00158q2 _ 0.000053q4,

(9.15c) (9.15d)

f’= qg-~’ -4 x 10

(9.15e)

PR=Po+P2Q2, Po=301b/in 2, P’2=Sx10 -41bs2/ins, -61bins4, M0=601bin, Ml=l.21bins, M4=1.453x10 MD=-1.82x10-slbins5, M6 =6.35x 10-111bins6.

(9.15f) (9.15g)

Theother parametersare taken as L -- 1 ft, p = 1.94 lb.s2/ft 2, Im = 20.05 lb’ft’s and IQ = 8310lb.s-~/ft ~. (The last numbercorresponds to a uniform channel of cross-sectional area 0.007 ft 2 and length 30 ft, whichby itself producesa load characteristic close to that given.) The simulation assumesthat fluid is not allowedto flow in the reverse direction, presumablythrough the use of a check valve. Therefore the flow remains at zero until the rotor has sufficient speedto overcomethe static back pressure of 60 psi. Thereafter, the flow accelerates very rapidly, and the equilbrium found previously is reached in less than 1.5 secondsfrom the time the motoris started. 9.3.5

Linear

Couplers

A linear static coupler, defined in terms of variables that are symmetricwith respect to the coupler, el R I e2.’-~’~--, q2 ~1

9.3.

NONCONSERVATIVE

can be represented Y:

COUPLERS

by an impedance

705

matrix

Z or an admittance

matrix

This element also is known as a two-port resistance field, explaining the traditional subscript f. In order to represent the linear static coupler by a more detailed bond graph, it is convenient to introduce a special two-port resistance element called the mutual resistance, which is designated here as R12: e~

R12 ~

e2

el

R1~4~,

=

e2 = R1241.

(9.17)

This appears to be very similar to the gyrator: e~ 0~

G ~

el

= G~2,

q2

e2 = G4i.

(9.18)

The gyrator, however, has anti-symmet~c power conventions on its two bonds, whereas the mutuM resistance has symmetric power conventions. This is a radical difference, since it makes the gyrator conserve energy and the mutual resistance consume the power P = e~ + e~2 = 2R~24~.

(9.19)

The mutual resistance is strictly symmetric with respect to its two ports; the gyrator is strictly anti-symmetric. The linear static coupler now c~ be represented as shown in part (b) of Fig. 9.17, from which you can see that

or

el -~ Rill q- (R12 q- G)O2,

(9.20a)

e~ -- R2~2+ (RI~. - G)~I,

(9.205)

I

Z= R12

R1 - G

RI~ + G] " R2

Three special categories of this coupler deserve attention. coupler, or coupler that satisfies reciprocity, has transfer and Z~ that equal one another. This happens if and only if G = 0,

(9.21) The reciprocal impedances Z12

(9.22)

so the gyrator is absent. The mutual resistance is inherently reciprocal, whereas the gyrator is not. Wholeclasses of systems, including ordinary electric circuits, are reciprocal (which shows why gyrators never are needed to model ordinary electric circuits).

706

CHAPTER 9.

MODELS WITH STATIC

COUPLERS

(a) symbol for two-port resistive field G (b) reticulation using mutual resistance element R 2

(c) reticulation using basic elements only

Figure 9.17: Representations of the general linear static

coupler

Symmetric Couplers are by definition unchanged if they are reversed endfor-end. Thus, not only are their transfer impedances equal to one another, the self-impedances Zll and Z22 also equal one another, or G = 0,

RI = Re.

(9.23)

Symmetrytherefore imposes reciprocity, but not vice-versa. The third category is the passive coupler, defined so that it never generates energy. The power of the general linear static coupler is 79 ----- elO~q- e_9~_9---- [R1~1+ (R12q- G)~_9]~1 + [/~2~2q- (~12-2 + 2R~z4102+ R.,4.~. = R~4~

(9.24)

This is never negative, regardless of the values of 41 and 42, if and only if 2 R~ _> 0,

n2 _> O, R12 _< ~

(9.25)

Violating the third condition gives 79 < 0 for ~1 = -42- Note that the presence of the gyrator has no effect on passivity, which is to be expected since the gyrator is inherently conservative. The bond graph of part (c) of Fig. 9.17 is equivalent to that of part (a), avoids the use of the mutual resistance. The equivalence can be established and the proper values of T and R1~ found by equating the power dissipation in the two models: 2 R~O~+ 2R12~1~2+ R.~O~= RI’0~ + R.~(T(I1 02) = (Ra’ + R.~T2)O~+ 2TR.~,~aO.~+ R~.(I~.. (9.26) 2Stated most succinctly, passivity applies if and only if the matrix Z is positive semidefinite.

9.3.

NONCONSERVATIVE

COUPLERS

707

The result is T = R12/R2, RI’ = RI - Rift/R2. Note that the conditions for passivity

(9.27a) (9.27b)

become simply

R~_> o, R2_> o,

(9.28)

which should be apparent from the bond graph.

9.3.6

Summary

A general static coupler can be characterized by two relations, each of which relates a different set of three of the four effort and flow variables. Each relation can be represented graphically as a family of curves. There are four causal interpretations, depending on which pair of variables is considered as independent, leaving the other pair as dependent. The determination of equilibrium and the unsteady behavior of models with static couplers has been illustrated. The four causal interpretations can be represented by 2 × 2 matrices of constants if the model is linear and unbiased. (There also are two acausal transmission matrices, as developed in Guided Problem 9.6.) Many linear couplers are reciprocal, which means that their two transfer impedances or two transfer admittances are equal. Examples include all two-port electric circuits comprised of linear components are examples. A linear coupler is symmetric, end-for-end, if it is reciprocal and its two self-impedances or two self-admittances are equal. A static coupler is passive if it cannot generate power. All these conditions can be represented as special cases of the bond-graph models given in Fig. 9.17. Dimensional analysis is a powerful tool for deducing what format or formats can be used to characterize a couplex static coupler, depending on the knowledge of what parameters and variables are relevant. Guided

Problem

9.6

This problem investigates the characteristics of a fairly general nonlinear coupler, and its matching to a source and load. A small boat is driven by a propeller attached directly to the drive shaft of an IC engine. The characteristics of a family of geometrically similar propellers are plotted in Fig. 9.18 part (a). Theefficiency of the propeller is ~/, its diameter is d, the shaft has momentMand angular velocity ~, the boat travels at velocity v and the water has density p = 1.94 lb s2/ft 4. The characteristic of the engine at full throttle is given in part (b) of the figure; the torque may be assumed be proportional to the percent throttle. The engine speed should not be allowed to exceed 450 rad/s in order not to produce excessive wear. The drag of the boat Fd, which includes a planing phenomenon,is given in part (c) of the figure. (a) Estimate the diameter of the propeller within the given family which gives maximumsteady speed of the boat, and determine this speed and the angular velocity of the shaft.

708

CHAPTER 9.

MODELS WITH STATIC COUPLERS

1.1 ~ ___ 100____9.0._M_~ 0.9 (a) propeller characteristics 0.7 0.5 0.3 0.06

I

I

I

I

I

0.08

0.10

0.12

0.14

0.16

0.18

120 (b) engine characteristics

80 M ft-lb 40

0

I

0

i

100

I

I

200

I

i

300 ~, rad/s

I

400

500

(c) drag force of boat

400 force, Fd lbs. 300 200 100 0

10

20

30 40 50 speed,v, ft/s

Figure 9.18: Guided Problem 9.6

9.3.

NONCONSERVATIVE

COUPLERS

709

(b) Superimpose on part (c) of the figure sketch-plots of the thrust force Ft vs. velocity for both full and one-half throttle, assuming the diameter found in part (a). (Identify three or four points for each curve. purpose is to enable investigations of other loading conditions, such as occur during acceleration.) Also, estimate the steady speed that results for one-half throttle. (c) Comment on whether the family of propellers described seems appropriate for the particular boat. (Alternative shapes are available with different ratios of the "pitch" to the diameter.) Also, might there be some merit in choosing a larger or smaller diameter propeller within the given family? Explain.

Suggested

Steps:

If the family of propellers described is close to have maximumspeed when the engine produces thing very close to it. Determine the associated ity. Also, the efficiency of the propeller should its maximum.

appropriate, the boat will maximumpower or sometorque and angular velocnot deviate radically from

Get a rough idea of how fast the boat might go by assuming that the efficiency is maximizedwhen the velocity is maximized. Do this by finding the corresponding value of ¢ and #. From # you can get the tentative diameter, and then from ¢ you can get the speed. Since you know the power and the efficiency you also can find the thrust force. Plot this point on the force-velocity diagram. Is the point near the actual drag-speed characteristic? If so, you are close to a solution; if the deviation is very great the proposed family of propellers likely is inappropriate. Choose an additional two or three values of ¢, and tabulate the corresponding values of #, d, v, r/and thereby the thrust force Ft. Plot these forces on the force-speed diagram and find the desired solution at the intersection with the drag curve. Use interpolation to find the diameter. Choose three or four points on the each of the two engine characteristics. For the chosen diameter, tabulate the consequent values of #, and thereby the corresponding values of ¢ and z/. This should give the points to plot on the force-speed diagram, and the steady-state speed for one-half throttle. Do the results suggest that a hysteresis phenomenonexists for some range of throttle positions? To answer the second question in part (c), consider the fuel consumption and the possible need to travel frequently at speeds slower than the maximum.Does this suggest a different strategy for optimization?

710

CHAPTER 9.

~

e~ 1

~

e2 LINEAR

MODELS WITH STATIC

COUPLER

e3 e4 ~ G ~.

COUPLERS

e5 0 ~.

R2 Figure 9.19: Cascade of static

Guided

Problem

elements for Guided Problem 9.7

9.7

This problem applies the acausal transmission matrix, introduced in Chapter 2, to the linear static coupler. It is particularly useful when one wishes to reduce a cascade comprising a coupler and one or more other elements. Recall that the transmission matrix is defined as follows; note in particular that unlike the causal matrices above, the power conventions of the two bonds are anti-symmetrical:

Find the four elements of the transmission matrix in terms of the four elements of the impedance matrix. Also, find the criteria for reciprocity and for symmetry in terms of the elements of the transmission matrix. Finally, find the overall transmission matrix for the cascade shown in Fig. 9.19. Suggested

Steps:

Expand the matrix relations for the impedance matrix and the transmission matrix into scalar algebraic equations. Change the sign of ~.) on one of these, so the equations refer to the same variables despite the different directions of the power conventions on the right-hand bonds. Compare the equations of step 1 to find the elements of the transmission matrix in terms of the elements of the impedance matrix. (You also might wish to do the inverse, and keep both results on file for future use.) To find the criterion for reciprocity in terms of the elements of the transmission matrix, define the determinant of the transmission matrix as A, evaluate it in terms of the elements of the impedance matrix, and substitute the condition for reciprocity. To find the criterion for symmetry, note that reciprocity must apply and that T~I --- T22. Alternatively, invert the transmission matrix, and change the signs of the two flows in the result. Since a symmetric system looks the same when inverted end-for-end, setting the new transmission matrix equal to the original transmission matrix gives the criteria for symmetry.

9.3.

NONCONSERVATIVE

COUPLERS

711

5. Find the tranmission matrices of the combinations of elements on either side of the coupler element in Fig. 9.19. 6. Multiply the three transmission matrices, in the same left-to-right order in which they appear, to get the overall transmission matrix. (Note that one could, if desired, convert the result back to impedance matrix form.)

PROBLEMS 9.23 Assume that the load of the system shown and analyzed in Fig. 9.14 (p. 701) and the associated text is a. nozzle located at some elevation above the pump, as in our water sprinkler system. Assumethat the nozzle is lowered to the same distance below the pump that originally it was above the pump. The length of the piping, etc.~ is unchanged.

(a) Find the new equilibrium operation point, assuming no pulley drive, using the induction motor. (b) Write linearized equations for the characteristics and the load about the operating point.

of the motor, pump

9.24 A pulley drive is placed between the induction motor and the pump of the system characterized in Fig. 9.14 (p. 701). Determine the pulley rati6 that maximizes the flow, and plot the corresponding equilibrium state. Losses in the pulley system may be neglected. Hint: Maximumflow means maximum power into the pump. 9.25 A propulsion system for a boat comprises an IC engine, ducts and an internal pump that takes water ingested in a forward-facing port and forces it out, at a higher velocity, from a backward-facing port. (Thus, no external propeller is needed, increasing safety and allowing operation in very shallow water.) The system has been analyzed to give the characteristics plotted on the next page. The engine is described, for the throttle position of interest, by the indicated engine torque-speed characteristic. It is connected to the pmnpby a gear pair, of undetermined ratio, that can be assumed to be frictionless. The pumpand duct system is described by two families of curves, one relating the pump-shaft torque, pump-shaft speed and craft speed, and the other relating the propulsion force, craft speed and shaft speed. The drag of the boat is represented by its own force-speed characteristic. Determine the maximumspeed that the boat can go, and the corresponding gear ratio. You might start by drawing a bond-graph model.

712

CHAPTER 9. MODELS WITH STATIC COUPLERS

600

~

500

200

thrust and drag vs. speed

7O 60 M 50 ft lbs 40 30 2O 10 400 ~, rad/s

500

600

pumptorque vs. pumpspeed, and engine torque vs. engine speed

9.3.

NONCONSERK&TIVE

9.26 The characteristics

e1

713

COUPLERS

of the components of the system e ~//r 2-PORT #1~...~.~. [1~’"-.-~0 e~._..,~__~ ~ 0 ~ 2-PORT #2 ~q~ q2

1-PORT are plotted below. Find the characteristics for the overall system. (It is suggested that you sketch with the aid of tracing paper.)

ec

. 0/l

j;2

e2=O

=2

/_.,-/-:7, . ~c for ~ ~,-~Jc for ------

¢C

9.27 A standard open-center symmetric hydraulic spool valve is connected to a symmetric hydraulic ram as shown on the next page. The pressure from the 2. power supply is fixed at 1500 psi, and the annular area of the ram is 2 in The differential pressure across the two components is P, the flow is Q and displacements of the spool and ram from their centered positions are y and x, respectively. The position of the spool depends on the balance of an applied force Fs and a spring-like flow-induced reaction force in the valve, which acts like a spring with a stiffness of 379 lb/in. The pressure-flow characteristics of the valve and the load are plotted below. (Note: The valve is shown at about two-thirds of full scale, whereas the ram is shownat only one-third of full scale.) (a) Draw a bond graph for the system, using the word element SPOOL VALVEfor the valve. Include the fact that Fs depends only on y (and vice-versa).

714

CHAPTER 9.

MODELS WITH STATIC

COUPLERS

(b) Find the equilibrium velocity ~ and force F for the spool force F8 0.68 lb. (c) Linearize the characteristics for small perturbations Fs* about the equilibrium. Find the resulting motion of the load if its effective mass corresponds to a weight of 1000 lbs and F~* = 0.1sin20t lbs, where t is in seconds.

pump

,,

to sump y=30×lO "~ in

X

-1500

-1000

~ -500

0

,-6 500 1000 1500 P1-P2, psi

9.28 For the linear coupler represented by the three impedance matrices below, (i)

Z= [21 ~] (ii)Z=

[20 ‘ (ii i)Z= [23

(a) Determine whether the systems are reciprocal, metric.

~]

passive and/or sym-

(b) If the systems are electric circuits, state whether they could comprise a network of resistors, or a network of resistors with a voltage amplifier or power supply.

9.3,

NONCONSERVATIVE

715

COUPLERS

(c) Find the parameters of the equivalent bond graph with a mutual resistance, as shownin part(b) of Fig. 9.17 (p. 706). (d) Find the parameters of the equivalent bond graphs with basic elements, as shown in part (c) of Fig. 9.17. 9.29 Linear two-port systems have been described in terms of the transmission matrix T, the impedance matrix Z and the admittance matrix Y. Following the causal definitions given in equation (9.8) (p. 699), an immitance matrix and adpedance matrix L also can be defined. (a) Relate the variables of a linear two-port element by means of (i) immitance matrix, and (ii) the adpedance matrix. (b) For each of the four combinations of two-port elements below, one of the four types of causal matrix simply sum to give the overall causal matrix of the same type. Determine which matrix applies in each case. ...,...4.2-PORT

#1 ~

-~--- 1 (i) -----,.-aii)

---.-~2-PORT

#2

1 .......~2-PORT #1 ~ ~2-PORT #2 ~

(ii)

0 ~ 2-PORT # 1 ~ 0 ~ ~2-PORT #2

~ 0 ..,....~2-PORT 0 ~

(iv)

"~2-PORT

#1 #2

(c) Various transformations relate the transmission and causal matrices. As an example, determine the elements of the immitance matrix in terms of the elements of the transmission matrix. Assumeall power conventions are directed from left to right.

SOLUTIONS

TO GUIDED

Guided

9.6

Problem

PROBLEMS

1. The maximum en.gine power occurs at or very close to its maximum allowable speed, at which ¢ = 450 rad/s, M= 88 ft-lb. The maximum efficiency of the engine is about r/-- 0.708. (Numerical values in this solution were determined by carefully scaling a blown-upcopy of the given plots.) 1000~1 2. ¢---- v-~=0.14; #_= 2 --0.688. de pdS~ For the conditions of maximum engine power, therefore, k,~" " ] = 0.688 x1.94 v = 0.14 x 0.799 x 450 = 50.3 ft/s.

x (450) 2. =0.799ft=9.59in.

716

CHAPTER 9.

MODELS WITH STATIC

COUPLERS

0.708 x 450 x 88 = 557 lb. 50.3 This point is plotted below, and is seen to be close to the load curve but not to lie onit. pointsfromstep 3: * ~ full throttle pointsfromstep 4: from step 2-~-~.~ 500 .......... point x

,x 400 force, F e lbs. 300 200

half throttle

100 0

0

10

20

30 40 50 speed,v, ft/s

The case above is given as the first row in the table below: ¢ 0.14 0.12 0.15 0.16

~ # = 0.224/d 0.688 0.835 0.600 0.505

0.7990 0.7686 0.8211 0.8499

50.34 41.50 55,42 61,19

0.708 0.687 0.685 0.639

557 656 490 414

Connectingthe four points on the plot showsa solution at about v = 54.4 ft/s, from whichFt ~_ 505 lb,d = 0.8166 ft, ¢ _~ 0.148, # ~ 0.617. Using d = 0.8166 ft mad choosing three different engine speeds for the two throttle conditions, ~, rad/s i M, ft-lb 450 88 44 400 98 49 350 3OO

v, ~/s F~, lbs

¢ 0.6171 0.3085 0.8698 0.4349 0.6202 0.8970

0.148 0.1735 0.1145 0.1661 0.1478 0.1104

0.694 0.512 0.675 0.599 0.695 0.661

54.4 63.76 37.40 54.25 42.24 27.05

505.2 159.0 707.5 216.4 308.1 416.8

These points also are plotted on the force-speed diagram above to give thrustspeed characteristics for the two throttle conditions. The equilibrium speed for half-throttle is about 44.3 ft/s. It can be seen that a similar characteristic for somewhatless than half throttle would give three equilibrium speeds; the outer two are stable and the center is unstable. This indicates a distinct hysteresis; no speed in the range betweenabout 25 to 35 feet per second could be maintained in steady-state.

9.3.

NONCONSERVATIVE

717

COUPLERS

Thefuel efficiency of the engineis greatest at distinctly less than full throttle. Whenthe consideration of miles per gallon of fuel is of considerable concern, therefore, one might prefer to emphasizeperformance at less than full speed. This implies shifting the speed at which the propeller has maximum efficiency to distinctly less than full speed, whichcan be done by using a smaller diameter propeller. Unfortunately, however, this can require the engine to exceed its desirable maximumspeed, causing excessive wear. A compromise could be sought, but it is hard to beat a propeller diameter for which maximum engine speed produces maximum boat speed. The only very practical way to increase et~iciency at the lower speeds would be to change the style of the propeller, which largely meanschanging its pitch-to-diameter ratio. (The given family of propellers has a nominalratio of 1:1.) Guided

Problem

9.7

1. Thetransmission matrix relation gives el = Tile~ + T12(1", ~1 = Tzlee + T22~ The impedancerelation, using the same ~ as above, is

(1) (2)

e~ = Z1~1+ Z~(-~)

(3)

ee = Z~Q~+ Z~e(-~e)

(4)

1 Zze . 2. Equation (4) can be rewritten ~ ~l = --e2Z~, + --Z~ q2 Comp~ing equations (2) and (5): T~ 1/ Z~; T~ = Z~/Z:~ Substitution of equation (5) into equation (3) gives

(5) (6)

Z:~. ] Z~ J which when compared to equation 3. Reciprocity

applies

(1) reveals

Z~= Z~: T~I = ~; T~ Z~ Z~z.

when 1 = Z~ = 1 (Z~Z~

z~

z~

~ ~ T~

_ Z~ Z~ T~ = T~,T~ - T~T:~ = a. ¯ Z~ Z~ Z~, 4. For symmetry, A = 1 and T~l = Tz2. This can be derived by inverting the transmission matrix to give

and changingthe signs of the two flows to give

The system is symmetricif ~d only if this transmission matrix relation, which describes right-to-left behavior, equals the original transmission matrix relation, which describes left-to-right behavior. The criterion therefore is A = 1 and T22 = T~, ~ given above.

718

9.4

CHAPTER 9.

Irreversible

MODELS WITH STATIC

COUPLERS

Couplers and Thermal Systems

An energy bond is a reversible coupler; energy can flow is either direction without degradation as well as loss. The same is true of transformers and gyrators, although this requirement has not been emphasized previously. The only irreversible elements we have considered are resistances. Energy "dissipated" in a resistance does not vanish, of course, it is transformed into thermal form, and thermal energy has yet to be considered. This omission is rectified in this section. Thermal energy is qualitatively different from the gravity, elastic, kinetic, electrostatic and electromagnetic energies considered thus far. Rather than being based on macroscopic coherent generalized displacements or velocities, it is based on the microscopic incoherent (random) motions of atoms and molecules. As a result, engineering devices can convert thermal energy only partly to the other forms of energy, as described by the second law of thermodynamics and the concept of irreversibility. On the other hand, the other forms are readily convertable to thermal energy. Since energy is conserved, as expressed by the first law of thermodynamics, this means that thermal energy has been excluded from the energy bookkeeping. That exclusion sometimes is unacceptable, particularly if you are interested in temperatures, or if temperature affects other key parameters. Irreversibilities are introduced here with the examples of heat conduction and friction. The energy fluxes in these cases are analogous to the other types of power flows in that they can be represented by simple bonds with a single effort and a single flow. The introduction of mass transfer complicates the representation, however; this subject is deferred to Chapter 12. 9.4.1

Effort

and

Flow

Variables

A quantity of heat is labeled as Q in most thermodynamicstexts; here the script Q is used to distinguish it from the volume flow rate of a fluid. The rate of heat transfer, which is an energy flux a with the dimensions of power, is given as ~. The first need is to identify the effort or generalized force associated with heat conduction. Stop a moment before reading the next paragraph and try to determine the proper physical variable. If you succeed your learning will be enhanced. 3Theterm flux implies per unit time, and also usually meansper unit area. In this text the latter is dropped;the energyflux refers to the rate of energytransfer for an entire macro bond.

9.4.

IRREVERSIBLE

COUPLERS AND THERMAL SYSTEMS

719

The correct answer is absolute temperature, which is indicated herein with the symbol 8, since T is already dedicated to the moduli of transformers. (8 is also used frequently in the literature to indicate temperature.) Temperature has precisely the characteristics of an effort: it is the traditional "force" driving the energy flux, and it is a symmetric scalar with respect to the control surface. The second needed identification is tougher. What variable is the flow or generalized velocity? This time you might like to get a pencil and some paper and recall the definition of entropy and the relationship between effort, fiow and power before reading on. The reward can be well worth the trouble. The flow is the ratio of the energy flux Q to the absolute temperature 0, or (dQ/dt)/O. However, recall that the change of entropy of a system due to a reversible transfer of heat ~Q is dS = --.

(9.29)

The flow is therefore the rate of entropy transfer, dS/dt, due to the heat conduction. This is called the entropy flux due to heat conduction, and is labeled ~, since the entropy gained by one system because of its contact with another system is lost by that other system. Thus the heat flux ~ is ~ = 0~.

(9.30)

The symbols 0 and ~ are entered in the updated table of efforts and flows given as Table 9.1. (A refinement appears in the use of the term a in the first two entries of the table. This is a coefficient equal or greater than 1 which can recognize the effect of non-square velocity profiles.) 9.4.2

Heat

Conduction

Heat conduction associated with a finite temperature drop is irreversible, and generates entropy. To see in simple terms how this happens, and to introduce a new entropy-producing bond graph element, consider the uniform slab of material shown in Fig. 9.20 through which a steady flux of heat passes. The heat flux entering one face equals the heat flux leaving the other, so that (9.31) which gives an entropy generation rate of (9.32) This is always non-negative, since the sign of ~1 is the same as.the sign of 01-02. In the linear special case, for example, 01~1

=02~2

= n(o1

-02),

(9.33)

720

CHAPTER 9.

Table 9.1 Effort

MODELS WITH STATIC COUPLERS

and Flow Analogies (Extended)

Fluid, incompressible general less thermal approximate micro-bond Mechanical, longitudinal general approximate micro-bond

generalized force or effort, e

generalized velocity or flow, 0

pu + P + apv2 /2 + pgz P + apv’2 /2 + pgz P any of above

Q Q Q v

Fc + pAv2/2 Fc o + pv2 /2

v =~ v =~ v =~

M Fs T e t?

~ v, V~ i ~

Mechanical, transverse rotation translation (shear) micro-bond Electric circuit Thermal, conduction

wherethe constant H is knownas a coefficient of heat conductance. This gives 2~2-~

=H(0~-02)

0102

’

(9.34)

which showsan irreversibility that grows with the square of the temperature difference. Entropy is generated throughout the slab. Energy storage and energy dissipation, whenthey occur, also are distributed spatially. Nevertheless, lumped modelsare most commonly used for thermal as well as other systems, for reasons of computationalsimplicity. Only spatial integrals of the extensive properties are represented in lumped models. The resistance element, ~R, is an example. The problem of approximating continuous phenomenaby lumpedmodels is discussed in more detail in Section 10.1, and distributed parameter models are discussed in Chapter 11. 9.4.3

The Irreversibility

Coupler

The energy-conservative but entropy-generating lumped general irreversible coupler is nowintroduced. Traditionally represented as el

~RS.’~-=--__~ 01

e2 q2

¯ this element often is called simply "the RS coupler." Whenapplied to heat transfer, as indicated in Fig. 9.20, el and e2 are the temperatures 01 and 0e, and 01 and 02 are the entropy fluxes ~1 and ~2, respectively.

9.4.

IRREVERSIBLE COUPLERS AND THERMAL SYSTEMS

721

equal heat fluxes in and out

temperature, 0O:

CONDUCTING

S~B O~,.-- RS O~ Figure 9.20: Steady heat conduction and the RS element

The RS coupler is defined partly as a static coupler that conserves but does not store energy; the energyfluxes ea01and e202are equal. In this sense it is the sameas the transformer element. It is very different in another sense, however; the transformer is used exclusively to represent reversible behavior while the RS coupler never represents reversibility except in the limit of zero energyflux. The transformer is reversible simplybecause it is never used whena variable entropy flux is involved. The same is true of gyrators. Therefore, the RS coupler is alwaysused in place of a transformeror a gyrator whenevereither or both ports have an essential entropy flux. In the specification of a particular RScoupler it is necessary to insure that the second law of thermodynamicsis satisfied, whichimplies that entropy never vanishes. In the case of heat conduction, the constitutive relation in equation (9.33) satisfies this requirement,as revealed the non-negative net entropy generation of equation (9.34).

9.4.4

Application

to Friction

You can replace an R element with an RS coupler whenever you wish your model to retain the energy converted to thermal forrn. A transformer or a gyrator cannot be so used, because the flow on the output bond is an entropy flux. A major application is the modelingof energy conversion to heat through friction, as the followingexampleillustrates.

722

CHAPTER 9.

MODELS WITH STATIC

COUPLERS

Example 9.10 A shaft is supported by a journal bearing having some friction. Heat conduction along the shaft is neglected, compared with the heat conduction through the bearing support to the frame. The bearing support offers significant resistance to heat flow, and the frame is at a knownenvironmental temperature

al

sistance

//////////0[////

R

Model the system with a bond graph, including the heat fluxes, and write the corresponding equations. The resistance of the traditional bond graph model with dissipation, showff’~bove right, can be approximated as a function of the form R(¢), or moPe’accurately of the form R(~, Oj). Solution: The bond graph opposite employs the element RS1 to represent the irreversible conversion of mechanical power to the thermal form 0j~j, where 0~ is the temperature of the journal. The element RS2 represents the subsequent heat conduction through the bearing suppport to the frame, which is at temperature Be. This cascade of elements details the reduction in the availability of the energy and its effect on the environment. The constitutive relation

M~

RS~

s~ R= R(~) implies that the frictional this case,

torque is independent of the thermal variables. In

2, 0e~e= 0~j = R~

so that if 8e and ~ are known, ~e can be calculated. Note that ~ never can be negative; the second law is satisfied. If a thermal conductance H between the journal and the environment (as in equation (9.33), p. 719) known, then 0j and ~j can be calculated also. If the friction in the journal depends significantly on the local temperature (as it well might since fluid viscosity depends on temperature), expressed as R = R(~, ~9~), the equations above still allow for the computation of the temperature 0j, the frictional torque, the heat transfer and the entropy fluxes.

9.4.

IRREVERSIBLE

723

COUPLERS AND THERMAL SYSTEMS

~

RS

RS

/

0

RS !

041 (b) bond graph

(a) spatial configuration Figure 9.21: Heat interaction interfaces

between regions separated by partially

insulating

Seh

"~

M IDEAL ~ MACHINE

Set Figure 9.22: Ideal heat engine (IHE)

9.4.5

Use of

0-

and

1-Junctions

A body can interact thermally with any number of other bodies, as suggested in Fig. 9.21. This is recognized by use of a 0-junction, or constant-temperature junction. On the other hand, a 1-junction rarely can be used for heat-conduction bonds, since this would imply a principle of conservation of entropy, in contradiction to the second law of thermodynamics. The only exception to this would be in the reticulation of a machine that is idealized as being reversible anyway. You might in fact wish to create a three-port element to represent an ideal heat engine (IHE), a favorite of thermodynamicists. Such an element is reticulated in Fig. 9.22. The port is mechanical and has no entropy flux. The other two are thermal with flows that are, identically, entropy fluxes. The requirement of reversibility forces these two entropy fluxes to sumto zero. Alternatively, this constraint can be shownexplicitly with the use of the 1-junction in the reticulation shown in part (b) of the figure. The thermodynamic efficiency of the IHE is

which agrees with the classical

Carnot result.

CHAPTER 9. MODELS }~qTH STATIC COUPLERS

724

S

So (a)

(b)

Figure 9.23: Thermalcharacteristic (equation (9.37)) An IHE with three or more thermal ports also can be defined; the sum of the entropy flows is zero. 9.4.6

Thermal

Compliance

The simplest modelfor thermodynamiccompliancerefers to a fixed mass; there is no convection. Also, work interactions with the environmentare neglected, whichimplies that either the volumeof the bodyis constant or its surrounding pressure is negligible. This leaves only a heat transfer interaction, and justifies use of the word "thermal" as opposed to the more general "thermodynamic." This thermal complianceis represented by a standard one-port compliance: ¯ ~C

S

A constitutive relation exists betweenS and 8, as it does betweenthe effort and generalized displacement of any one-port compliance. The most common assumptionis a constant specific heat, c, so that 8 ds = du + P d(1/p) = du = cdO

(9.36)

where u is the specific internal thermodynamicenergy and s is the specific entropy. Solvingfor 8, e( 0 S= S°)/rac O0

(9.37)

where mis the massof the compliancesubstance and the subscript 0 refers to an arbitrary reference state. This nonlinear relation is plotted in Fig. 9.23. The linearized model 8-8o=

(S-So)

or

A8= AS,

C=-~-o,

(9.38)

as shownin part (b) of the figure, can be used alternatively. This modelcorresponds to a specific heat that is proportional to absolute temperature, that is c = CS/m, (9.39)

9.4.

725

IRREVERSIBLE COUPLERS AND THERMAL SYSTEMS

cal g K

10° = W cm K 10~ =

bo

graphit~.~

/

-stainless

lO-3 A

i0-2

X

~tainless: steel i~rosilicat~

~ ~.~" glass

lO-S

°10

10 (a) specific heat

10 o 10

// 101

102 8, K 103

(b) thermal conductivity

Figure 9.24: Typical specific heats and thermal conductivities

Noblegases have constant specific heats, and the specific heats of liquids tend to be nearly independentof temperature. Also, the specific heats of most solids at moderate or high temperatures and of most other gases do not vary rapidly with temperature. These substances therefore agree or comeclose to agreeing with equation (9.37) for most practical purposes. Onthe other hand, the specific heats of solids at cryogenic temperatures decrease radically with decreasing temperature, even faster than the linear modelof equations (9.38) and (9.39). Sometypical behavior is given in part (a) of Fig. 9.24.

SoIne typical data for solids showingradical sensitivity to temperatures in the cryogenic range given in part (b) of Fig. 9.24. Suchdata is apt to be very sensitive to impurities and the crystaline form.

There is no such element as a thermal inertance. One consequenceis that purely thermal systems do not exhibit resonances.

726

CHAPTER 9.

MODELS WITH STATIC COUPLERS

Example 9.11 The case of a shaft with journal bearing with friction, given in Example 9.10, is now elaborated by including a path for heat conduction along the shaft, represented crudely by a heat transfer coefficient and a environmental temperature/93. The model also is made unsteady by including a lumped thermal compliance for the bearing. The bearing support has a fixed temperature 0.~. Represent this model with a bond graph. Then, annotate the graph with causal strokes and designations of the efforts and flows, leading finally to the associated differential equation(s). Solution:

The corresponding

bond graph is

RS~

ol

o

C With the added causal strokes and annotations,

RS~

TR~2/O(S)

I

0

3~ I RS

C The aSsociaged differential

d~ - O(S)

equation is (s ~ -s°I -Ha

/me 1- ~ ;

O = Ooe

In general, the thermal conductances ~ and Ha and the specific functions of the temperatures.

heat c are

9.4.

IRREVERSIBLE

COUPLERS AND THERMAL SYSTEMS

727

(1o) thermal storage

(a) heat conduction

Figure 9.25: Pseudo bond graphs for heat conduction

9.4.7

Pseudo

Bond

Graphs

for

Heat

Conduction

The assumptions that both the specific heats and the thermal conductivities of a system are invariant with temperature allow a striking computational simplification. This advantage exists, however, only if one substututes for the state variable S the state variable Q. Karnopp and Rosenberg (citation on page 16) represent this idea in the form of what they call a pseudo bond graph for heat conduction. A pseudo bond graph for heat passing through a thermal resistance is shown in part (a) of Fig. 9.25. The heat flow, Q, equals the temperature difference, 81 - 82, divided by the resistance, R. A pseudo bond graph for thermal energy storage is shown in part (b). The rate of change of temperature, {}, equals the difference between the heat flow in and the heat flow out, Q~ - Qa, divided by the thermal compliance, C, which in turn equals inc. Causal strokes can be applied as with true bond graphs, and differential equations written with relative ease even for fairly complex combinations of resistances and compliances. The bond graphs and the associated equations still can be used if the thermal resistances or compliances are fuctions of temperature, but the advantage over the true bond graph is lost. The designation pseudo recognizes that, unlike true bond graphs, the product of the effort, 8, and the pseudo-flow, ~, is not the energy flux. The pseudoflow itself is the energy flux. This poses a problem whenever a model for heat conduction, represented by a pseudo bond graph, is coupled to a model for a mechanical, fluid, electrical or hybrid system, represented by a true bond graph. Such a coupling might be represented by an ad hoc coupler defined to address the particular situation. One can write equations in terms of Q’s rather than S’s regardless of which type of graph is drawn. The author therefore prefers to avoid pseudo bond graphs, except when there is an isolated or an extensive region with constant thermal resistances and specific heats. Pseudo bond graphs nevertheless are quite commonin the literature.

728

CHAPTER 9. MODELS WITH STATIC COUPLERS Example 9.12 Convert the thermal conductance portion of the bond graph in the prior example to a quasi bond graph. Apply causal strokes, and annotate the graph variables assuminginvariant thermal conductivities and specific heat. Finally, write the correspondingstate differential equation(s). Solution:

dQ= Rl~p 2 1 Q d---~ R2 C’

1 Q 02 Ra C’

1 0~; C’ = mc; R2 = -~;

1 R3 = -~-~.

This is a simple first-order linear differential equation with constant coefficients. The associated time constant for transients is r = C’/(H1 + H2). Should the specific heats or the thermal conductances not be assumedas invariant, the advantageof using Q or 8 as the state variable vanishes. 9.4.8

Summary

Previous to this section thermal energy was treated as "dissipated," or useless, and accordingly it was not entered into the energy bookkeeping. Instead, it simply vanished in resistance elements. This approach is conventional in the elementary analysis of dynamicsystems, but fails whenevertemperature affects those properties of a systemwhichare of interest, or is of interest per se. The subject of this section is howto bookkeepthermal energy alone, or in conjuction with the other energies, in both static and dynamicsystems. Cases with coupled mass flow and thermal effects are deferred Chapter 12; heat conduction is the only modeof heat transfer considered here. The use of absolute temperature as the effort or generalized force for the transfer of thermal energy requires that the flow or generalized velocity be the associated entropy flux. Thermalenergy thereby becomesa generalized potential energy. There is no such thing as a generalized thermal kinetic energy. Thermalenergyis different from the other energies in that its convertibility to

9.4.

IRREVERSIBLE

COUPLERS AND THERMAL SYSTEMS

729

these other for~ns is restricted by the second law of thermodynamics. Heat conduction can be represented by the the RS static coupler element. The coupler is like a transformer or gyrator in that it conserves energy, but unlike them in that it allows irreversibilities and the associated generation of entropy. The RS coupler also can represent the irreversible flow of energy from the other forms into the thermal form, and is particularly useful in representing friction. Its constitutive relations depend on its particular application. Should you wish to model a thermal system by assuming that its specific heats are constant and that its heat conductions are proportional to its temperature differences, you should substitute the heats as the state variables. (You can alternatively use the associated temperatures, since in this case they are proportional to the heats.) This substitution can be represented in the form of a pseudo bond graph for heat conduction. This is not a true bond graph, however, because its energy fluxes are not the products of its efforts and flows, and does not mate with true bond graphs using conventional couplers. Guided

Problem

9.8

This guided problem illustrates the organization of steady-state problems involving heat engines and heat conduction provided by the methods presented in this section. The classical Carnot efficiency of the ideal heat engine operating between the temperatures ~ and 0L WaS confirmed in Fig. 9.22 and equation (9.35) (p. 723). In order to make the model conform more closely to a real system, interpose a thermal resistance with conductance H between the warmer heat reservoir and the ideal heat engine. Find how the resulting thermal efficiency is affected by the output power, 7Pout. Specialize the result for ~L/(~H= 0.5. Also, find the maximum7~o~,~ which is possible, and the associated thermal efficiency. SuggestedSteps: 1. Add an RS coupler to the bond graph model to represent the thermal resistance. Define symbols for all the efforts and flows, and place them on the graph. 2. Write a complete set of algebraic equations to represent the constraints of each of the elements in the model. Also, express the thermal efficiency as a function of appropriate parameters and variables. Treat 7~o~ as known. Make a list of the unknowns, and make sure their number equals the number of equations. 3. Eliminate all variables, solving for the efficiency. This is tricky; it may help to number the equations, and make a table in which the variables in each equation are indicated with check marks. 4. Verify that the thermal efficiency equals the Carnot efficiency in the limiting ’case "Po~,t = 0. If not, a mistake has been made.

730

CHAPTER 9.

MODELS WITH STATIC

COUPLERS

5. Make the indicated specialization ~L/~H = 0.5, and sketch-plot the efficiency vs. :Pout. The maximumpower and its associated thermal efficiency should become apparent.

Guided

Problem

9.9

This problem provides experience in the writing of differential equations when irreversibilities are present. It does not require mucheffort if the simulation part is omitted. The fading of brakes for vehicles on mountainous roads is a serious problem; the phenomenon of "runaway" trucks on long downhill runs is the most common result. Consider a truck weighing 100,000 lbs traveling down a long 6 percent slope. The coefficient of friction of the brake materials is -6°F-2 #0=0-25

~#=P0-#I(A8)

p~ =0.2x10

where A0 is the mean temperature of the brake material above the ambient. (Should the temperature become so high that this equation gives p < 0, the proper relation is# = 0.) The trucker attemps to keep the speed steady at 50 mph, and is successful for some time. The normal force squeezing the brake materials together cannot exceed eight times its initial value (when 58 = 0), however, and eventually the truck begins to accelerate. The effective mass of the heated material is 125 lbm, its specific heat is 0.30 Btu/lbm.°F (or 233.5 ft.lbf/lbm.°F), and the effective coefficient for heat transfer to the surroundings is 0.38 Btu/°F. Write state differential equations for a model of the system. Determine how long the speed remains constant, and the maximumlower speed for which the truck would never become a "runaway" on this slope. Optional: simulate the subsequent acceleration. Suggested

Steps:

1. Draw a bond graph for a simple model. Define symbols for the variables and the parameters. 2. Write the state differential is constant.

equations for the first phase, in which the speed

temperature above which the brakes will not prevent 3. Find the critical acceleration. Solve the equations of step 2 analytically (this .should be simple) to find the time at which this temperature is exceeded, and the acceleration phase begins. At this point one can also determine the critical speed below which the brakes would never have reached the critical temperature. 4. Write the differential eration phase.

equations in solvable form for the subsequent accel-

9.4.

IRREVERSIBLE

COUPLERS AND THERMAL SYSTEMS

731

5. The equations of step 4 are not easy to solve analytically. Solve them numerically using the MATLAB simulator or any other simulation program. Makesure the effective coefficient of friction never becomesnegative. Plot the resulting velocity as a function of time. PROBLEMS 9.30 A thin-walled spherical vessel, 0.5 m in diameter is full of hot water at 100°C when it is placed in an environment with an air flow at 0°C such that the mean local overall heat transfer coefficient is 150 kJ/hr.m2°C. Determine the temperature of the water as it cools over time. Note: For water, c = 4.18 3. kJ/kg.K and p = 997 kg/m 9.31 An ideal heat engine (IHE) operates between a hot reservoir of temperature 0H and a cold reservoir of temperature 0L. A thermal resistance with conductance H is placed between the IHE and the cold reservoir, with OL/OH= 0.5. Find how the resulting thermal efficiency is affected by the output power, Pout. Also, find the maximumPo,~t possible, and the associated thermal efficiency. Compareyour results with those of Guided Problem 9.8, which is similar except for the location of the thermal resistance. 9.32 Consider the slab of material below which exchanges heat with environments to the left and the right through thin partially insulating layers. The thermal conductivities can be assumed to be constant at H1 and H2, respectively. The specific heat of the slab can be assumed to be constant, also. RS~

0

RSz"-7--~"

(a) Write a differential equation for the state of the model in terms of the conventional bond-graph state variable S. The temperatures 01 and 05 are specified independently. (b) Repeat (a) substituting ~ as the state variable. You may wish represent your model with a pseudo bond graph. Which approach do you prefer? 9.33 A long piece of stainless steel of cross-sectional area A is used at cryogenic temperatures in the range 2 K to 50 K; typically, one end is muchcolder than the other. A simulation model consists of a series of alternating lumped thermal resistances and lumped thermal compliances for elements of length L. Choose a state variable and write a state differential equation for the ith such compliance using approximations for the thermal properties based on the data given in Fig. 9.24 (p. 725).

CHAPTER 9.

732

MODELS I~qTH

STATIC

COUPLERS

9.34 An assumed incompressible fluid at 120OFflows through a valve, dropping in pressure from 3000 psi to 1000 psi. The heat generated initially goes into the fluid, but ultimately goes to the environment at 70 °F. The density of the fluid is 0.8 × 10-4 lb.s:/in 4 and the specific heat is 0.50 BTU/lbm.°F. Find the heat generated, the temperature of the fluid immediately downstream, the entropy generation before the fluid cools, and the additional entropy generation due to the subsequent cooling. (a) Draw a bond graph, showing the two fluid ports and one thermal port, and two irreversible elements. Label all efforts and flows. (b) Find the heat generated. Note that 1 BTU=778.16ft-lb and that one pound mass weighs one pound force in one standard gravity. (c) Find the temperature fluid.

rise

and the temperature

of the downstream

(d) Find the entropy fluxes at each of the two temperatures.

SOLUTIONS TO GUIDED PROBLEMS Guided

Problem

9.8

1-2.

~7 = OH~H

(1)

o.8. = 0~-8

(2)

OH~H= H(O. - O~H)

(3) (4)

7~ o~ = (0’~ - OL)~ M IMA CHINE

eq. no. 1 2 3 4

~/ hH x x x x

OL

~[ x x

x x x

:

Equation (.1) shows that ~/ depends only on the unknown~H, and equation (4) allows S~ to be expressed in terms of unknown/9~t.Youstart therefore combiningequations (2) and (3) to eliminate and 0~. l;¥om equation (3), OL Substituted into equation (2), this gives ~ = On~H

9.4.

IRREVERSIBLE

COUPLERS AND THERMAL SYSTEMS

733

Substituting equation (1) into this result gives an equation whichcan be solved for the desired result, ~: Po~t = H OH-- OL . Therefore,

if we define p =

When~o~ = 0, this gives y = 1 - ~ which you know is correct. ~H ~

WhenOL/~h = 0.5, 1

~=~

The plot of this result reveals that the efficiency declines with increasing output power, at first slowly, and then more rapidly. The maximumpossible output power is 7aou~’: 0.0858HOH, which occurs whenthe efficiency is decreased to 29.3%.

0.5 ~ 0.4 0.3

0.1 0

Guided

Problem

0.0858

0.2

0.02 0.04 0.06 p 0.08

9.9

s, FW~i__~_,~ ~ Oo+4O I

Oo

C

~ = ~in - ~o~t, but use 0* as the state variable: Q~ = mcAO. ~i~ = Wsina&-- 100,000 x sin6 ° x (50 x 88/60) = 776,542 ft-lb/s, ~o~ = H 0". dS* ~2 76,542 0.38 Therefore, 0.’ dt --,,o~(Wsina~HO*) = 125 x 233,5 125 x 0.30 or (rs + 1)O* = (98.68s + 1)8" = 2591.7. 1 7 23. The brake overheats When~ ----/~ -- ~/~o so that ~o ---- ~(0~) from which the critical

temperature is 0~ = ~ --- 1045.8 °F. ~* -- 2591.7(1- e-~/~); specifically, 2591.7(1-~¢/~) --- - 1045.8 °F.

0.10

734

CHAPTER

9.

Therefore,--to_- 0.5167;tc

MODELS

WITH

STATIC

COUPLERS

= 50.99seconds.

T

3000 asymptote 2591.7 °F .....

At? °F

-.

7.t .......

2000 asymptote.- - "’"

10.8

ooo

~

tc=50.99 s 20

0 The critical

~ ~ ~ ~ ~ ~ -- ~

40

temperature

60

80

100 t, seconds

never would be reached

mph. Above the critical temperature, the drag force ~. creases with increasing temperature by the ratio Therefore,

the differential

if

120 1045.8 ~? < -- x 50 = 20.18 - 2591.7

produced by the brakes de~l~d = 8 - 6.4 x 10-6(0")

equations become

dO~* = 1__ {Wsina[8- 6.4 x 10-6(0")21~HO*} dt mc = 0.35813[8 - 6.4 x 10-6(0")~]~ - 0.010130", d~ W sin a. dt - ~g/g [1 - 8 + 6.4 x 10-6(A0)21 = 3.3631[--7

+ 6.4 × 10-6(A0)2].

If 0* > V/~//~a = 1118 °F, braking would cease altogether, d~ d-~ = 9 sin a = 3.363 ft/s 2. This state proached asymptotically.

so that

is never reached, however; it is ap-

A simulation starting at t = tc and 0* = 0~ is given below. Most of the runaway acceleration develops within a mere 10 seconds after the critical temperature is reached. " 200 speed,

150

100

~

AO-1000,

OF

50

~0.99’

dO ’ ~0 ’ 8;

’ 9~) ’ 100 t, seconds

Chapter 10

Energy Storage

Fields

Energy is distributed over space, or over a. field. In most of this text energy is approximated by functions of discrete numbers of variables. The associated process of field lumping is the subject of the first section below. It focuses on the lumping of the one-port resistance, compliance and inertance elements that have been assumed thus far. The name’ field is also applied, in a discrete sense, to resistance, compliance and inertance elements that have two or more ports. Multiport compliances represent energy stored by virtue of two or more energy fluxes and their associated displacements or efforts. These compliances are particularly important in modeling electromechanical and thermodynamic (or thermal-mechanical) components for which the fluxes are of different types. This is the subject of Section 10.2. Multiport inertances represent energy stored by virtue of two or more energy fluxes and their associated velocities or momenta, as discussed in Section 10.3. Here it is especially critical to distinguish energy storages that are truly dependent on two energy fluxes (called holonomic) from those that depend one energy flux and a remote generalized displacement (called nonholonomic). Engineers often assume linearity when faced with the complexities of multiport fields. Linearity is assumedin Section 10.1, but is not assumed in Sections 10.2 and 10.3. Linear multiport fields are discussed in Section 10.4, including a systematic glimpse of a wide variety of interdisciplinary phenomenaand devices.

10.1

Field

Lumping

This book emphasizes lumped models (with the exception of Chapter 11) described as compliances, inertances and resistances interconnected by a junction structure. Nevertheless, energy and the mechanisms that dissipate it actually are distributed over space. The process by which the energy storage or power dissipation over someregion or field of space is approximated in terms of a small number of variables is called field lumping. The state of a one-port field, for example, can be described in te~ms of a single variable: a characteristic gener735

736

CHAPTER 10.

ENERGY STORAGE FIELDS

alized displacement, velocity, force or momentum.In addition to their number of ports, fields can be categorized as scalar, vector or tensor, and also as nodic or non-nodic. This book has engaged implicitly and almost routinely in field lumping, without identifying it as such. In a typical example, the energy in a coil spring results from the torsion and twisting strain of the wire, which in turn results from the shear stress and strain at each point. To evaluate the "spring constant" k or the lumped compliance C = 1/k from first principles one can integrate the ~ strain energy over space and set the result equal to1 ~kxor 1gox2 . In another example, the resistance to a flow of a viscous fluid through a tube is associated with the integral of the power dissipation from the center of the tube to its periphery and from somewhat outside one end of the tube to somewhat outside the other end. Some of the basic concepts and methods are organized in this section, and they are extended particularly with regard to nodic vector fields. 10.1.1

Scalar

Fields

The scalar field is defined by the existence of a commoneffort, flow, momentumor displacement over the entire spatial extent of the field. The most familiar examples include translation of an incompressible solid or liquid, that is motion with a commondisplacement, velocity and momentumper unit mass. The result is overall or lumped inertance (e.g. mass), resistance, or, for motion in a gravity field, an effort (force) source as a degenerate or special type of compliance. Another familiar example is the compressibility field of a fluid in which the pressure is assumed to be uniform over the extent of the defined control volume. The result is a lumped compliance. A thermal field in which the temperature is assumed to be uniform over the extent of the defined control volume similarly results in a lumped compliance. A lumped resistance results from the assumption of uniform flow of a fluid through a homogeneouscylindrical porous plug with uniform pressures over its two ends. In all of these cases the lumped resistance, compliance or inertance equals a uniform property on a per unit mass or volume or length times the total mass or volume or length. 10.1.2

Rigid-Body

Vector

Fields

The gravity potential energy of a solid or fluid in rigid body in a uniform gravity field is expressed as the product of its mass and the elevation of its center of mass. This center is in fact defined so that the spatial integral of the potential energy, f gz dm, equals mgzc,~. Similarly, the kinetic energy of the body in planar motion is represented by two lumps, one with linear inertia (mass) and linear velocity and the other with rotational inertia and rotational velocity. Here, the center of mass and the rotational inertia J are defined so as to make this lumping work; that is, so that the kinetic energy is 7-=

lf~.d

~]v

1

m = "~mV;r

~

n + ~j~2.

(10.1)

10.1. 10.1.3

737

FIELD LUMPING The Role

of

Approximations

The approximation inherent in any definition of a lumped element is the assumption that one or at most a few commonvariables apply to all regions of the field. To the extent that a solid body is not rigid, for example, the very ideas of a center of mass fixed to the body and a constant rotational inertia become invalid. The spring compliance becomes an inaccurate description if the spring is deflected so fast that the axial force (or the torsion on the wire) is nonuniform along its lengt h due to the mass and the acceleration. The resistance of the tube to flow becomes nebulous if the dissipation of energy is affected by the acceleration of the fluid; this phenomenonindeed sometimes occurs because the velocity profile across the tube, which strongly affects the dissipation, can be ~narkedly altered by acceleration. Compliance, inertance and resistance fields can be coextensive, that is occupy the same space and involve the same matter at the same time. The lumping of these fields nevertheless is accurate if the fields do not interact with each other. The kinetic energy and the gravity energy of a truly rigid body, for example, do not interact. The compliance energy of the spring does not interact much with its inertance energy as long as the latter is muchsmaller than the former (or vice-versa!). The resistance field of the fluid-filled tube is not affected much by the coextensive inertance field if the latter has little energy, and the affect of the resistance field on the inertance field is limited in any case, permitting at least rough approximation. You must watch out for coupling between the inertance and compliance fields in the tube, however, which can produce wave-like behavior. The couplings of fields usually can be reduced to an acceptable degree if the spatial region of interest is reticulated into multiple fields of small extent. The solid body that is not quite rigid, for example, might be represented as having a single deflection variable with a single stiffness or compliance; this implies two inertance fields and one compliance field. If that model still is not good enough, the body could be modeled with several inertances with several overlapping compliances. In certain cases it might be appropriate to allow the fields to interact continuously; this is the subject of Chapter 11 on distributedparameter models. The whole business of modeling ultimately becomes an art which is strongly dependent on the objectives and skills of the modeler.

10.1.4

Nodic

Fields

A field is called nodic if the flow or generalized displacement therein has zero divergence. Physically, this means that the flow or displacement is like that of an incompressible fluid with no sources or sinks other than at boundary ports. Mathemetically, this "continuity" relation is stated by R or I field: C field:

~" ~l = 0, ~ ¯ q = 0.

(10.2a) (10.2b)

738

CHAPTER 10.

ENERGY STORAGE FIELDS

The rigid body motion discussed above corresponds to a nodic field. So does the incompressible flow of the liquid through a porous plug, and the motion of an elastic solid in pure shear, such as pure torsion. On the other hand, any kind of material dilation or compression violates the nodic condition, as does the thermal compliance. Nodic fields are generally incompatible with uniform effort, and exclude most fields described by tensors, such as complexstress-strain fields in solids. Fields of most commontypes can be analyzed numerically with a wide variety of finite-element and finite-difference software packages. The objective here is not to replace these packages, except in the simplest cases when a quick approxilnation can be made directly. Rather, the aim is a deeper understanding of the role and meaning of field lumping in the context of the modeling of complex systems. The focus here is on vector nodic fields. Vector fields are intermediate in complexity between scalar fields which require little exposition and tensor fields, which are discussed in Section 10.4.7. In particular, the focus is on fields in which the vector flow or displacement is in the direction of the gradient of the scalar effort or momentumand is uniformly proportional to it: R field:

£1----- -k Ve, q = -kVe,

(10.3a)

C field: I field:

~l = -k Vp.

(10.3c)

(10.3b)

Bold-face notation implies a vector field. For the R field, k is known as a conductivity, which is the reciprocal of a resistivity. For an electrostatic C field, k is the dielectric constant; for a magnetic C field it is the magnetic permeability. For an inertive I field, k is knownas the susceptivity, which is the reciprocal of the density. Substitution of equations (10.2) into (10.3) gives the well-known Laplace’s equation: RorCfield: Ifield:

V~e=O

or

O~e O~e ~x 2 +~+~z

~ =0,

(10.4a)

V2p=0

or

O~p 02p +~z2 02p ~+~y2 =0"

(10.4b)

Together with equations (10.2) these require that a set of equipotential (e or lines or surfaces form an orthonormal network with a set of flow or displacement (0 or q) lines or surfaces (across which no displacement or flow occurs). example with ports at the two ends of a channel with squar e cross-section is pictured in Fig. 10.1 part (a). The spacing between the flow or displacement surfaces is chosen to equal the spacing between the potential surfaces, resulting in an array of cubes-in-the-s~nall. The element can be represented by the bond graphs of part (b). In the special case where either e~ or e_~ is zero, the element becomes a simple oneport inertance, resistance or compliance. In general, the ratio of the integrated

10.1.

FIELD LUMPING

739

qor q e) or

potentiale2

surface 1 ~ potentiale1 or p~

(a) three-dimensional two-portfield el

e2 "R iil=(e~_e~)[

eI

e2

~ =el-e2~

,Io=P"

el-e2~

R

e1

e~

e~-e2= q/C

1

q l

C

(b) bond-graphmodels

//////(////// .~

0/2

el+e~

~ ~" "

I

/1111//11// (c) conductanceof planar squares

(d) sketchingmethodsfor planar fields

Figure 10.1: Two-and three-dimensional fields

or P2

740

CHAPTER 10.

ENERGY STORAGE FIELDS

flow or displacement through the channel to the difference between the efforts or momenta at the two ends is known as the field transmittance, ~-. For a resistive field this transmittance becomes the field conductance, G = l/R; for a compliance field it becomes the field compliance, C; for an inertance field it becomes the field susceptance, 1/I. 10.1.5

Planar

Vector

Nodic

Fields

A planar field is the special case with uniform depth, w. Consider first the special case of a square resistance planar field with flow from left to right, as shownin part (c) of Fig. 10.1. The conductance of this field is the conductivity, k, times the depth: G = kw. Divide this square into four equal squares, as shown. Each small square has one-half the flow and one-half the potential drop of the large square. Therefore, tl~e conductance of each small square, which equals the ratio of its flow to its potential drop, is the same as for the large square. This process of subdivision can be repeated. The same result would apply to the other transmittances (compliance and susceptance). The conclusion is that all squares in a planar vector nodic field have the same transmittance. A particular planar field is sketched in part (d) of the figure, The flow lines and the equipotential lines form an orthonormal net of "squares." Note that if each "square" were subdivided into four parts, these smaller regions would be closer to perfect squares. Therefore, the conductance of each subdivision is the same as every other, assuming they are drawn as nearly square as possible. There are ten "squares" in series for each stream tube, or sequence of squares from one port to the other; therefore, the transmittance of one stream tube is one-tenth that of one square, or kw/lO. There are three such stream tubes, so the overall transmittance is (3/lO)kw. In general, the overall transmittance of such a field is r = ~kw, (10.5) in which A, called the form factor, is the ratio of the number of squares in the direction transverse to the flow or displacement to the number of squares in the direction parallel to the flow or displacement; in the present example, A = 3/10. Whennecessary, it is correct to consider fractions of squares. A sophisticated analytical technique called conformal mapping is sometimes successful in mapping a curved region into a rectangular one, so that the form factor is determined analytically. Much more commontoday is the use of numerical computer routines for estimating the solution to Laplace’s equation, to the same end. Most popular among these are finite element methods. For rapid and intuitive approximate analysis, however, it is hard to beat trialand-error sketching of the field, particularly if it is planar, as suggested in the figure. Methods for sketching the orthogonal net of equipotential and flow or displacement lines are suggested in part (d) of the figure. If you use a pencil and a good eraser you usually can do a rather good job, and get a better grasp of the nature of the field than would result from using a computer solution. Note that lines drawn as diagonals of the "squares" also form an orthonormal set,

10.1.

FIELD LUMPING

741

and circles inscribed in the "squares" touch all four sides. Youstart by guessing either the flow/displacementlines or the equipotential lines; in the case shown, guessing the flow/displacementlines is probably easier. Further methodsto aid in estimation are given below. Estimates of the energy or dissipated powerstored in a field also can be deducedfrom the sketched solution. For a single square, the powerdissipated in the resistive case is ~(el - e2) = kW(el e~) 2 = (1/kw)(12, the energy st ored in the compliancecase is ½kw(el- e.2) ~ = (1/2kw)q~, and the energy stored in the inertance case is ½kw(pl-p2)2 = (1/kw)o"2. The potential differences, flows and displacements here refer to individual squares, of course, but are the same for all squares in a field. Therefore, these powerdissipations or energy storages are the samefor all squares, large and small. Thus, small squares have a higher power dissipation or energy storage density than large squares. Further, the powerdissipated or energy stored for an entire field equals the powerdissipated or energy stored per square times the total numberof squares. Usingpotential differences, flows and displacementswhichrepresent the entire field, the powerdissipated or the energy stored becomes resistance: compliance: inertance:

10.1.6 Estimating Upperand LowerBoundsfor Field Transmittances* Trial-and-error in the estimation of field transmittances maybe reduced, and upper and lower boundsof the form factor estimated, through the use of methods given by Paynter. ~ He applies the ThompsonPrinciple to give an estimate of T which,if done with reasonablecare, is on the low side of the actual value.’ He also applies the Dirichlet Principle similarly to give an estimate.on the high side. The ThompsonPrinciple uses an assumed distribution of the displacement or the flow. The total energy storage or powerdissipation associated with this distribution is estimated; errors in the assumedfield producean over-estimate in the energy or power. This happensbecause the actual flow or displacement distributes itself so as to minimizethe energyor powerin the field. Using the subscript T to indicate the estimate with an assumedresistance field, equation(10.6a) gives (~)~

(~)2

-

~H.M. Paynter, Analysis and Design o] Engineering section may be skipped without loss of continuity.

~0

- 7,

Systems,

M.I.T.

(10.7) Press,

1961.

This

742

CHAPTER 10.

ENERGY STORAGE FIELDS

showingthat Tr is indeed a lower boundfor ~-. The same result applies to the complianceand inertance fields. The Dirichlet Principle uses an assumeddistribution of the constant-potential lines (e or p). Again, the consequentfield wouldhave a greater energy or power dissipation than the actual field. Usingthe subscript D, the expression analogousto equation (10.7)

rD --= (el -- e:): > (el -- e:): -- (eL -- e2) =T,

(10.8)

showing that TD is indeed an upper bound for ~-. The Thompson-Dirichlet results can be summarizedas follows:

TT < T < TD,

(10.9a)

AT < A < AD.

(10.9b)

Example 10.1 Find a close overestimateof the inertance of a planar field of incompressible fluid of density p, betweenthe left and right boundaries as plotted below. The pressures along each boundaryare uniform, arid viscous effects are to be neglected.

Pl

10.1.

743

FIELD LUMPING

Solution: The Thompson principle, when applied approximately, gives the desired overestimate. First, streamlines are guessed; those shown below divide the field into four channels that each carry one-quarter of the total flow. This comprises the Thompsonestimate of the flow distribution. Each channel now is carefully subdivided into squares (estimated below using circles), without reference to the other channels:

"total numberof squares/circles: 79.9

If the original streamlines were chosen without error, the resulting equipotential lines for the four channels would line up; clearly, the estimate was not perfect. At this point the work could be erased and a more informed estimate attempted. If instead the decision is to proceed, the kinetic energy of the assumedfield is written

This recognizes that each square is supposed to have the same flow, 0/4. The Thompson susceptance therefore becomes 0~/2 rT -- "fT

16w pk

Here, k is the total number of squares, including fractions thereof; in the present case it is k = 79.9. It may be seen that errors in the original assumptions of the streamlines should always result, if care is taken, in too many squares, so therefore TT is an underestimate of the correct 7. The inertance estimate IT is the reciprocal of TT, so that I > IT = ~p ¯4.99

W

744

CHAPTER 10.

ENERGY STORAGE FIELDS

Example 10.2 Find a close underestimateof the inertance of the fluid field addressed in the above problem. Solution: The Dirichlet principle, applied approximately, gives the desired result. This starts with a guess of the equipotential lines. The potential drop is arbitrarily subdividedinto eighteen equal intervals. For each interval, squares or circles are drawnand counted; there are manyfractions of squares to count also, giving a total of about k -- 64.2: ///~////

~5"

3.,~’~3.~

3.8

3.7

2zZT . 3.6 3.3

total numberof squares/circles: 64.2 3.3

The kinetic energy of the assumedfield becomes

TD = ~

2(Api) i=l

with the result

TD TD-

(Pl

--P~)2/2

kp - 324w

The inertance I is the reciprocal of the transmittance or susceptance, T. Substituting the respective numbersof squares, k, gives I < ID p--. = 5.05

W

Note that the difference between this overestimate and the underestimate found in the previous exampleis little over 1~0.

