Energy Conversion Kenneth C. Weston Second Edition_2000.pdf

This book is dedicated to Engineering Students everywhere. Especially mine.

Preface to the Electronic Edition

In the Spring of 2000, Brooks/Cole (the most recent publisher of record) declared Energy Conversion out of print and returned the copyright to the author. Because of the unavailability of the films used in the printing of the book, the author decided to develop an electronic edition for the world-wide-web by revising a word processor draft to conform to the published text, converting it to Adobe Acrobat pdf form, with minor modifications and corrections where needed. This electronic textbook is offered in anticipation of the day when students will carry all of their texts and reference materials to class in a single electronic reader no larger than a book of a few hundred pages. Such readers are available now and notebook computers using the Acrobat reader could be used as such. It is in a spirit of conservation that the author hopes that students and other users will download chapters from the internet as they are needed and store them on their personal computers rather than printing them. With the cooperation and assistance of Dr. Dale Schoenefeld and Ms. Janet Cairns of the University of Tulsa, it was decided to provide the ebook from the university server. The author very much appreciates their participation and that of the University in this effort. Dr. Andrew Dykes again contributed to the updated Chapter 10; and my scholar-wife, Ruth, proof-read the entire book and helped me to avoid a multitude of problems with the text. I am, of course very grateful for their contributions to this non-profit venture. As I used the printed text as an instructor, before retiring, and as I worked on the electronic edition, I found a deeper appreciation of the quantity and quality of the efforts expended by the production editor of the printed version, Tad W. Bornhoft, and his associates. I came to admire and appreciate their work more and more. It was a job welldone. Thank you all very much. I am grateful to the instructors and students who have used Energy Conversion in the past. It is for them and future scholars that I am attempting to make this work more readily available. I hope that they find it useful and that it makes a pleasant and meaningful contribution to engineering education. Kenneth C. Weston December 2000

Preface to the First Edition

A solid grounding in heat and power has long been a characteristic expected of mechanical engineers. This text deals with energy conversion topics that should be well understood by all mechanical engineers. It is intended for use in an introductory three-semester-hour course in energy conversion, to follow first courses in thermodynamics and fluid mechanics and, where possible, heat transfer. No attempt is made to treat electrical motors, generators, and other conventional electric power equipment dealt with in electrical engineering power courses. Rather, it focuses, in the first six chapters, on the three predominant thermal power systems: the steam power plant, the gas turbine, and the reciprocating engine. These are considered the mainstream energy converters in which all mechanical engineers should be well grounded. The remaining five chapters provide a variety of choices for an instructor to select to round out a three-hour junior or senior level course. While the latter chapters depend on fundamentals appearing in the first six chapters as much as possible, they are organized and written to be independent of each other, so that they may be used in almost any order, or may be completely ignored. It is the author’s experience that many students have difficulty with the concepts of thermodynamics and fail to grasp its importance and power. A course in energy conversion, which applies the concepts of thermodynamics and introduces and analyzes the prime energy conversion devices, can be an eye-opener to those students who need concrete applications and are stimulated by them. At the same time, a good course in energy conversion can motivate further study of thermodynamics and the other engineering sciences. A number of textbooks are available for the study of energy conversion. Most of these books are best suited for advanced and graduate students because of an abundance of detail that may distract the undergraduate—fresh from first encounters with thermodynamics and fluid mechanics—from the fundamentals, from the major thrusts of system behavior and operation, and from engineering analysis. For this reason, the present text is an attempt at user-friendliness, which trades excessive technical detail for a fuller and easier development of mainline topics. Thus this text limits the amount of peripheral information presented and focuses on topics of current importance and those likely to play a significant role in the careers of the readers. It seeks to act as a convenient bridge between the engineering sciences and advanced courses and, at the same time, provide a useful terminal course for students pursuing other interests. The approach taken is to provide a brief review of fundamentals from thermodynamics and fluid mechanics and to immediately focus on an in-depth study of a major energy conversion system—the steam power plant—together with more advanced fundamentals needed for its understanding and analysis. The premise is that a thorough understanding of a major energy conversion system establishes a point of reference by which the student may

better appreciate and understand other systems. This is the major thrust of the first four chapters. The first chapter is a brief review and introduction to the text. Instructors may wish to assign the reading of Chapter 1 and a few problems for review at the first class meeting, and then answer questions and proceed to the important material of Chapter 2 in the second session. Chapter 2 starts with the Rankine cycle and examines the important refinements of the cycle, one at a time, culminating in the study of a typical cycle of a large modern steam power plant. Sections of the chapter may be assigned at a pace consistent with the students’ prior exposure to the Rankine cycle. The author usually assigns one or more problems at each class and spends one class each on the basic cycle, reheat, efficiencies and pressure losses, regeneration and feedwater heaters, combined mass conservation and First Law analysis with heaters, and two classes on the study of the power plant flowsheet. At this point the student should have the capability to interpret and analyze the flowsheet of any steam power plant presented to him or her. This may be an appropriate time for a first examination to assure that the attention of the student has focused on thermodynamic cycle and system analysis fundamentals before considering the basics of Rankine cycle implementation. With a good grasp of the cycles of steam power plants in hand, it is appropriate for the student to study some of the fundamentals of fuels and combustion in order to proceed further in understanding steam plant design and operation and to prepare him or her for later chapters. Thus the student should be easily motivated to the study of fuels and combustion in Chapter 3. To the extent that this and the preceding chapter are review for the student, the pace of coverage may be adjusted by the instructor in reading and problem assignments and class discussions. Chapter 4 completes the study of conventional steam power generation by following and analyzing the major flows in a steam plant: water, fuel, and gases, placing emphasis on the hardware and systems involved in these flows. A brief introduction to the fundamentals of engineering economy is included in this chapter, so that financial aspects of power plant operation may be considered. The latter material is not intended to replace a course in engineering economy. Instead, it provides basic information that may appear too late in some curricula for use in a junior-level energy conversion course. Chapter 4 concludes with a back-of-the-envelope type analysis of plant characteristics, which provides the student with an overview of the magnitudes of parameters associated with large steam plant operation and with an opportunity to reflect on the roles of thermodynamics, economics, and analysis in the plant design. The succeeding chapters provide opportunities to show the universality of many of the fundamentals and methods presented in the first four. The fifth chapter treats the important and exciting topics of gas turbines and jet propulsion. The characteristics of stationary gas turbines are studied and contrasted. The jet engine is studied as a form of gas turbine with special design and performance characteristics. Care is taken to provide the student with a conceptual understanding of the basic machine before delving into detailed analysis and alternate configurations. Advanced topics such as combined cycles, cogeneration, steam injection, polytropic efficiencies, and turbofan engines are deferred to Chapter 9, for those instructors who prefer not to treat these topics in an introductory energy conversion course.

As in other chapters, the instructor is largely in control of the level of the presentation by the choice and the pace of the assignments that he or she makes. One or several text sections may be assigned per class meeting, along with problems of appropriate number and difficulty for the students. At the end of Chapter 5, the instructor may wish to assign the development of a computer program or a spreadsheet for gas turbine analysis, or the use of an existing program or spreadsheet for optimization of a configuration for a given application, or for a preliminary design study of a gas turbine or steam plant. Although combined gas turbine and steam cycles are not considered until Chapter 9, some instructors may wish to deal with them at this point, and perhaps assign a design problem. Chapter 6 considers the phenomena and characteristics of the reciprocating engine and the engineering parameters describing its performance. The chapter develops and compares models of both Otto and Diesel cycles and explores some of the fundamental problems and aspects of their implementation in working engines. Chapter 7 complements Chapter 6 by providing a brief look at the Wankel rotary engine. While not a major player in energy conversion at this time, the rotary engine offers a unique opportunity for students to analyze and understand an entirely different, successful, and intriguing implementation of the Otto cycle. For those instructors wishing to substitute other materials, Chapter 7 may be omitted without concern for student comprehension of the remainder of the book. While refrigeration and air conditioning are not power generation technologies, they clearly involve energy conversion in important ways and represent an important mechanical engineering discipline. Chapter 8 briefly considers the fundamentals and hardware of this field briefly, giving a framework within which HVAC engineering may be understood and upon which engineering analyses may be built. The important topic of the analysis of moist air is introduced by treating moist air as a binary mixture of ideal gases and carried to the point where the student is introduced to HVAC system design using the psychrometric chart. Chapter 9 considers some of the important technological problems of the day and a selected collection of advanced energy conversion techniques that mechanical engineers use to deal with them. The treatments of most of these technologies are largely independent of each other, so that instructors may elect to use either the entire chapter or only those advanced topics that they consider most appropriate for their course. A notable exception is that the presentation on turbofans considers using either isentropic or polytropic efficiencies. It is recommended to cover the preceding section on polytropic efficiency before proceeding to turbofans. It is hoped that the study of this chapter will motivate students to investigate the selected topics in greater detail, aided by the bibliography. Instructors may wish at this point to assign more in-depth study using the library on one of these topics or on other advanced systems. Certainly, nuclear power is controversial. While the development of new nuclear fission plants in the United States is currently dormant, the subject remains important and is likely to become more important to mechanical engineers in the future. Some texts devote a great deal of space to nuclear power while others do not choose to treat it at all. The decision here was to take the middle ground, to provide a survey that allows mechanical engineers to place the subject in perspective with respect to other major energy conversion technologies. Chapter 10 thus focuses on relevant aspects of nuclear fission power without going into nuclear reactor physics and design in detail. The chapter may be studied at any time after

the first four chapters on steam power, or may be omitted completely if the instructor desires. The final chapter offers vistas on a number of promising technologies, some of which avoid the heat engine approach that dominates today’s modern power and propulsion systems. An introduction to electrical energy storage-battery technology provides a lead-in to fuel-cell energy conversion, now under intensive development as a flexible solution to a diversity of electrical power generation problems. Magnetohydrodynamics (MHD) is considered as a means of producing electricity directly from a hot fluid without the use of a turbine. The inspiration of the sun as a massive and eternal energy source continues to drive interest in solar energy, spurred by the success of small-scale photovoltaic conversion in watches and calculators, and of remote power applications. The chapter closes with a brief consideration of the use of hydrogen as a secondary energy source and the transition to a steady-state energy system. A course in energy conversion may serve as a terminal course in the thermal-fluidsenergy stem for those students pursuing other disciplinary interests. These students emerge with an appreciation of the purposes and methods of thermodynamics, of engineering hardware, and of important segments of industry. For others, the course may serve as a motivating lead-in to electives such as advanced thermodynamics, advanced fluids and heat transfer, refrigeration and air conditioning, solar energy, turbomachinery, gas turbines, and propulsion. At the University of Tulsa the course is offered to sixth semester students, allowing elective studies in the seventh and eighth semesters. The goal of this book is to provide a lucid learning tool for good students who may feel insecure in their understanding of the engineering sciences and their uses. An effort is made to tie theory, concepts, and techniques to earlier courses, which may not have been mastered. A special effort is made also to help the student become familiar with and appreciate the important hardware associated with energy conversion, through the presentation of numerous photographs and diagrams. The author hopes that this study of energy conversion will not only prepare the student for work in this field but will also provide motivation for further study of the engineering sciences because their usefulness in design and analysis is better appreciated. Numerous examples are provided in the text in connection with the more important topics. Photographs and schematic diagrams of modern equipment are included to aid the student in hardware visualization. Students should be encouraged to study these carefully as the useful learning tools that they are. As a further guide and aid to the student, terms which should be part of his or her vocabulary are italicized. Many acronyms and abbreviations are used in energy conversion as in other fields of endeavor. There is no point in delaying the inevitable. They are used sparingly but where appropriate in the text and are identified in their first occurrence and in the list of symbols. We are in the midst of a protracted transition from one system of engineering units, the English system, to an international system, the Systeme International d’Unites, usually called SI. It is essential that this and other texts provide experience in using both; therefore, an effort has been made to provide discussions and problems expressed in both the SI and English systems. The importance of units, not only as a tool of communication, but as an analytic tool warrants repeated emphasis. Using units for dimensional analysis may often be the

difference between obtaining a correct problem solution and falling into quantitative error. To emphasize the importance of units, equations in the text appear with the units of the dependent variable in both English and SI forms in brackets, as for example in the units of specific energy: [Btu/lbm | kJ/kg]. When an expression is dimensionless that fact is indicated by [dl] following the equation. The personal computer has become a valuable tool for practicing engineers and should be used by engineering students at every appropriate opportunity. The author has found spreadsheets to be particularly useful in a variety of activities including energy conversion. Spreadsheets are provided for the example problems presented in spreadsheet format in the text. This allows students to examine these calculations in as much detail as they wish, to conveniently repeat the calculations for alternate inputs, and to substitute alternate calculation details. Thus instructors may conveniently assign “what if” studies in these cases. A spreadsheet of data extracted from the JANAF tables is included to allow easy flame temperature and heat transfer calculations in connection with Chapter 3. While the spreadsheets were developed using Quattro Pro, the examples are provided in the WK3 format, so that they may be used with any spreadsheet that is compatible with this popular Lotus 1-2-3 format. Those unfamiliar with spreadsheets will find that a few hours with a modern spreadsheet and a good tutorial will enable its use in powerful ways. One of the characteristics of spreadsheets is that the new user can usually start solving problems at the first sitting. After using the tutorial, the new user should pick a modest problem of current interest to solve, or try a few problems that were previously completed by hand. It is wise for new users to select a spreadsheet that is widely used in their university or company or by their friends, so that it will be easy to exchange information with and learn from others. During the years of preparation of this book, many friends, colleagues, reviewers, and students have made constructive suggestions and offered ideas and encouragement. They are too numerous to thank individually here. It is, nevertheless, appropriate to express my appreciation to them and to acknowledge their invaluable contributions. It is particularly fitting, however, to recognize the special contributions of friend and colleague, Dr. Andrew A. Dykes, for both his insightful critique of the first draft and his substantive and continuing contributions to Chapter 10. Some of his expertise is written into the pages of that chapter; any deficiencies there are mine. I am greatly indebted also to industry for providing many of the photographs and illustrations that the reader will find in these pages. Sources are identified with the figures. Finally, I must acknowledge the patience and support of my family and the administration, faculty, and staff of the University of Tulsa, who helped ease the pain and maximize the joys associated with the preparation of this work.

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CHAPTER ONE FUNDAMENTALS OF ENERGY CONVERSION 1.1 Introduction Energy conversion engineering (or heat-power engineering, as it was called prior to the Second World War), has been one of the central themes in the development of the engineering profession. It is concerned with the transformation of energy from sources such as fossil and nuclear fuels and the sun into conveniently used forms such as electrical energy, rotational and propulsive energy, and heating and cooling. A multitude of choices and challenges face the modern energy conversion engineer. A few years ago major segments of the energy conversion industry were settled into a pattern of slow innovation. Most automobile manufacturers were satisfied to manufacture engines that had evolved from those produced twenty years earlier, some of which boasted 400 horsepower and consumed a gallon of leaded gasolene every eight or nine miles. Many electric power utilities were content with state-of-theart, reliable, fossil-fuel-consuming steam power plants, except for a few forward-looking, and in several cases unfortunate, exceptions that risked the nuclear alternative. Then came the oil embargo of the 1970s, high fuel prices, and threatened shortages. Also, the public and legislatures began to recognize that air pollution produced by factories, power plants, and automobiles and other forms of environmental pollution were harmful. International competitors, producing quality automobiles with smaller, lower-pollution engines, exceptional gas mileage, and lower prices shook the automobile industry. The limitations of the Earth's resources and environment started to come into clearer focus. These and other influences have been helping to create a more favorable climate for consideration, if not total acceptance, of energy conversion alternatives and new concepts. There are opposing factors, however. Among them are limited research and development funding due to budgetary constraints, emphasis on short-term rather than long-term goals because of entrepreneurial insistence on rapid payback on investment, and managerial obsession with the bottom line. But more open attitudes have become established. New as well as previously shelved ideas are now being considered or reconsidered, tested, and sometimes implemented. A few examples are combined steam and gas turbine cycles, rotary combustion engines, solar and windmill power farms, stationary and vehicular gas turbine power plants, cogeneration, photovoltaic solar power, refuse-derived fuel, stratified charge engines, turbocharged engines, fluidized-bed combustors, and coal-gasification power plants. We are living in a

2 rapidly changing world that requires continuing adaptation of old technologies and the development of new ones. Energy conversion engineering is a more stimulating, complex, and viable field today because of this altered climate. A Look Backward How did we get where we are today? A good answer requires a study of the history of science and engineering worthy of many volumes. Table 1.1 identifies a few pivotal ideas and inventions, some of them landmarks to energy conversion engineers, and the names of the thinkers and movers associated with them. Of course, the table cannot present the entire history of energy conversion engineering. Omitted are the contributions of Newton and Euler, the first rocket engine, the V-8 engine, the ramjet and fanjet. The reader could easily come up with many other glaring omissions and extend the table indefinitely. While the names of one or two persons are associated with each landmark achievement, most of these landmarks were the products of teams of unheralded individuals whose talents were crucial to success. Moreover, the successes did not occur in a vacuum, but benefited and followed from the advances and failures of others. Unknown or renowned, each engineer and his or her associates can make a contribution to the progress of mankind. Table 1.1 can only hint at how the persons, ideas, and events listed there relied on their predecessors and on a host of less well-known scientific and technological advances. A brief bibliography of historical sources is given at the end of the chapter. These works chronicle the efforts of famous and unsung heroes, and a few villains, of energy conversion and their struggles with ideas and limiting tools and resources to produce machines for man and industry. The historical progress of industry and technology was slow until the fundamentals of thermodynamics and electromagnetism were established in the ninteenth century. The blossoming of energy technology and its central role in the industrial revolution is well known to all students of history. It is also abundantly clear that the development of nuclear power in the second half of the twentieth century grew from theoretical and experimental scientific advances of the first half century. After a little reflection on Table 1.1, there should be no further need to justify a fundamental scientific and mathematical approach to energy conversion engineering. TABLE 1-1 Some Significant Events in the History of Energy Conversion ___________________________________________________________________________ Giovanni Branca Impulse steam turbine proposal 1629 Thomas Newcomen

Atmospheric engine using steam (first widely used Heat engine)

1700

James Watt

Separate steam condenser idea; and first Boulton and Watt condensing steam engine

1765 1775

3 Table 1.1 (continued) ___________________________________________________________________________ John Barber Gas turbine ideas and patent 1791 Benjamin Thompson (Count Rumford)

Observed conversion of mechanical energy to heat while boring cannon

1798

Robert Fulton

First commercial steamboat

1807

Robert Stirling

Stirling engine

1816

N. L. Sadi Carnot

Principles for an ideal heat engine (foundations of thermodynamics)

1824

Michael Faraday

First electric current generator

1831

Robert Mayer

Equivalence of heat and work

1842

James Joule

Basic ideas of the First Law of Thermodynamics; and measured the mechanical equivalent of heat

1847 1849

Rudolph Clausius

Second Law of Thermodynamics

1850

William Thompson (Lord Kelvin)

Alternate form of the Second Law of Thermodynamics

1851

Etienne Lenoir

Internal combustion engine without without mechanical compression

1860

A. Beau de Rochas

Four-stroke cycle internal combustion engine concept

1862

James C. Maxwell

Mathematical principles of electromagnetics

1865

Niklaus Otto

Four-stroke cycle internal combustion engine

1876

Charles Parsons

Multistage, axial-flow reaction steam turbine

1884

Thomas Edison

Pearl Street steam-engine-driven electrical power plant

1884

C.G.P. de Laval

Impulse steam turbine with convergent-divergent nozzle

1889

Rudolph Diesel

Compression ignition engine

1892

___

First hydroelectric power at Niagara Falls

1895

Albert Einstein

Mass-energy equivalence

1905

Ernst Schrodinger

Quantum wave mechanics

1926

Frank Whittle

Turbojet engine patent application; and first jet engine static test

1930 1937

4 Table 1.1 (concluded) ___________________________________________________________________________ Otto Hahn Discovery of nuclear fission 1938 Hans von Ohain

First turbojet engine flight

1939

J. Ackeret, C. Keller

Closed-cycle gas turbine electric power generation

1939

Enrico Fermi

Nuclear fission demonstration at the University of Chicago

1942

Felix Wankel

Rotary internal combustion engine

1954

Production of electricity via nuclear fission by a utility at Shippingport, Pennsylvania

1957 .

NASA

Rocket-powered landing of man on the moon

1969

Electricité de France

Superphénix 1200-MW fast breeder reactor – first grid power

1986

________________________________________________________________ Since energy conversion engineering is deeply rooted in thermodynamics, fluid mechanics, and heat transfer, this chapter briefly reviews those aspects of these disciplines that are necessary for understanding, analysis, and design in the field of energy conversion. 1.2 Fundamentals of Thermodynamics The subject of thermodynamics stems from the notions of temperature, heat, and work. Although, the following discussion makes occasional reference to molecules and particles, useful in clarifying and motivating concepts in thermodynamics, thermodynamics as a science deals with matter as continuous rather than as discrete or granular. System, Surroundings, and Universe We define a pure substance as a homogeneous collection of matter. Consider a fixed mass of a pure substance bounded by a closed, impenetrable, flexible surface. Such a mass, called a system, is depicted schematically in Figure 1.1(a). For example, the system could be a collection of molecules of water, air, refrigerant, or combustion gas confined in a closed container such as the boundary formed by a cylinder and a fitted piston, Figure 1.1 (b). A system should always be defined carefully, to ensure that the same particles are in the system at all times. All other matter which can interact with the system is called the surroundings. The combination of the system and the surroundings is termed the universe, used here not in a cosmological sense, but to include only the

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system and all matter which could interact with the system. Thermodynamics and energy conversion are concerned with changes in the system and in its interactions with the surroundings. State The mass contained within a system can exist in a variety of conditions call states. Qualitatively, the concept of state is familiar. For example, the system state of a gasmight be described qualitatively by saying that the system is at a high temperature and a low pressure. Values of temperature and pressure are characteristics that identify a particular condition of the system. Thus a unique condition of the system is called a state. Thermodynamic Equilibrium A system is said to be in thermodynamic equilibrium if, over a long period of time, no change in the character or state of the system is observed.

6 Thermodynamic Properties It is a fundamental assumption of thermodynamics that a state of thermodynamic equilibrium of a given system may be described by a few observable characteristics called thermodynamic properties, such as pressure, temperature, and volume. Obviously, this approach excludes the possibility of description of the condition of the molecules of the system, a concern that is left to the fields of statistical and quantum mechanics and kinetic theory. Nevertheless, it is frequently useful to think of thermodynamic phenomena in molecular terms. The Temperature Property. Temperature is a measure of the vigor of the molecular activity of a system. How can it be observed? A thermometer measures a system property called temperature when it is in intimate and prolonged contact (thermodynamic equilibrium) with the system. A mercury-in-glass thermometer, for instance, functions by thermal expansion or contraction of mercury within a glass bulb. The bulb must be in intimate thermal contact with the observed system so that the temperatures of the bulb and the system are the same. As a result of the equilibrium, elongation or contraction of a narrow column of mercury connected to the bulb indicates the temperature change of the system with which it is in contact. The Pressure Property. Another way to observe changes in the state of a liquid or gaseous system is to connect a manometer to the system and observe the level of the free surface of the manometer fluid . The manometer free surface rises or drops as the force per unit area or pressure acting on the manometer-system interface changes. Defining a State It has been empirically observed that an equilibrium state of a system containing a single phase of a pure substance is defined by two thermodynamic properties. Thus, if we observe the temperature and pressure of such a system, we can identify when the system is in a particular thermodynamic state. Extensive and Intensive properties Properties that are dependent on mass are known as extensive properties. For these properties that indicate quantity, a given property is the sum of the the corresponding properties of the subsystems comprising the system. Examples are internal energy and volume. Thus, adding the internal energies and volumes of subsystems yields the internal energy and the volume of the system, respectively. In contrast, properties that may vary from point to point and that do not change with the mass of the system are called intensive properties. Temperature and pressure are well-known examples. For instance, thermometers at different locations in a system may indicate differing temperatures. But if a system is in equilibrium, the temperatures

7 of all its subsystems must be identical and equal to the temperature of the system. Thus, a system has a single, unique temperature only when it is at equilibrium. Work From basic mechanics, work, W, is defined as the energy provided by an entity that exerts a force, F, in moving one or more particles through a distance, x. Thus work must be done by an external agent to decrease the volume, V, of a system of molecules. In the familiar piston-cylinder arrangement shown in Figure 1.1(b), an infinitesimal volume change of the system due to the motion of the piston is related to the differential work through the force-distance product: dW = Fdx = pAdx = pdV

[ft-lbf | n-m]

(1.1a)

dw = pdv

[Btu/lbm | kJ/kg]

(1.1b)

or

where p is the system pressure, and A is the piston cross-sectional area. Note that in Equation (1.1b), the lower case letters w and v denote work and volume on a unit mass basis. All extensive properties, i.e., those properties of state that are proportional to mass, are denoted by lowercase characters when on a unit mass basis. These are called specific properties. Thus, if V represents volume, then v denotes specific volume. Although work is not a property of state, it is dealt with in the same way. Also note that the English units of energy in Equation (1.1a) are given in mechanical units. Alternately, the British Thermal unit [Btu] may be used, as in Equation (1.1b). The two sets of units are related by the famous conversion factor known as the mechanical equivalent of heat, 778 ft-lbf/Btu. The student should pay close attention to the consistency of units in all calculations. Conversion factors are frequently required and are not explicitly included in many equations. For the convenience of the reader, Appendix A lists physical constants and conversion factors. When work decreases the volume of a system, the molecules of the system move closer together. The moving molecules then collide more frequently with each other and with the walls of their container. As a result, the average forces (and hence pressures) on the system boundaries increase. Thus the state of the system may be changed by work done on the system. Heat Given a system immersed in a container of hot fluid, by virtue of a difference in temperature between the system and the surrounding fluid, energy passes from the fluid to the system. We say that heat, Q [Btu | kJ], is transferred to the system. The system is

8 observed to increase in temperature or to change phase or both. Thus heat transfer to or from the system, like work, can also change the state of the matter within the system. When the system and the surrounding fluid are at the same temperature, no heat is transferred. In this case the system and surroundings are said to be in thermal equilibrium. The term adiabatic is used to designate a system in which no heat crosses the system boundaries. A system is often approximated as an adiabatic system if it is well insulated. Heat and Work Are Not Properties Mechanics teaches that work can change the kinetic energy of mass and can change the elevation or potential energy of mass in a gravitational field. Thus work performed by an outside agent on the system boundary can change the energy associated with the particles that make up the system. Likewise, heat is energy crossing the boundary of a system, increasing or decreasing the energy of the molecules within. Thus heat and work are not properties of state but forms of energy that are transported across system boundaries to or from the environment. They are sometimes referred to as energy in transit. Energy conversion engineering is vitally concerned with devices that use and create energy in transit. Internal Energy and The First Law of Thermodynamics A property of a system that reflects the energy of the molecules of the system is called the internal energy, U. The Law of Conservation of Energy states that energy can be neither created nor destroyed. Thus the internal energy of a system can change only when energy crosses a boundary of the system, i.e., when heat and/or work interact with the system. This is expressed in an equation known as the First Law of Thermodynamics. In differential form the First Law is: du = dq – dw

[Btu/lbm | kJ/kg]

(1.2)

Here, u is the internal energy per unit mass, a property of state, and q and w are, respectively, heat and work per unit mass. The differentials indicate infinitesimal changes in quantity of each energy form. Here, we adopt the common sign convention of thermodynamics that both the heat entering the system and work done by the system are positive. This convention will be maintained throughout the text. Thus Equation (1.2) shows that heat into the system (positive) and work done on the system (negative) both increase the system’s internal energy. Cyclic Process A special and important form of the First Law of Thermodynamics is obtained by

9 integration of Equation (1.2) for a cyclic process. If a system, after undergoing arbitrary change due to heat and work, returns to its initial state, it is said to have participated in a cyclic process. The key points are: (1) the integral of any state property differential is the difference of its limits, and (2) the final state is the same as the initial state (hence there is no change in internal energy of the system)

Šdu = uf – ui = 0 where the special integral sign indicates integration over a single cycle and subscripts i and f designate, respectively, initial and final states. As a consequence, the integration of Equation (1.2) for a cycle yields:

Šdq = Šdw

[Btu/lbm | kJ/kg]

(1.3)

This states that the integral of all transfers of heat into the system, taking into account the sign convention, is the integral of all work done by the system. The latter is the net work of the system. The integrals in Equation (1.3) may be replaced by summations for a cyclic process that involves a finite number of heat and work terms. Because many heat engines operate in cyclic processes, it is sometimes convenient to evaluate the net work of a cycle using Equation (1.3) with heat additions and losses rather than using work directly. Arbitrary Process of a System Another important form of the First Law of Thermodynamics is the integral of Equation (1.2) for an arbitrary process involving a system: q = uf – ui + w

[Btu/lbm | kJ/kg]

(1.4)

where q and w are, respectively, the net heat transferred and net work for the process, and uf and ui are the final and initial values of the internal energy. Equation (1.4), like Equation (1.2), shows that a system that is rigid (w = 0) and adiabatic (q = 0) has an unchanging internal energy. It also shows, like Equation (1.3), that for a cyclic process the heat transferred must equal the work done. Reversibility and Irreversibility If a system undergoes a process in which temperature and pressure gradients are always small, the process may be thought of as a sequence of near-equilibrium states. If each of the states can be restored in reverse sequence, the process is said to be internally reversible. If the environmental changes accompanying the process can also be reversed in sequence, the process is called externally reversible. Thus, a reversible process is

10 one that is both internally and externally reversible. The reversible process becomes both a standard by which we measure the success of real processes in avoiding losses and a tool that we can use to derive thermodynamic relations that approximate reality. All real processes fail to satisfy the requirements for reversibility and are therefore irreversible. Irreversibility occurs due to temperature, pressure, composition, and velocity gradients caused by heat transfer, solid and fluid friction, chemical reaction, and high rates of work applied to the system. An engineer’s job frequently entails efforts to reduce irreversibility in machines and processes. Entropy and Enthalpy Entropy and enthalpy are thermodynamic properties that, like internal energy, usually appear in the form of differences between initial and final values. The entropy change of a system, ªs [Btu/lbm-R | kJ/kg-K], is defined as the integral of the ratio of the system differential heat transfer to the absolute temperature for a reversible thermodynamic path, that is, a path consisting of a sequence of well-defined thermodynamic states. In differential form this is equivalent to: ds = dqrev /T

[Btu/lbm-R | kJ/kg-K]

(1.5)

where the subscript rev denotes that the heat transfer must be evaluated along a reversible path made up of a sequence of neighboring thermodynamic states. It is implied that, for such a path, the system may be returned to its condition before the process took place by traversing the states in the reverse order. An important example of the use of Equation (1.5) considers a thermodynamic cycle composed of reversible processes. The cyclic integral, Equation (1.3), may then be used to show that the net work of the cycle is: wn = Šdq = ŠTds

[Btu/lbm | kJ/kg]

This shows that the area enclosed by a plot of a reversible cyclic process on a temperature-entropy diagram is the net work of the cycle. The enthalpy, h, is a property of state defined in terms of other properties: h = u + pv

[Btu/lbm | kJ/kg]

(1.6)

where h, u and v are, respectively, the system specific enthalpy, specific internal energy, and specific volume, and p is the pressure.

11

Two other important forms of the First Law make use of these properties. Substitution of Equations (1.1) and (1.5) in Equation (1.2) yields, for a reversible process Tds = du + pdv

[Btu/lbm | kJ/kg]

(1.7)

and differentiation of Equation (1.6), combined with elimination of du in Equation (1.7), gives Tds = dh - vdp

[Btu/lbm | kJ/kg]

(1.8)

Equations (1.7) and (1.8) may be regarded as relating changes in entropy for reversible processes to changes in internal energy and volume in the former and to changes in enthalpy and pressure in the latter. The fact that all quantities in these equations are properties of state implies that entropy must also be a thermodynamic property. Because entropy is a state property, the entropy change between two equilibrium states of a system is the same for all processes connecting them, reversible or irreversible. Figure 1.2 depicts several such processes 1-a-b-c-2, 1-d-2, and a sequence of nonequilibrium states not describable in thermodynamic terms indicated by the dashed line (an irreversible path). To use Equation (1.5) directly or as in Equations (1.7) and (1.8), a reversible path must be employed. Because of the path independence of state property changes, any reversible path will do. Thus the entropy change, s2 – s1,

12 may be evaluated by application of Equations (1.5), (1.7), or (1.8) to either of the reversible paths shown in Figure 1.2 or to any other reversible path connecting states 1 and 2. The Second Law of Thermodynamics While Equation (1.5) may be used to determine the entropy change of a system, the Second Law of Thermodynamics, is concerned with the entropy change of the universe, i.e., of both the system and the surroundings. Because entropy is an extensive property, the entropy of a system is the sum of the entropy of its parts. Applying this to the universe, the entropy of the universe is the sum of the entropy of the system and its surroundings. The Second Law may be stated as "The entropy change of the universe is non-negative":

ÎSuniv $ 0

[Btu/R | kJ/K]

(1.9)

Note that the entropy change of a system may be negative (entropy decrease) if the entropy change of its environment is positive (entropy increase) and sufficiently large that inequality (1.9) is satisfied. As an example: if the system is cooled, heat is transferred from the system. The heat flow is therefore negative, according to sign convention. Then, according to Equation (1.5), the system entropy change will also be negative; that is, the system entropy will decrease. The associated heat flow, however, is into the environment, hence positive with respect to the environment (considered as a system). Then Equation (1.5) requires that the environmental entropy change must be positive. The Second Law implies that, for the combined process to be possible, the environmental entropy change must exceed the magnitude of the system entropy change. The First Law of Thermodynamics deals with how the transfer of heat influences the system internal energy but says nothing about the nature of the heat transfer, i.e., whether the heat is transferred from hotter or colder surroundings. Experience tells us that the environment must be hotter to transfer heat to a cooler object, but the First Law is indifferent to the condition of the heat source. However, calculation of the entropy change for heat transfer from a cold body to a hot body yields a negative universe entropy change, violates the Second Law, and is therefore impossible. Thus the Second Law provides a way to distinguish between real and impossible processes. This is demonstrated in the following example:

EXAMPLE 1.1

(a) Calculate the entropy change of an infinite sink at 27°C temperature due to heat transfer into the sink of 1000 kJ. (b) Calculate the entropy change of an infinite source at 127°C losing the same amount of heat.

13 (c) What is the entropy change of the universe if the aforementioned source supplies 1000 kJ to the sink with no other exchanges? (d) What are the entropy changes if the direction of heat flow is reversed and the source becomes the sink? Solution

(a) Because the sink temperature is constant, Equation (1.5) shows that the entropy change of the sink is the heat transferred reversibly divided by the absolute temperature of the sink. This reversible process may be visualized as one in which heat is transferred from a source which is infinitesimally hotter than the system:

ªSsink = 1000/(273 + 27) = + 3.333 kJ/K. (b) Treating the source in the same way:

ªSsource = – 1000/(273 + 127) = – 2.5 kJ/K. (c) Because the entropy change of the universe is the sum of the entropy changes of source and sink, the two acting together to transfer 1000kJ irreversibly give:

ªSuniverse = 3.333 – 2.5 = +0.833 kJ/K > 0 which satisfies the Second Law inequality (1.9). (d) A similar approach with the direction of heat flow reversed, taking care to observe the sign convention, gives

ÎSsink = (– 1000 )/(273 + 27) = – 3.333 kJ/K ÎSsource = (1000)/(273 + 127) = + 2.555 kJ ÎSuniv = – 3.333 + 2.5 = – 0.833 kJ/K. Thus we see that heat flow from a low to a high temperature reduces the entropy of the universe, violates the Second Law, and therefore is not possible. ____________________________________________________________________ Parts a, b, and c of Example 1.1 show that the entropy change of the universe depends on the temperature difference driving the heat transfer process:

ÎSuniv = Q(1/Tsink – 1/ Tsource) = Q( Tsource – Tsink) / Tsource Tsink Note that if the temperature difference is zero, the universe entropy change is also zero and the heat transfer is reversible. For finite positive temperature differences, ÎSuniv

14

exceeds zero and the process is ireversible. As the temperature difference increases, ÎSuniv increases. This exemplifies the fact that the entropy change of the universe produced by a process is a measure of the irreversibility of the process. For an isolated system, there is no change in the entropy of the surroundings. Hence the system entropy change is the entropy change of the universe and therefore must be non-negative. In other words, the entropy of an isolated system can only increase or at best stay constant. 1.3 Control Volumes and Steady Flows In many engineering problems it is preferrable to deal with a flow of fluid particles as they pass through a given region of space rather than following the flow of a fixed collection of particles. Thus, putting aside the system concept (fixed collection) for the moment, consider a volume with well-defined spatial boundaries as shown in Figure 1-3. This is called a control volume. Mass at state 1 enters at a rate m1 and leaves at state 2 with mass flow m2. If one mass flow rate exceeds the other, mass either accumulates in the volume or is depleted. The important special case of steady flow, in which no accumulation or depletion of mass occurs in the control volume, is considered here. In steady flow, the conservation of mass requires equal mass flows in and out, i.e., m1 = m2, [lbm /s, | kg /s]. If Q-dot is the rate of heat flow into the control volume and W-dot is the rate at which shaft work is delivered from the control volume to the surroundings, conservation of energy requires that the excess of inflowing heat over outgoing work equal the net excess of the energy (enthalpy) flowing out of the ports, i.e., [Btu/s | kJ/s]

(1.10)

15 where summations apply to inflows i and outflows o, and where other types of energy terms, such as kinetic and potential energy flows, are assumed negligible. For clarity, the figure shows only one port in and one port out. Kinetic and potential energy terms may be added analogous to the enthalpy termsat each port, if needed. Equation (1.10) may be the most important and frequently used equation in this book. Mastery of its use is therefore essential. It is known as the steady flow form of the First Law of Thermodynamics. It may be thought of as a bookkeeping relation for keeping track of energy crossing the boundaries of the control volume. The Second Law of Thermodynamics applied to steady flow through an adiabatic control volume requires that m2s2 $ m1s1, or by virtue of mass conservation: s2 $ s1

[Btu/lbm-R | kJ/kg-K]

(1.11)

That is, because entropy cannot accumulate within the control volume in a steady flow, the exit entropy must equal or exceed the inlet entropy. In steady flows, heat transfer can increase or decrease the entropy of the flow, depending on the direction of heat transfer, as long as the entropy change of the surroundings is such that the net effect is to increase the entropy of the universe. We will often be concerned with adiabatic flows. In the presence of fluid friction and other irreversibilities, the exit entropy of an adiabatic flow exceeds its inlet entropy. Adiabatic flows that have no irreversibilities also have no entropy change and therefore are called isentropic flows. 1.4 Properties of Vapors: Mollier and T-s Diagrams When heated, liquids are transformed into vapors. The much different physical character of liquids and vapors makes engines in which phase change takes place possible. The Newcomen atmospheric engine, for instance condensed steam to liquid water in a piston-cylinder enclosure to create a partial vacuum. The excess of atmospheric pressure over the low pressure of the condensed steam, acting on the opposite face of the piston, provided the actuating force that drove the first successful engines in the early eighteenth century. In the latter half of the eighteenth century, engines in which work was done by steam pressure on the piston rather than by the atmosphere, replaced Newcomen-type engines. Steam under pressure in reciprocating engines was a driving force for the industrial revolution for about two centuries. By the middle of the twentieth century, steam turbines and diesel engines had largely replaced the steam engine in electric power generation, marine propulsion, and railroad locomotives. . Figure 1.4 shows typical saturation curves for a pure substance plotted in temperature and entropy coordinates. A line of constant pressure (an isobar) is shown in which the subcooled liquid at state 1 is heated, producing increases in entropy, temperature, and enthalpy, until the liquid is saturated at state 2. Isobars in the

16

subcooled region of the diagram lie very close to the saturated liquid curve. The separation of the two is exaggerated for clarity. Once the substance has reached state 2, further transfer of heat fails to increase the system temperature but is reflected in increased enthalpy and entropy in a vaporization or boiling process. During this process the substance is converted from a saturated liquid at state 2 to a mixture of liquid and vapor, and finally to a saturated vapor at state 3. The enthalpy difference between the saturation values, h3 – h2, is called the enthalpy of vaporization or heat of vaporization. Continued addition of heat to the system, starting at state 3, superheats the steam to state 4, again increasing temperature, enthalpy, and entropy. Several observations about the isobaric process may be made here. Equation (1.5) and Figure 1.4 show that the effect of adding heat is to always increase system entropy and that of cooling to always decrease it. A similar conclusion can be drawn from Equation (1.10) regarding heat additions acting to increase enthalpy flow through a control volume in the absence of shaft work. A measure of the proximity of a superheated state (state 4 in the figure) to the saturated vapor line is the degree of superheat. This is the difference between the temperature T4 and the saturated vapor temperature T3, at the same pressure. Thus the degree of superheat of superheated state 4 is T 4 - T 3. In the phase change from state 2 to state 3, the temperature and pressure give no indication of the relative quantities of liquid and vapor in the system. The quality x is defined as the ratio of the mass of vapor to the mass of the mixture of liquid and vapor at any point between the saturation curves at a given pressure. By virtue of this definition, the quality varies from 0 for a saturated liquid to 1 for a saturated vapor.

17 Because extensive properties are proportional to mass, they vary directly with the vapor quality in the mixed region. The entropy, for example, varies from the entropy of the saturated liquid sl at state 2 to the saturated vapor entropy sv at state 3 in accordance with the following quality equation: s = sl + x(sv – sl)

[Btu/lbm-R | kJ/kg-K]

(1.12)

where s is the entropy per unit mass. Other extensive properties such as enthalpy and volume vary with quality in the same way. A variable closely related to the quality is moisture fraction (both quality and moisture fraction can be expressed as percentages). Moisture fraction, M, is defined as the ratio of the mass of liquid to the total mass of liquid and vapor. It can be easily shown that the sum of the quality and the moisture fraction of a mixture is one. A Mollier chart, a diagram with enthalpy as ordinate and entropy as abscissa, is much like the temperature-entropy diagram. A Mollier diagram for steam is included in Appendix B. An isobar on a Mollier chart, unlike that on a T-s diagram, has a continuous slope. It shows both enthalpy and entropy increasing monotonically with heat addition. Such a diagram is frequently used in energy conversion and other areas because of the importance of enthalpy in applying the steady-flow First Law. 1.5 Ideal Gas Basics Under normal ambient conditions, the average distance between molecules in gases is large, resulting in negligible influences of intermolecular forces. In this case, molecular behavior and, therefore, system thermodynamics are governed primarily by molecular translational and rotational kinetic energy. Kinetic theory or statistical thermodynamics may be used to derive the ideal gas or perfect gas law: pv = RT

[ft-lbf /lbm | kJ/kg]

(1.13)

where p [lbf /ft2 | kN/m2], v [ft3/lbm | m3/kg] and T [°R | °K] are pressure, specific volume, and temperature respectively and R [ft-lbf /lbm-°R | kJ/kg-°K] is the ideal gas constant. The gas constant R for a specific gas is the universal gas constant R divided by the molecular weight of the gas. Thus, the gas constant for air is (1545 ft-lbf /lb-mole-°R) / (29 lbm/lb-mole) = 53.3 ft-lbf /lbm-°R in the English system and (8.31 kJ/kg-mole-°K) / (29 kg/kg-mole) = 0.287 kJ/kg-°K in SI units. The specific heats or heat capacities at constant volume and at constant pressure, respectively, are: cv = (Mu / MT)v and

[Btu/lbm-°R | kJ/kg-°K]

(1.14)

18 cp = (Mh /MT)p

[Btu/lbm-°R | kJ/kg-°K]

(1.15)

As thermodynamic properties, the heat capacities are, in general, functions of two other thermodynamic properties. For solids and liquids, pressure change has little influence on volume and internal energy, so that to a very good approximation: cv = cp. A gas is said to be thermally perfect if it obeys Equation (1.13) and its internal energy, enthalpy, and heat capacities are functions of temperature only. Then and

du = cv(T) dT

[Btu/lbm | kJ/kg]

(1.16)

dh = cp(T) dT

[Btu/lbm | kJ/kg]

(1.17)

A gas is said to be calorically perfect if in addition to being thermally perfect it also has constant heat capacities. This is reasonably accurate at low and moderate pressures and at temperatures high enough that intermolecular forces are negligible but low enough that molecular vibrations are not excited and dissociation does not occur. For air, vibrational modes are not significantly excited below about 600K, and dissociation of oxygen does not occur until the temperature is above about 1500K. Nitrogen does not dissociate until still higher temperatures. Excitation of molecular vibrations causes specific heat to increase with temperature increase. Dissociation creates further increases in heat capacities, causing them to become functions of pressure. It can be shown (see Exercise 1.4) that for a thermally perfect gas the heat capacities are related by the following equation: cp = cv + R

[Btu/lbm-R | kJ/kg-K]

(1.18)

This relation does not apply for a dissociating gas, because the molecular weight of the gas changes as molecular bonds are broken. Note the importance of assuring that R and the heat capacities are in consistent units in this equation. Another important gas property is the ratio of heat capacities defined by k = cp /cv. It is constant for gases at room temperatures but decreases as vibrational modes become excited. The importance of k will be seen in the following example. EXAMPLE 1.2

(a) Derive an expression for the entropy change of a system in terms of pressure and temperature for a calorically perfect gas. (b) Derive a relation between p and T for an isentropic process in a calorically perfect gas. Solution

(a) For a reversible process, Equation (1.8) gives Tds = dh - vdp. Dividing by T and applying the perfect gas law gives ds = cp dT/T - Rdp/p.

19 Then integration between states 1 and 2 yields s2 - s1 = cp ln(T2 /T1) - R ln( p2 /p1) (b) For an isentropic process, s2 = s1. Then the above equation gives T2 /T1 = (p2/p1)(R/cp) But R/cp = (cp - cv )/cp = (k - 1)/k. Hence T2 /T1 = (p2 /p1)(k - 1)/k. ____________________________________________________________________ This and other important relations for an isentropic process in a calorically perfect gas are summarized as follows T2 /T1 = (p2 /p1)(k - 1)/k

[dl]

(1.19)

T2 /T1 = (v2 /v1)(k - 1)

[dl]

(1.20)

p2 /p1 = (v1 /v2)k

[dl]

(1.21)

These relations show that the ratio of heat capacities governs the variation of thermodynamic properties in an isentropic process. For this reason the ratio of heat capacities is sometimes called the isentropic exponent. 1.6 Fundamentals of Fluid Flow Almost all energy conversion devices involve the flow of some form of fluid. Air, liquid water, steam, and combustion gases are commonly found in some of these devices. Here we review a few of the frequently used elementary principles of fluid flow. The volume flow rate, Q [ft3/s | m3/s] at which a fluid flows across a surface is the product of the area, A [ft2 | m2], of the surface and the component of velocity normal to the area, V [ft/s | m/s]. The corresponding mass flow rate is the ratio of the volume rate and the specific volume, v [ft3/lbm | m3/kg]: m = AV/v = Q/v

[lbm /s | kg /s]

(1.22)

Alternatively the flow rate can be expressed in terms of the reciprocal of the specific volume, the density, , [lbm /ft3 | kg /m3]: m = AV, = Q,

[lbm /s | kg /s]

(1.23)

20 The first important principle of fluid mechanics is the conservation of mass, a principle that we have already used in Section 1.3. For a steady flow, the net inflow to a control volume must equal the net outflow. Any imbalance between the inflow and outflow implies an accumulation or a reduction of mass within the control volume, i.e., an unsteady flow. Given a control volume with n ports, the conservation of mass provides an equation that may be used to solve for the nth port flow rate, given the other n-1 flow rates. These flows may be (1) given, (2) calculated from data at the ports using Equation (1.22) or (1.23), (3) obtained by solving n-1 other equations, or (4) a combination of the preceding three. For isentropic flow of an incompressible (constant density, , ) fluid, the Bernoulli equation applies: p1 /, + V12/2 = p2 /, + V22/2 [ft-lbf/lbm | kJ/kg]

(1.24)

This is an invariant form, i.e. an equation with the same terms on both sides, p/, + V2/2. The subscripts identify the locations in the flow where the invariants are evaluated. The first term of the invariant is sometimes called the pressure head, and the second the velocity head. The equation applies only in regions where there are no irreversibilities such as viscous losses or heat transfer. The invariant sum of the two terms on either side of Equation (1.24) may be called the total head or stagnation head. It is the head that would be observed at a point where the velocity approaches zero. The pressure associated with the total head is therefore called the total pressure or stagnation pressure, po = p + ,V2/2. Each point in the flow may be thought of as having its own stagnation pressure resulting from an imaginary isentropic deceleration. In the event of significant irreversibilities, there is a loss in total head and the Bernoulli equation may be generalized to: or

p1 /, + V12/2 = p2/, + V22/2 + loss

[ft-lbf/lbm | kJ/kg]

(1.25a)

po1 /, = po2 /, + loss

[ft-lbf/lbm | kJ/kg]

(1.25b)

Stagnation pressure or head losses in ducts, such as due to flow turning or sudden area change, are tabulated in reference books as fractions of the upstream velocity head for a variety of geometries. Another example is the famous Darcy-Weisbach equation which gives the head loss resulting from fluid friction in a pipe of constant crosssection. 1.7 Compressible Flow While many engineering analyses may reasonably employ incompressible flow principles, there are cases where the compressibility of gases and vapors must be considered. These are situations where the magnitude of the kinetic energy of the flow

21 is comparable to its enthalpy such as in supersonic nozzles and diffusers, in turbines and compressors, and in supersonic flight. In these cases the steady-flow First Law must be generalized to include kinetic energy per unit mass terms. For two ports: [Btu /s | kJ/s]

(1.26a)

Care should be taken to assure consistency of units, because enthalpy is usually stated in thermal units [Btu /lbm | kJ/kg] and velocity in mechanical units [ft /s | m /s]. Another invariant of significance appears in Equation (1.26a). The form ho = h + V2/2

[Btu/lbm | kJ/kg]

(1.27)

is seen to be invariant in applications where heat transfer and shaft work are insignificant. The invariant, ho, is usually given the name stagnation enthalpy because it is the enthalpy at a point in the flow (real or imagined) where velocity approaches zero. In terms of stagnation enthalpy, Equation (1.26a) may be rewritten as [Btu/s | kJ/s]

(1.26b)

where conservation of mass with steady flow through two ports has been assumed. Writing dho = cp dTo with cp constant, we get ho2 - ho1 = cp(To2 - To1) Combining this with Equation (1.27), we are led to define another invariant, the stagnation temperature for a calorically perefect gas: To = T + V2/2cp

[ R | K]

(1.28)

The stagnation temperature may be regarded as the temperture at a real or imaginary point where the gas velocity has been brought to zero adiabatically. For this special case of a constant heat capacity, Equation (1.26b) may be written as [Btu /s | kJ/s]

(1.26c)

In both incompressible and compressible flows, the mass flow rates at all stations in a streamtube are the same. Because the specific volume and density are constant in incompressible flow, Equation (1.22) shows that the volume flow rates are the same at all stations also. However for compressible flow, Equation (1.23) shows that density change along a streamtube implies volume flow rate variation. Thus, while it is frequently convenient to think and talk in terms of volume flow rate when dealing with incompressible flows, mass flow rate is more meaningful in compressible flows and in general.