10.1.7 Three-Dimensional Vector Nodic Fields The three-dimensional equivalent of the square is the cube. Since a cube is the sameas a square of thickness equal to its other two edge lengths, defined as L, its transmittance is Vi ---kLi (10.10)

10.1.

745

FIELD LUMPING

The field comprises m stream tubes arrayed in a cluster instead of a sheet; fractions of such stream tubes can be included. The flow through each stream tube in a resistive field, and therefore through each cube, is the same. The product Ae~Li or ApiLi for the ith cube therefore is the same for all cubes. Choosing any particular stream tube, the total potential drop between the ports of the field therefore is (10.11) where n is the total number of cubes in the stream tube. The field conductance thus becomes G --

it

--

km

(10.12)

Doubling all dimensions (but keeping the same shape) doubles the conductance. Thus, it is traditional to represent the conductance as a product of the conductivity, k, a eharaeter|stlc length, L, which can be any dimension of the field, and a form factor, A: G = ALk

(10.13)

Considering compliance and inertance fields also, this result can be generalized to give T = ALk (10.14) Notice that, unlike for the planar field, the form factor depends on the choice of characteristic length. The energy dissipated or stored in each cube is proportional to the linear size of the cube, like the transmittance. Equations (10.6) apply to the three-dimensional field as well as the planar field. Rarely is it practical for the modeler actually to sketch cubes for a threedimensional field; the analysis above can be interpreted in more practical terms. In most cases the simplest and best approach is to estimate the flow lines so as to pekmit an estimate of the energy storage or power dissipation at each point in the field as a function of the total flow or displacement. This energy or power then is integrated over the field to give the total energy stored or power dissipated. The result is then equated to the proper power or energy of equation (10.6). Finally, the transmittance is deduced. This procedure tantamount to application of the Thompsonprocedure, and consequently gives a resulting 7 on the low side of the actual value. Starting with an estimate of the potential lines is an alternative: this application of the Dirichlet procedure gives an overestimate of T. 10.1.8

End Corrections

for

the

Inertance

of

Channels

A special case of a three-dimensional field has already been considered: the inertance of a slender fluid channel. The energy method given in Examples 3.3 and 3.4 (pp. 105-106) is essentially an application of the Thompsonapproach. All inertia in the local regions beyond the ends of the channel is neglected,

746

CHAPTER 10.

ENERGY STORAGE FIELDS

equivalent prismatic channels: .........

(a)

(c)

(b)

Figure 10.2: Types of channels however. Corrections to the inertance for these end effects are very important if the channel itself is very short; in the extreme case of a channel of zero length, which implies a slit or hole in a plate of zero thickness, the end regions comprise the entire field. This case is pictured in part (a) of Fig. 10.2. Part (b) shows case of a hole in an infinite baffle, for which part (a) is a special case. Part (c) shows a slender walled channel, which by virtue of the additional space around its ends gives less additional inertance. The end corrections frequently are important in unsteady hydraulics, and often are critical in acoustics. Beranek2 expresses the corrections as effective extensions of the length of the channel. For a circular hole of radius r in an infinite baffle he calculates an effective added length equal to 0.85r for each end. For a slender circular tube the length is 0.613r.

10.1.9 Multiport Vector Nodic Fields The linear vector fields considered thus far have two ports. Sometimes one wishes to model a linear field with three or more ports, which requires only a modest conceptual extension of what has been done already. An example of a three-port field is shown in Fig. 10.3 part (a). Assuminga resistive field (inertances could be readily substituted), a tee model, so called because of its shape, is shown in part (b), and a delta model, again named for its shape, shown in part (c). The first of two approaches starts by assuming that port 3 is blocked, and evaluates the resistance between the remaining ports 1 and 2. This resistance can be seen to equal R1 +R.9. Next, port 2 is assumedblocked, and the resistance /~2 q- R3 is computed. Again, the resistance R1 + R3 is computed in a similar manner. Finally, the individual resistances R1, R2 and R3 are deduced from the three sums. The second approach starts by assuming ports 1 and 2 are connected to the same effort, leaving a two-port field with conductance (1/R13) + (1/R23). ports 1 and 3 are similarly connected, leaving a two-port field with conductance (1/R12) + (1/R23). Connecting ports 2 and 3 together gives a two-port field 2Leo L. Beranek, Acoustics, NY, 1986.

2rid edition,

American Institute

of Physics,

Inc.,

New York

10.1.

FIELDLUMPING

747

port 2

e3

i

(a) field 2 0

i

|

2

I

1

RI2

R 3

3 R13

(b) tee model

(c) delta model 2

1

T

~1~

R b (d) one of manyother possible models Figure 10.3: Three-portresistive field

748

CHAPTER 10.

ENERGY STORAGE FIELDS

conductance (1/Rr)) + (1/R13). Knowledge of the three conductances permits the individual conductances or resistances to be computed. The tee model parameters can be converted to the delta model parameters, or vice-versa, if desired. There are many other equivalent three-parameter models for a three-port field. Part (d) of the figure shows one of these. Equivalence may be established most generally by equating the power dissipation or energy storage of the respective models. Access to m~ actual piece of hardware may permit three individual experiments to be carried out which determine the component parameters of a model, following the same logic as the analysis. Most experienced engineers employ a combination of analytical and experimental methods in modeling. The process of modeling from purposeful experiment has become known by the na~ne of experimental identification. 10.1.10

Summary

Field lumping is the essence of lumped modeling. The major source of error in a carefully executed lumped model results from neglect of couplings between different types of fields, which can be reduced at the expense of complexity by choosing smaller spatial domains for the fields. Scalar fields with a uniform effort are assumedfrequently and are relatively simple to understand and compute. Emphasis therefore has been placed on the next more difficult type of field, the linear nodic vector field. In the usual application of this field the vector generalized velocity or displacement is nodic, that is, acts like an incompressible mediumwithout internal sources or sinks. Si~nply organized nodic vector fields such as rigid body motion and displacement of a rigid body or a liquid in a gravity field can be addressed analytically, as you have seen previously. Special attention has been devoted to the case in which the effort or generalized momentumacts as a scalar potential; the flow is proportional to the gradient of this potential. The resulting field satisfies Laplace’s equation. Examples include resistance fields in which the flow of a fluid or of an electric current is impeded by a resistivity, compliance fields in which the displacement of electric charge is impeded by the dielectric phenomenonor the magnetic displacement is impeded by the permeability phenomenon and inertance fields in which the acceleration of a fluid is impeded by its inertance. Although precise analytic methods are available in some cases and computer methods are widely available, emphasis has been placed on approximate analytical and fieldsketching methods, since these methods impart an understanding which aids the design process more directly, and give quick estimates. Nodic vector fields with more than two ports can be treated as combinations of pairs of fields with two ports. The results can be organized in terms of bond graph equivalences. More complicated tensor fields such as are associated with the strain of an elastic solid are in general beyond the scope of this text, but simple approximate cases such as the classic bending of beams or torsion in rods

10.1.

~

749

FIELD LUMPING air -

water

coffer clam air

soil, permeability k

~water

~

...... soil, permeability 2k

Figure 10.4: Guided Problem 10.1 and coil springs can be accomodated readily, either through the standard development of their constitutive relations or through evaluation of their potential energy storage. See also the tensor fields in Section 10.4.7. Field lumping is one way to address systems which have coupled fields. This important topic is considered in Chapter 11, which focuses on distributedparameter models. Guided

Problem

10.1

As sketched in Fig. 10.4, a coffer dam holds back earth and water at an excavation site. The permeability (conductivity) of the soil to water seepage in the region to the left of the given dividing line is k; in the region to the right it is 2k. The area is undergirded by impermeable rock. Estimate the seepage rate under the coffer dam per running foot in terms of k. Suggested

Steps:

Photocopy the figure, perhaps doubling its size. Assumethat the dividing line between the two. soils is an equipotential surface. (The problem is in fact arranged so that this is a line of ~irtually uniform pressure.) Guess streamline above which half the water flows. Take each of the two regions above and below this streamline and sketch further estimated streamlines which separate their respective regions into two equal-flow sub-regions. This process of subdivision can be repeated further, but as a practical matter you might do this only for the soil on the left, and there only once more. Starting from the line dividing the two soils, sketch candidate constantpressure lines, attempting to form squares with the flow lines.

750

CHAPTER

10.

ENERGYSTORAGE

FIELDS

3. Critique your results, paying special attention to whether the aspect ratios of your would-be "squares" are close to 1:1 and the flow and equipoteno tial lines intersect perpendicularly to each other. Erase and re-sketch repeatedly until you are satisfied. (This is not getting as fancy as use of the Thompsonor Dirichlet principles, which together are directed at estimating error.) Pay special attention to the region im~nediately below the coffer barrier, where a near-singularity exists that produces very small "squares." 4. Find the form factors A for each of the two soil regions. 5. Find the conductances of the two regions in terms of k. Relate the total flow to the two conductances. (Note that the pressure drops add, and the flows are common.) 7. State a compatible set of units for k, Q and AP.

PROBLEMS 10.1 Estimate the resistance of the fuse link of uniform thickness t. The constitutive relation (equation (10.3a)) is J = -aVe, where a is the electrical conductivity.

10.2 Estimate the inertance referred to the velocity of the piston below. The system is axisymmetric and is filled with a fluid of density p. ////

//////

////

............. "---- l

.

:~:

l ~ open end (submerged)

10.1.

FIELD LUMPING

751

10.3 An impeller pump of the type used for cooling the steam in a power plant includes an inlet convergent section, a row of stator vanes, a row of rotating vanes, a second row of stator vanes and a diffuser section, with the rotating shaft contained within a fixed central column. (The axis drawn horizontally below actually is vertical.) The transient produced when the pump is started and the water rises from its initial level has the potential of producing an unacceptable waterhammereffect, so it is important to knowthe inertia of the water.

~_

~¢rdcal

sm’tor #1

_

0 20

40

60

80

inlet

100

(a) Estimate this inertia for the system shown as a function of the depth of the water, starting at its initial level at the bottom of the diffuser and ending when the water has reached the top of the diffuser. (b) Construct a calculable model that accomodates the bond-graph restriction to constant inertances. Define the moduli of all elements.

SOLUTION TO GUIDED PROBLEM Guided

Problem

10.1

1-3. The following sketch represents about ten minutes of draw-and-erase-and-drawagain: air water

-

~

coffer dam air

~ water

rOCK,perme;

-

-

¯

752 4.

CHAPTER 10.

ENERGY STORAGE FIELDS

On - . the left side, the formfactor is aboutA1= 3-~_~;on the right side, A.~= ~_3~ 8 4

5. The conductances are the transmittances T1 ---- kA1and v2 = 2kA2. 6. Q = ~AP -- (l/T1)

1 (1/7~)AP ----

kAP

3 - hr’ fte [Ap] = psi; [k] = hr ftpsi" 7. [Q]- fta/hr ft

10.2

Discrete

Compliance Fields

The potential energy stored in a system sometimes cannot be separated into terms V1 (ql), ]}2 (q2),..., ~,, (qn), but rather includes irreducible mixed V(ql, q2,..., q,~) that are called discrete compliance fields. Examples below include an electrical capacitance that varies with the displacement of a plate or of the meniscus of a dielectric liquid, the pressure-volume-heat flux relation of an ideal gas, and, later, the piezoelectric and piezomagnetic materials used in transducers. Such a field is represented by a multiport compliance element of the form

el

em C"~ ~

-"

The present discussion largely assumes general nonlinear relationships. Many compliance fields are modeled as linear, however; such fields are discussed in Section 10.4. 10.2.1

Generic

Relationships

The compliance field with stored potential

energy

V = V(ql, q2,...,

q,~)

(10.15)

dV ~ OP dqi dt cgqi dt"

(10.16)

has as its net input power or energy flux 7 )= ~ i=1

ei?ti-

i=1

Since this result applies regardless of the magnitude of the individual Oi, the key result is the theorem of virtual work: e~ = ~q~.

(10.17)

10.2.

753

DISCRETE COMPLIANCE FIELDS

This equation computes ei given the values of qi, i = 1, 2,..., consistent with integral causMity:

ql

m, and thus is

e2ti’2", e qm

It should be noted that the theorem of virtual work also applies directly to a one-port compliance; the partial derivative in equation (10.17) simply becomes an ordinary derivative.

10.2.2

Examples

with

Geometrically

Varied

Capacitance

Any capacitor that has its capacitance varied mechanically serves as an electromechanical transducer, that is, a device that transforms electrical energy to mechanical form or vice-versa. As a result, the force that the electric field places on the movable member can do work. Similarly, mechanical work done externally on the moveable memberaffects the electrical field. The device becomes a transducer that converts electrical energy to mechanical energy, and vice-versa. It may be useful either as a sensor (instrument transducer) or as an actuator (power transducer). Two examples are given below. The mechanical motion in the second case is actually the motion of a fluid. The further example of a capacitace microphone is given as a Guided Problem in Section 10.4. Example 10.3 The upper plate of a parallel-plate normal to the planes of the plates:

capacitor

is movable in the direction

Model the system with a bond graph, and relate the voltage e across the capacitor and the force F on the plate as functions of the gap width, x, and the electric charge, q. You. may assume that the capacitance is given by eA in whiche is the dielectric constant, A is the area of the plates, and fringing effects of the electrostatic field around its edges are neglected.

754

CHAPTER 10. Solution:

ENERGY STORAGE FIELDS

The bond graph is X

The electrostatic form

energy F can be represented

in the standard quadratic

1 .~ x 2 F = ~-~q" = 2-~Aq . Thus, from equation (10.17), the voltage e and the force F are OF q xq Oq - C- Ae’

e-

2eAe q2 OF "z Ox 2cA 2x Note that the applied force F and velocity/c are in the same direction, so the power product F~ is positive when mechanical power flows into the compliance.

Different substances have different dielectic constants, and thus the capacitance of a fixed-plate capacitor changes when different materials are movedinto the gap. This is the basis of the device analyzed in the following example, which could be used to sense the level of a liquid such as water. It also could be used as the heart of a pump which has no solid moving parts; the author has suggested this concept for use in microelectromechanical systems. Example

10.4

Two immiscible fluids with different dielectric constants, either or both of which are liquids, are separated by a meniscus that lies in the gap between two fixed plates of a capacitor:

meniscus diel[c~r~c’~d~’~ Q

~ ~1

I.

L

]fiat

plate electrodes

[ /(

~’-~e~e~t~i[~uid , .I depth

_ Q d~ension:

w

Model the system with a bond graph. Then, relate the voltage e across the capacitor to the electric ch~ge q and the position x of the meniscus, assuming the parameters L, g, w, el and e~ and neglecting the effects of electrostatic fringing. Also, relate the horizontal force F that acts on the meniscus, producing the pressure drop Pl - ~, to the same variables and parameters.

10.2.

DISCRETE

755

COMPLIANCE FIELDS

Solution: The bond graph below employs the force F and velocity k to describe the fluid within the gap. They are related to the more conventional pressures P1 and P2 and flow Q by the transformer, the modulus of which is the reciprocal of the area of the channel.

P2I - F e 1 ~---’-~T .-~-~ C"~.

T

P The meniscus will be assumed to be planar; this simplifies the model, and tends to be approximately ~orrect because both the surface tension and electrostatic forces encourage it. The capactor then can be viewed as two capacitors in parallel, one containing the first dielectric fluid and the other containing the second dielectric fluid. The total capacitance is the sum of the two. (Both have a commonpotential, and currents or charges that sum, and therefore they can be bonded to a common0-junction.) Thus, C = elwx + g

e2w(L - x)

9 where x and L - x are the respective lengths of the two slugs of fluid within the gap. Note that the channel has a rectangular cross-section. The potential energy becomes 1 q2 _ V = Y(q, x) = 2C(x)

gqZ/2 w[e2L " + (el - e2)x I

Therefore, OV gq e -- Oq w[ezL q- (,1

--

£2)X]’

F = O_~Y = -9(el - e2)q ~ __ OX22w[e2L d- (el -- £2)X]

w (~ __ e,~)e2" 29

Note that the equations correspond to the standard use of integral causality. 10.2.3

Thermodynamic Coupling Thermal Energies

Between

Mechanical

and

The temperature of a body is changed if the body is significantly compressed or expanded in size, and heat transfer to the surroundings is restricted. The resulting coupling between mechanical and thermal energies can be represented conveniently by a two-port compliance. This idea is now developed using a hydraulic accumulator as a case study. A hydraulic accumulator is based on the compression of a fixed mass of gas. In Section 3.6.3 (pp. 166-167) it was modeled crudely by a one-port compliance. It was noted, however, that unless the compression and expansion takes place either very slowly (isothermal conditions) or very fast (adiabatic conditions), heat.

756

CHAPTER 10.

transfer effects are fairly significant. and thermal energy fluxes:

ENERGY STORAGE FIELDS

A better model recognizes both mechanical

The volume flow rate of the hydraulic fluid is written as (-(/’); the minus sign results from the definition of V as the volume of the gas, not the liquid which displaces it, in contrast to the earlier treatment. Assumingan ideal gas of mass m, the conventional thermodynamic assumptions are PV = mRO,

(10.18a)

V =-’U :- mc.vO,

(10.18b)

in which ]; _= U is the internal energy, Cv is the specific heat at constant volume and R is the gas constant. The crudest approximation being made is that the heat flux, dQ/dt = 0~, is proportional to.the temperature difference between the temperature of the environment, 0e, and the mean temperature of the gas: dQ d--~ = H(O~- 0): The first

(10.19)

law of thermodynamics requires dY = dU = dQ - P dV.

(10.20)

Substituting equations (10.18) and (10.19) into equation (10.20), mcv dO = H ( Oe - O)dt - (mRO/V)dV.

(10.21)

This can be rewritten in the final desired form as dO H RO dV -(Oe - O) + ----. dt cv V dt rncv

(10.22)

In practice, dV/dt is treated as a causal input, that is a function of external variables~ and is integrated to give V. Equation (10.22) becomes the state differential equation for the compliance, giving 0. Finally, the pressure P is calculated using equation (10.18a). Problems in which a flow of gas enters a chamber are considered in Chapter 12, along with fancier equations of state including phase change. The simpler problem of the thermal exl~ansion of a solid or a liquid is considered in Section 10.4 below. The whole area of coupling between mechanical and thermal effects is important to the modeling and analysis of many nominally mechanical systems.

10.2.

DISCRETE COMPLIANCEFIELDS

757

~’--"F flat plates [ electrode ~ "--] 7 ...... " ]" .... "-..] -T .......... ~ i~el"e~t r"o"d~ [ I..__.__1 ~ x -----4 g depth dimension:w e

F

Figure 10.5: Guided Problem10.2

10.2.4

Summary

Potential energy storages whichare irreducibly a function of two or moregeneralized displacements can be represented by multiport compliances. If the stored energy is knownas an expressed function of these displacements,as in the examples given for variable capacitance, the efforts or generalized forces are readily computedas the gradient of the function in the direction of the respective displacements. This computationis consistent with integral causality, and thus can be incorporatedinto the writing of differential equations for a larger system into whichthe compliancefield is imbedded. Purther compliancefields are presented in Section 10.4 and Chapter 12. Guided

Problem

10.2

Themovableplate of a variable-plate capacitor, shownin Fig. 10.5, is supported mechanically so that it can movefreely in the direction parallel to the plane of the plates. (This concept has been used in some microelectromechanical systems.) Represent the resulting coupling betweenthe electric and mechanical variables by an annotated bond graph. Find approximate relations for this coupling,neglecting the fringing effects of the electrostatic field. Suggested Steps:

1. Write an approximateexpression for the electrostatic energy of the field, as functions of the displacementx and the electric charge q. The examples above should indicate howthis can be done. 2. Find the force F and voltage e by computingthe appropriate derivatives, as indicated by equation (10.17).

758

CHAPTER 10.

ENERGY STORAGE FIELDS

PROBLEMS 10.4 The slider-crank mechanismshownbelow lifts a weight, mg, vertically. The weights of the other membersmay be neglected by comparison. Thus the representation M mg T S~

¢

~

can be used, in whichT = T(¢). Alternatively, however,from the point of view of the shaft the device tooks like a rotary spring:

M lowestposition

(a) Writethe potential energyof the spring as a function of ¢, and determine the compliance relation M= M(¢). (b) Find a linearized modelvalid for small displacements from a nominal angle ¢ = ~, including evaluation of the linearized complianceC* and any other parameter(s). 10.5 A movable-plate capacitor has its movingmemberattached to a spring, as shownbelow. The spring exerts zero force whenx = xo. For the voltage e being constant, find any equilibrium positions x and determine their stability. You mayexpress the answerson a sketch-plot of the electric and spring forces, rather than algebraically. Also, find the minimum value of k necessary for stability at a particular value of the equilibrium x, and the associated ratio xo/x.

10.2.

759

DISCRETE COMPLIANCEFIELDS

10.6 Anelectrostatic motor, shownin cross-section below, has a fixed cathode of radius r, a rotor of thickness t comprising alternate 90° segmentsof high dielectric constant ki~eo and low dielectric constant kLe0, and an outer anode with four equal segments. A high constant voltage v0 is attached to terminals A-Aand a zero voltage to terminals B-B; every quarter revolution of the rotor the voltages are switched. The unit has thickness w. The clearances maybe assumedto be very small.

segmented

rotor

B~~

(smallclearances)~:~k r~~ segmentedanode A""~...JI.....~ / ~13 (a) Estimatethe potential energystored in the electric fields of the motor, valid for up to 90° of rotation. (b) Find the torque, M, generated by the motorin terms of the information given. (c) Find the electric current required by the motor. 10.7 The bladder of a hydraulic accumulator is filled with an open-cell solid foam of mass my and specific heat c~, intended to inhibit circulation of the compressedgas, and therefore reduce heat transfer to the walls and increase the reversibility of a compression-decompression cycle. The gas satisfies the perfect gas law PVg= mgROg;the gas has mass rng, specific heat Cv~, volumeV~and temperature ~ga. Heat transfer betweenthe gas and the foam can be assumedto satisfy Q = H(8~- 0f), whereH is a constant and/91 is the temperature of the foam. The volumeof the solid material in the foam does not change, although the voids expandin size so that the foamfills the entire space available. Drawa bond graph modelfor the situation of a changing volume, but with no heat transfer to the walls. Define variables and parameters as appropriate, and write a (set of) state differential equation(s). SOLUTION

TO

GUIDED

PROBLEM

Guided Problem 10.2 1. ]2= gq2 2ewx 2. x e= -~q OY = gq~w:

F

gq2

~

OY ewe.~ =--~ = 2~wz~ = 2g

760

10.3

CHAPTER 10.

Displacement

ENERGY STORAGE FIELDS

Modulated Inertances

The inertances considered thus far have a single port, and can be represented by an algebraic relationship between the flow and the generalized momentum. A complication introduced in Section 9.1 allows the generalized kinetic energy associated with an inertance to be affected by either a local or a remotedisplacment. This is done by interposing a variable transformer betweenthe inertance and the rest of the system, so that the inertance itself is constant. The modulation of these transformers, like any tranformers, must not involve a direct exchange of energy with a remote displacement or velocity. Such direct exchanges between generalized kinetic energy storage mechanismsand separate displacements nevertheless sometimes occur. They require the use of one or more multiport inertances. The strictly analogous situation to a multiport compliancethat that depends on more than one generalized displacement wouldbe a multiport inertance that depends on more than one generalized momenta.It turns out that this is not a very useful formulation, however.This section introduces, instead, a multiport inertance with generalized kinetic energythat dependsjointly on one generalized momentum and one or more separate displacements. Further, energy is directly exchangedwith these displacements. 10.3.1

Inertances

Dependent

on Holonomic

Constraints

A variable width flow channel with an inertive fluid flow is an exampleof a generalized kinetic energy that depends on a generalized momentum and on a remote displacement, namelythe width of the channel. The generalized kinetic energy in the major class of systems being considered is represented in the form T = T(pl, q~, q3,..., qm).

(10.23)

The variables in parentheses are treated as state variables. To be consistent with integral causality, they are the time integrals of the causal inputs to the energy-storage element. Accordingly, the mixedcausality employedis indicated by the causal strokes below:

q~emI~m’" q3 Only the first bond has generalized momentum as its input. The inertance can be said to refer to pl or Ol but at the sametime be modulatedby q2,.. ¯, q,~. Takingthe time derivative of equation (10.23) in the standard fashion,

dT

~

d-~=~-"e~(l~i=

0T

0T.

0T.

Op-~el + -~"~q~q2q~ +... +--Oq,~qm

(10.24)

10.3.

DISPLACEMENT MODULATEDINERTANCES

761

Therefore,

OT

(10.25a}

~1 = ~Pl

OT

ei = ~

i = 2,...,m

(10.25b)

Notethat if a displacementaffects the generalized kinetic energy, and that displacment is changed, powermust flow into or out from the inertance through the bond associated with that displacement or generalized velocity. The inertance is said to interrelate the variables of the various bonds by a holonomic constraint. The contrast to the nonholonomicconstraint is made below. There are holonomicconstraints that do not fit equation (10.23) and are 3beyondthe scope of this chapter. The examplebelow models the solenoid, which is the most commontype of electromechanicalactuator. It also can be used in the inverse wayto sense electrically the position of a mechanical member.Anotherexampleof a holonomic modulationof an inertance is given in GuidedProblem10.3. Example 10.5 A solenoid, pictured in cross-section below, can be described as a variablegeometryinductor. It comprises a coil with inductance dependent on the mechanicalposition of a movableferromagnetic prismatic core.

Represent the solenoid by a bond graph with the preferred causalities, and relate the causal output electric current i and mechanical force F to the input displacement x and voltage e. The dependence of the inductance on x, that is the function I(x), can be assumedas known. 3The approach here is a special case of the Hamiltonian 13 that may be available from the author.

method presented

in a Chapter

762

CHAPTER 10.

ENERGY STORAGE FIELDS

Solution: The local generalized displacement, q, is the time integral of the electric current, whichis the local flow. The local generalized momentum, p, is the integral of the voltage. Thegeneralized kinetic energycan be written in the form 2T = 2/~p in which x is a non-local displacement. The causal bond graph is

Substitution of the equation for ~- aboveinto equation (10.25) gives the results P

2[I(x)] 2 dx 2 dx In a particular design, the detailed task nowis to evaluate I(x). Withluck you could ~sign this job to a speciMist. 10.3.2

Inertances Constraints

Dependent

on Nonholonomic

Consider a constant inertance Ic driven through a transformer with modulus T(x) that depends on a remote displacement, x: IT f edt=~llc (l=p/TIc (lc=T(t=p/Ic

T= T(x)

Althougha variation in x changes the stored energy, the powerflow associated with the change equals the product of the generalzied velocity ~ and effort e exclusively; no generalized force must act on the displacement x itself. This kind of relationship is an exampleof a nonholonomlcconstraint. It is the sametype considered early in Section 9.1. By definition, a holonomicsystem has the same numberof generalized coordinates as degrees of freedom; a nonholonomicsystem has more generalized coordinates than degrees of freedom. The displacements 0, ~c and x are generalized coordinates, but the transformer ~ntroducesa constraint that reduces the degrees of freedom(the numberof geometrically admissible infinitesimal changes in displacement) without reducing the numberof generalized coordinates (the minimumnumberof displacements needed to specify the displacement state of the system). Specifically, Aqc is madedependent on Aq, while still requiring values of q, qc and x to describe the displacementstate. Observethat the generalized kinetic energy of the bond graph above can be written as

=

10.3.

DISPLACEMENT

MODULATED INERTANCES

763

This is the same as the following expression for the generalized kinetic energy of the holonomic variable-geometry inductor or solenoid of Example 10.3 above, T = ~I(x)i

2,

(10.27)

if the associations ~ = i and T~’(x)Ic = I(x) are made. Nevertheless, the bond graph above is incompatible with the bond graph of the solenoid, and the general holonomic relation of equation (10.23) cannot represent a nonholonomic system. An attempt to represent the variable-geometry inductor by the nonholonomic bond graph above, for example, would result in a system with no mechanical force on the moving member, which is obviously incorrect. Equation (10.23) expresses the energy in terms of the generalized momentum. It is not possible, in fact, to express the energy of the nonholonomic system above in terms of its generalized momentum,which is f e dt = f T(x)[9 dt (except in the trivial case in which x is a constant, which voids all coupling between the inertances and x.) Therefore, the two bond graphs themselves, or energy relations expressed in terms of generalized momentaas opposed to generalized velocities, are sufficient to distinguish the two types of constraints. The following example represents conceptually the nonholonomic coupling between a flywheel used to store energy for propelling a vehicle, and the vehicle drive wheel. It bears a superficial resemblance to a variable speed flywheel with a holonomic constraint given in Guided Problem 10.3, but the behaviors are dramatically different. Example 10.6 It is proposed to store kinetic energy for a bicycle or other vehicle in a flywheel. The flywheel is driven from a drive wheel by a transmission with a continuously adjustable speed ratio that is a knownfunction of the axial displacement of the flywheel, x:

x Adjustable Position

Rolling Contact (BearingNot Shown)

Part Of F"

764

CHAPTER 10.

ENERGY STORAGE FIELDS

Identify whether the system is holonomic, and modelit with a bond graph. Then, find the associated equations, assumingthat the momentMapplied to the drive wheeland the position x of the flywheelare specified. Solution: The bondgraph below represents the fixed inertia of the flyweel by. Ic. The speed ratio betweenthe angular velocities of the drive wheel (¢) and the flywheel/~)c) is represented by the modulusof the transformer, T(x). This transformer constrains infinitesimal displacments of qc without reducing the numberof generalized coordinates (three, for q, qc and x) necessary to describe the positional state of the system. M

Q~) ~:ic

The associated differential equation and output variable, as the annotations indicate, are

dp

1

dt - T(x) 1 ~- IcT(x)P, Note that adjustment of the friction drive requires zero essential force and energy, consistent with a nonholonomicmodulation and inconsistent with a holonomicmodulation. Minorforces due to the accelerations or friction in the adjustment mechanismcould, if desired, be modeledseparately. Heat transfer and thermal expansion can changethe modulusof an inertance. This change can be represented as a nonholonomiccontraint that modulates a transformer bonded to a fixed inertance. An exampleis offered in Guided Problem10.4. Heat transfer is possibly the most ubiquitous (and as such underrecognized) of all nonholonomicconstraints.

10.3.3

Summary

The bond graph formalism employedin this book requires that one-port inertances either be constants or at most functions of the local generalizedvelocity. A kinetic energy that dependspartly on a local displacementcan be represented by a displacement-modulatedtransformer in series with such an inertance. A kinetic energy that dependspartly on a non-local displacement can be accomodated the same wayif and only if the constraint is nonholonomic.Otherwise, it can be represented as a multiport inertance field, in ~nuchthe sameway as compliancefields are treated. The energy should be expressed as a function of 4the primary momentum and the remote displacement or displacements. 4In the reference cited in footnote 3 on page 761, the bond graph for the holonomicsystems considered in the present section becomesa special case of what is called a Hamiltonian bond graph.

10.3.

DISPLA

765

CEMENT MOD ULATED INERTANCES

Cylinderof Radiuso Rolling On Drive Wheel

Port Of Frame x DriveWheel.-----~I.~...%~,,.~

I|1 f.,~’J~. ~

Figure 10.6:..Guided

ble With

Problem 10.3

Holonomic constraints produce generalized forces on all bonds of the multiport inertance. Significant power is apt to flow through any of the bonds. Nonholonomicconstraints have no essential efforts associated with their influence on the kinetic energy through their modulation of the transformer. No power is required to effect this modulation, apart from any parasitic friction or other minor non-essential forces that you might wish to model separately. The contrast is illustrated in the comparison of the systems in Example 10.6 above and Guided Problem 10.3 below. Guided

Problem

10.3

A flywheel with adjustable inertia comprises two blocks, each with mass m/2, which move radially on rods welded perpendicularly to the shaft, as shown in Fig. 10.6. The blocks are held by cords that pass over p~ulleys; the centers of mass can be drawn inward or released outward with this cord. Model the system by a bond graph, define state variables, and find the state differential equations. The blocks may be idealized as point masses, all other inertias may be neglected, and all friction may be neglected. Suggested

Steps:

1. Define the adjustable radius, x, and the angle of the shaft, ~b, as the generalized displacements. Write the kinetic energy of the idealized system

766

CHAPTER 10.

ENERGY STORAGE FIELDS

heat flow

~ flow

in

~

insulation

~

~-~.

flow

out

Figure 10.7: Guided Problem 10.4 in terms of these displacements and associated generalized momenta. Note that the kinetic energy for radial velocity of the blocks can be expressed in a term separate from the kinetic energy for rotation. Drawa bond graph, apply causality, and identify the effort and flow variables. The torque on the shaft and the force on the cord may be considered as independent variables. Write the differential equations using the standard techniques, except that equation (10.25) above is used for the two-port inertance. Guided

Problem

10.4

The temperature-regulated valve shown in Fig. 10.7 is used with a highly unsteady flow, so the flow. inertance I as well as the flow resistance R are of interest. Consider that these parameters depend on the temperature of the expandable member in a known way. The thermal compliance of this member is C, which can be considered to be a constant. Its temperature is 0, the temperature of the environment is 0e, and there is a thermal conductance H between the two. Find differential equations which if solved would give the volume flow in response to the pressure drop AP and the environmental temperature. Neglect any heating or cooling of the expandable membercaused by the flow. Suggested

Steps:

1. Identify the state variables. 2. Formulate the key constraint holonomic.

and identify ,whether it is holonomic or non-

3. Draw a bond graph and label the variables. 4. Write the differential equations. (Since you are not given the forms I(O) and R(0) and are not asked to model these, the equations are not yet in a form ready for solution. This could be done as an extension to the problem, however.)

10.3.

DISPLACEMENT

767

MODULATED INERTANCES PROBLEMS

10.8 Identify whether the constraints in the five systems given in Problems 9.1, 9.2 and 9.3 (pp. 674-675) are holonomic or nonholonomic. 10.9 The electromechanical balance below has a position-dependent and inductor with values as given (which neglect fringing effects). is very small; x << d.

capacitor The motion

?rigid balance beam I ,~ .2 d+x

’TieA C(x)

R~~

coil fixed to frame ~ L(x)=Lo+aCr

(a) Model the system with a bond graph, and evaluate all moduli in terms of the given information. (b) Find the describing state differential

equations.

(c) For i = i0 sinwt, determine the frequency w for which the time-average of position x equals zero when no external force is applied to the beam. 10.10 A liquid-filled tank of uniform area AT is drained through a tube of uniform area AL and length L. Initially, a valve located a distance Lo downstream of the tank is closed, and the section of the tube downstream of this valve is empty. The valve is opened abruptly, giving a flow equal to a Ax/-~-fi, where zXP is the pressure drop across the valve and a is a constant. No other frictional pressure drop need be considered, and the liquid-air interface remains virtually normal to the axis of the drain line.

valve

area A L

(a) Model the system with a bond graph. (b) Find a solvable set of differential equations, which if solved would give the behavior of the system valid until the liquid-air interface reaches the downstream end of the tube.

768

CHAPTER 10.

ENERGY STORAGE FIELDS

10.11 Anincompressible fluid of density p flows through a long narrowchannel with width, x, that can be varied by the application of a force, F. Modelthe system. Find differential equations that interrelate the flow Q, the position x, the velocity 2, the force F and the pressures P1 and.Be. Viscosity and Bernoulli losses maybe neglected.The velocity of each fluid particle is virtually horizontal

(i.e. w~<< Q/wx).

~F

10.12 A cylindrical piston with dimensionsas shownis (i) supported submerged on a spring with stiffness k in an incompressibleliquid of density p, or (ii) floats on the liquid with the equilibrium position as shown.In both cases the cylinder is only slightly larger in diameter than the piston. Viscous effects maybe neglected.

m

area: A

area:A cylinder areas: Ac

(nominal)

P

(a) Find the effective mass, stiffness and natural frequencyin case (i), includingthe effect of the kinetic energyof the liquid in the annularregion.

(b) Definestate variables and find differential equationsfor case (ii) form suitable for solution.

(c) Linearize the equations of part (b), combine them to leave only position variable, and determinethe responseto an initial condition which is not far fromequilibrium.

10.3.

DISPLACEMENT

MODULATED INERTANCES

769

10o13 A dielectric fluid with density p and dielectric constant e is pumpedfrom a lower reservoir to a reservoir h higher. The environment is air, which has a dielectric constant e0 < e. This is accomplished by placing a ~roltage vo across a pair of plate electrodes of width w and separation d. As soon as the fluid reaches the top of the right-hand plate the voltage is discontinued, but inertia keeps the fluid moving upward for a while. The fluid that rises above the top of the right-hand plate falls back into an upper reservoir; the fluid remaining between the plates falls back to the lower reservoir. The cycle is then repeated. Notes: Parts (a)-(c) can be done independently of one another. Viscous wall-shear effects maybe neglected, but not all other resistive or inertial effects.

~upper reserv°ir il~lA

N1

lower reservoir,

(a) Model the system when the surface of the liquid is rising between the plates. Evaluate the parameters of your model, or alternatively give expressions for their energy in terms of defined state variables. (b) Model the system after the surface has broached the upper edge of the right-hand plate but before it has stopped rising. Give parameters as in

(a). (c) Modelthe system when the surface is settling Again, give the parameters.

downbetween the plates.

(d) If you have completed part (a), find a set of state differential equations to to describe that part of the cycle. Otherwise, find the equations for part (c) for part credit or part (b) for the least credit.

770

CHAPTER 10.

ENERGY STORAGE FIELDS

SOLUTIONS TO GUIDED PROBLEMS Guided

Problem

10.3

for the radial velocity is p~ = m~, 1. T = ½rn[(x~)2 ÷ ~o_]. However,the momentum and the momentum for the rotational velocity is pC = mx~. Therefore,

Op¢ OT OX

"~ rnx p~ 3ll~X

The differential equations are dt

dp~ dt dx 1 dt Note that ~ = p~/m. Guided Problem

~ mx

10.4

1. The entropy of the expandable member, S, and a momentumfor the flow, p, are proper state variables. The need for a constant inertance (to satisfy the requirementsof our schemefor writing differential equations) affects the choice of p, however, as noted below. 2. Both the resistance to the flow, R = R(8, Q), and the inertance, I = I(8), assumedto depend only on the temperature, 8, which is a direct function of S. Our schemefor writing differential equations requires that we employa constant inertance, however.This could be Io = 1(80), whichis connected to the flow Q a transformer with modulus T -~ T(~) -~ V/~/I(8o). But does this transformer represent a nonholonomicconstraint, which would deny any further coupling between the fluid flow and the entropy (or temperature)? The literal answer to this question is no. This follows from the observation that the generalized velocity TQis not integrable, that is f TQdt is a quasi-coordinate not representable by a function other than its being a time integral of a state function. Further, an accurate accounting of the kinetic energy of the fluid in the gap includes the effect of the velocity at whichthe gap opens or closes. This gives a direct force on the movingmemberdependant on both this velocity and the fluid flow. The problem is not included in the restricted class represented by equation (10.23).

10.4.

771

LINEAR MULTIPORT FIELDS

Nevertheless, the velocity at whichthe gap opensor closes is likely to be so small that its effects on the kinetic energy of the fluid are negligible. The change in the gap due to the strain of the expandable memberthat results from the fluidinduced force on it also is likely to be negligible. Therefore, the modelercan be quite justified to treat the coupling as thought it were nonholonomic.Nofurther coupling betweenthe thermal and fluid parts of the system is recognized by this approximation. This interpretation is consistent with the problem statement. The bond graph has two pieces:

1=I(Oo).

10.4

Linear

Multiport

Fields

Multiport inertance, compliance and resistance fields often are assumed to be linear. Linearity introduces special properties that allow the useful concepts of mutual inertance, mutual compliance, mutual resistance and certain standard bond graph equivalences to be employed. Applications considered below with a single power type include electric transformers that have two or more coils, and inertive rigid bodies that are actuated from more than one pivot point. These two cases will serve to introduce the linear inertance and resistance fields. Applications with two power types include the piezoelectric transducer (electric and mechanical powers, and electrostatic and strain energies) and the piezomagnetic transducer (electrical and mechanical powers, and magnetic and strain energies). These are used to introduce the linear compliance fields, which are finally applied to a general linear material with several coupled energy storage mechanisms.

10.4.1

Mutual Inertances and Resistances: the Real Electrical Transformer

A real transformer stores energy in the magnetic fluxes that pass through magnetically permeable material (most often iron) and are generated by the currents il and i2 in the two coils. The electric circuit symbol for the transformer, given in Fig. 10.8 part (a), includes two parallel lines that suggest the permeable core. The energy for the most general linear two-variable inertance may be written

772

CHAPTER 10.

ENERG-Y STORAGE FIELDS

e2

el

(a) circuit diagram symbol

(b) bond graph with mutual inertance

e1

ll lil

~

T ~

~

e212

0 -

e1

" T

(c) equivalem graph with transformer

i2’

(d) circuit graph equivalent

(e) reduction with perfect flux linkage

(f) added reduction for 12~ or ¢o-~o

el

~’~

Ts’----~

!?"~-T~ -------

0 ~.~ 0 li2’

(g) bond graph with resistances added

Figure 10.8: Non-ideal electric

(h) equivalent bond graph

transformer

o-i’

e2

10.4.

LINEAR MULTIPORT FIELDS

773

in the quadratic form 1 T = 1-I, i21 + I12ili., + -~Igi;.

(10.28)

For the transformer, I1 and I2 represent the inductances of the separate coils and I12 is knownas the mutual inductance Of the coils. (The letter I is used for inductance, rather ~han the more conventional L, to reduce the number of symbols.) It is this mutual inductance, due to the magnetic flux linkage through the commonpath for the two coils, that makes the transformer work. The currents il and i2 are appropriate flows or generalized velocities, as with the ideal trm~sformer. The inductances I1 and 12 then become common one-port inertances. It is convenient to represent I12 as a special two-ported mutual inertance, as shown in part (b) of the figure. Its respective efforts are el = I~2di.2/dtand e2 = lredi~/dt. A bond graph with a mutual inertance is awkward to implement causally in order to find the corresponding state differential equations. It serves very well, however, as a bridge between the energy expression and a corresponding bond graph that is indeed in a convenient form for application of the causal strokes and the writing of the differential equations. This latter bond graph is given in part (c) of the figure. It is a universal equivalent to the graph of part (b), assuming the parameters If and T are properly determined. For the equivalence to apply, the only. requirement is that the two bond graphs must represent the same energy in terms of their boundary variables. This new graph represents the energy 1 , .2 1_ .,2

(10.29)

which equals the energy of equation (10.28) if and only ~2" = i2 + Til,

(10.30a)

T = ~-2’

(10.30b)

I~ = I~ - T2I.~~ = 11- --.I~2 (10.30c) 1.2 You may verify this claim by direct substitution. Equation (10.30b) gives the junction structure of Fig. 10.8 part (c), which iucludes an ideal transformer. Equation (10.29) shows that two inertances I~ and I2 should be attached. This bond graph model can be converted back into the circuit diagram of part (d) the figure, which includes an ideal transformer, a series inductor and a parallel inductor. This diagram appears commonlyin textbooks for electrical engineers. Electrical engineers are taught how to write differential equations for circuits like this upon examination of the circuit diagram; you may accomplish the same result by working with the bond graph directly. The best possible performance of an electrical transformer results when the mutual inductance 112 is maximized. The theoretical limit of I12 corresponds

774

CHAPTER 10.

ENERGY STORAGE FIELDS

to perfect flux linkage between the two coils, l%r this value, I~ vanishes, and T has its maximumvalue consistent with the physical fact that I~ cannot be negative:

T= ~v~-~/I~.

I1.~ : ~

(10.31)

The corresponding bond graph and circuit diagram equivalent are shown in part (e) of the figure. The series inductance has been ~liminated, which becomes important if the device is to operate well at high frequency. The current passing through the inductor I2 in part (e) decreases as the frequency of excitation is increased or as the value of I2 is increased. If this current is sufficiently decreased, the energy ~ : 2 becomes negligible, the inertance I2 can be erased from the graph, and the d~evice virtually reduces to ~ ideal transformer as shownin part (f) of the figure, as is normally desired. The designer of a transformer that is to operate at 60 Hz would m~eIe large enough so that this condition is more or less satisfied at that frequency. He does this by adding iron to the core. This explains why power transformers are typically so heavy. Energy dissipation has been ignored thus far. Linear dissipation may be approximated by the quadratic expression P = R~z~ + 2R~2i~i: + R~i~,

(10.32)

where R~ and R: refer primarily to the electric~ resistances of the two coils and R~: is a mutual resistance associated largely with the eddy-current losses in the core. This description leads directly to the gener~ bond graph model of part (g) of the figure. The mutual resistance can be replaced by the s~e type of bond graph structure employed for the mutual inertance, however, giving the equivalent bond graph of part (h). The anMogousequations to (10.28) (10.29) are (10.33a) P = R~i~ + R~i~, i2 = i2 + TRil,

(10.33b)

R~

(10.33c)

R~ = R~ - T~Re = R~ - R~ Re

(10.33d)

Muchof the complexity of this model evaporates if R~ is neglected. 10.4.2

The Rigid

Inertive

Floating

Link

The commonutility of the bond graphs above is suggested by a direct analogy between the transformer without dissipation and a "floating link" of rigid material, pictured in Fig. 10.9, in which the two pivot points are movedvirtually in a direction fixed parallel to one another, and only within a small maximum displacement, so the axis drawn between them remains nearly perpendicular to the direction of the motions. The two velocities &l and ~e are analogous

10.4.

775

LINEAR MULTIPORT FIELDS

(a) link with pivot points F~

I2

I1

I1’

I2

(b) equivalent bond graphs (compare to Fig. 10.8)

(c) springs and dashpots added (same as vehicle modelof Fig. 5,15 (p. 322))

C~

C2

./R a

I~’

I2

(d) associated bond graph

Figure 10.9: Rigid inertive floating link

.R 2

776

CHAPTER 10.

ENERGY STORAGE FIELDS

to the two electric currents, and the inductance energies become analogous to the kinetic energies of the mass m in vertical translation and of the rotational inertia mr2 in rotation about the center of mass (cm): m(ab - 2) rn(b~" +2r2)" re(a2 + r2)’ I1-(a+5) ’ 12 = (a+b) -~ ’ I12 - 2 (10.34) (a+5) The bond graphs and equivalent circuits for the electric transformer (as given in Fig. 10.8) also apply to the floating link. The equivalent bond graph with the transformer includes the parameters 2mr 2ab - r I~ = a~ + r2 ; T - a2 + r:. (10.35) The analogy to maximizing the flux linkage occurs when the mass is concentrated at a point (the mass center, of course), reducing the rotational inertia (and the radius of gyration, r) to zero. The subsequent analogy to neglecting the current i2 by making I~. or the frequency large corresponds to making the mass so large or the frequency so high that the mass center does not moveappreciably. By not movingit acts as a pivot point; the floating link becomesan ideal lever, which is indeed properly modeled by an ideal transformer. This analogy can enhance your feeling for the behavior of a real electrical transformer. In parts (c) and (d) of the figure, springs C1 and C2 and dashpots R~ and are added to the pivot points of the floating link. This model could represent a vehicle with springs and shock absorbers at its front and rear axles. The resulting motion involves heave and pitch modes. 10.4.3

Multi-Coil

Transformer

with

Perfect

Flux

Linkage

Occasionally a system model will have most of the characteristics of linearity, but nevertheless contain a nonlinear characteristic. A three-coil electrical transformer such as shown in Fig. 10.10 is such a case. A perfect flux linkage is assumed, which means that the magnetomotive forces of the three coils sum together. The individual magnetomotive forces are proportional to the currents and the number of turns of the respective coils. The bond graph of part (b) the figure results; there is a single inertance/inductance. Thus far the system acts as though it were linear. If the currents are raised high enough, however, magnetic saturation occurs; there is a virtual limit to the magnetic flux and to the energy that can be stored. The constitutive relation can be plotted in the form of part (c) of the figure. Thus in dealing with electromagnetic components you can encounter nonlinear inertances, unlike in classical mechanics where they do not exist by definition. This nonlinear inertance can be accomodated fairly simply by means already developed, however, because it has only one port. 10.4.4

The Piezoelectric

Transducer

A piezoelectric crystal stores energy by virtue of both its mechanical stress or strain and its electric field or electric displacement (charge). The effects are

10.4.

777

LINEAR MULTIPORT FIELDS

il

"x

(b) bond graph

(a) schematic

P~g~etic

energy

;----~-R3

1

characteristic with saturation

TA+T~i~+T;~ Figure 10.10: Three-coil transformer with perfect flux linkage and saturation

coupled. It is supposed here that the stress or strain and the electric field or charge is imposed in only one direction, as shown in Fig. 10.11. The constitutive relation is further assumedto be linear: (10.36) The strain is S and the relative displacement of the two conductive surfaces is Ax = xoS, where xo is the thickness of the crystal; the tensile stress is a and the associated force is F = Aa, where A is the cross-sectional area; the electric field strength is E and the applied voltage is e = xoE; the electric displacement is D and the electric charge is q = AD. The mechanical compliance of the material is represented by the reciprocal of Young’s modulus, l/E, the electrical compliance (dielectric constant) by ~, and the piezoelectric coupling by de. The device is used as a transducer. This means that either a mechanical force or displacement is imposed and an electrical voltage or current is produced, or that an electrical voltage or current is imposed and a mechanical motion or force is produced. Such devices are used, for example, in crystal microphones and pressure transducers (to convert pressure or force o voltage) and in some vibration and valve control systems (to convert voltage to force and small motion). The behavior can be understood better in terms of bond graphs. The graph of part (b) of the figure shows three separate energy storage elements: C’¢ for strain energy, Cd for electric energy, and the mutual compliance C,,d for the

778

CHAPTER 10.

ENERGY STORAGE FIELDS

---~ x=.~+x*~(a) schematic ~xF0 "-’--"-C~d-’~----

0e ~ [

l

c~

q

C~=2/AE C~=A~/~

C~=d~

G

(b) bond graph wi~ mutual compli~ce ~--~- 0 ------~ x ~ ~ C~

T --~-~l ~ G

1q ~

T=Cd/C~a C~’=C~-Cof/Ca

(c) equivalent bondgraphfor computation

(d) equivalentcircuit diagram Figure 10.11: Piezoelectric transducer

10.4.

779

LINEAR MULTIPORT FIELDS

piezoelectric

or coupled energy. The stored energy can be written ~ + CadFe+ lzCdeU’z ~2 = ~C~F

(10.37)

It is awkwardto write equations directly from this graph, since there are three compliances but only two independent energy storages, so integral causality cannot be applied to all three compliances. This problem vanishes upon conversion to an equivalent bond graph with only two compliances, each with a single port. One of two possible versions is shown in part (c) of Fig. 10.11; the other has a mirror image structure, with the 1-junction on the left and the 0-junction on the right. This bond graph equivalence follows the same idea as the equivalence employed in Figs. 10.8 and 10.9 for the inertances. The relationships for the modulus of the new transformer and the modulus of the modified mechanical compliance C~ also are given in the figure. These are universal, and can be used whenever a graph of the form of part (b) of the figure is converted to a graph of the form of part (c). The equivalent circuit diagram is shown part (d). The bond graph or circuit diagram reveals potential limitations to piezoelectric transducers. At low frequency the compliance Cd can be seen to reduce the mechanical response to a voltage e, but not to current i -- 0. At high frequency the compliance C~ drains off flow, producing a break frequency above which the transducer largely fails to operate. This break frequency is raised as the mutual compliance Coa is increased, analogous to the effect of increasing the flux linkage in an electrical transformer or the concentration of mass at a single point in the floating link. In the theoretical limit of maximummutual compliance (not achieved with real crystals), C~ would vanish altogether, so that at sufficiently high frequencies the transducer would act just like an ideal transformer with modulusT. This objective, only partially realized in practice, is the ideal goal for the design of a piezoelectric transducer. 10.4.5

The

Thermoelastic

Rod

The thermoelastic rod stores energy by virtue of both its mechanical stress or strain and its temperature or entropy changes. Suppose that the stress or strain is imposed in only one direction,, as shownin Fig. 10.12. The constitutive relation is further assumedto be linear:

The strain, S, stress, a, and Young’s modulus, E, are the same as with the piezoelectric transducer: As = xoS and F = Aa. The change of temperature from some reference value is A0. The change of entropy per unit mass is As, so the total change in entropy is AS = pAxoAs. The thermal expansion coefficient, a, represents the thermoelastic effect, whereby not only does heating cause expansion, but stretching causes cooling.

780

CHAPTER 10.

]A

(a) systemand general bond graph (b) reticulated linearized At? ~ 0 model

~

ENERGY STORAGE FIELDS

~x r

[ ~ Coc~

0

l

F ~

T

~

0

Co=AE/L

cc =Ogpc

co ~

F

C~=I/Ltr

l

cc (c) equivalent bondgraph At? ~ 1 for computation

0

F ~

T= Cc/C~ Co= C~-C~/Cc

c~ Figure 10.12: Thermoelastic rod As with the piezoelectric transducer, the behavior can be understood better in terms of bondgraphs. The graph of part (b) of the figure showsthe analogous three separate energy storage elements: C~ for strain energy as before, Cc for thermal energy, and the mutual complia~aceC~c for the thermoelastic or coupled energy. The stored energy can be written in the same form as before,

v=

+c cF,O+ 2.Cc ,O

This graph is converted, as before, to the equivalent form shownin part (c) the figure, in order to permit integral causality. This graphreveals directly that 1

A0=

~-AS),

(lO.40a)

~cAS- ~F,

(10.40b)

1

demonstratingthat inwardheat flow reduces the tensile force (or increases compressive force) and that tensile force decreases temperature. ’ Equation(10.39) expresses the energy in terms of efforts. Alternatively, the energy could be expressed in terms of generalized displacements, ]? = y(Ax, AS),.

(10.41)

and the results of equation(lO.40) found in the traditional general manner Section 10.1: OV F = ~x" (10.42) A0= oA-~-~;

10.4.

LINEAR MULTIPORTFIELDS

781

Implementationof this approach employsthe inverse of equation (10.38). The initial bond graph with a mutual compliance has 1-junctions in place of 0junctions. See Problem10.14. This approach is often useful, as Guided Problems 10.5 and 10.6 are intended to demonstrate. 10.4.6

Piezomagnetic

(Magnetostrictive)

Transducer

A piezomagneticor magnetostrictive material is like a piezoelectric material except energy is stored in a magnetic field instead of an electrostatic field. Considerthe systemshownin Fig. 10.13 part (a), in whichthe magneticfield generated by a coil, and the core is madeof a magnetostrictive material which is fixed at one end. The constitutive relation is

=I where B is the magnetic flux density, H is the magnetizing force, ~ is the magnetic permeability coefficient and d, is the piezomagneticcoefficient. As with the piezelectric transducer, Ax = xoS and F -- Aa. The magnetomotive force Fm=zoH (which neglects the reluctance of the magnetic circuit other than the plunger) becomesanalogous to the voltage e, and the magnetic flux ¢ = AB becomesanalogous to the current 0- Fmis directly proportional to the actual applied electric current, however;specifically, Fm= 4pNil. Thus, the bond graph on part (b) of the figure is analogous to that of part (b) the preceding two figures for the piezoelectric transducer and thermoelastic rod, except that the coil is represented by the addedgyrational coupling with modulus G = 4pN. Similarly, the equivalent bond graph of part (c) contains only two energy storage elements, which are causally independent. To simplify this result, the 1-junction with its compliancecan be dualized to leave a 0junction with an inertance; the transformer thus becomesa gyrator and viceversa, as shownin part (d). Finally, the values of the gyrator and inertance moduli can be modifiedto allow the modulusof the transformer to becomeunity, so the transformer can be replaced by a simple bond, as shownin part (e). This is the simplest representation possible, unless one or more of the remaining effects are neglected. Note that, unlike the piezoelectric transducer and the thermoelastic rod, the piezomagnetictransducer is essentially gyrational. 10.4.7

Generalized

Linear

Media

The piezoelectric, thermoelastic and piezomagneticmedia considered aboveare special cases of moregeneral media.First, they were restricted to unidirectional strain and stress only. In general, there are 3 normal and 3 shear stresses and strains; the elastic behaviorcan be represented by a symmetric6×6array of elastic coefficients, leading to 6 one-port compliancesand 15 mutualcompliancesfor the 21 independentcoefficients. Second,adding directionality to the dielectric and magnetic permeability coefficients gives two 3 x 3 arrays with 3 two-port and 6 mutual compliances each. Finally, thermal expansion also can be added,

782

CHAPTER 10.

ENERGY STORAGE FIELDS

F

(a) schematic F ~ 0 "--~-~Cu-’~’~--

:~H e 0 ~ G ~

C~=.~/EA C2=AIz’:2

G =4 zN (b) bond graph with mutualcompliance

Cl

C2

(c) equivalent bondgraphwith transformer x~--~-

0 ~-I G’I---~,.-

0 ~- T ’~--

T’=G

(d) bond graph wi~ dualized region x

~

~ q

I~’=C~

(e) final reducedbondgraph Figure 10.13: Piezomagnetic(magnetostrictive) transducer

10.4. Table 10.1

783

LINEAR MULTIPORT FIELDS Sub-~natrix coefficients for energy storage in a mediumwith elastic, electric, magnetic and thermal energies Sij

8E,H,O ’ ~j~!

H~O dij~ TH8

Bm

mink T,E.O #ink

(symmetric)

p.~

Ek

p~.E

Hk

pcE,H,T/

o

AO

Thesubscripts i, j, k, l, m refer to Cartesian directions. sub-matrix E.H,O 8ijkl

and name of coeff, elastic

e,~kT’S’O dielectric pT,E,O magnetic permeability mk pcE’H’T/o specific heat °diHj~ piezoelectric

sym~netric? dimensions independ, coeff. yes

6× 6

21

yes

3× 3

6

yes

3× 3 1×1

6

not applic. no

6×3

18

1

dE,O

piezomagnetic

no

6×3

18

~Ej,H T,O

t~hermal expansion

no

6 ×1

6

mink

magneto-dielectric

no

3×3

9

p~H

pyro-electric

no

3×1

3

pTm’E

pyro-magnetic

no

3 x1

3

cI

energystorage

yes

13 × 13

91

~j~

to give an additional one-port compliance. The couplings between these four different types of fields can be described by 6 dielectric coefficients, 18 piezoelectric coefficients, 18 piezomagnetic coefficients, 6 thermal expansion coefficients, 3 pyro-electric coefficients and 3 pyro-magnetic coefficients, all representable by mutual compliances. The complete result is a symmetric 13×13 array of coefficients, as shown with matrix tensor notation in Table 10.1. Stress is represented by the symbol T, temperature increments by A0, and entropy increments by Aa. The 13 diagonal elements are representable by one-port compliances, and the 78 independent offdiagonal elements are representable by mutual compliances. This gives the bond graph of Fig. 10.14, in which the 78 bonds interconnecting the 13 zero-junctions implicitly represent the mutual elements. Fortunately in engineering one always vastly simplifies this extremely complicated general case. Probably no real material exhibits significant effects for all the coefficients or even a majority of them, and if it did the material itself or at least most of its effects would be ignored.

784

CHAPTER 10.

ENERGY STORAGE FIELDS

C~3~yy

~,~ ~x \C -~

Note: The78 lines connectingthe 13 0-junctions represent mutual compliances (~ C ~).

Figure 10.14: Compliancefield for energy storage in a volumeof a substance 10.4.8

Reticulation

Of General

Multiport

Fields

Breedveld5 has developed a congruence canonical form for representing a linear n-port C, I or R field with n one-port C, I or R elements interconnected with a "weighted" junction structure comprising 0 and 1-junctions and (n2 n)/2 transformers. Althoughnot unique, his structure appears to be as simple as any. The two-port case is identical to that employedrepeatedly above and inside the rear cover. The three-port case is shownin Fig. 10.15 part (c) for the compliancefield and Fig. 10.16 part (b) for the inertance field. The dual junction structures also can be used; these are shownin Fig. 10.15 part (b) and Fig. 10.16 part (c). (The options shown in the parts (b) are somewhat easier to manipulate. The two cases shownin the parts (b) are perfect duals of one another, as are the two cases shownin the parts (c).) The graphs not symmetricwith respect to the three ports; there are therefore six options for each field. The two-port fields also are not symmetric, as you have seen, and possesses two options. Integral causality is shownin each case; differential causality also maybe used. Thesefield reticulations are helpful becausethey exposethe essential transformational couplings betweenthe parts. If, for example, the effort el in Fig. 10.16 part (b) is madean independentinput (changing the causality of its bond and the bondfor C1), its contribution to e~ is simplyTa~e~and its contribution to e2 is simply T~le~. Also, state variables maybe associated with each energy storage element and the correspondingstate differential equations can be found by the standard methods. 5p.c. Breedveld, "Decomposition of multiport elements in a revised multibond graph notation," J. I~ranklin Institute, v. 318 n. 4 pp. 253-273, 1984.

10.4.

785

LINEAR MULTIPORTFIELDS

I ’C

C12

(a) bondgraph with mutualcompliances

(b) equivalent bondgraph(dual of Breedveldstructure)

Figure 10.15: Reticulation of a general linear three-port C field

786

CHAPTER10.

ENERGYSTORAGEFIELDS

V I

112

[]13

112

1"13]

12

123[ q ! /v3J

]23

(a) bondgraph with mutualinertances

T31 =113/13 Z21 = (123~13-I1313)/(~213-I132)

(b) equivalent bondgraph(due to Breedveld)

Figure 10.16: Reticulation of a general linear three-port I field

10.4: 10.4.9

LINEAR MULTIPORTFIELDS Further

787

Equivalences

Someof the bond graphs with mutual elements above e~nploy 1-junctions, while others employ0-junctions. Not infrequently, the most convenient energy expression is in terms of displacementinstead of effort, or momentum instead of flow. Similarly, the most convenient energy dissipation expression might be in terms of effort rather than flow. As a result, the type of junction to which the expression corresponds is switched. Equivalent graphs without the mutual elementsare given inside the rear cover for all cases. Determinationof these equivalences is straightforward. Anexampleis developed step-by-step in GuidedProblem10.5 below. 10.4.10

Summary

The energy stored or dissipated in linear models is represented by a sum of terms, someof which are proportional to the square of an effort, momentum, displacementor flow variable and others of whichare proportional to the product of a pair of variables. Those terms of the energy associated with each separate variable are directly representable, as before, by standard single-ported compliances, inertances and resistances. Those associated with a pair of variables are directly representable by special double-portedor mutualcompliances, inertances or resistances. It is possible to write state differential equations directly frombondgraphs that contain both self and mutual energy storage elements, but the numberof independentdifferential equations is less than the numberof elements, requiring someuse of differential causality. The recommended alternative is to construct equivalent bond graphs which contain only single-ported compliancesand inertances, plus ideal transformers. Theseequivalent graphs not only permit simple interpretation in the form of differential equations, but also reveal in familiar terms the inherent transformational structure of the coupled storage mechanism. The process of workingfrom the stored energy expression through a bond graph with mutual elements and then to an equivalent graph without these kinds of elementsand finally a set of state differential equationscan be simpler than the apparently more direct methodsthat have been employedpreviously. Guided

Problem

10.5

Find the general bond graph which is equivalent to that shown below, but contains no mutual element and can be given integral causality. The parameters C1, Ca and C1~ are constants. (This graph is used in Guided Problem 10.6 below. Note its difference from the graphs of Figs. 10.11 and 10.13.

C~

788

CHAPTER10.

ENERGYSTORAGEFIELDS i

rigid 13 diaphragm perforated ~ electrode electrode ~ ~ d+x

Figure 10.17: Capacitance microphoneof Guided Problem 10.6

Suggested Steps: 1. Write the total energy stored in the graph elements in terms of ql and q2. 2. Propose a graph with either of the forms 1.

T

0

Ca ~ 0 ~------

Cb T ~ 1.

Ca and also find its energyin terms of ql and q.~. 3. Equatethe two energy expressions for arbitrary values of ql and q2, so as tofind the constant parameters Ca, Cb and C12, 4. Applyintegral causality to the newequivalent graph. Guided Problem 10.6 A capacitance microphoneis b~sedon a parallel plate capacitor: one electrode is a diaphragm of effective area A and mass m which moves with a uniform displacement, x, while the other electrode is rigid and perforated to allow free passageof air, as shownin Fig. 10.17. The meanseparation of the plates is d, to give a capacitance of C = ¢A/(d+x). The effective stiffness of the diaphragmis

10.4.