22 A measure of the compressiblity of a flow is often indicated by a Mach number, defined as the dimensionless ratio of a flow velocity to the local speed of sound in the fluid. For ideal gases the speed of sound is given by a = (kp/,)½ = (kRT)½

[ft /s | m /s]

(1.29)

Compressible flows are frequently classified according to their Mach number: M=0 0<M<1 M=1 M>1

Incompressible Subsonic Sonic Supersonic

Studies of compressible flows show that supersonic flows have a significantly different physical character than subsonic and incompressible flows. For example, the velocity fields in subsonic flows are continuous, whereas discontinuities known as shock waves are common in supersonic flows. Thus the student should not be surprised to find that different relations hold in supersonic flows than in subsonic flows. 1.8 Energy Clasification Energy exists in a variety of forms. All human activities involve conversion of energy from one form to another. Indeed, life itself depends on energy conversion processes. The human body, through complex processes, transforms the chemical energy stored in food into external motion and work produced by muscles as well as electrical impulses that control and activate internal functions. It is instructive to examine some of the processes for transformation between technically important forms of energy. Table 1.2 shows a matrix of energy forms and the names of some associated energy converters. Table 1.2

Energy Transformation Matrix

From:

To:

Thermal Energy

Mechanical Energy

Electrical Energy

Chemical Energy

Furnace

Diesel engine

Fuel cell

Thermal Energy

Heat exchanger

Steam turbine

Thermocouple

Mechanical Energy

Refrigerator, heat pump

Gearbox

Electrical generator

Nuclear Energy

Fission reactor

Nuclear steam turbine

Nuclear power plant

The table is far from complete, and other energy forms and energy converters could readily be added. However, it does include the major energy converters of interest to mechanical engineers. It is a goal of this book to present important aspects of the design, analysis, performance, and operation of most of these devices. One of the major criteria guiding the design of energy conversion systems is

23 efficiency. Each of the conversions in Table 1.2 is executed by a device that operates with one or more relevant efficiencies. The following section explores some of the variety of definitions of efficiency used in design and performance studies of energy conversion devices. 1.9 Efficiencies Efficiency is a measure of the quality of an operation or of a characteristic of a device. Several types of efficiencies are widely used. It is important to clearly distinguish among them. Note that the terms work and power are equally applicable here. The efficiency of a machine that transmits mechanical power is measured by its mechanical efficiency, the fraction of the power supplied to the transmission device that is delivered to another machine attached to its output, Figure 1.5(a). Thus a gearbox for converting rotational motion from a power source to a device driven at another speed dissipates some mechanical energy by fluid and/or dry friction, with a consequent loss in power transmitted to the second machine. The efficiency of the gearbox is the ratio of its power output to the power input, a value less than one. For example, a turboprop engine with a gearbox efficiency of 0.95 will transmit only 95% of its power output to its propeller.

24 Another type of efficiency that measures internal losses of power is used to indicate the quality of performance of turbomachines such as pumps, compressors and turbines. These devices convert flow energy to work (power), or vice versa. Here the efficiency compares the output with a theoretical ideal in a ratio [Figure 1.5(b)]. The resulting efficiency ranges from 0 to 1 as a measure of how closely the process approaches a relevant isentropic process. A turbine with an efficiency of 0.9, will, for example, deliver 90% of the power of a perfect (isentropic) turbine operating under the same conditions. This efficiency, sometimes referred to as isentropic efficiency or turbine efficiency, will be considered in more detail in the next chapter. Another form of isentropic efficiency, sometimes called compressor efficiency (or pump efficiency), is defined for compressors (or pumps). It is the ratio of the isentropic work to drive the compressor (or pump) to the actual work required. Because the actual work required exceeds the isentropic work, this efficiency is also less than to 1. A third type of efficiency compares the magnitude of a useful effect to the cost of producing the effect, measured in comparable units. An example of this type of efficiency compares the net work output, wn, of a heat engine to the heat supplied, qa, to operate the engine. This is called the to (th = wn /qa) [Figure 1.5(c)]. For example, the flow of natural gas to an electrical power plant provides a chemical energy flow rate or heat flow rate to the plant that leads to useful electric power output. It is known from basic thermodynamics that this efficiency is limited by the Carnot efficiency, as will be discussed in the next section. Another example of this type of efficiency as applied to refrigerators and heat pumps [Figure 1.5(d)] is called the coefficient of performance, COP. In this case the useful effect is the rate of cooling or heating, and the cost to produce the effect is the power supplied to the device. The term "coefficient of performance" is used instead of efficiency for this measure of quality because the useful effect usually exceeds the cost in comparable units of measure. Hence, unlike other efficiencies, the COP can exceed unity. As seen in Figure 1.5(d), there are two definitions for COP, one for a refrigerator and another for a heat pump. It may be shown using the First Law of Thermodynamics, that a simple relationship exists between the two definitions: COPhp = COPrefr + 1. 1.10 The Carnot Engine On beginning the study of the energy conversion ideas and devices that will serve us in the twenty-first century, it is appropriate to review the theoretical cycle that stands as the ideal for a heat engine. The ideas put forth by Sadi Carnot in 1824 in his “Reflections on the Motive Power of Heat” (see Historical Bibliography) expressed the content of the Second Law of Thermodynamics relevant to heat engines, which, in modern form (attributed to Kelvin and Planck) is: “It is impossible for a device which operates in a cycle to receive heat from a single source and convert the heat completely to work.” Carnot’s great work also described the cycle that today bears his name and provides the theoretical limit for efficiency of heat engine cycles that operate between two given temperature levels: the Carnot cycle. The Carnot cycle consists of two reversible, isothermal processes separated by two

25

reversible adiabatic or isentropic processes, as shown in Figure 1.6. All of the heat transferred to the working fluid is supplied isothermally at the high temperature TH = T3, and all heat rejected is transferred from the working medium at the low temperature TL = T1. No heat transfer takes place, of course, in the isentropic processes. It is evident from Equation (1.5) and the T-s diagram that the heat added is T3(s3 - s2 ), the heat rejected is T1 (s1 - s4 ), and, by the cyclic integral relation, the net work is T3(s3 - s2 ) + T1 (s1 - s4 ). The thermal efficiency of the Carnot cycle, like that of other cycles, is given by wn / qa and can be expressed in terms of the high and low cycle temperatures as :

26

because both isothermal processes operate between the same entropy limits. The Carnot efficiency equation shows that efficiency rises as TL drops and as TH increases. The message is clear: a heat engine should operate between the widest possible temperature limits. Thus the efficiency of a heat engine will be limited by the maximum attainable energy-source temperature and the lowest available heat-sink temperature. Students are sometimes troubled by the idea of isothermal heat transfer processes because they associate heat transfer with temperature rise. A moments reflection, however, on the existence of latent heats–e.g., the teakettle steaming on the stove at constant temperature-makes it clear that one should not always associate heat transfer with temperature change. It is important here, as we start to consider energy conversion devices, to recall the famous Carnot Theorem, the proof of which is given the most thermodynamics texts. It states that it is impossible for any engine operating in a cycle between two reservoirs at different temperatures to have an efficiency that exceeds the Carnot efficiency corresponding to those temperatures. It can also be shown that all reversible engines operating between two given reservoirs have the same efficiency and that all irreversible engines must have lower efficiencies. Thus the Carnot efficiency sets an upper limit on the performance of heat engines and therefore serves as a criterion by which other engines may be judged. 1.11 Additional Second-Law Considerations The qualitative relationship between the irreversibility of a process and the entropy increase of the universe associated with it was considered in section 1.2. Let us now consider a quantitative approach to irreversibility and apply it to a model of a power plant. Reversible Work Instead of comparing the work output of the power plant with the energy supplied from fuel to run the plant, it is instructive to compare it with the maximum work achievable by a reversible heat engine operating between the appropriate temperature limits, the reversible work. It has been established that any reversible engine would have the same efficiency as a Carnot engine. The Carnot engine provides a device for determining the reversible work associated with a given source temperature, TH, and a lower sink temperature,TL.

27 Irreversibility The irreversibility, I, of a process is defined as the difference between the reversible work and the actual work of a process: I = Wrev - Wact [Btu | kJ] It is seen that the irreversibility of a process vanishes when the actual work is the same as that produced by an appropriate Carnot engine. Moreover, the irreversibility of a non-work-producing engine is equal to the reversible work. It is clear that I $ 0, because no real engine can produce more work than a Carnot engine operating between the same limiting temperatures. Second-Law Efficiency We can also define a “second-law efficiency,” II, as the ratio of the actual work of a process to the reversible work: [dl] This is an efficiency that is limited to 100%, as opposed to the thermal efficiency of a heat engine, sometimes referred to as a “first-law efficiency,” which may not exceed that of the appropriate Carnot engine. Note that an engine which has no irreversibility is a reversible engine and has a second-law efficiency of 100%. A Power Plant Model Let us consider a model of a power plant in which a fuel is burned at a high temperature, TH, in order to transfer heat to a working fluid at an intermediate temperature, TINT. The working fluid, in turn, is used in an engine to produce work and reject heat to a sink at the low temperature, TL. Figure 1.7 presents a diagram of the model that shows explicitly the combustion temperature drop from the source temperature, TH, to the intermediate temperature, TINT, the actual work-producing engine, and a Carnot engine used to determine the reversible work for the situation. The Carnot engine has an efficiency of C = 1 - TL /TH and develops work in the following amount: Wrev = WC = QIN - QC = CQIN

[Btu|kJ]

Suppose that the engine we are considering is a Carnot engine that operates from the intermediate source at TINT. Its efficiency and work output are, respectively, I = 1 - TL /TINT and Wact = QIN – QI = I QIN . The irreversibility, I, of the power plant is then

28

I = Wrev - Wact = (C - I )QIN = [1 - TL /TH - ( 1 - TL /TINT)]QIN = TLQIN ( TH - TINT ) /TINTTH

[Btu|kJ]

and the second-law efiiciency is:

II = Wact /Wrev = (1 - TL/TINT ) / (1 - TL/TH)

[dl]

Note that when TINT = TH, the irreveribility vanishes and the second-law efficiency becomes 100%. Also, when TINT = TL, the irreversibility is equal to the reversible work of the Carnot engine and the second- law efficiency is zero. The latter condition indicates that a pure heat transfer process or any process that produces no useful work causes a loss in the ability to do work in the amount of Wrev . Thus the reversible work associated with the extremes of a given process is a measure of how much capability to do work can be lost, and the irreversibility is a measure of how much of that workproducing potential is actually lost. The following example illustrates these ideas. EXAMPLE 1.3

Through combustion of a fossil fuel at 3500°R, an engine receives energy at a rate of 3000 Btu/s to heat steam to 1500°R. There is no energy loss in the combustion process. The steam, in turn, produces 1000 Btu/s of work and rejects the remaining energy to the surroundings at 500°R. (a) What is the thermal efficiency of the plant? (b) What are the reversible work and the Carnot efficiency corresponding to the source

29 and sink temperatures? (c) What is the irreversibility? (d) What is the second-law efficiency? (e) What would the irreversibility and the second-law efficiency be if the working fluid were processed by a Carnot engine rather than by the real engine? Solution

(a) The thermal efficiency, or first-law efficiency, of the plant is I = Wact/QIN = 1000/3000 = 0.333 or 33.3%. (b) The relevant Carnot efficiency is 1 - 500/3500 = 0.857, or 85.7%. The engine’s reversible work is then CQIN = 0.857(3000) = 2571 Btu/s. (c) The plant irreversibility is 2571 - 1000 = 1571 Btu/s. (d) The second-law efficiency is then I /C = 0.333/0.857 = 0.389, or 38.9% or Wact /Wrev = 1000/2571 = 0.389, or 38.9%. (e) The Carnot efficiency corresponding to the maximum temperature of the working fluid is 1 - 500/1500 = 0.667 or 66.7%. The second-law efficiency for this system is then 0.667/0.857 = 0.778, compared with 0.389 for the actual engine. The actual work produced by the irreversibly heated Carnot engine is 0.667(3000) = 2001 Btu/s, and its irreversibility is then I = 2571 -2001 = 570 Btu/s. ___________________________________________________________________ In summary, the thermal efficiency, or first-law efficiency, of an engine is a measure of how well the engine converts the energy in its fuel to useful work. It says nothing about energy loss, because energy is conserved and cannot be lost: it can only be transformed. The second-law efficiency, on the other hand, recognizes that some of the energy of a fuel is not available for conversion to work in a heat engine and therefore assesses the ability of the engine to convert only the available work into useful work. This is a reason why some regard the second-law efficiency as more significant than the more commonly used first-law efficiency. Bibliography and References

1. Van Wylen, Gordon J., and Sonntag, Richard E., Fundamentals of Classical Thermodynamics, 3rd ed. New York: Wiley, 1986. 2. Balmer, Robert, Thermodynamics. Minneapolis: West, 1990. 3. Cengel, Yunus A., and Boles, Michael A., Thermodynamics. New York: McGrawHill, 1989.

30 4. Faires, Virgil Moring, Thermodynamics, 5th ed. New York: Macmillan, 1970. 5. Silver, Howard F., and Nydahl, John E., Engineering Thermodynamics. Minneapolis: West, 1977. 6. Bathie, William W., Fundamentals of Gas Turbines. New York: Wiley, 1984 7. Wilson, David Gordon, The Design of High Efficiency Turbomachinery and Gas Turbines. Boston: MIT Press, 1984. 8. Anderson, John D., Modern Compressible Flow. New York: McGraw-Hill, 1982. 9. Anderson, John D., Introduction to Flight. New York: McGraw-Hill, 1978. 10. Chapman, Alan J. and Walker, William F., Introductory Gas Dynamics. New York: Holt, Rinehart, and Winston, 1971. Historical Bibliography

1. Barnard, William N., Ellenwood, Frank E., and Hirshfeld, Clarence F., Heat-Power Engineering. Wiley, 1926. 2. Bent, Henry, The Second Law.NewYork: Oxford University Press, 1965 3. Cummins, C. Lyle, Jr., Internal Fire, rev. ed. Warrendale, Penna.: Society of Automotive Engineers, 1989. 4. Tann, Jennifer, The Selected Papers of Boulton and Watt. Boston: MIT Press, 1981. 5. Carnot, Sadi, Reflexions Sur la Puissance Motrice de Feu. Paris: Bachelor,1824. 6. Potter, J. H., “The Gas Turbine Cycle.” ASME Paper presented at the Gas Turbine Forum Dinner, ASME Annual Meeting, New York, Nov. 27, 1972. 7. Grosser, Morton, Diesel: The Man and the Machine. New York: Atheneum, 1978. 8. Nitske, W. Robert, and Wilson, Charles Morrow, Rudolph Diesel: Pioneer of the Age of Power. Norman, Okla: University of Oklahoma Press, 1965. 9. Rolt, L.T. C., and Allen, J. S., The Steam Engine of Thomas Newcomen. New York: Moorland Publishing Co., 1977. 10. Briggs, Asa, The Power of Steam. Chicago: University of Chicago Press, 1982.

31 11. Thurston, Robert H., A History of the Growth of the Steam Engine. 1878; rpr. Ithaca, N.Y.: Cornell University Press, 1939.

EXERCISES

1.1 Determine the entropy of steam at 1000 psia and a quality of 50%. 1.2 Show that the moisture fraction for a liquid water-steam mixture, defined as the ratio of liquid mass to mixture mass, can be written as 1 - x, where x is steam quality. 1.3 Write expressions for the specific entropy, specific enthalpy, and specific volume as functions of the moisture fraction. Determine the values of these properties for steam at 500°F and a moisture fraction of 0.4. 1.4 Show that for a thermally perfect gas, cp - cv = R. 1.5 During a cyclic process, 75kJ of heat flow into a system and 25kJ are rejected from the system later in the cycle. What is the net work of the cycle. 1.6 Seventy-five kJ of heat flow into a rigid system and 25 kJ are rejected later. What are the magnitude and sign of the change in internal energy? What does the sign indicate? 1.7 The mass contained between an insulated piston and an insulated cylinder decreases in internal energy by 50 Btu. How much work is involved, and what is the sign of the work term? What does the sign indicate? 1.8 Derive Equation (1.8) from Equation (1.7). 1.9 Use Equation (1.8) to derive an expression for the finite enthalpy change of an incompressible fluid in an isentropic process. If the process is the pressurization of saturated water initially at 250 psia, what is the enthalpy rise, in Btu /lbm and in ft-lbf / lbm, if the final pressure is 4000 psia? What is the enthalpy rise if the initial pressure is 100 kPa and the final pressure is 950 kPa? 1.10 Sixty kg /s of brine flows into a device with an enthalpy of 200 kJ/kg. Brine flows out of the other port at a flow rate of 20 kg /s. What is the net inflow? Is the system in steady flow? Explain. 1.11 Use the steam tables in Appendices B and C to compare the heats of vaporization at 0.01, 10, and 1000 psia. Compare the saturated liquid specific volumes at these pressures. What do you conclude about the influence of pressure on these properties?

32 1.12 Using the heat capacity equation for nitrogen: cp = 9.47 - 3.47 x 103/T + 1.16 x 106/T2 where cp is in Btu / lb-mole and T is in degrees Rankine (From Gordon J. Van Wylen and Richard E. Sonntag, Fundamentals of Classical Thermodynamics, 3rd ed. New York: Wiley, 1986). Compare the enthalpy change per mole of nitrogen between 540°R and 2000°R for nitrogen as a thermally perfect gas and as a calorically perfect gas. 1.13 Use Equation (1.7) to derive Equation (1.20) for a calorically perfect gas. 1.14 Use Equation (1.7) to derive a relation for the entropy change as a function of temperature ratio for a constant-volume process in a calorically perfect gas. 1.15 Use Equation (1.8) to derive a relation for the entropy change as a function of temperature for an isobaric process in a calorically perfect gas. 1.16 A convergent nozzle is a flow passage in which area decreases in the streamwise direction (the direction of the flow). It is used for accelerating the flow from a low velocity to a higher velocity. Use the generalized form of the steady-flow First Law given in Equation (1.26a) to derive an equation for the exit velocity for an adiabatic nozzle. 1.17 Derive an equation for the pressure drop for a loss-free incompressible flow in a varying-area duct as a function of area ratio. 1.18 Two units of work are required to transfer 10 units of heat from a refrigerator to the environment. What is the COP of the refrigerator? Suppose that the same amount of heat transfer instead is by a heat pump into a house. What is the heat pump COP? 1.19 A power plant delivers 100 units of work at 30% thermal efficiency. How many heat units are supplied to operate the plant? How many units of heat are rejected to the surroundings? 1.20 A steam turbine has an efficiency of 90% and a theoretical isentropic power of' 100 kW. What is the actual power output? 1.21 Thomas Newcomen used the fact that the specific volume of saturated liquid is much smaller than the specific volume of saturated steam at the same pressure in his famous "atmospheric engine." Calculate the work done on the piston by the atmosphere if steam is condensed at an average pressure of 6 psia by cooling in a tightly fitted piston-cylinder enclosure if the piston area is 1 ft2 and the piston stroke is 1 ft. If the process takes place 10 times a minute, what is the power delivered? Discuss what can

33 be done to increase the power of the engine. Describe the characteristics of a Newcomen engine that would theoretically deliver 20 horsepower. 1.22 Expand Table 1.2 to include solar and geothermal sources. 1.23 Twenty pounds of compressed air is stored in a tank at 200 psia and 80/F. The tank is heated to bring the temperature to 155/F. What is the final tank pressure, and how much heat was added? 1.24 Ten kilograms of compressed air is stored in a tank at 250 kPa and 50/C. The tank is heated to bring the air temperature to 200 °C. What is the final tank pressure, and how much heat was added? 3

1.25 An 85-ft tank contains air at 30 psia and 100/ F. What mass of air must be added to bring the pressure to 50 psia and the temperature to 150/F? 1.26 A 20-L tank contains air at 2 bar and 300K. What mass of air must be added to bring the pressure to 2 bar and the temperature to 375K? 1.27 Air enters a wind tunnel nozzle at 160/F, 10 atm, and a velocity of 50 ft/s. The entrance area is 5ft2. If the heat loss per unit mass is 10 Btu/lbm and the exit pressure and velocity are, respectively, 1.5 atm and 675 ft/s, what are the exit temperature and area? 1.28 Air enters a wind tunnel nozzle at 90°C, 250 kPa, and a velocity of 40 m/s. The entrance area is 3 m2. If the heat loss per unit mass is 7 kJ/kg and the exit pressure and velocity are, respectively, 105 kPa and 250 m/s, what are the exit temperature and area? 1.29 Air enters a wind tunnel nozzle at 160/F, 10 atm, and a velocity of 50 ft/s. The entrance area is 5ft2. If the heat loss per unit mass is 8 Btu/lbm and the exit pressure and temperature are, respectively, 1.25 atm and 120/F, what are the exit velocity and area? 1.30 Air enters a wind tunnel nozzle at 90/C, 250 kPa, and a velocity of 40 m/s. The entrance area is 3 m2. If the heat loss per unit mass is 5 kJ/kg and the exit pressure and temperature are, respectively, 120 kPa and 43°C, what are the exit velocity and area? 1.31 Sketch a Mollier diagram showing the character of three isotherms and three isobars for a calorically perfect gas. Label each curve with a value in SI units to show the directions of increasing temperature and pressure. Explain how the diagram would differ if the gas were not calorically perfect.

35 CHAPTER TWO Fundamentals of Steam Power 2.1 Introduction Much of the electricity used in the United States is produced in steam power plants. Despite efforts to develop alternative energy converters, electricity from steam will continue, for many years, to provide the power that energizes the United States and world economies. We therefore begin the study of energy conversion systems with this important element of industrial society. Steam cycles used in electrical power plants and in the production of shaft power in industry are based on the familiar Rankine cycle, studied briefly in most courses in thermodynamics. In this chapter we review the basic Rankine cycle and examine modifications of the cycle that make modern power plants efficient and reliable. 2.2 A Simple Rankine-Cycle Power Plant The most prominent physical feature of a modern steam power plant (other than its smokestack) is the steam generator, or boiler, as seen in Figure 2.1. There the combustion, in air, of a fossil fuel such as oil, natural gas, or coal produces hot combustion gases that transfer heat to water passing through tubes in the steam generator. The heat transfer to the incoming water (feedwater) first increases its temperature until it becomes a saturated liquid, then evaporates it to form saturated vapor, and usually then further raises its temperature to create superheated steam. Steam power plants such as that shown in Figure 2.1, operate on sophisticated variants of the Rankine cycle. These are considered later. First, let’s examine the simple Rankine cycle shown in Figure 2.2, from which the cycles of large steam power plants are derived. In the simple Rankine cycle, steam flows to a turbine, where part of its energy is converted to mechanical energy that is transmitted by rotating shaft to drive an electrical generator. The reduced-energy steam flowing out of the turbine condenses to liuid water in the condenser. A feedwater pump returns the condensed liquid (condensate) to the steam generator. The heat rejected from the steam entering the condenser is transferred to a separate cooling water loop that in turn delivers the rejected energy to a neighboring lake or river or to the atmosphere.

36

As a result of the conversion of much of its thermal energy into mechanical energy, or work, steam leaves the turbine at a pressure and temperature well below the turbine entrance (throttle) values. At this point the steam could be released into the atmosphere. But since water resources are seldom adequate to allow the luxury of onetime use, and because water purification of a continuous supply of fresh feedwater is costly, steam power plants normally utilize the same pure water over and over again. We usually say that the working fluid (water) in the plant operates in a cycle or undergoes of cyclic process, as indicated in Figure 2.2. In order to return the steam to the high-pressure of the steam generator to continue the cycle, the low- pressure steam leaving the turbine at state 2 is first condensed to a liquid at state 3 and then pressurized in a pump to state 4. The high pressure liquid water is then ready for its next pass through the steam generator to state 1 and around the Rankine cycle again. The steam generator and condenser both may be thought of as types of heat exchangers, the former with hot combustion gases flowing on the outside of water-

37

filled tubes, and the latter with external cooling water passing through tubes on which the low- pressure turbine exhaust steam condenses. In a well-designed heat exchanger, both fluids pass through with little pressure loss. Therefore, as an ideal, it is common to think of steam generators and condensers as operating with their fluids at unchanging pressures. It is useful to think of the Rankine cycle as operating between two fixed pressure levels, the pressure in the steam generator and pressure in the condenser. A pump provides the pressure increase, and a turbine provides the controlled pressure drop between these levels. Looking at the overall Rankine cycle as a system (Figure 2.2), we see that work is delivered to the surroundings (the electrical generator and distribution system) by the turbine and extracted from the surroundings by a pump (driven by an electric motor or a small steam turbine). Similarly, heat is received from the surroundings (combustion gas) in the steam generator and rejected to cooling water in the condenser.