LINEAR MULTIPORT FIELDS

789

k, which includes any effect of compression of the air in the volume behind the diaphragm. The voltage applied is ~ + e*, where ~ is constant; the associated electric charge is ~ + q* where ~ is constant. Write differential equations which model the dynamics for small changes x << d, e* << ~ and q* << ~. Suggested

Steps:

Write an expression for the energy stored (Y), including both the electrostatic energy and the elastic energy. Employas the variables the generalized displace~nent pair V (equal to Ax) and q. Note that when x = 0 the diaphragm has a strain deflection, which you could call -5, due to the charge ~.

Calculate the pressure P* -- 0Y/DV and voltage e* + ~ = O)?/Oq. Note that the constant terms in the first derivative must cancel, and the constant terms in the second derivative equal ~.

Complete the linearization started in step 2 by retaining only first order terms in V and q; the higher order terms are relatively small and may be neglected. Represent the results by the bond graph below. Evaluate the compliances as functions of the constant parameters.

4. Add the inertia of the diaphram to this graph. Find its value through use of its kinetic energy.

5o

Apply the bond graph equivalences found in Guided Problem 10.5 to eliminate the mutual compliance in favor of a transformer. Apply integral causality to the graph. Write state variable differential equations for the system, and relate the causal outputs to the state variables. P

790

CHAPTER 10.

ENERGY STORAGE FIELDS

PROBLEMS 10.14 A yoke shown below is rigidly attached to a rotating shaft at a right angle. A small mass m slides frictionlessly on a small rod attached to the yoke as shown; the mass is loosely constrained by springs with a net spring rate k. The yoke has momentof inertia I~ about the shaft.

shaft

(a) Write an e.xpression which approximatesthe system, using ¢ and 2 as the velocity variables.

kinetic

energy of the

(b) Find a linearized model for small displacements of z, identifying self a.nd mutual inertances as well as a compliance. Drawa corresponding bond graph. (Note: When~ is large, the nonlinear behavior of this system can be significant for fairly modest displacements. Problems of this type are addressed in the reference cited in footnote 3 on page 763.) (c) Find an equivalent bond graph with no mutual inertance. (d) Find the natural frequency of the model when the momentin the shaft is zero. 10.15 Oil-cooled transformers tend to be noisy. (You may have heard them buzzing on telephone poles.) Eddy current and hysteresis losses are proportional to the square of the rate of change of the magnetic flux. Magnetostrictive action of the nickel/iron core causes a dilational fluid flow Q. There is an effective compressibility of the oil, and the acoustic pressure P acts against an acoustic impedance with resistance R.~ and inertance I3. It is proposed that these effects can be represented in a summary fashion in the bond graph model on the next page. (a) Identify qualitatively and briefly the meanings of the various parameters (other than R2 and I3 which are already identified). (b) Identify what you consider the most glaring omission in this model. (c) Simplify the given graph in preparation for part (d), defining any elements in terms of what is replaced. Apply causal strokes, identify input variables, and relate the output pressure P to the state variables. (This part can be done independently of parts (a) and (b).)

10.4. LINEAR

791

MULTIPORT FIELDS

(d) Write in solvable form the set of state differential sponding to your graph of part (c).

~small /’-""oil (I

el

I ~ I

)/

volume of air (neglect r-,.’x for most purposes)

cooling

tubes

R~

(neglect)

all parameters assumedconstant

li 1

equations corre-

~Cl2~

1

I1

~

0

C 3

~

1

~

1

12

10.16 You are asked to evaluate the feasibility of basing a solar engine on the thermal expansion of a metal, avoiding some of the problems associated with fluids. Since the strain and percent change in absolute temperature are small, linearized characteristics can be used to model the behavior. Consider a unit volume of metal, for example in the shape of a rod, with uniform state, which undergoes a thermodynamic cycle by being alternately exposed to and shaded from sunlight. (a) Represent the strain, thermoelastic and thermal energies of the metal in the form associated with a linear model. Draw the corresponding bond graph, and convert this graph into an equivalent graph with only two energy storages, one associated with the strain, S, and the other with the changes in entropy, (fa. (b) Make the metal undergo a Carnot cycle, with the processes 1 to below. Find the corresponding stress (T) - strain diagram, and the thermal efficiency of the cycle. (c) Define the ratio of the actual thermal expansion coefficient to the maximum theoretically possible as e. Evaluate e for two or three metals for which you can find data (such as aluminum, iron, copper or magnesium), and compare to an ideal gas (such as air or a monatomic gas). (d) Find the ratio of the energy converted per cycle to the maximum energy stored during the cycle (at the end of process 1) in terms of

792

CHAPTER 10.

ENERGY STORAGE FIELDS

(e) Assume that the frictional loss per cycle is a fixed fraction, ], of the maximum strain energy. Find an expression for the mechanical e~ciency i~ terms of e and f. (f) Draw"practical conclusionsabout the feasibility of solar engines based on the expansionof metals from the results of the steps above.

10.17 A magnetic pick-off transducer comprises a U-shaped magnetic path including a permanent magnet with magnetomotiveforce (mmf) F0, a coil with N turns and an electrical resistor. Whenthe bar of magnetically permeable material (such as iron) passes by at a distance w and velocity ~:, as shownbelow, magnetic flux surges through the path, producing a pulse in the voltage e. The magnetic energy is f F de, in which the total mmfis F = Fo - Ni and the magnetic flux, whichis conserved around the loop, is ¢ = BA; the flux per unit area is B = pH = # dF/ds where s is distance along the path, and A is cross-sectional area.

(a) Relate ¢ to F as a function of the position x. All reluctance in the magnetic path maybe neglected in comparison to the reluctance of the air gap, where # = #0 and fringing maybe neglected. (b) Modelthe system with a bond graph. Note that the magnetic energy can be represented as a compliance. (c) WhenR is infinite so that i vanishes, find e and the force F~ which acts on the bar. (d) WhenR is arbitrary, find equations whichif combinedor solved would give e and F~.

10.4.

793

LINEAR MULTIPORT FIELDS

SOLUTIONS TO GUIDED PROBLEMS Guided

Problem

10.5

2. The two graphs are mirror images of one another. For the left-hand graph, 1T 2 Y=~-Tq’+~b(

1

1 ~ ~-~+~bb q~+-~bbq~q’~+~bqg 1(1 T2) ~

qI+q2)2=~

3. Equating the three respective coefficients gives, directly, 1

2T

1

1

T

C2 from which T = Ct2’

Guided

Problem

1

1

Ca = 1 - C~C~/C~2’

Cb = C~

10.6

1 [V .~)~ x. v =d+x ~(O;)+~-~(~ - ~)~- d+V/A,_ 7~-~~q+ q*?+ 1. ~ ~-~ = 2--e-~ (q + 2~q* + q,2) + 01.’--= OY d + VIA ..... Canceling the const~t terms, 1 ..~ P=~q*+~q-+~V

k

d , ~V+ e* = ~q + eA ~

1

Neglecting the higher-order terms in q* and V, and comparing to the bond graph, ~

.

k

.

1

1 ,

P= ~.~_,q+7~v= c-77~.~v" +~t. e* = ~..-~q* +cA,z Therefore, Cv = ~ ;

~-~qq*+ d Cq= e.’-~;

V] Cvv = "~ ~

794

CHAPTER

4.

I

10.

ENERGY

STORAGE

:.IlQ ~ 1 . 2 z = ~mx ; therefore,

FIELDS

I = m

l ~’Cvq-~-"--li__~--Z~-~iq 5.

1 P (~p/l" ~ 1 _l’-~--r:~__ Tl~--"."eo V/C~ -Q= V

q/C,~

c; whereT-6. The state equatiom are

~ =q

c~ C9 _ Ad, Cyq ~ v~ables

are

Cv k/A = Cv’ _ ~ 1 - CvCqC~q 1 - kd~AZ /~ no~d ~ the

~aph above,

~om w~eh the

d; . -~ = P-~vc~, TGq dV

1

~ d---~ =-i

dqdt = i + TQ = I + ~P and the causal outputs are Q --- l~- = -~p;

1 . e* = ~-q

~e~ntiM

Chapter

11

Introduction to DistributedParameter Models All physical systems are distributed in space, so it is natural to seek models that also are distributed in space. A distributed-parameter model has at least two independent variables, and possibly three or four. If the model is dynamic, one of these is time. The others are spatial, for example the Cartesian coordinates x, y and z. The lumped dynamic models considered exclusively thus far have one independent variable -- time -- leading to ordinary differential equations. The presence of two or more independent variables requires that distributed parameter models be represented mathematically by partial differential equations. A distributed-parameter model of a physical system is not necessarily more accurate than a lumped model of the same system. It also is not necessarily more difficult to handle than the lumped model. Distributed-parameter models should be thought of as alternatives which not infrequently produce superior results for a given degree of complexity. Nonlinear distributed-parameter models are commonly addressed with numerical simulation, like lumped models. Rather than delve into the complexities of simulation with two independent variables, however, this book assumes linearity and taps directly into the powerful and revealing analytical consequences of the property of superposition. Certain nonlinear boundary conditions will be treated, nevertheless. The models considered employ only one spatial variabl~. The restriction to one-dimensional models is not as severe as might appear, since two and threedimensional effects often can be represented. Further, complex systems often can be viewed as networks of one-dimensional distributed-parameter models joined at simple junctions. 1 Most phenomena which occur in two and three dimensions also occur in one dimension, however, and areintroduced here. 1Asecondchapter including morecomplexmodelsthan are treated in this bookhas been written and maybe available fromthe author. 795

796

CHAPTER 11.

DISTRIBUTED-PARAMETER M~

M~

MI

C (a) shaft

MODELS M~

!

(b) spring model

(c) inertia model

Figure 11.1: Rotating shaft and low-order lumped models

11.1

Wave Models Conditions

with

Simple

Boundary

The simplest and most commonly assumed distributed-parameter model is the pure bilateral-wave-delay model. Physical structures often so modeled include the rotating shaft, the rod with longitudinal motion, the acoustic tube, the stretched string with lateral vibrations, and the electic coaxial cable. The discussion below focuses on the Simple wave propagation associated with this model. Attention is restricted to steady-state boundary conditions, apart from an initial excitation. Methods capable of handing dynamic boundary conditions are discussed in Section 11.4. 11.1.1

Comparison

of

Lumped

and

Distributed

Models

Consider the cylindrical shaft of Fig. 11.1 part (a), which is constrained to rotate about its axis. Perhaps one end is attached to a motor and the other end to a load. If the torque is high but the angular accelerations are low, you might use the simple lumped model given in part (b) of the figure, in which C represents the flexibility (compliance) of the entire shaft. If, on the other hand, the torque is low and the angular accelerations high, the simple alternative lumped model of part (c) might be appropriate, in which I represents the inertia (inertance) of the entire shaft. For either the inertia or the compliance of the shaft to be negligible, as they are assumed to be in the two models, respectively, the frequencies of excitation must be low. For high excitation frequencies, both the torque and angular accelerations become large, and therefore both compliance and inertance are significant. If both the compliance and inertance of the shaft are important, one of the alternative lumped models shown in part (a) of Fig. 11.2 might suffice. For quite high frequencies, however, waves can be seen to propagate back and forth along the shaft, and the model of part (b) might be proposed. This model has seven lumps, and is not particularly simple to handle analytically or computationally. Further, it does a rather poor job of representing the wave propagation. More lumps would improve the representation, but at the expense of additional complication. As shown below, it is possible to characterize the shaft with a distributed-

11.1

WAVE MODELS WITH SIMPLE

BOUNDARY CONDITIONS

797

(a) symmetric third-order models M: 1 __..._.~ 0 .~_.~ 1 _...___~ 0 __....~ 1 ...~.~ 0 .~_._~ 1 M..~

I/6

C/3

I/3

C/3

I/3

C/3

I/6

(b) symmetric seventh-order model Figure 11.2: Higher-order lumped models for rotating

shaft

parameter model that has only two parameters, the wave travel time from one end to the shaft to the other, and the ratio of the angular velocity changes in these waves to the associated torque changes, called the characteristic or surge admittance. This model could be represented by a new kind of bond graph element, the pure bilateral-wave-delay element: M1 " M~ ~1 D ¢2 Analysis and simulation with this delay model can be considerably simpler than with a multi-lumped model, as may be seen intuitively and as will be demonstrated below. Further, the model is more accurate. Or is it, always? Nowconsider that the shaft is extremely small and comprises a perfect single crystal with lattice planes perpendicular to the axis. Further, presume that the frequencies of excitation are exceedingly high, near the natural frequencies of the individual atoms within the lattice. The lattice vibrations in this periodic structure 2 modif~v the gross wave behavior markedly, changing the propagation. velocity and introducing at least one absorption band. To model this behavior, you must return to a lumped model, using point masses for the atoms and lumped springs for the interatomic forces. You have seen an example in which a lumped model is simpler than a distributed model for the same system, and a counter example in which a distributed model is simpler than a lumped model. You also have seen an example in which a distriSuted model is more accurate than a lumped model, and a counter example in which a lumped model is more accurate than a distributed model. The most appropriate type of model to use tends to change as the scale of the system is increased or decreased dramatically; there could be several reversals if the change is over a truly enormous range. The same observation 2SeeLeonBrillouin, WavePropagationin Periodic Structures, first edition McGraw-Hill, 1946; secondedition, DoverPublications, 1953.

798

CHAPTER 11.

DISTRIBUTED-PARAMETER

MODELS

Oe_dx

c~ Figure 11.3: Bond-graph model for an infinitesimal like element

segment of a simple wave-

applies to changes in the frequency of interest. There are no simple rules that dictate the choice of a lumped or a distributed model. Modeling, again, becomes partly an art, based on knowledge, experience and intuition. Beware of the term distributed-parameter systems, since all physical systems are distributed over space. The term distributed-parameter model is clear, however. To a mathematician, nevertheless, a system is merely a set of equations. For example, a matrix partial differential equation often is called a distributed system.

11.1.2

The Pure Bilateral-Wave-Delay

Model

The rotating shaft is well represented by a simple wave-like model, as found in Example 11.1 below. Other examples that can be so approximated are given in Examples 11.2-11.5. The commonmodel for these cases can be represented by the bond graph of Fig. 11.3, which describes an infinitesimal segment of the structure having length dx; the independent position variable is x. If I’ and C’ are the inertance per unit length and the compliance per unit length, respectively, the inertance Idx and the compliance Caz are Id~ = I’

dx; Cd~ = C’ dx.

(11.1)

ExampIe 11.1 Identify the generic variables e and 0 and the parameters I ’ and C~ for a uniform elastic rotating shaft. Solution." For the shaft~ e represents the moment,M, and 0 represents the angular velocity, ~, as before:

The momentdue to inertia is the integral over the area of the radius times the acceleration times the density, and the momentof inertia I ~ is the ratio

11.1

WAVE MODELS WITH SIMPLE BOUNDARYCONDITIONS

799

of this momentto the angular acceleration, or I’ -’- f~ r(rO~/c~t)p(2rcrdr) 4_ p~ra

-

o$/ ot

The shear stress on a transverse element at a radius r is Ou in which /~ is the shear modulus. The momentthat this stress produces on the shaft can be found by integrating the product of the stress and the momentarm (i.e. the radius, r) over the area of the shaft, whichhas radius a:

47rpa

0¢ 2 Ox The complianceis the ratio of the twist in the shaft to the moment,or

f Ox Jo

C’- O¢/Ox = .2 Example 11.2 Represent the approximatebehavior of the longitudinal vibrations of a rod (or any slender prism) by the pure bilateral-delay model, defining the variables e and ~ and the parameters I and C. Commenton the accuracy of this modelas comparedwith the torsional shaft of Example11.1 Solution: Longitudinal vibrations in a rod or any slender prism can be approximatedin terms of the axial force, F, as the effort, and the longitudinal velocity, v = 0, as the flow:

~

~ BX

The inertance is simply the cross-sectional area, A, times the density; the complianceis the reciprocal of Young’smodulus,E, times the area, or: I’ = pA;

C’ = 1lEA.

The Poisson’s ratio, however,produces small transverse expansions and contractions. This implies the existence of small transverse velocities and associated kinetic energy. The gradient of the expansionsalso producessmall stresses in the transverse direction, and consequentstrains. As a result of these neglected secondaryeffects, the pure bilateral wavedelay modelis not nearly as accurate as it is for the twist of a shaft, whichhas no comparable effects. Note that the slenderer the rod or prism, the more accurate the model.

800

CHAPTER 11.

DISTRIBUTED-PARAMETER MODELS

Example 11.3 Represent the approximate behavior of acoustic waves in a tube by a pure bilateral-delay model,identifying the efforts and flows and the inertance and compliance. Comment on any limitations in the model. Solution: Acoustic wavesin a rigid tube are almost the same as the longitudinal wavesin the rod or prism of Example11.2, except that one usually uses the pressure perturbations as the effort and the volumeflow rate as the flow:

As a result, the inertance and complianceare I’ = p/A; C’ = A/3. In place of the Young’smodulusthere is the bulk modulus,8- The difference is that fluid does not expand!aterally, unlike the solid, and as a result the secondary effects of the solid are missing in the fluid. On the other hand, however,a real fluid has viscous effects whichintroduce dissipation that is missing in the pure bilateral wavemodel.

Example 11.4 Represent the approximatebehavior of the lateral vibrations of a stretched string by a pure bilateral-delay model, identifying the effort and flow and the inertance I’ and the complinaceC’. Solution: The string is presumedto be flexible enoughand stretched tight enoughfor its tension, T, to be considered uniformand constant. The effort is the lateral force, which equals the tension times the slope Oy/Ox, where y is the lateral deflection. This slope is presumedto be small. The flow is the lateral velocity Oy/Ot. ~+/~-Txe

dx q + d~-qdx

The product of the effort and the flow is the propagated power, as always. The inertance becomesthe mass per unit length, pA, and the compliance becomes lIT.

11.1

WAVE MODELS WITH SIMPLE BOUNDARYCONDITIONS

Table 11.1 Examples Modeled by the Pure Bilateral

rotating shaft

M

801

Wave Delay

~ v

rod, longitudinal F P tube, acoustic Q Oy Oy string, lateral T-~x Ot coaxial cable

e

i

Example 11.5 Represent the approximatebehavior for the propagation of electromagnetic wavesalong coaxial cable, whichis the simplest case of a pair of conductors. Identif~v the effort and the flow. Either computeor find in a reference the inertance I ~ ~. and the complianceC Solution." The variables for planar electromagnetic wavestraveling along a pair of conductorsare the voltage difference and the current. The compliance is the capacitance per unit length, and the inertance is the inductance per unit length.

e+~--ed~ dx ---~. The capacitance and inductance can be found in nearly any eleruentary text on electromagnetic theory to be C’ = 2~re ln(ro/ri)

I’ = ~ ln(ro/ri).

Here e and # are the dielectric constant and magnetic permeability, respectively, for the annular material, and ri and ro are the radii of the inner conductorand the inner radii of the outer conductor, respectively.

The results of the five examplesabove are assembled in Table 11.1, plus further consequencesof the modelsas discussed in the following subsection.

802

CHAPTER 11.

DISTRIB

UTED-PARAMETER MODELS

-to

~>e ÷ 0

Xo

x

Figure 11.4: Arbitrary traveling wave

11.1.3 Analysis of the Pure Bilateral-Wave-Delay Model It is convenientto employdifferential causality, as shownin Fig. 11.3 (p. 798). Solvingfor e and (1, 0e cO o e = e - ~xdX--Idx--~, 0e 0(1 (1 = (1 + -~x dx + Cdx

(ll.2a) (11.25)

Withthe help of equation (11.1) (p. 798) these give the differential equations

Oe

., 04

0(1 _C, COe

-5-~=

(1i.3)

Or"

These equations can be combinedto eliminate either e or (1. The latter case gives ’02e = I’C (11.4) 2" ~x~= - "~02(t -5~

at

There are two general solutions to this equation, whichcan be written as e+ = e+(x - ct);

e_ = e_(x + ct).

(11.5)

The terms in parenthesis represent the arguments of the functions. Imagine a pulse-like function e+, such as shownin Fig. 11.4. The peak occurs when the argumentx - ct equals x0. Therefore, if the function is plotted as a function of x, instead, the peak occurs whenx = xo + ct. As time advances this peak movesto the right at the constant velocity c, whichis knownas the wavespeed or celerity. The shape of the waveremains unchanged.The function e_ differs in that it represents a wavetraveling to the left with the samewavecelerity c. The complete solution is the sum of these two waves, which are transparent to one another. To prove that these functions are indeed the solution, and to evaluate the wavecelerity, you can substitute the proposedsolution into the differential equation (11.4). Thefirst and secondderivatives of each functiori :hie’indilated with

11.1

WAVE MODELS VVITH SIMPLE BOUNDARYCONDITIONS

803

respect to its argument(x- ct) or (x + ct) by e+~, e+" and e2, e2’, respectively. Therefore,

0%= eJ’ + e_";

O’~e

")2Ot Ox Withthis notation the substitution gives

c’~(e~/’ +eft).

e+" +e_" = I’C’ c’2(e-~’ +eft),

(11.6)

(11.7)

whichestablishes the correctness of the solution and gives the wavecelerity as

The consequentwavecelerities of the various physical systems considered thus far are included in Table 11.1. The bilateral wavesof effort e+ and e_ implywavesof flow, 0, also: gl =O+(x- ct) + ~l-(X +ct)

(11.9)

It is necessary to understand these flows, because the end conditions usually are expressedin terms of flow as well as effort. Substitution of only the righttraveling wavesinto equation (11.3b) gives

OJ= C’ce+’.

(11.10)

Thus 0+ and e+ are proportional to one another, with a ratio called the characteristic admittance,

~+ ~J - C’ c =

(11.11)

Substitution of the left-traveling wavesinto equation (11.3b) gives ~_’ = -C’c e_’,

(11.12)

so that

[

~-~-- = -Ye. I

(11.13)

The difference betweenthe left and right-travelling waves,therefore, is sign. This follows directly from the fact that ~ is defined as positive to right. The characteristic admittancesfor the specific cases considered in are entered into Table 11.1. The reflections of wavesat boundaries can be determinedfrom the relations, whichresult directly from the equations above:

the minus fromleft

gl=Ye(e+- e_),.

(11.14)

Fig. 11.3 following

804

CHAPTER 11.

DISTRIBUTED-PARAMETER

(a) t=O

(b) t= T/4

(c) t=T/2

(d) t=3T/4

(g) t=3T/2

(h) t=7T/4

~

key:

MODELS

string position ~ ~ q÷ wave ......

(i) t=2T

q. wave

~ .......

Figure 11.5: Plucked stretched string The inverse of these relations also is useful:

~(e- 4/~;). 11.1.4

Fixed

and

Free

Boundary

(11.15)

Conditions

A procedure for treating fixed and free boundary conditions is now developed, starting with the case study of a musical instrument with a stretched string that is clamped (fixed) at each end. As shown in Fig. 11.5, the string is plucked at location one-quarter of the distance from one end to the other. At the instant of release, the lateral velocity, ~, is zero for all x. Equations (11.15) require e+ = e_ = e/2 everywhere at this instant. Recall that e is the lateral force, which equals the tension T times the Slope of the string, Oq/Ox.. Therefore, Oq+/Ox = Oq_/Ox = (Oq/Oz)/2. The displacements q, q+ and q_ equal the integrals of these slopes over x, so you can conclude that, at the first instant at every location, q+ = q_ = q/2.. The bilateral waves are shown by the dashed line in part/a) of the figure. As the waves q+ and q_ propagate, relected waves are continuously generated at the ends of the string in such a manner that the end conditions are satisfied. At each end ~ = 0 and q = 0, which from equations (11.22) requires e+ = e_, which in turn requires q+ = q_. (Note that if q+ is the wave incident on

11.1

WAVE MODELS WITH SIMPLE

BOUNDARY CONDITIONS

805

the right end, q_ is the reflected wave there; if q_ is the wave incident on the left end, q+ is the reflected wave there.) The consequent wave pattern and total string position is shownin parts (b) - (i) of the figure for the times t = T/4, T/2, 3T/4, T, 5T/4, 3T/2, 7T/4 and 2T. It is helpful to include the virtuM wave shapes beyond the ends of the string in order to anticipate the pattern of the reflected waves which satisfy the boundary condition q = 0. The position of the string at t = 2T is identical to its position at t = 0, indicating a cycle time of 2T: Free boundary conditions can be treated in the same manner as fixed boundary conditions, except it is e rather than q or 0 which is zero. Thus, for example, at each end e+ = -e_ = 0/21~. This means that the reflected wave still has the same amplitude as the incident wave, but its sign is inverted. The.difference is illustrated by a compression wave in a tube approaching either a blo.cked or an open end. The blocked end produces a reflected compression wave, whereas the open end produces a reflected rarefaction wave. Fourier

11.1.5

Analysis

with

Fixed

or

Free

Boundary

Con-

ditions Functions of t.ime, t, are represented by a Fourier sum of sine and cosine waves in Section 7.1.1. With distributed parame.ter models there is a second independent variable, position, x; often it is useful also to represent a function of ¯ by a Fourier series. Continuing with the case study of the stretched string, with its fixed ends, Fig. 11.6 shows that the components of displacement q(x) can be represented by the sine series = q,~ sin(n~rx/L).

(11.16)

This is similar to the sine series for an odd periodic function of time, except that the interval is 0 < x < L rather than -T/2 < t < T/2. As a result, the lowest or fundamental harmonic has one-half a wavelength over the interval, rather than a whole wavelength of the periodic function of time. This is ~cceptable because, unlike before, the function does not have to repeat for intervals outside of its length, L. Values of qn can be found to match any given continuous function q(x) that satisfies q(O) = q(L) = 0. Multiplying both sides of equation (11.16) sin(m~x/L) and integrating over the interval 0 < x < L, in standard Fourier fashion, q(x)sln

~0 L

.

---f

(mTrx)~

T- = q,,

~ L

sin

(n Trx)__~

si n\(m~rx~L

]dx

re=l,2,....

(11.17) As before, the integral on the right side is non-zero only when m ~ n, and then it equals Lqn/2. Therefore,

806

CHAPTER 11.

DISTRIBUTED-PARAMETER MODELS

exact deflection

note: deflectionsare greatly exaggerated x=L x=O x=L

x--O

(a) first three harmonics

(b) sumof first 12 harmonics

d full-wave/~ravel:,~ (competecycle),l I" half-wave travel:~2/2---~2/2 ~ (sum reverses) quarter-wavelengthtravel:~-2/4--~2/4-~ (sumcancels)

".,~,~~:’:., : ....... .,..::’.~2’:’:.,.:.’..’...:.-., first harmonic: L=2/2~i.----- L -~ secondharmonic:L=A .~, third harmonic:L=3A/2 ~,

L

.:’. .... ,-

’-~ L

,~

(c) right and left-traveling harmonicswith standing-wavesums Figure 11.6: Fourier sine series decompositionof initial string displacement

11.1

WAVE MODELS WITH SIMPLE

BOUNDARY CONDITIONS

q~ =-~2 foL q(x) sin(n~rx/n) dx. For the initial

807

(11.18)

position of the string as shownin Fig. 11.6, q(x) = 4qox/L; = 4q0(L-

0 < x < L/4, x)/3L;

L/4 < x < L.

(11.19)

Substitution of this function into equation (11.18) gives, after cancelation several terms, q~_ 32 (nTr) (11.20) q0 3(n7~) 2 sin ~ . The respective values of the first 12 harmonies are 0.764, 0.270, 0.085, 0, -0.031, -0.030, -0.016, 0, 0.009, 0.011, 0.006, 0. Part (a) of the figure shows the first three harmonics and their sum. Part (b) shows the sum of the first twelve harmonics, which is a reasonable approximation to the complete straight-line V shape. Each harmonic produces a pair of bilateral propagating sinusoidal waves of one-half the amplitude of the harmonic itself. These waves all propagate at the constant wave celerity, and the sumof each pair satisfies the end conditions. Part (c) of the figure is intended to show how the propagation of the componentpairs of traveling waves sum to produce standing waves with the periods Tn = 2L/cn. The sinusoid depicted with a solid line represents the left and right-traveling waves of some harmonic at a moment when they are superimposed. The dashed lines represent the waves after they have propagated a distance of one-quarter of a wavelength. They sum to zero, so the standing wave has advanced by onequarter of a cycle. The dotted lines represent the traveling waves after they have propagated a distance of one-half a wavelength. They sum to exactly the negative of the original standing wave, and therefore the standing wave has advanced by one-half cycle. The cycle is completed when the left- and righttraveling waves have moved a whole wavelength. The w-avelength of the nth harmonic of the stretched string is 2L/n, and the wave speed is c; therefore, its frequency is 1/T~ = nc/2L. Musicians call the second harmonic3 the first overtone; a frequency doubling is precisely one octave. The frequency of the third harmonic, by the same reasoning, is triple that of the first, and so forth. For example, if the fundamental frequency is 256 Hz (middle C), the first overtone is at 512 Hz (the next higher), and the second overtone is at 1024 Hz (G above that). (If you raise dampers for this higher C and G on a piano, by holding down their keys, and then strike middle C, these higher strings will resonate. Their reverberation is audible if the middle C is quickly damped. This phenomenon does not occur for non-harmonic notes.) 3Someauthors identify as the first harmonicwhat is called here the secondharmonic, and the secondharmonicwhat is called here the third harmonic,etc. Their use of the word "harmonic"becomesessentially the sameas the present use of the word"overtone."

808

CHAPTER 11.

DISTRIBUTED-PARAMETER MODELS

compression rarefaction

@~

compression

rarefactio@

Figure 11.7: Wavesin the hodographplane The quality of a musical tone is affected by its pattern of overtones, plus any randomness that mayoccur due to phenomenasuch as turbulence-induced vibrations of an air column. Different instruments produce different patterns. The plucked string itself is affected by the location of the plucking; were it to be plucked in the center, the second harmonicwouldbe noticeably absent, which maybe considered undesirable. This affects the way guitars and harps are played. 11.1.6

The Hodograph

Plane

Sometimesthe terminations at either end of a pure bilateral delay line can be expressedas static sources or resistances, that is by effort-flow characteristics. These cases are, in general, nonlinear and morecomplicated than the fixed and free boundary conditions considered above. They are addressed here for the simplest kind of wave: a discontinuity betweentwo otherwise steady states such as shownin Fig. 11.7. Suchlongitudinal wavesin rods and fluid tubes have four types: compressionto the right, compressionto the left, rarefaction to the right, and rarefaction to the left. "Compression"and "rarefaction" can be generalized for other mediato meanan increase or decrease, respectively, in the effort, e. It is convenientto represent the twoconditions on either side of such a wave graphically, as points on a mapwith the generalized flow as the ordinate and the generalized effort as the abcissas. Straight line segments drawnbetweenthe two states separated by a simple wave discontinuity must have slopes of =kYc in order to satisfy equations (11.10) and (11.12). The vertical and horizontal componentsof the line segment represent the rightward-traveling waves0+ and e+, respectively, if the slope is positive. If the slope is negative, they represent the leftward-travelling waves0_ and e_, respectively. A mapusing these coordinates is knownas a hodograph plane. The arrow drawn near the middle of the line segmentspoints toward the newstate, which is sweepingawaythe old state as the wavepropagates.

11.1

WAVE MODELS WITH SIMPLE

BOUNDARY CONDITIONS

809

Example 11.6 An acoustic tube is closed at its left end and open on its right end:

At any one time, presume there are only two states in existence, separated by a single wave that is traveling to the left or to the right, as suggested in the drawing. Show on a hodograph plane the sequence of waves that comprise a cycle. Solution: The abcissas of the hodograph plane is the characteristic of the left end, and the ordinate is the characteristic of the right end. Start with, say, a rarefaction wave traveling to the left. This is represented in the hodograph plane below by the line segment in the first quadrant. The arrow is pointing toward the new state that is sweeping out the old state, at the left (and upper) end of the line segment. ~ = Q] characteristic characteristic~’~k ~ for left end right e~ I ",~. \for "~

waves

0 Whenthe wavereaches the left end of the tube, it reflects; what had been the new state becomes the old, and the new state is found where the new surge characteristic intersects with the proper boundary characteristic. Since the newwave is traveling to the right, its line segment has a positive slope, and appears in the second quadrant. The new state is at the left (and lower) end of this line, toward which the arrow points. After this wave reaches the open right end of the tube, it reflects again, producing a compression wave with a new line segment (third quadrant). The wave that completes the cycle (fourth quadrant) is produced after this second leftward-traveling wave reaches the closed end of the tube. The states repeat themselves after every two full round trips of the wave. The oscillation never dies out, and has a period of 4T, where T is the time for a wave to propagate from one end to the other.

810

CHAPTER 11.

DISTRIBUTED-PARAMETER

MODELS

Example 11.7 An acoustic tube is blocked at both ends. It cannot contain only one wave discontinuity, since the flow is zero at both ends. It can, however, contain two such waves that move symmetrically, one of compression and the other of rarefaction. In this case the pressure at the center of the tube remains constant. As suggested in the drawiag below, the tube then can be analyzed as two half-tubes, each of which has a zero flow at one end and a constant pressure at the other end. Show the hodograph analysis, assuming some compatible initial state, and find the period of a cycle.

Solution: This is the same as Example 11.6, except that since the time for a wave to travel from one end of the half-tube to the other is T/2, the period is 2T. The hodograph analysis is as follows: 4=Q[ characteristic characteristic for center

o

0 Example 11.8 In a more interesting case, a hydraulic tube of length L = 125 in has constant pressure of 2000 psi maintained at its right end by a large accumulator (tank with compressed gas inside). The left end also has an accumulator, with pressure of 3000 psi. An adjustable valve is located immediately to the right of this accumulator, however, which when open gives a pressure drop proportional to the square of the flow and equal to 1000 psi when the flow rate is 20 in3/s. The hydraulic fluid has a density of p = 0.8 × 10-4 4lb s/in and a bulk modulus of B = 200,000 lb/in 2. Two different cross~sectional areas of the tube are to be considered: A = 0.04 in 2 s. and A = 0.10 in

WNaccumulator and valve Po~’P~ 3000 psi (constant)

rigid line, area A

accumulator

~

P2 2000 psi (constant)

Find the pressures P~ and the P’2 and flows Q~ and Q2 at the two ends of the tube as functions of time, following a initial quiescent state with the valve closed and a subsequent instantaneous opening.

11.1

VfAVE

MODELS WITH SIMPLE

BOUNDARY CONDITIONS

811

Solution: The first step is to draw the pressure-flow characteristics for the two ends of the tube in the hodograph plane (Q vs. P). The right-end characteristic is a vertical line at 2000psi. Initially, the valve at the left end is closed, so that there is no flow and the pressure everywhere in the tube is 2000 psi. After the valve is opened, the pressure-flow characteristic at the left end of the tube (just to the right side of the valve) is as plotted below:

characteristic

waves

for

forA

=0.10

right

en~~.

in2---:*’--/ 1600

~ , i

2000

P, psi

~ 3000

The pressure drop across the valve equals 3000 - P1, and the flow is proportional to the square root of this pressure drop, producing the parabolicshaped characteristic. The surge admittances for the smaller and the larger tube areas are, respectively, :ks = A/x/~ = 0.0100 and 0.0250 inS/lb.s. A wave starts propagating from the left end of the tube toward the right the momentthe valve opens. The slope of the line segment in the hodograph plane that connects the two states on either side of the wave therefore must have the positive slope ~"~. Further, the old state being swept out has zero flow and 2000 psi, and the new state must be on the characteristic for the left boundary. This defines the new state, which is labeled L1. When the wave reaches the right side and reflects back, the original state is gone and the new state on the right side, labeled R2, lies at the intersection of the characteristic for the right side and the surge characteristic with negative slope 1~ that passes through state L1. The process continues until the equilibrium is reached. The flows at the two ends are plotted below:

. ........

Q,30~Iin3/s | 2hi_ ~

. :. -_-.~ .... i .. L ’ .....

-------- .....

:"~---7~"~ "’~’--’~:(~’in2

2A=0.10in

I oL ...........

L.-" A -- ...... !::-/" I lumpedQ ............................ 0

"

5

10

15 t, ms

20

812

CHAPTER 11.

DISTRIBUTED-PARAMETER

MODELS

Note that the case with the smaller area approaches the equilibrium flow quickly, and without overshooting. The larger area produces an overshoot and an oscillation, although the oscillation decays fairly rapidly. Sometimes one wishes to design such a system to minimize oscillations. It is instructive to compare this solution to the one that results if the compressibility of the fluid is neglected, leaving a lumped model with the same boundary characteristics at each end but just a fluid inertia in between. 2, For the line area of 0.04 in P1 = P’). +____PL dO, _: dQ 2000 + 0.25 A dt dt’ Q1 ----"

/3000 - P1 20v ]"~’~ ¯

Combining these equations leads to 2, d~Q = 4000 - 10Q dt which has the solution Q = 20 tanh(200 t). 2 The line area of 0.10 in changes the coefficient 200 s -1 to 500 s -~. Both solutions are also plotted above. It can be seen that the more precise wave solutions zig-zag about the lumped solutions, a typical occurance. 11,1.7

Summary

Spatially distributed phenomenaare the stuff of elasticity, plasticity, fluid mechanics, heat and mass transfer and electromagnetics. Most conventional methods and associated computer packages are restricted to static or steady-state problems. They often e~nploy finite element or finite difference techniques. In the analysis of complex engineering systems these tools can be used to aid in field lumping, that is the approximation of a phenomenon distributed over a spatial field by discrete or lumped models. This text employs lumped models exclusively, except for the present chapter. The more difficult or less obvious field lumping situations have been avoided. Sometimes you may find it more practical to engage a specialist to carry out needed field lumping, rather than to become a universal expert yourself. The dynamic interadtion of inertance, compliance and resistance fields can be particularly troublesome. The behavior at low frequencies usually can be deduced by lumping these fields separately, even when they are coextensive. For high frequencies, however, partial differential equation models may be more practical. This subject has been introduced via a special class of systems: onedimensional models exhibiting bilateral waves of constant wave speed. Examples include models of the rotating shaft, the translating rod, the acoustic behavior of fluid in tubes, the vibrating string and the electrical transmission line. These

11.1

WAVE MODELS WITH SIMPLE

BOUNDARY CONDITIONS

813

are characterized by two constant parameters: the wave speed or celerity and the surge impedance, which is the ratio of the effort of a waveto its flow. One approach is to plot the effort or the displacement or flow versus position, and to decompose it into bilateral waves. The reflections of these waves are particularly easy to represent for fixed and free terminations. The waves also can be decomposedinto Fourier components, resulting in an infinite set of natural frequencies. Purely sinusoidal behavior can be analyzed for unforced conservative systems and forced systems with static or dynamic boundary conditions by assuming a particular effort or flow and tracing it for one complete spatial cycle, noting that the pure wavelike mediummerely advances or delays the phase by the angle wL/c. Transient behavior has been addressed for the case in which the terminations of the wavelike system can be represented by steady-state characteristics. A hodograph plane, or plot of flow versus effort, is used. Waves are represeated by line segments having a slope equal to plus or minus the inverse of the surge 4impedance.

Guided

Problem

11.1

Twosimple bilateral wave media with the same wave celerity but different surge or characteristic admittances, Ys~ and Yst, are joined. They might be, for example, acoustic tubes of different diameter, or slender torsion shafts of different diameter. A wave approaches the interface from the side with surge admittance Y~. Find the ratios of the amplitudes of the reflected and transmitted waves to the incident wave. Suggested

Steps:

1. Define the incident wave as (e+)~, the reflected wave as (e_)~ and transmitted wave as (e+)t. Note that there is no wave (e-)t. 2. Find the effort e~ and flow ~ in the incident memberas functions of the incident and reflected waves. 3. Find the effort ee and flow ~t in the transmittance of the transmitted wave.

memberas functions

4. The efforts and flows of the two membersmust be equal at the interface. Therefore, equate the respective results of steps 2 and 3. 5. Solve the two equations from step 4 simultaneously for the reflection ratio (e-)i/(e+)i and the transmission ratio (e+)t/(e+)~. 6. As a check, make sure that when Y,~ = Y~t, which implies no discontinuity at all, the reflection ratio is zero and the transmission ratio is unity. 4Ageneralization of this approach,called the methodof characteristics, is givenin the reference cited in the footnoteon page795.

814

CHAPTER 11.

P0 = 1000t

DISTRIBUTED:PARAMETER MODELS

rigid fluid line, area A

~¢~ P = 0

Figure 11.8: System for Guided Problem 11.2 Guided

Problem

11.2

A rigid hydraulic tube connects a hydraulic chamberwith a constant pressure of P0 = 1000psi to a valve with Bernoulli-type characteristics whichis connected to a virtually zero-pressure region, as shownin Fig. 11.8. The valve is opened abruptly. Choosethe area of the tube such that the duration of the transients -4 in the pressure and the flow is minimized.The density of the fluid is 0.8 x 10 lb.s2/in 4, its bulk modulusis 200,000 psi, and the equilibrium desired flow is Qo = 30 in3/s. Suggested Steps: 1. Sketch the pressure-flow characteristics of the terminations of the fluid line in the hodographplane. Numbersaren’t needed at this point; it is sufficient to scale the plot using the symbolsP0 and Q0. Note the final equilibrium state. 2. Sketch the initial waveon the hodographplane, choosing an arbitrary surge impedance(inverse slope of the line segment) and taking care get its proper direction of propagation. Then continue with the second or reflected wave. 3. It is possible for the surge impedanceto be selected such that the second waveleaves the entire tube at the desired final flow, Qo. Showgraphically howthis is done. 4. Evaluate the pressure and flow behind the initial wavein terms of Po and Qo. Find the surge impedancein terms of these parameters. 5. Relate the surge impedanceto the density, bulk modulusand tube area, and solve for the area as desired. PROBLEMS 11.1 The strings for the lowest register on a piano are wrappedwith heavy wire in order to increase their effective massper unit length without increasing the tension force. Determinethe effect on the wavespeed and resonant frequency of quadrupling the effective mass in this manner. Whatmusical interval does this represent?

11.1

WAVE MODELS WITH SIMPLE BOUNDARYCONDITIONS

815

11.2 One of the most commonstandard steel hydraulic tubing has an outer diameter of 0.5 inch and a wall thickness of 0.049 inch. Computethe wavespeed and the characteristic admittanceif this tubing is used to propagate power(a) with a hydraulic oil having a bulk modulusof 200,000 psi and a specific weight density of 50 lb/ft 3, (b) with air at 100 psig and 70°F, (c) as a push/pull and (d) as a rotating shaft. The Young’sand shear modulifor steel are 30 × 106 psi and 11 × 106 psi, respectively, and its weight density is 0.283 lb/in 3. The effect of expansion of the tube caused by internal pressure maybe neglected. 11.3 A virtually incompressible fluid of density p passes slowly through a tube with thin, radially compliant walls with meanradius r, thickness t, Young’s modulusE and a Poisson’s ratio of essentially zero. Determinethe compliance and inertance per unit length, the wavespeed and the characteristic admittance. 11.4 The properties of coaxial cables most commonlyspecified are the capacitance per unit length and the "impedance,"which meanssurge or characteristic impedance. A particular typical cable is listed as having 67.2 pf/m (or ## f/m) and 75 ft. Determinethe inductance per unit length and the wavespeed; represent the latter as a fraction of the speed of light (3.00 x l0s m/s). 11.5 Placing thumb tacks on the hammersof a piano produces a "honky- tonk" sound. Explain the phenomenonqualitatively in terms of the harmonics or overtones. 11.{}A string in a musical instrument is plucked at its midpoint. (a) Sketchthe positions of the string for every one-quarterof a wave-travel time, similar to the exampleof Fig. 11.5. (b) Find the harmonic content of the resulting motion. Do you expect this is moreor less desirable than the result given in the text for plucking the string at the location one-quarter of the distance from one support to the other? Why? 11.7 A rigid tube through which fluid wavespropagate bifurcates at a Y junction into two tubes with the same properties. Determine the reflection and transmission of a simple wavediscontinuity which is incident on this junction. Hint: Note the similarity to GuidedProblem11.1. 11.8 A rod 100 in. long and 0.375 in. in diameter is suspended horizontally from aboveby two wires so that it is free to movein every direction except the vertical. The rod, which is observed to weigh 3.13 lbs, is then struck on one end by a hammer.Subsequent vibrations, monitored by a small accelerometer mountedon the other end, reveal harmonics at 505 Hz and multiples thereof. Determine the Young’smodulusof the material.

816

CHAPTER 11.

DISTRIBUTED-PARAMETER

MODELS

11.9 The system described and analyzed in Example 11.8 has reached its equilibrium state at 20 in3/s when the valve opening is suddenly changed. Find the resulting waves, and plot the flows as a function of time.

(a) The valve is shut completely. (b) The valve is closed partly to give an equilibrium flow of 5 in3/s, and then is shut completely. (c) Generalize from the results the qualitative rapid closure of the valve.

effect of slow as opposed

SOLUTIONS TO GUIDED PROBLEMS Guided

Problem

11.1

waves: ~(e_)~

(e+),

interface 2. Fromequation (11.21), ei = (e+)i + (e-)i

Oi= Yod(e+)i 3. Similarly, et = (e+)t

6 = l;~(e+)~ 4-5. (e+)t = (e+)~ + (e-)i

~(e+)~= c,[(e+)~+ (e-)d= Yc~[(~+)~ Therefore, (Y c~ + Yct)(e-)~ = (Y~ whichgives the reflecgion ratio (e-)i (e+)i and the transmission ratio

(~-)~

(e+),

If Y~t = :t~, the above results give (7+)/ = 0 and (7+)~ = 1, as expected.

11.2 Guided

ONE-DIMENSIONAL Problem

817

MODELS

11.2

1-2. neglecting wall shear: finalstate locus at right end Q~X~ locusat left end initialstate P

Po

P

In this ideal case, l~=--=---=---=(1/2)Q° 2Qo2 30 0.020inS/lb.s (3/4)P0 3P0 3 1000 50 but Yc = Alva, so A = I~fl = V/0"8 x 10-4 x 200, 000 2= 12.5 in

in 5 lb.sin 4~ ~ lb lb.s~/ in

11.2

One-Dimensional

Models

The pure. bilateral-wave-delay model is a very special one-dimensional model. General linear one-dimensional models are represented mathematically in this section. Subcategories are described by the power number and whether local or global symmetry or anti-symmetry apply. Analysis of these models is deferred to later sections. 11.2.1

General

Formulation

The operator S =- O/Ot is employed to represent differentiation with respect to time, t, as before. The model is represented in general with the following matrix equation which is first-order in the spatial variable x: I OY -A(x,S)y

+ B(x)u(x,t).l~xx

=

(11.22)

The state variables are represented by the vector y = y(x, t). Whenthe square matrices A and B are not functions of x, the model is said to be uniform in x.

818

CHAPTER 11.

DISTRIBUTED-PARAMETER

MODELS

The example of a uniform rod in torsion illustrates this form. If the state variables are chosen to be the torque or moment, M, and the angular velocity, ~, and no excitation exists for the region of x under consideration (this does not include the ends of the rod), equation (11.22) becomes either (11.23a) or equivalently

[:].

(11.23b)

The rotational mass momentof inertia per unit length is I, and the rotational compliance per unit length is C. Without the operator notation the equations become OM O~ (11.24a)

Ox -

0~_ Ox

cOM Ot ’

(11.24b)

or equivalently OM T0~’¢

Ox -

(11.25a)

0¢_ CM. (11.25b) Ox If the Laplace transform is taken of these equations with respect tO the independent variable t, there results the respective forms

dx

Cs

(11.26a)

(11.26b)

dx .

The derivative operator S can be seen to correspond to the Laplace notation and variable s except for the initial-condition terms, which usually are not of interest. Both notations serve the useful purpose of converting partial differential equations into ordinary differential equations. An alternative approach to equation (9.22) employs a Heaviside operator :D = O/Ox, or the corresponding Laplace operator: 0y _ A(x,T~)y B(x)u(x,t).

Ot The particular

(11.27)

example of the torsion rod becomes, for example,

(11.2s)

11.2

819

ONE-DIMENSIONAL MODELS

There is no clear preference between the alternative forms in this particular case. Both are used in the literature. The first form is used exclusively in the remainder of this book, however, because it allows functions A(S) which are irrational (employingsquare roots, for example), very usefully expandingthe domainof the modelsand permitting important three-dimensional effects to be included. The pair of equations (11.24a) and (11.24b) or (11.25a) and (11.25b) combinedto give 0’2 M O~ M (11.29) ~ - ICS~-M = IC -~. Since this form of the modeldescribes only Mand not ¢ or ~ it can be called incomplete. Completeness, of course, has nothing to do with exactness, which never applies to a model. Addingthe equation 0"2¢ ICS2¢ 2Ox

IC

Ot

(11.30)

does not confer completeness,either, since the relation betweenMand ~b is left undescribed. For general purposes, then, the matrix modelsare preferred.

11.2.2

One-Power Models

The torsion rod is an exampleof a one-powermodel; a transverse cut at any location defines a single effort and its conjugateflow, the product of whichequals the instantaneous power. The connection betweenthe two sides of the cut can be represented by a single bond. Wheneverpractical, the effort is defined as symmetric with respect to the cut, while the flow and the product of the two are anti-symmetric, oriented in either the plus-x or the minus-xdirection. This is precisely as described in Chapter 2. The state vector comprisesthe effort on top and the flow below, or in general the symmetricvariable Y8on top and the anti-symmetric variable Ya below:

[Ys]

Y = Ya "

(11.31)

The general 1-powermatrix A is partitioned with the following notation:

In the usual case of effort and flow variables, the dimensionsof Z are those of the gradient of effort divided by flow; Z is called the series impedanceper unit length. The dimensions of Y are those of the gradient of flow divided by effort; Y is similarly called the shunt admittance per unit length. The symbols Wand X, which.experience muchless interest and use for a reason youwill see shortly, are chosenhere simply becauseof their juxtaposition in the alphabet to Y and Z.

820

CHAPTER 11.

DISTRIBUTED-PARAMETER

MODELS

Occasionally one assumes that the effort or the flow of a one-power model is itself a constant, leaving only a single state variable. This is called a degenerate one-power model. Only W or X then exists. 11.2.3

Symmetric

One-Power

Models

A locally symmetric model has the same differential equation if written in the lninus-x direction as it does if written in the plus-x direction. This is a consequence of the symmetry of a slice of the system of infinitesimal length dx. The symmetric variable appears the same regardless of the +x or -x perspective, but the anti-symmetric variable has its sign changed. Therefore, the differential equations in the two perspectives become, respectively,

-~x ya 0

-

:Y

Ys

(Ii.33a)

Ya ’ Ys

(11.33b)

Local symmetry requires the two square matrices to be equal. Thus, the model exhibits local symmetry if and only if W = X = 0.

(11.34)

A locally symmetric system is globally symmetric if it is locally symmetric and is uniform in x. Note that the uniform rod in torsion satisfies this condition, but a non-uniform rod usually does not, despite its local symmetry. Since all degenerate one-power models include only W or X, they are necessarily anti-symmetric. 11.2.4

Multiple-Power

Models

A transverse cut through a non-degenerate multiple-power model reveals two or more pairs of effort and flow variables. Each pair can be represented by its own bond; there is a bundle of parallel bonds. Each bond represents one of the powers. If one of the powers is not coupled to the others, it can be separated therefrom, reducing the complexity of the remaining model. It will be assumed that such separations have been made. The state variables need not be effort and flow variables, as noted above; each conjugate pair of variables nevertheless includes one that is symmetric with respect to the transverse cut and one that is anti-symmetric. All the symmetric variables are collected together into a vector, Ys, and all the anti-symmetric variables are collected into a second vector, Ya. To be as orderly as possible, the elements in the two vectors are placed in the same order. Thus, for example, the second asymmetric element is the conjugate state variable paired to the second symmetric element. Finally, the two vectors are assembled into the overall state vector:

11.2

ONE-DIMENSIONAL

MODELS

[y81

Y = Ya "

821

(11.35)

The state differential equation can be partitioned, like the one-power model above, to give square matrix elements W, X, Y and Z:

The identification of locally symmetric and anti-symmetric cases given above for one-power models can be generalized directly to multiple-power models, with the result that symmetry exists if and only if Wand X vanish. In other words,

represents the general symmetric model. Solutions to the state differential equations are given in the following sections, starting with the simpler cases. The basic approach and several of the 5results have been published by the author. 11.2.5

Summary

The distributed-parameter models addressed in this chapter are linear and employ only one spatial variable as an independent variable. Nevertheless, certain two and three-dimensional phenomena can be included if the first-order cannonical form of equation (11.31) is used. This form defines a square matrix in which the dynamics is represented by use of the derivative operator S or the Laplace derivative operator s. The models can be categorized according to the number of bonds necessary to represent the power which flows across any transverse cut. The state of a non-degenerate n-power model therefore is represented by a state vector with 2n elements. The symmetric variables are collected in the first subvector component of the state vector, and the asymmetric variables, ordered to match, form the balance of the state vector. This partitioning defines four submatrices of A. One advantage of this scheme is that local or global symmetry is immediately apparent, corresponding to the vanishing of the upper-left and lower-right submatrices. Guided

Problem

11.3

Find a state differential equation modeling gravity waves with wavelength considerably greater than the mean depth. Assume a horizontal bottom and wave heights considerably smaller than the mean depth. 5F.T. Brown,"A Unified Approachto the Analysis of UniformOne-DimensionalDistributed Systems,"ASME Transactions J. of Basic Engineering, v 89 n 2 pp 423- 432, June 1967.

822

CHAPTER 1 I.

Suggested

DISTRIBUTED-PARAMETER

MODELS

Steps:

Choose state variables, one symmetric and the other asymmetric. These could be efforts and flows, but need not be. Presume that the velocity of the water is largely horizontal and uniform over the depth, which follows from the long wavelength assumption. 2. Find the continuity relation between the state vertical control surfaces dx apart can be helpful,

variables.

3. Find the momentumrelation between the state variables. horizontal contol volume can help.

Drawing two

Drawing a small

4. Assemble your results in the canonical form.

Guided

Problem

11.4

The static bending of a uniform slender beam is presented in elementary textbooks by the model dM/dx = F, ~/dx ~ = (1/EI)M and dF/dx = -W(x), where Mis the bending moment,F is the shear force, ~ is the lateral deflection, W(x) is the applied loading per unit length, E is Young’s modulus and I is the area momentof inertia of the cross-section. Introduce dynamics into this model with the simplest possible assumptions, and present it in the canonical form given above. The resulting model is known as the Bernoulli-Euler beam. Check for the symmetryof your result. Solutions will be sought in Section 11.5. Suggested

Steps:

Draw an element of the beam. Label the variables

appropriately.

Power is propagated along the beam by virtue of transverse force and motion and rotatibnal moment and motion. Identify two pairs of symmetric amd anti-symmetric state variables, and collect them into a state vector y. 3°

Place the given differential

equations into the canonical form given above.

The simplest possible representation of the dynamics includes the inertia for lateral velocity but overlooks the inertia for rotational velocity. Write the corresponding differential equation, place it in the cannonical form and complete the matrix A accordingly. 5°

Identify the component submatrices W, X, Y and Z. The model is known to be symmetric, so your Wand X should be zero.

11.2

ONE-DIMENSIONAL

823

MODELS PROBLEMS

11.10 A comxial cable (with inductance and capacitance) has a series resistance per unit length R~. Find the elements of the associated matrix A. 11.11 Repeat the above problem, adding a shunt, length.

conductance

G~ per unit

11.12 The fluid in an acoustic tube has a viscosity #. Find the elements of the associated matrix A assuming a resistance consistent with laminar steady flow, as can be deduced from equation (2.14) in Section 2.3 (p. 44). 11.13 A fluid-filled tube with series inertance I t per unit length and shunt compliance C’ per unit length leaks through pores in its walls; the shunt conductance is G~ per unit length. Evaluate the matrix A. 11.14 An acoustic muffler comprises a long central tube with numerous radiM holes leading to otherwise sealed chambers, as pictured below. four holes of diameter dh evenly spaced around periphery for each segment

(a) Construct a bond-graph model of one segment of this discrete system, using the pressure and the axial flow as the state variables. The orifices may be modeled as inertances, and the chambers as compliances; energy dissipation may be neglected. (b) Evaluate the moduli of the bond graph elements of part (a). orifices may be assumedto be through a sheet of zero thickness; see Section 10.1.8 (pp. 745-746). (c) Convert the descrete model into an approximately equivalent distributed-parameter modeli give the elements of the matrix A.

11.15 Identify whether or not the models for the five preceding problems are locally or globally symmetric. (You need not solve the problems to answer this question.)

824

CHAPTER I1.

SOLUTIONS

TO GUIDED

Guided

11.3

Problem

DISTRIBUTED-PARAMETER

MODELS

PROBLEMS

1. The choice here is the depth of the water, y (let ]"0 be the meandepth), and the horizontal velocity, v. 2.

~, ,

~ y(~)

,v(x+dx)

face

= y(x)v(x) - y(x + dx)v(x ~ttdx ~--yo ~xdxOV = -~---~(yv)dx _ VO~xxd x

~ ~y(~+dx)

Oy The mean velocity will be assumedto be zero, so that ~- = - y0~xx At the

bottom:

P(x)~

P(x+dx)

If the dynamicheads are neglected (due to the zero meanvelocity), Oy Ov = pg-~xdX = -(pdx)-~

~xdX

4 ~z Guided

=- S/yo Problem

11.4

2. The transverse displacement, r/, is symmetric, whereasthe transverse force, F, is anti-symmetric; these can form the first conjugate pair of variables since the associated power is f O~/Ox. The moment, M, is symmetric, and the angle OrlOXis asymmetric; these can form the second pair since the associated power is MO2rl/OxOt.The resulting state vector is T. [~l, M, F, ~grl/~gx] ~/ 0

0

=-

0 0

o

~ -01

+

o

0 O,]/(9x -1/EIm This is valid only for the static case.

Orl/Ox

-

00 WOO(m)

M per unit length times the F 4. The net lateral force per unit length equals the mass lateral acceleration, or OFdxo---~= (pAdx)-~02~(omitting the effect of the external loading, W(x)).

11.3

825

WAVE PROPAGATION Entering this into the equation above gives

1[ ] i 0

PAos~

Z=

11.3

1

(This is used in Section 11.5, equation (11.120).)

0

-1/EIm 0 ’

-l/El,,

; W= X =

.

Wave Propagation

The concepts of phase and group velocities, wave number, attenuation factor and dispersion are introduced. This is done in the context of degenerate onepower models, in particular the case study a heated incompressible fluid flowing at a constant velocity through a tube of uniform area. The waves in this case are propagated unilaterally in the downstream direction, simplifying the consideration. Frequency and transient reponses are calculated for three assumed wall conditions. 11.3.1

Simplest

Model:

Pure

Transport

The simplest model for heated fluid flowing in a tube presumes a slug-flow velocity profile, as shownin Fig. 11.9, with negligible mixing of the fluid. Axial heat transfer is neglected, and the walls are assumed to be adiabatic. The behavior of the system is clear before any equations are written: temperature disturbances propagate, without distortion, at the velocity v of the fluid. This fact will be expressed mathematically. The differential equation that describes the system can be written directly. This approach is inadequate to address more difficult problems, however, so a more powerful method will be used. Often when mass transport is involved, it is best to emply the substantial or material derivative d/dt which describes the behavior of a fixed quantity of mass: d 0 0 dt - Ot + v. A = -~ + V~x. insulated walls velocity v

Figure 11.9: Pure transport of heated fluid in tube

(11.38)

826

CHAPTER 11.

DISTRIBUTED-PARAMETER

MODELS

The form on the far right is the scalar version for one spatial dimension. Since the temperature 8 of any particle of fluid in pure transport is constant, (11.39)

d-~- = 0, or, in the canonical operator form of equation (11.2) (p. 817),

Ox

-

$8. v

The solution of this equation can be found readily derivative in operational form: 8(x, t) -- e-TSs(o, t) = 8(0, t -T -- x/v.

(11.40) by retaining

the time

(ll.41a) (ll.41b)

The operator e -TS is the pure delay operator, and e-Ts is its Laplace counterpart. T is the delay time associated with the transport velocity. Equation (11.41) also can be generalized slightly to give 8(x2,t) = 8(x~, t T~I),

(11.42)

where T2~ is the time delay required for a fluid particle to traverse the distance x2 - xl. This solution corresponds to what you should anticipate. The response to the sinusoidal input disturbance 8(0, t) = Re jwt 8o(O)e

(11.43)

can be found by assuming a solution of the form 8(x, t) = Re 8o j~t. (x)e Uponsubstitution comes

into the differential

(11.44)

equation, this frequency response be-

80(z)_ e_~k~, 80(0) k = ~/v -~ 2~r/A.

(11.45) (11.46)

The same result follows most directly from substituting j~ for S in equation (11.41). The symbol k is employed here in its traditional role as wave number, which is defined as the ratio of the frequency of a propagating sinusoid to the phase velocity, Vp. This also equals the ratio of 2;r divided by the wavelength, A. The phase velocity is defined as the velocity at which a sinusoidal wave propagates. In the present case, the phase velocity equals the transport velocity: Vp = v.

11.3 11.3.2

827

WAVE PROPAGATION First Modification: stant-Temperature

Thermal Leakage Environment

to

a Con-

Attenuation can be introduced into the above problemby allowing heat to flow through the walls to an environmentthat has a constant temperature over time, thoughit mayvary as a function of x. Specifically, it will be assumedthat the thermal conductivity-per-unit-length, G, is constant, and that all thermal storage in the walls or in the thin boundarylayer associated with the conductivity is negligible. Thus, equation (11.39) is modifiedto become Afcf~dO(x, t) =-G[8(x,t)

(11.47)

- 8e(x)],

in which Af is the area of the tube, cf is the specific heat of the fluid and 81 is the temperature of the environment, This equation represents the actual thermal energy transferred per unit length. Usingequation (11.38), it gives Ox

v cot

(8 - 81);

(11.48a)

~-~=_AIcI /G,

(11.48b)

whereTf is a time constant for heat transfer to or fromthe fluid. If the temperature8(0, t) is held constant for a while, a steady-state solution is approachedthat will be rather complicatedif 81 is a complicatedfunction of x. Tosimplify matters it is useful to distinguish the steady-state solution 8o (x) from an incrementalunsteady solution 81 (x, t): O(x,t) = 00(x) + 0~ (x,

(11.49)

Twoequations result, one that can be solved for Oo(x), 000 ..... Ox

1 81), r~v

(11.50)

and one whichcan be solved for 8~ (x, t): O~l _Ox (S+81"v l/r~)

(11.51)

The solution of the latter is alwayssimple because it is independentof 81 (X, t) e-TSe-T/r" 8( 0, t) = e-T/r~ 0(0 , t - - x/v ).

(11.52)

Thetwosolutions are essentially identical, of course, if toe is uniformin position. This result reveals that the sole effect of the conductanceis an exponential attenuation of the temperaturefor increasing x. As illustration, the responseto an upwardfollowed by a smaller downwardchange in the upstream temperature is plotted in Fig. 11.10.

828

CHAPTER 11.

DISTRIBUTED-PARAMETER MODELS

0(0,0

I

I

I

I

0.5 1 1.5 (a) input disturbance

~:

0 0.5

2

of propagation 1

1.5G~

v 2

(b) responseat time Figure 11.10: Step responses of heated flow in tube, first modification perfect insulation wall wi~ ~e~al compliance possibleinsulating spacers

Figure 11.11: Heated flow in tube, second modification 11.3.3

Second Walls

Modification:

Thermal

Compliance

in

the

Consider finally that the previously time-invariant environmentaltemperature becomesthe time-varying temperature of the wall of the tube, 0~,(x,t). The conductances G and G~ refer to the film coefficient associated with the thin boundarylayer, which is presumedto be independent of position. The exterior of the wall is assumedto be adiabatic, or perfectly insulated. The wall itself is assigned a uniform thermal compliance Au, cw per unit length. Axial heat transfer in the wall is neglected, as emphasizedin Fig. 11.11 by the probably hypothetical placementof insulating spacers. (Inclusion of axial heat transfer would require a two-powermodel.) These assumptions might apply reasonably well to the hot water pipes in your residence, particularly whenyou turn on the faucet in the morningand wait for the water to becomehot.

11.3

WAVE PROPAGATION

829

Equation (11.47) is modified only by the substitution of the variable wall temperature for the constant environmental temperature: A fcf dO(x, ~ = t)-G [0(x, t) - 0.,(x,

t)].