38

At the start of the twentieth century reciprocating steam engines extracted thermal energy from steam and converted linear reciprocating motion to rotary motion, to provide shaft power for industry. Today, highly efficient steam turbines, such as shown in Figure 2.3, convert thermal energy of steam directly to rotary motion. Eliminating the intermediate step of conversion of thermal energy into the linear motion of a piston was an important factor in the success of the steam turbine in electric power generation. The resulting high rotational speed, reliability, and power output of the turbine and the development of electrical distribution systems allowed the centralization of power production in a few large plants capable of serving many industrial and residential customers over a wide geographic area. The final link in the conversion of chemical energy to thermal energy to mechanical energy to electricity is the electrical generator. The rotating shaft of the electrical generator usually is directly coupled to the turbine drive shaft. Electrical windings attached to the rotating shaft of the generator cut the lines of force of the stator windings, inducing a flow of alternating electrical current in accordance with Faraday's Law. In the United States, electrical generators turn at a multiple of the generation frequency of 60 cycles per second, usually 1800 or 3600 rpm. Elsewhere, where 50 cycles per second is the standard frequency, the speed of 3000 rpm is common. Through transformers at the power plant, the voltage is increased to several hundred thousand volts for transmission to distant distribution centers. At the distribution centers as well as neighborhood electrical transformers, the electrical potential is reduced, ultimately to the 110- and 220-volt levels used in homes and industry.

39 Since at present there is no economical way to store the large quantities of electricity produced by a power plant, the generating system must adapt, from moment to moment, to the varying demands for electricity from its customers. It is therefore important that a power company have both sufficient generation capacity to reliably satisfy the maximum demand and generation equipment capable of adapting to varying load. 2.3 Rankine-Cycle Analysis In analyses of heat engine cycles it is usually assumed that the components of the engine are joined by conduits that allow transport of the working fluid from the exit of one component to the entrance of the next, with no intervening state change. It will be seen later that this simplification can be removed when necessary. It is also assumed that all flows of mass and energy are steady, so that the steady state conservation equations are applicable. This is appropriate to most situations because power plants usually operate at steady conditions for significant lengths of time. Thus, transients at startup and shutdown are special cases that will not be considered here. Consider again the Rankine cycle shown in Figure 2.2. Control of the flow can be exercised by a throttle valve placed at the entrance to the turbine (state 1). Partial valve closure would reduce both the steam flow to the turbine and the resulting power output. We usually refer to the temperature and pressure at the entrance to the turbine as throttle conditions. In the ideal Rankine cycle shown, steam expands adiabatically and reversibly, or isentropically, through the turbine to a lower temperature and pressure at the condenser entrance. Applying the steady-flow form of the First Law of Thermodynamics [Equation (1.10)] for an isentropic turbine we obtain: q = 0 = h2 – h1 + wt

[Btu/lbm | kJ/kg]

where we neglect the usually small kinetic and potential energy differences between the inlet and outlet. This equation shows that the turbine work per unit mass passing through the turbine is simply the difference between the entrance enthalpy and the lower exit enthalpy: wt = h1 – h2

[Btu/lbm | kJ/kg]

(2.1)

The power delivered by the turbine to an external load, such as an electrical generator, is given by the following: Turbine Power = mswt = ms(h1 – h2)

[Btu/hr | kW]

where ms [lbm /hr | kg/s] is the mass flow of steam though the power plant.

40 Applying the steady-flow First Law of Thermodynamics to the steam generator, we see that shaft work is zero and thus that the steam generator heat transfer is qa = h1 – h4

[Btu/lbm | kJ/kg]

(2.2)

The condenser usually is a large shell-and-tube heat exchanger positioned below or adjacent to the turbine in order to directly receive the large flow rate of low-pressure turbine exit steam and convert it to liquid water. External cooling water is pumped through thousands of tubes in the condenser to transport the heat of condensation of the steam away from the plant. On leaving the condenser, the condensed liquid (called condensate) is at a low temperature and pressure compared with throttle conditions. Continued removal of low-specific-volume liquid formed by condensation of the highspecific-volume steam may be thought of as creating and maintaining the low pressure in the condenser. The phase change in turn depends on the transfer of heat released to the external cooling water. Thus the rejection of heat to the surroundings by the cooling water is essential to maintaining the low pressure in the condenser. Applying the steady-flow First Law of Thermodynamics to the condensing steam yields: qc = h3 – h2

[Btu/lbm | kJ/kg]

(2.3)

The condenser heat transfer qc is negative because h2 > h3. Thus, consistent with sign convention, qc represents an outflow of heat from the condensing steam. This heat is absorbed by the cooling water passing through the condenser tubes. The condensercooling-water temperature rise and mass-flow rate mc are related to the rejected heat by: ms|qc| = mc cwater(Tout - Tin) [Btu/hr | kW] where cwater is the heat capacity of the cooling water [Btu/lbm-R | kJ/kg-K]. The condenser cooling water may be drawn from a river or a lake at the temperature Tin and returned downstream at Tout, or it may be circulated through cooling towers where heat is rejected from the cooling water to the atmosphere. We can express the condenser heat transfer in terms of an overall heat transfer coefficient, U, the mean cooling water temperature, Tm = (Tout + Tin)/2, and the condensing temperature T3: ms|qc| = UA(T3 - Tm)

[Btu/hr | kJ/s]

It is seen for given heat rejection rate, the condenser size represented by the tube surface area A depends inversely on (a) the temperature difference between the condensing steam and the cooling water, and (b) the overall heat-transfer coefficient. For a fixed average temperature difference between the two fluids on opposite sides of the condenser tube walls, the temperature of the available cooling water controls the condensing temperature and hence the pressure of the condensing steam.

41 Therefore, the colder the cooling water, the lower the minimum temperature and pressure of the cycle and the higher the thermal efficiency of the cycle. A pump is a device that moves a liquid from a region of low pressure to one of high pressure. In the Rankine cycle the condenser condensate is raised to the pressure of the steam generator by boiler feed pumps, BFP. The high-pressure liquid water entering the steam generator is called feedwater. From the steady-flow First Law of Thermodynamics, the work and power required to drive the pump are: wp = h3 – h4

[Btu/lbm | kJ/kg]

(2.4)

and Pump Power = mswp = ms(h3 – h4)

[Btu/hr | kW]

where the negative values resulting from the fact that h4 > h3 are in accordance with the thermodynamic sign convention, which indicates that work and power must be supplied to operate the pump. The net power delivered by the Rankine cycle is the difference between the turbine power and the magnitude of the pump power. One of the significant advantages of the Rankine cycle is that the pump power is usually quite small compared with the turbine power. This is indicated by the work ratio, wt / wp, which is large compared with one for Rankine cycle. As a result, the pumping power is sometimes neglected in approximating the Rankine cycle net power output. It is normally assumed that the liquid at a pump entrance is saturated liquid. This is usually the case for power-plant feedwater pumps, because on the one hand subcooling would increase the heat edition required in the steam generator, and on the other the introduction of steam into the pump would cause poor performance and destructive, unsteady operation. The properties of the pump inlet or condenser exit (state 3 in Figure 2.2) therefore may be obtained directly from the saturated-liquid curve at the (usually) known condenser pressure. The properties for an isentropic pump discharge at state 4 could be obtained from a subcooled-water property table at the known inlet entropy and the throttle pressure. However, such tables are not widely available and usually are not needed. The enthalpy of a subcooled state is commonly approximated by the enthalpy of the saturated-liquid evaluated at the temperature of the subcooled liquid. This is usually quite accurate because the enthalpy of a liquid is almost independent of pressure. An accurate method for estimating the pump enthalpy rise and the pump work is given later (in Example 2.3). A measure of the effectiveness of an energy conversion device is its thermal efficiency, which is defined as the ratio of the cycle net work to the heat supplied from external sources. Thus, by using Equations (2.1), (2.2), and (2.4) we can express the ideal Rankine-cycle thermal efficiency in terms of cycle enthalpies as:

42

th = (h1 – h2 + h3 – h4)/(h1 – h4)

[dl]

(2.5)

In accordance with the Second Law of Thermodynamics, the Rankine cycle efficiency must be less than the efficiency of a Carnot engine operating between the same temperature extremes. As with the Carnot-cycle efficiency, Rankine-cycle efficiency improves when the average heat-addition temperature increases and the heatrejection temperature decreases. Thus cycle efficiency may be improved by increasing turbine inlet temperature and decreasing the condenser pressure (and thus the condenser temperature). Another measure of efficiency commonly employed by power plant engineers is the heat rate, that is, the ratio of the rate of heat addition in conventional heat units to the net power output in conventional power units. Because the rate of heat addition is proportional to the fuel consumption rate, the heat rate is a measure of fuel utilization rate per unit of power output. In the United States, the rate of heat addition is usually stated in Btu/hr, and electrical power output in kilowatts, resulting in heat rates being expressed in Btu/kW-hr. The reader should verify that the heat rate in English units is given by the conversion factor, 3413 Btu/kW-hr, divided by the cycle thermal efficiency as a decimal fraction, and that its value has a magnitude of the order of 10,000 Btu/kW-hr. In the SI system of units, the heat rate is usually expressed in kJ/kW-hr, is given by 3600 divided by the cycle efficiency as a decimal fraction, and is of the same order of magnitude as in the English system. It is evident that a low value of heat rate represents high thermal efficiency and is therefore desirable. EXAMPLE 2.1

An ideal Rankine cycle (see Figure 2.2) has a throttle state of 2000 psia/1000°F and condenser pressure of 1 psia. Determine the temperatures, pressures, entropies, and enthalpies at the inlets of all components, and compare the thermal efficiency of the cycle with the relevant Carnot efficiency. Neglect pump work. What is the quality of the steam at the turbine exit? Solution The states at the inlets and exits of the components, following the notation of Figure 2.2, are listed in the following table. The enthalpy and entropy of state 1 may be obtained directly from tables or charts for superheated steam (such as those in Appendices B and C) at the throttle conditions. A Mollier chart is usually more convenient than tables in dealing with turbine inlet and exit conditions. For an ideal isentropic turbine, the entropy is the same at state 2 as at state 1. Thus state 2 may be obtained from the throttle entropy (s2 = s1 = 1.5603 Btu/lbm-R) and the condenser pressure (1 psia). In general, this state may be in either the superheatedsteam region or the mixed-steam-and-liquid region of the Mollier and T-s diagrams. In the present case it is well into the mixed region, with a temperature of 101.74°F and an enthalpy of 871 Btu/lbm.

43 The enthalpy, h3 = 69.73 Btu/lbm, and other properties at the pump inlet are obtained from saturated-liquid tables, at the condenser pressure. The steady-flow First Law of Thermodynamics, in the form of Equation (2.4), indicates that neglecting isentropic pump work is equivalent to neglecting the pump enthalpy rise. Thus in this case Equation (2.4) implies that h3 and h4 shown in Figure (2.2) are almost equal. Thus we take h4 = h3 as a convenient approximation. State 1

Temperature (°F)

Pressure (psia)

Entropy (Btu/lbm-°R)

Enthalpy (Btu/lbm)

1000.0

2000

1.5603

1474.1 871.0

2

101.74

1

1.5603

3

101.74

1

0.1326

69.73

4

101.74

2000

0.1326

69.73

The turbine work is h1 – h2 = 1474.1 – 871 = 603.1 Btu/lbm. The heat added in the steam generator is h1 – h4 = 1474.1 – 69.73 = 1404.37 Btu/lbm. The thermal efficiency is the net work per heat added = 603.1/1404.37 = 0.4294 (42.94%). This corresponds to a heat rate of 3413/0.4294 = 7946 Btu/kW-hr. As expected, the efficiency is significantly below the value of the Carnot efficiency of 1 – (460 + 101.74)/(460 + 1000) = 0.6152 (61.52%), based on a source temperature of T1 and a sink temperature of T3. The quality of the steam at the turbine exit is (s2 – sl)/(sv – sl) = (1.5603 – 0.1326)/(1.9781 – 0.1326) = 0.7736 Here v and l indicate saturated vapor and liquid states, respectively, at pressure p2. Note that the quality could also have been obtained from the Mollier chart for steam as 1 - M, where M is the steam moisture fraction at entropy s2 and pressure p2. __________________________________________________________________ Example 2-2 If the throttle mass-flow is 2,000,000 lbm/hr and the cooling water enters the condenser at 60°F, what is the power plant output in Example 2.1? Estimate the cooling-water mass-flow rate.

44 Solution: The power output is the product of the throttle mass-flow rate and the power plant net work. Thus Power = (2 × 106)(603.1) = 1.206 × 109 Btu/hr or Power = 1.206 × 109 / 3413 = 353,413 kW. The condenser heat-transfer rate is msqc = ms ( h3 – h2 ) = 2,000,000 × (69.73 – 871) = – 1.603×109 Btu/hr The condensing temperature, T3 = 101.74 °F, is the upper bound on the cooling water exit temperature. Assuming that the cooling water enters at 60°F and leaves at 95°F, the cooling-water flow rate is given by mc = ms|qc| / [ cwater(Tout – Tin)] = 1.603×109 /[(1)(95 - 60)] = 45.68×106 lbm/hr A higher mass-flow rate of cooling water would allow a smaller condenser coolingwater temperature rise and reduce the required condenser-heat-transfer area at the expense of increased pumping power. ____________________________________________________________________ 2.4 Deviations from the Ideal – Component Efficiencies In a power plant analysis it is sometimes necessary to account for non-ideal effects such as fluid friction, turbulence, and flow separation in components otherwise assumed to be reversible. Decisions regarding the necessity of accounting for these effects are largely a matter of experience built on familiarity with the magnitudes of the effects, engineering practices, and the uses of the calculated results. Turbine In the case of an adiabatic turbine with flow irreversibilities, the steady-flow First Law of Thermodynamics gives the same symbolic result as for the isentropic turbine in Equation (2.1), i.e., wt = h1 – h2

[Btu/lb | kJ/kg]

(2.6)

except that here h2 represents the actual exit enthalpy and wt is the actual work of an adiabatic turbine where real effects such as flow separation, turbulence, irreversible internal heat transfers, and fluid friction exist.

45

An efficiency for a real turbine, known as the isentropic efficiency, is defined as the ratio of the actual shaft work to the shaft work for an isentropic expansion between the same inlet state and exit pressure level. Based on the notation of Figure 2.4, we see that the turbine efficiency is:

turb = (h1 – h2 )/(h1 – h2s )

[dl]

(2.7)

where h2s, the isentropic turbine-exit enthalpy, is the enthalpy evaluated at the turbine inlet entropy and the exit pressure. For the special case of an isentropic turbine, h2 = h2s and the efficiency becomes 1. Note how state 2 and the turbine work change in Figure 2.4 as the efficiency increases toward 1. The diagram shows that the difference between the isentropic and actual work, h2 – h2s, represents work lost due to irreversibility. Turbine isentropic efficiencies in the low 90% range are currently achievable in well-designed machines. Normally in solving problems involving turbines, the turbine efficiency is known from manufacturers’ tests, and the inlet state and the exhaust pressure are specified. State 1 and p2 determine the isentropic discharge state 2s using the steam tables. The actual turbine-exit enthalpy can then be calculated from Equation (2.7). Knowing both p2 and h2, we can then fully identify state 2 and account for real turbine behavior in any cycle analysis. Pump Work must be supplied to a pump to move liquid from a low pressure to a high pressure. Some of the work supplied is lost due to irreversibilities. Ideally the remaining effective work to raise the pressure is necessarily less than that supplied. In order for

46

the efficiency of a pump to be less than or equal to 1, it is defined in inverse fashion to turbine efficiency. That is, pump efficiency is the ratio of the isentropic work to the actual work input when operating between two given pressures. Applying Equation (2.4) and the notation of Figure (2.5), the isentropic pump work, wps = h3 – h4s, and the pump isentropic efficiency is

pump = wps /wp = (h4s – h3)/(h4 – h3)

[dl]

(2.8)

Note the progression of exit states that would occur in Figure 2.5 as pump efficiency increases for a fixed inlet state and exit pressure. It is seen that the pump lost work, given by h4 – h4s decreases and that the actual discharge state approaches the isentropic discharge state. States 4 and 4s are usually subcooled liquid states. As a first approximation their enthalpies may be taken to be the saturated liquid enthalpy at T3. More accurate approximations for these enthalpies may be obtained by applying the First Law for a closed system undergoing a reversible process, Equation (1.8): Tds = dh - vdp. For an isentropic process it follows that dh = vdp. Because a liquid is almost incompressible, its specific volume, v, is almost independent of pressure. Thus, using the notation of Figure 2.5, integration with constant specific volume yields h4s = h3 + v3 ( p4 – p3 )

[Btu/lbm | kJ/kg]

where a knowledge of state 3 and p4 determines h4s.

47 Using Equation (2.8), and without consulting tables for subcooled water, we can then calculate the pump work from wp = v3(p3 – p4)/p

[ft-lbf/lbm | kN-m/kg]

(2.9)

Note that the appropriate conversion factors must be applied for dimensional consistency in Equation (2.9). EXAMPLE 2.3

Calculate the actual work and the isentropic and actual discharge enthalpies for an 80% efficient pump with an 80°F saturated-liquid inlet and an exit pressure of 3000 psia. Solution

From the saturated-liquid tables, for 80°F, the pump inlet conditions are 0.5068 psia, 48.037 Btu/lbm, and 0.016072 ft3/lbm. Using Equation (2.9), we find that the pump work is or

wp = [0.016072(0.5068 – 3000)(144)]/0.8 = – 8677 ft-lbf / lbm wp = – 8677/778 = – 11.15 Btu/lbm.

Note the importance of checking units here. The actual discharge enthalpy is h4 = h3 – wp = 48.037 – (–11.15) = 59.19 Btu/lbm. and the isentropic discharge enthalpy is h4s = h3 – p wp = 48.037 – (0.8)(– 11.15) = 56.96 Btu/lbm. ____________________________________________________________________ EXAMPLE 2.4

What is the turbine work, the net work, the work ratio, and the cycle thermal efficiency for the conditions of Example 2.1 if the turbine efficiency is 90% and the pump efficiency is 85%? What is the turbine exit quality? Solution

By the definition of isentropic efficiency, the turbine work is 90% of the isentropic turbine work = (0.9)(603.1) = 542.8 Btu/lbm. By using Equation (2.9), the isentropic pump work is [(0.01614)(1 – 2000)(144)] / 778 = – 5.97 Btu/lbm.

48 The actual pump work is then – 5.97/.85 = – 7.03 Btu/lbm and the work ratio is 542.8/| – 7.03| = 77.2 The cycle net work is wt + wp = 542.8 – 7.03 = 535.8 Btu/lbm.. Applying the steady-flow First Law of Thermodynamics to the pump, we get the enthalpy entering the steam generator to be h4 = h3 – wp = 69.73 – (– 7.03) = 76.76 Btu/lbm. The steam-generator heat addition is then reduced to 1474.1 – 76.76 = 1397.3 Btu/lbm. and the cycle efficiency is 535.8/1397.3 = 0.383. Study of these examples shows that the sizable reduction in cycle efficiency from that in Example 2.1 is largely due to the turbine inefficiency, not to the neglect of pump work. From Equation (2.6), the true turbine exit enthalpy is the difference between the throttle enthalpy and actual turbine work = 1474.1 - 542.8 = 931.3 Btu/lbm. The quality is then x = (h2 – hl)/(hv – hl) = (931.3 – 69.73)/(1105.8 – 69.73) = 0.832. Thus the turbine inefficiency increases the turbine exhaust quality over the isentropic turbine value of 0.774. ____________________________________________________________________ 2.5 Reheat and Reheat Cycles A common modification of the Rankine cycle in large power plants involves interrupting the steam expansion in the turbine to add more heat to the steam before completing the turbine expansion, a process known as reheat. As shown in Figure 2.6, steam from the high-pressure (HP) turbine is returned to the reheat section of the steam generator through the "cold reheat" line. There the steam passes through heated tubes which restore it to a temperature comparable to the throttle temperature of the high pressure turbine. The reenergized steam then is routed through the "hot reheat" line to a low-pressure turbine for completion of the expansion to the condenser pressure. Examination of the T-s diagram shows that reheat increases the area enclosed by the cycle and thus increases the net work of the cycle by virtue of the cyclic integral, Equation (1.3). This is significant, because for a given design power output higher net work implies lower steam flow rate. This, in turn, implies that smaller plant components may be used, which tends to reduce the initial plant cost and to compensate for added costs due to the increased complexity of the cycle. Observe from Figure 2.6 that the use of reheat also tends to increase the average temperature at which heat is added. If the low-pressure turbine exhaust state is superheated, the use of reheat may also increase the average temperature at which heat is rejected. The thermal efficiency may therefore increase or decrease, depending on specific cycle conditions. Thus the major benefits of reheat are increased net work,

49

drying of the turbine exhaust (discussed further later), and the possibility of improved cycle efficiency. Note that the net work of the reheat cycle is the algebraic sum of the work of the two turbines and the pump work. Note also that the total heat addition is the sum of the heat added in the feedwater and reheat passes through the steam generator. Thus the

50 thermal efficiency of the reheat cycle is: (h1 – h2) + (h3 – h4) + (h5 – h6)

th = -----------------------------------

[dl]

(2.10)

(h1 – h6) + (h3 – h2)

Relations such as this illustrate the wisdom of learning to analyze cycles using definitions and applying fundamentals to components rather than memorizing equations for special cases such as Equation (2.5) for the efficiency of the simple Rankine cycle. Note that the inclusion of reheat introduces a third pressure level to the Rankine cycle. Determination of a suitable reheat pressure level is a significant design problem that entails a number of considerations. The cycle efficiency, the net work, and other parameters will vary with reheat pressure level for given throttle and condenser conditions. One of these may be numerically optimized by varying reheat pressure level while holding all other design conditions constant. Reheat offers the ability to limit or eliminate moisture at the turbine exit. The presence of more than about 10% moisture in the turbine exhaust can cause erosion of blades near the turbine exit and reduce energy conversion efficiency. Study of Figure 2.6 shows that reheat shifts the turbine expansion process away from the two-phase region and toward the superheat region of the T-s diagram, thus drying the turbine exhaust. EXAMPLE 2.5

Reanalyze the cycle of Example 2.1 (2000 psia/1000°F/1 psia) with reheat at 200 psia included. Determine the quality or degree of superheat at the exits of both turbines. Assume that reheat is to the HP turbine throttle temperature. Solution

Referring to Figure 2.6, we see that the properties of significant states are the following: State

Enthalpy (Btu/lbm)

Temperature (°F)

Pressure (psia)

Entropy (Btu/lbm-°R)

1

1000.0

2000

1.5603

1474.1

2

400.0

200

1.5603

1210.0

3

1000.0

200

1.84

1527.0

4

101.74

1

1.84

1028.0

5

101.74

1

0.1326

69.73

6

101.74

2000

0.1326

69.73

51 Properties here are obtained from the steam tables and the Mollier chart as follows: 1. The enthalpy and entropy at state 1 are read from the superheated-steam tables at the given throttle temperature and pressure. 2. State 2 is evaluated from the Mollier diagram at the given reheat pressure and the same entropy as in state 1 for the isentropic turbine expansion. 3. Reheat at constant pressure p3 = p2 to the assumed throttle temperature T3 = T1 gives s3 and h3. Normally, T3 is assumed equal to T1 unless otherwise specified. 4. The second turbine flow is also specified as isentropic with expansion at s4 = s3 to the known condenser pressure p4. 5. The condenser exit (pump entrance) state is assumed to be a saturated liquid at the known condenser pressure. 6. Pump work is neglected here. The steady-flow First Law then implies that h6 = h5, which in turn implies the T6 = T5. The turbine work is the sum of the work of both turbines: (1474.1 - 1210) + (1527 - 1028) = 763.1 Btu/lbm. The heat added in the steam generator feedwater and reheat passes is (1474.1 - 69.73) + (1527 - 1210) = 1721.4 Btu/lbm. The thermal efficiency then is 763.1/1721.4 = 0.443, or 44.3%. Both the net work and the cycle efficiency are higher than in the simple Rankine cycle case of Example 2.1. From the Mollier chart in Appendix B it is readily seen that state 2 is superheated, with 400 - 381.8 = 18.2 Fahrenheit degrees of superheat; and state 4 is wet steam, with 7.4% moisture, or 0.926 (92.6%) quality. Thus the first turbine has no moisture and the second is substantially drier than 0.774 quality value in Example 2.1. ____________________________________________________________________ Reheat is an important feature of all large, modern fossil-fueled steam power plants. We now consider another key feature of these plants, but temporarily omit reheat, for the purpose of clarity. 2.6 Regeneration and Feedwater Heaters The significant efficiency advantage of the Carnot cycle over the Rankine cycle is due to the fact that in the Carnot cycle all external heat addition is at a single high

52 temperature and all external heat rejection at a single low temperature. Examination of Figures 2.2 and 2.6 shows that heat addition in the steam generator takes place over a wide range of water temperature in both the simple and reheat Rankine cycles. Presumably, the Rankine-cycle thermal efficiency could be improved by increasing the average water temperature at which heat is received. This could be accomplished by an internal transfer of heat from higher-temperature steam to low-temperature feedwater. An internal transfer of heat that reduces or eliminates low-temperature additions of external heat to the working fluid is known as regeneration. Open Feedwater Heaters Regeneration is accomplished in all large-scale, modern power plants through the use of feedwater heaters. A feedwater heater (FWH) is a heat exchanger in which the latent heat (and sometimes superheat) of small amounts of steam is used to increase the temperature of liquid water (feedwater) flowing to the steam generator. This provides the internal transfer of heat mentioned above. An open feedwater heater is a FWH in which a small amount of steam mixes directly with the feedwater to raise its temperature. Steam drawn from a turbine for feedwater heating or other purposes is called extraction steam. Feedwater heaters in which extraction steam heats feedwater without fluid contact will be discussed later. Consider the regenerative Rankine-cycle presented in Figure 2.7. The steam leaving the high-pressure (HP) turbine is split with a small part of the mass flow extracted to an open FWH and the major part of the flow passing to a low pressure (LP) turbine. The T-s diagram shows that steam entering the FWH at state 2 is at a higher temperature than the subcooled feedwater leaving the pump at state 5. When the two fluids mix in the FWH, the superheat and the heat of vaporization of the extraction steam are transferred to the feedwater, which emerges with the condensed extraction steam at a higher temperature, T6.. It is assumed that all streams entering and leaving the FWH are the same pressure so that the mixing process occurs at constant pressure. The T-s and flow diagrams show that heat from combustion gases in the steam generator need only raise the water temperature from T7 to T1 rather than from T5 when extraction steam is used to heat the feedwater. The average temperature for external heat addition must therefore increase. Despite the reduced flow rate through the lowpressure turbine, we will see by example that the thermal efficiency of the steam cycle is improved by the transfer of energy from the turbine extraction flow to the feedwater. The analysis of cycles with feedwater heaters involves branching of steam flows. In Figure 2.7, for example, conservation of mass must be satisfied at the flow junction downstream of the high-pressure-turbine exit. Thus, assuming a mass flow of 1 at the HP turbine throttle and a steam mass-flow fraction, m1, through the feedwater heater, the low-pressure-turbine mass-fraction must be 1 - m1. Note that the latter flow passes through the condenser and pump and is reunited with the extraction flow, m1, in the FWH at state 6, where the exit-flow-rate fraction is again unity.

53

It will be seen later that it is common for more than one FWH to be used in a single power plant. When more than one FWH is present, mass flows m1, m2...mn are defined for each of the n FWHs. Conservation of mass is used to relate these flows to

54 condenser flow rate and the reference throttle flow rate. This is accomplished by taking a mass flow of 1 at the high-pressure-turbine throttle as a reference, as in the case of a single FWH discussed above. After solving for each of the thermodynamic states and FWH mass fractions, actual mass flow rates are obtained as the products of the known (or assumed) throttle flow rate and FWH mass-flow fractions. The function of feedwater heaters is to use the energy of extraction steam to reduce the addition of low-temperature external heat by raising the temperature of the feedwater before it arrives at the steam generator. Feedwater heaters are therefore insulated to avoid heat loss to the surroundings. Because the resulting heat loss is negligible compared with the energy throughflow, feedwater heaters are usually treated as adiabatic devices. In order to avoid irreversibility associated with unrestrained expansion, constant pressure mixing of the streams entering the FWH is necessary. Returning to Figure 2.7, this implies that the pressures of the feedwater at state 5 and at the FWH exit state 6 are chosen to be the same as that of the extraction steam at state 2. Note that, as with reheat, the inclusion of a FWH also introduces an additional pressure level into the Rankine cycle as seen in the T-s diagram. In the figure, the extraction pressure level, p2, is another parameter under the control of the designer. The extraction mass flow rate, m1, is in turn controlled by the designer’s choice of p2. The mass-flow rate is determined by the physical requirement that the feedwater entering the FWH at state 5 increase in temperature to T6 through absorption of the heat released by the condensing extraction steam. This is accomplished by applying the steady-flow First Law of Thermodynamics, using appropriate mass fractions, to the insulated open FWH: q = 0 = (1)h6 – m1h2 – (1 – m1 )h5 + 0

[Btu/lbm | kJ/kg]

Every term in this equation has dimensions of energy per unit throttle mass, thus referring all energy terms to the mass-flow rate at the throttle of the high-pressure turbine. For example, the second term on the right is of the form: FWH Extraction mass Enthalpy at state 2 Enthalpy at state 2 -------------------------- × -------------------------- = ---------------------Throttle mass FWH Extraction mass Throttle mass Similarly, the structure of the third term on the right has the significance of Pump mass Enthalpy at state 5 Enthalpy at state 5 ---------------- × --------------------- = ---------------------Throttle mass Pump mass Throttle mass

55 Solving for the extraction mass fraction, we obtain m1 = (h6 – h5) / (h2 – h5)

[dl]

(2.11)

For low extraction pressures, the numerator is usually small relative to the denominator, indicating a small extraction flow. The T-s diagram of Figure 2.7 shows that increasing the extraction pressure level increases both h6 and h2. Thus, because the small numerator increases faster than the large denominator, we may reason, from Equation (2.11), that the extraction mass-flow fraction must increase as the extraction pressure level increases. This conforms to the physical notion that suggests the need for more and hotter steam to increase the feedwater temperature rise. While such intuitions are valuable, care should be exercised in accepting them without proof. The total turbine work per unit throttle mass flow rate is the sum of the work of the turbines referenced to the throttle mass-flow rate. Remembering that 1 - m1 is the ratio of the low-pressure turbine mass flow to the throttle mass flow, we obtain: wt = (h1 – h2 ) + (1 – m1 )(h2 – h3 )

[Btu/lbm | kJ/kg]

(2.12)

The reader should examine the structure of each term of Equation (2.12) in the light of the previous discussion. Note that it is not important to remember these specific equations, but it is important to understand, and be able to apply, the reasoning by which they are obtained. For a given throttle mass flow rate, mthr [lbm/s | kg/s], the total turbine power output is given by mthrwt [Btu/s | kW]. We see in Figure 2.7 that the heat addition in the steam generator is reduced, due to extraction at pressure p6 = p2, by about h7 – h5 to qa = h1 – h7

[Btu/lbm | kJ/kg]

(2.13)

At the same time, the net work also decreases, but more slowly, so that the net effect is that the cycle efficiency increases with increased extraction.

EXAMPLE 2.6

Solve Example 2.1 (2000 psia /1000°F/1 psia) operating with an open feedwater heater at 200 psia. Solution

Referring to Figure 2.7, we find that the properties of significant states are:

56 State

Enthalpy (Btu/lbm)

Temperature (°F)

Pressure (psia)

Entropy (Btu/lbm-°R)

1

1000.0

2000

1.5603

1474.1

2

400.0

200

1.5603

1210.0

3

101.74

1

1.5603

871.0

4

101.74

1

0.1326

69.73

5

101.74

200

0.1326

69.73

6

381.8

200

0.5438

355.5

7

381.8

2000

0.5438

355.5

States 1 through 4 are obtained in the same way as in earlier examples. Constant pressure mixing requires that p5 = p6 = p2, the extraction pressure level. State 6, a pump entrance state, is assumed to be a saturated-liquid state as usual. Subcooledliquid states are approximated, as before, consistent with the neglect of pump work. The extraction mass fraction obtained by applying the steady-flow First Law of Thermodynamics to the FWH, Equation (2.11), is m1 = (355.5 – 69.73)/(1210 – 69.73) = 0.251. The net work (neglecting pump work) by Equation (2.12), is then wn = (1474.1 – 1210) + (1 – 0.251)(1210 – 871) = 518.1 Btu/lbm This may be compared with the simple-cycle net work of 603.1 Btu/lbm. The heat added in the steam generator by Equation (2.13) is qa = h1 – h7 = 1474.1 – 355.5 = 1118.6 Btu/lbm. The resulting cycle efficiency is th = 518.1/1118.6 = 0.463, or 46.3%, a significantly higher value than the 42.94% for the corresponding simple Rankine cycle. Note, however, that the LP-turbine exhaust quality is the same as for the simple Rankine cycle, an unacceptable 77.4%. This suggests that a combination of reheat and regeneration through feedwater heating may be desirable. We will investigate this possibility later after looking at closed feedwater heaters. _____________________________________________________________________ Closed Feedwater Heaters We have seen that feedwater heating in open feedwater heaters occurs by mixing of extraction steam and feedwater. Feedwater heating also is accomplished in shell-and-

57 tube-type heat exchangers, where extraction steam does not mix with the feedwater. Normally, feedwater passes through banks of tubes whereas steam condenses on the outside of the tube surfaces in these heaters. Such heat exchangers are called closed feedwater heaters. Pumped Condensate. Closed feedwater heaters normally are employed in two configurations in power plants. In the configuration shown in figure 2.8, condensate is pumped from the condenser through the FWH and the steam generator directly to the turbine along the path 4-5-8-9-1. Ideally, p5 = p1 assuming no pressure drop in the FWH and steam generator. Note that if m1 mass units of steam are extracted from the turbine for use in the FWH, only 1 - m1 units of feedwater pass throught the condenser, pump, and the tubes of the FWH. The condensed extraction steam (condensate) emerging from the FWH at state 6 is pumped separately from p6 = p2 to throttle pressure p7 = p1, where it becomes part of the steam generator feedwater. The pumped condensate at state 7 thus mixes with the heated feedwater at state 8 to form the total feedwater flow at state 9. Constant pressure mixing ( p7 = p8 = p9) is required at this junction to avoid losses associated with uncontrolled flow expansion. The enthalpy of the feedwater entering the steam generator can be determined by applying the steady-flow First Law of Thermodynamics to the junction of the feedwater and FWH streams: h9 = (1 – m1 )h8 + m1h7

[Btu/lbm | kJ/kg]

As in the open FWH analysis, the extraction mass fraction depends on the choice of intermediate pressure p2 and is obtained by applying the steady-flow First Law of Thermodynamics to the feedwater heater. Throttled Condensate. The second closed FWH configuration is shown in Figure 2.9 where the FWH condensate drops in pressure from p6 = p2 through a trap into the condenser at pressure p7 = p3 = p4. The trap allows liquid only to pass from the FWH at state 6 in a throttling process to state 7. As usual, it is assumed that the throttling process is adiabatic. The T-s diagram shows that the saturated liquid at state 6 flashes into a mixture of liquid and vapor in the condenser with no change in enthalpy, h7 = h6. For this configuration, the closed FWH condensate mass-flow rate is equal to the extraction mass-flow rate. As a result, conservation of mass applied to the condenser shows that the mass-flow rate leaving the condenser and passing through the pump and FWH tubes is the same as the throttle mass-flow rate. The throttled-condensate, closed feedwater heater is the preferred configuration in power plants, because it is unnecessary for each FWH to have a condensate pump.

58

59

60 EXAMPLE 2.7

Rework Example 2.1 (2000 psia/1000°F/1 psia) with reheat and a closed feedwater heater with extraction from the cold reheat line and FWH condensate throttled to the condenser. Both reheat and extraction are at 200 psia. Assume that the feedwater leaving the FWH is at the temperature of the condensing extraction stream. Draw appropriate T-s and flow diagrams. Solution

Referring to the notation of Figure 2.10, verify that the significant the thermodynamic state properties are: State

Enthalpy (Btu/lbm)

Temperature (°F)

Pressure (psia)

Entropy (Btu/lbm-°R)

1

1000.0

2000

1.5603

1474.1

2

400.0

200

1.5603

1210.0

3

1000.0

200

1.84

1527.0 1028.0

4

101.74

1

1.84

5

101.74

1

0.1326

69.73

6

101.74

2000

0.1326

69.73

7

381.8

200

0.5438

355.5

8

101.74

1

__

355.5

9

381.8

2000

__

355.5

Applying the steady-flow First Law of Thermodynamics to the FWH, we obtain: 0 = h9 + m1h7 – m1h2 – h6 + 0 which, solved for m1, yields: m1 = ( h9 – h6 )/( h2 – h7 ) = (355.5 – 69.73)/(1210 – 355.5) = 0.3344 The total net work per unit of mass flow at the throttle of the HP turbine is the sum of the specific work of each of the turbines adjusted for the HP turbine throttle mass flow: wn = h1 – h2 + (1 – m1)( h3 – h4 ) = 1474.1 – 1210 + (1 – 0.3344)(1527 – 1028) = 596.2 Btu/lbm

61

62 As in the earlier examples in this series, pump work has been neglected. The heat addition per unit HP-turbine-throttle mass is the sum of the heat addition in the main pass and reheat pass through the steam generator, the latter as adjusted for the reduced mass flow. Thus the steady-flow First Law of Thermodynamics yields qa = h1 – h9 + (1 – m1 )( h3 – h2 ) = 1474.1 – 355.5 + (1 – 0.3344)(1527 – 1210) = 1329.6 Btu/lbm The thermal efficiency of the cycle is wn / qa = 596.2 / 1329.6 = 0.448, or 44.8%. The Mollier chart shows that the discharge of the first turbine (state 2) has 20 degrees of superheat and the second turbine (state 4) 7.4% moisture, or a quality of 0.926. ___________________________________________________________________ In the above calculation it was assumed that the feedwater temperature leaving the FWH had risen to the temperature of the condensing extraction steam. Since the FWH is a heat exchanger of finite area, the feedwater temperature T9 usually differs from the condensing temperature of the extraction steam T7. If the surface area of the FWH is small, the feedwater will emerge at a temperature well below the extraction-steam condensing temperature. If the area were increased, the feedwater temperature would approach the condensing temperature. This aspect of FWH design is reflected in the parameter known as the terminal temperature difference, TTD, defined as TTD = Tsat - Tfw

[R | K]

where Tfw is the temperature of the feedwater leaving the tubes and Tsat is the condensing temperature of the extraction steam in the closed FWH. In Figure 2.10, for instance, Tfw = T9 and Tsat = T7. Thus, if the TTD and the extraction pressure are known, the true FWH exit temperature may be determined. An application of the TTD will be considered in a later example. Table 2.1 summarizes, for comparison, the results of the calculations for the several plant configurations that we have considered. The reader is cautioned that since these calculations have not accounted for turbine inefficiency, the thermal efficiencies are unusually high. While the efficiency differences with respect to the simple cycle may seem insignificant, they are of great economic importance. It must be realized that hundreds of millions of dollars may be spent on fuel each year in a power plant and that capital costs are equally impressive. As a result, the choice of cycle and design characteristics are of great significance. Some further improvement in net work and efficiency could be shown by selecting extraction and reheat pressure levels to maximize these parameters.

63 Table 2.1 Comparison of Rankine Cycle Modifications Net Work (Btu/lbm)

Efficiency %

Heat Rate (Btu/kW-hr)

Turbine Exit Quality

Simple cycle

603.1

42.9

7956

0.774

Reheat cycle

763.1

44.3

7704

0.926

One open FWH

518.1

46.3

7371

0.774

One closed FWH and reheat

596.2

44.8

7618

0.926

Multistage Extraction It has been shown that increases in cycle efficiency may be accomplished in a steam power plant through regeneration via the feedwater heater. Large steam power plants typically employ large numbers of feedwater heaters for this purpose. Multistage extraction refers to the use of multiple extractions to supply steam to these feedwater heaters. Earlier discussions of examples involved extractions taken only from the flows between turbines. However, the number of extractions is not limited by the number of turbines. In fact, large turbines are designed with several extraction points through which steam may be withdrawn for feedwater heating and other purposes. Assigning Extraction-Pressure Levels. Given n feedwater heaters, it is necessary to assign values to the n associated extraction pressures. For preliminary design purposes, the extraction-pressure levels assigned may be those that give equal feedwater temperature rises through each heater and through the steam generator to the boiling point. Thus, for n heaters the appropriate temperature rise is given by

Topt = ( Tsl – Tcond )/( n + 1)

[R | K]

(2.14)

where Tsl is the temperature the saturated liquid at the throttle pressure and Tcond is the temperature the feedwater leaving the condenser. The corresponding steam condensing temperature in the ith heater is then Ti = Tcond + ( i )Topt = Tcond + i ( Tsl – Tcond )/( n + 1)

[R | K]

(2.15)

where i = 1, 2..., n. Steam tables may then be used to evaluate the corresponding extraction-pressure levels. It is, of course, possible and sometimes necessary to assign extraction-pressure levels in other ways.

64 EXAMPLE 2.8

Evaluate the recommended extraction-pressure levels for single heater for the 1000° F/2000 psia throttle and one psia condenser that have been used throughout this chapter. Solution

The feedwater temperature rise to establish an appropriate extraction-pressure level for a single heater for a plant such as that shown in Figures 2.7 through 2.9 is (Tsl – T4 )/2 = (635.8 – 101.74)/2 = 267.05°F where Tsl was evaluated at p1 = 2000 psia. This would make T6 = 101.74 + 267.05 = 368.79°F and the corresponding extraction pressure level p6 = p2 = 171 psia, using the saturated-steam tables. ____________________________________________________________________ At this point we have the tools necessary to evaluate the performance and penalties associated with a given configuration. The following example examines the gains that follow from the use a single feedwater heater and the sensitivity of the thermal efficiency to the assigned feedwater temperature rise. EXAMPLE 2.9

Consider a single open feedwater heater operating in a Rankine cycle with a 2000 psia saturated-vapor throttle and a 1 psia condenser. Evaluate the thermal efficiency as a function of feedwater temperature rise. Compare the temperature rise that maximizes the thermal efficiency with the results of Equation (2.14). Solution

Utilizing the notation of Figure 2.7 and taking the throttle state as a saturated vapor, we get the results that are summarized in spreadsheet format in Table 2.2. (This table is a direct reproduction of a Quattro Pro spreadsheet used in the analysis. Care should be taken if this spreadsheet is used for "what if" studies, because it is dependent on manual entry of thermodynamic properties. To explore other cases, appropriate properties must be obtained from steam tables or charts and inserted in the spreadsheet. Despite this drawback, the spreadsheet provides a convenient means of organizing, performing, and displaying calculations.) Details of the methodology are given in the right-most column. It is seen that the net work drops, as expected, as more extraction steam is used to heat the feedwater. Figure 2.11 shows the percentage increase in thermal efficiency as a function of the feedwater temperature rise for this case. Over a 9% increase in thermal efficiency is achieved with feedwater temperature rises between 200/F and 300°F. Thus the prediction of Topt = 267°F using Equation (2.14) in Example 2.8 is clearly in this range. _____________________________________________________________________

65

Example 2.9 shows that improved thermal efficiency is achieved over a broad range of feedwater temperature rise and therefore extraction pressure. This gives the designer freedom to assign extraction-pressure levels so as to make use of existing designs for feedwater heaters and turbines without severely compromising the efficiency of the plant design. Calculation Methodology. Once the extraction- and reheat-pressure levels are established for a cycle with multistage extraction, and once throttle and condenser conditions, turbomachine efficiencies, and FWH terminal temperature differences are known, significant state properties should be determined. Symbols for extraction mass-fraction variables should be assigned for each heater and related to other unknown flows using mass conservation assuming unit mass flow at the highpressure-turbine throttle. The steady-flow First Law of Thermodynamics should then be applied to each of the FWHs, starting with the highest extraction pressure and progressing to the lowest-pressure FWH. Analyzing the heaters in this order allows each equation to be solved immediately for a mass fraction rather than solving all of the equations simultaneously. Important performance parameters such as thermal efficiency, net work, and work ratio may then be evaluated taking care to account properly for component mass flows. The following example illustrates this methodology.

66

67 EXAMPLE 2.10

Consider a power plant with 1000/F/2000-psia throttle, reheat at 200 psia back to 1000/F, and 1-psia condenser pressure. The plant has two closed feedwater heaters, both with terminal temperature differences of 8/F. The high-pressure (HP) heater condensate is throttled into the low-pressure (LP) heater, which in turn drains into the condenser. Turbomachine efficiencies are 0.88, 0.9, and 0.8 for the HP turbine, the LP turbine, and the boiler feed pump, respectively. Draw relevant T-s and flow diagrams and evaluate FWH mass fractions, thermal efficiency, net work, and work ratio. Solution

The notation used to study this plant is shown in Figure 2.12. The pertinent thermodynamic properties and part of the analysis are presented in the spreadsheet given in Table 2.3. The earlier-stated caution (Example 2.9) about using spreadsheets that incorporate external data applies here as well, because changing parameters may require changes in steam-table lookup values. To start the analysis we first determine the extraction-pressure levels. The ideal FWH temperature rise is given by ( Tsl – T7 )/3 = ( 635.8 – 101.74)/3 = 178.02°F where the saturation temperature is evaluated at the HP-turbine throttle pressure of 2000 psia. The corresponding extraction condensing temperatures and extractionpressure levels are 101.74 + 178.02 = 280/F $ p9 = p5 = 49 psia and 101.74 + (2)(178.02) = 457.8°F $ p12 = p2 = 456 psia where the extraction pressures have been evaluated using the saturated-steam tables. After the entropy and enthalpy at state 1 are evaluated, the enthalpy h3s at the HPturbine isentropic discharge state 3s is determined from s1 and p3. The HP-turbine efficiency then yields h3 and the steam tables give s3. The entropy and enthalpy at the HP-turbine extraction state 2 may be approximated by drawing a straight line on the steam Mollier diagram connecting states 1 and 3 and finding the intersection with the HP-extraction pressure P2. This technique may be used for any number of extraction points in a turbine. Once the hot reheat properties at state 4 are determined from the steam tables, the LP-turbine exit and extraction states at 6 and 5 may be obtained by the same method used for the HP turbine.

68

69

The determination of the FWH condensate temperatures and pressures at states 9 and 12 have already been discussed. The temperatures of the heated feedwater leaving the FWHs may be determined from the terminal temperature differences: T11 = T9 - TTD = 281 - 8 = 273°F T14 = T12 - TTD = 457.5 - 8 = 449.5°F Recalling that the enthalpy of a subcooled liquid is almost independent of pressure, we note that the enthalpies h11 and h14 may be found in the saturated-liquid tables at T11 and T14, respectively.

70 The pump discharge state 8 is a subcooled-liquid state, which may be approximated in the same way as in Examples 2.3 and 2.4. Thus h8s = h7 + ( p8 – p7 )v7 = 69.7 + (2000 – 1)(144)(0.016136)/778 = 75.7 Btu/lbm and h8 = h7 + ( h8s – h7 )/p = 69.7 + (75.7 – 69.7)/0.8 = 77.2 Btu/lbm The pump work is then wp = h7 – h8 = 69.7 – 77.2 = – 7.5 Btu/lbm The extraction mass-flow fractions designated m1 and m2 relate other flows to the unit mass flow at the high-pressure-turbine throttle. For example, the condensate flow rate from the LP heater at state 10 is given by m1 + m2. The steady-flow First Law of Thermodynamics may now be applied to the heaters. For the HP FWH: 0 = m1h12 + (1)h14 – m1h2 – (1)hll may be rewritten as m1 = ( h14 – hll )/( h2 – h12 )

[dl]

This and the T-s diagram show that the HP extraction-flow enthalpy drop from state 2 to state 12 provides the heat to raise the enthalpy in the feedwater from state 11 to state 14. Also, for the LP FWH: 0 = (1)h ll + (m2 + m1 )h9 – (1)h8 – m2h5 – mlhl3 becomes m2 = [ m1( h9 - h13 ) + h11 – h8 ]/( h5 – h9 )

[dl]

This and the T-s diagram show that the discharge from the HP FWH at state 13 aids the mass flow m2 in heating the LP FWH flow from state 8 to state 11. The values of m1 and m2 are evaluated at the bottom of spreadsheet in Table 2.3.

71 With all states and flows known, we may now determine some plant performance parameters. The turbine work referenced to the throttle mass-flow rate is easily obtained by summing the flow contributions through each section of the turbines: wt = h1 – h2 + (1 – ml )( h2 – h3) + (1 – ml )( h4 – h5) + (1 – ml – m2 )( h5 – h6)

[Btu/lbm | kJ /kg]

The net work is then wt + wp, and the heat added in the steam generator is the sum of heat additions in the feedwater pass and the reheat pass: qa = h1 - h14 + (1 - ml )( h4 - h3)

[Btu/lbm | kJ/kg]

These parameters and the work ratio are evaluated in Table 2.3. ____________________________________________________________________ Example 2.10 shows that a good thermal efficiency and net work output are possible with the use of two feedwater heaters despite taking into account realistic turbomachine inefficiencies. The high work ratio clearly demonstrates the lowcompression work requirements of Rankine cycles.