(11.53)

Equatingthe heat efflux from the fluid to the heat influx to the wall, A or

wc~,--~-,= G(0- 8,,,), 1 ~ - 1 + ~dS;

(11.54) (11.55a)

~, ~ A~c~,/G, (11.55b) where 7wis a time constant for heat transfer to or from the wall. The Combination of equations (11.53) and (11.55) with equation (11.38) can be written --

= -W0;

(11.56a)

v ~ 1 +~WS " Note that the use of the symbolWis consistent with its use in Section 11.2. The modelis of necessity anti-symmetric. The steady-state solution is trivial: a constant uniformtemperaturefor everything. The solution of equation (11.56) for the incremental unsteady temperature is O~(x,t) = e-~w0~(0,t) =e-~/~’ e-~SeZ/r~’(~+r~S)o~ (0,

(11.57)

This is rather complicated. The simplest case is the frequency response, which can be found by employingthe substitution ~ S ~ j~: ~ 0~ , = Re 0~0e 010(x~ = e_W(~)~= e_(j~+~)~, 0~0(0)

(11.58a) (11.58b) (11.58c)

. =

1(

( l.SSe)

Here, k is the wavenumber, as before, and ~ is knownas the attenuation factor. All phase information is contained in the wavenumber,which is traditionally plotted against frequencyas shownin Fig. 11.12. For re~ons that will 6The rigor of this approach can be demonstrated, but its application can overlook the possibility that the response to an old transient disturbance still lingers. This problem occurs mostly in conservative systems, where transient responses do not decay.

830

CHAPTER 11.

DISTRIBUTED-PARAMETER MODELS

0.4 3

0.3 asymptote

k 0.2

1 / 0.1

0

0

1

2 (r.,v)k

0

4

0

2

(a) dispersionrelation (dimensionless)

4

86( r..v)k

10

(b) attenuation per radian

10 dashed

(c) log scales for parts (a) and(b)

lines

are

/

asymptrltlJ 1.(

0.2

1.0

( r..v)k

10

Figure 11.12: Frequencybehavior for heated flow in tube, second modification

11.3

831

WAVE PROPAGATION

become apparent later, equation (11.58c) is known as a dispersion relation. The phase velocity is precisely the slope of the chord drawn from the origin to the point of interest, or Vp

w w~ k 2~r

v - ’ 1 + r~v/7l(1 + ~:~’-~w2)

(11.59)

At very high frequencies this phase velocity approaches the fluid transport velocity. This should be anticipated, for at high frequencies the temperature of the wall would not fluctuate appreciably, due to its thermal compliance and the brevity of each half-cycle. Thus the problem becomes virtually the same as the earlier problem with constant environmental temperature. At very low frequencies, on the other hand, the temperatures of the wall and the adjacent fluid track each other nearly perfectly, so that the thermal resistance of the boundary layer becomes inconsequential. The phase velocity is less than the transport velocity because the wall as well as the fluid is alternately heated and cooled: L v__~p = 1 (11.60) 1+ ~o v At intermediate frequencies the phase velocity changes continuously from its lower to its upper asymptotic values. The attenuation of sinusoidal waves is represented exclusively by the attenuation factor, which is plotted in part (b) of the figure. At very high frequencies the attenuation, like the phase velocity, approaches that of the earlier problem with a constant environmental temperature. At lower frequencies the attenuation is reduced~ approaching zero at zero frequency and infinite wavelength. Transient responses are harder to deduce from equation (11.57). The first factor on the right clearly represents an attenuation that is exponential in position, and the second factor represents a pure delay corresponding to the transport velocity. The meaning of the third and final factor, called a percolation operator, is not immediately apparent, however. It can be evaluated by employing the one-sided time Laplace transform of the equation, presuming zero initial conditions. Using a bar over a symbol to represent its Laplace transform, E1 (X, 8) = e-T/r~e-Tse

T/r~ (l+rs)~ 1 (0,

8).

(11.61)

The percolation term can be simplified by noting the following property of Laplace transforms, taken from Table 7.1 (p 511): f_.-~[f(a + s)] = e-~t£-’[f(s)].

(11.62)

Here f is any function and a is any constant. Thus, after both sides of equation (11.61) are multiplied by eT%the anti-transform becomes

t+ x/v) =

s-

(11.63)

Whent?~ is a unit impulse, ~(0, s) = 1, which is not a function of s all, so ~ (0, s - 1/Tw) = 1 also. The inverse transform within equation (11.62)

832

CHAPTER 11.

0"8 t /T/~=

0

0.25/

DISTRIBUTED-PARAMETER

MODELS

case

~0

~e ~L~S: ]~p~]~e ~d ~ep ~e~po~es fo~ ~e~ed ~o~ ~ ~ube, second ~od~c~t~o~ becomes, from transform pair #44 of Appendix C with b = 1,

(11.64) in which Io and I1 are modified Bessel functions. The impulse response here is stated as the time derivative of the step response. Both responses, with the other terms in equation (11.64) also accounted for, are plotted in Fig. 11.13. Note that the impulse and step terms propagate at the velocity of the fluid, but decay rapidly. The remainder of the responge, which is its bulk except close to the source, propagates more slowly. This observation agrees with the earlier conclusions for the propagation of sinusoidal disturbances.

11.3.4

Dispersion and Absorption

A non-sinusoidal propagating wave is distorted by the dependencies of both the phase velocity and the attenuation factor on frequency. These two effects can be separated, at least conceptually, by first considering the hypothetical situation of the actual varying phase velocity but zero attenuation, and then later the hypothetical situation of constant phase velocity but the acutal attenuation. The first case is called dispersive bt~t non-absorptive; the energy is propagated without loss but at a range of speeds depending on the frequency. To see how dispersion produces distortion, consider the following particular

11.3

833

WAVE PROPAGATION first ~erm

(a) signal of equation Oj(O,O= O0(sinWot +~sin 3~Oot) all cases:r..= ~3 ~oo=I /2 ~,, (b) signal 0.424of the longer wavelength. downstream,with attenuationneglected

secona term sum

"-’~second

term~

(c) signal 0.848of the longer wavelength downstream, with attenuationincluded Figure 11.14: Attenuation and dispersion of heat flow in tube disturbance, which contains two sinusoidal components: 1 01(0, t) = 00 (sin wot+-~ sin 3w0t).

(11.65)

This is plotted in Fig. 11.14 part(a). The two componentspropagateat different speeds. At regular intervals in position the wavepattern at x = 0 reappears, but elsewhere the pattern is different. The pattern midwaybetween these intervals, correspondingto a relative phase shift of 180° of the higher frequency components,is shownin part (b) of the figure. If three or more irrationally related frequencycomponentswere present, the waveprofile at the origin never wouldbe repeated. To see hownon-constant absorption produces distortion, presumethat both frequencycomponentsof equation (11.64) are propagatedwith different attenuation factors. As shownin part (c) of the figure, the ratio of the two components changes as the wavepropagates downstrean~;ultimately one virtually vanishes comparedwith the other, leaving a nearly pure sine wave. The particular case showncorrespondsto the separated dispersion and attenuation of the heat transport problemabove, the property r~o/Tf being set at unity and wobeing set at 1/2rw. Anyperiodic signal can be represented by a Fourier series, and most aperiodic signals by a Fourier integral. Knowledge of the wavenumberand attenuation factor as functions of frequencythen can be used to construct the response

834

CHAPTER 11.

DISTRIBUTED-PARAMETER MODELS

to these signals. 11.3.5

Group

Velocity*

Engineers often are interested in narrow-band signals. Examplesinclude AM and FMmodulation in communication, and bursts of non-sinusoidal disturbances in various media. The velocity at which a modulation or a burst or a group propagates is not generally equal to the phase velocity for any frequency within its frequency band. Indeed, the group velocity, defined as the velocity at which a disturbance propagates in the asymptotically limiting case of zero bandwidth,can be radically different from the phase velocity. To see this, consider a wavecomprising two frequency componentsw + A~/2 and w - Awl2 in which Awis very small, and the corresponding wave numbers k + Ak/2 and k- Ak/2: 0 = 0oRe

{e j[(~+Aw/2)t-(k+Ak/2)x]

-t- e j[(w-Aw/2)t-(k-Ak/2)x]

}

= Re[ej(A~tl~.-Ak x/e) + e-J(A
(11.66)

This result is plotted in Fig. 11.15. Thefirst cosine term represents an envelope of the response with a long wavelength;the secondcosine oscillates at the mean frequency of the two components. You may have experienced an analogous phenomenon in the "beating" of two musical tones that are at slightly different frequencies; musicians often tune their instruments to a knownstandard by eliminating the beating. The two propagation velocites represented by equation (11.66) can be found by holding the respective phase angles in parenthesis constant. The high- frequencywavewithin the envelope propagates at the phase velocity

~

(11.67)

and the envelope propagates at the velocity

very = --. Ak

(11.6s)

The group velocity vg is defined as the limit of this velocity as Aw-~ 0, or as the wavelengthapproachesinfinity: v~ = L ~-~o

Ak dk"

(11.69)

Returning to the examplegiven in Fig. 11.12 (p. 830), you can see that this velocity is the local slope of the dispersion plot of w as a function of k, whereas the phase velocity is the slope of the chord drawnfrom the origin.

11.3

WAVE PROPAGATION

835

AAAAAAAA

~°~°~ VVVVVVVVV This sine wave

AAAAAAAAAA ~o~o~ VVVVVVVVVVV plus this... ¯.,.

envelope

~.-.

Ill ~ p°siti°n /Ill / ~’" "’,V i/ equalsthis. Figure 11.15: Conceptof group velocity

836

CHAPTER 11.

DISTRIBUTED-PARAMETER MODELS

Dispersion implies a difference betweenVp and Vg. If vg > Vp, as with the exampleof the heated fluid travelling downthe tube, the dispersion is called anomolous,whereas if vg < vp it is called normal. These terms originated in the field of optics wherenormaldispersion is the usual case; the non-technical implication of the terms is misleading in manyother media and structures, including the particular examplehere. Had a wave-train been chosen with a more complicated composition within the same narrow band, the envelope of the wave would have been more complicated, but the short-wavelength oscillation with the envelope wouldremain. In the limit of vanishingly small bandwidth, this complicated envelope would travel without distortion at the samegroup velocity as the simpler envelope. For a somewhatwider frequency band the envelope distorts slowly as it propagates at virtually the same speed. Thus the group velocity is an important property of a distributed mediumor structure. Its importance is enhancedfurther when,in the following section, the velocity at whichenergy propagates through conservative distributed-parameter systems is seen to equal the group velocity. 11.3.6

Summary

The use of the derivative (Heaviside) and Laplace operator notations have been illustrated for a particular degenerate one-powermodel. The simplest version demonstrates unilateral wavepropagation without energy absorption or dispersion; the more general versions demonstrate both absorption and dispersion in their unilateral waves. Dispersion meansthat the velocity at which a sinusoid propagates, called the phase velocity, dependson its frequency, and the shape of non-sinusoids therefore is continuously being distorted. The wavenumberk(w) of a sinusoid equals 27~/A, where A(w)is the wavelength; the phase velocity equals o~/k. The group velocity equals do~/dk. The group velocity is the speed at which the envelope of a narrow-band wave propagates. The propagation of complicated waveforms can be calculated either by Fourier decomposition or through the use of Laplace transforms, which are typically of non-polynomial type. PROBLEM 11.16 Surface wavesin water of density p and surface tension T in a narrow tank of uniform depth h can be shownto obey the dispersion relation

where9 is the acceleration of gravity. (a) "Deepowaterwaves"are said to exist if increasing the depth has very little effect on the phase or group velocities. Determinean approximate

11.4.

ONE-POWER SYMMETRIC

837

MODELS

depth necessary for this condition, as a. function of the wavelength, and find the phase velocity as a function of the wavelength and the parameters. Also, find the wavelength and frequency for water at room temperature (p = 1000) kg/m3, T = 0.075 N/m) below which surface tension is more important, and above which gravity is more important. (b) The motion of "shallow water waves" penetrates to the bottom without any decrease in amplitude, unlike deep-water waves which have negligible motion at the bottom. (See Guided Problem 11.3, pp. 821,824.) Repeat (a) for this case. (c) Find the ratio of the group velocity to the phase velocity for four cases: shallow-water gravity waves, shallow-water capillary waves, deepwater gravity waves and deep-water capillary waves. (d) Describe qualitatively what you would would expect if you throw stone into a deep pond and observe the ripples. Note that the disturbance is an impulse in both space and time, and therefore contains all wavelengths equally. Suggestion: It is helpful to find and plot the phase and group velocities as functions of the wavelength. Phenomenato consider also include the attenuation due to spreading and the nonlinear effects of the shorter wavelengths.

11.4

One-Power

Symmetric

Models

The fundamental concepts of propagation operator, characteristic impedance and causal and transmission matrices are introduced or extended is this section to apply to general one-power symmetric models. Emphasis is placed on the propagation of waves and on boundary-value problems. Applications are given to heat conduction, and to pure wave-like behavior as represented by the uniform rod in torsion. Finally, exponentially tapered systems are. analyzed. 11.4.1

Wave

Behavior

The general symmetric uniform one-power model with no internal from Section 11.2.3 (p 820),

excitation

is,

0

Substitution

of the second componentequation into the first Ox~" - YZys or O( ~x,/,~-~x)2 0"2ys O~-ys = ys.

The general solution

becomes

gives (11.71)

838

CHAPTER 11.

DISTRIBUTED-PARAMETER

in which the propagation operator 7(S) and characteristic Yc(S), also known as the surge admittance, are defined

~(S)

c(s) = rSVP,

MODELS

admittance

(11.73)

and Ysl and ys2 are any functions of time. Equations (11.70), (11.72) and (11.73) are employedbelow repeatedly. The special case of 1’~ for the pure bilateral wave delay has been given in Section 11.1 (Equation (11.11), p 803). The reciprocal of the characteristic admittance, known as the characteristic impedance or the surge impedance, often is used instead. For frequency response calculations, the substitution S -~ ja~ gives, as before, a wave number k and attenuation factor c~: [3’(jw) Thus the phase velocity

jk + ~ . (11

.74)

becomes a~ w I Vp - -£ - Im ~(jw)" 1

(11.75)

These results also are of considerable general utility. Example

11.9

Find the propagation operator, characteristic admittance and phase velocity for the torsion rod of Example11.1 (p. 798). kbu may use the information for Cj and I j given in that example. Also, observe the nature of the propagating waves from the results. Solution: The symmetric and asymmetric variables are, respectively, y~ = M and ya = ~. From Example 11.1 or Table 11.1 (p. 801), Z(S) = I’S = (7rr4p/2)S, Y(S) : C’S = (2/~rr4#)S, in which r is the radius of the rod, p is its mass density and # is its shear modulus. With these substitutions, equations (11.73) give the propagation operator and characteristic admittance as follows:

-- c,/ s -Yc= C~_

2

~r4v~. .

11.4.

ONE-POWER SYMMETRIC MODELS

839

The ]~ agrees with equation (11.11) (p. 803). Equation (11.75) then that the phase velocity is a constant (consistent with equation (11.8), 803): vp-

Im3,(j~o)

- ~ =

As a result, the operator e-Tx represents a pure time delay, and the operator e7z represents a pure time advance. Waves travel at the same speed in both directions; equation (11.72) reveals that the entire behavior can represented as the sum of such right and left-traveling waves. The model is non-dispersive, and the group.velocity equals the phase velocity. Further, Yc is a constant, so waves of ¢ propagate in phase with the corresponding waves of M.

Example 11.10 Find the propagation operator, characteristic admittance, phase velocity and attenuation factor for heat conduction. Assume one spatial dimension with no internal heat source or sink, for which the differential equations can be written 8 Note that this uses the heat flux q as the asymmetric ~ariable ya, rather than the flow variable which is the entropy flux, largely because it is more traditional. The thermal conductivity is ke, and C is the specific heat; both are assumed to be invariant. Also, observe the nature of the propagating waves from the results, including a comparison of the group velocity with the phase velocity. Solution: Only one energy storage mechanism is present, as indicated by the single time derivitive operator S =- O/Ot in the given differential equation. Equations (11.73) give the propagation operator and characteristic admittance

= = v%CS.

The phase velocity Vp and attenuation Letting S -~ ja~, ~/(jw)

factor a are defined for sine waves.

= Cv@-~/ko = jk + c~;

k =a =

The traditional dispersion plot of frequency as a function of wave number (with the constant coefficient C/ko for the ordinate) is given at the top of the following page.

840

CHAPTER 11.

DISTRIBUTED-PARAMETER MODELS

4 3 ko 2 1

Cv °o

0.5

1’.o

?.5

k The phase and group velocities equal the slopes of the chord (w/k) and of the curve (dw/dk), respectively: Vp = -~ =

--

,

dw vg - dk - 2Vp" Heat conduction behaves very differently from the pure bilateral-wavedelay model. Since energy cannot slosh back and forth between different storages, the system is non-oscillatory in the usual sense, and possesses monatonic(continuously increasing or decreasing) reponses to step disturbances. It is highly dispersive and absorptive. The equality of k and c~ implies that the envelopeof the decayingsinusoidal response for unilateral waves(that is either 01 or O: alone), as shownbelow, falls off by the factor e -2~ per wavelength. 1.13 0.8 responsesto sinusoidal disturbance atx= 0

0.6

lead ¯ / envelope

0.4 0.2 0 -0.2’

T

3-

~

T

The wavelengthis inversely proportional to the square root of frequency. The heat flux is not in phase with the temperature variations, but leads by

11.4.

ONE-POWER SYMMETRIC

841

MODELS

a constant 45°, which is the phase angle of ~"~. The ratio of the heat flux variations to the temperature variations increases with the wave number and therefore with the frequency, as can be seen b’y the plot of the magnitude of 1~ given below:

magnitude of

1

1

characteristic

’~

admittance 0 0

Example

0.5

1~.0

k~

1~.5

11.11

Continue the example of the infinite rod above by finding the temperature at a location x to the right of the point where an impulse disturbance in temperature is imposed. Also, find the reponse when the disturbance is a step in time. The Laplace transform table given in Appendix C may be used. Solution: You can examine the rightward propagation of transient waves by setting the amplitude of the leftward waves in equation (11.72) (p. 837) to zero. The Laplace transform of the response is

0(x,s) The responses to the unit impulse (02(s) = 0o) and step (02(8) = Oo/s) given in Appendix C by transform pairs #38 and #39, respectively: impulse response : step response : v

O(x,

t)

0(x,t)

--

-

O~ 1,/-~q~i_3[2p_Cx2[4kot

~ 2 V ~rko

=00erfc

Here "erfc" stand8 for the complementary error function, which is ~videly tabulated. These results are plotted below. Note that a single dimensionless curve characterizes each case completely; it can be scaled for any distance or time. Notice also that the tails of the responses are extremely long. This feature of many distributed-parameter models cannot be modeled with great accuracy using only a few lumped elements.

842

CHAPTER 11.

DISTRIBUTED-PARAMETER MODELS

step response,"~o

/Cx~\ ~

4~

I

5

I

6

X

The differential equation given for heat conduction in ExampleI1.10 above, and also used in Example11.11, also is knownas the diffusion equation. This classical equation describes the diffusion of the concentration of a component through a mixture. Specifically, 0 becomesthe concentration per unit length and q becomesthe flow of the componentof interest. Withthis interpretation, ko becomesthe permitivity and C becomesunity. 11.4.2

Energy

Velocity

in

Conservative

Media*

The time-averaged power propagated through a non-absorptive mediumby a sinusoidal wavecan be respresented as the product of the time-averagedenergy density and the velocity of energy propagation, re: ~ = (T + V),~.

(il.80)

The overbars imply the time averaging. For non-dispersive media this velocity equals the phase velocity, vp, which also equals the group velocity, vg. For dispersive media, however,the energy velocity equals the group velocity, which can differ markedlyfrom the phase velocity: (11.81) ~ This relation does not apply in the presenceof dissipation, for whichthe concePt of an energy velocity has nebulous meaning. For example, the group velocity

11.4.

843

ONE-POWER SYMMETRIC MODELS

7 in a dissipation band is known for light traveling through a material medium to exceed the speed of light in a vacuum,c. Energy, of course, never travels faster than c. A proof of equation (11.81) nowwill be given. Since the mediumis conservative, the series impedanceZ(jw) and shunt admittance Y(jw) must be pure imaginary functions. Therefore, Z(S) and Y(S) can be represented Z(S) = IS =IoS + I1S3 -t- I2S5 + I3S7 +’", Y(S) CS= CoS+ C~S3+ C’~S5+ C3S~+’".

(11.82a) (11.82b)

Consider the traveling wave e = e0 sin(wt - kx),

(11.83a)

0 = 00 sin(wt - kx).

(11.83b)

The series expansions require /Co -- C1032 ~- C2034 - C3~d6 ~- "-"

0°=1~e°; ]{:=V ~°:/-~’°+/-~--’/-~-~-::’-

’

(11.84)

and the mean power becomes 1 P = e~ = e0~0sin2(wt- kx) = ~eoqo = 1yveS’2

(11.85)

The plan is to solve equation (11.80) for Ve, so the time-averaged energy density, P + T is needed next. To find ~, the complianceC is excited with an effort e(t) that, at the initial time, equals zero and has zero derivatives m dme/dt for all m_< n, wheren is twice the largest subscript of interest in the expansion for C given in equation (9.82b). For larger time it is required that e -~ e0 sin wt;

t -~ oo.

(11.86)

Sucha function is illustrated in Fig. 11.16. Its details for small values of time greater than zero nre immaterial; only the conditions at t = 0 and as t -~ c~ matter. The flow ~ becomes n d~+1 e ~ = ~ C~ dt~+. (11.87) ~ This allows the potential energy to be written as td~+~e P(t) =te(idt ~- ~f C~ ]o e d-~JY L

(11.88)

i----O

7Leon Brillouin, Wave Propagation and Group Velocity, Academic Press, New York, 1960. Brillouin shows that equation (11.81) applies when C = C(w) is a slowly varying function of w, and stroagly implies that it does not hold otherwise. The proof herein shows that it applies for any real functions C = C(w) and I = l(w).

844

CHAPTER 11,

DISTRIBUTED-PARAMETER

MODELS

eo

-e o

Figure 11.16: Function used to find the mean potential

energy in the compliance

c = c(~) The term for i = 0 can be integrated directly. The term for i = 1 can integrated by parts. The term for i = 2 can be integrated by parts twice, and so on for larger values of i. The result is l:(t)

= Coe 2 + C~ e dt 2 2 \ dt ] J + C2 e dt 4 dt dt "~ + ~ \-~-~ +C3 edt 6 dt dt 5 + dt "z dt a 2 ~,dt 3] J +""

]

(11.89)

Substituting the values of e(t) for t = 0 and t --~ ~c, and noting the time averages for the square of sine waves, there results 1 _., 2. ~ = ~C eo,

C* =- Co - 3Clw~ + 5C2w4 - 7C3~v6 +’" .

(11.90)

A similar process gives the mean kinetic energy 1 , .~ ~ = -~I qS;

I* -- Io - 3I~w~ + 5I~4 - 713~v6~ +..’.

(11.91)

The substutution of equations (11.85), (11.90) and (11.91) into (11.80), that the phase velocity is Vp -- 1/v/~, gives ve _ 2/Vp Vp C*/Yc + I*Yc

_ (C*/C)

2 + (I*/I)"

(11.92)

In the nondispersive case C = Co and I = I0, C* = C and I* -- I confirming that ve = The general dispersion relation k = wv/-f~

= w/v~

(11.93)

gives the group velocity dw v~ - dk - 1 v~

1 w dv," v~

dw

(11.94)

11.4.

ONE-POWER SYMhlETRIC

MODELS

845

However,

(11.95) so that equation (11.94) yields + 2-~wdI V p 1 + ~-~w~wdC d--~

(11.96)

+

(1+

The terms in parentheses can be evaluated through the use of the expansion for C and I; comparison with the definitions of C* and I* show that they are identical, respectively, to C*/C and I*/I. Therefore the right sides of equations (11.92) and (11.96) are equal, proving that ve = vg as was to be shown. 11.4.3

Boundary-Value

Problems;

Transmission

Matrices

The response of distributed-parameter systems has been computed thus far in terms of waves. More commonly, however, you are confronted with certain end conditions and wish to find the remaining end conditions, with little interest in the internal b.ehavior. For the problem with hot water transport in Section 11.3, only one possible input existed -- the inlet temperature -- and the remaining end condition was in fact a single wave. For the torsion rod and the heat conduction or diffusion problem, however, or in fact any non-degenerate one-power model or indeed any general line~tr two-port system, four different combinations of independent and dependent end conditions (at x = 0 and x = L) have direct~ causal meaning. E~ch of these can be represented by a causal matrix, as with lumped models. Placing the independent or input variables on the right and the de ~endent or output variables on the left, these are admittance matrix : impedance matrix : (Z adpedance matrix : immitance matrix :

(11.97)

Only two more combinations of the variables exist: transmission matrix : inverse of T :

[

y~(O)] = T [ys(L)] pa(O) [ya(L) J yo(L) ] T-~

[yo(0)]"

(11.98)

846

CHAPTER 11.

DISTRIBUTED-PARAMETER

MODELS

P2

~P~ ~___.~2-port N

=T~

;

=T~

= T~ note:

;

;

same form of result

~P~

tbr

=Tc

;

etc.

T~ = T~-T~ .Tc --. 4-ports,

6-ports,

etc.:

~

Figure 11.17: Cascade of two-port elements These last two relations are acausal in the sense that in a physical experiment the two variables on the right sides cannot be varied independently; as you have seen almost from the beginning, it is necessary for one variable to be specified from the right and another to be specified from the left. These causalities also have meaning in the assembly of equations for numerical simulation, as you have seen.

Transmission matrices nevertheless have one very useful property: if two or more two-port devices are connected end-to-end, the transmission matrix of the overall cascade equals the product of the individual transmission matrices. Recall that if the cascade is oriented left-to-right, the order of the matrix factors corresponds to the order of the corresponding elements, as shown in Fig. 11.17. The reverse order- applies for the inverse transmission matrices. In practice it is convenient to compute the transmission matrix representation of a distributed-parameter model first, and then convert it to one of the causal forms if necessary. The conversion formulas, which can be derived readily, are V= TI~ G= ~1~

-TI~ -Tx2

;

;

H= ~22

Z=~

1

-Tie

-T21

T = [T~l[Tll T:~T~2] ,’ A ~ detT.

’ ’ (11.995)

The transmission matrix for the general model of equation (11.70) (p 837) c~ be found by evaluating Ys and y~ ~ given in equation (11.72) at x = and x = L in terms of Ys~ (t) and ys2(t), and inverting one of these relations to

11.4.

847

ONE-POWER SYMMETRIC MODELS

eliminate Yst and Ys2. The result, utility:

r

yo(0)] =,~l~sinhr.

which you should verify, is of considerable

F=yL.

coshF(1/ ’S)sinhr]r , [ya(L s(L/] i ;

transmission

matrix,

(11.100)

T

Applications now will be illustrated.

Example

11.12

A slab of conducting material of thickness L is insulated on its left surface and excited on its right surface by a sinusoidally varying temperature: insulating

~

specified

boundary q(O) =0

temperature O(L) =

Find the temperature at the left side of the slab as a function of frequency, and plot its real (in-phase) part imaginary (90° out-of phase) parts. (Use a complex plane for a frequency response is knownin the controls literature, including Chapter 8, as a Nyquist plot.) Solution: The given information is in the form suitable mittance matrix H: q(L)]

= H

[e(n)

= H

for using the im-

e(n)

Only H~2 and H2.) are needed, which from equation (11.99) H,~. = A/T~.~ = sechr; The following identities lations:

H~.~ = -T2I/T.),~ -= -]~ tanh F.

are particularly

useful for frequency response calcu-

sinh((~ ÷ jfl) = sinh ~ cos ~ ÷ j cosh ~ sin fl, cosh((~ + jf~) -- cosh ~ cos B + j sinh ~ sin ~. ~om the solution to Problem 11.11 (p. 841), F = j~

+ ~;

~ = kL = L~.

848

CHAPTER 11.

DISTRIBUTED-PARAMETER MODELS

Therefore, the ratio of the temperatureat the left to the excitation temperature on the right is ~(O, ja~) ~(L,j~) = sechr 1/coshF = cosh f~cos~ + jsinhflsinFt sechf~see f~ e -j tan- 1 (tanh ~ tan f~) X/1+ (tanh f~ tan This result is plotted below, using the complexplane or Nyquist coordinates. Note that the phase lag in the response doesn’t reach 180° until a frequency at whichthe amplitude has decayedto a small value. =

Im

5zr/4 3z/2 = D ’ 0.2 5z/6# 3 z/4’l

’ 0.4

’ 0.6

t z/12

I

O(O, jta) -0’ 2z/3~ 7n-/12 ~

0 ~ Re

, 0.8

/

/~16

-0.6

Example 11.13 Consider a torsion rod of length L, for which P = ",/L = (L/VB)S

=

TS; T

= L/vp =

Lv’--C-] = LX/-~.

Here, T is the time a wave takes to propagate from one end of the rod to the other. The rod is terminated in a linear ideal rotary dashpot with a constant resistance R, so that M(L) R~(L):

IM@~ ......

x=O

rotational ~_~ lumped dashpot x=L

A constant torque M(0) is suddenly applied at the left end. Plot the consequent angular velocity ~(0) at this driving point as a function of time for the cases R]~ = 0, 1/3, 1 and 2.

11.4.

ONE-POWER SYMMETRIC

Solution: The transmission matrix relation

[~I((00))]

849

MODELS is

[ =}’~. sinh (1/1~)sinhTS][ TS cosh TS J t R.~(L)]. ¢(L) [ coshTS

Solving for ~(0),

TS+coshTS .1 + a .... 4¢0) Y ~ RYcsinh cosh TS + sinh~¢’./~ M(0) = } ~ ~ C~ =

]%(1

+ 2ae -2TS

+ 2a2e -4TS a~

+ 2a3e -6TS

+’"

.)M(0),

11 +

The successive operators on the right side of the equation represent pure delays of, respectively, zero, 2T or one round-trip delay, 4T or two round-trip delays, 6T or three round-trip delays, etc. The amplitude of each successive term or wave is attenuated more than its predecessor if R > 0, since then a < 1. The requested responses are plotted below. Note that for the special case of R = 1/~), that is when the resistance of the termination equals the surge impedance, no waves at all are reflected from the termination; a = 0. It is as if the rod were infinitely long.

6 Y~M(O) 4

2

RYe= 1

1 0o

2

4

6

RY~ = 2

8 t/T 10

12

Back in the heyday of analog electronics it was very difficult to achieve pure time delays of electrical signals, and a mechanical torsion rod with the reflectionless termination sometimes was employed, with appropriate input and output electromechanical transducers, to do the job. Today, important analogs to this ideal termination remain in the design of hydraulic and acoustic systems.

850

CHAPTER 11. Example

DISTRIB

UTED-PARAMETER

MODELS

11.14

A tosion rod is free at its left end and is terminated by a (rather stiff) rotary spring at its right end:

.... ~M~~

~

x=O

rotational spring lumped

x=L

This system has no damping, and possesses an infinity of resonant frequencies. Determine these frequencies as functions of the spring compliance Ct, the characteristic admittance ~ for torsion of the rod and the wave delay time, T. Also, compare the result when the spring is very compliant to that of a lumped model that neglects the flexibility of the rod. Solution:

For the lumped termination

comprising the spring,

Since M(0) -- 0, the transmission relation

becomes

[ ~0) = ] I[YcsinhTS c°shTS (i/Yc)sinhTS1 coshTS

J I~(L’)/C~S1 [ ¢(L)

For a non-trivial solution to exist, the first scalar equation requires coth TS = -CtS/Yc. This can happen only at the resonant or natural

frequencies

coth T ja~i = -j cot Tcai = - Ct ja~i / Yc, or

cot Tcai= Ctcai/Yc.

The two sides of this transcendental equation are plotted on the next page for three values of the parameter Ct/TY~, which represents the ratio of the complance of the spring termination to the total rotational compliance of the rod. Solutions are represented by the intersections. When this parameter is very small (i.e very stiff termination), the natural frequencies are seen to approach (n - 1/2)n, n = 1, 2 3..., which is the limit for the shaft clamped at the right end. Whenthe parameter is very large (i.e. very flexible termination), the first natural frequency becomes so low that the higher natural frequencies likely become unimportant. Since cot Ta~l ~ litton, the first natural frequency approaches 1//x/~-L~t, which is the natural frequency of the lumped mass-spring model in which the rod is the rotational inertia, (IL), and the spring is the termination spring (Ct).

11.4.

851

ONE-POWER SYMMETRIC MODELS

4 3 2 I 0 -1 Virtually identical analyses to the last two examples above can be carried out for analagous systems represented by pure wave delay models, such as those highlighted in Section 11.1. 11.4.4

Exponentially

Tapered

Systems*

The simplest behavior for a non-uniform locally symmetric system occurs when there is an exponential taper: Z = ZoeaX; -a Y x. = l~be

(11.101)

Note that the product YZ is independent of x. Such a model can represent, for example, a torsion rod or a push rod or an acoustic tube for which the diameter is exponentially tapered; the nominal wave speed of the mediumis independent of x. Systems with other tapers might well be approximated by a sequence of exponential segments. Elimination of y, in favor of y~ gives d~y_~£~ 2dx = ZoYo ys + a~x " Assumption of the particular

(11.102)

solution y~ = e’Zy~o

(11.103)

gives the following eigenvalues and general solution: #a,2 = ~ =k b; 0

Zoya(z, t) =

b =

+ ZoYo;

(ll.104a)

ebx

-

(11.104b) Equation (11.104b) can be inverted to solve for y~l(t) and y~2(t) as functions of y~(x, t) and Zoy~(x, t). The result then can be substituted back into the special

852

CHAPTER 11.

DISTRIBUTED-PARAMETER

MODELS

case of equation (11.104b) for which x = 0. The final result is the transmission matrix relation ZoYa(O,t) T= |

a [coshbx+-~sinhbx [

1 ~ sinh bx! 1 [e-’ ~~ a coshbx- ~-~sinhbxJ

Z0]~ . ,. ~- s~naox

(ll.105a)

.ZoYa(X,t) ]

eax/~ 0)] t~/~

(ll.105b) The right-hand matrix factor in equation (ll.105b) represents a transformer action with transformer modulus e-ax/2. The left-hand term represents both the delay that would occur in the absence of the taper and the distortion produced by the taper. The distortion is large for low frequencies and long wavelengths and small for high frequencies and short wavelengths; note that the distortion coefficient a/2b --+ 0 as w -~ oc. It is the transformer action that makes water waves grow as they approach the beach, that concentrates the sound in the historical hearing tubes, and makes the tool mounted at the small end of an exponentially tapered rod used in ultrasonic machining vibrate with a much larger amplitude than the large driven end. These examples can be modeled by the pure bilateral wave model with nominal wave celerity Co - 1/v/~-~ = constant and nominal wave number ko = w/co = ,k0/2~r. For frequency response, b= 3~ ° 1.

(11.106)

For frequencies below aco/2 this is a real number; waves do not propagate, and the distortion is considerable. For frequencies above this critical frequency, on the other hand, b is an imaginary number, waves propagate and grow or decay according to the transformer action. The distortion coefficient in equation (11.105b) becomes a 1 2b y/1 - (2w/aco)2

--

1

-V/1 - (2ko/a) 2 V/1 - 2(4~r/Aoa)

(11.107)

This reveals that little distortion exists for aAo< < 4~r. For example, if the taper is a factor of two in one wavelength, aA0 = in 2 and a/2b = j0.0552, which is fairly small; shorter wavelengths produce yet lesser distortion. 11.4.5

Summary

The bilateral waves in symmetric one-power models propagate in a similar way to the unilateral waves in degenerate one-power models. The information on the speed and attenuation of waves is contained in the propagation operator, % which can be employed as a temporal or Laplace operator or as a complex function of frequency. A wave of the anti-symmetric variable is related to the corresponding wave of the symmetric variable by the characteristic or surge

11.4.

ONE-POWER SYMMETRIC MODELS

853

P=Posin~ot ] area A, fluid properties p. fl .~

I ,

L

q transducer volume

Figure 11.18: System for Guided Problem11.5 impedance, Yc, which if not constant also can be treated as a temporal or Laplace operator or as a complexfunction of frequency. Energypropagates at the group velocity whenthere is no dissipation. Boundary value problems are most commonlyaddressed starting with the transmission matrix that interrelates the states at the two ends. Matrices for elementsthat are cascadedcan be multiplied directly to give an overall transmission matrix. This acausal matrix can be converted, if desired, into any of four causal matrices, dependingon whichboundaryvariables are specified and which are to be found. Both operator and frequency approaches and both uniform and exponentially tapered mediahave been illustrated. The pure bilateral wavemodelis based on an inertance per unit length and a complianceper unit length. A variety of systems can be approximatedby this model, as detailed in Section 11.1; the exampleof a torsion rod has been given here. The modelfor one-dimensional diffusion or heat transfer is based on a complianceper unit length and a resistance per unit length. Guided

Problem

11.5

Capacitive-type pressure transducers typically have a significant internal volume,V, which introduces fluid compliance. Often they are placed at the end of an instrument line, as shownin Fig. 11.18. Presumethat the pressure of interest is Po sin wt and exists at the opposite end of the line from the transducer. Also presumethat the line is rigid and has a length L and an internal area A, the density and bulk modulusfor the fluid are p and ~, and the volumehas no gas bubbles and is virtually rigid mechanically. Find the ratio of the measured pressure to the pressure P. Suggested Steps: 1. Represent the systemby a bilateral pure delay line with a terminal compliance. Relate the complianceto the volumeand the fluid properties. 2. Find the transmission matrix as a function of S -~ jw. 3. Substitute the relation for the flow into the transducer volume,expressed as a function of the pressure there, into the transmission matrix relation.

854

CHAPTER 11.

DISTRIBUTED-PARAMETER

MODELS

Then, use this relation to find the frequency transfer function between the two pressures. Report the result as a magnitude ratio and a phase angle.

Guided

Problem

11.6

A flexible shaft of radius r = 1.0 in., length L = 40 in., shear modulus p = 11.5 x 106 psi (steel) and weight density pg = 0.283 lb/in 3 is driven at its left end by an applied moment M = Mo sin wt, Mo = 200 in-lb, and is terminated at the right end by a flywheel of the same material having a radius of 5.0 in. and a thickness of 0.10 in.; a viscous damping momentwith resistance R = 2000 in-lb-s is applied to the flywheel. Find the driving-point admittance ~)o(O)/Mo and the transfer admittance ~o(L)/Mo as a function of frequency. Plot their magnitudes and phase angles for frequencies up to 6000 Hz. Suggested

Steps:

1. Write the relation angular velocity.

between the moment applied to the flywheel to its

2. Write the transmission matrix relationship between the two ends. Substitute the known applied momentat the left end and the result from step 1 at the right end. 3. Solve for the desired admittances. At this point your work ought to be in terms of symbols. 4. Evaluate the parameters nmnerically, substitute into the results of step 3, and plot. Linear axes are suggested rather than the usual Bode axes in problems dominated by traveling waves. PROBLEMS 11.17 A uniform linearly elastic tube contains water and has an effective inertance to fluid motion I’, compliance C~ and resistance R~, all per unit length, which can be assumed to be constants. In addition, the tube is pierced with manysmall holes to produce a constant leakage conductance, G~, per unit length. (The resulting model is the same as in Problem 11.13, p. 823.) (a) Define appropriate variables and find linear describing differential equations in standard form, neglecting the time-mean velocity and other second-order or nonlinear effects. (b) Showthat for given values of ~, C~ and R~, aparticular va lue of G~ causes the characteristic impedance to be constant. Find this value, and investigate its affect on wave propagation (phase velocity, group velocity, dispersion, distortion, attenuation).

11.4.

ONE-POWER SYMMETRIC

MODELS

855

(c) Showthat with the condition of part (b) a particular nation, Re, eliminates wave reflections. Also, find this

resistive

termi-

11.18 A steel torsion rod 500 in. long is driven at the left end and terminated at the right end by a linear viscous damper with coefficient 0.07 in..lb.s. (The properties of steel are noted in Guided Problem 11.6 above.) The left end starts at rest and is given a step change in angular velocity to 1800 rpm. Find and plot the momentin the rod at the driving end and the angular velocity at the right end as functions of time for rod diameters of (a) 0.250 in. (b) 0.375 11.19 A 0.250 in. torsion rod of the preceding problem is unconstrained at the left end and terminated at the right end by a linear rotary spring with rate 7.74 in lb/rad. Find the eigenvalue equation for the system and determine the first four natural frequencies. 11.20 The torsion rod of the preceding problem is driven sinusoidally from the left end. Determine the ratio of the moment to the angular velocity at the driving end. 11.21 A long torsion rod of length L, wavespeed c and characteristic impedance Zc is clamped at its left end and terminates in a flywheel of rotational inertance I~, at its right end. (a) Find the eigenvalue equation for the resonant frequencies. (b) Approximatethe first three natural frequencies in the special case for which the rotational inertance of the entire rod equals (c) Evaluate the lowest eigenvalue when the rotational inertance of the rod approaches zero. Then compare this result with that for a lumped model that ignores the inertance of the rod in favor of its compliance. 11.22 A coaxial cable is terminated in a parallel combination of a resistance and a capacitance. The value of the resistance equals the surge impedance, so that if the capacitance were absent there would be no wave reflections and no standing waves. The termination can be corrected at any given frequency for the presence of the capacitance by placing an open-ended stub of the same cable in parallel with the load, as shown below. load terminal end of coaxial cable

856

CHAPTER 11.

DISTRIBUTED-PARAMETER

MODELS

The proper length of the stub, L, depends on the load capacitance, C. This example of "stub matching" is practical, since most coaxial cables are used only for a very narrow band of frequencies. (Note: Stub matching has been used to prevent reflections and resonances due to compliance and inertance in fluid transmission lines and acoustic ducts, also.) (a) Consider the stub by itself, with zero current at its open end. Find the ratio of the voltage to the current at driven end in terms of the length of the stub and its surge impedance and wave speed. (b) Place the impedance from part (a) in parallel with the actual load. the resulting impedance equal to the surge impedance to find the proper length of the stub.

11.23 The muffler of Problem 11.14 (p. 823) has length L = 2 ft, and joins together two infinitely long tubes with the same diameter di as the perforated inner tube within the muffler. The resonant (Helmholtz) frequency of all the side chambers is w, = 2~r x 100 rad/s, and the ratio of the outer to the inner diameter is (do/di) 10. A si nusoidal wa ve wi th fr equency w, wave spe ed c = 1100 ft/s and pressure amplitude P+ is incident to the muffler at one end, and a consequent wave PL emerges at the other end. Approximating the muffler as a uniform continuous structure, find and plot the magnitude ratio as a function of f~ = w/w,~. Commenton the performance of the muffler. 11.24 A magnetostrictive driver oscillates the large end of a steel exponentially tapered ultrasonic machine tool at the frequency 30kHz. A motion amplification of 40 is desired, and the large end of the taper has a diameter of one inch. Make reasonable choices for the length and diameter at the small end of the taper. The taper should not be unnecessarily long.

SOLUTIONS

TO

Guided Problem

11.5

1.

P~

D Pt

C

[ cos Tw 2. T= [(1/Z~)sinhTw

GUIDED

PROBLEMS

C = V/3 ZccosTw sinh TwJ" ]

T= L/c=

Lv/~;

3. Qto = CjwPto The first equation in the transmission relation becomes P0 = [cos Tw+( Cjw ) j Zc sin Tw] Pto so that Pto= Po/(cosTw - CwZc sinTw); Pt = Ptosin~t P,o---The magnituderatio therefore is -~0

(cosTw-~wZcsinTw)l

11.4. ONE-POWER

SYMMETRIC

857

MODELS

and the phase angle is 0° when Pto/Po > 0 and 4-1800 when Pto/Po < O. Guided

Problem

11,6

1. M(L) = (Ifs ÷ R)~(L), or Mo(L) = (jlfw ÷ R)~o(L), where If is the moment of inertia of the flywheel. 2. [~o(O)j=

3. ~o(L) Mo

[(j/Zc)sinTw

~°(L) cosTw J [II’~+R]

1 (Iijw + R) cos Tw+ jZ~ sin Tw

~o(0~ = (Iijw + R)(j/Z~) sin Tw+ cos T~ Mo (Iijw + R) cos Tw + jZ~ sin Tw With MATLAB it is not necessary to write expressions for the real and imaginky p~ts sep~ately, or otherwise simplify the equations above. ~.

1 ~ 1 0.283 ~ = ~ = !(~.~)~)~ .(0.~)(~) ~ ~. = 0.0n0 ~.i...~ 2 2 386 R = 2000lb.in..s. 0.283 L ~ =40 386~11.5~ 10 ~ = 3.194 x 10-~ seconds T=~=L ~ a ra ~(1.0) ~ /0.283~ Z¢ = ~ = ~/~11.5 ~ 10 = 144.2

lb.in..s.

5O Mois 200lb in

4O

~o(O)...... 3O 20[

0

t

~

2

4

5

Tea

10

12

858

CHAPTER 11.

.angle 2

DISTRIBUTED-PARAMETER

of (~o(L)/Mo

MODELS

(drops~

~ angle of ~o(O)/M o

2

11.5

Multiple-Power

4

6

8

10 To)

12

Models

Systems such as flexible fluid-carrying tubes, parallel and counter-flow heat exchangers and beams with lateral vibrations can be represented by one-dimensional distributed-parameter models with more than a single power. The structure of such models has been given in Section 11.2.4 (pp. 820-821). The general boun.dary-value problem is addressed here, followed by some examples. (Wave behavior and internal excitations are discussed in the reference cited in footnote 1 on page 795.) 11.5.1

Symmetric and Models

and

Anti-Symmetric

Variables

A conjugate pair of symmetric and anti-symmetric variables have been employed in modeling one-power one-dimensional models. Multiple-power models, except in degenerate cases, employ a conjugate pair of symmetric and anti-symmetric variables for each power. Whenthese are chosen as true power-factoring variables (efforts and flows) the sum of the products of the respective symmetric and anti-symmetric variables is the total power. This choice is not mandatory, however, as before: the variables can be merely proportional to the efforts and flows or their time derivatives or integrals, as long as they satisfy the criteria of symmetry and anti-symmetry. The symmetric variables are collected together in a vector, Ys, and the anti-symmetric variables are collected together in a second vector, Ya~ employing the same order with respect to the powers being conveyed. Thus, repeating equation (11.35), Ys ] " y = y~

(11.108)

The example of a Bernoulli-Euler beam has been given in Guided Problem 11.4 (pp. 822, 824). The general differential equation is, repeating equation (11.36),

11.5.

MULTIPLE-POWER

MODELS

859

Y~ + Bu(x, Whenthe model is locally symmetric, W= X = 0. In this case, the transsmission matrix can be shown to be

= LY(~) -~ ~i~h ~

co~h ~

]’

(11.110)

The matrix elements of this partitioned matrix can be found in terms of the matrix products ZY and YZ and their identical eigenvalues. This can be accomplished using Sylvester’s theorem or the Cayley-Hamilton theoremS; the example of the lateral vibration of the slender beam is carried out below. Note that for the one-power case, this equation reduces to the familiar form of equation (11.100) (p. 847). Another advantage of the ordering of the variables for the symmetric case results if you wish to examine the propagation of waves, or use wave-scattering variables. The characteristic values of A are desired. The coefficients of the odd-powered terms in the characteristic polynomial vanish, as can be seen from Bocher’s formulas, t° Further, since trace(ZY) m = trace(YZ) m

m = 1, 2, 3 ....

.,

(11.111)

the 2n (or less) nonzero eigenvalues of A can be shown to be equal to the square roots of the n (or less) nonzero eigenvalues of ZYor YZ. This relationship exists regardless of whether Z and Y are square (partial degeneracies). As result, matrix and numerical manipulations are kept to a minimum.

11.5.2

Case flow

Study of a Degenerate Heat Exchanger

System:

A Counter-

A totally degenerate model, for which all symmetric or all asymmetric variables have been purged, has A = X or A = W, respectively, and cannot be sym~netrical. The simple counterflow heat exchanger shown in Fig. 11.19 can be represented by such a degenerate two-power uniform one-dimensional model. The degeneracy results from converting the otherwise variable fluid velocities into assumed constant parameters, v~ and ve, leaving the temperatures 0t and 0e as the only state variables. To simplify matters it is assumed that the thermal compliance of the fluids per unit length, C~ and C2, are so much larger than the compliances of the walls that the latter can be neglected. The thermal conductivity between the two tubes, per unit length, is G. The outer walls are SSeefootnote5 (p. 821)for a referenceto a derivation. 9See for exampleL.A. Pipes, Matrix Methods]or Engineering,Prentice-Hall, Inc, EnglewoodCliffs, NJ, 1963. ~Oibid.

860

CHAPTER 11.

DISTRIBUTED-PARAMETER

MODELS

0~(0) ,.v~" .~. ~, ,. 1 k\\\\\\\\\\\\~ b--~x 0 Figure 11.19: Counterflow heat exchanger assumed to be perfect

d [01] ~xx[ J02

insulators.

=_

The model becomes

vl GF C2 v2

[

Clvl G/C2 + S 02 v2

"

(11.112)

The Cayley-Hamilton method of evaluating a function of a square matrix will be illustrated for the transmission matrix exp(-WL). The eigenvalues W are defined as #1 and ~. A function of an n x n matrix can be expressed ag a polynomial in the matrix of order n - 1. Since Wis second order, e -wL = a0I + alW.

(11.113)

The problem is to find the coefficients ao and al. According to the CayleyHamilton theorem, scalar equations corresponding to equation (11.113) apply, in which an eigenvalue is substituted for W. Therefore, e -~L =- ao + al]ll,

(11.114a)

e-N2L ~--- a0 -1- alP2.

(ll.l14b)

Inverting this result gives the coefficients:

(11.115) The causal inputs are 01 (0) and O~(L). The transfer function between the latter and the output 02(0), for example, is the lower-right element of the transmission matrix exp(-WL), or T~_2, which T’22 ----

a0 + al ~/~2 ~---

~

[(u,. - w2’,)e-"’L - (m- w2~)e-"~L] ¯ (11.116)

The eigenvalues are #1 = 11 -/2; 1

[

11=-~ S

(~2

-

#2 --- ll + 12, V~)

G

G

+ C~v2 C-~,’

(ll.l17a) (11.117b)

11.5.

MULTIPLE-POWER

MODELS

12 =

’~ G CiC2v, v2’

l’~

+

13 = ~ S With ~hese substitutions

i

861

the transfer

-’IL e 0_~(0) =- Ta2: ~[1.,.

O~(L)

(11.117c)

+ C~v-~. +

(ll.l17d)

"

function reduces to . cosh(/2L)- lu sinh(/.,L)]._

(11.118

Should the frequency response be desired, the substitution S ~ jw gives the result directly. Wor~ble transfer functions for the direct evaluation of transient responses are harder to achieve, although approximations combining the ratios of polynomials and pure delays can be made to apply at high frequencies, and ratios of polynomials without pure delays at low i~equencies. Fourier decomposition might be the best route in complicated cases.

11.5.3

Case Study of a Symmetric Model: The BernoulliEuler Beam

The Bernoulli-Euler model of the slender beam with lateral vibration has been introduced in Guided Problem 11.4 (p. 822,824). It assumes the classical relation between the bending momentand the curvature of the centerline of a slender beam, as presented in elementary strength-of-materials courses, assuming a shear-free rotation of each differential element. It relates the gradient of the momentto the shear force. Finally, it restricts inertial forces to the product of the mass per unit length and the lateral acceleration. With the mass density, Young’s modulus, cross sectional area, and area momentof inertia of the cross section being defined respectively as p, E, A and I,~, the state differential equation becomes, ll repeating from the solution to Guided Problem 11.4, d ~x

I° °

0 =- pAs "2 [ O~/OxJ 0 -

0 0 1/EI~

-1 0 0

"

(11.119)

O~/Ox

Note that W and X vanish, verifying the local symmetry of the beam or, assuming the symmetry is known, serving ~ a partial check of the model. ~ ~ A somewhatbetter model, known~ the Timoshenkobeam,results from including the effects of rotational inertia and shear-induceddeflections. This modelis described by the following, in whichG is the shear modulusand k2 = Ga2/[E/(1- ~2)] wherep is Poisson’s ratio anda2 is the Timoshenko shear coefficient: d __

0

= ~

0

~

k2GA EIm

o

pls2+k2G~ Elm

.o

’

?j

862

CHAPTER 11.

DISTRIBUTED-PARAMETER

MODELS

The four eigenvalues of A, which equal the square roots of the eigenvalues of ZY or YZ, are (_pAS2. ~ ~/4 (11.120) P = \ Elm It is convenient to label the principal eigenvalue, which is positive real if S ~ jw, as #~. This allows the following nondimensionalization of the state differential equations:

d

[~

d-~/~

0

0

-1

=-"1 -1

0

0

"i~ I~ [.

~!

(11.121)

GO~ ~ The transmission matrix can be found using equation (11.110), with each quadrant evaluated using the Cayley-Hamilton method as illustrated above. (The reader might well check one of the quadrants.) The result is

~

/

1

~2

kl

~3

~4

/

~

(11.~22a)

kX ~ eoSh~l(Z - zo) + cos~l(Z - z0); k~ ~ cosh ~(z - zo) - cos~(x c ~o); ka ~ sinh ~(z - zo) + sin ~1 (z - zo); (11.122b) k4 ~ sinh~x (z - zo) - sin ~l(Z - zo). You now may substitute whatever boundary conditions are imposed, and solve for the remaining boundary variables. For an example, see Guided Problem 11.7 below. 11.5.4

Summary

Boundary-value problems for one-dimensional distributed-parameter models have been addressed generally. Advantages accrue from segregating the symmetric and asymmetric elements in the vector of state variables. If the model is locally symmetric, two of the four quadrants of A vanish, the eigenvalues can be found from treating a matrix of half the dimension, and the transmission matrix is simplified. The evaluation of functions of matrices using the Cayley-Hamilton theorem has been illustrated. The example of a simple counterflow heat exchanger typifies a totally degenerate two-power model, and the example of the Bernoulli-Euler beam typifies a 2multi-powered locally symmetric model) I’)A fuller understanding of the Bernoulli-Euler example is given in the reference cited in footnote 1 on page 795, where a new kind of wave is shown to decay spatially while propagating no power.

11.5.

MULTIPLE-POWER MODELS

Guided

Problem

863

11.7

Estimate the bending natural frequencies of a slender beam of length L that is cantilevered at one end and free at the other end. Suggested Steps:

Since the beam is slender, the Bernoulli-Euler model likely is acceptable except for the higher modes that have short wavelengths. Identify which of the boundary variables in equation (11.122a) are zero. With this information find a 2 × 2 matrix function of #1L which has a zero determinant. Write the characteristic plify.

equation, and use trigonometric identities

to sim-

Represent the resulting equation as one simple trigonometric function of #IL equal to another. Then, sketch-plot the two functions and estimate their intersections, which represent the respective modes. 5. Solve for the modal frequencies in terms of the parameters of the beam.

PROBLEMS 11.25 A slender uniform beamis cantilevered at one end and is free at the other end, apart from the attachment of a compact mass equal to the total mass of the beamitself. (a) Find the first three resonant frequencies of the system in terms of the dimensions and material properties of the beam. (b) Compare the first resonant frequency to the frequency found from simple analysis that ignores the dynamics of the beam altogether, instead considering it as a pure lateral spring. Showthat the results agree in the limit in which the mass of the beam becomes insignificant compared to the mass of the end. (c) Estimate the effective lumped mass and compliance of the beam using field luInping technique valid when this mass is small comparedto the mass at the end. Determine the corresponding natural frequency. Compare this result with the distributed-parameter answer. Is this an improvement to the estimate of part (b)?

864

CHAPTER 11.

DISTRIBUTED-PARAMETER

MODELS

11.26 Hydroelectric turbines are fed with water through long tunnels, called "penstocks," bored through the rock and lined with reinforced concrete so as to be quite rigid. A sudden reduction in the flow admitted to the turbine causes a large waterhammer wave to propagate upstream and subsequently to reverberate. In order to limit the pressures of these waves, the installation of a relatively small flexible air hose throughout the length of the penstock is proposed. The compliant hose is kept under enough pressure to prevent its collapse. It contains frequent porous plugs to restrict the axial flow of air.

(a) Construct an equivalent bond graph model or lumped circuit for element Ax of the system. Do not include nonlinear or relatively unimportant phenomena. Hints: Start with bond graphs or equivalent circuit diagrams for the water-filled tunnel and the air hose as though they were not coupled. Neglect the inertia of the air relative to the resistance to flow from the porous plugs, which can be assumed to be continuously distributed. Then add the coupling due to the radial flexibility of the tube walls, noting that the sum of the cross-sectional areas of the water and the air is constant. Relate the parameters of your model to the density and bulk modulus of the water, the mean absolute pressure and the ratio of specific heats of the air, the cross-sectional areas of the tunnel and the tube, and the compliance and resistance of the tube. The latter two may be left unrelated to physical parameters for purposes of this problem. (b) Write a differential equation model for the system. Hints: Identify the symmetric and anti-symmetric variables, and write the partial differential equation in the recommended form. Note which elements of the A-matrix represent the coupling between the tunnel and the tube. Are the elements of W and X indeed zero? (c) Find the eigenvalues for wave propagation in terms of frequency and the parameters of your model. Plot the resulting phase velocity and attenuation factors for the special case in which the compliances of the water, the air and the wall of the tube are equal. Hints: Form the matrix product ZYor YZ, and find its eigenvalues. The eigenvalues of interest are the. square-roots of these. Whenthe three compliances are equal, the eigenvalues can be characterized in terms of two time constants (an I/R and an RC time constant), allowing the definition of di~nensionless eigenvalues and frequency. Thus a single plot for each attenuation factor and phase velocity can be made. (d) Discuss the effectiveness of this solution. Hint: Note the effect of the air tube on the surge impedance of the system, particularly fdr infinite frequency or abrupt surges, as well as the propagation characteristics. Ought the air tube be made larger, or more flexible?

11.5.

MULTIPLE-POWER

MODELS

865

SOLUTION TO GUIDED PROBLEM Guided

Problem

11.7

1.-2. Cantilevered at left end (x = 0): r/(0) (O~/Ox)(O) = l~ree at right end: M(L) = F(L)

Therefore,

M/121EIm F/#~EIm 0

1 k’~ ka k3 k4 = "~ k3 k4 k3 ~2 ~ ~=0 k~

(1/#l)(O#/Ox

(ll~)(O,yO~)

~:+L

kl ~ cosh~L + costaL, ks ~ sinhp~L + sinp~L, k4 ~ sinh #IL - sinp~L. 3. The determinant of this square matrix must equal zero, or k~ - k3k4 = O. expanding: cosh2 #1L+2cosh #IL cos #lL+cos2 #~ L-sinh ~ #~L+sin~ #~L = 0. Useof the identities coshe x - sin~ x -- 1; cos~ x + sin2 x = 1 reduces this result to 1 + coshp~LcOs#lL= O. 1 4.-5. The result above can be cast as cos#~L = coshg~L’ These functions can be plotted (a quick sketch will do) to get approximate answers:

firsl foureigenvalues 0

0

2

4

6

8 #~L 10

For i _> 4, #IL -~ (i - .5)~r, so that wi

12

(i - .5)2~r ~" ~ L2 V ~-

To resolve the first three eigenvalues accurately, you can use the MATLAB function fzero. Start with the m-file funct±ony=beam(x) y=l+cosh(x). *cos(x)

866

CHAPTER 11.

DISTRIBUTED-PARAMETER

MODELS

which you can save as beam.m.Then, the commandxzero=fzero (’ beam’, 1.6) gives the zero whichis closest to 1.6, namely1.8751. The guess 5 for the second zero produces 4.7124, and the guess 8 for the third zero produces 7.8548. The respective frequencies are proportional to the squares of these numbers: w~ =(#liL) ’2 ~/EI,~/pA.

Chapter

12

Thermodynamic

Systems

Irreversibilities were introduced in Section 9.4 with the examples of heat conduction and friction. The energy fluxes in these cases are analogous to the other power flows considered, in the sense that they can be represented by simple bonds with a single effort and a single flow. Mass transfer complicates the energy flux, however; a new convection bond is introduced in Section 12.1 to address this. The bond has two efforts and a single flow. It represents the entropy fluxes as well as the mass flows, and thereby accounts for the thermodynamic availability as well as the conservation of energy. Implementation for heat interactions and junctions for convection bonds is given in Section 12.2. Steady-state analysis can be carried out at this point, as illustrated in Section 12.3, analogous to the steady-state analysis of simple systems in Chapter 2 and parts of later chapters. Thermodynamicenergy storage is a potential energy representable by a compliance field. For purposes of lumped modeling, the kinetic energy associated with coherent velocites can be represented by separate inertances. Lumpedthermodynamiccompliance fields are presented in Section 12.4 This is followed in Section 12.5 by a detailed consideration of practical ways to represent thermodynamic state functions for real substances with computational models, foregoing table look-up alternatives. Examples given include simulations of the unsteady behavior of an air compressor and a refrigeration cycle. The chapter ends in Section 12.6 with an introduction to systems with unsteady chemical reaction.

12.1

The Convection Bond and Compressible Flow

The mechanical energy in the incolnpressible flows considered thus far is decoupled from the thermal energy, and the latter has been discounted as "dissipation." For compressible fiow there is no decoupling, since a change in temperature changes the density of the fluid and therefore its velocity, etc. As a result, it becomes necessary to bookkeep both the mechanical and thermal forms of 867

868

CHAPTER 12.

THERMODYNAMIC SYSTEMS

energy. Sometimesyou might opt to keep track of thermal energy and power even whenyou presumeincompressible flow. A new kind of compoundbond, called the convection bond, is introduced below.This leads to a generalization of the two-port irreversible or RS element, and later to a special heat interaction element and generalizations of the 1 and 0-junctions. Ultimately, thermodynamicenergy is represented by a multiport C element, and is extended to accomodatechemical reaction. 12.1.1

Flow Through

a Port

Consider a micro-port of area dA through which a compressible or incompressible fluid is flowing. The volumeand massflow rates, respectively, are dQ = v dA, dm= pv dA,

(12.1a) (12.1b) wherev is the fluid velocity and p is its mass density. The total energy flux whichcrosses a fixed control surface that cuts this micro-port is the surn of a "flow-work"powerand a transport of energy. The flow-workpoweris dT~l -= Pv dA= Pdrh, (12.2) P in whichP is the local absolute pressure. This is the samepowerthat, a piston with force P dA and velocity v wouldtransmit. The energy transport comprises the transport of internal energy (per unit mass: u), kinetic energy (per unit mass: v’2/2) and gravity energy (per unit mass: gz), assuming the absence of capillarity and electric and magneticfield effects: dT)t = (u + v~-/2 + gz)drh.

(12.3)

Thetotal energy flux is therefore d7) = dT) l + d7)t = + u + -~ + gz, dA.

(12.4)

The total powerfor a complete flow channel or port of cross-sectional area A is the central interest. Equation(12.4) is integrated across the channel evaluate this power. The simplest .assumptions are that P, p, u, v and z are constants over the cross-section. In this case, the integral becomes 7) =

+ u + --~ + gz ~h.

(12.5)

Withone exception these assumptionsare usually acceptable, particularly if the area of the cross-section is not large as in pipe flow, etc. The exceptionis the velocity, v. If the meanvelocity is defined as ~, the integral can be written as 7)=

+ u + c~-~ + g z rh

= (P + pu + c~p~ + pgz)Q,

(12.6)

12.1

THE CONVECTION BOND AND COMPRESSIBLE

Table

12.1

Effort

and Flow Analogies

FLOW

(Extended)

generalized force or effort, e

generalized velocity or flow, 4

Fluid, incompressible: general less thermal approximate micro-bond

pu + P + apf~2 /2 + pgz P’+ apf;2 /2 + pgz P any of above

Q Q Q v

Fluid, compressible: general approximate micro-bond

ho + gz h either above

rh pv

Mechanical, longitudinal: general approximate micro-bond

Fc + pAv2 /2 Fc a + pv2 /2

v = v = v =

Mechanical, transverse: rotation translation (shear) micro-bond

M Fs ¯ T

vs v~.

e

i

Electric circuit:

869

Thermal, conduction: where a depends on the shape of the velocity profile. The lower right-hand term in parentheses appears as the effort in the expanded list of effort and flow analogies given in Table 12.1, while Q appears as the flow; this is the usual choice for incompressible flow if thermal energy is to be included in the bookkeeping. Thetransportof kinetic energyis f-~pv ~ 2vdA =½v~p~2Q. Sinceq = ~A,this gives f v3 dA (12.7) a~3A A flat or slug-like profile, with v = ~, gives a = 1. A parabolic profile in a round tube, such as occurs in fully developed laminar flow, gives a = 2. Nearly all other cases are intermediate between these values; turbulent flow in most channels is apt to give a value closer to 1. " The enthalpy, h, is defined as Pip + u. The stagnation enthalpy, h0, is h + ~aw. This is the enthalpy that the stream would have if its velocity were slowed reversibly until the kinetic energy term (½ag’~) became negligible. Therefore, the power can be written in the condensed form 7~ = (ho + gz)~. The gravity term is omitted below, since in most instances it has little significance. It can be reintroduced easily.

(12.8) or no

870

CHAPTER

12.

THERMODYNAMIC

SYSTEMS

The density of a compressible flow may change from point to point along a streamline, particularly if it is in a channel of varying area, even if the flow is steady. In this instance the volume fiow rate Q = ~h/p varies from point to point also. This discourages the choice of Q as the flow or generalized velocity, unlike with assumed incompressible flow. The mass flow rate, rh, on the other hand, is constant along a channel if the flow is steady. Thus, rh is chosen as the flow variable when compressibility is assumed. It also can be used when the flow is considered incompressible, if you so prefer. Equation (12.7) indicates that, neglecting gravity effects, the proper effort or generalized force to be paired with ~h is the stagnation enthalpy, ho, giving ho These effort and flow variables are added to Table 12.1. Several authors have used this representation, but it contains a very significant ambiguity that demands resolutfon. An explanation follows.

12.1.2

The Convection

Bond

The conjugate pairs of effort and flow M and ~, F and ~, P and Q, and e and i describe work interactions. The conjugate pair 0 and ) describes a heat interaction. The problem with the pair h0 and ~n is that it describes a sum of a work interaction and a heat interaction, with unspecified proportions. Heat ~ and work are not universally convertable. The meaning thus is non-unique. Complete description of the interaction requires two variables in place of the single h0, assuming that no electrical, magnetic, capillary, chemical or other special effects exist. Recall the state postulate in thermodynamics: the state of a pure substance in the absence of these special effects is specified uniquely by two independent intensive variables. The most commonlyused pair is temperature and pressure, although these are not independent in the two-phase liquid-vapor region. Other commonlyused variables include density (or specific volume), quality (in the liquid-vapor two-phase region), enthalpy, entropy the Gibbs function. Discussion of the Gibbs function is deferred to Sections 12.4 and 12.6, where it will be used in connection with thermal and chemical energies. The available enthalpy, h~, and the unavailable enthalpy, h~, are a conceptually satisfying pair of efforts or generalized forces, but they are ~ not so computationally convenient and are not emphasized. 1 The same is true when the internal

energy u is allowed to appear in the effort

for incom-

pressible flow, as it does in one row of Table 12.1. Recall, however, that in past chapters u was dropped with the explanation that it is decoupled from the work, and conversion from work to heat was labeled as "dissipation." 2 Availability is defined as the useful work potential of the fluid, which is the work delivered when the fluid undergoes a reversible process from its current state to the state (temperature and pressure, for example) of the environment. The available enthalpy, more conventionally known as the stream availability, is shown in elementary thermodynamics textbooks to be ha = (ho - h~) - O’(s - s*),

12.1

THE CONVECTION BOND AND COMPRESSIBLE

FLOW

871

The choice of which pair of intensive variables to use in describing the state of a pure fluid substance is ultimately a practical matter which depends on computational requirements and the availability of state data. For most purposes the most convenient pair is Po, ho. The fact that a convection bond has two independent variables to describe the effort while a single variable describes the flow is indicated by adding a dashed line on the effort side of the bond: P0, ho Note that P0 is not directly an effort, because the product P0rh is not an energy flux. a Rather, h0 is the proper effort, and the value of P0 qualifies the character of this effort. Knowledgeof P0 and h0 is sufficient to define other stagnation state properties, including density. Whencombined with the mass flow rate (rh) they are sufficient also to define the velocity and all the actual (non-stagnation) state properties. The choice of Po and ho to describe the effort is based partly on considerations of causality, which become critical when unsteady behavior (Sections 12.4 and 12.5) is addressed. One of these, together with rh, must form a bilateral causal pair. The enthalpy h0 is physically causal only in the actual direction of the flow, unlike the usual effort variables on simple bonds. It could be said that the causality of enthalpy moves with the flow; it reverses wheneverthe flow reverses. The pressure P0, fortunately, suffers no such limitati~)n. Therefore, P0 and the mass flow rh form the needed bilateral causal pair, like the P0 and volume flow Q of a bond for incompressible flow,; the traditional causal stroke is used to indicate the causalities of P0 and rh only. For example, the causal stroke in the example Po, h0 indicates that rh is specified from the left and P0 is specified from the right. No indication of the causality of h0 is indicated or needed. The convection bond is compatible with the previously existing stable of bond graph elements, although certain logical generalizations are needed. In particular, two new primitive elements are necessary. Certain additional combinations of primitive elements also will be defined as new macro-elements, for purposes of computational convenience. The result will be a comprehensive although simple scheme for representing lumped models of thermofiuid and hybrid systems. 12.1.3

The RS Element

for

Fluid

Flow

The RS element is used in Section 9.4 to represent a thermal resistance and a mechanical resistance with thermal power included. Its application is now where the asterisk (.) indicates the state of the environment. The product hath is the theoretical maximumrate of work the flow could produce. The balance of the enthalpy, ho - ha, is the unavailable enthalpy, hu. The energy flux hurh represents only part of a heat interaction. 3When it is desired to break h0 into two actual components or efforts, the Gibbs function, g, and the product Os appear to be the best choice, as discussed in Section 12.6.

872

CHAPTER 12.

THERMODYNAMIC

SYSTEMS

extended to the flow of a fluid through a valve, or any other device, that drops its pressure without any heat interaction or storage of energy: /~01, h0~_ RS P02, ho The stagnation enthalpy and the total power remain constant, but the total pressure drops, i.e. if rh > 0, then P01 > P02. The unavailable enthalpy increases at the expense of the available enthalpy. Further, the rate of entropy generation, which is the difference between the entropy effiux and the entropy influx, must be positive: :~g = (s2 - sl)rh > (12.9) The fundamental property relation 0 ds = dh - dP/p

(12.10)

often is helpful in computing an entropy change. It is awkwardto carry the subscript 0 wherever it applies, so this subscript is dropped below, and stagnation properties are assumed. The actual calculations are made the same way you treated flow through a valve in your study of thermodynamics. Usually, the enthalpy and the pressures are known, so all other properties can be computed. The flow rh would have to be computed from knowledge of the mechanics of the flow, expressible functionally as

rh = fn(P,,P2,hl).

(12.11)

Finding such functions in general is beyond the scope of this text, but two cases are used repeatedly. The first is incompressible inviscid flow through an idealized orifice, for which Bernoulli’s equation model gives ~h = dm x/2p(P1

- P~),

(12.12)

where Amis the effective minimumarea of the orifice or the vena contracta. The second is compressible flow of an ideal and inviscid gas with constant specific heats through a converging orifice, which is given in Guided Problem 1.1 below. Sometimes the mass flow rate, the upstream enthalpy and one of the two pressures are specified and the other pressure is to be found. This also is readily accomphshedin most cases once the details of equation (12.11) are established. 12.1.4

Summary

The state of a pure substance is specified by any two independent intensive variables. Specification of the state of a flowing fluid requires a third variable to describe the rate of flow. The mass flow rate is chosen for this third variable, since in steady flow it is invariant along a channel of non-uniform area. The two intensive variables describe the effort associated with the flow. For most purposes the stagnation enthalpy and the stagnation pressure are chosen..The stagnation enthalpy is the true effort; its product with the mass flow rate gives

12.1

THE CONVECTION BOND AND COMPRESSIBLE

FLOW

873

the overall energy flux (neglecting any gravity effects). The pressure serves not only to properly qualify this effort, but also acts as the causal effort variable. A convection bond is distinguished from a simple bond by the addition of a dashed line to its effort side. Flow through a restriction or throttle drops the stagnation pressure and generates entropy but leaves the stagnation enthalpy unchanged. This irreversible process is represented by a convection-bond adaptation of the RS element. Guided

Problem

12.1

In this problem you will compare the flows of an inviscid incompressible fluid (as given by equation (12.12)) and an inviscid compressible ideal gas through the same simple, smoothly convergent nozzle. The latter is given in textbooks on compressible flow as

[A,nP1/k ---~

2 ]{k+l)/(k-1)

V [~--~J

P.2

’

if ~ S r,

where P1 and 01 refer to the upstream stagnation state, R is the gas constant and k is the ratio of spedfic heats, cp/cv, whi& is ~sumedconstant. P2 refers to the absolute static pressure at the throat, which equals the downstreamstagnation pressure since the walls expand abruptly (without "pressure recovery") downstream. The dimensionless mass flow rate, as determined by the above equation, is plotted as a function of P2/P~ in Fig. 12.1 for air (k = 1.4). For pressure ratios below the indicated critical ratio, the flow is independent of the downstre~ pressure, and is said to be choked. (a) Place equation (12.12) %r incompressible flow into the same %rm, plot the result for air directly onto the graph in Fig. 12.1%r comparison. Observe the nature of the error. (b) Expand the compressible-flow function for small (first-order) differences in the pressure drop, &P = ~ - P~, and show that this approximation agrees with equation (12.12). {c) Find the rate of entropy generation for the ideM g~. (d) (Optional.) Find the reduction in the available enthalpy across nozzle, ha~ - ha2. Assume the temperature of the environment is the same as 01, and the pressure of the environment is the same as P2.

874

CHAPTER 12. 1.0 rh A,~P~/ , ~-~

THERMODYNAMIC

SYSTEMS

for ideal gas (k= 1.4)

0.5

0

0.5

p2/p I

1.0

0

Figure 12.1: Inviscid compressible flow through a nozzle

Suggested

Steps:

1. Factor out from equation (12.12) the term A,~P1/x/-~, noting that P1 = pRO1. The result is a simple function of P’2/P1. Roughly plot this into Fig. 12.1 to compare this relation for incompresible flow to that for compressible flow. 2. Replace P2 in the equation given for compressible flow by P~ - Ap. Get each of the pressure-ratio terms under the square root into the form (1 aAP)~, which for small Ap can be approximated as 1 - xaAP. Simplify and compare, again using P = pRO if necessary. 3. To find the rate of entropy generation, start by finding 02, knowing that h = h(O) for an ideal gas, and hi = h2 for a steady-flow nozzle. Then employ equation (12.10), noting also the ideal gas law. 4. Use footnote 2 and the results above to find the reduction in the available enthalpy. PROBLEMS 12.1 Consider incompressible flow in a long uniform tube of area A in which the velocity profile can be approximated as parabolic. (a) Verify that the factor (12.7), equals

a in Table 12.1, as computed using equation

(b) The inertance of the fluid per unit length, relative to the volumeflow, can be written as 7p/A. The factor 7 is not in general equal to a, even though both factors are based on the same profile. Develop a formula for 7 in terms of the velocity profile v(r), where r is radius, and show that the parabolic case "r -- 4/3.

12.1

THE CONVECTION BOND AND COMPRESSIBLE

875

FLOW

12.2 Air flows at 100 psig through a channel with a 10-foot change in elevation at velocities (i) 10 ft/s and (ii) 1000 ft/s and at temperatures of (a) 70°F (b) 470°F. Compare the magnitudes of the four terms that contribute to the effort in equation (12.6). Commenton the significance of uncertainty regarding the factor a. (You will need some data on the thermodynamic state. It is important that you evaluate the efforts relative to some reasonable reference environmental state, which can be taken as 0 psig and 70°F.)

SOLUTION Guided

TO GUIDED

Problem

PROBLEM

12.1

1. incompressible approximation:

=

=

=

As shownin the plot below, this approximationis fairly accurate for (P21P~) 0.9, but increasingly overestimates the flow for smaller pressure ratios. ~h

1.0f incompressible / approximation [ideal gas (k=1.4)".~",~.

°; 2. For smallAP/P~,

°.5 2,p, ~ P~I

Therefore,

= 1=

~]

=1

k P~

1-~

=1

A~P~ _ __~ [(1 2

~]

-

k

Pa

k

~)]

This approximationis idengical ~o the incompressible result of step 1. a. Since h~ = h~ and h = h(O), 0~ = 01 (constan~ ~em~erature). ds_dhdP dP 0 p8 =O-R~;integrated, thisgives~2=s~+Rln Finally,

~g = ~(se - s~) = ~Rln

4. h~ = h~ - h* -~(s~ - s*) = -O~(s~ h~ = h2 - h * -8~(s~ - s*) = -~(s2 -

(P~)

876

CHAPTER 12.

THERMODYNAMIC SYSTEMS

04

note:

P, z h

of

P, h ~e

m

P P, h e hz ~ HS --~ ~

m

m

(a) special reversible element

m

(b) general irreversible element

RS o P~,hz

--~ m

h~ RS~P2 HS m

P2, h2 m

~

(c) addedirreversibilities fromfluid friction andheai conduction Figure 12.2: Heat interaction elements Therefore, h~ - h~ = 01(s~ - sl) =R01In ~-~

12.2 Heat Interaction

and Junctions

The RS element applied to the flow of a fluid has a pressure drop but no heat transfer. Anotherelement with two fluid ports is nowintroduced that has just the inverse, heat transfer (through a third port) but no pressure drop. It can used to represent an ideal heat exchanger. There are two versions: a special case that is reversible, and a general case that is not. In the secondcase the irreversibility is kept to a theoretical minimumconsistent with the boundary conditions. The 0 and 1-junctions are used with convection bonds in Mmostthe same way as they are used with simple bonds. The presence of two effort variables for a convection bond introduces complications, however. The merging of two fluid streams, for example,is in general an irreversible process, and therefore cannot be represented by a 0-junction alone. 12.2.1

The Reversible

Heat

Interaction

Element

This element, abbreviated as the H element, is shownin Fig. 12.2 part (a). There are convectionbonds for the inlet and outlet flows and a simple bondfor the heat conductedto or fromthe fluid. Reversibility requires, in addition to the absenceof friction, that heat is transferred across no morethan an infinitesimal temperature difference. Thus this element is very special: the temperatures for the three bonds must be virtually the same. There are two ways this can occur. In the first case, only an infinitesimal amountof heat is transferred. This

12.2.

HEAT INTERACTION

877

AND JUNCTIONS

describes a micro-element: an infinite number of such elements would be needed to represent a significant heat interaction. In the second case, the entering and leaving fluids are liquid-vapor mixtures, and since the pressure doesn’t change the two fluid temperatures would be the same even if a rather large quantity of heat is transferred. It is still necessary, however, for the temperature at the heat conduction port or bond to be no more than infinitesimally different from the fluid temperature. The element therefore implies infinite thermal conductivity for the heat conduction path. For the first case in which only an infinitesimal amountif heat is transferred, the entropy flux becomes infinitesimal, i.e. ~ -+ d~, ~2-~ ~ d~, h.~-hl ~ dh, etc. The conservation of energy requires rh dh = O d~.

(12.13)

Reversibility requires that the change of entropy flux of the fluid, m ds, equals the entropy flux of the heat transfer:

Combining these relations

/ads =d~.

(12.14)

dh=Ods.

(12.15)

gives

The thermodynamic identity dh = dP/p + ~ ds, when compared to equation (12.15),

(12.16)

gives the important result dP = 0,

(12.17)

proving that the input and output pressures are identical. 12.2.2

The

General

Heat

Interaction

Element

An ideal heat exchanger usually has a varying temperature along its length, and can be represented by this more general element, abbreviated as the HS element and shown in Fig. 12.2 part (b). This element is irreversible, since the two fluid boundary temperatures are unequal. The conduction-bond temperature, labeled 0out, is defined to equal whichever of ~ or t?2 represents the temperature of the downstream or outlet flow. (The flpw could move in either direction.) Additional irreversibilities can be added in the form of RS elements. In Fig. 12.2 part (c), the element RSI on one of the fluid ports represents fluid friction with its pressure drop, and the element RSo on the heat conduction port represents resistance to conduction with its temperature drop. The HS element can be understood better by decomposing it into an infinite cascade of infinitesimal H elements with coupled RS elements, as shown in Fig. 12.3. The first conclusion 4 is that the stagnation pressure Pin equals the

878

CHAPTER

12.

THERMODYNAMIC

m

Figure 12.3: elements

SYSTEMS

m

Decomposition

of an HS element

into

infinitesimal

H and RS

stagnation pressure Pout, since for each H element, dP = 0. The second conclusion is that the irreversibility and its associated entropy generation results from the assumption of a single bond with a single temperature for the heat interaction with the environment. Since this temperature is at one end of the interval between 0in and t?out, the heat fluxes through the various RS elements and through any added external RS element are all in the stone direction, in or out. Assuming the fluid is not in the two-phase region, only the specific heat of the fluid, cp, is needed to represent the H and HS elements computationally. 4The sophisticated reader familiar with the classical Rayleigh line solution mayinitially wince at this result. The Rayleigh line assumes zero wall shear, for which heat transfer indeed forces both the static and stagnation pressures to change along the flow path whenever the Mach number is .significantly different from zero. The same result would follow from the present analysis, moreover, if the static temperature were used in place of the stagnation temperature in the equations for energy balance and reversibility. Proper use of an entropy balance and proper use of momentumbalance are redundant (see Guided Problem 12.2). Thus, momentumbalances are not needed in energy-based modeling, although sometimes they are convenient, particularly for finding the forces of constraint. The question thus becomeswhether the static or the stagnation temperature should be used for the modelof a reversible heat exchanger. The author considers the stagnation temperature to be far more useful, since the walls of a real adiabatic channel are at that temperature. The Rayleigh model corresponds, as it were, to walls which moveat the same velocity as the bulk fluid. With stationary walls, irreversibility demandsthat wall shear result from heat transfer. In cooling, for example, the molecules that give up part of their kinetic energy to the walls also impart an impulse force to them. A real heat exchanger has additional wall shear associated with the friction irreversibility. This effect can be represented by placing an RS element in series with the H or HSelement, as shown above. Conventional analysis of a channel with heating or cooling and wall shear uses a friction factor for the latter, resulting in a wall shear that is essentially independent of the heat transfer but a pressure drop that is not independent. The model recommended here is rather the reverse: in the simplest case, the pressure drop is independent of the heat transfer but the wall shear is not. The pressure rise in cooling indicated by the Rayleigh line never even becomesa theoretical possibility, and the analyst is not mislead.

12.2.

879

HEAT INTERACTION AND JUNCTIONS

Since the total pressure is constant, dh = CpdO.

(12.18)

Thusif Cp is assumedto be constant over the temperature range 01 to 02, where the subscripts 1 and 2 refer to the inlet and outlet, respectively, h2 -- hi -=Cp(02--

(12.19)

The conservation of energy requires (12.20)

(h2 - hl)~t -- ~82.

In practice one usually knowshi, P and rh and either ~ or 382. (Other cases are discussed later.) If ~0e is known,equation (12.20) is the only computation necessary. If ~ is known,you can combineequations (12.19) and (12.20) solve for 02: ¯ (12.21) 8e -

1-

Therefore, he = hi + (~/~)81 1 - ~/Cpfn"

(12.22)

It is necessaryin this case to find 81, whichis a state function of hi and P. Therate of entropy generation, definedas ~g, ma~vbe of interest, particularly if you wishto understandthe irreversibility or loss of availability. Thebalance of entropy fluxes is (12.23) s2~ = s~ + ~ + ~. Solving the aboverelations for ~ gives ~=Cp~

~-l-ln =

12.2.3

~

E-1

.

(12.24)

The HRS Macro Element

The RS element in a modelof a real heat exchanger likely generates more entropy than the HS element with its ~g from equation (12.24). The combination of the HSelement with a conductionRS elementis so nearly universal that it is practical to represent it with a macroelement, called here the HRSelement. As shownin Fig. 12.4, the HRSelement allows the boundary temperature for the heat conduction, 8c, to be muchdifferent from either 81 or 8.~. Further, 8¢ can be taken as an assumedor independent variable, in place of ~ or the product 82~ as assumed above. The parameters now are Cp and the thermal conductance of the RS element, which is defined as H. The conservation of energygives the results h~. = h~ + 8c~c/¢n, (12.25)

880

CHAPTER 12.

m

m

THERMODYNAMIC SYSTEMS

m

m

Figure 12.4: The HRSmacro element 1 - O1/Oc ~c = 1/H + 1/cpfa’ ~g ~- Cp~Z ~2 --

[01

1 - In

(0~)] +H (0~" =-0"9)2

(12.26) (12.27)

The first term in the equation for the entropy generation, ~e, comesdirectly from equation (12.24) for the HS componentwithin the HRSelement, and the second term, for the RS component, comes from equation (9.34) (p. 719). before, it is necessary to use property data to find 0~ and 0~, from knowledgeof the enthalpies and the pressure, before equation (12.27) can be evaluated. Also as before, the total pressures at the inlet and outlet are equal, i.e. P2= P1. EXAMPLE12.1 A flow of 2801bm/secof exhaust gas from a gas turbine enters a water boiler at 800T. Saturated liquid water (already preheated) enters at a pressure 500 psia and temperature of 467T. Subsequent pressure drops can be neglected. If the effective integrated heat transfer coefficient is 300Btu/s.T(independent of the flow rate), at what rate should the water be pumpedin so it leaves as saturated vapor? The specific heat of the gas is 0.280Btu/lbmR, and the enthalpy of vaporization of the water is 756Btu/lbm.Organize your work with a bondgraph. Also, find the rate of entropy generation. Solution: Youcan start by drawing a bond graph comprising an HS element, an RS element, an H element ~d bonds, and labelling the input and output variables. Then, as shownon the right below, combine the HS and RS elements to form an HRSelement: Pg, hg~ Pg, gas: ~ ~ ~ "7-HS~ "-:" -- -r-rhg 0~l,~ mg

RS P~., hw2 water:-- Pw, ~ --hw~ .--- n ---:-- -7-/n w

m w

12.2.

881

HEAT INTERACTION AND JUNCTIONS

The conduction entropy flux at the water temperature can be found using equation (12.26): :~c= 1 -(l/H) 091/0u, = 1 - (800+ 460)/(467 + 460) + (1/0.260 x 280) (1/cpgmg) (1/300)

= 21.04 Btu/R s

The heat transfer nowcan be found, followed by the flow of the water and the temperatureof the effiux gas: Q = 0w~c= (467 + 460)21.04 : 19,510 Btu/s ~hu, =

;~ = 25.8

lbm/sec

O~e = 0~ - Q/cv~m~ = 800 - 19150/0.260(280) = 532°F Finally, you can find ~,: ~ (0~ - 0~) = 0.260(280)

L~460

k~ 460]J

~ (532 - 467) + 300 (532 + 460)(467 + 460) = 3.~0 Btu/R s 12.2.4

The 0-Junction

for

Convection

Bonds

This junction is defined such that all bonds joined to it represent a common thermodynamicstate; both the enthalpy and the pressure are common.As with simple-bonded0-junctions, energy is neither dissipated nor stored; the powers on the respective bonds, with signs indicated by the power-conventionhalfarrows, sumto zero. A mixture of simple and convection bonds is not allowed for any 0-junction. The junction represents the mergingand/or dividing of separate convection channels. The dividing of a flow from one into two or more channels is the simplest case, since all the flows necessarily have a common pressure and enthalpy. As sho~vnin Fig. 12.5, this is the samestructure as a junction for an incompressible flow represented by simple bondsand a 0-junction. (In both cases the dynamicpressures for the separate channels can be different, requiring special treatment; these differences are neglected here.) 12.2.5

Merging

Streams:

the

0S Junction

The merging of two or more streams is inherently more complicated. In general, the merging streams will have the same pressure (again neglecting any

882

CHAPTER

12.

THERMOD}WAMIC

SYSTEMS

Figure 12.5: Use of a 0-junction to represent bifurcating flows

Figure 12.6: Mergingflows with irreversibility

differences in the dynamic pressure or head), but they will not have the same enthalpy or temperature. Consequently, a zero-junction alone cannot represent this merging except for the very special case in which the’ streams happen to have a commonstate. The merging of two streams at different temperatures is irreversible. Mixing with heat transfer takes place, and entropy is generated. A bond graph can represent the merger of two streams by separating the heat transfer operation from the mixing operation. As shown in Fig. 12.6 part (a), an HS element is used to equalize the two temperatures which, since the two pressures are constrained to be equal, equalizes the two states. (Whentwo streams at different pressures are to merge, at least one of them must first pass through a resistance to equalize their pressures; for incompressible flow this resistance is a 1-junction with a shunt R element, and for compressible flow it is an RS element.) The 0-junction then can be used to represent the physical mixing. If you don’t know in which directions the three flows may be heading, three HS elements can insure proper accounting of the irreversiblility, as shown in part (b). The HS element(s) attached between outflow bonds simply have 5no effect, since the associated temperature differences are zero. The combination shown in-Fig. 12.6 part (b) occurs so frequently that the macro 0S junction is created to represent it. The 0S junction, shown in part (c), applies for any number of joined bonds and channels. This, the second the macro elements, has the added advantage of simplifying computation. The 5This is one reason for defining the effort on the thermal conductionbondof HSelements as the temperatureof the outlet fluid stream.

12.2.

HEAT INTERACTION

883

AND JUNCTIONS

P__.£~,m_hl~,~ l~P2’hz’ s ~ m m

P~,h, s~_v_~..1S~P~’h" s2 m m

(a) reversible junction

(b) irreversible junction

Figure 12.7: Reversible and irreversible

1-junctions

conservation of energy gives

hour-- ~ rhinhin

(12.28)

The entropy generation can be deduced from the HS elements of the more primitive graph. Equation (12.26) applies for each HS element therein, if the subscripts are adjusted accordingly.

12.2.6

The 1-Junction and 1S-Junction with Convection

1-junctions are defined to force a commonflow on all attached bonds. Since the convection bond has only a single flow variable, unlike its effort variables, the problem experienced with the 0-junction does not exist. Two types of 1junctions are useful, nevertheless. The basic 1-junction is reversible; the entropies of the convection bonds are equal. The 1S-junction is irreversible; the pressures on the convection bonds are defined as being equal. In both cases, shown in Fig. 12.7, simple bonds as well as convection bonds can be joined to the junction. This is the essential way in which you tie together parts of systems represented by simple bonds with other parts represented by convection bonds. The vertical bonds in the examples of Fig. 12.7 are simple bonds; they have no enthalpy or pressure or thermodynamic state associated with it, but rather scalar potentials equal in magnitude to the enthalpy difference ~ hi - h2. An ideal (lossless, reversible) motor or turbine operating.with a compressible flow can be represented by a combination of the 1-junction and a simple-bonded ideal machine, as shown in Fig: 12.8 part (a). Losses are added in the bond graphs of parts (b) and (c). In part (b), the losses are represented in a summary fashion; in part (c), losses are detailed according to their physical origin. The power convention arrows shown give positive moduli for the various elements when the machine is acting as a turbine. The arrows can be left as is when the machine acts as a compressor, although you probably would rather retain postive moduli by inverting the arrows on certain bonds. In both cases, the IDEAL MACHINE element could be replaced by a simple transformer if overall behavior 6Thescalar potential on the simplebondcan also be said to equal the difference of the available enthalpies, h~l - ha2. This followsfromthe inherent reversibility of the junction, whichrequires the unavailableenthalpies, hul and hu2, to be equal.

884

CHAPTER 12.

IDEAL MACHINE

IDEAL MACHINE

h2 ~P~’h~ ! ~--.~P2’ m

THERMODYNAMIC SYSTEMS

0

/\

m

(a) ideal

h,.._...~_,~. m

m

m

s~

(b) with knownadiabatic efficiency

1

~ RSf,.

(c) withirreversibilities IDEAL MACHINE PI,hl~.. O__~ 1 =--~--~RSI~-~’-OS-=--~--~HS

(d) unspecified

--

P2 h2 ~

P~, ~ PUMP or P~, h~ --~ TURBINE ~

Figure 12.8: Pumpor turbine for compressible flow

12.2.

HEAT INTERACTION

AND JUNCTIONS

885

resembles that of a positive-displacement machine; the effects of mechanical and fluid friction and fluid leakage are represented elsewhere. Were the model of part (b) 100%efficient, the transformer modulus T would be unity and the elements 1S and Ts would be removed. The state of the fluid leaving the compressor, particularly its enthalpy he --- h2, would be determined by the requirement that its entropy equal sl and its pressure equal P2. The product of the enthalpy drop (for a turbine) and the mass flow would equal the output mechanical power. If the IDEALMACHINE is represented by a transformer of modulus Tin, the torque on the shaft would be the product of the enthalpy drop and Tin, and the speed of the shaft would equal rh/T,~. Thus, the modulus Tm would be the ratio Dive, where D is the volumetric displacement per radian of shaft rotation and v~ is the specific volume of the fluid entering the machine. An actual machine will suffer mechanical losses due to mechanical friction, fluid losses due to fluid friction (i.e. pressure drops through orifices, etc.) and fluid losses due to internal leakage. Heat interactions with the surroundings also can produce irreversibilities, but the present discussion assumes that these are being treated separately, which is relatively easy to do. Instead, the overall machine is treated as being adiabatic. In the summary model of part (b), the considered overall losses are described conventionally by the adiabatic efficiency of th~ machine, written here as *~ad. This efficiency is the ratio of the output power to the input power; for the turbine, this means the ratio of the actual turbine work to the isentropic work, or hi - h2 ?~ad-- hl - hc

(12.29)

The irreversibility occurs exclusively within the IS element. Following the efforts from the 1-junction through the two transformers to the IS-junction, you can see that 1 T-~s(h~ - hc) = h2 - hc = (-h~ - h2) + (h~ -

(12.30)

Substitution of equation (12.29) gives the result (12.31)

~ = 1 + r~aa.

An additional condition must be specified in order to determine the individual values of T and Ts. This condition depends on the nature of the machine. A postive-displacement-like motor can be described partly by its volumetric ettlciency, r~v, that is defined as the ratio of the actual shaft speed to the speed it would have if there were no fluid leakage or slippage. In this case the effort and flow on the bond between the 0-junction and the transformer Tm equal (h~ - h,.)/~v and rh, rh , respectively. Thus~(T - 1ITs) = rl~, and T=fl-2-~ ; ~,~d

T~ -

1 r~v(1/r~d - 1)

(12.32)

886

CHAPTER 12.

THERMODYNAMIC

SYSTEMS

The model shownin part (c) of Fig. 12.8 details the causes of lossesin a turbine, allowing the adiabatic efficiency to vary depending on the state. Leakage passes through the element RSi. Were the machine a pump or compressor, the 0S element would be placed on the inlet side rather than the outlet; you might also prefer to reverse some of the power convention half-arrows. For a positivedisplacement machine, the simple-bonded ideal machine would be a transformer with constant modulus. The element RS~n represents mechanical friction, and the element RSI represents frictional losses in the fluid. Note that the heat generated by friction is carried off in the fluid. For most dynamic machines the 7siinple-bonded ideal machine does not resemble a fixed-modulus transformer, but nevertheless can be defined by.a transformer hedvily modulated by two independent variables: either Mor ¢ and either hi - h2 or rh. In some cases it may be better to employ overall eInpiricM relations regarding the behavior of a fluid machine, and leave the bond graph model in the general form of part (d) of the figure. Most of these matters are beyond the scope this text. Nevertheless, the model of part (b) is used in the example below, and the model of part (c) is given within a case study in Section 12.3. Guided Problem 12.2 below serves in part as preparation for the case study. EXAMPLE 12.2 A piston compressor has a volumetric displacement D per radian, and is knownto give an adiabatic efficiency of 80% and a volumetric efficiency of 95%. Drawa bond graph of the system, give the moduli of its transformers, and find an expression for the output enthalpy in terms of variables that could be found from knowledge of the state of the inlet fluid and the outlet pressure. Solution: The bond graph of Fig. 12.8 applies, except that the power flows in the opposite direction through four of the bonds, because that graph deals with a turbine instead of a compressor. The power direction arrows on these bonds have been reversed in order to keep their moduli positive:

G(h~-h~)l~/G 0

/\ r rs 7The simple-bonded ideal dynamicmachine maybe represented better by a modulated gyrator. The element RS~is not needed. Details for ]’2ulerian turbomachineswith incompressible flow have been given by H.M.Paynter, "The Dynamicsand Contro! of Eu|erian Turbomachines,"d. DynamicSystems, Measurementand Control, ASMETrans., v. 94 n. 3 pp. 198-205,1972.

12.2.

HEAT INTERACTION AND JUNCTIONS

887

The flow on the bond below the ideal positive displacementtransformer T,~ wouldbe #t if the volumetric efficiency were 100%.The flow actually is larger, specifically rh/~?v, since the shaft has to rotate faster to supply the leakage as well as the through-flow.Since the conservationof energyrequires the total mechanicalpowerto equal the total changein fluid energy increase per unit time, the effort on this bond is ~,(h2 - hi). The modulusof is the ideal ratio of the mass flow pumpedto the shaft speed; this equals the volumetric displacementper radian, as in an incompressible-flow pump, times the density of the flow that enters the compressor,whichis l/v1. Thedefinition of the adiabatic efficiency (the inverse of equation(12.29)),

(the inverse of that given in equation (12.29)) gives the workingformula h2 - hC~ad (~ld--1) hl’ where hi is given and hc is found from its knowentropy Sc = sl and known pressure Pc = P2. Further, the effort side of the transformers T and Ts give T- ~,,(h2 - h~) _ ~ he - hi ~ad’

Ts- ~(h~ - h~) _(h2 - ~(h~ - hl) _ h~ - hc h~) - (he -h~) --1----2~d" The relations for the flow sides of the transformers confirm these results. As a partial check, note that whenboth efficiencies equal unity, T -~ 1 and Ts ~ ec. This wouldeliminate the path for Ts, which it should.

12.2.7

Summary

Heat conduction has been represented previously by the simple-bonded RS element. A meansalso is needed to transfer heat betweena flowing fluid and a solid. This transfer is represented by the H, HS and HRSelements. All three have two convection bonds, representing the inlet and outlet flows, and one conduction bond. The H element is reversible; all three bonds have the same temperature, whichimplies either that the heat is transferred at an infinitesimal rate or that the fluid enters and leaves in the two-phaseregion. The irreversible HSelement allows the fluid to enter and exit at different temperatures, but like the H element assigns the temperatureof the fluid efflux port to the conduction port also. Additional thermal resistance to heat conduction can be represented by bonding a conduction RS element to the conduction port. The HRSmacro element represents this combination. Causal strokes can be added to the bonds, consistent with the constraints, and the causal output variables computedas functions of the causal input vari-

888

CHAPTER 12.

z~choked nozzle, P,,O, --~ th I

~.~ ~.~area A ’~J mixing chamber

]

THERMODYNAMIC SYSTEMS

~ , motor, displacement D

m~ 2 +m

Figure 12.9: The configuration of GuidedProblem12.3 ables and the parameters of the element. The rate of entropy generation, which never is negative, also can be computedas a measureof the irreversibility. 1-junctions, in which all bonds have a commonflow, can be used to join systems described using simple bonds with systems described using convection bonds. This allows the 1-junction to represent the reversible heart of a fluid machine. The net power flow on a single simple mechanical bond of the 1junction equals the net power flow on its convection bonds. A transformer on this simple bond represents the ratio of the mass flow and the mechanical generalized velocity. Friction and leakage losses can be addedwith proper use of R, RS, HS and additional junction elements. Irreversibilities also can be represented by use of the 1S-junction, which has a pair of convection bonds with equal pressures rather than equal entropies. 0-junctions require identical efforts on all their bonds, and therefore cannot be used with mixed simple and convection bonds, since convection bonds have two effort variables rather than one. A convection0-junction can directly represent a. flow which is being divided betweentwo or more output channels. The mergingof two streams, however,is an irreversible process. It can be ~nodeled by adding HS bonds to a convection 0-junction, or by employing the macro equivalent 0S junction. Guided

Problem

12.2

A knownmass flow rhl of air at a knowntemperature 01 is augmented in a mixing chamberby a mass flow of air rh~, as shownin Fig. 12.9. This second mass flow first passes through a simple nozzle of throat area A with known upstream temperature 02 and a knownpressure P2, which is high enoughrelative to P1 to produce choked flow. The flow leaving the mixing chamber passes through a positive displacement motor with knownvolumetric displacement per radian, D, relative to its inlet.’ Thetorque on the shaft, M, is known.Thexnotor can be assumedto behave isentropically, and has a knownexhaust pressure, Pe (which should be atmospheric pressure). The air maybe approximated as ideal gas with invariant specific heats. Youare asked to find relations for the pressure P1, massflow ~h2, shaft speed ~ and outlet temperature, 0e.

12.2.

HEAT INTERACTION

Suggested

AND JUNCTIONS

889

Steps:

1. Represent the system model with a bond graph and label all the pertinent variables on the bonds~ recognizing when different bonds have the same generalized efforts or flows. Place causal strokes on all bonds on the graph, consistent with the given information. This graph should guide you through the following steps. 2. Find the transformer modulus for the motor in terms of known quantities except for an unknown temperature and pressure. 3. Use the result of step 2 to relate the ratio of the inlet and outlet temperatures of the motor to the known shaft moment, M. 4. The ratio of the inlet to outlet temperatures of the motor also is related to the knownratio of inlet to outlet pressures, since the machine is assumed to be isentropic. Determine this relationship. 5. Combinethe results of steps 3 and 4 to find the inlet pressure, P16. Find the mass flow rate through the nozzle, vh2, using the choked flow relationship given in Guided Problem 12.1. 7. Find the temperature of the flow leaving the mixing chamber. 8. Find the outlet temperature, Oe. 9. Find the shaft speed, ~. PROBLEMS 12.3 A complicated distributed-parameter model is required to represent the dynamics of a counterflow heat exchanger. For most modeling purposes a much simpler steady-state model suffices, however. You are asked to find the basic equations for such a model. The mass flows in the two directions are the same; so are the specific heats. There is a thermal conductivity G per unit length between the two streams. Longitudinal heat conduction may be neglected, as may pressure drops.

890

CHAPTER 12.

THERMODYNAMIC SYSTEMS

(a) Drawa bond graph representing an element of the heat exchanger dx long. The complete heat exchangerrepresents an infinite cascade of such micro elements. Define variables on your graph. (b) Write equations whichgive the output enthalpies in terms of the input enthalpies, the massflow and fixed parameters. This is your basic model. (c) Write an expression ibr the rate of entropy generation. (This need be in a simple form, but could contain integrals.) (d) Introduce frictional pressure drops to your model. Does this affect your answers to both parts (b) and (c)? 12.4 An open feedwater heater for a steam power plant mixes a small amount of steam tapped from an interstage of the turbine with cold water pressurized by a pump, to give hot water. The pressures in the mixing process may be approximatedas constant, but as a practical matter the flow of steam must be regulated by a valve, introducing someadditional irreversibility to what you likely considered in an introductory thermodynamics course. In .practice, also, the state of the output hot water will not be saturated liquid, unlike what you mayhave assumedpreviously, but will lie in the compressedliquid region. It is desired to develop a computational model that can readily adapt to small changesin the operating states. (a) Drawa bond graph that models the valve and the mixer, using RS and 0S elements. Reticulate the 0S element into a combination of HS and 0 elements. (b) Write expressions for the changes in the enthalpies of the water and the vapor in terms of the temperature downstreamof the valve Or, the temperatures of the cold and hot water 0c and ~h, the specific heats of the vapor and the liquid (cp~, and %t) and the enthalpy of vaporization, hyg. SOLUTION

TO

Guided Problem 12.2

GUIDED

PROBLEM

CASE STUDY WITH QUASI-STEADY

R03 RO3 1 (b P1D (o 3 _~) p~Q3 = p3D = P~D - P~D

2. T- ~n, ~ -~ 3. M

891

FLOW

R 1 =(T"h3 - he) = Cp(0~ T - ~4) _ (1 - l/k) no~

(~ - ~/~ 4. Isentropic flow: ~ 5. Equate ratios

0~/0~ : ~ =1 ( Pe ) (~-~/~)

Therefore, M = DP~[1- (p~/p~)(~-~/k)]

~ - ~/k

6.

~=~V

1

DP~-1/kM

which can be solved for P~.

kk+l]

7. ~omequation (12.28), 0a -

~0~ + ~0~

(Note that the air leaving the nozzle h~ the same sgagnation enthalpy as the air entering, and thus also h~ ~he same temperature, 0~.) 8. Substitute tge P~ found ~oma numerical evaluation in step 5 into the equation from step 4 go find 0~. ROa, . + ~) (All desired results nowcan be evaluated.)

12.3

Case

Study

with

Quasi-Steady

Flow

The double-rotor two-lobed compressor pictured in Fig. 12.10, often called a "Roots blower," is a very simple positive-displacement machine that is used to increase gas pressures up to a factor of about two) It typically runs dry with no oil mist. Metal-to-metal friction is avoided without excessive leakage by using a small clearance between the two rotors, and between the rotors and the housing. External gears provide the necessary synchronization. In position (A), the chamberon the right side of rotor A is still in communicationwith the inlet port. By position (B) the chamber has been isolated from both the inlet and the outlet, but remains at constant volume and thus at constant pressure. In position (C) a surge of high-pressure fluid flows back into the chamber, charging it to the outlet pressure. In position (D) the fluid that was in the chamber being pushed out, or discharged through the outlet.

12.3.1

Bond Graph Model

The bond graph of Fig. 12.11 can be used to model the time-average behavior of the compressor. It is adapted from Fig. 12.8(c), which is primarily for a mosstaged Rootscompressorssometimesproduceair pressures in excess of 100 psi.

892

CHAPTER 12.

THERMODYNAMIC SYSTEMS

Figure 12.10: Operatingcycle of a two-impeller straight-lobe rotary compressor

’

p2,

h2@

S~ m

convection bond number 1 pressure P~ enthalpy hI temperature O~ mass flowrate rh

2

3

4 5 P4 P2 P~ P2 h2 h3 h,~ h4 O~ 04 02 04 rh rh+rhi ~+~hi th+rhi

6 7 8 P2 P~ P~ h4 h4 h4 04 O~ 04 rh rhi th~

Figure 12.11: Bondgraph for the rotary compressor

CASE STUDY WITH QUASI-STEADY

893

FLOW

tor rather than a pump, by reversing the power-convention arrows through the transformer to avoid confusion, and by interchanging the 0S and 0-junctions since the leakage flow through the RSs element is from the right (outlet) the left (inlet). The inlet bond is labeled with the subscript 1 and the outlet bond with the subscript 2. Other bonds are labeled with circled numbers which also are used as subscripts. The table below the graph explicitly identifies constraints between bond variables which are implicit in the definitions of the various elements; the number of variables is reduced from the outset by using these identifications. The information which follows will be interpreted to quantify the elements T, RSIm, RSi and RSI, and to predict the mass flow, outlet temperature and shaft torque as a function of the shaft speed, inlet conditions and outlet pressure. The machine displaces D = 25.0/2~r fta/rad (its overall size is little over a cubic yard), draws in atmospheric air (P1 = 14.7 psia, 01 = 70°F, R 0.3704 psia.fta/lbm.R, k = 1.4) which can be considered an ideal gas, and normally operates in the range 200-600 rpm. The torque loss due to gear and bearing friction is Mo= 173 ft.lb plus the fraction ] = 0.05 of the balance of the moment. The fluid leakage around the rotor from the outlet to the inlet is controlled mostly by viscosity, since the leakage passages are relatively slender and long. As a result, the leakage is almost proportional to the pressure drop; it is determined to be 51 scfm for each psi. ("scfm" means standard cubic feet per minute, i.e. volume flow at atmospheric pressure and temperature.) This is all the quantitative information that is needed. The transformer modulus equals the volumetric displacement per radian times a density. Since the given displacement D corresponds to the density in the inlet chamber, the density P3 should be used, giving T = paD = P1D -~3’

(12.29)

from which rh + rhi

M = M0 + (1 + f)T(h4

=T~= P~D~ R03 ’

(12.30)

ha) = Mo+ ( 1 f)P iD(h4

=

..

ha) RO.~ RID(04/03

-

:77g

1)

(12.31)

The fluid 1-junction implies the isentropic relation 04 --

03

h4 _ (k-I)lk ¢P4~

ha

(12.32)

894

CHAPTER 12.

12.3.2

THERMODYNAMIC SYSTEMS

Irreversibilities

The mechanicalfriction energy is converted to heat in the element RSfm rate

at

the

1 + f ’ (12.33) causing the air to increase its enthalpy and temperature in the HS element according to: 0,2 h~ (Mo + fM)~ (1 - 1/k)(Mo + fM)~ - = 1 + - 1 + 04 h4 (1 + f)h4rh (l + f)RO4#t

(12.34)

The leakage flow through the element RS~ is given in the form 51 ft3/sec; a = ~-~

rhi = p~a(P2- P~);

P~ = ~-~.

(12.35)

The mixing of this flow with the air from the intake increases the temperature and enthalpy according to equation (12.28): 03 01

h3 _ dn + (h4/h~)dni rh + r~i hi

__

_-

~h + (04/0~)~i rh +

(12.36)

As a pumping chamber begins to communicate with the discharge port, which is at a lower pressure, flow surges from the chamberto the port. This presents an inherent irreversibility representable as a frictional or throttling loss. The steady-flow equivalency for this irreversibility occurs in the bondgraph element RSI. It can be approximated by the classic problem in which a constant-volume chamber,with initial conditions indicated by the subscript 0, is charged through a restriction by a mass mc with conditions indicated by the subscript c, until equilibrium is reached at mass mo + mc and conditions indicated by the subscript f. The conservation of energy for the chamberas a control volume requires AE =hcmc, or (~0 + m~)uf -- moUo= m~hc.

(12.37)

Since du = c~dOand dh = c~d0, this gives mock,(01- 0o) m~(cpO~ - cv0f).

(12.38)

The enthalpies h~ and hI are equal, so that 0~ = 0I. Solving for the ratio of the temperatures and noting the definition of k, this gives 0I _

1

0-~- 1 - l} - 1).~/.~o The ideal gas relation requires mc +mo _ Pf _ P~Oo mo

Po PoO]

(12.39)

(12.40)

CASE STUDY }VITH QUASI-STEADY FLOW

895

The final state, indicated by the subscript f, is the state on bonds5, 6 and 7, so that 0~, = 04 and Pf = P2. The initial state, indicated by subscript i, is the state on bond 3, so that 00 = 03 and P0 = P1. Combiningequations (12.39) and (12.40) nowgives 04 = 1 + (k - 1)P2/P1 (12.41) ’ where P] and Pi are the corresponding pressures. 12.3.3

Computation

of Results

Equations (12.30), (12.31), (12.32), (12.34), (12.35), (12.36) and prise seven equations in the seven unknowns ~h, ’#ti, 0_%03, 04, P4 and M. The variables P1, P2, 01 and ~ and the parameters D, a, M0, f, R and k are assumedto be given. Equation(12.35) gives ~hi directly. Substitution of equation (12.41) into (12.31) gives the torque on the shaft: M= Mo+ (1 + f)D(P2 - P1).

(12.42)

The mass flow rh can be found by solving equations (12.30) and (12.41) for and 04, respectively, and substituting the results into equation (12.36) to give the quadratic #t 2 [a(P.-~-P1)

¯ a(p2D---_C-p1)k[l+(k-1)-~11-k] =0. (12.43)

1 ~n

](~-~i)-~

Equation(12.34) nowcan be solved for 02, with the help of equation (12.41): 02= . PID~ [1 +(kkR(rh+ ~hi)

1-:-~P2] (fM + M0)(1 1/ k)~b + )/-’1J (1 + f)Rrh

(12.44)

The efficiency, ~, of the machineis of special interest. Efficiency is defined as the ratio of the powerthat wouldbe required to achieve the same massflow rate and output pressure if the machinewere reversible, to the actual power required. In the reversible machinethe balance of mechanicaland fluid powers requires Rm01 ~02 reversible power= (h2 - hl)rh ]- --i-~/k\~ - " 1)

(12.45)

The isentropic relation of equation (12.32) applies, with 02 = 04 and 01 = 03. Thus, R’/YtO

1

(l-l/k> ] [(P2~

~ = (1 - 1]k)M$ [\PlJ - 1 . (12.46) The massflow, shaft torque, inlet temperature and efficiency for the given parametersare plotted in Fig. 12.12 for two shaft speeds. Youmust pay careful attention to the units for the various variables in carrying out this type of computation. The results closely resemble data from an actual Roots compressor.

896

CHAPTER 12.

1.4

I

I

I

THERMODYNAMIC SYSTEMS

I

I

T~ +200, °F, 200 rpm h+10, Ibm/s, 400 rpm 1.2

1.0 +200, °F, 400 rpm

0.8

400 rpm r/, 200rprr

0.6 rh+ 10, lbm/s, 0.4

0.2

0

0

fl-lb

2

4

6

8

10 Pz-P~,psi

12

Figure 12.12: Modelresults for the rotary compressor

12.4.

THERMODYNAMIC COMPLIANCE

897.

The entropy generation in the elements RSIm, HS, RSf, RS{ and 0S can be computed and compared. This information could aid a designer interested in potential improvements. The irreversibility in RSf, for example, is inherent in the design and could not be removed, for example, whereas the others could be mitigated. The tools now at your disposal should allow you to represent and reduce most lumped engineering models for the steady-state behavior of thermodynamic enginering systems. Consideration of models of unsteady or dynamic systems is next. This will have a side benefit of allowing equilibria to be calculated without solving numerous simultaneous algebraic equations, as to some extent had to be done above.

12.4

Thermodynamic

Compliance

The coherent kinetic energy of a substance is represented by inertance elements. The incoherent kinetic energy of the molecules plus the energy due to intermolecular forces is the thermodynamic energy. From the classical macroscopic viewpoint, taken here, this energy can be represented as a function of generalized displacements. Thus it becomes a potential energy described in bond graphs by compliance elements. In particular, the thermodynamic energy of a pure substance in a control volume, neglecting the effects of capillarity and electric and magnetic fields, can be changed by heat transfer, mass transport and the work of volume change.

12.4.1 Application of a Simple ThermalCompliance The simplest model for thermodynamic compliance refers to a fixed is no convection. Also, work interactions with the environment are which implies either that the volume of the body is constant or its pressure is negligible. This leaves only a heat transfer interaction. erate thermodynamic compliance was called a thermal compliance 9.4.6 (pp. 724-726), and it was represented by a standard one-port

mass; there negl’ected, surrounding This degenin Section compliance:

The thermal regenerator pictured in Fig. 12.13 illustrates a use of the simple thermal compliance in the presence of fluid flow. This device comprises a porous mass m~ of metal or ceramic which is heated or cooled by fluid passing through the pores. In a typical application to a thermal engine, the material is heated by hot exhaust gases during one portion of the cycle, and in turn heats cool inlet gases during another portion of the cycle. The mass of the gas within the pores is probably small enough to allow neglect of its thermal capacity. A bond graph model for the regenerator is given in part (b) of the figure. This model assumes that the entire mass of metal or ceramic is at a uniform

898

CHAPTER

12

THERMODYNAMIC

SYSTEMS

C IP, h

,,_,

2P, h

(a) schematic

(b) bond graph

Figure 12.13: Thermal regenerator

temperature, 8. Considering the HRS element and the causal stroke on the compliance, the differential equation become~(from equation (12.26), p. 880)

dt

1/H + 1/cp~n’

(12.47)

in which 01 is the known temperature of the inlet gas, 81 =01(h~),

(12.48)

H is the heat transfer coefficient, Cp is the specific heat of the gas, and rh is the mass flow. From equation (9.37) (p. 724), 8c (s-S t?c °) = /m’c, 8oe

(12.49)

in which c is the specific heat of the mass mr. The enthalpy of the outlet gas is, from equation (12.25) (p. 880), h2 --- hi - -=- -- h~ rn

gn/H + 1/Cp"

(12.50)

The flow through the regenerator is driven by a small pressure drop. This likely could be represented by a single RS element placed on either side of the HRS element; one on each side would improve the accuracy of the model slightly by making it symmetric. These elements do not change the enthalpies. For the whole regenerator, then, the independent variables are P1, P3 and h~. The system is first order. 12.4.2

Causality

The introduction of compliance leads to differential equation models; as always, causal strokes are very helpful in constructing such models. Convection bonds have two effort variables; the choice used here, as noted in Section 12.1.2 (pp 870-871), is P, h (the stagnation pressure and enthalpy) and rh. Recall that the enthalpy h is causal physically in the actual direction of the flow. (There is a special exception with theoretical rather than practical

12.4.

899

THERMODYNAMIC COMPLIANCE

significance. 9) Since at any point in a simulation this direction is kn~,wn, no further specification of its causality is needed. This leaves the other effort variable, P, which together with the flow, rh, form a bilateral causal pair just like the effort and flow of a simple bond. The direction of the causality is indicated on a bond graph by the usual causal stroke. Thus, for example, h,P indicates that rh is specified from the left and P is specified from the right. The elements introduced earlier in this chapter impose various constraints on causality. The convection RS element imposes the same mass flow on both bonds, therefore allowing three possible causal combinations:

The thermal conduction bond of the H and ttS elements must have ~ as its causal input, since the temperature is constrained by the specification of the enthalpy on the convection bond with the incoming flow. This leaves only two possible causalities for these elements:

..~--._~/-/S

-~.=_~

On the other hand, the HRS macro element allows the conduction temperature to be set independently; either Oc or ~c could be the causal input:

~

HRS

~

~

HRS

~

Note that the causality of the convection bonds on these graphs can be reversed to give two more possibilities. The pressure of the 0 and 0S junctions can be specified by only one bond, just like the simple 0-junction.

12.4.3

General Thermodynamic Compliance

The more general thermodynamic energy of a system or control volume in which the volume and/or the mass can change is now addressed. The state of the substance will be assumed to be homogeneous throughout the control volume. If its volume, V, changes by an amount dV, the system does work PdV on its surroundings. If mass dm enters, the system energy increases by h din, where 9For a e-element with a convection bond, the enthalpy h is causally determinedby the elementitself regardlessof the direction of rh. In practice, however,the C-elementis replaced by the C~S-element, describedin Section 12.4.4 below,whichdoes not suffer this anomaly.

900.

CHAPTER

12

THERMODYNAMIC

SYSTEMS

h is the stagnation enthalpy. If heat is conducted across the control surface into the control volume, the energy increases by 0 dsc where dsc is the inward entropy displacement due to the heat. Denoting the total energy by )~, the first law of thermodynamics requires dF hrh dt

+ 0~c

P~.

(12.51)

This form suggests a three-port compliance element with one convection bond for the mass transport and two simple bonds for the heat and work, respectively:

I

h,P rh

-P V

If the practice with simple compliances is followed, the generalized displacements are used as the state variables. The time integral of the mass flow ~ is the total mass, m; the integral of I? is the volume, V. The integral of the conduction )c is only part of the entropy of the system, however, since additional entropy is t.ransported along with the mass flow rh. Thus the implied third state variable is the total entropy of the system, S: S =/(~c The total compliance energy therefore

+ s~h)dt. becomes a function

V = ]?(S, V, m),

(12.52) of S, V and m: (12.53)

from which d~ c9~ dS c9~ dV c9~ drn + + dt OS dt OV dt Orn dt (12.54) A comparison of equations (12.51) and (12.54) reveals 1° 0=

O(~)

(12.55a)

~°Thedefinition of Gibb’sfunction g_~ h-Os allows equation(12.55c)to be recast OV

12.4.

901

THERMODYNAMIC COMPLIANCE

-p= ),

0(~1~

(12.55b) ~’,rn

In practice it is simpler to use the specific quantities u ~

F m

1 V ~ p m

s

S ~, m

(12.56)

so that u = u(s, p),

(12.57)

and equations (12.55) are replaced O=O(~s) p

P h = u + -P

(12.58)

Although this computational scheme is ideal theoretically , its direct implementation assumes the availability of analytic state functions of the form of equation (12.57). They are not generally available, however. As a result, it necessary to guess the entropy and iterate until the error is sufficiently small. This iterative method is not presented here in favor of an alternative procedure based on a different set of state variables. This new procedure is aided by the definition of a particular macro element, described next. 12.4.4

The

CS Macro

Element

The general thermodynamic C element virtually always occurs in combination with other primitive elements. As a practical matter, a particular combination of elements called the CS (macro) element can be employed as the basic building block for dynamic models. This approach is used to eliminate the iteration associated with the primitive C element, by using temperature, 0, as a state variable in place of S. The other two state variables, m and V, are unchanged. Consider a fluid-filled chamber with an input port and an output port, such as the cylinder of an engine shown in Fig. 12.14. A bond graph model is shownin part (b) of the figure. The input flow ~h~, which enters at the lower left, is likely at a different temperature from the fluid in the cylinder, so an HS element is needed to equalize the temperature. The resulting entropy flux ~mis combined with whatever external entropy flux is associated with heat conduction (~q at the upper left) to give ~0. The output flow rh2 is at the lower right, and the mechanical bond for volume change is at the upper right.

902

CHAPTER 12 THERMODYNAMIC SYSTEMS

Oi, P th~ (a) schematic

m (b) bondgraph RS

h~,P -~ h~,P ~----~---~- CS ~ m~_plkf, m~

(c) macroelement Figure 12.14: Fluid chamberwith interactions

12.4.

903

THERMODYNAMIC COMPLIANCE

This combination of elements occurs repeatedly in the modeling of thermofluid systems. Usually, a thermal conductance RS element is added to the upper 0-junction to recognize conduction resistance and allow temperature causality for the heat conduction. Provision often is wanted to allow the material flows to reverse direction, also. The result is an expanded combination, shownin part (c) of the figure. This is the combination designated as the CS macro element. The causal strokes on both the detailed and the macro representations imply that 1;~ and ~ = rhl - rh~ are integrated to give the state variables V and m, respectively. The causal strokes on the primitive C element do not dictate S as the third state variable, although they suggest its use. In the associated differential equation, which is consistent with equation (12.52), ~ = ~ + s~, ~

+

~

+ (~n - ~o~t)s.

(12.59)

The subscripts "in" and "out" refer to either bonds 1 or 2, depending on the direction of the flow. The causal thermal input at the top could be either ~, ~ = 0~, or ~, so no causal stroke is shown. The un-elaborated CS element, on the other hand, does not by itself suggest that S should be a state variable. It allows the temperature, 0, to substitute, which is precisely what is needed to allow the practical computation of thermodynamicstate properties from available analytical state functions (d~duced from empirical data). The other two state variables (m, and when the volume varies, V) remain. There are two commonlyavailable forms for empirically-based state formulas of a single-phase pure substance. The most widely available are "P-v-T" formulas of the form IF= P(v,

O),J

that are used in conjunction with specific heat relations 1Iv = 0) using the form

(12.60) at zero density (p

These %rmul~ are closely ~sociated with the fundamental experiments. Some commonsubstances, including water, ammoniumand certain refrigerants, are available in the alternative %rmof the Helmholtz kee energy, ¯ ~ U - ~S, as a function of density and temperature:

Note that both equations (12.60) with (12.61) and (12.62) employ the same independent variables (v and 0), and therefore the two forms can be treated similarly. Expressions for the properties of pressure, entropy, internal energy, and enthalpy can be deduced from either of these forms, as will be shown in Section 12.5. Thus, for example, equations of the form u = u(v, ~)

(12.63)

904

CHAPTER

12

THERMODYNAMIC

SYSTEMS

can be deri~ed, where v = 1/pm. Since v = V/m, this equation requires

~m

fi=~vv

+NO.

The conservation of total energy, U, can be expressed as ~7 = rn~ + u~ = Q - PI;" + ~ h.i~i.

(12.65)

i

The substitution of equation (12.64) into equation (12.65) gives, after manipulation including use of the definition h ~ u + Pv, ~ the major result

dt

.~Ou/OO~ + i~(hi - h)~ + P + ~

.

(12.66)

Equation (12.66) applies also to an incompressible substance (approximation of a liquid or solid) if the term Ou/O0(in the denominator) is set equal to the specific heat, c, and the term Ou/Ovis set equal to zero. To treat the saturated mixture region, formulas of the form [ P =- P~t = Psat(O)

]

(12.67)

have been deduced for many substances from the basic relations of equations (12.60) and (12.61) or (12.62). Using the thermodynmnic identities h = ho - (sg - s)O, sg-s=(v~-v)

(@),

h = u + Pv, (the second being the Clapeyron relation), urated vapor, gives

(12.67a) (12.67b) 12.67c)

where the subscript g refers to sat-

- dP~t \(v - v). u = u~ + Psat - 0--~-) o

(12.68)

To evaluate the term Vg, state formulas of the form [Vg =

Vg(0)

(12.69)

also have been deduced for many substances, based also on the fundamental relations. The term u0 can be deduced in the form ~o = ua(vg, O)

(12.70)

11The development from equations (12.60) through (12.72), and Figs. 12.22 and 12.23 the application to a refrigeration system, are taken from the author’s paper "Non-Iterative Evaluation of Multiphase Thermal Compliances in Bond Graphs," which was submitted for publication to the Proceedings of the Institution of Mechanical Engineers PartI for publication in late 2001, with permission from Professional Engineering Publishing, Ltd.

12.4.

905

THERMODYNAMICCOMPLIANCE

by combiningequation (12.69) with equation (12.63). The time derivative of equation (12.68) nowcan be written

-

- ~ Psat ~(v~ - v)&

(12.71)

Substituting this result into equation (12.65) and solving for ~ gives the desired differential equation for the time derivative of the state variable, 0:

denom=m \Ov~ + P~at-#--~-)

-~- +-~---"--~--tv9

¯

(12.72)

The temperature (8) and pressure (P) are nominally assumed to be form over the volume, consistent with lumpedmodeling; for mixedsaturation states, u, h and s need not be uniform, but for examplecould be affected by gravity separation. Moregeneral approximationsare possible, but can lead to computationalinstabilities, and are beyondthe scope of the presentation here. 12.4.5

Computations

for

the Ideal

Gas

A simple very special case is the ideal gas with assumedconstant specific heats, for which s-s0=c, ln ~-0 -Rln . (12.73) The subscript 0 refers to any convenient reference state. Thus, s is knownin terms of v and 8, and no iteration is needed to carry out the schemein which S is a state variable. In place of equation (12.58) (p. 901) there results O=O°exp [ s-s°+ c~R ln(vo/v)

] ,

P = RS/v,

(12.74a) (12.745)

h = c,(0 - 00) Pv.

(12.74c)

These equations can be used to address the following case study. 12.4.6

Case

Study:

A Piston-Cylinder

Compressor

A one-cylinder air compressor, shownin Fig. 12.15, has an assumedsinusoidal piston motion. For simplicity of presentation, the inlet and outlet valves are assumedto open to their full extent wheneverthe pressure on their left side exceeds that on their right side, and to be closed otherwise, thus passing flow

906

CHAPTER

Figure 12.15: Piston-cylinder

12

THERMODYNAMIC

SYSTEMS

compressor and model

only in the rightward direction, like ideal check valves. Whenopen, the pressure drops across them are assumed to be represented by the equations given in Guided Problem 12.1 (p. 873). In the bond-graph model shown in part (b) the figure, these valves are represented by the two convection RS elements. The magnitude of the friction force between the piston and the cylinder is modeled as a constant, giving an equivalent pressure, PF, that heats the piston and cylinder walls, which are assumed to be uniform in temperature. Heat flows between the gas in the cylinder and the metal of the piston and cylinder, with an assumed constant heat conductance associated with a thin thermal boundary layer. A separate larger constant heat conductance is applied between the walls and the surroundings. The inlet and outlet pressures are assumed to be given and constant. The internal energy of the fluid in the cylinder and the thermal conductance of the boundary layer is represented by the CS element. The thermal compliance of the metal is represented by a C element. The frictional conversion of mechanical power to thermal energy is represented by a simple-bonded RS element; the fact that thermal fluxes to the C element can come from both the fluid and the friction, at the same temperature, is represented by a 0-junction. The synchronization of the motions of the valves and the piston is indicated by dashed lines. The causal strokes to the CS element produce the first-order

state differential

12.4.

907

THERMODYNAMIC COMPLIANCE

equations dV d’-~- = V1sin(~)t); dm -- = rhl - ~2. dt

V(0) = V0,

(12.75a) (12.75b)

In this example, the third state variable for the CS element is chosen to be S, rather than 8 (either would be a good choice), so the third differential equation becomes, from equation (12.59) (p. 903),

The final state differential equation is associated with the energy storage in the walls of the cylinder, namely the C element:

W= V~-I -He I-V~ + 8---[Before these can be evaluated, the variables rhl, ~h2, 81, 8, 8c, ~1 and s need to be related to the state variables. The mass fluxes are associated with the RS elements. The flow ~h~ is set at zero when P > P~, and ~h2 is set at zero when P < P2. Otherwise, assuming unchoked flow (this condition can be checked after simulation and corrected if necessary), from Guided Problem 12.1:

(12.76b) P: = pROa.