2.7 A Study of a Modern Steam Power Plant Modern steam power plants incorporate both reheat and feedwater heating. A flowsheet for the Public Service Company of Oklahoma (PSO) Riverside Station Unit #1, south of Tulsa, is shown in Figure 2.13. This natural-gas-burning plant was sized for two nominal 500-megawatt units. Several other plants in the PSO system have similar unit flowsheets, including a coal-burning plant. Note the flowsheet coding W, H, F, and A for flow rate in lbm/hr, enthalpy in Btu/lbm, temperature in °F, and pressure in psia, respectively. The steam generator, not shown on the flowsheet, interacts through the feedwater and steam lines on the right-hand side of the diagram. The high pressure turbine throttle is at 1000°F and 3349 psia and has a mass-flow rate of 2,922,139 lbm/hr. This type of unit is called supercritical, because the pressure in the main steam line to the HPturbine throttle exceeds the 3208.2-psia critical pressure of steam. Note that a large fraction of the HP-turbine mass-flow rate enters the cold reheat line at 630 psia and is reheated to the intermediate-pressure (IP) turbine throttle conditions of 1000°F and 567 psia. Most of the steam flow through the IP turbine passes through the crossover at 186 psia to the double-flow low-pressure (DFLP) turbine. The term double-flow refers to the fact that the incoming flow enters at the middle, splits, and flows axially in opposite

72

73 directions through the turbine. This causes the large axial force components on the blades and shaft to oppose each other so that the resultant axial thrust is small and does not necessitate heavy thrust bearings. The combined HP and IP turbines are similarly configured. The plant is equipped with six closed FWHs and one open FWH (the deaerator). Note that the condensate of each of the closed feedwater heaters is throttled to the next lowest pressure FWH or, in the case of the lowest-pressure heater, to the condenser. The extraction steam for the four lowest-pressure FWHs flows from the DFLP turbine. Extraction steam for the highest pressure FWH is provided by the HP turbine, and the IP turbine supplies heater HTR1-6 and the open feedwater heater identified as the deaerator. The deaerator is specially designed to remove non-condensable gases from the system, in addition to performing its feedwater heating duties. The feedwater starts at the "hot well" of the condenser on the left of the diagram, enters the condensate pump at 101.1°F and 2"Hg abs., and starts its passage through the FWHs. Note that the feedwater increases in temperature from 102.1° to 180°, 227.2°, 282.7°, and 314.4° in passing through the 4 lowest pressure FWHs. The feedwater from the deaerator is pumped to 405 psia by the booster pump and subsequently to 3933 psia by the boiler feed pump (BFP). The BFP exit pressure exceeds the HP-turbine throttle pressure of 3349 psia in order to overcome flow losses in the high pressure heater, the boiler feed line, the steam generator main steam pass, and the main steam line, all of which operate at supercritical pressure. The boiler feed pump turbine (BFPT) shown in the upper left of the diagram supplies the shaft power to drive the BFP at the lower right. The BFPT receives steam from an extraction line of the DFLP turbine and exhausts directly to the condenser. The reader should study Figure 2.13 thoroughly in the light of the preceding discussions of reheat and feedwater heating. It is particularly useful to consider the flow rates with respect to mass and energy conservation. Mastery of this flow sheet will make it possible to quickly understand flowsheets of other major power plants.

Example 2.11

Verify that the steam generator feedwater flow rate satisfies the conservation of mass into all the feedwater heaters shown for the Riverside Unit #1 in Figure 2.13. You may neglect all flows of less than 2000 lbm/hr. Solution

The shell side of the low pressure heater, labeled HTR1-1, receives condensate from heaters 2, 3 and 4 as well as steam entering from the LP turbine. The total condensate from the low-pressure heaters into the condenser are:

74 Source Flow rate, lbm/hr _______________________________________________ Condensate from HTR1-4 75,005 Extraction steam into HTR1-3

125,412

Extraction steam into HTR1-2

102,897 ---------303,314

Total condensate into HTR1-1 Extraction steam into HTR1-1 Total condensate leaving HTR1-1

157,111 --------460,425

The feedwater flow rate through the four low-pressure heaters (the condenser condensate pump flow rate) is the sum of the flows into the condenser: 460,425 + 162,701 + 1,812,971 = 2,436,097 lbm/hr. An easier approach to evaluating this flow rate is by imagining a control volume around the entire left side of the diagram that cuts it in two parts between the deaerator and HTR1-4 and through the crossover steam line. Because these are the only points where the control volume is penetrated by large mass flows, the two flows must be equal. Consequently the crossover mass-flow rate of 2,434,357 lbm/hr agrees very well with our above calculation of the feedwater flow rate into the deaerator. Now, observing that the boiler feedwater all flows from the deaerator through the booster pump, we sum all of the flows into the deaerator: Feedwater into deaerator

2,434,357

Steam to deaerator

148,321

Steam to HTR1-6

107,661

Steam to HTR1-7 Total feedwater into HTR1-7

222,876 ----------2,913,215 lbm/hr

This compares well with the tabulated value of 2,922,139 lbm /hr to the steam generator. Accounting for the small flows should improve the agreement. ______________________________________________________________

75 2.8 Deviations from the Ideal - Pressure Losses It is evident from study of Figure 2.13 that there are significant pressure drops in the flows through the steam generator between the HP FWH and the HP-turbine throttle and in the reheat line between the HP and IP turbines. While we have neglected such losses in our calculations, final design analysis requires their consideration. A first attempt at this may be made by applying a fractional pressure drop based on experience. Two per cent pressure drops through the main steam and feedwater lines and a 3.7% loss through the steam generator would, for instance, account for the indicated 14.8% loss from the boiler feed pump to the HP turbine. In the final analysis, of course, when realistic values are available for flow rates and properties, known fluid mechanic relations for pressure drop may be employed to account for these losses. Bibliography and References

1. Anon., Steam, Its Generation and Use. New York: Babcock and Wilcox, 1978. 2. Singer, J. G., (Ed.), Combustion/Fossil Power Systems. Windsor, Conn.: Combustion Engineering, 1981. 3. Wood, Bernard, Applications of Thermodynamics, 2nd ed. Reading, Mass.: AddisonWesley, 1981. 4. Li, Kam W., and Priddy, A. Paul, Powerplant System Design. New York: Wiley, 1985. 5. El-Wakil, M. M., Power Plant Technology. New York: McGraw-Hill, 1984. 6. Skrotzi, B. G. A. and Vopat, W. A., Power Station Engineering and Economy. New York: McGraw-Hill, 1960. EXERCISES 2.1 An ideal Rankine-cycle steam power plant has 800-psia saturated steam at the turbine throttle and 5-psia condenser pressure. What are the turbine work, pump work, net work, steam generator heat addition, thermal efficiency, maximum cycle temperature, and turbine exit quality? What is the Camot efficiency corresponding to the temperature extremes for this cycle? 2.2 A Rankine-cycle steam power plant has an 800-psia/900/F throttle and 5-psia condenser pressure. What are the net work, turbine work, pump work, steam generator

76 heat addition, thermal efficiency, and turbine exit quality? What is the Carnot efficiency corresponding to the temperature extremes for this cycle? 2.3 Solve Exercise 2.2 for the cases of (a) an 85% efficient turbine, (b) an 85% efficient pump, and (c) both together. Tabulate and discuss your results together with those of Exercise 2.2. 2.4 Solve Exercise 2.2 for the case of (a) 1000/F throttle, (b) 2000-psia throttle, (c) 2-psia condenser, and (d) all three changes simultaneously. Make a table comparing net work, quality, and thermal efficiency, including the results of Exercise 2.2. What conclusions can you draw from these calculations? 2.5 Sketch coordinated, labeled flow and T-s diagrams for the ideal Rankine cycle. Tabulate the temperatures, entropies, pressures, enthalpies, and quality or degree of superheat for each significant state shown on the diagram for a throttle at 1000 psia and 1000/F and a condenser at 5 psia. Determine the net work, heat added, thermal efficiency, heat rate, and heat rejected in the condenser. If the power plant output is 100 megawatts and the condenser cooling-water temperature rise is 15 Rankine degrees, what is the steam flow rate and cooling-water flow rate? Neglect pump work. 2.6 Consider a simple Rankine cycle with a 2000-psia/1100/F throttle and 1-psia condenser. Compare the thermal efficiencies and net work for cycles with a perfect turbine and one having 86% turbine isentropic efficiency. Assume isentropic pumping. 2.7 A boiling-water reactor operates with saturated vapor at 7500 kPa at the throttle of the high-pressure turbine. What is the lowest turbine exit pressure that ensures that the turbine exit moisture does not exceed 12% if the turbine is isentropic? What would the lowest pressure be if the turbine isentropic efficiency were 85%? 2.8 Consider a steam plant with a single reheat and a single open feedwater heater that takes extraction from the cold reheat line. Sketch carefully coordinated and labeled T-s and flow diagrams. If the throttle is at 1000/F and 3000psia, the condenser is at 1 psia, and reheat is to 1000/F at 400 psia, what is the extraction mass fraction, the heat rate, and the thermal efficiency? The turbine efficiency is 89%. Neglect pump work. 2.9 A Rankine-cycle power plant condenses steam at 2 psia and has 1000/F and 500 psia at the turbine throttle. Assume an isentropic turbine. (a) Tabulate the temperature, pressure, entropy, and enthalpy of all states. Determine the quality and moisture fraction for all mixed states. (b) Calculate the heat transferred in the condenser and the steam generator and the turbine work, all per unit mass. What is the thermal efficiency? (c) Calculate the pump work. What is the ratio of turbine to pump work?

77 (d) What is the turbine work and thermal efficiency if the turbine efficiency is 85%? Include pump work. 2.10 For throttle conditions of 1000/F and 1000 psia and a condenser pressure of 2 psia, compare the net work, thermal efficiency, and turbine discharge quality or degree of superheat for a simple cycle and two reheat cycles with reheat to 1000/F at 50 and 200 psia. Tabulate your results. Sketch a single large, labeled T-s diagram comparing the cycles. Turbine isentropic efficiencies are 85%. 2.11 Consider a regenerative Rankine cycle with a 1000/F and 500-psia throttle, 2-psia condenser, and an open feedwater heater operating between two turbines at 50 psia. Turbine efficiencies are 85%. Neglect pump work. (a) Draw labeled, coordinated T-s and flow diagrams. (b) Determine the fraction of the throttle mass flow that passes through the extraction line. (c) Calculate the turbine work per unit mass at the throttle. (d) Calculate the cycle efficiency, and compare it to the simple-cycle efficiency. 2.12 Consider a 1120°F, 2000-psia, 10-psia steam cycle with reheat at 200 psia to 1000/F and a closed feedwater heater taking extraction from a line between two turbines at 100psia. The FWH condensate is throttled to the condenser, and the feedwater in the FWH is raised to the condensing temperature of the extraction steam. (a) Draw labeled T-s and flow diagrams for this plant. (b) Tabulate the enthalpies for each significant state point. (c) What is the extraction fraction to the FWH? (d) What are the net work and work ratio? (e) What are the thermal efficiency and the heat rate? 2.13 A turbine operates with a 860/F, 900-psia throttle. Calorimetric measurements indicate that the discharge enthalpy is 1250 Btu/lbm at 100 psia. What is the isentropic efficiency? 2.14 An ideal Rankine cycle has 1000-psia saturated steam at the turbine throttle. The condenser pressure is 10psia. What are the turbine work, steam generator heat addition, maximum cycle temperature, turbine exit quality, and Carnot efficiency corresponding to the temperature extremes of the cycle? Neglect pump work. 2.15 Assume that the extraction mass-flow rate to FWH #7 in Figure 2.13 is not known. Calculate the FWH extraction mass fraction (relative to the HP-turbine throttle flow) and the extraction mass-flow rate. Compare the extraction-steam energy loss rate with the feedwater energy gain rate.

78 2.16 Compare the inflow and outflow of steam of the DFLP turbine in Figure 2.13, and calculate the percentage difference. Calculate the power output of the DFLP turbine in Btu/hr and in kW. 2.17 Calculate the power delivered by the PSO Riverside Unit #1 boiler feed pump turbine, BFPT. Based on the feedwater enthalpy rise across the BFP, determine its power requirements, in kilowatts. What fraction of the plant gross output is used by the BFPT? 2.18 Without performing a detailed analysis of the FWHs, determine the PSO Riverside Unit #1 feedwater flow rate from heater number 4 to the deaerator. Explain your methodology. 2.19 Total and compare the inflows and outflows of mass and energy to the PSO Riverside Unit #1 deaerator. 2.20 Rework Example 2.4 neglecting pump work. Repeat your calculations for an 80% efficient pump. Compare and comment on the significance of accounting for pump work and turbomachine efficiency. 2.21 For a 1080/F, 2000-psia, 5-psia Rankine cycle with 85% turbine efficiency and 60% pump efficiency: (a) Compare the actual net work and the isentropic turbine work and the isentropic net work. (b) Calculate the actual heat transfer and work for each component, and evaluate the cyclic integrals of Q and W. (c) Compare the real cycle efficiency with that for the ideal Rankine cycle. 2.22 For a 1080/F, 2000-psia, 5-psia Rankine cycle with 85% turbine efficiency and 60% pump efficiency, evaluate the effect of a single reheat to 1080/F at 500 psia on: (a) Heat addition in the steam generator. (b) Work of each turbine, total turbine work, and net work. Compare the net work with the cyclic integral of the external transfers of heat. (c) Cycle efficiency and heat rate. (d) Quality or degree of superheat at the exit of the turbines. Draw labeled flow and T-s diagrams. 2.23 Consider a 1080/F, 2000-psia, 5-psia Rankine cycle with 85% turbine efficiency and 60% pump efficiency. Compare the simple cycle with the same cycle operating with a single reheat to 1080/F at 1000 psia with respect to: (a) Heat addition in the steam generator. (b) Work of each turbine, total turbine work and net work, condenser heat rejection, and cyclic integral of heat added.

79 (c) Cycle efficiency. (d) Quality or degree of superheat at the exit of the turbines. Draw labeled flow and T-s diagrams. 2.24 For a 1080/F, 2000-psia, 5-psia Rankine cycle with 85% turbine efficiencies and 60% pump efficiencies and using a single open feedwater heater operating at 500 psia: (a) Draw labeled and coordinated flow and T-s diagrams. (b) Evaluate the feedwater heater mass fraction. (c) Evaluate heat addition in the steam generator, work of each turbine, total turbine work, and net work, all per pound of steam at the HP-turbine throttle. (d) Evaluate condenser heat transfer per unit mass at the HP-turbine throttle. (e) Evaluate cycle efficiency and heat rate. Compare with simple-cycle efficiency. (f) Evaluate the cyclic integral of the differential heat addition, and compare it with the net work. 2.25 Consider a 1080/F, 2000-psia, 5-psia Rankine reheat-regenerative cycle with perfect turbomachinery and a closed feedwater heater taking extraction from the cold reheat line at 500 psia. FWH condensate is pumped into the feedwater line downstream of the feedwater heater. Assume that the enthalpy of the feedwater entering the steam generator is that of the saturated liquid leaving the FWH. (a) Draw coordinated and labeled flow and T-s diagrams. (b) Determine the extraction mass fraction, the net work, and the total heat addition. (c) Determine the thermal efficiency and heat rate. (d) Determine the superheat or quality at the turbine exhausts: 2.26 Taking the reheat-pressure level as a variable, plot net work, thermal efficiency, and turbine exhaust superheat and/or moisture against reheat pressure for the conditions of Example 2.5. Select a suitable design value based on your analysis. 2.27 Solve Example 2.6 for 1200/F throttle substituting a closed FWH for the open heater. Consider two cases in which the FWH condensate is (a) throttled to the condenser, and (b) pumped to throttle pressure. 2.28 Solve Example 2.6 using the method for assigning extraction-pressure levels given in the subsection of Section 2.6 on multistage extraction systems. 2.29 Solve Example 2.7 using the method for assigning extraction-pressure levels given in the section on multistage extraction systems, and determine by trial and error the reheat-pressure level that maximizes the thermal efficiency. 2.30 Solve Example 2.7 with the extraction condensate from the closed FWH pumped ahead to the feedwater-pressure level.

80 2.31 Solve Example 2.6 for 900/F throttle temperature with the open FWH replaced by a closed FWH where the feedwater is (a) throttled to the condenser, and (b) pumped into the feedwater line downstream of the FWH. 2.32 Compare the work and exhaust quality of 90% efficient turbines with 2500-psia throttle pressure and 1000/F and 1200/F throttle temperatures exiting to a 2-psia condenser. 2.33 Draw a large T-s diagram showing the states associated with the important flows of the PSO Riverside Unit #1 (Figure 2.13). 2.34 A Rankine-cycle steam power plant has 5-MPa saturated steam at the turbine throttle and 25-kPa condenser pressure. What are the net work, steam generator heat addition, thermal efficiency, heat rate, maximum cycle temperature, and turbine exit quality? What is the Carnot efficiency corresponding to the temperature extremes for this cycle? 2.35 A Rankine-cycle steam power plant has a 5-MPa/450/C throttle and 10-kPa condenser pressure. What are the net work, steam generator heat addition, thermal efficiency, heat rate, and turbine exit quality? What is the Carnot efficiency corresponding to the temperature extremes for this cycle? 2.36 Solve Exercise 2.35 for the cases of (a) an 85% efficient turbine, (b) an 85% efficient pump, and (c) both together. What conclusions may be inferred from your results? 2.37 Solve Exercise 2.35 for the case of (a) a 550/C throttle, (b) a 15-MPa throttle, (c) a 5-kPa condenser, and (d) all three changes simultaneously. What conclusions can you draw from these calculations? 2.38 Sketch coordinated, labeled flow and T-s diagrams for the following Rankine cycle. Tabulate the temperatures, entropies, pressures, enthalpies, and quality or degree of superheat for each significant state shown on the diagram for a throttle at 10 MPa and 550/C and condenser at 5 kPa. Determine the net work, heat added, thermal efficiency, and heat rejected in the condenser. If the power plant output is 100 megawatts and the condenser cooling water temperature rise is 15/Rankine, what is the steam flow rate and cooling-water flow rate? Neglect pump work. 2.39 Consider a Rankine cycle with a 20MPa/600/C throttle and 3-kPa condenser. Compare the thermal efficiencies and net work for cycles with a perfect turbine and one having 86% turbine isentropic efficiency.

81 2.40 Consider a steam plant, with a single reheat and a single open feedwater heater, that takes extraction from the cold reheat line. Sketch carefully coordinated and labeled T-s and flow diagrams. If the throttle is at 550/C and 15 MPa, the condenser is at 5 kPa, and reheat is to 3 MPa and 550° C, what are the extraction mass fraction, work ratio, and thermal efficiency? The pump and turbine efficiencies are 82% and 89%, respectively. 2.41 A Rankine-cycle power plant condenses steam at 10 kPa and has 550/C and 5 MPa at the turbine throttle. Assume an isentropic turbine. (a) Tabulate the temperature, pressure, entropy, and enthalpy of all states. Determine the quality and moisture fraction for all mixed states. (b) Calculate the heat transferred in the condenser and steam generator and the turbine work, all per unit mass. What is the thermal efficiency? (c) Calculate the pump work. What is the ratio of turbine to pump work? (d) What is the turbine work and thermal efficiency if the turbine efficiency is 85%? Include pump work. 2.42 For throttle conditions of 550°C and 5 MPa and a condenser pressure of 10 kPa, compare the net work, thermal efficiency, and turbine discharge quality or degree of superheat for a simple cycle and two reheat cycles. Consider reheat to 500/C at (a) 4MPa and (b) 1 MPa. Tabulate and compare your results. Sketch a large, labeled T-s diagram for a reheat cycle. Turbine efficiencies are 85%. 2.43 Consider a regenerative Rankine cycle with a 600/C and 4-MPa throttle, a 5-kPa condenser, and an open feedwater heater at 500 kPa. Turbine efficiencies are 85%. Neglect pump work. (a) Draw labeled, coordinated T-s and flow diagrams. (b) Determine the fraction of the throttle mass flow that passes through the extraction line. (c) Calculate the turbine work per unit mass at the throttle. (d) Calculate the cycle efficiency, and compare it with the simple-cycle efficiency. (e) Calculate the heat rate. 2.44 Consider a 600/C, 15-MPa steam cycle with reheat at 2 MPa to 600/C and extraction to a closed feedwater heater at 600 kPa. The FWH condensate is throttled to the condenser at 5 kPa, and the feedwater in the FWH is raised to the condensing temperature of the extraction steam. Neglecting pump work: (a) Draw labeled T-s and flow diagrams for this plant. (b) Tabulate the enthalpies for each significant state point. (c) What is the extraction fraction to the FWH? (d) What is the net work? (e) What is the thermal efficiency?

82 (f) What is the heat rate? 2.45 A turbine operates with a 600°C, 7-MPa throttle. Calorimetric measurements indicate that the discharge enthalpy is 3050 kJ/kg at 0.8 MPa. What is the turbine isentropic efficiency? 2.46 A pressurized water-reactor nuclear power plant steam generator has separate turbine and reactor water loops. The steam generator receives high-pressure hot water from the reactor vessel to heat the turbine feedwater. Steam is generated from the feedwater in the turbine loop. The water pressure in the reactor is 15 MPa, and the water temperature in and out of the reactor is 289°C and 325°C, respectively. The plant has one turbine with a single extraction to a closed FWH with condensate throttled to the condenser. Throttle conditions are 300/C and 8 MPa. The extraction and condenser pressures are 100 kPa and 5 kPa, respectively. The reactor-coolant flow rate is 14,000 kg/s. Assume no heat losses in heat exchangers and isentropic turbomachines. Neglect pump work. (a) What is the rate of heat transfer from the reactor in MWt? (b) Draw coordinated flow and T-s diagrams that show both loops. (c) Determine the extraction mass fraction of the throttle flow rate. (d) Determine the cycle net work, heat rate, and thermal efficiency. (e) Calculate the steam flow rate. (f) Assuming the electrical generator has 97% efficiency, calculate the power output, in MWe (electric). 2.47 Perform an optimization of the extraction pressure of a Rankine cycle with a 2000-psia saturated-vapor throttle, a 1-psia condenser with a single closed feedwater heater, as in Example 2.9. Compare the optimum extraction temperature given by Equation (2.14) with your results. 2.48 Prepare an optimization study of thermal efficiency with a table and plot of net work and thermal efficiency as a function of reheat pressure level for Example 2.5. Discuss the selection of reheat pressure for this case. How does the reheat pressure used in Example 2.5 compare with your results? 2.49 Solve Exercise 2.25 for reheat and extraction at 200 psia. Compare the extraction mass fraction, net work, thermal efficiency, heat rate, and turbine exit conditions with those of Exercise 2.25. 2.50 Rework Exercise 2.25, accounting for 90% turbine efficiencies and a 10/F terminal temperature difference. 2.51 A 1000/F/2000-psia-throttle high-pressure turbine discharges into a cold reheat line at 200 psia. Reheat is to 1000/F. The low-pressure turbine discharges into the

83 condenser at 0.5 inches of mercury absolute. Both turbines are 90% efficient. Design the cycle for the use of three feedwater heaters. Draw coordinated T-s and flow diagrams. State and discuss your decisions on the handling of the feedwater heater design. 2.52 A steam turbine receives steam at 1050/F and 3000 psia and condenses at 5 psia. Two feedwater heaters are supplied by extraction from the turbine at pressures of 1000 psia and 200 psia. The low-pressure heater is an open FWH, and the other is closed with its condensate throttled to the open heater. Assuming isentropic flow in the turbine and negligible pump work: (a) Sketch accurately labeled and coordinated T-s and flow diagrams for the system, and create a table of temperature, pressure, and enthalpy values for each state. (b) What are the extraction flows to each feedwater heater if the throttle mass flow rate is 250,000 pounds per hour? (c) How much power, in kW, is produced by the turbine? (d) Compare the thermal efficiency of the system with the efficiency if valves of both extraction lines are closed. (e) What is the heat rate of the system with both feedwater heaters operative? 2.53 Apply the steady-flow First Law of Thermodynamics to a single control volume enclosing the two turbines in Example 2.7. Show that the same equation is obtained for the turbine work as when the work of individual turbines is summed. 2.54 Apply the steady-flow First Law of Thermodynamics to a single control volume enclosing the two turbines in Example 2.10. Show that the same equation is obtained for the turbine work as when the work of individual turbines is summed. 2.55 Resolve Example 2.7 with 4% pressure drops in the main steam pass and reheat pass through the steam generator. Make a table comparing your results with those of the example to show the influence of the losses on plant performance. Calculate and display the percentage differences for each parameter. Assume turbine throttle conditions are unchanged. 2.56 Draw labeled and coordinated T-s and flow diagrams for a steam power plant with 1000°F / 3000-psia / 2" Hg absolute conditions, assuming isentropic turbomachinery. The plant has reheat at 500 psia to ll00/F. The plant has the following feedwater heaters: 1. A closed FWH with extraction at 1000 psia and pumped condensate. 2. A closed FWH at 400 psia with condensate throttled into the next-lowestpressure FWH. 3. An open FWH at 20 psia. Define mass fraction variables. Show mass-flow variable expressions, with arrows indicating mass fractions along the various process paths on the T-s and flow diagrams.