(12.76c)

Temperatures are unchanged across convection RS elements, since enthalpy is a function of temperature. The other two temperatures are given by the standard relations for constant specific heats: 8 = 8o exp

Is- so1+[~ln(p/po)

(12.77a)

80 exp

(12.77b)

’

cv

8c=

[~]

,

The entropies so and Sco are defined at the arbitrary reference temperature and may be set equal to zero. The density Po also is defined at that temperature and the pressure P1; the density p is m

P = V’

(1~.78,~)

908

CHAPTER 12 THERMODYNAMIC SYSTEMS volumeof cylinder x

104, m3

air temperature.-’ 100, K

4 2

l\..v "z,

\ - , "/, \ .:.,-

/,

0 pressure, bars

-2

i~

-4 -6 -8

0 net massflow x 100, kg/s (positive:inlet; negative:outlet)

-10 -12

0

O.b2

O.b4 ’

O.b6 ’ O.b8 ’ 0.10 time, seconds Figure 12.16: Simulationresults for piston-cylinder compressor

Similarly, S s = --. (12.78b) m Results of a particular simulation are plotted in Fig. 12.16. Sucha simulation can give good estimates of the amountof flow compressedand the energy efficiency of the process. It can showhowthe valve areas affect this performance, and what timing of the valves ought to be used. It can predict howhot the parts will becomedue to friction, and howboth friction and heat transfer betweenthe gas and the metal and the surroundings affect the performance. Morerefined questions can be addressed by elaborating the modelto include profiles of the valve openingas a function of the position of the piston, substituting a morerealistic slider-crank geometrical constraint, improvingthe friction model, adding an output tank, applying a torque rather than a velocity and adding a modulated inertia (including, presumably,the effect of a flywheel), etc. Noneof these .modifications are beyondthe scope of what you have done before, although as a systems engineer you might wish to consult with specialists to establish some of the parametersof the model,such as friction and heat transfer coefficients. The inlet and output pressures are 1.0 bar and 4.0 bar, respectively. The maximum and minimumvolumes of the cylinder are 0.1 and 0.6 liters, respectively. The shaft turns at 185rad/s, the inlet air is at 293K, the effective areas of both valves are 4.0 cm2, the heat transfer coefficient H and He are 0.012 and 0.120 Joules/m2/K, respectively, the massof the walls is 1.0 kg and their specific heat is 0.45 Joules/kg K. Theinitial temperatureof the walls is 744 K, close to equilibrium under the given circumstances of an assumedfriction force of 10 N. (Moreexternal cooling would be desirable.) The initial condition 3, the cylinder is minimumvolumewith the air at 440 K and density 3.15 kg/m

12.4.

THERMODYNAMIC COMPLIANCE

909

hout-hin~ hin, Pin, Sin

hout Pideal Sin

m

t hout, Pout

m

Figure 12.17: Simple model for the joining of the convection and mechanical bonds of a quasi-steady compressor

which is close to the steady-state values for that part of the cycle. The plotted results actually modeled the air as a mixure of nitrogen, oxygen and argon with very accurately modeled properties, as described in Section 12.5.3 below. This fancier simulation is detailed in Guided Problem 12.4 at the end of the next section. (It adapts equation (12.66) with temperature as state variable rather than equation (12.75c) which uses entropy. Using entropy with the more accurate model of the air would have required repeated iteration in the evaluation of state properties.) The assumption of an ideal gas with constant specific heats, as in the equations above, is considerably simpler to implement, and does not produce results different enough to justify a separate plot. 12.4.7

Treatment

of

a Quasi-Steady-State

Fluid

Machine

The piston-cylinder system above treats a fluid machine as a device that goes through a cycle for each revolution of its shaft. Often the cycle time for one revolution is a minute fraction of the time required for a major change to occur in the thermodynamic properties of a system outside of the fluid machine. An example, to be detailed in Section 12.5.6 is a refrigeration system. The shaft makes thousands of revolutions while the refrigerant masses and temperatures gradually change during start-up. If these changes are the center of interest, rather than the rapid changes over each rotation of the shaft, it makes sense to employ a quasi-steady model for the compressor. Quasi-steady models for turbines and compressors were introduced in Section 12.2.6 (pp. 883-887), and the example of a Roots blower was developed in Section 12.3 (pp. 891-897). Consider the simple model of a compressor shown in Fig. 12.17, in which the irreversibilites are restricted to those of an equivalent virtual series flow restriction that is represented by the element RS. Internal and external leakage is neglected, and any mechanical friction losses are assumed to occur to the right of the ideal compressor element T,~, and not to affect the flow. The transformer modulus is the volumetric displacement of the machine per radian of. rotation times the specific volume of the upstream fluid. The mass flow is computed as the product of the shaft speed and this modulus, and the torque is computed as the the product of the difference hour - hi,~ and this modulus. The enthalpy hi,~

910

CHAPTER 12 THERMODYNAMIC SYSTEMS

is causal from the upstream part of the system, but howis hour determined? The 1-junction is a reversible as well as conservative element. Therefore, the entropies s of the upstream and downstreamflows are the same. Further, the pressure Pideal can be found from knowledgeof rh and Pout, the latter being a result of the state of the compliance element. Therefore, the state of the fluid for the convectionbondto the right of the 1-junction is to be found from knowledgeof its entropy and pressure. This is awkward,since the available state functions assume knowledgeof the temperature and the specific volume (or density) instead. Example12.2 in Section 12.2.6 (pp. 886-887) gives an alternative model, which the volumetric and adiabatic efficiencies of the compressorare assumed to be known.In this case, it was found that (12.79) hour=hc/~ad-- (1/~c,d -- 1)hin, in which hin is known,and he corresponds to a fluid state having the known entropysin and knownpressure Po~,t. Poutis different from(less than) the Pide~,t of Fig. 12.17, but the task of finding hc is essentially the sameas the task of finding hour; only the simple step of equation (12.79) is addedto find hour. This situation generally requires iteration: one starts with a guess of the proper temperature and specific volume(probably using the most recent known values), and computes the associated entropy and pressure. A comparison of these values with the actual values producesa seconditeration. The iterative procedure may be expedited by makinguse of the gradients OP/Ov, OP/O0,Os/OPand Os/O0, all of which can be computed as functions of v and 0. A Newton-Raphsoniteration becomes

i+l= 0 i-[OslOv OslO0

si

’

in which the thermodynamicidentity 8s/Ov =- OP/O0can be used to reduce computation.Since the partial derivatives are usually nearly constant over the narrowrange of consideration, only one such iteration is usually necessary to reduce the errors to within acceptable limits. Equation (12.80) does not work under all circumstances, however. The entropy of a saturated vapor depends largely on temperature, and the second or later iterations maywork better if the terms Os/Ov=- OP/O0in equation (12.80) are set equal to zero. The process, including the iteration, is illustrated in Example12.5 (on p. 926) and in the case study given in Section 12.5.6 (on p. 935). EXAMPLE 12.3 The compressorof Example12.2 (pp. 886-887) operates on an assumedideal gas with inlet temperature0~,~ and a pressure Pi,~ and outlet pressure Pout. Find the outlet enthalpy, ho,t, as is necessary in simulating a systemthat includes such a compressor. Solution: The plan is to find the temperature 0c, use this to find hc, and then employequation (12.79) to find hout. The state c has the sameentropy

12.4.

911

THERMODYNAMICCOMPLIANCE

as the inlet, and an isentropic process satisfies cp/cv, and Pv = RO. Therefore,

Pvk

=constant, where k =

Fromequations (12.74b) and (12.74c) (p. h~ = cv(Oc- 00) + Pcvc = (c, + R)Oc- C~0o= c~0~- c~0o, where 0o is the reference temperature for which s = 0. Finally, equation (12.79) gives

~a--~ [\’~-/~ ]

- 1 + 1 - ~-0o.

EXAMPLE 12.4 The compressorof Fig. 12.17 operates on ideal gas, and the RS element is based on a virtual orifice of effective area A. The flow throughthis orifice can be assumednot to be choked. Developa procedurefor finding ho~,t given Po,~t, ~, D and the inlet conditions. (This procedure is morecomplexthan that found in the previous example;you are not asked to carry it out, which in practice wouldbe done numerically.) Solution: The mass flow is given by rh = (D/V/n)~, and the pressure PideaL must be found from the orifice relation

L\ Pidea;

]

\ Pideal

The isentropic process across the 1-junction requires (as in equation (12.32), p. 894, for the case study of a rotary compressor),

Substituting ~his into ~he previous eqna~iongive~ a re~nl~ in ~e fo~m ~ = ~(~de~, ~n, Po~, constants). Since everythingin this equation except 8i~ea~ is known,it can in principle be solved for 0id~, after which the answeris hour : hideat

~ CpOideal -- CvO0.

In practice, 0id~ must be found iteratively, a Newton-Raphsonprocedure.

presumablythrough the use of

912

CHAPTER 12 THERMODYNAMIC SYSTEMS

Figure 12.18: Approximatelumpedmodel of a channel with unsteady flow 12.4.8

Fluid

Inertance

With

Compressible

Flow

The incoherent kinetic energy of fluid molecules has been included in the fluid compliance.The coherentkinetic energy of a fluid is represented in an inertance. The use of stagnation properties for the pressure, enthalpy and temperature suppresses most problems associated with fluid inertia in steady flow. In unsteady flow, the kinetic energy of a compressible flow is muchless apt to be significant than ihe kinetic energy of an incompressibleflow, since its density typically is less and its thermodynamicor potential (compliance) energy dominates. This is fortunate, since if the fluid in a channel has both significant kinetic and potential energies, wavepropagation results. Lumpedmodelingis then apt to becomeeither inaccurate or quite complex, as has been shownin Chapter 11. A lumped model comprising alternate C and I elements with 0 and 1junctions nevertheless can be constructed, as shownin Fig. 12.17. The greater the numberof C and I elements, the more accurate the model. Note that the bonds between the I elements and the 1-junctions are simple bonds; the only efforts thereon are the changesin Pip due to fluid acceleration. Note also that the nominal 3-port C element is treated here as a degenerate 1-port element, consistent with constant volumeand reversibility. This allows the methodsof Chapter 11 to apply, approximately. The ultimate computational method is finite-element analysis. This method, once fully developed, will allow analysis in two and three dimensions, including phase changes, etc. 12.4.9

Pseudo Bond Graphs for Thermofluid Systems

Compressible

The convection bond graph is a new development. At the time of this writing, the major bond graph representation for compressible thermofluid systems in the literature was an extension of the pseudo bond graph described in Section 9.4 (p. 727). The concept was proposed by Karnopp.12 A pseudo bond graph for a variable volumewith heat and massfluxes followedby a fluid restriction is shownin part (b) of Fig. 12.19; a convection bondgraph for the samemodel shownin part (c). The term pseudorecognizes that the products of the indicated 12Karnopp, Dean C., "State Variables and Pseudo Bond Graphs for Compressible Thermofluid Systems," Trans. ASME, J. Dynamic Systems, Measurement and Control, 101(3), Sept 1979.

12.4.

913

THERMODYNAMIC COMPLIANCE F (a) schematic

m~ a E~

~ ~/ ~, k

B rn2 ~

1.---~R

~ m 2 E3

P’ hl~ CS P’ h’.~RS P~’ h~ mj m2 k m2 S

S (b) pseudo bond graph

(c) convection bond graph

Figure 12.19: Comparison of pseudo and convection bond graphs

efforts and flows are not the actual energy fluxes, unlike true bond graphs. The energy flux into the compliance is /~, which includes the mechanical power -/~m = -P1? that is communicated through the modulated flow source Sy and the diagonal bond, not the horizontal bond with flow ~. Karnopp writes, "This signal interaction is the price we pay for connecting a pseudo bond graph with a true bond graph." The purpose of the graph is to aid in the writing of state differential equations, not particularly to explicate the energy fluxes. The causal strokes indicate that the state variables for the compliance are the mass m of the fluid therein, the total energy E of this mass, and the volume V. Dividing m and E by V gives the density, p and the specific energy, u, which are sufficient to define the thermodynamic state. The implication of the graph is that the pressure, P, and the temperature,/9, are deduced from the knowledge of p, u. The available state formulas, as noted in Section 12.4.4 above (p. 903), unfortunately, use 0 in place of u as an independent variable. Thus, the same kind of iteration needed for dealing with p, s as independent variables must be employed. It is largely for this reason that pseudo bond graphs are not emphasized in this book. Their relative complexity and awkwardness in joining with other parts of a model represented by true bond graphs also contributes to this decision. Pseudo bond graphs for non-compressible media do however have an advantage over true bond graphs when dealing with the special case of constant specific heats and constant thermal conductivities, as pointed out in Section 9.4.

914 12.4.10

CHAPTER

12

THERMODYNAMIC

SYSTEMS

Summary

The thermodvnamicenergy in a relatively uniform region of space is represented by the compliance element C. For the special case of a virtually incompressible solid or liquid body of fixed identity, its entropy suffices as the only independent variable needed. Thus the compliance has only one port, entropy is the generalized displacement, entropy flux is the generalized velocity, and temperature is the generalized effort. The generalized energy of heat flux, represented by a simple bond, is the product of the temperature and the entropy flux. This topic was addressed in Section 9.4. A body of a pure substance with fixed identity but changing density also has its volume as its second generalized displacement, and rate of change of volume as its corresponding generalized velocity. The associated effort is minus its pressure. Thus a second simple bond is added to the compliance. Whenthe mass of the pure substance is no longer fixed, that. is when there is a control volume with mass entering or leaving, a third generalized velocity is added: the rate of change of mass. This requires a convection bond with enthalpy as its power-factor effort. Irreversibilites due to heat transfer and fluid mixing can be represented in a bond graph by adding HS, 0 and conduction RS elements to the C element. This combination is so commonly useful it is represented by the CS macro element. Irreversibilites due to throttling can be appended by adding convection RS elements. Causal strokes can be applied to the bonds; on the effort side of a convection bond, the causality refers to the pressure. Causally, enthalpy and entropy are causal in the direction of the flow, so no special causal mark for them is needed on the bond graph. The thermodynamic energy of a pure substance can be represented, on a continuum basis per unit mass, as a function of any two independent variables. The most natural pair is specific volume (or its reciprocal, mass density) and entropy. All other continuum intensive variables, such as temperature, pressure and enthalpy, can be deduced therefrom. This scheme works well for ideal gases, but for more accurate models of most substances it requires an awkwardcomputational iteration. It is better to substitute temperature for the entropy as the state variable (making entropy a derived variable). The associated differential equations for a CS element have been found, but their evaluation depends on the details of the representation of the properties of the substance. This is the subject of the next section. The joining of a fluid flow to a mechanical machine such as a compressor or turbine involves a 1-junction across which the entropy is common.Knowledgeof entropy rather than enthalpy requires an iterative procedure for which a strategy is given. In some cases a 1S junction is helpful to represent the irreversible aspects of the coupling. The pair of convection bonds for this junction have the same pressure. The inertia of a compressible flow in a channel can be approximated by one or more combinations of a 1-junction and a simple-bonded inertance, as with incompressible flow. Unsteadiness plus compressibility can result in significant

12.4.

THERMODYNAMIC COMPLIANCE

915

Outlet Inlet Valve

Figure 12.20: Piston-cylinder compressors of Guided Problem 12.4 Reprinted courtesy o] Parker-Hannifin Corporation

gradients in the mass flow along the channel, however, possibly requiring an excessive number of cascaded pairs of inertance and compliance elements (interconnected with junctions) for adequate accuracy. In these cases, which involve wave motion, distributed parameter and finite-element models such as those presented in Chapter 11 may be more appropriate. Guided

Problem

12.3

Piston-cylinder compressors (see Fig. 12.20) often have two stages with an interstage cooler to improve the efficiency of the process. You are asked to develop a model including a set of state differential equations for this system. Makeassumptions similar to those for the single-stage compressor modeled above, and add an externally insulated load tank. P~eplace the air with helium for simplicity. Apply a momentto a crank shaft rather than a particular motion. You are not asked to work out the details of the geometry, however; simply note general transformer relations and a general modulated inertia. Suggested

Steps:

1. Draw a bond graph for your model. It is suggested that you ignore the volumeof fluid within the intercooler. Apply causality.

916

CHAPTER

12

THERMODYNAMIC

SYSTEMS

2. Label the graph with the appropriate variables, noting that enthalpies remain constant across convection RS elements and pressures remain constant across CS and HRS elements. Write three differential equations for each CS element (two for fixed volume), one differential equation of each one-port C element and two differential equations for the variable inertia element, in terms of convenient variables.

4. Relate the "convenient variables" of step 3 to the state variables. This requires appropriate constitutive relations for the non-storage elements, consistent with the indicated causality, plus state relations for an ideal gas. PROBLEMS 12.5 Show that, when equation (12.66) (p. 904) is applied to an ideal (Pv R8 and constant specific heats) for an isentropic expansion, the classical formula Pvk .= constant results. 12.6 A fixed-volume chamber containing an ideal gas is connected to a cylinder with piston of area Ac to form a catapult that accelerates a mass me. Draw a bond graph, define state variables, and write state differential equations. Neglect all effects of friction and heat transfer. 12.7 Repeat the above problem, placing a flow restriction between the chamber and the cylinder.

of effective

area Ar

12.8 Repeat Problem 12.6 when the chamber is charged with a saturated steamwater mixture. Assume the needed properties of the fluid are known; do not evaluate the associated partial derivatives in equation (12.72) (p. 905). 12.9 Repeat Problem 12.7 when the chamber is charged with a saturated steamwater mixture. Gravity separation allows only vapor to enter the orifice. Assume the needed properties of the fluid are known; do not evaluate the associated partial derivatives in equations (12.66) and (12.72). (pp. 904, 12.10 Repeat Example 12.3 (p. 910), considering rather than a compressor.

the machine to be a turbine

12.11 Solve Example12.4 (p. 911) in the case where the virtual orifice is known to be choked (implying rather large losses). (See Guided Problem 12.1, p. 873, for the orifice flow equation.)

12.4.

THERMODYNAMIC COMPLIANCE

917

SOLUTION TO GUIDED PROBLEM Guided

Problem

12.4

1-2. P~h4 P~hz P~ h~ P~ h3 P~h~ Po ho P~ ho P~ h~ SOURCEm, ~ RS~ CS~HR~ RS~ CS~ RS~ CS~ LOAD

MOTOR CS1 thermodynamic eompli~ee, mixing and heat conductance, cylinder ghermodyn~ie compliance, mixing and cylinder b ghermodynamiccompliance and mixing, g~ in sgorage gank flow resisgance, inlet valve flow resistance, interstage valve RSa flow resistance, outleg valve interstage heat exchanger wigh conducgance "Se environmental thermal reservoir thermal compli~c~, walls of cylinder a thermal compli~ce, wMlsof cylinder b thermal compliance, walls of tank inertance of moving p~ts (modulated) frictional "dissipation," cylinder frictional "dissipation," cylinder b slider-crank constraint, piston slider-crank constraint, piston b 3. g~, cylinder a:

dV~ = T~(¢)~ dt dmm dt dt

cylinder b:

dt dmb dt dS~ dt

g~ in

918

CHAPTER

THERMODYNAMIC

SYSTEMS

d~Ttc

gas, tank:

walls, cylinder a: walls, cylinder b: walls, tank: inertance: therefore: also:

12

-- = rh3 - ~h4 dt dS_~ = -~cc + cp(1 - 03/04)~h3 + ~s~ dt dS~ = ~ + Rio [T.~[ dt dSc~ = ~ + Pf~[T~[ dt O~b ¯ dSc~ = ~cc dt

d~ d~ _ 1 [M-

[

~dlq~

2-

Pfa~T~ + Pfb]n~]

dt

~ = $

There ~e therefore 13 state v~iables (~, ~, ma, rob, mc, S., Sb, S~, Sca, &~, &~, ~, ¢). The p~ameters T.(¢), Tb(¢), Cp, PI. and P~ (p res sures required to balance friction), lq(¢) and M~e presumed known. This leaves 17 other v~iables for which algebraic relations ~e needed: ~, ~, S~,~, ~, s~, s3, s4, P~, P2, ~, ~2, ~3, ~4, Oca and Oc~. 4. 10 state relations (R, po, ~o, m~., m~b, c., c, SO, S~o, S~o and. S~¢o are presumedknown;the l~t four are probably set at zero): P~ = p~R~ = m~RO~/V~ ~2 = p3ROa = m~ROa/B~

Sl = &/m.; s~ = &/m~;s, = S~/m~

=Ooex ] L mcac o] ~

mcbC

The outlet valve of cylinder a and the inlet valve of cylinder b ~e redundant in this model, since the energy of the small m~sof g~ in the intercooler is ~ neglected. Thus there ~e three valve relations (k, At(C), A~(¢), A3(¢) 0o are ~sumed known):

12.5.

EVALUATION

919

OF THERMODYNAMIC PROPERTIES

]ntercooler relation (E, c~ ~d 8~ ~e ~su~edk~ow~;gSe~bse~ceof compliancecan be criticized): ~ = ~ + .

~ =

.,~%’

Three conduction relations

(:In) + (:/~)

(H~, Hb ~nd Hc ~e ~sumed known):

This completes17 equations. Newv~iables P3 and8cc have beenintroduced, whichhavethe following state relations (mcc~ndSccO~e ~sumed to be known; the l~tter probablyis set at zero): Pa = pRe~ = m¢RedV¢ L

12.5

Evaluation

of

mccC J

Thermodynamic

Properties

The thermodynamic properties of a wide variety of substances are available in analytical forms that represent fits to experimental data. The most widely available form is that of the "P - v - T" relation and the specific heat relation, as given in equations (12.60) and (12.61), repeated below as equations(12.81) (12.82). The next most widely available form is that of the Helmholtz relation, equation (12.62) (equation (12.94) below). This section presents practical by which either of these forms can be used to calculate other thermodynamic properties such as internal energy, entropy, enthalpy, etc. The state is assumed implicitly to be in thermodynamic equilbrium. Illustrations and details are given for oxygen, nitrogen, air and a commonrefrigerant. A case study gives the complete simulation routine for the start-up of a refrigeration system.

12.5.1

The Most Commonly Available for State Properties

Thermodynamicstate functions were originally largely as "P-v-T" relations of the form

Analytical

developed and are still

Form available

lP = or P = = 1/p,J plus specific heat relations at p = 0 of the form [c~° = I c~°(0).

(12.82)

The first consideration below is the calculation of specific entropy, s, from these functions. It uses the density p rather than its reciprocal, the specific volume, v (but could be transformed to use v).

920

CHAPTER 12.

THERMODYNAMIC SYSTEMS

gas /saturated mixture\ liquid

p,_eo _ A P Figure 12.21: Path of integration in the evaluation of entropy or internal energy The fundamental "T ds" relation

=

+

(12,s3)

with substitutions for the definition of c,, and the identity Ou

OP

(12.84)

gives ds = dO- ~5.

P"

(12.85)

The entropy is defined to be zero at somearbitrary state Po, 80, whichoften is taken at the triple point with 100%liquid (e.g. 0°C for liquid water). Equation (12.85) must be integrated along any path from this state to the desired state p, 8 in order to get the desired s. Since c, is knownonly for zero density, integration is chosen along the path shownin Fig. 12.19 for which temperature change occurs only at zero density. Addingand subtracting a term -R In p from equation (12.84), this integral becomes s -- fo: ~dO-R fPdP+ fOl [pR-(~)o]odP.]oo-~ ]o

Note that the second integrand also is integrated from p to 0 on the lower leg and from 0 to p on the upper leg, but these two (infinite) terms precisely cancel each other and thus are not represented. The remaining term can be integrated outright. The right-most integral minus the term R lnpo is a constant, and is represented here as -s0. Thus, 0

fp

(12.87)

12.5.

EVALUATION

OF THERMODYNAMIC PROPERTIES

921

This is the desired result, the implementation of which is discussed below. The integrand of the second integral vanishes for the special case of an ideal gas, leaving a formula which may be familiar. The corresponding equation for the specific internal energy, u, can be found by starting with the expression u = u(O, l/p), taking the derivative, substituting equation (12.84), and integrating as was done for s, to give

The P-v-T relations apply only in the single phase region. This includes, on its boundary, both the saturated vapor and saturated liquid states. These states can be identified from a relation in the form of equation (12.81a) because, at any particular temperature, they are the only two states having the same pressure. It is nevertheless awkwardto have to cNculate the saturation states this way every time they are needed. Fortunately, equations of the forms

ps=;s(0), are widely available. The pressure, P, is given directly as a function of the density or specific volume. The specific enthalpy, h, can be computed from its definition h = u + Pv=_ u + P/p.

(12.90)

The plan for evaluating s in the two-phase or saturation mixture region starts with finding sg, namely the value of s when p = pg at the desired temperature, by substituting the density p~ from equation (12.89b) as the limit of integration in equation (12.87). The balance of the answer, s - s~ (which is negative) theri added. This final term is based on the Clapeyron relation s - s~

OP

(12.91)

which is derived in most thermodynamics textbooks. [Psat

= Psat(O)

]

Equations of the form (12.92)

have been deduced from the P-v-T relations and are widely available to expedite the evaluation of the right side of equation (12,91). Changes in enthalpy in the two-phase region associated with changes in density at constant temperature simply equal the changes in entropy times the temperature. Thus, in the two-phase region, h = hg + (s - s~)O.

(12.93)

922

CHAPTER 12.

THERMODYNAMIC SYSTEMS

W.C.Reynolds13 has collected and interpreted data on manysubstances in the form of equations (12.81), (12.82), (12.89a) and (12.92). Stewart has done the same for manyregrigerants and several other substances. These equations are generally very complex, involving manycoefficients with many significant digits. Relations of the form of equation (12.89b) are conspicuously missing from Reynolds,but are included in Stewart, whoalso carries out the integrations to give the entropy (equation (12.87)) and enthalpy (equation (12.90) with equations (12.8t) and (12.88) substituted) for specific substances.

12.5.2 Helmholtz Analytical Formfor State Properties The properties of a few substances, including water, ammonium and certain commonrefrigerants, are available in the form of the Helmholtz free energy @-- U - 0S as a function of the density and the temperature:

: = ¢(p, 0). ]

(12.94)

The arguments of this formulation are the same p, 0 use in the ’P-v-T’ and c°~ formulation. The Helmholtz form is more convenient, however, since the entropy, internal energy and pressure are given by simple derivatives (without the confusion of constants of integration):

o~ r= 1/0

(12.9a)

Thesereplace equations (12.87), (12.88) and (12.81), respectively. Application to water is given in Section 12.g.6 below. Application to R12, R22~5 and R123is given by various a’uthors.

12.5.3 Application to Gases The specific heats c°v of noble gasses such as argon, helium and neon are constants. For nitrogen and oxygen, Reynoldsgives

o

7

cv = Z GiOi-4 ÷ a~e~°(\e~/O: )"

"

(12.96)

13William C. ]~eynolds, ThermodYnamic Properties in SI, Department of Mechanical Engineering, Stanford University, Stanford CA, 1979. ~4R.B. Stewart et al., ASHRAE Thermodyanrnic Properties of Refrigerants, American Society of Heating, Refrigeration and Air-Conditioning Engineers, 1988. 15Fluid Phase Equilibria, v 80 (1992), Elsevier Science Publishers B.V., Amsterdam.

12.5.

923

EVALUATION OF THERMODYNAMICPROPERTIES

in which the coefficients are constants included in the MATLAB file Gasda~a.m that appears in Appendix D. For the pressure-volume-temperature relation, Reynolds gives P =pRO+ p’~

AIO +

A~O~/~ + ~ i=3

AiO 3-i

+ p3 ~ Ai07-i ~

i=6

12

~-~ + pSA~3 + p6(A~4/0 + A~5/~2) + p7At6/O + p4 ~ A~O i=10

pS(A~/O+ A~s/O2) + p9A~9/O~ + [p~(A~o/O2 + A2~/03) p~(A~2/O~ + A~/O4) + p7(A~/O~ + A~/O3) + pg.(A26/~~ + + p~(A28/O ~ + A~/e ~) + p~(A~o/e~ + A~/e ~ -~ + p~ A~2/Oa)]e

A27[04)

(12.97) This relation also applies to hydrogen, methane, oxygen, nitrogen and air; the coefficients for oxygen,nitrogen and air are given in gasdaga.m. For temperatures well above the saturation state, equation (12.97) can approximatedas an ideal gas, whichmeansneglecting all terms beyondthe first on its right side. Thesecondintegral in equation (12.87) also vanishes, leaving for the entropy

~/0 +Gs e~/0_l

~/0o e~/°o-1

+ ~ ~ In ~Z~) 0 00

+ so. (12.98)

The internM energy becomes, from equation (12.88),

+ ~,(0-0o)+ ~(0~ -o~)+~(0~-0~) G~

[:

4

-

1]

e~/oo

(12.99)

+ uo.

The inclusion of more than the first term in equation (12.97) augmentsthe relations for entropy and internal energy as follows: ~ s=s~-p.

.4~ A~+2~

0A4

(

;~ (

-~p~( A~ 2A5) 0a

As 0~

~l~h.+;~ (~ ~l~h~ al~

~ ~kA~°-~/

~ko

+~]+~o~

2~)

924

CHAPTER 12.

U =Uh + p *

+

+ A3 + "~- +

p7 (2~

~.

THERMODYNAMIC SYSTEMS

+ "-~ AT + --~- + -~-/

+3Als~+pS3A~°+~

o ~ ~) [Ae2+5X~3~ [3A2s

4A29~

~

[3A2o+4A~

~

t~

[3A2~

4As5~

(3Aso

4A3~

~) [3A~ 5A~2~

1 (1-e-~°~) Wi+~ ~+~ =~-0

e-~0

5A~v~ (12.101a) (12.101b)

l
(12.101c)

The term sm is the entropy of mixture, related to ~he reference state. Equations (12.98)-(12.99) for an ideal gas at low (literally ~ero) pressure are evalutated in the MATLAB function file ga~.~ given in AppendixD. The derivatives Os/OTand O~/OTalso are evaluated, to aid in the evaluation of the terms in equation (12.80) and similar endeavors. G~eous Nr usuNly c~ be treated ~ a m~ture of nitrogen, oxygen and argon. ~he entropy and internal energy can be eMculated by summingthe contributions of each component,using equations (12.98) and (12.99) as com~uted using gaa .m. Since equation (12.97) applies to the g~ mixture of air, the corrections of equations (12.100)-(12.101) can be applied directly to the mixture rather than to the individual components.Proper constants sm and ~ for the mixture should be added, re~lacing the individual s0 and u0. A function m-file a~r.m, given in AppendixD, uses the equations above and their derivatives to compute P, u, h = u + P/p, s, Ou/OT, Ou/Ov, Os/OT, Os/Ov ~ OP/OTand OP/Ov,~ functions of 0 and p. Note that the first two partial derivative terms are needed to evaluate the differentiM equations for temperature, as given by equation (12.66), and the additional derivatives can be very useful in evaluating a state whena ~air of variables other than 0 and 0 (or v = l/p) are known, such ~ P and ~, for they allow the terms of a Newton-Raphson iteration to be

12.5.

EVALUATION

925

OF THERMODYNAMIC PROPERTIES

evaluated analytically, giving rapid convergence. This procedure is discussed in Section 12.4.7 (p. 909) above; the derivatives are used in equation (12.80). that air.m can be used for pure nitrogen or oxygen, also, by setting the mass weighting coefficients Fo, Fn and the gas constant t~ appropriately. A separate script file spech.m computes the specific heat of the air mixture, given the temperature, needed for estimating the flow through a restriction. EXAMPLE 12.5 A piston-cylinder compressor draws in atmospheric air (temperature 293 K, density 1.205 kg/m3) and charges an insulated tank from its initial atmospheric state. The tank has volume 0.1 m3. The compressor is driven at 1750 rpm, has a volumetric displacement of 1 × 10-5 m3/rev, an adiabatic efficiency of 80%, and a volumetric efficiency of 95%. Simulate and plot the resulting pressure and temperature in the tank for 60 seconds. Solution: The system can be represented by the bond graph below: Sy

M _

Two state variables are needed, one for the mass of air in the tank and the other for its temperature. Calling the shaft speed ~, the volumetric displacement per radian D and the specific volume of the atmospheric air ~in, the first differential equation becomes dm dt The second differential

0.95Vin

equation is adapted from equation (12.66) (p. 904):

d~ 1[ dt - mOu/O0

( d~v~]dm hour-h+ P + dv ]]

~

The enthalpy of the air leaving the compressor, hour, is found from equation (12.67). The other terms in the differential equation are fluid properties in the tank that depend on the state variables m and ~. They can.be determined by running the routine air.m with the appropriate coefficients given by gasda~a, m. A function file that computes the differential equations and the master program follow: ~ction ~=compdi~e(t,x) % diff. equations ~ charging from atmospheric air g~oba~ V DPH0 D v~n sin bin vc TC

fo~ compressor/~k

926

CHAPTER

12.

THERMODYNAMIC

SYSTEMS

gasdata; v=V/x(1); Z specific volume of air in tank [P]=air(x(2),l/v); Z call Calf’ just to get the pressure, ~ Then call Cair’ to get the properties of state c: [Pc,uc,hc,sc,dudTc,dudvc,dsdTc,dsd~c,dPdTc,dPdvc3=air(Tc,1/vc); E=(P/Pc-l)’2+(sin/sc-l)~2; ~ self-scaled error index n=O; ~ start a count of the number of iterations while E>le-8 ~ iterate until error is acceptably small if n==O ~ that is, for the first iteration only M=[dPdvc dPdTc;dPdTc dsdTc]; ~ matrix for Newt.-Raphson q=M\[Pc-P;sc-sin]; vc=vc-q(1); ~ improved estimate, spec. volume, state Tc=Tc=q(2); ~ improved estimate, temperature, state else ~ following the suggestion on page 910 vc=vc+(P-Pc)/dPdvc; Tc=Tc+(sin-sc)/dsdTc; end ~ Now the revised estimate of the state c is found: [Pc,uc,hc,sc,dudTc,dudvc,dsdTc,dsdvc,dPdTc,dPdvc] =air(Tc,i/vc); E=(P/Pc-l)~2+(sin/sc-l)’2; ~ new (reduced) n=n+l end hout=hc/.8-(1/.8-1)*hin; Z from equation (12,67) f(1)=DPHO*D/vin*.95; Z first differential equation f(2)=((hout-h+(P+dudv)*v)*f(1))/x(1)/dudT; second diff. eqn. f(3)=P; ~ to enable plotting of P f=f’; t ~ to see the progress of the simulation on the screen ~ comptank.m, master program for simulation of an air ~ compressor charging a tank from atmospheric air ~ x(1) is the mass of air in the tank; x(2) is its temperature global V DPHO D vin bin sin vc Tc V=.I; DPHO=I750*2*pi/60; D=le-4; vin=I/l.205; T0=293; vc=l; Tc=300; ~ non-critical trial values to get started gasdata ~ imports the many coefficients that describe air [P,u,hin,sin]=air(TO,1/vin); ~ finds the needed hin and sin [t,x]=ode45(’compdife’,[O 60],[V/vin TO 0]); k=size (t); i=k(1); ~ for use below to find the pressure, for j=l:i-i dt(j)=t(j+l)-t(j); ~ values of time increments P(j)=(x(j+~,3)-x(j,3))/dt(j) Z P is time deriVativeof tl(j)=(t(j+l)+t(j))/2; center times of time intervals end plot(t,x(:,2),tl,P/lO00).~ plots the temperature and pressure

12.5.

927

EVALUATION OF THERMODYNAMICPROPERTIES

The plot that results fromthe final instructions aboveis given below(annotations added). The high temperature reached showsthe importance of the omitted heat transfer. 4000

,

,

,

3500 3000 2500 2000 1500

temperature,K_.~.~._. ~ooO~oo

~

,~

~’0

30

,0

~’0

time, seconds

60

The model or air used above above is dry. Atmospheric air normally is considered a mixture of dry air and water vapor. The ratio of the massof water vapor to the mass of the dry air is called the absolute or specific humidity, TP~

~ =v .

(12.102)

Assuminga perfect gas, the specific humidity also can be expressed in terms of the total pressure, P, and the partial pressures of its components,Pa and P~,: ~z= Pv/R~, Pa/Ra -0"622~ = p_p~. p~ 0.622P~

(12.103)

Whenthe partial pressure of the water, P,,, equals the saturation pressure of water, Pg, the air is said to be saturated; it can hold no more water. The relative humidity, ¢, is the ratio of the actual massof water in the air to this maximum, or ¢ = my _ P~, _ m~ P~ (0.622

wP + ~)Pg

(12.104)

Should the temperature of a given mixture cool beyond the temperature at which the relative humidity reaches 1.00, or 100%,knownas the dew point, the excess water condensesout. Extensive properties such as the total entropy and enthalpy equal the sumsof the contributions from the various componentsof the mixture, including water vaporand, whenit exists, liquid water. A treatment for water is given in Section 12.5.5.

928 12.5.4

CHAPTER 12. Application

THERMODYNAMIC SYSTEMS

to Refrigerants

Refrigerants changephase in their useful applications. Manyrefrigerants, including R-12 (the "freon" workhorse that is nowoutlawed for new equipment due to its alleged effect on ozone in the upper atmosphere), R-22 (a somewhat less offensive halogenated hydrocarbonthat is scheduled to be phasedout over the next 30 years) and R-134a(an emergingpopular refrigerant 16 that does not attack ozone) have been characterized by the Martin-Houequation of state, in whichthe specific volumev is the reciprocal of the density, p: p = R~ ~ 1 (Ai + Bi~ +~ 2 (v -b) v~-b ~

+

Cie -K0/~) -k

A6 -t-

B6~ e+ K~/~ C6 e"V(l+ ~v) ce

(12.105) The saturation pressure for most refrigerants is given by the single equation F3 3lnPsat =F1 + F~. -~- + ~ + F~lnO + F~0 + Fo02 + FTO + Fs (F9 - 8)ln[(F9 - tg)F~o],

(12.106)

in which logi0 sometimesis substituted for ln. The density of the saturated liquid for most refrigerants is given by 5 Pl = Z DiX(i-l)/3

+ D6X1/2

"4-

D7X2; X =- 1 - O/Pc.

(12.107)

i~1

The density of the satm:ated vapor for most refrigerants is given by Pg=pc(O/Oc)E=sexp[~ki=lo " Ei+11Xi/3]

(12.108)

The specific heat for manyof these refrigerants, including R-12 and R-22, has been given in the form 4

0 Cv = Z GiOi-1

+ G50-2

-t-

G604.

(12.109)

A slight variation has been used for R-134a: 2 + C~4~ ~ + Cv~/~. %o= ’Cp° _ R = Cv~- R + Cv.,. 8 + Cv3~

(12.110)

To compute the entropy, a procedure similar to that used to give equation (12.87) (p. 920) can be used. Instead of adding and subtracting the ~6Properties of R-134a are given by D.P. Wilson and R.S. Basun in "Thermodynamic Properties of a New Stratospherically Safe Working Fluid - Refrigerant 134a," ASHRAE Transactions v. 94 Pt. 2 (1988) pp. 2095-2118.

12.5.

EVALUATION OF THERMODYNAMICPROPERTIES

929

-R lnp, however, R ln(v - b) is added and subtracted, the term pR in the integrand is replaced by R/(v - b), and dp/p2 becomes-dv. Note also that the partial derivative of P with respect to ~ for constant v must be computedfrom equation (12.105). The entropy for the vapor states of substances described by equations (12.105) and (12.109) s=JRln(v-b)+

J6A.,

Bi (v_b)~_~_j 1- i

av ae

5 Ci(v - b)~ - i i~2

+ G3~/2 + G~3/3 - G~/(20 ~) + G~a/4 + ~

(12.111)

and for substances described by equations (12.105) (with B~= 0) and (12.110) is s =Rln ( (V -~:~ - ~ l - iB~ (v _ b )~~ ~ C~(v - ~-i b) +~Ke-Ke/e~ i 1 ~ 8 ~ C~ + Cp3~ + Cp4 3 0 + l;

+Cp~lnO+Cp~O (12.112)

whereY is a constantof integration that includes the datumstate entropy, s0. Entropystates for the saturatedvapor,s~, are calculatedusingthe specific volume,v~. Entropystates for a saturated mixture then can be computed from the C]apeyron relation, s=s~-(ve-v)

~

(12.113)

in whichdPsa~/dO is foundby differentiating equation(12.106). The enthalpy for the vapor states of substancesdescribed by equations (12.105)and(12.109) 5

h = ~ J[A~(v- ~)~1-~/(i_ ~)1jA~/(u~.~) i=2

-~°/°~ ~ C~(v b)(1-~} + J(1 + KO/O~)e ~ + JPv+alO i=2

+ G~O~/2+ GaOl/3 + G40~/4- G~/O+ G~O~/5+ X, and for substances describedby equations(12.105)and (12.110) h

~ Ai(v - ~-~ b) i 1 + Pv - RO i~2

(12.114)

930

CHAPTER 12. -~

(

1 -1-

If e--KO/O¢

~)

THERMODYNAMIC SYSTEMS

;-_. ~ S ECi(v-b)(1-il

-1- CplO -}-

Cp202/2

+ Ca~3/3+ C4~4/4+ C,s ln0 + X,

(12.115)

where X is the constant of integration that includes the datumstate enthalpy. In the two-phase saturated mixture region, the enthalpy is computedfrom h = ha - O(sa - s).

(12.116)

Reynoldsgives the coefficients for evaluating equation (12.107) for the density of the saturated liquid. Fromthe Clapeyronrelation, s f = sa - -~ ua -

,

so that hI = ha - O(sg - sf),

(12.118)

and the quality becomes (12:119)

sa - s f" Thesecalculations are included in the master function file propre~ .m. Parameters for R-22 and R-134a are given in AppendixD. 12.5.5

Application

to Water

The Helmholtz free energy for water has been given by Keenanet a117 in the form (12.120a)

¢ = ¢0(0) + R0[ln p + pQ(p, T)], 6

¢o = E Ci/Ti-1 + C7 ln0 + Cs lnO/T

(12.1205)

i=1

Aij (p - pay)i-’ -E~ +e

jQ ~" = (v -- To) E(T -- raj) j=l

i-~ Aijp

i=1

7 ~ 1000/0.

(12.120c) (12.120d)

Althoughthe International Association for the Properties of Water and Steam (IAPWS-IF1997) h~ replaced this modelas the international standard for cal~s culations in the powerindustry with a vastly morecomplexset of formulations, 17J.H. Keenan et al, Steam Tables, Int. ed. 2d ed, Wiley, New York, 1978. ~SWolfgang Wagner, Properties of Water and Steam, Springer-Verlag, 1998.

OF THERMODYNAMIC PROPERTIES

931

the Keenan model should suffice for most purposes. Substitution (12.95) and (12.92) gives

into equations

12.5.

EVALUATION

P=pRO

l+pQ+p

2 ~p

,

,

s=_R[lnp+pQ_pT(O~p) h = RO pT

]p

d¢OdO,

OQ + l + pQ + p~ ~ P

(12.121b)

+ dT

The partial derivatives above are best carried out anMytically. The saturated mixture region can be addressed by using equations (12.116)(12.119), which apply to any substance. The needed relations for the saturation pressure, density of the saturated liquid and approximate density of the saturated vapor are

p~ = p~ ~ + D~[~- 0/0~)~/~ , p~ =~)

exp

Ei

1-

(~.~b) .

(12.122c)

Coe~cients for equations (12.120) and (12.122) ~re given in the program ~aterdat.m at the end of Append~ D. The equations are evaluated in the program prop~at, which calls the program ste~.m. Equation (12.122b) due to Reynolds (footnote 13 p. 922), and equation (12.122c) is due to the thor. This l~t equation is accurate within 0.06% compared to the core model, producing modest errors in the derived properties. It could be improved.

12.5.6

Case Study: a Refrigeration

Cycle

A crude model for the dynamics of a refrigeration cycle is given in Fig. 12.21. The evaporator is represented by the compliance element CS1, the heater (which in practice assures that the compressor does not pmnpliquid) by element the normally superheated inlet region of the condenser by elements CS3 and and the normally saturated-mixture region of the condenser by CS5. Thermal compliances for the metal shells of these components are omitted, for simplicity, but in practice could easily be added. Knowntemperature reservoirs are assumed for the hot and cold sides, and the heater power is given. An orifice, represented by the element RS~, is assumed to accomplish the expansion process. Elements RS~, RS4 and RS5 also are treated as virtual orifices (with

932

,CHAPTER 12.

THERMODYNAMICSYSTEMS

hot thermalreservoir (a) schematic

/ expansion

condenser ~ valve ~ .~mpressor evaporator ~ heater

cold thermalreservoir (b) convectionbondgraph

3CSs~RS45~ T . m45

Sen

CS4~RS34~CS /1145 m34

P,,h.~orh~[m,~ RSs~

m34

p3, h3nlrh23.

1S ~_

ds ~RS ~ CS~ O~ l ’1

S,~

h~ q~

S,~

T~=D/v2; T= q~/q~ ; Ts= ~/(1-~)

Figure 12.22: Simplemodelof a refrigeration cycle

12.5.

EVALUATION

OF THERMODYNAMIC PROPERTIES

933

muchlarger throat areas) to represent small frictional flow losses. The compressor is driven by a constant-speed shaft, and the moduli T and Ts are set to give a constant adiabatic efficiency, as described in Example12.2 (pp. 886887) and Section 12.4.7 (pp. 909-911). The refrigerant is R-12, the properties which are given in the computer file "dataR12" in Appendix D. The evaporator is assumed to emit only saturated vapor, because of gravity separation. The condenser emits only saturated liquid whenever there is any, also assuming a gravity separation, but when the chamber is in the superheated region it emits vapor at its xnean state. The assumed parameters of the system are listed in Table 12.1. Table 12.1 Parameter

values,

refrigeration

parameter

symbol value

mass of refrigerant specific heat ratio volumetric displacement, compressor shaft speed,compressor adiabatic efficiency, compressor volume, chamber 1 volume, chamber 2 volume, chamber 3 volume, chamber 4 volume, chamber 5 heat cond. coefficient, chamber 1 heat cond. coefficient, chamber 3 heat cond. coefficient, chamber 4 heat cond. coefficient, chamber 5 temperature of cold reservoir temperature of hot reservoir heater power area, expansion valve area restriction, chambers 1 to 2 area restriction, chambers 3 to 4 area restriction, chambers 4 to 5

MT K D DPH0 Eta V1 V2 V3 V4 V5 CH1 CH3 CH4 CH5 TEL TEH HH A51 A12 A34 A45

cycle

6 kg 1.3 2.305 × i0 -4 m3/rad 28 rad/s 0.75 30.0060 m 30.0003 m 30.0006 m 30.0006 m 30.0060 m 622 W/K

4.07 W/K 13.895

a49.3 w/~ 260 K 295 K 588.4 W ’~ 1.246 x 10-6 m 2.385 x 10-4 2m 3.762 x 10-4 2m 3.790 x 10-4 m

The state of the system before start-up has a uniform pressure throughout. Since the evaporator is at the temperature of the cold reservoir, this means that all liquid is located there; it was blown through through the expansion valve when the c6mpressor was last shut off. After start-up, however, considerable refrigerant slowly migrates through the compressor to the condenser, which finally enters the saturated mixture region and approaches the steady-state equilibrium. The temperatures are seen to approach their equilibrium values more quickly.

934

CHAPTER

12.

THERMODYNAMIC

SYSTEMS

The main MATLAB program for the simulation, called refgmst, is given below. It defines the parameters, including running dataR12, establishes initial conditions and calls the basic MATLAB integrator ode23. The state variables comprise the refrigerant ~nasses in each of the first four chambers (the mass in the fifth equals the known total mass minus the first four masses) and the temperatures in each of the five chambers. ~ refgmst.m,master program for simulatingrefrigeratorstart-up ~ Five lumped thermal compliancesassumed: one for the evaporator, Z one for the heater, and three for the condenser. ~ The state variablesare: ~ x(1) is the mass in the evaporator ~ x(2) is the mass in the heater ~ x(3) is the mass in the condenser,superheatedinlet section ~ x(4) is the mass in the condenser,.superheated exit section ~ The mass in the condenser,nominal saturated mixture section, ~ is m5 which equals the total mass minus the sum of the above. ~ x(5) is the temperaturein the section with mass x(1) Z x(6) is the temperaturein the sectionwith mass x(2) ~ x(7) is the temperaturein the section with mass x(3) ~ x(8) is the temperaturein the.sectionwith mass x(4) ~ x(9) is the temperaturein the section with mass ~ The compressorruns at a constant speed DPHO, and has a fixed ~ adiabaticefficiencyEta. ~ The expansion comprises flow through an orifice, either liquid ~ (Bernoulli’sequation)or unchoked compressibleflow. global MT K C D DPHO EO0 TEL TEH Vc Tc Eta global HH V1 V2 V3 V4 V5 A12 A34 A45 A51CHI CH3 CH4 CH5 A12=2.385e-4; A34=3.762e-4; A45=3.79e-4; A51=1.246e-6; C=.35506; Vl=6e-3; V2=.3e-3; Y3=.6e-3;~Y4=.6e-3; VS&6e-3; Eta=.75; HH=588.4; vh=.099513; vg=.08539; x2=V2/vg; x3=V3/vh; h4=V4/vh; x5=VS/vh; xl=(MT-x2-x3-x4-xS); K=l.3; C=.355; D=2.305e-4; DPH0=28; E00=le-12; CH1=622; CH3=4.07; CH4=13.895; CH5=849.3; TEL=260; TEH=295; MT=6; Tc=260; Vc=.08; ~ values to start the simulation dataR12 ~ gets the coefficientsthat describe the refrigerant xO=[xl x2 x3 x4 TEL TEL TEH TEH TEH]; ~ initial conditions [t,x]=odefBs(’refrigde’,[O 14],x0); ~ integratorfor stiff systems The bulk of the programming for the refrigerator resides in the function m-file refrgde, which gives the differential equations. This program starts by computing the five specific volumes from the mass state variables. It then communicates these values and the corresponding five temperatures to the routine propre~, and receives back all the needed (and some unneeded) derivative state variables for each of the five chambers. The mass flow rate through the compressor (dm23) then can be computed:

12.5.

EVALUATION

OF THERMODYNAMIC PROPERTIES

935

function f=refrgde(t,x) Z differential equations for refgmst.m global MT E C D DPHO EO0 TEL TEH Vc Tc Eta global HH V1 ~2 V3 ~4 V5 AI2 A23 A34 A45 A51CH1CH2 CH3 CH4 CH5 m5=MT-x(1)-x(2)-x(3)-x(4) ~ mass in chamber vl=Vl/x(1);v2=V2/x(2);v3=V~/x(3);v4=V4/x(4);v5=V5/m5; [Pl,hl,hgl,sl,sgl,vfl,vgl,hfl,dsdvl,dsdTl,dudvl,dudTl,dPsl,d2Psl, dvdgl,dsgl,dugdvl,dugdTl,dPdvl,dPdTl]=propref(vl,x(5)); [P2,h2,hg2,s2,sg2,vf2,vg2,hf2,dsdv2,dsdT2,dudv2,dudT2,dPs2,d2Ps2, dvdg2,dsg2,dugdv2,dugdT2,dPdv2,dPdt2]=propref(v2,x(6)); [P3,h3,hg3,s3,sg3,vf3,vg3,hf3,dsdv3,dsdT3~dudv3,dudT3,dPs3,d2Ps3, dvdg3,dsg3,dugdv3,dugdT3,dPdv3,dPdT3]=propref(v3,x(7); [P4,h4,hg4,s4,sg4,vf4,vg4,hf4,dsdv4,dsdT4,dudv4,dudT4,dPs4,d2Ps4, dvdg4,dsg4,dugdv4,dugdT4,dPdv4,dPdt4]=propref(v4,x(8)); [Pl,h5,hg5,sS,sgS,vf5,vg5,hf5,dsdv5,dsdT5,dudv5,dudTS,dPs5,d2Ps5, dvdgS,dsgS,dugdvS,dugdT5,dPdvS,dPdT5]=propref(v5,x(9)); dm~=D*DPHO/v~~ the flow pumped by the compressor; The program continues with the tricky job of finding the enthalpy hideal ~hc that would exist downstream of the compressor were that machine operating reversibly. As described in Section 12.4.7 (p. 910), the entropy and the pressure of this state equal s2 and Pa, respectively; a Newton-Raphsoniteration with full evaluation of derivatives in the appropriate state regime (superheated or saturated mixture) is carried out. The actual enthalpy h3n of the outflow ~om the compressor, resulting ~oman adiabatic efficiency of 75%, is then computed: ~ Start by using values of Vc and Tc from the previously ~ determinedstate as the first trial: [Pc,hc,hgc,sc,sgc,vfc,vgc,hfc,dsdvc,dsdTc,dudvc,dudTc,dPsc,d2Psc, dvdgc,dsgc,dugdvc,dugdTc,dPdvc,dPdTc]=propref(Vc,Tc)); E=(Pc-P3)’2*le-12+(sc-s2)’2~le-4 ~; weighted error value vc=Vc; while E>E00 ~ iterate to get vc and Tc for Pc=P3 and sc=s2 M= [dPdvcdPdTc; dsdvcdsdTc]; q=M\[Pc-P3; sc-s2] vc=vc-q(1) Tc=Tc-q(2); [Pc,hc,hgc,sc,sgc,vfc,vgc,hfc,dsdvc,dsdTc,dudvc,dudTc,dPsc, d2Psc,dvdgc,dsgc,dugdvc,dugdTc,dPdvc,dPdTc]=propref(vc,Tc)); E=(Pc-P3)^2,le-l~+(sc-s2)’2~le-4; end h3n=hc/Eta+(l-1/Eta)*h2;~ compressorenthalpy (eqn. (~2.67)) The gravity separation that sometimes takes place at the outlet of the condenser is then recognized. Next, the program orifice is called repeatedly to compute the flows through the four actual or virtual fixed orifices, including the expansion valve. (This program, given in Appendix D, is more general than

936

CHAPTER

12.

THERMODYNAMIC

SYSTEMS

is needed for the present simulation; it permits the inflow to be a saturated mixture or a compressed liquid as well as a saturated liquid or a superheated vapor.) This leads directly to the differential equations for the rates of change of mass in chambers 1 - 4, that is the first four state variables (note that the rate of change in the fifth chamber equals minus the sum of the other four): if v5vg2 f(6)=(HH+CH2*(TEL-x(6))+(hgl-h2)*dml2+(P2+dudv2)*v2*f(2)) /dudT2/x(2); else den=x(2)*((dugdv2+P2-x(6)*dPs2)*dvg2+dugdT2 -x(6)~d2Ps2~(vg2-v2)); f(6)=(CHl~(TEL-x(6))+(hgl-h2)~dml2+x(6)~dPs2~v2~f(2))/den; end f(7)=(CH3~(TEH-x(7))+(h3n-h3)~dm23+(P3+dudv3)~v3~f(3))/dudT3/x(3); f(8)=(CH4~(TEH-x(8))+(h3-h4)~dm34+(P4+dudv4)~v4~f(4))/dudT4/x(4); if v5>=vg5 f(9)=(CH5*(TEH-x(9))+(h4-h5)*dm45-(h5e-h5)*dm51 +(PS+dudvS)*vS*f5)/dudT5/mS; else

12.5.

EVALUATION

OF THERMODYNAMIC PROPERTIES

937

heat in asymptote 5 heat, power, ratio, 4

heat

~

/__~.~ //

¯ 350 temperature, K

emperamre, chamber 3

/"

chamber

5

2

’

-

300

hot reservo"~

~

fref~iger~t mass, eh..~._~_~r5, kg,% 0 0

asymptote

/

~ tem~rat~e, ’

mass5 325

4

temperate,

cola reservoir"

chamber 1 \ 6

8

I0

12 4 time, seconds

Figure 12.23: Start-up of the refrigeration

system

den=m5*((dugdv5+P5-x(9)*dPs5)*dvgS+dugdT5-x(9)*d2PsS*(vgS-vS)); f(9)=(CHS*(TEH-x(9))+(h4-hS)*dm45-(hSe-hS)*dmS1 +x(9)*dPsS*vS*fS)/den; end Vc=vc; Selected results are plotted in Fig. 12.23. Temperature equilibrium is approached long before mass equilibrium; only the former occurs (virtually) in the 14 seconds shown. The condenser reaches saturation state at about three seconds, and holds there for awhile, rapidly alternating between discharging liquid and vapor. After eight seconds the discharge is pure liquid. Were less refrigerant used, mass equilibrium would occur sooner, and the blow-down process following shut-down also would be faster. 12.5.7

Application

to

the

Liquid

Region

There are two different approaches to evaluating properties in the liquid region. The first uses either the ’P - v - T’ or the Helmholtz formulation, which may or may not apply in this region for a particular substance, depending upon the data used in its generation. The second approach uses a simple model including a specific heat and a bulk modulus. The first approach is most apt for states near the critical point, where the specific heat and bulk modulus are

938

CHAPTER 12.

THERMODYNAMIC

SYSTEMS

far from constant. The second approach is usually quite satifactory for lower temperatures. Direct use of a ’P-v-T’ equation can give an inaccurate pressure, since in the liquid region pressure is extremely sensitive to small changes in density or specific volume. For a discussion and procedure to increase accuracy, see 19 Reynolds. The simpler model gives the pressure as the sum of the saturation pressure at the same temperature plus a pressure difference due to compression with a constant bulk modulus, ~: (12.123)

P = P~r+ ~ln (vsa,~.

The internal energy, u, equals the internal energy of the saturated liquid at the same temperature, defined ~ ui~, plus the work done in compressing the liquid: u=uyo+

Pdv=u]o+~

v-vi+vln

~ .

(12.124)

The entropy is unchanged in an assumed reversible process, and the enthalpy can be found from its definition h = u + Pv. The term Ou/O~, necessary for differential equation for temperature (equation (12.66), p. 904), is assumedto a constant specific heat, c, values of which usuMlyare available. The values of cv and c, are assumed to be the same, which is exactly true only for incompressible substances. The value of the bulk modulus may not be so readily available, but it can be estimated from data for saturated liquid. If the internal energy at the saturated state having the same specific volume, v, (rather than ui0 , which is at the same temperature) is defined as ui~ , the internal energy is approximately u = ufv + c(0 - 0i~),

(12.125)

where 0]~, is the temperature of that saturated state. Equating the two equations for the internM energy given by equations (12.124) and (12.125) gives, Mter little algebra, fl = 2[u]~, - ui8 ~ + c(O - O]v)]

v(1- v/v]o)

A sample calculation uses tabular data for refrigerant tures of -20°C and 20°C:

(12.126) R12 at the tempera-

- 54,440 + 977(20 - 720) Z =2117, ~~ 720 ~~~0~) = 0.689 ~ ~0" P~ = ~6,000 psi. (~2.127) This value is almost half that of water, showing that the liquid is nearly incompressible for most purposes. As a practical matter, a simulation involving occ~ional compressed liquid states with this bulk modulus Coul~ have differential equations so "stiff" that solution is significantly impeded. Little error ~gSee footnote 13 (p. 922).

12.5.

EVALUATION

OF THERMODYNAMIC PROPERTIES

939

would be introduced by the expedient of decreasing the assumed bulk modulus. One could also recognize that, in reality, the effective bulk modulus is apt to be significantly lower than the theoretical value, because of the effects of the expansion of the container due to pressure-induced stresses, and. the microscopic vapor bubbles that usually abide in surface cracks. 12.5.8

Considerations

of

Reversing

Flows

The differential equations for temperature include terms for the enthalpy fluxes entering and leaving the control volume. Whena flow reverses, the differential equation itself needs to be changed, since the source of one or more fluxes changes. The switching can be implemented simply through the use of i/statements. Sometimes these//statements can result in a false simulation, however. This occurs when the different evaluations of a flow rate within a single time step, as organized by the Runga-Kutta integration algorithm, produce different signs. This would not be meaningful physically. You are urged to keep a vigil for this phenomenon, and take special corrective steps should it occur. The first telltale sign may be a distinct sensitivity of a result to the step sizes (or error index) used. You also can display the running values of a flow on the screen, and directly witness a rapidly alternating sign. A simple remedy for this phenomenon is to put the the flow rates or the enthalpy fluxes themselves through a simple smoothing filter. The chamber has a residence time, so the model is reasonable as long as the assumed time constant for such a filter is shorter than this time. For the time constant to be effective, on the other hand, it should be larger than the time steps. This procedure does not guarantee total elimination of the phenomenon,but it radically curtails any errors than might otherwise accrue. 12.5.9

Summary

The thermodynamic properties of substances are measured in terms of pressurevolume-temperature relations and specific heat relations, the latter at asymptotically small pressures. The independent variables are temperature and specific volume (or density). Pressure, enthalpy, entropy and internal energy are treated as derivative properties. The data for a wide variety of pure substances is available in the form of complex algebraic equations, with manycoefficients but far less data than required by a table of properties. This permits direct computation to replace table look-up with its interpolation. The data for several commonsubstances have been converted to a Helmholtz formulation, from which the derivative properties can be deduced in a similar but somewhat simpler manner. This section has presented the necessary basic analytical equations. Applications to ideal gases, with air as a specific example, multiphase pure substances with refrigerants as a specific example, and compressed liquids have been discussed. MATLAB coding is given in Appendix D for the cases of air (or its components oxygen or nitrogen), refrigerant R-12 and water.

940

CHAPTER 12.

THERMODYNAMIC SYSTEMS

The assumption of thermodynamicequilibrium overlooks certain highly complex dynamicprocesses that occur in very rapid phase change. The bubble nucleation and growththat limits the rate of flashing (rapid boiling) is an example. Examplesof an air compressorand a refrigeration system have been used.to illustrate the completeprocess of writing differential equations and evaluating the various terms, including MATLAB coding. Guided

Problem

12.4

The simulation equations for the air piston compressor presented in Section 12.4.6 (pp. 905-909) assumes an ideal gas with constant specific heats. Carry out a moreprecise simulation by representing the air as a mixture of nitrogen, oxygenand argon, as described in Section 12.5.3 (pp. 922-924). The script files gasdata.m and spech.m and the function file a±r.m, all of which appear in Appendix D, may be used. Suggested Steps: 1. Define a set of state variables, using the bondgraph given in Fig. 12.15 (p. 906) as a guide. Use the temperatures of the gas and the walls, rather than their entropies that were used in Section 12.4.6. Usingthe gas temperature, in particular, precludes the need for iteration in dealing with the properties. 2. Writethe correspondingset of state differential equations. 3. Codethe differential equationsin a function m-file that calls on the available property files as necessary. 4. Write a master programthat calls the integrator ode23s, which in turn calls your function file for the differential equations. The master program can call the file gasdata.monce, and transfer the results via a ’global’ statement. System parameters and initial conditions are given on the samepage as the results plot. 5. Carry out a simulation for the same 0.1 seconds shownin Fig. 12.16 (p. 908), and plot the results. The results should be the same, since this is the procedure used to producethat figure. PROBLEMS 12.12 A tank charged with oxygen discharges into the atmosphere through nozzle of effective area 0.0001 m3. The tank has a volumeof 0.5 m3 and is equipped with a check valve that prevents reverse flow. The initial pressure of the gas is 5.0 MPaabsoute (about 720 psi gage), and its temperature is equilibrium with the 300 K surroundings, to give a specific volumeof 0.01516 mU/kg. Heat flows into the tank with the conduction coefficient 1.0. W/K. Simulate the discharge for a period of 120 Seconds.Plot the resulting massand

12.5.

EVALUATION

OF THERMODYNAMIC PROPERTIES

941

temperature as a function of time. Check to make sure that the oxygen remains in the gaseous phase. (The condition at the moment the check valve closes is the greatest threat to this assumption; you could merely print out the running specific volume on the screen and compare its values at that instant to those in a table or plot of properties.) 12.13 The refrigerator simulated in Section 12.5.6 is running at equilibrium when it is shut off. You are asked to find the mass distribution and the temperatures until equilibrium is reached. To simplify the model and simulation, ignore all volumes and masses except for the principal condenser and evaporator (which contain almost all of the refrigerant anyway). Since the compressor not running, its flow is zero; the problem is essentially a blow-downfrom one tank to another through a fixed orifice. You may assume that the two tanks have the same volume of 0.006 m3, and start with equal masses of 3 kg. The condenser starts at 304.3K and the evaporator at 251.5K. 12.14 Model.a steam catapult for use on an aircraft carrier as a single variable chamber that acts on a piston of area 0.024 m~ to push an aircraft of mass 16000 kg. The aircraft engine is producing a thrust of 100,000 N. The chamber initially contains saturated mixture at 600 K with specific volume 0.002 m3/kg. Assumethe take-off is so fast that no water enters or leaves the chamber, and there is negligible heat transfer. Assumean equilibrium flashing process, and neglect friction. Simulate the take-off for the length of the catapult, which is 100 m. Determine the necessary initial size of the chamber if the final velocity is to be 55 m/s. (Note: Problem 12.8, p. 916, is a start.) 12.15 A fluid chamber is normally in the saturated-mixture region, but sometimes enters the compressed-liquid region. Show what coding changes or additions would be needed, using the coding for the refrigeration cycle in Section 12.5.6 as reference. 12.16 The flow dml2 normally entering a chamber 2 from chamber 1 and the flow din23 normally leaving chamber 2 to enter chamber 3 may reverse direction. Assume that chamber 2 is in the saturated mixture region, and that the port from chamber 1 is above the level of the liquid and the port to chamber 3 is below that level. Write code for the differential equation for the temperature of chamber 2. SOLUTION Guided

Problem

TO

GUIDED

PROBLEM

12.4

1. The first two state variables can be the same as before: the volume of the cylinder, V, and the massof the air therein, m. Thenext two state variables are chosen as the temperature of the air in the cylinder, 8, and the temperature of the walls of the cylinder, ~. (Alternatively, you could stick with the entropy of the walls of the cylinder, So, but for the air the use of equation(12.66) (p. 904), which precludes the need for iteration, demandsthe use of its temperature.)

942

CHAPTER

12.

dt dO ~ H(Oc-O)+(h~-h)~,l+ -dt mOu/O8

THERMODYNAMIC

SYSTEMS

P+ ’

~ = F ¯ I~1 + H(~- ~). dt The flo~ through the valves, ~ and ~2, ~e given by equations (12.76) (p. 907, not reproduced here), which are approximations since they ~sume a const~t value for the specific heat. This value, for the mixture of air, is computed in a script m-file Spacheat.m which is given in Appendix D. function f=aircompd(t,x) ~ derivatives for simulation of X air compressor ~ x(1) is the volume of the cylinder ~ x(2) is the mass of the air in the cylinder ~ x(3) is the absolute temperature of that mass ~ x(4) is the absolute temperature of the cylinder wall ~ dml and dm2 are the mass flows into and out from the ~ cylinder, respectively global Pl T1 H1 P2 Cm C Hc Hc0 Dphi PF V1 al R E1 E2 A0 T rho=x(2)/x(1); ~ density of the [P,u,h,s,dudT,dudv]=feval(’air’,x(3),rho); ~ finds air properties T=x(3); specheat; ~ finds cvm (slightlyvarying specific heat) k=R/cvm+l:E3=2/k; E4=l+l/k; A=al*sqrt(2*k/(k-1)/K); Prl=P/P1; Pri=P2/P; ~ pressure ratios across valves if Prl
12.6.

SYSTEMS WITH CHEMICAL REACTION*

943

T=T1; gasdata; specheat; kl=l+R/cvm; El=2/kl; E2=l+l/kl; gO=al*sqrt(2*kl/(kl-l)/K); [t,x]=ode23(~aircompd’,[O .1],[VO mO 440 744 0]);

Theplots of Fig. 12.16 (p. 908) result fromthe followingpost-simulationMATLABprogram: tl=t(l:size(t)-l);Z time vector shortened one P=diff(x(:,5))./diff(t); Z P is the derivativeof dm=diff(x(:,2))./diff(t); Z mass flow is the derivativeof plot(t,x(:,l)*le4,t,x(:,3)/lOO,tl,P/leS,tl,dm*lO0) Z plots volume, temperature,pressure and mass flow, respec%ively,as scaled

12.6

Systems

with Chemical

Reaction*

The dynamicsof chemical reactions ("chemical kinetics") depends on transport phenomenasuch as mixing and diffusion mechanismsas well as the local properties of composition, temperature and density. This section introduces the subject, with emphasis on lumpedmodels, which implies the ideal case of well 2° mixedreactants. Muchof the bondgraph notation used is that of the author. 12.6.1

Energy

of a Pure

Substance

The thermodynamicenergy of a pure substance has been represented by a threeported compliance, as shownin Fig. 12.24 part (a), where~q is the conduction entropy flux. Recalling equations (12.53), (12.55) and footnote 10 (pp. 900-901), Y = Y(S, V, m),

(OY)

0 = OS v,,~

_p = 0(_~=~..),

(12.128)

(12.129a) (12.129b)

S,m

h = Os + g, g --

(12.129c) .

(12.129d)

V,S

Whenchemical reactions are involved, focusing on the Gibbsfunction g is especially helpful. This can be done in a bondgraph by separating the terms 0s and g with a 1-junction, as shownin part (b) of the figure. The bondwith effort and flow ~h is then connected to a bondwith effort 8 and flow srh by an ideal 2°F.T. Brown, "Convection Bond Graphs," Journal of the Franklin Institute, 1991, pp. 871-886.

w 328 n. 5/6,

944

CHAPTER 12 THERMODYNAMIC SYSTEMS 8& T.__s. ~Ie~v ,h=g+~s~ -p , t=--==~.’~.’-.’-~ ¯ (a) previousversion, pure substance

~

.h=g+&~. g ~_ _p l ~ C ~m V

(b) reticulated version, pure substance

(c) C element, more ~ (d) vector bond version one species of (c) Figure 12.24: Thermodynamicenergy storage transformer having modulus s. This moduluscma change in time, but may not changedrastically. The flow srh is the convection entropy flux due to the mass transport into (or out of) the system. Whenthis is added to the conduction entropy flux ~q through the use of a zero junction, the result is the total entropy flux, ~. The product Os~his the energy flux directly associated with the convection entropy flux. The product grh is the total "free energy" flux which conveys work but no entropy whatsoever. Note that its Bondand the bond for the effort Os are simple bonds. 12.6.2

Energy of Multiple

Species

Whenmore than one substance is present, equation (12.128) is modified V = V(S, V, ml, m2, m3,...),

(12.130)

so that 0= 0(~_~)

,

(12.131a)

-P= O(’-~V)s

,

(12.131b)

O(~-~m/)

(12.131c)

The C element of part (c) of the figure represents this relationship, in which each ~i is a partial Gibbs function known~ a chemical potential, although the constraints betweenS ~d ml, ..., mnare not as yet shown.

12.6.

945

SYSTEMS WITH CHEMICAL REACTION*

A bundle of bonds of similar type is represented widely in the literature a vector bond, drawn with two solid parallel lines:

by

This notation permits the graph of part (c) of Fig. 12.24 to be represented shown in part (d). The vector nature of ~ and ~ is acknowledged through the use of bold-faced type. 12.6.3

Stoichoimetric

Coefficients

A chemical reaction of arbitrary

and Reaction

Forces

complexity can be described by li~P ~- ~ ~ Pi,

i=1

(12.132)

i=1

where fii is the molar density and l/~ and li~’ are the stoichoimetric coetilcients for the ith species. For example, 3H ~ He + H gives li~ = 3, li~ = 1, li~2 = 0, li~2 = 1. Similarly, 2NO + 02 ~ 2NO.~ It02 : 1, l/tt02 -~ O, I ItN02 giveslifo= 2, UN --~ 0, l/NO Or!~- O, I :’’ = 2. Thefirst consideration here is a reaction in a closed chamber (no mass flows across its boundaries) of volume V, which could vary in time. A cylinder of an engine with its valves closed is an example. The molar densities are thus

rn~ ~i = V Mi ’

(12.133)

where the Mi are the respective molar masses. A reaction rate 7~ is defined such that for each species, rhi = Mi(li~’ - l/~)~:

(12.134)

Note that the signs are chosen so that if 7~ is positive, the reaction proceeds from left to right. Note also that the stoichoimetric coefficients are constrained to satisfy the conservation of mass n

=o.

946

CHAPTER 12 THERMODYNAMIC SYSTEMS

//o.-

TI = Mj(v~"-v~’) ~ ml

~ = M~(~"-v~’)

~ 1/...~.-.-r:~: m~

:~

T~=M~(v~’ ’- 3 ’)

(a) simple bonds

(b) vector bond

Figure 12.25: Stoichoimetric coefficients and reaction force The constraints of equation (12.134) are represented by the bondgraph of Fig. 12.25, in which part (b) gives a vector-bond-graphshorthand for part (a). effort e, knownas a reaction force or chemical al~nity, results naturally: (12.136) i=l

i-~l

The reaction force will be evaluated for the ideal gas and the ideal solution. For the ideal gas, Pi = ~i - 0gi, (12.137) in which the enthalpy ~i is a function of temperature only, while ~ is also a function of pressure. Further, the T ds relation 0 ds = dh - v dP

(12.138)

mandates OP~)o

so that, assumingalso the equation of state Pivi = #~ + RO

(12.139)

0 Pi = ROi,

_o___ Pi #~ + RO In

,

(12.140)

in which/2/0is the value of/2~ at temperature0 and reference pressure Po, which usually is taken as one atmosphere.A similar result applies to the ideal solution:

#~=#/0+ RO in

.

(1~.141)

HerePi is the molardensity of the ith species of solute or solvent, and p~0is the value of/2i at temperature0 and concentration t~0.

12.6.

SYSTEMSWITH CHEMICALREACTION*

947

The reaction force or affinity for the ideal gas nowbecomes u i -ui tt~+R~ln = _ E(,~ , --oROln[~(~) " - ~)~

/

(12.142)

The product e~ is an energy flux that generates entropy, nnd is a me~sureof irreversibility. 12.6.4

Chemical

Equilibrium

For chemicalequilibrium at temperature 0, no entropy ~s generated even in the presence of a virtual displacement5~, so that ~ = 0. Thus,

ln/~f’ ~¢I

~(~ , -o

/

(12.143)

wherethe subscript ¢ imitates equilibrium, With the definitions K~ ~ ~l~) K.e

,

~tE)

(12.144a)

’

(12.144b)

substitution of equation (12.143) into (12.142) gives ~ = -R~(InK -lnK~) -RO ln (~)

(12.145)

or

K~ = exp

(12.146)

- .

The subscript g h~ been dropped from K~ ~d K~e here, because the same result follows for the ideal solution if one substitutes the followingK~and K~, respectively: K~ ~ ~(~(~’-’~} ~o/ K~ ~--~(~(~7-G)

’

(12.147a) (12.147b)

K~ (and Kse and K~) is knownus an equilibrium constant. It is a readily calculablg function of temperature based on the thermodynamicproperties of

948

CHAPTER

12

THERMODYNAMIC

the components. For the ideal gas, for example, inverting gives 1 E(u~ _ u,i,)p Kge = exp ~-~

SYSTEMS

equation (12.143)

(12.148)

Knowledgeof the equilibrium constant permits calculation of the partial pressures or concentrations of the components at equilibrium. In general this is awkward, as the direct computation of the equilibrimn of any type of system often is awkward because of the simultaneous nonlinear algebraic equations. The approach here, as with non-chemical systems, is to simulate the behavior in time. Equilibrium is achieved ultimately in the absence of disturbances; if the details of the dynamics are of no interest, convergence to the equilibrium state can be hastened by assuming artificially high reaction rates. 12.6.5

Reaction

Rates

The actual reaction rates are given at least approximately by the law of mass action, which specifies that for a one-step reaction 7~ = V kf fil

L

i=l

-

~’’ \pO]

~ ,

--

(12.149)

b

in which kf and kb are the forward and backward specific reaction-rate coefficients, respectively. The reaction rate vanishes at equilibrium, so that the ratio of these two coefficients is

where,as before, ~he subscrip~e designates ~he equilibrium condition. As no~ed, employin~~he definition of K~ from equ~kion{1~.147a), ~here results ~ = VhyH(~ 5 (1The same form results

K~

{12.151)

%r the ide~ g~,

(12.1~2) and the subscripts s and g may be removed. Like Ke, k.f is a function of temperature which depends on the particular reaction. Normally it is given the form -E/ kf RO = .BOO~e

(12.153)

12.6.

SYSTEMS WITI-I

949

CHEMICALREACTION*

T~

T tt

~Ca -P

Figure 12.26: Reaction in a closed container E is called the activation energy; the Arrhenius factor e -E/RO is the fraction of the total numberof molecularcollisions betweenthe reacting chemicalspecies for whichthe energy is sufficient to induce a reaction. The balance of the right side is proportional to the numberof collisions. Along with the frequency factor B and the quantity a, E is a constant parameter determined by the nature of the reaction. They are commonlydetermined empirically; a is most frequently ~et equal to 0, =k½or 1. Substitution of equation (12.146) into (12.151) for the solution or (12.152) for the ideal gas reveals, respectively, ~ / = Vkfi=lLOo

- e-~/n°

,

(12.154a)

~=~kP0] ~ as the constitutive relation whichdetermines the reaction rate ~ as a function of the reaction force or affinity, the temperature and the densities or partial pressures of the componentspecies. Note that the factor in square brackets vanishes for ~ = 0, equals e/ROfor values thereof very small comparedto unity (near equilibrium), and reaches unity asymptotically for large values of e/RO.

12.6.6 Models of Reactions Without Mass Flows The energy flux eT~ is converted irreversibly to heat, muchlike mechanical workis converted to heat by friction. Like the friction example,therefore, this conversion is represented in our bondgraph by an RS element, as shownin Fig. 12.26. The generation of entropy is determinedby the required conservation of energy: ~7~-= 0~. The boundarybondon the top of the graph permits.,heating or cooling of the system. The boundarybond on the right permits the volume to change,as in the cylinder of an engine. It has been assumedtacitly that the chemicalreaction is a one-step process. Mostimportant reactions are not. For example, the reaction 2NO + 2H~ ~ N~ + 2H~O taken withv~o=2, U~o =0,H2 v’ =2, =O,u" =l,u’H~O =0, Ha u" =O,u’ N2 N2 u{}2o= 2 gives an incorrect reaction rate if interpreted with the equations above.

950

CHAPTER 12 THERMODYNAMIC SYSTEMS

Figure 12.27: Multi-step reaction in a closed container Twoseparate reactions in fact take place, for whichthe results apply separately: 2NO+ He ~ N2 + H2G~2 HeO2 + He ~ 2HeO The secondreaction takes place muchfaster .than the first; often it is assumed to be in equilibrium. Anynumber of elemental reactions can be accomodated without such an approximation,, however, by implementing the bond graph of Fig. 12.27. One reaction rate T/j and corresponding element RSj is used for each elemental reaction. The actual reaction rate in a combuster, for example, can be muchslower than the equations abovewouldgive, because the properties are not uniform over the volumedue to diffusion and mixing phenomena.In such cases empirically determinedrate coefficients might be used. Frequently, also, reactions occur so quickly that it matters little whatcoefficients are used, as long as they are not too small. In these cases it is the rate of the massflows into and out of the system that controls the dynamics. 12.6.7

Models

of Reactions

With

Mass Flows

A systemwith two in-flows and one out-flow is shownin the bondgraph of Fig. 12.28. The HS elements are employed as in a non-reacting CS element. The rates of change of masses in the control volumeare ~hv, and the mass fluxes undergoingreaction are rhr; in the absence of in-flows or out-flows these would be the same. The heart of the graph - the C and the loop with the two zero junctions, the two transformers and the RS element - is not affected by the mass influxes or effiuxes. Causal strokes are applied in the usual mannerto direct the routine writing of state equations. The addition of further mass fluxes can be accomodateddirectly. The output bond with a solid line between two parallel dashed lines is a generalization of the convection bond for two or more componentspecies. An

12.6.

SYSTEMS WITH CHElVlICAL REACTION*

951

Figure 12.28: Single-step reaction with two in-flows and one out-flow absence of internal diffusion is assumedin the definition. The numberof state variables equals two plus the numberof species present, or five in the example shown, including the three mass flows, pressure and one convected property. The causal stroke applies to the massfluxes and the pressure. Note that this bond can be reduced to a simple convection bond if the species are assumed to retain their proportions subsequently due to an absence of further chemical reaction or phase change. The system as represented can be defined as a single macro-elementfor computational purposes. A complicated system could include several st/ch macros. Distributed-parameter systems, such as a single spherical lump of coN, might be represented by several such macro-elements. The discussion above plus the examplesgiven in Section 12.5 for non-reacting systems indicate howthis can be done, but the detailed developmentof software is left as unfinished business. 12.6.8

Summary

The energy of a system comprising a well-mixed pure substance has been represented as a function of its total entropy, total volumeand total mass. When a mixture of potentiMly reactive substances is involved, the only difference is that the massesof the individual species are needed. The effort on the convection bond remains the total enthalpy, but this is decomposedinto the thermal term Osand the free energy term ~-~i #i, where#i is the partial Gibbsfunction (O~/Omi)v,s,m~e,.The stoichoimetric coefficients that constrain the changes in the composition, assuminga knownreaction, can be represented by a family of transformers that relate the individual #i and rhi to a common reaction rate and reaction force (~). The product ~7~represents the irreversible generation of heat and entropy. The assumption 7~ = 0 allows the conditions for chemical equilibrium to be deduced. A dynamicmodelfor a single reaction in a closed container is completed

952

CHAPTER 12 THERMODYNAMIC SYSTEMS

by adding an RS element to regulate the rate of the reaction and provide a location for the entropy generation. The most commonlyassumed model has been presented: the law of ~nass action. Multi-step reactions can be modeled by introducing separate families of transformers, each with its ownRS element to control the rate of its particular reaction. Influx and effiux of reactants and product can be treated in the samewaythey have been for non-reacting systems. The assumption of well-mixedspecies ma~vnot apply in practice except for small spatial domains. Individual trays in a distillation columnoften are designed to makethe assumptiona fair estimate. At the other extreme, a combustion process usually is controlled largely by complexturbulence and diffusion phenomenathat are well beyond the scope of this introduction to chemical kinetics.

Appendix

A

Introduction

to MATLAB

1 is a commercial interactive software package that is widely used for MATLAB numeric computation, data analysis and visualization. Its basic data element is an array that requires no dimensions, making it easy to use. It is especially well suited for the analysis of linear models. It performs symbolic math, although this feature is not used explicitly in this book. The most recent version at the time of writing is Version 6 (Release 12). Innumerable applications of MATLAB appear in a large number of "Toolboxes" that are marketed separately. The Student Edition is claimed to be full-featured regarding the basic MATLAB; several of the toolboxes also are available for use with this edition, with some restrictions. Some of the commands in this book may require the Control Systems Toolbox or the Signal Processing Toolbox, depending on the version of MATLAB being used. The on-line help within Version 12 of MATLAB is extensive. A wide 2selection of texts present applications of MATLAB. Most of the use of MATLAB in this book regards simple-to-use commands described in a largely self-contained wayin the following sections: Topic

Sections

nonlinear simulation transfer function transformations linear simulation roots,.factorization, eigenvalues, eigenvectors step, impulse responses; partial fraction expan. Bode plots discrete convolution Fourier analysis root locus, Nyquist and Nichols plots thermodynamic properties and system simulation

3.7.2, 5.1.4 5.3.6, 6.2.3, 6.2.6 5.3.7, 5.4.8 6.1.1, 7.3.8, B.2.5 6.1.9, 6..3.1, 6.3.2 6.2.2, 6.2.7 6.4.3 7.1.4 8.1.6, 8.3.2-8.3.5 12.5.6, Appendix D

1Aproduct of The MathWorks,Inc., 24 Prime Park Way,Natick, MA01760. 2A list of hundreds of MATLAB-based books is maintained on the MathWorksHome Page, w~.raa~h~o~:l~s.c~a., including MathWork’s ownLearning MATLAB, Release ~2, F~. B. Magrabet al, An Engineers Guide to MATLAB, Prentice Hall, 2000, and A. Biran and M. Breiner, MATLAB 5 .for Engineers, 2nd Ed., Addison-Wesley,1999. 953

954

APPENDIX A brief

Scalar

A

primer on the ruidments of MATLAB follows.

Calculations

MATLAB can be used like a simple calculator. For example, to evaluate (5/7r 2) sin(3) you enter, in response to the MATLAB prompt (>>), >> 5/pi’2*sin(3)

0.0715 The answer is placed in a. default variable "arts", and is placed on the screen as shown. Expressions are evaluated from left to right with power operation having the highest order of precedence, followed by multiplication and division, and trailing with addition and substraction. The elementary math functions treated are given in Table A.1

Variables Variables, such as the x, y used in the table above, can be defined by expressions. For example, the same answer as above results from >> omega=3 ; ¯ >> T=I ; >> x=5/pi ~ 2. s in (omega*T) except that it is stored in the variable x. The semicolons at the ends of the first two statements inhibits printing of their numerical values. The subsequent typing of >> omega,T,x givestheresponse omega = T= 1 X----

0.0715 Thevariables thathavebeendefinedat any pointin yoursession willbe displayed" if youenter >>who Your variables are omega T x Theinstruction <whos>alsogivesdetails regarding thenatureof thevariables. Variables are case sensitive, must start with a letter, and are limited to 19 characters.

INTRODUCTION Table A.1 abs(x) acos (x) acosh(x) angle (x) asin(x) asinh (x) atan(x) atan2(x,y) atanh (x) ceil(x) conj (x)

cos(x) cosh(x) exp (x) fix(x) floor(x) imag(x) log(x) loglO(x) real(x) rem(x,y) round(x) sign(x) sin(x) simh sqrt(x)

tan(x) tanh(x)

TO MATLAB

955

Elementary Math Functions absolute value or magnitude of complex number inverse cosine inverse hyperbolic cosine angle of complex number inverse sine inverse hyperbolic sine inverse ta.ngent four quadrant inverse tangent inverse hyperbolic tangent round towards plus infinity complex conjugate cosine hyperbolic cosine ~) exponential (e round towards zero round towards minus infinity complex imaginary part natural logarithm common logarithm complex real part remainder after division of x/y round towards nearest integer signum (sign) function; returns 1, 0 or sine hyperbolic sign square root tangent hyperbolic tangent

Complex Numbers If you enter >> a=2; b=3; c=4; >> y= (-b+sqrt (b^2-4*a*c))/2/a; the response to typing is y= -0.’~500

+ 1.1990i

where i is the unit imaginary number. Imaginary numbers can be entered by using the identifier or <j >, as in standard mathematical notation: p=5+2i; q=6-3j*sin(2) Starting with Version 4, MATLAB does not require an asterisk between a number and the i or j. Other predefined variables are (for co) and (for 0/0).

956

APPENDIX

A

Arrays and Matrices A row array can be entered as r=[l 2 3]; and a column array as either of the forms >> c=[1;2;3]; >> c=[1 2 3]’; The prime (’) denotes transpose. Entry of a general array is illustrated A=[1 2 3;4 5 6] A= 1 2 3 4 5 6 Elements in matrices may be imaginary or complex. When arrays are added, substracted, or multiplied with <+>, <->, <*>, they are treated as conventional matrices. On the other hand, array multiplication, array division and array power, indicated by the commands <.*>, <./>, <.’>, simply perform the indicated operation on the respective elements in the array, ~. e. p=[4 5 6]; >>p. *q

q=[3 2 1];

= allS

12 10 >>p./q

6

= 8_*1S

1. 3333 2. 5000 >>p.-q

6. 0000

= axis

64

25

6

The dot and vector products are given by, respectively, >> p*q’ = 8.D.S

28 >>x=p’*q = X

12 15 18

8 4 10 5 12 6

The commandsize(A) returns a two-element vector stating the numbers of rows and columns in the matrix A, respectively. Thus, if A is any 2 × 3 matrix, size(A) returns the vector 2 3, size(A, 1) returns the 2 and size(A,2) returns the 3. Further, the statement y=x(size (A, 1),

INTRODUCTION

TO MATLAB

957

in which x is the 3 × 3 matrix defined above, returns the row vector y= 15

10

5

The size argument indicates the second row of x, and the wild card (:) indicates all columns. To save time, the special matrices illustrated below are Offered. AssumingA has been defined as a 3x3 matrix and B as a 2x4 matrix, >>zeros (size = D/IS

(A))

0 0 0 0 0 0 0 0 0 >>ones(size(B)) = aiis 1 1 1 1

1 1 1 1 >>eye(size(A)) D/IS=

1 0 0 0 1 0 0 0 The linear

matrix equation Ax = b,

with A and b known, can be solved for x two ways, assuming A is square and nonsingular: >>x=inv (A)*b >>x=A\b; For example, if A=

5 8

;

either commandwill give X---

1. 0000 2. 0000 3. 0000 The determinant of A is readily found, also: >>det = ans

(A)

b=

32 , 23

958

APPENDIX

A

The use of the backslash operator \ is preferred. It employs LUfactorization, which is a modification of Gaussian elimination and produces greater accuracy with fewer internal multiplications. When there are more equations than unknowns, so A has more rows than columns, the command/~\b gives the least squares solution. This can be extremely useful. The command±nv(A)*b, on the other hand, produces an error message.

Evaluating

and Plotting

Functions

The array operations on vectors are very useful in evaluating and plotting functions. As an example, the time response

x = (1/cos¢)e-;~’*

cos(~/~-

~2 wnt + ¢);

¢= - tan-l(~/v~-

with ~" = 0.2 and Wn- 2~r and 101 values of t evenly spaced between 0 and 4, can be evaluated as follows: >>z=0.2;om=2*pi; >>omd=sqrt(l-z’2)*om; phi=-atan(z/sqrt(l-z’2)); >>t=[0:.04:4]; >>x=exp(-z*om*t).*cos(omd*t+phi)/cos(phi); Notice that only one array operation is needed; this operation is unnecessary for multiplying or dividing a scalar and an array. The 101 resulting values of t and x are stored in the row arrays t and x, and will be displayed by entering these variables. You can secure a plot of x versus t as shown at the top of the next page: >>plot(t,x) >>xlabel(’t’);ylabel(’x); >>title(’homogenous responsewith zero intial velocity’) Semilog and log-log plots to the base 10 also can be made; the commandsare <semilogx(x,y) >, <semilogy(x,y) >, and . The command draws grid lines, and the command places the word "text" (or your substitute) at (pl,p2) in screen coordinates where (0.0,0.0) is the lower left and (1.0,1.0) is the upper right of the screen.

INTRODUCTION

959

TO MATLAB homogeneous response withzeroinitiol velocity

t 0.8 0,6 0.4 x

0.2 0 -0.2 -0.4 -0.6

’ 0.5

1

1.5

2

2.5

3

3.5

t

Fitting

Curves

to

Data

The characteristic of~a source, load, compliance, etc. often is observed experimentally, and you wish to represent it with a polynomial expansion for use in simulation, or perhaps just for curve plotting. This is an example of a task for which the MATLAB help feature, particularly in Version 6, is a sufficient guide. You find the information by clicking on "MATLAB Functions Listed by Category," then "Mathematics," then "Polynomial and interpolation functions," then "Polynomials," and finally "polyf it polynomial curve fitting." You learn that the command [p] =polyf it (x,y, fits the coefficients of a polynomialp(x) of degree n to the data, p(x(i)) to y(i), in a least squares sense. The result is a row vector of length n + 1 containing the polynomial coefficients in descending powers: p(x) = plx ’~ + p~x"-1 + ...

+ Pnx + Pn+~.

A complete example is given. You also learn how to use the command f=polyval (p, z) to compute the values of the polynomial for values of a vector z, which could be considerably longer than the vector x. The illustration includes use of the plot commandfor both the original data points and the resulting polynomial curve. You also are told that the algorithm used by polyfit makes use of the backslash operator, \, as described above. The meaning and use of further optional output arguments of polyfit also are described.

960

Control

APPENDIX

A

Flow Commands

Statements can be executed conditionally form is as follows:

with the ±f command. The simplest

±f condition statement; action statements; end; A more complicated form is as follows: if condition statements; action statements; elseifcondition statements; else action statements; end; The for commandallows statements to be repeated a specifed number of times. It is most commonlyused relative to indices, and often is used in conjuction with the ±f command. An example using for, if, else and follows: n=5; for i = l:n for j = l:n if j > i A(i,j) = 2; elseif j == i A(i,j) 1; else A(i,j) = 0; end; end; end; The resultisthematrix

A=

122 0 1 2 001 000

,

(which is small enough to have been entered directly with less fuss.)

Script Files When interactions with MATLAB are not particularly short or are repetitive, it is desirable to write programs off-line. Files executable by MATLAB are called M-files, since they must have the extension .m. The toolboxes such as the signals and systems toolbox and the symbolic math toolbox that may come

INTRODUCTION

TO MATLAB

961

with the student version comprise a collection of Mofiles; these subprograms represent most of the ultimate power of MATLAB. A script is one type of M-file. Scripts are ordinary ASCII text files and can be created off-line with nearly any text editor and stored on your disk. You will need to make sure that MATLAB has a path to the location of your file, which you might transfer to or write in the location C: terap. This can be accomplished within MATLAB by clicking on f±le and then set path. The procedure from here depends on the version of MATLAB.Back in the CommandWindow of MATLAB, you can open an existing m-file or write a new one after clicking file. The MATLAB Editor/Debugger should appear, and you should save the file to the location C:temp (or whatever location you have placed in the path). Subsequent transfers between the Editor/Debugger and the CommandWindow are executed merely by clicking on their windows. Don’t forget to save any changes you make in a file before attempting to run it. A script file can be run merely by typing its name, without extension or location, into the Command Window, and pressing Return. Test the system by entering the following program: x=[1:360] ; z=exp (j*pi*x/180) plot (imag (z), real (z)) If you name the program myprog, m and type "myprog" in response to the MATLABprompt, you should see an ellipse; the imaginary part of the argument is plotted versus the real part. The plot can be annotated to your taste. The subsequent command plot(x,imag(z)) plots one cycle of a sine wave. Data

Files

The command save ]~lename x,y,A writes the content of the variables x, y A into a binary file with the name filename .mat. If no variables are specified, all workspace variables are saved. If no filename is specified, the default file nameis matlab .mat. The command load ]ilename x, A retrieves Function

the indicated data. The same defaults apply. Files

There are two types of M-files: the script files described above, and function files. A function file is identified by the the word " function" at the beginning

962

APPENDIX

A

of its first line, followed by a statement in equation form that includes the name of the file. The example given in Section 3.7.3 starts with functionf = pendulum(t,x) Unless special provisions are made, the arguments t ,x are the only variables communicated from the main progra~n to the function. The variable f is the only variable returned to the main program. All other variables defined in the function file are hidden from the main program. If more than one output variable is to be computed, the list should be placed in square brackets. As an example, function

[y,x,t]

= step(num,den)

(see Section 6.1.9) is the first line of one of the hundredsof function M-files built into MATLAB. The variable names y, x, t can be changed when the function is called. Functions can be evaluated through the use of the command [yl .....

yn] = feval(’]unction

file

name’ ,xl .....

xn)

The variables xl ..... xn are communicatedto the function file, and the variables yl ..... yn are communicated back to the calling file (or the command window). The left side can be omitted if no returns are intended.

Communication Between Files An M-file can call another M-file. The statement in Section 3.7.3 It,x]= ode23(’pendulum’ ,O, [O 3],[0 pi/18]); calls the built-in function M-file ode23, which in turn calls the user-written function M-file pendulum. In this example, the parameter values L, W and I are redefined every time pendulumis called. This is not efficient. Repetitive definitions or calculations can be eliminated through the use of global variables. Those functions that declare a particula~ name as a global variable, and the base workspace if it makes the same declaration, share a single copy of that variable. The simulation described in Section 3.7.3 becomes more efficient, therefore, if you write a script M-file named "pendulumsim" and a shortened function M-file as follows: Y. ScriptM-file ’pendulumsim’; global L W I Y. global variables L = 1; W = 1; g = 32.2; I=L^2*W/(3*g); It,x] = ode23(’pendulum’,[0 3],[0 pi/18]); plot(t,x(: ,2).180/pi) xlabel ( ’time,seconds ’ title(’Angleof swinging pendulum’)

INTRODUCTION

963

TO MATLAB

function f = pendulum(t,x)~ Function M-file computes global L W I ~ global variables ~ The two differentialequationsfollow: f(1) = -W*(L/2)*sin(x(2)); f(2) = x(1)/I; f=f’; The entire computation and display follows from entering the name of the script M-file into the commandwindow in response to the coInmand prompt. The use of global also allows variables to be communicatedfrom the function file to the commandwindow or the script file, which can expedite data output. Sometimes you will like to plot the derivative of a state vector, but the function f’ is not communicated to the controlling script file. Further, the function M-file evaluates f several times per time interval; a plot of all these points would display undesired and misleading scatter. The best you can do is to plot the differences between successive values of the state variable, divided by the corresponding differences in time. The simplest way to accomplish this is by using the MATLAB function diff. For example, to plot the moment on the pendulum above, namely dxl/dt, you could enter the command plot(t(l:size(t)-l),diff(x(:,l)/diff(t)) Note that if the vector x(:, 1) contains n elements, the vector diff(x(:,l) contains n - 1 elements. The vector for the abcissas of the plot, which is the time t, must be shortened by one element for the plot routine to work. This explains the use of the size (t) command. PROBLEMS A.1 Use MATLAB to find the magnitude numbers (a) 3- 4j, (b) 0.5 + A.2 Find the real and imaginary parts e-’75j=, (b) j(-5=/4).

and phase angles

of the complex

of the following complex numbers:

Give MATALB coding for the following (but do not run): (a) If d < 3, set r = 0; otherwise, set r = 2d. (b) If x- y > 2, print x and (c) If the natural logarithm of x is between 2 and 10, increment the index m by one.

964

APPENDIX A

A.4 Enter the following matrices into MATLAB:

A=

1 ; .

B=

1

;

C=

-1 0

;

b=

.

-1. (a) Using MATLAB, compute the matrices AB, BC’, and ((A’C) (b) Using MATLAB, solve the equation Bx = b for A.5 Plot the characteristic of the hydraulic source as given in Problem2.15 (p. 55). On a second graph, plot the powerproduced by this source as a function of flow. A.6 The IC engine in Problem 2.16 (p. 55) has the nonlinear torque-speed characteristic plotted in Figure 2.39. (p. 75). (a) Read a few values of the shaft speed and the associated torque, and plot these points using the ’o’ argument, of the plot command.Holdthis plot. (a) Makeand plot three polynomial approximations to this characteristic: (i) a linear model(equation); (ii) a second-order(quadratic) (iii) a third-order (cubic) model. Plot these polynomialsas essentially continuous curves on the same axes as the data points from part (a). (b) Locate the points for whicheach of the modelsin part (b) give maximumpower. Hint: Type "help max" in the commandwindow.

Appendix

Classical

B

Vibrations

Longbefore the state-space methodsfor modelingand analyzing dynamicsystems given in this textbook were promulgated,engineers addressed the relatively narrowclass of modelscomprisingvibrating inertias interconnected by springs. Important applications were made to structures and machines. The methods that evolved are simple and insightful, and still have relevence today. On the other hand they do not extend your capabilities, and therefore maybe considered optional. The approach in the tradition of dynami c systems, emphasizedin this book, treats linear modelswith the first-order matrix differential equation ~ = Ax + Bu(t).. (B.1) The approachin the tradition of classical vibrations, on the other hand, treats linear modelswith the second-order matrix differential equation Mi~ + C~ + Kx = F(t). (B.2) This latter approach has the advantage of dealing with matrices and vectors with half the dimensionof those used in the correspondingfirst-order models. Also, Mand K and often C are symmetric matrices, further simplifying the analysis, whereasA generally is asymmetric. The purpose of this appendixis to explain and illustrate the application of equation (B.2). This is done for modelswith two or moredegrees of freeedom(of fourth or higher order), since the differences for modelswith one degree of freedomare not significant. Usually it is assumedthat the matrix Mis invertable, whicheliminates first, third and higher odd-orderedmodelsfrom consideration, a serious restriction that motivates the tradition of dynamicsystems to use equation (B.1) instead.

B.1 Models With Two Degrees

of Freedom

The vibration absorber analyzed in Example6.9 (pp. 416-418) is repeated Fig. B.1 with the displacementsof the centers of mass, Xl and x~_, indicated as 965

966

APPENDIX

B.

Figure B.I: System with vibration the principle variables. of motion ml

q-

0

Direct application

m~ 0

~1

m3 5~

[ kl

-~

[ -k2

CLASSICAL

VIBRATIONS

absorber

of Newton’s law gives the equations

k2

k~

x~

The first square matrix in this classical representation is known as the mass matrix; no dynamical coupling exists in this example since this matrix is diagonal. The second matrix is known ~ the stiffness matrix; the non-diagonal terms here ~e said to impose static coupling. Were there no such coupling, the problem could be treated as two independent single-degree-of-freedom problems. B.1.1

Normal

Mode

Vibration

In the unforced case (Fe = 0) the model can oscillate

Substitution

sinusoidally:

of these assumed motions into equation (B.3) gives

The determinant of the square matrix must vanish, giving the characteristic equation A =_ (ml + m~)m3w4 - [(kl

+ k~)m3 + k,~(ml + m~)]w2 + klk.2 = 0. (B.6)

This equation is quadratic in terms ofw’-), giving two real roots. The two natural frequencies are the two square-roots of these roots, which are called wl and w2 below, and are presumed to be positive.

B.1.

MODELS WITH TWO DEGREES OF FREEDOM

967

The sinusoidal motionsassumedabovedo not represent the general solutions, but rather are special cases called normal modes. If you substitute w = Wl or w = w~. into equation (B.5) you get the amplitude ratios

(~xlo (Xlo~

~ : x_~o/~ k~ + k2 - (ml + rn2)w~

~ m3~ ~

k.2 = x20/~: kl + k2 - (ml + m2)w~

k2

’

(B.7a)

(B.Tb)

Thusthe amplitudesof the two displacementshave a fixed ratio that is different for the two modes, and along with the respective frequencies serves to define the modal motions. The general unforced motion comprises a sum or superposition of the two modalmotiofls: Xl = Xll

sin(wit + ¢1) + x~2sin(~v2t +

x2 = x21sin(wit +

(~1)

Xll \ x~° | xl0 } , ~lX:21;

-[" X12

x22

(B.8a)

sin(w2t+ ¢2) X22

(B.8b) (B.8c)

Thus there are two independent undetermined amplitudes, which can be taken to be x12 and x2e, and two independent phase angles, ¢1 and ¢_~. These amplitudes and angles are determinedby the initial conditions Xl(0), 5:1 (0), x2 and 2(0). Unlike the general analysis above, the results of Example6.9 are specialized to the case of ma -- 0.2(m~+ m2) and ko_ -- 0.2k~. In this case, the natural frequencies are w~ -- 0.80109worad/s and w~ -- 1.24830wotad/s, where wo = v/k~/(ml +ra~). The modal shape ratios are (x~o/X2o)~ = -2.209 and (xlO/X2o)~: -6.791. Th is means th at if thesyst em is s et oscillating at 0.80109worad/s, the larger massexhibits simple sinusoidal motion 180° outof-phase with the smaller mass, and with an amplitude 2.209 times larger than that of the smaller mass. At 1.24830w0,the motionis similar except the ratio of amplitudes is 6.791. In general, the unexcited motionrepresents a sumof these two modesand two frequencies, with proportions to suit the initial conditions. B.1.2

Forced

Harmonic

Motion

The particular solution of the system described by equation (B.3) corresponding to the forcing function Fe = m~r~2 coswt (B.9) has the sameform as the special solutions of equation (B.4),

968

APPENDIX B. CLASSICAL VIBRATIONS

although w is nowthe forcing frequency rather than a natural frequency. Substitution of this solution into equation(B.3) gives

A

k2 J

(B.11)

where A is the determinant defined in equation (B.6). The result for Xl(t) corresponds to the G(S) used in Example6.9, but allows any set of physical parameters. In the numerical examplediscussed above and plotted in Example6.9, the motion of xl ceases whenthe excitation frequency is precisely V~’2/m3= wo. This is the purpose of the vibration absorber.

B.2 Higher-Order B.2.1

Modal

Models

Motions

The general linear undampedand unexcited vibration problem with n degrees of freedom can be represented by the matrix equation M~ + Kx = 0,

(B.12)

in which Mis the inertia matrix and K is the stiffness matrix. The matrices Mand K are symmetric. Premultiplying equation (B.12) by -1, ii

+ Ax = 0, A = M-1K.

(B.13)

As with the two-degree-of freedommodel, special modalsolutions of this equation exhibit simple in-phase sinusoidal motions at the frequencies wi, where i = t,...,n. The assumption x(t) xisinwit,

(B.14)

wherex~ is a vector of constants and ~9i is a constant, gives uponsubstitution into equation (B.12) (A - o~I)xi = 0. (B.15) The characteristic equation

~ -- IA- ~II = 0

(B.16)

gives n eigenvalues w~. Substitution of an eigenvalue into equation (B.13) gives the proportions of the corresponding eigenvector, xi. That is, the eigenvector can have any magnitudebut the ratios of its membersare fixed. The eigenvector describes the modalshape of the vibration at the special frequency wi.

B.2.

HIGHER-ORDER

969

MODELS

It is convenient to fix the length of the eigenvector xi at somevalue, and then represent smaller or larger versions of the same eigenvector as cixi, where ci is the appropriate coefficient. This practice is e~nployed below. Sometimes, the first element in x~ is set equal to 1, and the other elements scaled accordingly. At other times, particularly in standard software packages such as MATLAB that handle the extra computation easily, the length of the eigenvector is set equal to 1; this length is the square-root of the scalar product, x~xi, where the prime (’) means transpose. This second option, unlike the first, handles the special case in which the first element in xi is identically zero. Either scheme can be applied to the results below. A model given an initial position corresponding to a particular ~node shape, with zero velocity, or an initial velocity corresponding to the modeshape with zero position, Will experience the motion of that modeexclusively. An arbitrary initial condition, on the other hand, likely excites all the modes, some more than others. The eigenvectors possess a very important property that now will be established. For x equalling the ith eigenvector, equations (B.12) and (B.14) w~Mxi = Kx~.

(B.17)

Premultiplying both sides by the transpose of x~, 2i wi x~Mx~-- x~gxi.

(B.~8)

Taking the transpose of this equation, noting the general theorem that the transpose of a product of three conformable matrices equals the products of the respective inverse matrices in reverse order, and noting that Mand K are symmetric, w~x’~Mxj = x’~Kxj (B.19) Equation (B.18) also can be written with the roles of i and j reversed: w~xiMxj ---Equation (B.20) is now subtracted

x~gxj.

(B.20)

from equation (B.19) to yield

(w~ - w~)xiMxj =

(B.21)

As long as wi and wj are two different natural frequencies, it follows that x~iMxj-- O.

(B.22)

By the same approach it can be shown that x~Kx/= 0.

(B.23)

It is said that the eigenvalues are orthogonal with respect to M and K. On occasion, an eigenvMue is repeated. In this case there are two eigenvectors associated with a commoneigenvalue; no unique mode exists corresponding to this frequency.

970

APPENDIX B. CLASSICAL VIBRATIONS

Finally, whenj = i the terms corresponding to the left sides of equations (B.22) and (B.23) do not vanish. Instead, we define the generalized mass, /~fi, and the generalized stiffness,/(i, as follows: Mi = x~Mxi, Ki = x~iKxi. B.2.2

The Initial

Value

(B.24a) (B.24b)

Problem

The general solution to the initial value problemis given by the sumof all its modalcomponents,each with its modalamplitude ci and phase angle ¢i, or its componentsof amplitudes ai and hi: x = Exicisin(wit+¢i)

=

i=1

xi(aisinwit+bicoswit).

(B.25a)

i=l

bi ) ¢i = tan -~ ~

(B.25b)

The initial conditions are x(0) = b~xi,

(B.26a)

i=1 :~(0)

= E coiaixi.

(B.26b)

i=1

The solution of equation (B.25) is completed by evaluating ai and bi, i = 1,.-.,n, as functions of x(0) and ~(0). This is accomplished by multiplying both sides of equations (B.26) by x}M: x}Mx(0) = E bix}Mxi

(B.27a)

i=1 n

x M(0) = ia x}Mxi i=l

In view of the orthogonality relation of equation (B.22) and the definition equation (B.24a), the right sides of equations (B.27a) and (B.27b) equal, spectively, bjMj and wjajMj. Therefore, the desired result is bi = ~,-~/x~gx(0), 1 ai = --x~M~(O).

(B.28a) (B.28b)

B.2. B.2.3

HIGHER-ORDER Forced

971

MODELS

Response

The coupled set of differential F(t) is written

equations in the presence of a forcing function M~ + Kx = F(t).

(B.29)

The solution is expedited by tranforming the variables so as to achieve a set of uncoupled differential equations, one for each mode. The complete solution then becomes the sum of the individual modal solutions. The transformation is written x = Py. (B.30) To make this idea work, P has to be a very special matrix, called the modal matrix. The n columns of P are the n eigenvectors, xi: (B.31)

P = [x~ x2 ...xn].

Note that the transposed matrix P’ comprises the same eigenvectors arranged, instead, in rows. Substitution of equation (B.30) into equation (B.29), and subsequent ~ multiplication by P’, gives P’MP~ + P’KPy = P’F.

(B.32)

The element on the ith diagonal element of P~MPis identically Mi, and all offdiagonal elements are identically zero. Similarly{ the element on the ith diagonal element of P~KPis identically Ki, and off-diagonal elements are identically zero. These results follow directly from the orthogonal properties and definitions given above. Thus, equation (B.32) represents n uncoupled second-order differential equations Mi~/i + Ki~/~ = [P’F(t)]~; i = 1,...,n. (B.33) Oncethese equations are solved, the desired response x(t) is found by substitution into equation (B.30). B.2.4

Modal

Damping

The equation of motion with linear

damping introduced

M~ + C~ + Kx = F(t).

becomes (B.34)

Use of the modal matrix P and transformation x = Py based on zero damping gives P’MP~} + P’CP:9 + P’KPy = P’F(t). (B.35) 1Theparallel use of the modalmatrix in Section 7.3 for the first-order model~ -- Asemploys the inverse matrix p-1 in place of the present transposematrix P’. Onlyin the present case is the matrix A. generally symmetric;as a consequenceof this and the normalizationof the component eigenvectorsin P, there results P-1 = P’. Herein,P’ is used, since it is easier to find than P-~.

972

APPENDIX B. CLASSICAL VIBRATIONS

The matrices P~Mpand P~KPare diagonal, as before, but in general P~CPis not; the result is modalcoupling due to damping. In manycases, nevertheless, P~CPis indeed diagonal, and the modelis said to have proportional damping. This happens when C = aM + ~K

(B.36)

for any values of the constants a and ft. The non-zero diagonal elements are [P’CP]i = aMi + ~3Ki,

(B.37)

so that the separate modesare described by ~liiji + (aMi+ ~3Ki)~+ K0’i : [P’ F(t)]i,

(B.38)

or

’~;Yi + fli

(B.39a)

+ Yi = [P’F(t)]i,

wi = X/’-~i/M~; ~i = -~

+/3wi ¯

(B.39b)

Manysystems are lightly damped. Light damping does not produce much modalcoupling, even if equation (B.36) does not apply. Nevertheless, damping can be of critical importancein limiting the amplitudeof resonances. It is often reasonable, therefore, to approximate dampingby a proportional model. You should understand, however, that the approach of the dynamicsystems tradition~ whichis based on matrix first-order differential equations allows, perfect modaldecoupling regardless of the form of the linear damping. B.2.5

Example

Using

MATLAB

A ten-story building is shakenby horizontal groundmotion; the resulting motion is sought. The building, pictured in part (a) of Fig. B.2, is modeledby ten masseseach equal to m= 5 × 105 kg, ten springs for lateral displacements each equal, to k = 1 × 109 N/m, and ten corresponding dampingcoefficients each equal to c = 1 × l0T N s/m. The ground is assumedto have infinite stiffnesses in translation and rotation. The equation of motionfor the first floor is miil + 2cicl + 2kxl = c~2 + kx2 + c~.o ÷ kxo.

(B.40)

For the top floor the equation is

(B.4~)

mY~lo+ c5:~o+ kx~o = c~:9 +kxg, and for floors 2 < i < 9 it is m~,i

+ 2c’,~i

+ 2kxi

: c(~i+l

+ xi-1)

+ k(Xi+l

.-b

xi-1).

(B.42)

B.2.

HIGHER-ORDER

973

MODELS 10 9 8 floor

7 6 5 4 3 2 1 0

-0.4

(a) structure

0

0.4

(b) modes

Figure B.2: Ten-story building and lowest-frequency The matrices

M, C and K therefore

mode shapes

become

M=mI

(B.43a) -1 2 -1 0

K=k

0 0 0 0 0 C = (c/k)g.

0 0 0 0 0

0 -1 2 -1 0 0 0 0 0 0

0 0 -1 2 -1 0 0 0 0 0

0 0 0 -1 2 -1 0 0 0 0

0 0 0 0 -1 2 -1 0 0 0

0 0 0 0 0 -1 2 -I 0 0

0 0 0 0 0 0 -1 2 -i 0

0 0 0 0 0 0 0 -I 2 -1

c=leT;

K=k*[2 -1 0 0 0 0 0 0 0 0;-1

(B.43b)

(B.43c)

Thisin~rmationcanbeenteredintoaMATLABm-filebu±ld.mas~llows: m=5e5; k=le9; M=m*eye(10);

0 0 0 0 0 0 0 0 -i I

2 -1 0 0 0 0 0 0 0;

0 -I 2 -i 000000;00 -i 2 -I 0000 O; 000-12-10000;0000-12-1000; 00000-12-100;000000-12-10; 0000000-12-i;00000000-11];

974

APPENDIX B.

CLASSICAL

VIBRATIONS

C=c/k*K; A=inv(M)*K; [P,W]=eig(A); GM=P’*M*P;GK=P’*K*P;GC=P’*C*P;GF=P’(I,:)’; Notice the commandsfor the inverse of a matrix and ~r the modal matrix, P, and eigenvalue matrix, W. The matrix Wis square and diagonal. If upon running MATLAB you respond to the prompt as follows, >>build >>diag(W)’ the responseis 1.0e+003 * Columns 1 through7 2.0000

1.0678

Columns 8 through 6.4940

7.3050

0.3961

0.0447

3.1099

4.2989

5.4614

iO 7.8223

The lowest-frequency mode is the fourth in the order chosen by MATLAB, the second lowest mode is the third, and the third lowest modeis the second. 2 The frequency of the losest mode, in Hz, can be found by entering EDU>> sqrt (W(4,4))/2/pi 8/23 =

1. 0638 The eigenvec~ors corresponding to these three modes can be found by entering >> [P(:,4)P(:,3) P(:,2)] 0.0650 0.1894 -0.2969 0.1286 0.3412 -0.4352 0.1894 0.4255 -0.3412 0.2459 0.4255 -0.0650 0.2969 0.3412 0.2459 0.3412 0.1894 0.4255 0.3780 0.0000 0.3780 0.4063 -0.1894 0.1286 0.4255 -0.3412 -0.1894 -0.4063 0.4352 -0.4255 These modeshapes are plotted in part (b) of Fig. B.2. 2Theorderofthe

modesvariesdependingontheversion

of

MATLAB.

B.2.

HIGHER-ORDER

MODELS

975

The elements Mi, Ki and C~ of the generalized mass, spring and damping vectors are the diagonal elements on the square diagonal matrices GM,GKand ~¢. The values for the lowest three modes can be displayed as follows: >> [GM(4,4)GM(3,3) 8_n.S =

1.0e+005 * 5. 0000 5. 0000 5. 0000 >> [GC(4,4) GC(3,3) GC(2,2)] a.D.S =

1. Oe+O06* O. 9.234 1.9806 5. 3390 >> [GK(4,4)GK(3,3)GK(2,2)] D/2S =

l. Oe+O08* O. 2234 1. 9806 5. 3390 >> [GF(4) GF(3) GF(2)] allS =

0.0650 0.1894 -0.2969

The information above can be assembled as follows to give the equations of motion for the first three modes: 5 × 105#1 + 0.2231 × 1069~ + 0.2234 × 10Syl = 0.0650(109k0 + 107Xo) 5 x 105~)o. + 1.9806 x 106~2 + 1.9806 x 10Sy2 = 0.1894(10950 + 107x0) 5 x 105#3 + 5.3390 x 106#3 + 5.3390 x 10Sy3 = -0.2969(109~0 + 107xo) (B.44) Thee actual displacements at the various floor levels are 10

xi = xo + Z PijYj

(B.45)

The behavior of the modes as a function of frequency can be seen from Bode plots; the lowest three modesare given in Fig. B.3. The first of these plots will be displayed as a result of following commands: >> nl=[0.0650e9 0.0650e7] >> dl= [5e5 0.2234e6 0.2234e8] >> bode(nl,dl) >> hold

976

APPENDIX B. CLASSICAL VIBRATIONS 5o

-..50 -~ 10

I I IIIIIII

I I IIIIIII

I I IIIIIII

I

I I flllll

I

I I Illlll

I

I I IIIIII

I .I

I IIIIII

I

I ,llliH

I

I

I I I I ! ,

IIIIIII I I I IIIIII~..,~-~"~~ I IIIIII I I I_~J.,I-~1"I"~~IIIII I IIIII~~ I I IIIIII I l~i~lllll I IIIIII I I I IIIIII ! ! ::~__~-,~] ~ I II IIIII ,,~illl I I IIIIIII 10-2 lO-t 100

I I I I ! ,

I

I I I1~111,, I 1611111 I I J,,~lll~l,,,../

Ill~k.l~~l I IIIIII I IIIIII I IIIIII I IIIIII I IIIIII I I IIIIIII 101

Frequency (rod/sec)

I,,,,’N~P’~LIIII .~ I I I I I I I I Ii !i I

I IIIIII I I IIIII I I IIIII I I IIIII I I IIIII ~ I 1,111 10

II

°. note: Thephaseofy~shouldbe shiftedby 180 100

I I

I Jllllll I~.L..LLIIIII i i iiiii1~,---1--i i Iiiiii

I

I I IIIIII

I I I Illlll

’, I I’,;;;’,’, I I i iiiiii

’, I ’~.~I’.I’.~LY~I IIIIII i i i1~1111 -~1~1 iiiiii

I

I

I IIIIIII

I

I1|111

It, I~[ IIIII

JALIIII__L.J_LIJII.U___L_L.I_LIIUL_J_J._ALI __ ~. -lOO___I_J. I I I IIIIII I I I IIIIII I I I IIIIII I I IIIIII I

_ I IIIIII I I I IIIIII I I I IIIIII I I I IIIIII I I IIIIII I I I IIIII I I I IIIIII I I I IIIIII I I I IIIIII I I IIIIII I I I IIIII TiTI---~-i-T TITRr--i--f--, ~ T~T~T--r-i-i-I-~ T,T -200 "---I-~ 7 7 Tl-rlT--F-i-f-i7 10-3 10-2 10-1 100 101 10 Frequency(rod/sec)

Figure B.3: Bodeplots for the three lowest vibrational modesof the building The commandhold allows the subsequent Bode plots to be superimposed on the first. Repeatingthis command releases the plots. Note that if the frequency content of an excitation does not exceed the resonant frequency of the third mode,there is little need to include morethan the three modesin the solution.

B.2.6

Mode Reduction

The response of a high-order modelis found routinely using digital computation, as the above examplesuggests. If the order is very high or there are a large numberof cases to address, however, the computation can be costly. If the computation is done on-line in a controller, time and complexity become important. Moreover, often only a few modes are important, most often the lowest-frequency modes. These cases are candidates for modereduction. The essential idea is to employ the reduced modal matrix of dimensions n x mcomprising the m < n (usually m << n) eigenvectors of interest: Xll

X21

Xml

(B.46) Xln

X2n

Xmn

In the exampleabove the three lowest-frequency modesas given could be used.

B.2.

HIGHER-ORDER

MODELS

977

The vectors y and P~x become of reduced length 3, and the matrices P~MP, P~CP and P’KP become of reduced size 3 × 3. The analysis is carried out as before, and the transformation x = Py restores the full displacement vector corresponding to the neglect of the missing modes. In the example, the result of this reduced calculation would be identical to the truncation of equation (B.45) at i = 3 and the use of equations (B.44). MATLABwas not directed to use the reduced matrices P~MP, P~CP and P~KPsimply because the full computation required an insignificant amount of computer time. Were the order much higher and the matrices M, C and K fuller, however, or the calculation repeated a large number of times, the saving might be worthwhile. PROBLEMS B.1 A torsional system has three identically and three identical disks with inertance I.

(a) Find the mass and stiffness

compliant shafts with stiffness

K

matrices.

(b) Using MATLAB, find the natural

frequencies

and mode shapes.

(c) A torque M(t) is applied at the free end. Find the modal equations of motion, and relate the displacement of the free end to these modes. (d) Present a Bode plot for the response of the free end when M(t) is a sinusoidal function. B.2 The ten-story building considered above is buffeted by wind rather than an earthquake. The force on each floor is sinusoidal, due to the shedding of vortices, with an amplitude of 1 kN and frequency of one cycle in three seconds. Find the displacement amplitude of the top floor. Note: Only the lowest frequency modeis very significant, so consider only two modes.

Appendix

C

Laplace

Transform Pairs

The following is a highly selective list. For more extensive pairs, see, e.g., Paul A. McCollum and Buck F. Brown, Laplace Trans]orm Tables and Theorems, Holt, Rinehart and Winston, Inc., NewYork, 1965 a~d Ruel V. Churchill, Operational Mathematics, 2nd ed., McGraw-Hill, New York, 1958. See also Table 6.1 for selected basic relations. Note: n is a positive integer. f(t)

F(s)

1

~(0)

1

2

~(t)

-

3

t

28

tn--1

1

4

1

(n- 1)! 1

5

8~-a

6

re-at

7

1 (n - 1)! tn-le-at

2(s+a)

979

n(8 + a)

980

APPENDIX

f(t)

F(s)

8

sin wt

w 2w s2 +

9

cos wt

82 2+ w

10

sinh wt

11

cosh wt

12

1(1 - -at) e a

8,

s(s+a)

13

1 (e_at_e_bt) b-a

(s+a)(s+b)

14

1 (be_bt_ae_at) b-a

(s+a)(s+b)

15 i6

~-:~(11_

17

-~(at

e_at_ ate_at - 1+ -at)

18

e -at

19

e-at COS o2t

20 21

~e-( ~t sin

w,~v/~

2 s(s+a) s 2 (s + a)

sin wt

2 (8-4-a)2+w

s+a 2(s-4-a)2-4-w - (2t

1V/~_ ~2e ( smn(w~X/]-- (2t ¢ = tan -~

22

s(s+a)(s+b)

l[l+a--~b(be-at-ae-bt)]ab

s 2 + 2(w.~s + w.2 s ~ + 2(w~s

~-

s(s ~ + 2~s + ~2)

C

981

LAPLACE TRANSFORM PAIRS

/(t) 022

23

1 - coswt

24

wt - sin wt

25

sin wt - wt cos wt

26

1 -- t sin wt 2w

27

t cos wt

s(s2+022) 023

28

1 02~ -- 0212

82 2_ 02 2(82-~-~d2)

(cos w~t," cos.~t)

(82 ~t- ~d12)(82 .~- 0222)

~-~ (sin wt +wt cos wt) 1

1

e_at

(b- a)(c-

~a (b-a)(c-a)

-bt e.

a) + (a- b)(c1

~ (a - e)(b31

~ (s~+w~)

28

29 30

2) s~(s2+02

e_Ct

b (a-b)(c-b)

~-at

(s+a)(s+b)(s+c) e-bt

C

(a-c)(b-c) 32

A e-~t + ~/(b - a) ~ + w~ e_at sin(wt - ¢) .4 = (b - a)~ + ~

33

e_Ct

(s + b)[(s + a +02u]

¢ = tan-~

Ae-ct + Be-at sin(wt + ¢) -c

(s+a)(s+b)(s+c)

1 / a 2 2+ w

(a - c)2 + 022B= ~V(c- a)2 + ~-~

2] (s+c)[(s+a) 2 +w

982

APPENDIX C

I(t) 34

le-at

sinh

1 (s + a)2 2-- w 1

wt

1

35

1 s~/~

36 ta--1

37

e-°’z (a > 0)

38 39

1

r(a) a > 0 (F is Gammafunction)

erfc

a

(complementaryerror function)

-e

-- a >_ 0

8

40 41

Le-a2/4t

--~e J0(2v/-~) (Bessel function)

~- a > O le-a/~ 8

42 43 44

1 --a/s

v~COS 2V~ 1 cosh 2v~

[t~\/|_~](b-1)/2ib__l(2V~) (modified Bessel fn.) s--gel ~/~. b>0

Appendix

D

Thermodynamic Data Computer Code

and

Complete programs that find the state variables corresponding to a specified density and entropy are given below for dry air as a mixture of nitrogen, for oxygen and argon, for refrigerant R-12, and for water. These files, along with programs from Example 12.5 (pp. 925-926), Section 12.5.6 (pp. 934-936), Guided Problem 12.4 (pp. 942-943), may be downloaded from the author’s page at the website www.lehigh.edu/~inmem. The data necessary for treating refrigerants R-22 and R-134Aalso are given below.

D.1

Programs nents

and Data for

Air and Compo-

The following files are discussed in Section 12.5.3: function [s,dsdT,u,dudTS=gas(gl,g2,g3,g4,gS,g6,gT,gS,b,sO,T0g,u0, T) ~ Evaluates the entropies s and dsdT and the internal energies u ~ and dudT of an ideal (zero-pressure) gas, given the temperature. ~ Based on equations (12.98) and (12.99). y=b/T; y0=b/T0g; ey=exp(y)-l; ey0=exp(y0)-l; T2=T*T; T3=T2*T; T4=T2*T2; T0~=T0g*T0g; TO3=TO2*TOg; T04=TO2*T02; s=-gl/3*(1/T3-1/TO3)-g2/2*(1/T2-1/TO2)-g3*(1/T-1/TOg); s=s+g4*log(T/TOg)+gS*(T-TOg)+g6/2*(T2-TO2)+gT/3*(T3-T03); s=s-gS*(y/ey-yO/eyO+y-yO-log(ey/eyO))+sO; dsdT=gl/T4+g2/T3+g3/T2+g4/T+gS+g6*T+gT*T~-gS/T*(y/ey)~*(l+ey); u=-gl/2*(1/T2-1/TO2)-g2*(1/T-1/TOg)+g3*log(T/TOg)+g4*(T-TOg)+uO; u=u+gS/2*(T2-TO2)+g6/3*(T3-TO3)+gY/4*(T4-TO4)-gS*b*(1/ey-1/eyO); 983

984

APPENDIX

D.