84 Write equations for conservation of energy for the FWHs that allow you to solve easily for the mass fractions in terms of known state enthalpies and other mass fractions. Indicate a solution method for the mass fractions that involves simple substitution only. 2.57 A pressurized-water nuclear-reactor steam generator has separate turbine and reactor loops. The steam generator linking the two loops cools high-pressure hot water from the reactor vessel and transfers the heat to the turbine feedwater producing steam. The water pressure in the reactor is 2250 psia, and the water temperatures in and out of the reactor are 559°F and 623°F, respectively. The plant has one turbine with a single extraction to an open FWH. Throttle conditions are 555/F and 1100 psia. The extraction and condenser pressures are 100 psia and 1 psia, respectively. The reactorcoolant flow rate is 147,000,000 lbm/hr. Assume no heat losses in heat exchangers and isentropic turbomachines. (a) What is the rate of heat transfer from the reactor, in Btu/hr and in MWt? (b) Draw coordinated flow and T-s diagrams that show both loops with states in their proper relations with respect to each other. (c) Determine the extraction mass fraction relative to the throttle flow rate. (d) Determine the cycle net work. (e) What are the cycle thermal efficiency and heat rate? (f) Calculate the turbine-steam flow rate. (g) Assuming the electrical generator has 100% efficiency, calculate the turbine power, in Btu/hr and in MWe. 2.58 Determine the efficiencies of the boiler feed pump and boiler feed pump turbine of the PSO Riverside Station Unit 4/1 (Figure 2.13). 2.59 A Rankine cycle with a single open feedwater heater has a 1040/F and 550-psia throttle. Extraction from the exit of the first turbine (assumed isentropic) is at 40/F of superheat. The second turbine has an efficiency of 85% and expands into the condenser at 5 psia. (a) Draw matched, labeled T-s and flow diagrams. (b) Accurately calculate and tabulate the enthalpies of all significant states. Neglect pump work. (c) What is the feedwater-heater mass fraction relative to the mass flow at the first throttle? (d) What is the quality or degree of superheat at the condenser inlet? (e) What are the net work, thermal efficiency, and heat rate? (f) Estimate the feedwater-heater condensate pump work and its percentage of turbine work.

85

CHAPTER 3 FUELS AND COMBUSTION

3.1 Introduction to Combustion Combustion Basics The last chapter set forth the basics of the Rankine cycle and the principles of operation of steam cycles of modern steam power plants. An important aspect of power generation involves the supply of heat to the working fluid, which in the case of steam power usually means turning liquid water into superheated steam. This heat comes from an energy source. With the exception of nuclear and solar power and a few other exotic sources, most power plants are driven by a chemical reaction called combustion, which usually involves sources that are compounds of hydrogen and carbon. Process industries, businesses, homes, and transportation systems have vast heat requirements that are also satisfied by combustion reactions. The subject matter of this chapter therefore has wide applicability to a variety of heating processes. Combustion is the conversion of a substance called a fuel into chemical compounds known as products of combustion by combination with an oxidizer. The combustion process is an exothermic chemical reaction, i.e., a reaction that releases energy as it occurs. Thus combustion may be represented symbolically by: Fuel + Oxidizer Y Products of combustion + Energy Here the fuel and the oxidizer are reactants, i.e., the substances present before the reaction takes place. This relation indicates that the reactants produce combustion products and energy. Either the chemical energy released is transferred to the surroundings as it is produced, or it remains in the combustion products in the form of elevated internal energy (temperature), or some combination thereof. Fuels are evaluated, in part, based on the amount of energy or heat that they release per unit mass or per mole during combustion of the fuel. Such a quantity is known as the fuel's heat of reaction or heating value. Heats of reaction may be measured in a calorimeter, a device in which chemical energy release is determined by transferring the released heat to a surrounding fluid. The amount of heat transferred to the fluid in returning the products of combustion to their initial temperature yields the heat of reaction.

86 In combustion processes the oxidizer is usually air but could be pure oxygen, an oxygen mixture, or a substance involving some other oxidizing element such as fluorine. Here we will limit our attention to combustion of a fuel with air or pure oxygen. Chemical fuels exist in gaseous, liquid, or solid form. Natural gas, gasoline, and coal, perhaps the most widely used examples of these three forms, are each a complex mixture of reacting and inert compounds. We will consider each more closely later in the chapter. First let's review some important fundamentals of mixtures of gases, such as those involved in combustion reactions. Mass and Mole Fractions The amount of a substance present in a sample may be indicated by its mass or by the number of moles of the substance. A mole is defined as the mass of a substance equal to its molecular mass or molecular weight. A few molecular weights commonly used in combustion analysis are tabulated below. For most combustion calculations, it is sufficiently accurate to use integer molecular weights. The error incurred may easily be evaluated for a given reaction and should usually not be of concern. Thus a gram-mole of water is 18 grams, a kg-mole of nitrogen is 28 kg, and a pound-mole of sulfur is 32 lbm. _____________________________________________________________________ Molecule Molecular Weight ------------------------------------------C 12 28 N2 O2 32 S 32 H2 2 _____________________________________________________________________ The composition of a mixture may be given as a list of the fractions of each of the substances present. Thus we define the mass fraction, of a component i, mfi, as the ratio of the mass of the component, mi, to the mass of the mixture, m: mfi = mi /m It is evident that the sum of the mass fractions of all the components must be 1. Thus mf1 + mf2 + ... = 1 Analogous to the mass fraction, we define the mole fraction of component i, xi, as the ratio of the number of moles of i, ni, to the total number of moles in the mixture, n: xi = ni /n

87 The total number of moles, n, is the sum of the number of moles of all the components of the mixture: n = n1 + n2 + ... It follows that the sum of all the mole fractions of the mixture must also equal 1. x1 + x2 + ... = 1 The mass of component i in a mixture is the product of the number of moles of i and its molecular weight, Mi. The mass of the mixture is therefore the sum, m = n1M1 + n2M2 + ..., over all components of the mixture. Substituting xin for ni, the total mass becomes m = (x1M1 + x2M2 + ...)n But the average molecular weight of the mixture is the ratio of the total mass to the total number of moles. Thus the average molecular weight is M = m /n = x1M1 + x2M2 + ... EXAMPLE 3.1

Express the mass fraction of component 1 of a mixture in terms of: (a) the number of moles of the three components of the mixture, n1, n2, and n3, and (b) the mole fractions of the three components. (c) If the mole fractions of carbon dioxide and nitrogen in a three component gas containing water vapor are 0.07 and 0.38, respectively, what are the mass fractions of the three components? Solution

(a) Because the mass of i can be written as mi = niMi , the mass fraction of component i can be written as: mfi = niMi /(n1M1 + n2M2 + ..)

[dl]

For the first of the three components, i = 1, this becomes: mf1 = n1M1/(n1M1 + n2M2 + n3M3) Similarly, for i = 2 and i = 3: mf2 = n2M2/(n1M1 + n2M2 + n3M3) mf3 = n3M3/(n1M1 + n2M2 + n3M3)

88 (b) Substituting n1 = x1 n, n2 = x2 n, etc. in the earlier equations and simplifying, we obtain for the mass fractions: mf1 = x1M1/(x1M1 + x2M2 + x3M3) mf2 = x2M2/(x1M1 + x2M2 + x3M3) mf3 = x3M3 /(x1M1 + x2M2 + x3M3) (c) Identifying the subscripts 1, 2, and 3 with carbon dioxide, nitrogen, and water vapor, respectively, we have x1 = 0.07, x2 = 0.38 and x3 = 1 – 0.07 – 0.038 = 0.55. Then: mf1 = (0.07)(44)/[(0.07)(44) + (0.38)(28) + (0.55)(18)] = (0.07)(44)/(23.62) = 0.1304 mf2 = (0.38)(28)/(23.62) = 0.4505 mf3 = (0.55)(18)/(23.62) = 0.4191 As a check we sum the mass fractions: 0.1304 + 0.4505 + 0.4191 = 1.0000. ________________________________________________________________ For a mixture of gases at a given temperature and pressure, the ideal gas law shows that pVi = niúT holds for any component, and pV = núT for the mixture as a whole. Forming the ratio of the two equations we observe that the mole fractions have the same values as the volume fraction: xi = Vi /V = ni /n

[dl]

Similarly, for a given volume of a mixture of gases at a given temperature, piV = niúT for each component and pV = núT for the mixture. The ratio of the two equations shows that the partial pressure of any component i is the product of the mole fraction of i and the pressure of the mixture: pi = pni /n = pxi EXAMPLE 3.2

What is the partial pressure of water vapor in Example 3.1 if the mixture pressure is two atmospheres?

89 Solution

The mole fraction of water vapor in the mixture of Example 3.1 is 0.55. The partial pressure of the water vapor is therefore (0.55)(2) = 1.1 atm. _____________________________________________________________________ Characterizing Air for Combustion Calculations Air is a mixture of about 21% oxygen, 78% nitrogen, and 1% other constituents by volume. For combustion calculations it is usually satisfactory to represent air as a 21% oxygen, 79% nitrogen mixture, by volume. Thus for every 21 moles of oxygen that react when air oxidizes a fuel, there are also 79 moles of nitrogen involved. Therefore, 79/21 = 3.76 moles of nitrogen are present for every mole of oxygen in the air. At room temperature both oxygen and nitrogen exist as diatomic molecules, O2 and N2, respectively. It is usually assumed that the nitrogen in the air is nonreacting at combustion temperatures; that is, there are as many moles of pure nitrogen in the products as there were in the reactants. At very high temperatures small amounts of nitrogen react with oxygen to form oxides of nitrogen, usually termed NOx. These small quantities are important in pollution analysis because of the major role of even small traces of NOx in the formation of smog. However, since these NOx levels are insignificant in energy analysis applications, nitrogen is treated as inert here. The molecular weight of a compound or mixture is the mass of 1 mole of the substance. The average molecular weight, M, of a mixture, as seen earlier, is the linear combination of the products of the mole fractions of the components and their respective molecular weights. Thus the molecular weight for air, Mair, is given by the sum of the products of the molecular weights of oxygen and nitrogen and their respective mole fractions in air. Expressed in words: Mair = Mass of air/Mole of air = (Moles of N2 /Mole of air)(Mass of N2 /Mole of N2) + (Moles of O2/Mole of air)(Mass of O2 /Mole of O2)

or

Mair = 0.79 Mnitrogen + 0.21 Moxygen = 0.79(28) + 0.21(32) = 28.84 The mass fractions of oxygen and nitrogen in air are then mfoxygen = (0.21)(32)/28.84 = 0.233, or 23.3% and mfnitrogen = (0.79)(28)/28.84 = 0.767, or 76.7%

90 3.2 Combustion Chemistry of a Simple Fuel Methane, CH4, is a common fuel that is a major constituent of most natural gases. Consider the complete combustion of methane in pure oxygen. The chemical reaction equation for the complete combustion of methane in oxygen may be written as: CH4 + 2O2 Y CO2 + 2H2O

(3.1)

Because atoms are neither created nor destroyed, Equation (3.1) states that methane (consisting of one atom of carbon and four atoms of hydrogen) reacts with four atoms of oxygen to yield carbon dioxide and water products with the same number of atoms of each element as in the reactants. This is the basic principle involved in balancing all chemical reaction equations. Carbon dioxide is the product formed by complete combustion of carbon through the reaction C + O2 Y CO2. Carbon dioxide has only one carbon atom per molecule. Since in Equation (3.1) there is only one carbon atom on the left side of the equation, there can be only one carbon atom and therefore one CO2 molecule on the right. Similarly, water is the product of the complete combustion of hydrogen. It has two atoms of hydrogen per molecule. Because there are four hydrogen atoms in the reactants of Equation (3.1), there must be four in the products, implying that two molecules of water formed. These observations require four atoms of oxygen on the right, which implies the presence of two molecules (four atoms) of oxygen on the left. The coefficients in chemical equations such as Equation (3.1) may be interpreted as the number of moles of the substance required for the reaction to occur as written. Thus another way of interpreting Equation (3.1) is that one mole of methane reacts with two moles of oxygen to form one mole of carbon dioxide and two moles of water. While not evident in this case, it is not necessary that there be the same number of moles of products as reactants. It will be seen in numerous other cases that a different number of moles of products is produced from a given number of moles of reactants. Thus although the numbers of atoms of each element must be conserved during a reaction, the total number of moles need not. Because the number of atoms of each element cannot change, it follows that the mass of each element and the total mass must be conserved during the reaction. Thus, using the atomic weights (masses) of each element, the sums of the masses of the reactants and products in Equation (3.1) are both 80: CH4 + 2O2 Y CO2 + 2H2O [12 + 4(1)] + 4(16) Y [12 + 2(16)] + 2[2(1) + 16] = 80 Other observations may be made with respect to Equation (3.1). There are 2 moles of water in the 3 moles of combustion products, and therefore a mole fraction of water in the combustion products of xwater = 2/3 = 0.667. Similarly, xCarbon dioxide = 1/3 = 0.333 moles of CO2 in the products. There are 44 mass units of CO2 in the 80 mass units of products for a mass

91 fraction of CO2 in the products, mfcarbon dioxide = 44/80 = 0.55 Likewise, the mass fraction of water in the products is 2(18)/80 = 0.45. We also observe that there are 12 mass units of carbon in the products and therefore a carbon mass fraction of 12/80 = 0.15. Note that because the mass of any element and the total mass are conserved in a chemical reaction, the mass fraction of any element is also conserved in the reaction. Thus the mass fraction of carbon in the reactants is 0.15, as in the products. Combustion in Air Let us now consider the complete combustion of methane in air. The same combustion products are expected as with combustion in oxygen; the only additional reactant present is nitrogen, and it is considered inert. Moreover, because we know that in air every mole of oxygen is accompanied by 3.76 moles of nitrogen, the reaction equation can be written as CH4 + 2O2 + 2(3.76)N2 Y CO2 + 2H2O + 2(3.76)N2

(3.2)

It is seen that the reaction equation for combustion in air may be obtained from the combustion equation for the reaction in oxygen by adding the appropriate number of moles of nitrogen to both sides of the equation. Note that both Equations (3.1) and (3.2) describe reactions of one mole of methane fuel. Because the same amount of fuel is present in both cases, both reactions release the same amount of energy. We can therefore compare combustion reactions in air and in oxygen. It will be seen that the presence of nitrogen acts to dilute the reaction, both chemically and thermally. With air as oxidizer, there are 2 moles of water vapor per 10.52 moles of combustion products, compared with 2 moles of water per 3 moles of products for combustion in oxygen. Similarly, with air, there is a mass fraction of CO2 of 0.1514 and a carbon mass fraction of 0.0413 in the combustion products, compared with 0.55 and 0.15, respectively, for combustion in oxygen. The diluting energetic effect of nitrogen when combustion is in air may be reasoned as follows: The same amount of energy is released in both reactions, because the same amount of fuel is completely consumed. However, the nonreacting nitrogen molecules in the air have heat capacity. This added heat capacity of the additional nitrogen molecules absorbs much of the energy released, resulting in a lower internal energy per unit mass of products and hence a lower temperature of the products. Thus the energy released by the reaction is shared by a greater mass of combustion products when the combustion is in air. Often, products of combustion are released to the atmosphere through a chimney, stack, or flue. These are therefore sometimes referred to as flue gases. The flue gas composition may be stated in terms of wet flue gas (wfg) or dry flue gas (dfg), because

92 under some circumstances the water vapor in the gas condenses and then escapes as a liquid rather than remaining as a gaseous component of the flue gas. When liquid water is present in combustion products, the combustion product gaseous mass fractions may be taken with respect to the mass of flue gas products, with the product water present or omitted. Thus, for Equation (3.2), the mass of dry combustion products is 254.56. Hence the mass fraction of carbon dioxide is 44/254.56 = 0.1728 with respect to dry flue gas, and 44/290.56 = 0.1514 with respect to wet flue gas. In combustion discussions reference is frequently made to higher and lower heating values. The term higher heating value, HHV, refers to a heating value measurement in which the product water vapor is allowed to condense. As a consequence, the heat of vaporization of the water is released and becomes part of the heating value. The lower heating value, LHV, corresponds to a heating value in which the water remains a vapor and does not yield its heat of vaporization. Thus the energy difference between the two values is due to the heat of vaporization of water, and HHV = LHV + (mwater /mfuel)hfg

[Btu/lbm | kJ/kg]

where mwater is the mass of liquid water in the combustion products, and hfg is the latent heat of vaporization of water. Air-Fuel Ratio It is important to know how much oxygen or air must be supplied for complete combustion of a given quantity of fuel. This information is required in sizing fans and ducts that supply oxidizer to combustion chambers or burners and for numerous other design purposes. The mass air-fuel ratio, A/F, or oxygen-fuel ratio, O/F, for complete combustion may be determined by calculating the masses of oxidizer and fuel from the appropriate reaction equation. Let’s return to Equation (3.2): CH4 + 2O2 + 2(3.76)N2 Y CO2 + 2H2O + 2(3.76)N2

(3.2)

The A/F for methane is [(2)(32) + (2)(3.76)(28)]/(12 + 4) = 17.16 and the O/F is 2(32)/(12 + 4) = 4. Thus 4 kg of O2 or 17.16 kg of air must be supplied for each kilogram of methane completely consumed. Of course it is possible, within limits, to supply an arbitrary amount of air to a burner to burn the fuel. The terms stoichiometric or theoretical are applied to the situation just described, in which just enough oxidizer is supplied to completely convert the fuel to CO2 and H2O. Thus the stoichiometric O/F and A/F ratios for methane are 4.0 and 17.16, respectively. If less than the theoretical amount of air is supplied, the products will contain unburned fuel. Regardless of the magnitude of A/F, when unburned fuel remains in the products (including carbon, carbon monoxide, or hydrogen), combustion is said to be incomplete. Because air is virtually free and fuel is expensive, it is usually important to burn all of the fuel by using more air than the theoretical air-fuel ratio indicates is needed. Thus most burners operate with excess air.

93 The actual air-fuel ratio used in a combustor is frequently stated as a percentage of the theoretical air-fuel ratio % theoretical air = 100(A/F)actual /(A/F)theor

(3.3)

Thus, for methane, 120% of theoretical air implies an actual mass air-fuel ratio of (120/100)(17.16) = 20.59. Excess air is defined as the difference between the actual and the theoretical air supplied. Accordingly, the percentage of excess air is % excess air = 100[(A/F)actual – (A/F)theor ]/(A/F)theor

(3.4)

Thus, for methane, 120% of theoretical air implies % excess air = (100)(20.59 – 17.16)/17.16 = 20%. Note also that combining Equations (3.4) and (3.3) yields the following general result: % excess air = % theoretical air – 100%

(3.5)

Again, the excess air percentage is 120% – 100% = 20%. Table 3.1 shows examples of ranges of excess air used with certain fuels and combustion systems. The air/fuel parameters just discussed emphasize the amount of air supplied to burn a given amount of fuel relative to the theoretical requirement. An alternate approach considers a given amount of air and indicates the mass of fuel supplied , the fuel-air ratio, F/A, which is the inverse of the air-fuel ratio. A measure of how much fuel is actually supplied, called the equivalence ratio, is the ratio of the actual fuel-air ratio to the theoretical fuel-air ratio:

M = (F/A)actual / (F/A)theor = (A/F)theor / (A/F)actual = 100/( % theoretical air) Thus 100% theoretical air corresponds to an equivalence ratio of 1, and 20% excess air to M = 100/120 = 0.833. When the equivalence ratio is less than 1, the mixture is called lean; when greater than 1, it is called rich. This section has dealt with the application of combustion chemistry or stoichiometry applied to methane gas. Other fuels for which a reaction equation such as Equation (3.1) or (3.2) is available may be treated in a similar way. Before considering more complex combustion problems, it is appropriate to investigate the nature and description of the various types of fossil fuels.

94

3.3 Fossil Fuel Characteristics Most chemical fuels are found in nature in the form of crude oil, natural gas, and coal. These fuels are called fossil fuels because they are believed to have been formed by the decay of vegetable and animal matter over many thousands of years under conditions of high pressure and temperature and with a deficiency or absence of oxygen. Other fuels such as gasoline, syngas (synthetic gas), and coke may be derived from fossil fuels by some form of industrial or chemical processing. These derived fuels are also called fossil fuels. Coal Coal is an abundant solid fuel found in many locations around the world in a variety of forms. The American Society for Testing Materials, ASTM, has established a ranking system (ref. 3) that classifies coals as anthracite (I), bituminous (II), subbituminous (III), and lignite (IV), according to their physical characteristics. Table 3.2 lists

95

seventeen of the many United States coals according to this class ranking. Coal is formed over long periods of time, in a progression shown from left to right in Figure 3.1. The bars on the ordinate show the division of the combustibles between fixed carbon and volatile matter in the fuels. “Fixed carbon” and “volatile matter” indicate roughly how much of the fuel burns as a solid and as a thermally generated gas, respectively. It is seen that the volatile matter and oxygen contained in the fuels decrease with increasing age. Peat is a moist fuel, at the geologically young end of the scale, that has a relatively low heating value. It is not considered a coal but, nevertheless, follows the patterns of characteristics shown in the figure. Peat is regarded as an early stage or precursor of coal. At the other extreme, anthracite is a geologically old, very hard, shiny coal with high carbon content and high heating value. Bituminous is much more abundant than anthracite, has a slightly lower carbon content, but also has a high heating value. Subbituminous coal, lignite, and peat have successively poorer heating values and higher volatile matter than bituminous. Coal is a highly inhomogeneous material, of widely varying composition, found in seams (layers) of varying thickness at varying depths below the earth's surface. The wide geographic distribution of coal in the United States is shown in Figure 3.2.

96

According to reference 1, the average seam in the United States is about 5.5 ft. thick. The largest known seam is 425 ft. thick and is found in Manchuria. Coal Analyses It is often difficult to obtain representative samples of coal because of composition variations from location to location even within a given seam. As a result there are limits on the accuracy and adequacy of coal analyses in assessing coal behavior in a given application. Before discussing the nature of these analyses, it is important to establish the basis on which they are conducted. Coal contains varying amounts of loosely held moisture and noncombustible materials or mineral matter (ash), which are of little or no use. The basis of an analysis helps to specify the conditions under which the coal is tested. The coal sample may be freshly taken from the mine, the as-mined basis. It may have resided in a coal pile for months, and be analyzed just before burning, the as-fired basis. It may be examined immediately after transport from the mine, the as-received basis. Exposure to rain or dry periods, weathering, and separation and loss of noncombustible mineral matter through abrasion and the shifting of loads during transport and storage may cause the same load of coal to have changing mineral matter and moisture content over time. It is therefore important to specify the basis for any test that is conducted. Published tabulations of coal properties are frequently presented on a dry, ash-free, or dry and ash-free basis, that is, in the absence of water and/or noncombustible mineral matter. Coal ranking and analysis of combustion processes rely on two types of analysis of coal composition: the proximate analysis and the ultimate analysis. The proximate analysis starts with a representative sample of coal. The sample is first weighed, then raised to a temperature high enough to drive off water, and then reweighed. The weight

97

loss divided by the initial weight gives the coal moisture content, M. The remaining material is then heated at a much higher temperature, in the absence of oxygen, for a time long enough to drive off gases. The resulting weight-loss fraction gives the volatile matter content, VM, of the coal. The remainder of the sample is then burned in air until only noncombustibles remain. The weight loss gives the fixed carbon, FC, and the remaining material is identified as non-combustible mineral matter or ash, A. The proximate analysis may be reported as percentages (or fractions) of the four quantities moisture, ash, volatile matter, and fixed carbon, as in Table 3.2, or without ash and moisture and with the FC and VM normalized to 100%. Sulfur, as a fraction of the coal mass, is sometimes reported with the proximate analysis. The proximate analysis, while providing very limited information, can be performed with limited laboratory resources. A more sophisticated and useful analysis is the ultimate analysis, a chemical analysis that provides the elemental mass fractions of carbon, hydrogen, nitrogen, oxygen, and sulfur, usually on a dry, ash-free basis. The ash content of the coal and heating value are sometimes provided also. Data from a dry, ash-free analysis can be converted to another basis by using the basis adjustment factor, 1 - A - M, as follows. The mass of coal is the mass of ultimate or proximate analysis components plus the masses of water (moisture) and ash: m = mcomp + mash + mmoist

[lbm | kg]

98 Dividing through by the total mass m and rearranging, we get the following as the ratio of the mass of components to the total mass: mcomp / m = 1 – A – M

[dl]

where A is the ash fraction and M is the moisture fraction of the total coal mass. A component of a coal analysis may be converted from the dry, ash-free basis to some other basis by forming the product of the component fraction and the basis adjustment factor. Thus an equation for the wet and ashy volatile matter fraction in the proximate analysis may be determined from the dry, ash-free proximate analysis by using VMas-fired = (Mass of combustibles/Total mass)VMdry,ashfree = ( 1 - A - M ) VMdry,ash-free

(3.6)

where A and M are, respectively, the ash and moisture fractions for the as-fired coal. Here the as-fired (wet, ashy) mass fraction of volatile matter is the product of the dry, ash-free mass fraction and the basis adjustment factor. Fixed carbon, heating values, and components of the ultimate analysis may be dealt with in a similar way. Table 3.3 gives proximate and ultimate analyses for a number of United States coals on a dry basis. Another extensive tabulation of the characteristics of American and world coals is given in Appendix E. EXAMPLE 3.3

If the as-fired moisture fraction for Schuylkill, Pa. anthracite culm characterized in Table 3.3 is 4.5%, determine the as-fired proximate and ultimate analysis and heating value of the coal. (Culm is the fine coal refuse remaining from anthracite mining.) Solution

The FC, VM, and ash contents are given in Table 3.3. Because ash is already present in the analysis, the appropriate adjustment factor is 1 - A - M = 1 – 0.0 – 0.045 = 0.955. Using Equation (3.6) and the data from Table 3.3, we get VMas-fired = (0.955)(8.3) = 7.927 FCas-fired = (0.955)(32.6) = 31.133 Aas-fired = (0.955)(59.1) = 56.411 Mas-fired = 4.500 Check Sum = 99.971 Heating valueas-fired = (0.955)(4918) = 4697 Btu/lbm Similarly, the as-fired ultimate analysis is 32% C, 1.15% H2, 4.87% O2, 0.57% N2, 0.48% S, 56.44% ash, and 4.5% moisture, with a checksum of 100.01. _____________________________________________________________________

99

100

101

102

As a solid fuel, coal may be burned in a number of ways. Starting with the smallest of installations, coal may be burned in a furnace, in chunk form on a stationary or moving grate. Air is usually supplied from below with combustion gases passing upward and ash falling through a stationary grate or dropping off the end of a moving grate into an ash pit. A wide variety of solid fuels can be burned in this way. Though all furnaces were onced fired manually, today many are fired by or with the assistance of mechanical devices called stokers. Figure 3.3 shows a spreader stoker, which scatters coal in a uniform pattern in the furnace, the finer particles burning in suspension in the rising combustion air stream while the heavier particles drop to the grate as they burn. The particles that reach the grate burn rapidly in a thin layer, and the remaining ash drops off the end into the ash pit. This type of combustion system has been in use for over fifty years for hot water heating and steam generation. In large installations, coal is crushed to a particular size, and sometimes pulverized to powder immediately before firing, to provide greater surface exposure to the

103

oxidizer and to ensure rapid removal of combustion gases. Because of the wide variation in the characteristics of coals, specialized types of combustion systems tailored to a specific coal or range of coal characteristics are used. Natural Gas Natural gas is a mixture of hydrocarbons and nitrogen, with other gases appearing in small quantities. Table 3.4 shows the composition of samples of natural gases found in several regions of the United States. For these samples, it is seen that the gases contain 83-94% methane (CH4), 0-16% ethane (C2H6), 0.5-8.4% nitrogen and small quantities of other components, by volume. The ultimate analysis shows that the gases contain about 65-75% carbon, 20-24% hydrogen, 0.75-13% nitrogen, and small amounts of oxygen and sulfur in some cases. The higher heating values are in the neighborhood of 1000 Btu/ft3 on a volume basis and 22,000 Btu/lbm on a mass basis. In regions where it is abundant, natural gas is frequently the fuel of choice because of its low sulfur and ash content and ease of use.