THERMODYNAMIC

DATA

AND

CODE

dudT=gl/T3+g2/T2+gS/T+g4+gS*T+g6*T2+gT*T3+gS*exp(y)*(y/ey)’2;

function [P,u,h,s,dudT,dudv,dsdT,dsdv,dPdT,dPdv]=air(T,rho) ~ Computes the indicated functions for air as a mixture of nitro~ gen, oxygen and argon, given the values of temperature and ~ density. "n", "o" and "a" designate nitrogen, oxygen and argon, ~ respectively. Based on equations (12.97), (12.100) and (12.101). global Gln G2n G3n G4n G5n G6n GTn G8n Bn SOn TOn T2n T3n T4n UOn global Glo G2o G3o G4o G5o G6o G7o G8o Bo SOo TOo T2o T3o T4o UOo global Cva SOa TOa UOa Sm R Fn Fo eyOn eyOo global Gamma A1 A2 A3 A4 A5 A6 A7 A8 A9 AIO All AI2 AI3 AI4 AI5 global AI6 AI7 AI8 AI9 A20 A21 A22 A23 A24 A25 A26 A27 A28 A29 A30 global A31 A32 ~ Note that dsdv=dPdT (an identity), so one of these (presumably ~ dsdv) can be removed. They are left as a check; the coding is ~ different. r=rho; r2=r*r; r3=r*r2; r4=r2*r2; r5=r2*r3; r6=r3*r3; rT=r3*r4; rS=r4*r4; rq=r4*r5; rlO=r5*r5; rll=r5*r6; r12=r6*r6; r13=r6*rT; r14=rT*rT; er=exp(-Gamma*r2); W=ones(6); dW=ones(6); W(1)=I/2/Gamma*(l-er); dW(1)=r*er; for i=1:5 W(i+l)=(-r^(2*i)/2*er+i*W(i))/Gamma; dW(i+l)=(r’(2*i)*(Oan~aa*r-i/r)*er+i*dW(i))/Samma; end T2=T*T; T3=T*T2; T4=T2*T2; T5=T2*T3; T6=T3*T3; yn=Bn/T; ynO=Bn/TOn; yo=Bo/T; yoO=Bo/TOo; eyo=exp(yo)-l; eyn=exp(yn)-l; P=r*R*T; p=p+r2.(AI.T+A2*sqrt(T)+A3+A4/T+AS/T2)+r3*(A6*T+AT+A8/T+Aq/T2); P=P+r4,(AIO,T+AII+AI2/T)+r5*AI3+r6*(AI4/T+AI5/T2)+rY*AI6/T; P=P+r8*(AIZ/T+AI8/T2)+rq*AIq/T2+(r3*(A20/T2+A21/T3); p=p+r5.(A22/T2+A23/T4))*er+(rT*(A24/T2+A25/T3)+rq*(A26/T2+A2?/T4) P=P+rlI*(A28/T2+A29/T3))*er+rI3*(A30/T2+A31/T3+AS2/T4)*er; [sn,dsdTn,un,dudTn] =feval(’gas’,Gln,G2n,G3n,G4n,G5n,G6n,GTn,G8n,Bn,SOn,TOn,UOn,T); [so,dsdTo,uo,dudTo] =feval(’gas’,Glo,G2o,G3o,G4o,G5o,G6o,GTo,G8o,Bo,SOo,TOo,UOo,T); u=Fn*un+Fo*uo+(l-Fn-Fo)*(Cva*(T-TOa)+UOa); u=u+r.(A2/2.sqrt(T)+A3+2*A4/T+3*A5/T2)+r2/2*(AT+2*A8/T+3*Aq/T2); u=u+r3/3.(AII+2*AI2/T)+r4/4*AI3+rS/5*(2*AI4/T+3*AI5/T2); u=u+r6/6*2*KI6/T+rT/7*(2*AIT/T+3*AI8/T2)+r8/8*3*AIq/T2; u=u+W(1)*(3*A20/T2+4*A21/T3)+W(2)*(3*A22/T2+5*A23/T4);

D.1

PROGRAMS AND DATA FOR AIR

AND COMPONENTS

u=u+W(3),(3,A24/T2+4*A25/T3)+W(4)*(3*A26/T2+5*A27/T4); u=u+W(5),(3,A28]T2+4*A29/T3)+W(6)*(3*A30/T2+4*A31/T3+5*A32/T4); h=u+P/r; s=-R*log(r)+Sm+Fn*sn+Fo*so; s=s+(1-Fn-Fo)*(Cva*log(T/TOa)+SOa); s=s-r*(AI+A2/2/sqrt(T)-A4/T2-2*AS/T3)-r2/2*(A6-AS/T2-2*Ag/T3); s=s-r3/3*(AlO-A12/T2)+r5/5*(A14/T2+2*A15/T3)+r6/6*A16/T2; s=s+r7/7*(A17/T2+2,A18/T3)+rS/8*2*A19/T3+W(1)*(2*A20/T3+3*A21/T4); s=s+W(2)*(2*A22/T3+4*A23/T5)+W(3)*(2*A24/T3+3*A25/T4); s=s+W(4)*(2*A26/T3+4*A27/T5)+W(5)*(2*A28/T3+3*A29/T4); s=s+W(6)*(2*A30/T3+3*A31/T4+4*A32/T5); dudT=Fn*dudTn+Fo*dudTo+(1-Fn-Fo)*Cva; dudT=dudT+r*(A2/4/sqrt(T)-2*A4/T2-6*AS/T3)-r2/2*(2*AS/T2+6*A9/T3); dudT=dudT-r3/3*2*A12/T2-r5/5*(2*A14/T2+6*A15/T3)-r6/6*~*A16/T2; dudT=dudT-r7/7*(2*A17/T~+6*A18/T3)-r8/8*(6*Aig/T3); dudT=dudT-W(1)*(6*A20/T3+12*A21/T4)-W(2)*(6*A22/T3+20*A23/T5); dudT=dudT-W(3)*(6*A24/T3+12*A25/T4)-W(4)*(6*A26/T3+20*A27/T5); dudT=dudT-W(5)*(6*A28/T3+12*A29/T4); dudT=dudT-W(6)*(6*A30/T3+12*A31/T4*20*A32/T5); dudr=A2/2*sqrt(T)+A3+2*A4/T+3*A5/T2+r*(AT+2*AS/T+3*A9/T2); dudr=dudr+r2*(All+2*A12/T)+r3*A13+r4*(2*A14/T+3*A15/T2)+rS*A16/T; dudr=dudr+r6*(2*A17/T+3*A18/T2)+r7*3*A18/T2; dudr=dudr+dW(1)*(3*A20/T2+4*A21/T3)+dW(2)*(A22/T2+5*A23/T4); dudr=dudr+dW(3)*(3*A24/T2+4*A25/T3)+dW(4)*(3*A26/T2+5*A27/T4); dudr=dudr+dW(5)*(3*A28/T2+4*A29/T3); dudr=dudr+dW(6)*(3*A30/T2+4*A31/T3+5*A32/T4); dudv=-r2*dudr; dsdT=Fn*dsdTn+Fo*dsdTo; dsdT=dsdT+(1-Fn-Fo)*Cva/T; dsdT=dsdT+r*(A2/4/T^(1.5)-2*A4/T3-6,AS/T4)-r2/2,(2,A8/T3+6,A9/T4); dsdT=dsdT-2/3*r3*A12/T3-rS/5*(2*A14/T3+6*A15/T4)-r6/3*A16/T3; dsdT=dsdT-rT/7*(2*A17/T3+6*A18/T4)-rS/8*6*A19/T4; dsdT=dsdT-W(1)*(6*A20/T4+12*A21/TS)-W(2)*(6*A22/T4+20*A23/T6); dsdT=dsdT-W(3)*(6*A24/T4+12*A25/TS)-W(4)*(6*A26/T4+20*A27/T6); dsdT=dsdT-W(5)*(6*A28/T4+12*A29/T5); dsdT=dsdT-W(6)*(6*A30/T4+12*A31/T5+20,A32/T6); dsdr=-R/r-(AI+A2/2/sqrt(T)-A4/T2-2*AS/T3)-r*(A6-AS/T2-2*A9/T3); dsdr=dsdr-r2*(AlO-A12/T2)+r4*(A14/T2+2*A15/T3)+r5,A16/T2; dsdr=dsdr+r6*(A1Z/T2+2*A18/T3)+rT*2*A19/T3; dsdr=dsdr+dW(1)*(2*A20/T3+3*A21/T4)+dW(2)*(2,A22/T3+4,A23/T5); dsdr=dsdr+dW(3)*(2*A24/T3+3*A25/T4)+dW(4)*(2,A26/T3+4,A27/TS); dsdr=dsdr+dW(5)*(2*A28/T3+3*A29/T4); dsdr=dsdr+dW(6)*(2*A30/T3+3*A31/T4+4*A32/TS); dsdv=-r2*dsdr; dPdT=r*R+r2*(AI+A~/2/sqrt(T)-A4/T2-2*AS/T3)+r3*(A6-AS/T2-2,A9/T3); dPdT=dPdT+r4*(AIO-A12/T2)-r6*(A14/T2+2*A15/T3)-rT,A16/T~;

985

986

APPENDIX

D.

THERMODYNAMIC

DATA

AND

CODE

dPdT=dPdT-r8*(A17/T2+2*A18/T3)-r9*2*A19/T3; dPdT=-r3*(2*A20/T3+3*A21/T4)*er-(r5*(2*A22/T3+4*A23/T5); dPdT=dPdT-rT*(2*A24/T3+3*A25/T4))*er-(r9*(2*A26/T3+4*A27/TS); dPdT=dPdT-rll*(2*A28/T3+3*A29/T4))*er; dPdT=dPdT-r13*(2*A30/T3+3*A31/T4+4*A32/TS)*er; dPdr=R*T+2*r*(AI*T+A2*sqrt’(T)+A3+A4/T+AS/T2); dPdr=dPdr+3*r2*(A6*T+AT+AS/T+A9/T2)+4*r3*(AIO*T+All+A12/T); dPdr=dPdr+5*r4*A13+6*rS*(A14/T+A15/T2)+Y*r6*A16/T; dPdr=dPdr+8*rT*(A17/T+A18/T2)+9*rS*A19/T2; dPdr=dPdr+(3*r2-2*Gamma*r4)*(A20/T2+A21/T3)*er; dPdr=dPdr+(5*r4-2*Gamma*r6)*(A22/T2+A23/T4)*er; dPdr=dPdr+(Y*r6-2*Gamma*rS)*(A24/T2+A25/T3)*er; dPdr=dPdr+(9*rS-2*Gamma*r10)*(A26/T2+A27/T4)*er; dPdr=dPdr+(ll*rlO-2*Gamma*r12)*(A28/T2+A29/T3)*er; dPdr=dPdr+(13*r12-2*Gamma*r14)*(A30/T2+A31/T3+A32/T4)*er; dPdv=-r2*dPdr;

~ script file gasdata.m: data for ideal gasses. "n", "o" and "a" ~ designate nitrogen, oxygen and argon, respectively. global Gln G2n G3n G4n G5n G6n GTn G8n Bn SOn TOn T2n T3n T4n UOn global Glo G2o G3o G4o GSo G6o G?o G8o Bo SOo TOO T2o T3o T4o UOo global Cva SOa TOa UOa Sm Fn Fo eyOn eyOo g Ser global Gamma A1 A2 A3 A4 A5 A6 A7 A8 A9 AIO All AI2 global AI6 AI7 AI8 AI9 A20 A21 A22 A23 A24 A25 A26 A27 A28 A29 A30 global A31 A32 E Gamma=5.97105475117183e-6; R=287.0686; AI=I.5562309840913?E-I; A2=l.25288666202326el; A3=-2.92541568638838e2; A4=4.29432480725523e3; A5=-5.58450959675108e5; A6=3.92054480883008e-4; AT=-4.4098564188134?e-2; A8=5.8638?178724129e-4; Ag=?.97411385439405e4; AlO=9.88045320906742e-9; All=2,97999237261289e-4; A12=-6.817830409590?Oe-2; A13=2.02551630992042e-?; A14=-l.6272428184949?e-7; A15=-l.O6340143152999e-4; A16=~.51428501875049e-lO; AlT=-l.?O388092279449e-13; A18=5.91103444646786e-ll; Alg=-l.O5363473794348e-14; A20=-Z.32?326SlI96979e4; A21=-5.42674649S24748eS; A22=-4.489354S6142735e-I; A23=2.81453138446295e2; A24=-8.83132042?91851e-?; A25=-l.32229814838386e-5; A26=-2.16521865046609e-12; A2?=-l:47835008246593e-9; A28=-6.9321984930150Ie-19; A29=6.06?43598768355e-l?; A30=-3.20538718135891e-24; A31=-4.73178337355130e-23; A32=3.83950822306912e-22; Gln=-218203.473713518; G2n=lO157.3580096247; G3n=-165.50472165724; G4n=?43.17599919043; G5n=-5.1460562354SO25e-3; G6n=5.18S47156760489e-6; G?n=-l.O5922170493616e-9;

D2.

PROGRAMS AND DATA FOR REFRIGERANT

987

R-12

G8n=298.389393363817; Bn=3353.4061; S0n=3288.2374; T0n=63.15; U0n=196622.81; Glo=-1294427.11174062; G2o=59823.1747005341; G3o=-897.850772730944; G4o=655.236176900400; G5o=-l.13131252131570e-2; G6o=3.49810702442228e-6; GTo=4.21065222886885e-9; G8o=267.997030050139; Bo=2239.18105; S0o=3271.4925; T0o=54.34; U0o=228267.4; Cva=312.192; S0a=2270.67;T0a=83.8;U0a=149355.4; Sm=174.519; Fn=0.7553; Fo=.23146;Ser=0.001; T2n=T0n*T0n; T3n=T0n*T2n; T4n=T2n*T2n; T2o=T0o*T0o; T3o=T0o*T2o; T4o=T2o*T2o; y0o=Bo/T0o; y0n=Bn/T0n; ey0o=exp(y0o)-l; eyOn=exp(y0n)-l; g=Fn*[Gln G2n G3n G4n G5n G6n GTn]; g=g+Fo*[Glo G2o G3o G4o GSo G6o GTo];

~ file specheat.m;computes the specific heat of air, ~ given its temperature.Oxygen and nitrogen components based on ~ equation(12.96). global Gln G2n G3n G4n G5n G6n GTn G8n Bn Fn T global Glo G2o G3o G4o G5o G6o GTo G8o Bo Fo Cva cvn=Gln/T\’{}3+G2n/T/T+G3n/T+G4n+G5n*T+G6n*T*T+GTn*T\’{ }3; cvn=cvn+G8n*exp(Bn/T),(Bn/T/(exp(Bn/T)-l))\’{ cvo=Glo/T\^{}3+G2o/T/T+G3o/T+G4o+G5o~T+G6o,T~T+GTo*T\’{ }3; cvo=cvo+G8o~exp(Bo/T)~(Bo/T/(exp(Bo/T)-l))\’{ cvm=Fn~cvn+Fo~cvo+(l-Fn-Fo)*Cva;

D.2 Programs and Data for Refrigerant

R-12

The following files are discussed in Section 12.5.4: ~ program dataRl2.m~ data defining model of refrigerantR-12 global A2 A3 A4 A5 B2 B3 B4 B5 C2 C3 C4 C5 G1 G2 G3 G4 G5 global E6 Ell E13 E22 E23 E24 F1 F2 F3 F4 SO UO TO TC R B Ck global D1 D2 D3 D4 D5 D6 D7 ~c VC Pe A2=-91.6210126;A3=.I01049598;A4=-5.74640225e-5; A5=O; B2=.0771136428;B3=-5.67539138e-5;B4=O; B5=4.08193371e-ll; C2=-1525.24293;C3=2.19982681;C4=0; C5=-1.66307226e-7; DI=558.0845400;D2=854.4458040;D3=O; D4=299.40771103;DS=O; D6=352.1500633; DT=-50.47419739;.Rc=4.616*120.91; G1=33.89005260; G2=2.507020671; G3=-.003274505926; G4=l.641736815e-6; GS=0; E6=.788904825e-5; El1=-.1900330949; E13=-5.947813469; E22=-512.6536438; E23=1142.99428; E24=-690.5751515; F1=93.3438056; F2=-4396.18785; F3=-12.4715223; F4=.0196060432; S0=894.48764; U0=169701.87; T0=200; TC=385.1Z; R=68.7480;

988

¯APPENDIX

B=4.06366926e-4;

Ck=5.475;

D.

THERMODYNAMIC DATA AND CODE

VC=.0017004;

function[P,h,hg,s,sg,vf,vg,hf,dsdv,dsdT,dudv,dudT,dPs,d2Ps,dvg,dsg dugdv,dugdT,dPdv,dPdT]=propref(v,T) ~ end ~ Program that finds propertiesof refrigerantR12 given v, T. ~ Calls program onephase.mbelow, and uses equations ~ (12.106)-(12.108), (1~.1~3) and (~2.~6)-(12.1~9). global DID2 D3 D4 D5 D6 D7 T02 T03 T04 Ck TC A2 A3 A4 A5 global C1 C2 C3 C4 C5 G1 G2 G3 G4 G5 E6 Ell El3 E22 E23 E24 global RI Rc R B UO T2 T3 T4 ft ftO ft2 ft3 dataRl2 T2=T*T; T3=T*T2;T4=T2*T2; TO2=TO*TO;TO3=TO*T02;TO4=TO2*T02; ft=Ck/TC*exp(-Ck*T/TC); ftO=Ck/TC*exp(-Ck*TO/TC); ftl=ft*TC/Ck; ft2=(I+Ck*T/TC)*exp(-Ck*T/TC); ft3=-ft*Ck*T/TC; vOl=l.369-B;vO2=vOl*vOl;vO3=vOl*v02;vO4=vO2*v02; x=I-T/TC; if x<.Ol ~ supercriticaltemperature(with some allowance) [P,h,s,dudv,dudT,dsdv,dsdT,dPdv,dPdT]=onephase(v,T); else ~ subcritical temperature y=E6*x’(-5/~)+Ell+E13*x’(2/3)+E22*x^(ll/3)+E23*x’4; y=y+E24*x’(13/3); vg=l/~c*exp(-y); if vg<=v ~ superheated [P,h,s,dudv,dudT,dsdv,dsdT,dPdv,dPdT]=onephase(v,T); else ~ saturated mixture or compressedliquid x2=x’(1/3); x3=x~*x2; x5=x2*x; rf=Dl+D2*x2+D3*x3+D4*x+D5*xS+D6*sqrt(x)+DT*x*x; ~ density of liquid phase vf=I/rf; X=(v-i/rf)/(vg-I/rf); ~ quality of mixture vlg=vg-B; v2g=vlg*vlg;v3g=vlg*v2g;v4g=v2g’2;v5g=v2g*v3g; P=exp(FI+F2/T+F3*Iog(T)+F4*T); hg=GI*(T-TO)+G2/2*(T2-TO2)+G3/3*(T3-TO3)+G4/4*(T4-T04)-; hg=hg-G5*(i/T-i/TO)+(A2+ft2*C2)/vlg+(A3+ft2*C3)/2/v2g; hg=hg+(A4+ft2*C4)/3/v3g+(A5+ft~.C5)/4/v4g+P*vg+UO; dPs=P*(-F~/T~+F3/T+F4); d~Ps=P~(~*F~/T3-F3/T2)+dPs~(-F~/T~+F3/T+F4); sg=GI.log(T/TO)+G~.(T-TO)+G3/2.(T~-TO~)+G4/3.(T3-T03) sg=sg-G5/~*(1/T~-l/TO~)+~log(vlg/vO1)+B2~(1/vO1-1/vlg); sg=sg+B3/~*(i/vO2-1/v~g)+B4/3,(1/vO3-1/v3g); sg=sg+B5/4*(I/vO4-1/v4g)+C~(ft/vlg-ftO/v01);

D2.

PROGRAMS AND DATA.FOR

REFRIGERANT

R-12

989

sg:sg+C312,(ftlv2g-ftO/vO2)+C4/3*(ft/v3g-ftO/v03); sg=sg+CS/4,(ft/v4g-ftO/v04)+915.1; ug=GI,(T-TO)+G2/2,(T2-TO2)+G3/3*(T3-TO3)+G4/4*(T4-T04); ug=ug-GS,(llT-11TO)+(A2+C2*ft2)/vlg+(A3+C3*ft2)/v2g/2; ug=ug+(A4+C4*ft2)/v3g/3*(A5+C5*ft2)/v4g/4+UO; hg=ug+P*vg; sm=(vg-v)*dPs; s=sg-sm; sf=sg-(vg-i/rf~dPs; h=hg-T*sm;~ enthalpyif saturated mixture sf=sg-dPs*(vg-I/rf); ~ entropy of saturated liquid hf=hg-T*(sg-sf);~ enthalpy of saturated liquid u=h-P*v;~ internal energy if saturated mixture dvg=l/TC*vg*(-5/3*E6*x’(-8/3)+2/3*E13*x^(-1/3)); dvg=dvg+vg*(ll/3*E22*x’(8/3)+l]TC*(4*E23*x’3+13/3)); dvg=dvg+vg*E24*x’(lO/3); dsg=(R/vlg+B2/v2g+B3/v3g+B4/v4g+BS/vSg)*dvg; dsg=dsg-ft*(C2/v2g+C3/v3g+C4/v4g+C5/v5g)*dvg; dsg=dsg-Ck/TC*ft*(C2/vlg+C3/2/v2g+C4/3/v3g+CS/4/v4g); dsg=dsg+G1/T+G2+G3*T+G4*T2+GS/T3; dugdv=-(A2+C2*ft2)/v2g-(A3+C3*ft2)/v3g-(A44C4*ft2)/v4g; dugdv=dugdv-(AS+C5*ft2)/v5g; dugdT=GI+G2*T+G3*T2+G4*T3+GS/T2; dugdT=dugdT+ft3*(C2/vlg+C3/2/v2g+C4/3/v3g+C5/4/v4g); dsgdv=R/vlg+B2/v2g+B3/v3g+B4/v4g+B5/vSg; dsgdv=dsgdv-ft*(C2/v2g+C3/v3g+C4/v4g+CS/vSg); dsdT=dsg-dvg*dPs-(vg-v)*d2Ps; dPdv=O; dPdT=dPs; dsdv=dPs; end end function [P,h,s,dudv,dudT,dsdv,dsdT,dPdv,dPdT]=onephase(v,T) ~ Program that finds the propertiesof vapor phase refrigerant ~ given its specific volume and temperature. ~ Called by program propref.mabove. ~ Based on equations(12.105), (12.109),(12.111) and (12.114). global B2 B3 B4 B5 TO2 TO3 TO4 TC TO A2 A3 A4 A5 Cl C2 C3 C4 global C5 G1 G2 G3 G4 G5 K B UO T2 T3 T4 ft fro ft2 ft3 Ck ftl=ft*Tg/Ck; vOl=l.369-B;vO2=vOl*vOl;vO3=vOl*v02;vO4=vO2*v02; vl=v-B; v2=vl*vl; v3=vl*v2;v4=v2*v2; v5=v2*v3;v6=v3*v3; s=GI*Iog(T/TO)+G2*(T-TO)+G3/2*(T2-TO2)+G4/3*(T3-T03); s=s-G5/2*(i/T2-1/TO2)+R*log(vl/vOl)+B2*(I/vOl-i/vl); s=s+B3/2*(i/vO2-1/v2)+B4/3*(i/vO3-1/v3);

990

APPENDIX

D.

THERMODYNAMIC

DATA AND CODE

s=s+B5/4*(1/vO4-11v4)+C2*(ft/vl-ftO/vO1)+C3/2*(ft/v2-ftO/v02); s=s+C4/3*(ft/v3-ftO/vO3)+CS/4*(ft/v4-ftO/v04)+915.1; P=R*T/vl+(A2+B2*T+C2*ftl)/v2+(A3+B3*T+C3*ftl)/v3; P=P+(A4+B4*T+C4*ftl)/v4+(AS÷BS*T+CS*ftl)/v5; h=GI*(T-TO)+G2/2*(T2-TO2)+G3/3*(T3-TO3)+G4/4*(T4-T04); h=h-GS*(1/T-1/TO)+(A2+ft2*C2)/vl+(A3+ft2*C3)/2/v2; h=h+(A4+ft2*C4)/3/v3+(AS+ft2*CS)/4/v4+P*v+UO; dPdT=R/vl+(B2-C2*ft)/v2+(B3-C3*ft)/v3+(B4-C4*fZ)/v4; dPdT=dPdT+(B5-C5*ft)/vS; dPdv=-R*T/v2-2*(A2+B2*T+C2*ftl)/v3-3*(A3+B3*T+C3*ftl)/v4; dPdv=dPdv-4*(A4+B4*T+C4,ftl)/v5-5*(AS+BS*T+C5,ftl)/v6; dudv=-(A2+ft2,C2)/v2-(A3+ft2*C3)/v3-(A4+ft2*C4)/v4; dhdv=dudv-(A5+ft2*CS)/v5+P+v*dPdv; dudT=GI+G2*T+G3*T2+G4*T3+GS/T2; dhdT=dudT+ft3*(C2/vl+C3/2/v2+C4/3/v3*CS/4/v4)+v*dPdT; dsdv=R/vl+B2/v2+B3/v3+B4/v4+B5/v5-ft*(C2/v2+C3/v3+C4/v4+CS/vS); dsdT=-Ck/TC*ft*(C2/vl+C3/2/V2+C4/3/V3+C5/4/v4)+G1/T+G2+G3*T; dsdT=dsdT+G4*T2+GS/T3; function[dm] =orifice(A, Pu, Pd,vu,vd, vfu,vfd,vgu,vgd,Tu,Td) ~, Computesthe flow throughan orifice based on equationp. 873. ~ global C K r=Pd/Pu; if r<=l dmvap=A*C*Pu*sqrt ( (r" (2/K)-r"(1+I/K))/Tu) if Tu>381 dm=dmvap; elseif vu>=vgu dm=dmvap; else xu= (vu-vfu) / (vgu-vfu) dmliq=A*sqrt (2/vfu*(Pu-Pd)) dm=xu*dmvap+ (1-xu)*dmliq; end else r=l/r; dmvap=-A*C*Pu*sqrt ( (r~ (2/K)-r^ (I+1/K))/Td) if vd>=vgd dm=dmvap; else xd=(vd-vfd) / (vgd-vfd) dmliq=-A*sqrt (2/vfd*(Pd-Pu)) dm=xd*dmvap+ (l-xd)*dmliq; end end

D.3.

DATA FOR REFRIGERANT R-134A

D.3

Data for Refrigerant Equation of state equation (12.94)

R b A2 B2 C2 A3 B~ C3 A4 A5 B5 C5 K

= = = = = = = = = = = = =

-5 0.814881629 x 10 -3 0.3455467 × 10 -0.1195051 0.1137590 -3 × 10 -1 -0.3531592 × 10 -3 0.1447797× 10 -7 -0.8942552 × 10 -2 0.6469248 × 10 -s -0.1049005 x 10 -0.6953904 -1 × 310 -14 0.1269806 × 10 -l° -0.2051369 × 10 5.475

991

R-134A

Ideal gas heat capacity equation (12.99) Cpl = Cp2 = C~3 =

-2 -0.5257455 × 10 -2 0.3296570 x 10 -7 -0.2017321× 10

C~4 =

0.0

C~5 =

0.1582170x 102

saturated liquid density equation (12.96) D1 D2 D3 D4 D5

= = = -=

0.5122 × 103 ~ 0.819.6183× 10 0.1023582x 104 -0.1156757x 104 0.7897191x 103

All parametersI above and below have units compatible with the critcal parameters below; pressure is in kPa, density in kg/m3, energy in kJ/kg and temperaturein K. Coefficients not listed are zero. Critical Parameters: Critical pressure: Pc = 4067 kPa; Critical volume: vc = 1/512.2 m3/kg Critical temperature: 0c = 374.25 K; Constants X and Y: not specified Saturation pressure equation (12.95) F1 F2 F~ F5 Fs F9

= = = = = =

F10 =

Saturation vapor density equation (12.97)

0.248033988× 102 -0.3980408 × 104 -1 -0.1336296 × 10 -4 0.2245211 × 10 0.1995548 0.3748473× 103

not given

1.000

D.4 Data for Refrigerant

R-22

Critical parameters: Critical pressure: Pc = 4977 kPa; Critical volume: vc = 0.16478 L/mol Critical temperaure: 0c = 369.16 K; Molecular weight.= 86.469 1Taken from D.P. Wilson and R.S. Basun, "Thermodynamic Properties of a NewStratospherically Safe Working Fluid - Refrigerant 134a,"ASHRAETransactions v. 94 Pt. 2 (1988) pp. 2095-2118.

992

APPENDIX D. THERMODYNAMIC DATA AND CODE

entropy constant: .Y = 46.8883 J/mol-K; enthalpy constant: X = 25990.02 J/mol All parameters2 have units compatible with the critical parameters above; pressure is in kPa, density in mol/L and temperature in K. Coefficients not listed are zero. The value of J in equations (12.100) and (12.102) is 1000. equation of state ideal gas heat capacity equation (12.94) equation (12.98) -2 R = 0.83136998378x 10 -1 b = 0.10796166646x 10 A2 = -0.87466122549 B = 0.87054394311× -3 10 2 C2 = -0.88533742184 x 10 A3 = -1 -0.18939998642 x 10 -3 0.14890632572x 10 B3 = 0.16091656668x 10 C3 = -1 = 0.13524290185x 10 A4 B4 = -4 -0.37996244181× 10 C4 = 0.0 X 10 A5 = -0.11768746066 -2 -5 0.30463870398× 10 B5 = -0.58307411777× 10 C5 = -2 s0.94001896962x l0 A6 = Be - = -0.20757984460x 104 C~ = 0.0 K = 4.2 u = 0.10155456431x 103 c = 0.0 saturation pressure equation (12.95) with log10 F1 F2 F3 F4 Fs F~ F~ Fs F9 Flo

= = = = = = = = = =

G1 G2 G3 G4 G5 G5

= = = = = =

0.10183212 2× 10 0.14697171 -4 -0.76354723x 10 0.0 0.28754121 x 105 0.0

saturated liquid density equation (12.96) D1 D2 D3 D4 D5

= = = = =

0.6068821 x 10 0.1012108 2× 10 0.6807772 x 10 -0.4129719 × 10 0.3792696 x 10

saturated vapor density equation (12.97)

0.25189356867x 102 E7 -0.21362184178 × 104 En 0.0 El2 -0.78610312200 x 10 E14 -2 0.39436902792x 10 E20 0.0 E21 0.0 E22 0.44574670300 E23 -3 0.38116666667× 10 E24 1.8 E25

= = --

= = = = = = =

-4 -0.1680024749 x 10 0.1344723347 -0.2845461873 x 10 0.1292966614 0.3106835616x 104 -.01189422522 x 105 s0.1914759823x l0 -0.1472081631 x 105 0.4527640563x 104 ~ 0.1306817379x 10

2TakenfromR.B. Stewart et al., ASHRAE Thermodynamic Properitesof Re]rigerants, American Societyof Heating,RefrigerationandAir-Conditioning Engineers,1988.

D.5.

PROGRAMS

AND DATA

D.5 Programs

FOR WATER

and Data for

993

Water

The coefficients Ci and Aij for the Helmholtz free energy (equation (12.120), 930) are given in the script M-file below, and the properties are computed using this data in the two subsequent function M-files.

~ file waterdat.m, data for the properties of water, from ~ Keenan (footnote 17, p. 930); units in cgs global C1 C2 C3 C4 C5 C6 C7 C8 R E tc global All AI2 AI3 AI4 AI5 AI6 AI7 A21A22 A23 A24 A25 A26 A27 global A31 A32 A33 A34 A35 A36 A37 A41 A42 A43 A44 A45 A46 A47 global A51 A52 A61 A62 A67 AT1 A72 A81 A82 global Agl A92 A93 A94 A95 A96 A97 global AIOI AI02 AI03 AI04 AI05 AI06 AI07 global D E1 E2 E3 E4 E7 EIO El3 El4 El5 F PC rc CI=1857.065; C2=3229.12; C3=-419.465; C4=36.6649; C5=-20.5516; C6=4.85233; C7=46; C8=-I011.~49; R=.46151; E=4.8; tc=1.54491~; AII=29.492937; AI2= -5.1985860; A13=6.8335354; A14=-0.1564104; A15=-6.3972405; A16=-3.9661401; A17=-0.69048554; A21=-132.13917; A22=7.7779182; A23=-26.149751; A25=A24=-0.72546108; 26.409282; A26=15.453061; A27=2.7407416; A31=274.64632; A32=-33.301902; A33=65.326396; A34=-9.2734289; A35=-47.740374; A36=-29.142470; A37=-5.1028070; A41=-360.938~8; A42=-16.254622; A43=-26,181978; A44=4.3125840; A45=56.323130; A46=29.568796; A47=3.9636085; A51=342.18431; A52=-177.31074; A61=-244.50042; A62=127.48742; A71=155.18535; A72=137.46153; A81=5.9728487; A82=155.97836; A91=-410.30848; A92=337.31180; A93=-137.46618; A94=6.7874983; A95=136.87317; A96=79.847970; A97=13.041253; AI01=-416.05860; A102=-209.88866; AI03=-733.96848; AI04=I0.401717; AI05=645.81880; AI06=399.17570; AI07=71.531353; D=[3.6711257 -28.512396 222.65240 -882.43852]; D=[D 2000.2765 -2612.2557 1829.7674 -533.50520]; El=f3; E2=-1.393188; E3=-6.418663; E4=18.46092; E7=-35.25809; EI0=219.5021; E13=-2199.413; E14=3765.631; E15=-1809.692; F=[-741.9242 -29.72100 -11.55286 -0.8685635]; F=[F 0.1094098 0.439993 0.2520658 0.05218684]; PC=22.088; rc=0.3170;

function [P,h,vg,dvg,dudT,dudv,dugdT,dugdvg,dPs,d2Ps] Z cont. =propwat(v,T) Z end Z Computes the properties of water sufficient for equations Z (12.66) and (12.72), (pp 904, 905). Irrelevant outputs

994

APPENDIX

D.

THERMODYNAMIC

DATA

AND

CODE

~ be spurious. The program requires that the script file ~ waterdat.m be run first. Also canoutput quality X, ~ specific volume of liquid vf, entropies s and sg, enthalpy ~ hg and energy ug. Easy to add some others, including u, uf, ~ sf, hr. Note that units are different from those for ~ refrigerant RI2; pressures are in MPa, energy densities are ~ in ~ kJ/kg), and specific volumes are in m’3/kg. global C1 C2 C3 C4 C5 C6 C? C8 R E tc global All AI2 AI3 AI4 AI5 AI6 AI7 A21 A22 A23 A24 A25 A26 A27 global A31 A32 A33 A34 A35 A36 A37 A41 A42 A43 A44 A45 A46 A47 global A51 A52 A61 A62 ATI A72 A81 A82 global Agl A92 A93 A94 A95 A96 A97 global AIOI AI02 AI03 AI04 AI05 AI06 AI07 global D El E2 E3 E4 E7 EIO El3 El4 El5 F PC rc t=lOOO/T; Tc=lOOO/tc; x=l-tc/t;rho=i/v; if x<.Ol ~ supercritical temperature (with some allowance) [P,h,dudT,dudr]=steam(rho,T); dudv=-dudr/v/v; else ~ subcritical temperature y=E2*x’(I/3)+E3*x’(2/3)+E4*x+EY*x*x+ElO*x’3+E13*x’4; y=y+El~*x’(13/3)+E15*x’(14/3); vg=.OO3155*(t/tc)’(E1)*exp(-y); ~ specific volume of saturated vapor at T if v<=vg ~ saturated mixture or compressed liquid xl=x’(i/3);x2=xl*xl; xO=[xl x2 x x*xl x,x2 x*x x*x*xl x*x*x2]; rf=rc*(l+D*xO’)*lO00; vf=i/rf; ~ specific volume of saturated liquid at T if v>=vf ~ saturated mixture X=(v-vf)/(vg-vf); ~ quality of mixture tO=.65-.Ol*T+2.7316; tO2=tO*tO; tO3=tO~*tO; tO4=tO2*t02; tOS=tO3*t02; tO6=tO3*tO3; tl=[1 tO tO2 tO3 tO4 tO5 tO6 tO4*t03]; t2=-.Ol*[O 1 2*tO 3,t02 4,t03 5,t04 6,t05 7,t06]; t3=.O001*[O 0 2 6*tO 12,t02 20,t03 30,t04 42,t05]; P=PC*exp(.OI*(Tc/T-1)*F*tl’); sa turation pr essure dPs=P*.OI*(-Tc/T/T*F*tl’+(Tc/T-1)*F*t2’); ~ derivative of Psat with respect to T d2Ps=P*.OI*(2*Tc/T’3*F*tI’-2*Tc/T/T*F*t2’; d2Ps=d2Ps+P*.Ol*(Tc/T-l)*F*t3’)+dPs’2/P; ~ second derivative of Psat with respect to T dvg=(-El/T+i/Tc/3*(E2/x2+2*E3/xl+3*E4+6*ET*x))*vg; dvg=dvg+i/Tc/3*vg*(9*ElO*x*x+12*El3*x’3); dvg=dvg+i/Tc/3*vg*(13*El4*x’(lO/3)+14*ElS*x-(ll/3)); ~ derivative of vg with respect to T

D.5.

PROGRAMS AND DATA FOR WATER

995

[P,hg,sg,dudT,dudr]=steam(1/vg,T); ug=hg-P*lOOO*vg; ~ internal energy of vapor component sm=(vg-v)*sg; h=hg-T*sm; ~ enthalpy of mixture s=sg-sm; ~ entropy of mixure dugdT=dudT-dudr/lOOO/vg/vg*dvg; ~ derivativeof ug with respectto T dugdvg=dugdT/dvg; ~ derivativewith respect to vg else ~ single phase (vapor or liquid), use steam.m [P,h,dudT,dudr]=steam(rho,T); dudv=-dudr/v/v; end end end

function [P,h,s,dudT,dudr]=steam(rho,T) ~ computesproperties ~ of single-phasewater. Requires that script file waterdat.m ~ be run first. Intended as sub-program for propwat.m. ~ If used directly,note some use of cgs units. ~ Units of energy are J/g = kJ/kg. global Cl C2 C3 C4 C5 C6 C7 C8 R E tc global All AI2 AI3 AI4 AI5 AI6 AI7 A21 A22 A23 A24 A25 A26 A27 global A31 A32 A33 A34 A35 A36 A37 A41 A42 A43 A44 A45 A46 A47 global A51 A52 A53 A54 A55 A56 A57 A61 A62 A63 A64 A65 A66 A67 global AT1 A72 A73 A74 A75 A76 A77 A81 A82 A83 A84 A85 A86 A87 global A91 A92 A93 A94 A95 A96 A97 global AIOI A102 AI03 AI04 AI05 AI06 AI07 t=lOOO/T;r=rho/lO00;~ density in g/cm’3 tl=t-2.5;t2=tl*tl;t3=tl*t2; t4=t2*t2;t5=t3*t2; rO=r-.634; rO2=rO*rO; rO3=rO*r02; rO4=rO2*r02; rO5=rO3*r02; rO6=rO3*r03; rOT=rO4*r03; rl=r-1; r2=rl*rl; r3=rl*r2; r4=r2*r2; r5=r3*r2; r6=r3*r3; rY=r4*r3; EO=exp(-E*r); tcc=t-tc; al=All+A21*rO+A31*rO2+A41*rO3+A51*rO4+A61*rOS+ATl*r06; al=al+A81*rO7+EO*(Agl+AlOl*r); a2=A12+A22*rl+A32*r2+A42*r3+A52*r4+A62*r5+A72*r6+A82*r7; a2=a2+EO*(A92+AlO2*r); a3=AI3+A23*rI+A33*r2+A43*r3+EO*(A93+AIO3*r); a4=A14+A24*rI+A34*r2+A44*r3+EO*(A94+AIO4*r); aS=AIS+A25*rI+A35*r2+A45*r3+EO*(A95+AIOS*r); a6=AI6+A26*rI+A36*r2+A46*r3+EO*(A96+AIO6*r); aT=AIT÷A27*rI+A37*r2+A47*r3+EO*(A97+AIO7*r); Q=al+tcc*(a2+t1*a3+t2*a4+t3*aS+t4*a6+tS*aT); dal=A21+2*A31*rO+3*A41*rO2+4*ASl*rO3+5*A61*rO4+6,ATl*r05; dal=dal+7*A81*rO6+EO*(AlO1-E*(A91+A10I*r));

996

APPENDIX

D.

THERMODYNAMIC

DATA AND CODE

da2=A22+2*A32*rl+3*A42*r2+4*A52*r3+5*A62*r4; da2=da2+6*A72*r5+7*A82*r6+EO*(AlO2-E*(A92+AI02*~)); da3=A23+2*A33*rI+3*A43*r2+EO*(AIO3-E*(A93+AIO3*r)); da4=A24+2*A34*r1+3*A44*r2+EO*(AIO4-E*(A94+AIO4*r)); da5=A25+2*A35*rI+3*A45*r2+EO*(AIO5-E*(A95+AIO5*r)); da6=A26+2*A36*rI+3*A46*r2+EO*(AIO6-E*(A96+AIO6*r)); daT=A27+2*A37*rI+3*A47*r2+EO*(AIOT-E*(A97+AIOT*r)); dQdr=dal+~cc*(da2+~l*da3+~2*da4+~3*da5+~4*da6+~5*daT); d~d~=a2+(~l+~cc)*a3+~l*(~l+2*~cc)*a4+~2*(~l+3*~cc)*a5; dQd~=d~d~+~3*(~l+4*~cc)*a6+~4*(~l+5*~cc)*aT;~ pO=Cl+C2/t+C3/t/t+C4/t’S+C5/t’4+C6/t’5+CT*log(T)+C8*log(T)/t; dpOdT=(C2/Z+2*C3/t’2+3*C4/t’3+4*C5/t’4+5*C6/Z^5+CT)/T; dp0dT=dp0dT+C8/t*(log(T)+l)/T; dpOtdt=Cl-C3/t/t-2*C4/t’3-3*C5/t’~-4*C6/t’5+C?*log(T)-CT-CS/t; P=r*R*T*(l+r*Q+r*r*dQdr); Z in N/cm’2 = MPa u=R*T*r*t*dQdZ+dp0Zdt; s=-R*(log(r)+r*Q-r*t*d~dt)-dp0dr; h=u+P/r; d2Odt~=~*aS+(4*tl+~*tcc)*a4+(2*tl*(tl+S*tcc)+4*t~)*aS; d2Qdt2=d2Qdt2+(3*t2*(~l+4*~cc)+5*t3)*a6; d2Qdt2=d2Qdt2+(4*t3*(tl+5*tcc)+6*t4)*aT; d2pOtdZ2=2*C3/t’3+6*C4/t’4+12*C5/t’5+20*C6/t’6-CT/t+C8/~/t; dudT=-R*r*t*t*d2Qdt2-t/T*d2p0tdt2; d2~dtdr=da2+(Zl+tcc)*da3+tl*(tl+2*tcc)*da4+t2*(~l+3*tcc)*da5; d2Qdtdr=d2Qdtdr+t3*(tl+4*tcc)*da6+~4*(tl+5*tcc)*daT; dudr=lO00*R*(dQdt+r*d2~dtdr);

Index

Accumulator, hydraulic 166-167, 755756, 759 Activation energy 949 Actuator, hydraulic 63-64, 69, 72 Admittance,characteristic or surge 797, 803, 838 shunt, per unit length 819 Aircraft arresting engine 257, 671-672, 679-680, 681 Amplifiers, electronic 684-685, 690-693 Amplitude ratio 410 Analog computer 174 Analog-to-digital converter 493 Analogies, effort and flow 38-39, 720, 869 Analyzer, dynamic or spectrum 493494, 500-501,507 Arrhenius factor 949 Arresting engine for aircraft 257, 671672, 679-680, 681 Atter~uation factor 829

for poles and zeros at origin 418419, 424 for real poles and zeros 418-422 for right-half plane zeros 418-420, 440-441,445-446, 458-459 Bond 21 activated 570, 657, 682-692 convection 867, 870-871 micro 36-38 vector 945 Bond graph 4, 7 equivalences of 242-253 pseudo 363, 727-728, 912-913 reduction of 247-253 word 21 Brake, mechanical 46-47 thermal runaway of 730, 733-734 Break frequency 421 Breedveld, P.C. 16n, 784n, 785-786 Briener, M. 953n Brenan, K.E. 321n Brillouin, Leon 797n, 843n Brown,F.T. 249n, 560, 795, 821, 858, 859n, 904, 943 Bulk modulus 95

Balance, electromechanical 767 Basun, R.S. 928n Beam, Bernoulli-Euler model of 822, 824-825, 861-863, 865-866 Timoshenko model of 861 Beating phenomenon404, 408 Beranek, Leo L. 746 Biran, A. 953n Block diagrams 345-346 Block, floating 96-97, 99, 154, 159 Boat, thrust and drag 25-26, 28-29, 5253, 56, 367, 797-709,711-712, 715-717 Bode plots 411-413, 419-439, 444-445 for combinedpoles and zeros 424441 for complexpoles and zeros 422.424

Cable, coaxial 801,855-856 Cadsim Engineering 134n Campbell, S.L. 321n Cannonical form, congruence 784 Capacitor, capacitance 94 variable geometry753-755, 757, 758, 759 Cauchy theorem 631 Causal strokes 135, 281-282, 871,899 Causality 134-143, 281-302, 898-899 with activated bonds 686-688 admittance 135 of compliancesand inertances 137138 997

998 with dependentenergy storages 309313, 326, 328-329 differential 136, 137, 309-313 for differential equations 139-141, 281-303, 309-331 of effort and flow sources 134-135 impedance 136 integral 136, 137 of junctions 135-139 with meshes 321-327 of over-causal models282, 309-313, 320-327, 328-329 of under-causal models 282, 309, 313-320, 329-331 for thermodynamicelements 898899 of transformers and gyrators 292295 Celerity (of wave) 802 Cellier, F.E. 16 Channel, fluid, end corrections for inertance 745-746 Characteristic 20-22 Characteristic admittance and impedance 797, 838 Characteristic equation and polynomial 390 Characteristic length 745 Characteristics, methodof 813n Chargeor displacement, electrical 32, 94 Chemicalaffinity 946 Chemical equilibrium 947-948 Chemical kinetics 943-952 Chemical potential 944 Chemical reactions 949-951 Chokedflow 873-875 Circuits 191-200, 205-209 electric 192-194,205-207,690-’692, 696-697 fluid 194-195, 198-200,207-209,221223, 225-228 mechanical 195-198 regenerative hydraulic 208-209 Coffer dam749-750, 751-752 Coil device, translating 68, 70, 72 Complexplane 409-410, 424-425, 497498, 510, 588-591, 593-599, 607, 616, 619, 629-650 Compliance 91, 93-99, 162-166

INDEX chordal 163-164, 170-171 electrical (capacitance) 94, 753755, 757-759 fluid, compressiblity 95-96, 166167, 919, 922 fluid, gravity 94-95, 108, 164-165 generalized 93-94, 162-164, 752 incremental or tangential 362-363 mechanical(gravity) 99, 100, 108, 172 mechanical (spring) 92-93, 162163~ 182-184 multiport 752 thermal 724-726, 897 thermodynamic 897-916 thermodynamic C-element 900901 thermodynamic CS-element 9019O5 Compressor 883-887, 891-897 " piston-cylinder 905-909, 915-916, 917-919, 925-927, 940, 941943 Conductance 119 Conductivity 738 Conductors, electric 31-32 Conformal mapping 740 Constraints, electrical 116-117,120-121 fluid 117-118, 120-121, 207-210 geometric 63, 115, 122, 213-219 holonomic 760-762, 765-768 nonholonomic 762-764 Constitutive relation, 93 Control (automatic) 583-650 bang-bang 611-612 design by pole placement for 611 digital 621-622, 647-648 with disturbances 592-593 dynamic compensation for 605-650 feedback 587-650 frequency-response methods for 629-650 modeling for 7-8 open-loop 583-587 optimal 584-587 position 584-590, 592, 594, 606608, 610-612, 615-620 suboptimal 586 Control mass, surface, volume18 Controllab Products B.V. 134n

INDEX Controller, phase lag 618-620, 642-643 phase lead 615-618, 640-642, 649650, 653-656 phase lead-lag 620-621 process or three-mode 614-615,644, 645-647 proportional 587, 639 proportional-plus-derivative (PD) 610-613, 644 proportional-plus-integral (PI) 606610, 644 proportional-plus-integral-plusderivative (PID) 614-615,644, 645-647 Control mass, volume and surface.18 Convolution 470-481,509-510 continuous 477-479, 480-481 discrete 471-476 integral 477, 509-510 Coulombfriction 46-47 Coupled and uncoupled behavior of elements 135-139 Couplers, irreversible 718-731 linear 704-707 passive 706 reciprocal 705 symmetric 706 nonconservative 697-711 CS macro element 901-905 Cylinder, hydraulic 63, 84-85, 207-209 Damping,critical 153 proportional 972 Dampingratio 149-153, 155 Dashpots 44-46, 300-302~308 DASSL321 Dauphin-Tanguy, G. 16 Decibel scale 412 Decrement, logarithmic 158 Degrees of freedom 416, 965 Delta model 746-747 Derivative, material or substantial 825 Derivative operator 345 Derivative time 610 Design process 1-5 Determinant, graph 563 Dewpoint 927 Differential-algebraic equations (DAE’s) 309, 313-321 Diffusion equation 842

999 Dimensional analysis 700-702 Dimensions, treatment of 10-12 Dirichelet principle 741-742 Dispersion and absorption of waves 831836 anomolous and normal 836 Displacement(variable) 30n, 32, 34, 92, 93 Displacement (volumetric) of pump motor, 64, 885 Distributed-parameter models 795 boundary problems in 845-851 comparison to lumped models 796-798 degenerate one-power 820 exponentially tapered 851-852, 856 globally symmetric one-power 820 locally symmetric one-power 820 multiple-power 820-821,858-863 one-dimensional 817-822 one-power 819-820, 837-854 uniform 817 Disturbances, response to 592-593 Drive, chain and sprocket 61 cam659-662, 670, 676, 678, 680 epicyclic gear 219 friction 721-722, 674 mechanical modulated 657-658 rolling contact 59-60, 674 Scotch yoke 676, 677 sheave (pulley) 60, 77-80, 216-218, 227 sheave (pulley), variable 674 timing belt 61 toothed or gear 61, 73, 219, 688 Ear, drain tube for 202-203, 334, 447, 575, 579-580 Efficiency, adiabatic 885 volumetric 885 Effort 30-39, 718-719 Eigenvalues and eigenvectors 528-529, 538, 540-544~966-967, 968 from MATLAB 543-544, 923-976 matrix of 529 Energy, internal, evaluation of 921 Enthalpy, stagnation 869 available and unavailable 870-871, 883n

1000 Entropy, evaluation of 920 Entropy flux 719 Equilibrium 17, 23-24, 258-271, 368377 meta-stable 371, 373 with resistances and compliances 262-263 unstable 17, 23-25, 265-269, 373374, 424 Equilibrium constant 947 Error, steady-state 588 Euler integration 179 Existence and uniqueness theorem 390 Exponential inputs 395-396 Feedback, unity 587-588 Feedwater heater 890 Field, compliance, conductanceand susceptance of 740 discrete compliance752-757, 784785 discrete inertance 760-762, 765-766, 786 discrete multiport 771-789 discrete resistance 697-705 energy storage 735 field, nodic 737-750 multiport vector 746-748 planar vector 740-744 three-dimensional vector 744746 rigid body 736 scalar 736 transmittance of 740 Field lumping 735-750 Final value theorem 519 Flow, in bond graph 30-39, 718-719 of incompressible fluid 33-34 coefficient (discharge) of Flywheel, adjustable 763-766, 770 Force, generalized 30-31, 38. Form factor 740 Fourier analysis 485-501 for wavelike behavior 805-808 Fourier series 486-491,497-501 Fourier series pair 488-489 Fourier transform 491-499, 507-508 Frequency response 408-446, 546-548 for undamped models 413-418 Friction, by RSCoupler 721-722

INDEX Gain 411 path 563 Gain margin 636-637 Gas, ideal 905 Gawthrop, P. 15 Gibbs function 870, 900n, 943 Ground-effect machine 316-319, 329331,335, 337-341 Gyrator, abstract 59 cascaded 73-74 connecting source to load 81-82, 213-214, 246-247 coupled to R, C or I elements 246247 ¯ mechanical67-68 transducing 68-70 H element 877 Half-arrows (power convention) 21, Heat conduction 719-721,827-832,839842, 847-848, 877-881 Heat engine, ideal 723, 729, 731, 732733 Heat exchanger, counterflow 859-861, 876-881,889 Helmholtz free energy 903, 922, 930 Helmholtz resonator 448 Hindmarsh, A.C. 321n Hodograph plane 808-812, 814 Holonomicconstraints 760-762, 765 Homogeneoussolution 146-147, 151155, 390-393 HRS element 879-881 HS element 877-879 Humidity,absolute or specific and relative 927 Hydraulic systems 3-5, 77-80, 117-118, 199, 207-210, 221-223, 225229, 230, 231, 239, 240, 275277, 300, 307 bleed-off, meter-in and meter-out 198-200, 205, 270-271, 280281 wave behavior in 810-812 Hysteresis 52 IC engine 74-77, 273-274 Identification, experimental 436-439, 444-446, 748

INDEX Impedance,characteristic or surge 838 self 706 series, per unit length 819 source and load 50 transfer 705 Impulse function 393, 496, 513 responses to 399-403, 459-462 Inductance 104-105, 168, 761-762 mutual 772-773 Inertance 91,102-108, 168-171 chordal 171 compressible flow 912 dependent 664-669 displacement modulated 760-766 electrical (inductance) 104-105,108, 168-169, 761-762 fluid channel 105-106, 912 generalized 103, 168-171 mechanical(mass and inertia) 102104, 108 mutual 771-776 nonlinear 168-171 rotational 104 Inertia, mass momentof 104 Inertia matrix 968 Initial value problem398-399, 515-516, 527, 970 Initial value theorem 518 IRC models 122-123 Jet, fluid 274-275 Jumpeffect 182-184, 187-189 Junction, generic 91, 110-111, 122-124, 2O5 0-junction 110-111, 118-122, 123, 124, 166-168, 723, 881 0S-junction 881-883 1-junction 111-118, 123, 124, 166168, 723, 883 1S-junction 883-886 Junction structure, 205 Karmakar, Ranjit 16 Karnopp, D. 15, 727, 912 Keenan, J.H. 930 Kierzenka, J.A. 321n Kinetic energy 102-103, 169-171 Laplace transform 459, 463-464, 507522 definition 510

1001 derivative relations 510-512 inverse 510 one-sided 508 of output variables 514-517 of singularity functions 512-513 tables 461, 511,979-982 two-sided 508 Laplace’s equation 738 Leakage, internal and external 210 Learning styles 8-10 Lever or link, floating 218, 774-776 Limit-cycle 259, 266-268, 269, 286-291, 374-377 Linearity 341-354 and differential equations 343-345 differentiation and integration properties with 401 stationary, non-stationary 344 Linear media, generalized 781-783 Linearization 358-381 and bond graphs 359, 360, 363 of compliance 362-363 of differential equations 368-371 for one variable 361-363, 378-379 of resistance 358-361 stick-slip example374-377 for two variables 365-367, 380-381 Load, bond graph 21-22 LODSI 321 Loop rule 560-576 Loops, flat 567 Loops, open 569 Lorens, C. 561 Lorens Simulation S.A. 134n Machine, compressible fluid, 883-887, 909-911 ideal 58-70, 205-219, 657-659 Magnetostrictive transducer 781, 856 Magnitude ratio 410 Magrab, B. 953n Margolis, D. 15 Mass, generalized 970 Massaction, law of 948 Mass matrix 966 MathWorks,Inc., The 953n MATLAB953-963 arrays and matrices 956-958 Bodeplots 413, 433, 975 classical vibrations 972-976

1002 complex numbers 955 control flow commands960 convolution 474-476 curve fitting 959 data files 961 eigenvalues, eigenvectors 392,543544, 974-975 factorization, pole-zero 425-426 function files 175, 961-962 Fourier analysis 497-498 global variables 962 impulse 401-403 M-files 175, 960-963 Nichols charts 639, 643, 646, 654, 655 Nyquist plots 630, 635 Pade approximants 440n partial fraction expansion460-461, 462, 467, 469 plotting 958-959 responses to steps, impulses 401403 root locus 595-599, 604-605, 608, 627-628 roots 392 scalar calculations 594 script files 175, 960-961 simulation, linear 349-354, 375-377, 528 simulation, nonlinear 175-178, 925927, 934-937, 962-963 step 401-403 with thermodynamic data 983-996 transfer function equivalences 348349, 517 variables 954 Matrix, admittance and impedance 705 transmission 82-83, 250-252, 845852 Matrix exponential 352, 527, 542 ¯ Matrix methods 526-553 Mesh stub 569 Meshes, bond graph 247-250, 321-327 even, odd, neutral 249, 569-570 Meta-stability 371 Microphone, capacitance 788-789, 793794 Minimumand nonminimum phase lag 434 Minsky, Marvin vi

INDEX Modal damping 971-972 Modal matrix 533, 540 Modalvariables, shape 534, 536, 968, 973-974 Mode~normal 967 of motion 528 Modereduction 976-977 Modeling,and engineering science. 5-6 for control 7-8 guidelines for 220-223 languages for 6-7 misconceptions in 223-224 process of 2-5 Momentum,generalized 103 MOSFET53-54, 57 Motion, constraints on 213-219 lateral 35-36 longitudinal 32-33 rotational 34-35, 216-219 Motor, DCelectric 68-69, 81-82, 210212, 228-229, 233, 234, 241 capacitor start 26-27, 29 electrostatic 759 hydraulic 66, 209-210, 659, 698 induction 18-21, 26-27, 29, 378379, 386-387 synchronous 43 Muffler, acoustic 823, 856 Mukherjee, Amalendu 16 Nailer, hydraulic 305-306 Natural frequency 149-150, 154, 391 damped 151~ 391 Nichols charts 638-648 Nyquist plots 629-637, 654, 847-848 Nyqui~tstability criterion 630-635,64965O Nonholonomicconstraints 762-764 Nonlinearity, essential 364=365 Norton equivalent 242-244 Nozzle flow 46-47, 358-360, 873-875 Operator notation 345 Order of models, differential equations 139, 295-298 first-order 145-148 increase due to modulation 661664 minimum-order 296 second-order 148-155

INDEX Orifice flow 46-47, 358-360, 873-875 Orthogonal functions 488 Over-causal models 282, 309-313, 328 Overshoot 588 Pade approximants 440 Partial fraction expansion 459-467,517518 Particular solution 390, 397-398 Passivity 48, 245-246, 269 Paynter, Henry vi, 7, 106, 111, 741, 886n Pendulum172, 177-178, 962-963 coupled 550-551,552-553,554, 556558, 559-560 inverted, control of 623-624, 627628, 649-650, 653-656 Penrose, Rogervi, 8 Penstock, hydraulic 160, 407, 864 Percolation operator 831 Periodic excitations and responses 486491 Permeability 749 Perturbations 360, 361 Petzoid, L.R. 321n Phase angle, shift or lag 409-410, 434 Phase margin 636-638 Phase plane 290, 291,371-372 Phasors 410 Pipes, L.A. 859n Piston-and-cylinder 63, 207-209, 905909, 915-919, 925-927 Pitch surface, radius, diameter 61 Planing of boat 52-53, 367 Plant, example of 584-585 Polar plots for frequency response 629637 Pole placement, design by 611 Pole-zero excess 436 Poles 418, 424 Port, energy 19, 21 Potential energy 94, 169-171 Power 19, 30 Power-law waveforms 396 Propagation operator 838 Propeller drive 707-709, 711-712, 715717 Pseudo bond graphs 727-728, 912-913 Pulley system 216-218 Pump, displacement of 64-65

1003 dielectric 769 dynamicor impeller 48, 67, 697704, 751 fixed positive displacement 63-66, 209-210, 884-885 variable positive displacement 659 Pure-delay model and operator 439, 826, 839 Rayleigh line 878n Ram,hydraulic 63, 84-85, 89, 207-209, 407 Rampwaveform or function 394 Reaction force, chemical 946 Reaction rates, chemical 948 Reciprocity 705 Reduction, steady-state models, 250252, 259-262, 263-265 Refrigeration cycle 931-937 Regenerator, thermal 897-898 Reichelt, M.W.321n Residues 464, 508 Resistance (R element), chordal linear 43-46 linear biased 44, 47 linearized or tangential 360 mutual 705, 774 nonlinear 46-48 Resistivity 738 Reticulation 18 Reversing flows 939 Reynolds, W.C. 922, 930, 931 Robustness 636 Rocker 667-669, 678-679 Rod, longitudinal distributed 799, 801, 851-852 torsional distributed 798-799,838839, 848-852, 854, 857 Root-locus plots 588-591,593-599, 604605, 607, 616, 619, 627-628, 632 Roots blower 891-897 Roots of characteristic equation 390393 Rosenberg, R. 15, 134n, 727 Routh-Hurwitzstability criterion 600 RS element 720-723, 871-872 Runge-Kutta integration 178-181 S-plane 424-425,508, 548-549, 588-591,

1004 593-599, 607, 616, 619, 629650 Sample-data, sample-and-hold 621-622, 647-648 Saturation, magnetic 168-169 Saturation element 611, 612 Servomechanism, hydraulic 713-714 Settling time 588-590 Shaft, rotational waves 796-799, 801 variables 19-21 Signal-flow graphs 560-567 Similitude, Eulerian 702 Simulation, analog 174 MATLAB 175-178, 349-352, 962963 numerical 175-181 piston-cylinder compressor907-908, 925-927 refrigeration cycle 931-937 Runge-Kutta 178-181 Singularity waveforms393-395, 512-513 Sinusoidal waveforms395-396 Sinks, effort or flow 41-43 Shampine, L.F. 321n Slider-crank mechanism660-661, 662666, 672-673, 681-682, 758 Smith, L. 16 Snubber, mechanical 300-302, 308 Solenoid 761-762 Source 18, 41-43 :.. effort and flow 41-43 Source-load synthesis 22-25, 48-54 Spectral analysis, digital 493-499 Springs 92-93, 162-163, 265, 553 Sprinkler system 21-24, 77-80 Stability 17, 23-27, 258-259, 261-262, 265-269, 371-377, 600, 630635 Stability, relative 636-638 State, thermodynamicequations of 903, 919, 922 for gases 922-925, 983-987 for liquids 937-938 for refrigerants, 928-930, 987-993 for water 930-931,993-996 State postulate 870 State space 134, 174, 344-349 527, 533 trajectories in 290-291,371-372 transition to/from scalar form 346349

INDEX State transition matrix 527 State variables 140, 142, 174, Stationary vs. nonstationary systems 344 Steady-state 17, 22, Steady-state error 588 Steady-state response 409 Step waveformor function 393 responses to 147-149, 152-153, 401403, 462-463 Stewart, R.B. et al 922 Stick-slip instability 265-268, 286-291, 374-377, 385 Stiffness, generalized 93, 171,970 Stiffness matrix 966, 968 Stoichoimetric coefficients 945 String, stretched, wave motions 800801, 804-808 Superposition 148-149, 156, 341-343 Surge admittance 797, 801, 838 Susceptance, generalized 122, 171 Susceptivity 738 Synthesis, source-load 22-27, 48-54, 213214 System 18 Tachometer 68 Tank, liquid storage 94-95, 164-165,182, 185-186, 299, 358-361, 362363, 529-535, 566-567, 601602 Tee model 746-747 Tetrahedron of state 106-107 Thermal expansion coefficient 779 Thermal systems 718-731,867-952 control of 609-610 Therr~odynamicproperties, evaluation of See State, thermodynamic equations of Thermoelastic rod 779-781 Thevenin equivalent 242-244 Thoma, J. 16 Thompsonprinciple 741, 743-745 Three-tank system 529-535, 601-602, 604-605 Time constant 145, 391 Time-delay operator 439, 514,826, 839 Time-derivative operator 3.45, 512 Time, integral 606 Time-integral operator 512

INDEX Toilet 689-690, 696 Tool chatter 277-278 Transducer 62, 68 magnetic pick-off 792 piezoelectric 776-779 piezomagnetic (magnetostrictive) 781-782, 856 Transfer function 344-345,409, 508,515 expansion 459-467 factorization 424-426 frequency 409 matrix 345, 515 Transformers, abstract 58-59 cascaded 72-73 connecting source to load 74-81, 247-248 electric 62, 74, 771-774, 776-777, 790-791 mechanical 59-61 modulated 657-672 electrical 674, 675 hydraulic 675 locally 659-661, 664-667 remotely 658-659 multicoil 776-777 oil-cooled 790-791 transducing 62-67 Transient responses, general 463-464 Transient solutions 151 Transistor 55 Transmission, vehicle, hydraulic 694695 Transmission matrix 82-83, 250-252, 845-852, 859 Tube, acoustic 800, 801,809-812, 851, 853-854, 856 Tube, hot fluid wave propagation in 825-832 Two-tank system 299, 306 Type-zero, -one, -two, -minus-one models 435-436 U-tube 126-127, 133,143,145,157 Under-causal models313-328, 329-331 algebraic reduction for313-316 commerical software for 321 differentiation methodfor 316-319, 329-331, 337-341 non-zero virtual energy storage

1005 methodfor 320, 329-331,340341 Undetermined coefficients, method of 397-398 Units, treatment of 12-15 Valve, adjustable hydraulic 365-366, 380-381, 713-714 relief 200 reverse check 275 spool 380-381,388, 713-714 Variables, 17 linearized or perturbation 360-361 state 140 Vehicle drive 74-77, 694-695 Vehicle dynamics 321-326,539-545,694695 Velocity, angular 19 of energy propagation 842-845 generalized 30-31, 39 group 834-836 phase 826 Vibration absorber 182-184, 187-189, 416-418, 453, 505, 965-968 Vibration isolator 131, 144, 158, 285, 370-371, 414-415, 442, 447448, 454-455, 552, 558-559 Vibration of building 453-454, 554-555, 972-976 Vibrations, classical 965-977 Virtual work, theorem of 752 Viscometer, drum 125-126, 133, 156, 160 Viscous damper, untuned 453 Wagner, Wolfgang 930n Water skiing 56 Wavedelay model, bilateral 797, 798804 Fourier analysis for 805-807 Wave number 826 Wavepropagation 825-836 pure transport model of 825-826 Wavespeed 801, 803, 826, 839 Wavetravel time 439, 797 Wavelength 826 Waves,dispersion and absorption 832836 gravity or surface in liquid 821822, 824, 836-837

1006 in one-power symmetric models 837-842 reflected and transmitted 813, 816 Wheatstone bridge 193-194 Wilberforce spring 553 Wilson, D.P. 928n

INDEX Z-transform 622 Zeros 418-426 at origin 418-419, 435 in left-half plane 418-420,434-436, 445-446 in right-half plane 418-420, 434436, 439-442, 445-446, 458459