104 EXAMPLE 3.4

Determine the molecular weight and stoichiometric mole and mass air-fuel ratios for the Oklahoma gas mole composition given in Table 3.4. Solution

Equation (3.2), CH4 + 2O2 + 2(3.76)N2 Y CO2 + 2H2O + 2(3.76)N2

(3.2)

shows that there are 2 + 2(3.76) = 9.52 moles of air required for complete combustion of each mole of methane. Similarly for ethane, the stoichiometric reaction equation is: C2H6 + 3.5O2 + (3.5)(3.76)N2 Y 2CO2 + 3H2O + 13.16N2 where 2 carbon and 6 hydrogen atoms in ethane require 2 CO2 molecules and 3 H2O molecules, respectively, in the products. There are then 7 oxygen atoms in the products, which implies 3.5 oxygen molecules in the reactants. This in turn dictates the presence of (3.5)(3.76) = 13.16 nitrogen molecules in both the reactants and products. The reaction equation then indicates that 3.5(1 + 3.76) = 16.66 moles of air are required for complete combustion of one mole of ethane. In Table 3.5, the molecular weight of the gas mixture, 18.169, is found in the fourth column by summing the products of the mole fractions of the fuel components and the component molecular weights. This is analogous to the earlier determination of the average air molecular weight from the nitrogen and oxygen mixture mole fractions. The products of the mole fractions of fuel components and the moles of air required per mole of fuel component (as determined earlier and tabulated in the fifth column of Table 3.5) then yield the moles of air required for each combustible per mole of fuel (in the sixth column). Summing these, the number of moles of air required per mole of fuel yields the stoichiometric mole air-fuel ratio, 9.114. The stoichiometric mass A/F is then given by the mole A/F times the ratio of air molecular weight to fuel molecular weight: (9.114)(28.9)/18.169 = 14.5. Table 3.5 Calculations for Example 3.4 i

Mi

xi

xiMi

Moles air per mole i

Moles air per mole fuel

Methane

16

0.841

13.456

9.52

(0.841)(9.52) = 7.998

Ethane

30

0.067

2.010

16.66

(0.067)(16.66) = 1.116

CO2

44

0.008

0.351

0.0

Nitrogen

28

0.084

2.352

0.0

1.000

18.169

Totals

moles air /mole fuel = 9.114

105

Liquid Fuels Liquid fuels are primarily derived from crude oil through cracking and fractional distillation. Cracking is a process by which long-chain hydrocarbons are broken up into smaller molecules. Fractional distillation separates high-boiling-point hydrocarbons from those with lower boiling points. Liquid fuels satisfy a wide range of combustion requirements and are particularly attractive for transportation applications because of their compactness and fluidity. Table 3.6 gives representative analyses of some of these liquid fuels. Compositions of liquid and solid fuels, unlike gaseous fuels, are usually stated as mass fractions. 3.4 Combustion Reactions and Analysis Mechanism of Combustion Details of the mechanics of combustion depend to a great extent on the fuel and the nature of the combustion system. They are sometimes not well understood and are largely beyond the scope of this book. There are, however, certain fundamentals that are useful in dealing with combustion systems. The chemical reaction equations presented here do not portray the actual mechanism of combustion; they merely indicate the initial and final chemical compositions of a reaction. In most cases the reactions involve a sequence of steps, leading from the reactants to the products, the nature of which depends on the temperature, pressure, and other conditions of combustion. Fuel molecules, for instance, may undergo thermal cracking, producing more numerous and smaller fuel molecules and perhaps breaking the molecules down completely into carbon and hydrogen atoms before oxidation is completed. In the case of solid fuels, combustion may be governed by the rate at which oxidizer diffuses from the surrounding gases to the surface and by the release of combustible gases near the surface. Combustion of solids may be enhanced by increasing the fuel surface area exposed to the oxidizer by reducing fuel particle size.

106 The following simple model illustrates the effect.

Example 3.5 is, of course, an idealized example. In reality, the reacting surface area of solid fuels is usually much larger than the spherical surface area implied by their size. We have seen that, for combustion to occur, molecules of oxidizer must affiliate with fuel molecules, an action enhanced by the three T’s of combustion: turbulence, time, and temperature. Chemical reactions take place more rapidly at high temperatures but nevertheless require finite time for completion. It is therefore important that burners be long enough to retain the fuel-air mixture for a sufficiently long time so that combustion is completed before the mixture leaves. Turbulence, or mixing, enhances the opportunities for contact of oxidizer and fuel molecules and removal of products of combustion. A flame propagates at a given speed through a flammable mixture. It will propagate upstream in a flow of a combustible mixture if its flame speed exceeds the flow velocity. If a fixed flame front is to exist at a fixed location in a duct flow in which the velocity of the combustion gas stream exceeds the propagation speed, some form of flame stabilization is required. Otherwise the flame front is swept downstream and flameout occurs. Stabilization may be achieved by using fixed flameholders (partial

107 flow obstructions that create local regions of separated flow in their bases where the flame speed is greater than the local flow velocity) or by directing a portion of the flow upstream to provide a low-speed region where stable combustion may occur. Each combination of oxidizer and fuel has been seen to have a particular stoichiometric oxidizer-fuel ratio for which the fuel is completely burned with a minimum of oxidizer. It has also been pointed out that it is usually desirable to operate burners at greater than the theoretical air-fuel ratio to assure complete combustion of the fuel and that this is sometimes referred to as a lean mixture. Occasionally it may be desirable to have incomplete combustion, perhaps to produce a stream of products in which carbon monoxide exists or to assure that all the oxidizer in the mixture is consumed. In that case a burner is operated at less than the stoichiometric air-fuel ratio with what is called a rich mixture. There are limits to the range of air-fuel ratios for which combustion will occur called limits of flammability. Here the density of the mixture is important. The limits of flammability around the stoichiometric A/F are reduced at low densities. If combustion is to occur reliably in mixtures at low densities, it is necessary to closely control the air-fuel ratio. Combustion Analysis of Solid Fuels In the determination of the air-fuel ratio and flue gas composition for the combustion of solid fuels, it is important to account for the ash and moisture in the fuel in the as-fired condition. In the following analyses, all of the elements of the reactants in the fuel and oxidizer are assumed to be present in the flue gas products except for the ash, which is assumed to fall as a solid or flow as molten slag to the furnace bottom. Nitrogen and oxygen are present in many solid fuels and should be accounted for in predicting the flue gas composition. While both carbon monoxide and oxygen may be present in combustion products at the same time because of imperfect mixing of combustibles and oxygen in some instances, we will assume for prediction of the flue gas composition that perfect mixing occurs such that no carbon monoxide is present when excess air is supplied. EXAMPLE 3.6

A coal with a dry, ash-free composition of 0.87 C, 0.09 H2, 0.02 S, and 0.02 O2 is burned with 25% excess air. The as-fired ash and moisture contents are 6% and 4%, respectively. (a) What are the stoichiometric and actual air-fuel ratios? (b) What is the flue gas composition? Solution

(a) Before performing combustion calculations, it is necessary to convert coal composition data to an as-fired basis. The ratio of as-fired to dry, ash-free

108

109

mfj = (kg j / kg fg) = (kg j / kg coal) / (kg fg / kg coal)

110

111

112

113

114

115

(Note: The reference conditions of the 1985 JANAF Thermochemical tables used differ slightly from those of preceding editions.) Heats of formation are usually determined based on statistical thermodynamics and spectroscopic measurements. By definition, heats of formation are zero for all elements in the standard state. Hence, from the steady-flow First Law of Thermodynamics, the heat transferred in a formation reaction of a compound created from elements in the standard state is the heat of formation for the compound, as in the hydrogen-to-water example just mentioned. Heat Transfer in a Chemically Reacting Flow Consider now the combustion problem in which fuel and oxidizer flow into a control volume and combustion products flow out. The steady-flow First Law of Thermodynamics applied to the control volume may be written as Q = Hp – Hr + Ws

[Btu | kJ]

(3.7)

where Q is heat flow into the control volume, Ws is the shaft work delivered by the control volume, and the enthalpies, H, include chemical as well as thermal energy. The subscripts r and p refer to the reactants entering and products leaving the control volume, respectively. The enthalpy Hp is the sum of the enthalpies of all product streams leaving the control volume. A similar statement applies to Hr for the entering reactant streams.

116 The individual enthalpies may each be written as the product of the number of moles of the component in the reaction equation and its respective enthalpy per mole of the component. For example, for k products: Hp = n1h1 + n2h2 +...+ nkhk

[Btu | kJ]

(3.8)

where the n’s are the stoichiometric coefficients of the chemical equation for the combustion reaction, and the enthalpies are on a per-mole bases. EXAMPLE 3.9

Write an equation for the enthalpy of the products of complete combustion of octane in oxygen. Solution The balanced equation for the complete combustion of octane is

C8H18 + 12.5O2 Y 8CO2 + 9H2O The mole coefficients, 8 and 9, of the products are stoichiometric coefficients that yield Hp = 8h(CO2) + 9h(H2O)

[Btu | kJ]

per mole of octane consumed. ___________________________________________________________________ The enthalpy of any component of the reactants or products may be written as the sum of (1) its enthalpy of formation at the standard temperature, To, and standard pressure, and (2) its enthalpy difference between the actual state and the standard state of the components. Thus, for each component: h(T) = hf (To) + [h(T) – h(To)]

[Btu /mole | kJ /mole]

(3.9)

where it is assumed that the sensible gas enthalpy difference is independent of pressure. Sensible enthalpies (those that depend on temperature but do not involve reactions or phase change) relative to the standard reference state are given in Appendix D. Thus, returning to the formation reaction for the combustion of hydrogen in oxygen at the standard state to produce water, as discussed in the preceding section, we see that the steady-flow First Law of Thermodynamics becomes Q = (1)hf, H2O – (1)hf, H2 – (0.5)hf, O2 = –103,996 – 0 – 0 = – 103,996 Btu/lb-mole with water in the vapor phase as the product of combustion of one mole of H2. Here the sensible enthalpy differences are zero, because both the products and the reactants are at the standard state. Note that because the stoichiometric coefficients of both

117 hydrogen and water are the same in this reaction, the resulting heat transfer may be interpreted as either per mole of water formed or per mole of hydrogen consumed. If, instead, liquid water is the combustion product, the heat transfer is given by Q = – 122,885 – 0 – 0 = – 122,885 Btu/lb-mole of H2O. The difference between the two cases, 18,919 Btu/lb-mole H2O, is equivalent to 18,935/18 = 1,051.9 Btu/lbm of water, the enthalpy of vaporization of water. This result compares with the enthalpy or latent heat of vaporization of water at 77°F, 1050.1 Btu/lbm, given in the steam tables. With either liquid or gaseous water as the product, the heat transfer term for the control volume is negative, indicating, in accordance with the usual sign convention, that the heat flows from the control volume to the surroundings. The two calculations above illustrate the fact that the heat transfer in a formation reaction at the standard state is the heat of formation of the compound created. EXAMPLE 3.10

What is the enthalpy of water vapor at 1800°R and 1 bar? What is the heat transfer in the formation reaction of water from hydrogen and oxygen if the products are at 1800°R and the reactants are (a) at the standard state, and (b) at 900°R? Solution

The heat of formation of water vapor at the standard state of 298.15K. (536.7°R) and one bar is – 103,966 Btu/lb-mole. The enthalpy of water vapor at 1800°R is the sum of the heat of formation at the standard state and the sensible enthalpy difference of water vapor between 536.7°R and 1800°R. Thus: Hp = – 103,966 + 11,185 = – 92,781 Btu/lb-mole of water. (a) In this case, the reactants, oxygen and hydrogen, have zero enthalpies because they are in the standard state and, as elements, their heats of formation are zero. Thus the heat transferred is – 92,781 Btu/lb-mole, or 5154.5 Btu/lbm of water. (b) For reactants at 900°R , Appendix D gives hH2(900) – hH2(536.7) = 2530 Btu/mole of H2 and hO2(900) – hO2(536.7) = 2617 Btu/mole of O2. The enthalpy of the reactants is then Hr = (1.0)(2530) + (0.5)(2617) = 3838.5 Btu/lb-mole H2O. The heat transferred is then: Q = Hp – Hr = – 92,781 – 3,838.5 = – 96,619.5 Btu/lb-mole of water, or Q = – 96,619.5 / 18 = – 5,367.8 Btu/lbm of water.

118 Thus more heat must be transferred from the control volume to form water vapor at 1800°R if the reactants are at 900°R than if they are in the 536.7°R standard state. Combustion Flame Temperature In many combustion problems, the reactants enter the combustor near room temperature and products emerge at an elevated temperature. The temperature of the products flowing from the control volume depends on the energy released in the combustion reaction and heat gain or loss through the boundary of the control volume. The resulting combustion product temperature is sometimes called the flame temperature. EXAMPLE 3.11

Methane and stoichiometric air enter a combustor at the standard state. Using a spreadsheet, calculate the heat transfer as a function of the exit (flame) temperature of the products of complete combustion. Assume the water in the products remains a vapor. Solution

The reaction equation for complete combustion of methane in air is: CH4 + 2O2 + (2)3.76N2 Y CO2 + 2H2O + 7.52N2 The enthalpy of the products at temperature T and of the reactants at the standard state is Hp = (1)hf, CO2 + (1)[hCO2 (T) – hCO2 (537)] + (2)hf, H2O + (2)[hH2O (T) – hH2O (537)] + (7.52)[hN2 (T) – hN2 (537)] Hr = hf, CH4 = – 32,189.6 Btu/lb-mole of methane, where the heats of formation of elemental nitrogen and oxygen are zero and the heat of formation of water is for the vapor phase. Writing the enthalpy differences as )h’s and applying the steady-flow First Law of Thermodynamics, we get Q = Hp – Hr = – 169,184 + )hCO2(T) + (2)(– 103,966) + (2) )hH2O(T) + 7.52 )hN2(T) – (– 32,189.6) = – 344,926.4 + )hCO2(T) + (2))hH2O(T) + 7.52 )hN2 Btu/lb-mole of methane. This function is tabulated in the spreadsheet in Table 3.10 and plotted in Figure 3.4 using values of enthalpies at the temperature T from the the JANAF tables. Negative values of Q indicate that heat must be rejected from the control volume to maintain product effluent temperature below about 4200°R. Beyond 4200°R, the CO2,

119

N2, and H2O outflow carries more energy than is released in the control volume by chemical reaction; hence, heat must flow into the control volume to achieve the resulting high exit temperatures. Thus the final flame temperature clearly depends on the chemical composition of the flow and on the consequent control volume heat transfer. ____________________________________________________________________

120 Heat of Combustion and Heating Value The heat of combustion, or enthalpy of combustion, of a fuel is defined as the energy transferred during a steady-flow process in which the fuel is completely burned and where the products are returned to the temperature and pressure of the reactants. It will be seen that the enthalpy of combustion evaluated at the standard state may be determined from the heats of formation. The heat of combustion of hydrogen has, in fact, been determined in a preceding section that examined the formation reaction for water. The negative of the enthalpy of combustion of a fuel burned in air is usually referred to as the heating value of the fuel. When water in the combustion products is condensed, the heat of vaporization of the water adds to the chemical energy released, and the resulting heating value is called the higher heating value, HHV. Recall also that the heating value obtained when the product water stays a vapor is called the lower heating value, LHV. The difference between HHV and LHV has been illustrated in the previous section for the formation reaction of water resulting from the combustion of hydrogen. For methane, note also that the heat of combustion, and thus the magnitude of the lower heating value, appears in the value of Q in the top row of table 3.10, since there the combustion products are at the reference temperature. EXAMPLE 3.12

Illinois no. 6 raw coal has the following dry mass composition: 61.6% C, 4.2% H2, 9.7% O2, 1.3% N2, 4.6% S, and 18.5% ash. Using heats of formation, determine the higher and lower heating values, in kJ / kg, of the as-fired coal with 10% moisture, and compare them with the heating value in Table 3.3. Solution

To adjust the composition for 10% moisture, the factor 1 – A – M becomes 1 – 0 – 0.1 = 0.9. The resulting moist coal composition is given in the following table. It was seen earlier that the heat of reaction of hydrogen in its standard state, and thus its heat of combustion, is the heat of formation of its product of combustion. The reaction equation H2 + 0.5O2 Y H2O shows that one mole of hydrogen produces one mole of water. Thus, from Table 3.9, the heat of formation of steam, – 241,826 kJ per kg-mole of water formed or per kgmole of hydrogen burned, is also the heat of combustion of hydrogen in the standard state. Thus the hydrogen contributes 241,826/2 kJ per kg of hydrogen in the coal. The total energy released by the hydrogen in the coal is then 241,826/2 times the mass fraction of hydrogen in the coal, as shown in the following table. Similar arguments may be made for hydrogen with product water in the liquid phase and the carbon and sulfur components of the coal.

121 Element i

Dry mfi

Wet mfi

Heat of Combustion

kJ/kg coal

(241,826)(0.0378)/2 =

4,570.5 (v)

(285,830)(0.0378)/2

5402.2 (l)

(393,522)(0.5544)/12 =

18,180.7

(296,842)(0.0414)/32 =

384.0

H2: For LHV

0.042

0.0378

For HHV C

0.616

0.5544

O2

0.097

0.0873

N2

0.013

0.0117

S

0.046

0.0414

Ash

0.185

0.1665

H2O

_____

0.1000

0.999

0.9991

23,135.2 (v) 23,966.9 (l)

Thus the lower and higher heating values of the coal are 23,135.2 and 23,966.9 kJ/kg, respectively. Table 3.3 lists a heating value of dry Illinois no. 6 raw coal of 11,345 Btu/lbm. The corresponding heating value for the wet coal is (0.9)(11,345) = 10,210.5 Btu/lbm. This corresponds to (10,210.5)/(0.43) = 23,745 kJ/kg. ____________________________________________________________________ Adiabatic Flame Temperature The results of Example 3.11, tabulated in Figure 3.10, show that for a given air-fuel mixture there is a unique product temperature for which the control volume is adiabatic. This temperature is known as the adiabatic flame temperature. It can be determined as in Example 3.11, or it may be calculated from the steady-flow First Law of Thermodynamics by setting Q = 0. The resulting First Law equation for the adiabatic flame temperature, designated T*, becomes: Hp (T*) = Hr (T r)

[Btu | kJ]

(3.10)

where the reactants are at the temperature Tr . The enthalpy terms depend on the individual enthalpies of the components as functions of temperature. Thus a trial-anderror solution is required using data on heat of formation from Table 3.9 and the enthalpy tables in Appendix D. Given a known Tr , the adiabatic flame temperature may also be obtained as the intercept (Q = 0) on a graph of Q versus temperature, T, such as Figure 3.4. Adiabatic Flame Temperature for Solid Fuels

122 As a final example, we will determine the combustion products, heat of combustion, and adiabatic flame temperature for a solid fuel specified by its ultimate analysis. The solution is presented in a spreadsheet in which enthalpies are tabulated as a function of temperature for the relevant chemical components as given in the JANAF tables.

123

124 The number of moles of excess O2 in the flue gas per pound of coal may also be obtained from the excess air-fuel ratio: [(A/F)actual – (A/F)theor ](0.233)/32 = (8.288 – 6.375)(0.233)/32 = 0.0139 The number of moles of flue gas nitrogen is also given by [(A/F)actual ](1 – 0.233)/28 + d = 8.288(0.767)/28 + 0.000357 = 0.227 With the actual balanced reaction equation known, the mole coefficients may then be used to write an equation for the enthalpy of the products per unit mass of coal. Because all the reactants are assumed to be elements at the reference temperature, the enthalpy of the reactants is zero. By setting Q = Hp = 0, we can solve this First Law equation for the adiabatic flame temperature by trial and error. However, with a spreadsheet, it is convenient to calculate the heat transfer at each temperature-enthalpy data point and determine the adiabatic flame temperature by inspection and, if additional precision is required, by explicit interpolation. The heat transfer equation is shown on the spreadsheet, and the values of the flue gas mole coefficients and heats of formation are shown above the appropriate JANAF enthalpies to which they relate. It is seen that the adiabatic flame temperature for combustion with 30% excess air is about 2110K, or 3800°R. Note also that the heat of combustion of the coal, 8506.7 Btu/lbm of coal, may be read from the spreadsheet at the JANAF table reference temperature. This is possible because the heat of reaction is independent of the amount of excess air employed in the reaction. _____________________________________________________________________ The spreadsheet for Example 3.13 is easily modified to compute and plot the composition of the flue gas as a function of the percentage of excess air. Figure 3.5 shows that the mole fractions of excess oxygen and of nitrogen increase while the fractions of other products decrease. Excess oxygen measured from a flue gas sample is commonly used as a measure of excess air in adjusting the air-fuel ratio of combustion systems. 3.6 Molecular Vibration, Dissociation, and Ionization The temperature of a gas is a measure of the random translational kinetic energy of molecules. The simple kinetic theory of the heat capacity of a gas predicts heat capacities that are independent of temperature and determined by the number of degrees of freedom of the molecules. The kinetic theory is usually regarded as applicable at low pressures and moderate and high temperatures, conditions at which collisions between molecules are rare.

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For monatomic gases, kinetic theory predicts an internal energy per atom of u = 3kT/2 and an enthalpy of h = u + pv = 3kT/2 + kT = 5kT/2 Here k, the Boltzmann constant, is the ideal-gas constant per molecule, which can be calculated from the universal gas constant, ú, and Avogadro’s number of molecules per mole. Thus k = ú/No = 8.32/(6.025 × 1023) = 1.38 × 10 -23 J/K From their definitions, Equations (1.14) and (1.15), this leads to cv = 3k/2 and cp = 5k/2 per atom and to a heat capacity ratio of k = cp /cv = 5/3 = 1.667. The concept of equipartition of energy assumes that the energy of a particle is equally divided among its various degrees of freedom. Each mode of energy storage of a molecule is assigned an energy kT/2. The theory then represents the internal energy as nkT/2, the enthalpy as (n + 2)kT/2, and the heat capacity ratio as k = (n + 2)/n, where n is the number of modes of energy storage, or degrees of freedom, of the molecule. For atoms with three translational and no rotational degrees of freedom, n = 3 (an atom is presumed to be a point and to have no rotational kinetic energy

126 because it has zero radius); and these relations reduce to the findings described in the preceding paragraph. The simple kinetic theory suggests that a diatomic molecule has five degrees of freedom: three degrees of translational freedom and two modes having significant rotational kinetic energy. For this case we obtain u = 5kT/2, h = 7kT/2, and k = 7/5 = 1.4, in agreement with experiment for oxygen and nitrogen at moderate temperatures and densities. At higher temperatures, diatomic molecules start to vibrate, adding additional degrees of freedom that reduce the heat capacity ratio k below 1.4. As the temperature increases and the collisions between molecules become more vigorous, molecules not only vibrate but they start to be torn apart, each forming separate atoms, the process known as dissociation. The energy required to break the bonds between atoms in molecules is called the dissociation energy. At low temperatures, few molecules have sufficient kinetic energy to provide the energy needed to cause dissociation by collision. At higher temperatures, when more molecules have energies exceeding the dissociation energy, the chemical equilibrium shifts to a composition in which there are more atoms and fewer molecules in the gas. This trend continues as temperature increases. At still higher temperatures, when particle kinetic energies exceed the ionization energy of the gas, outer electrons are separated from atoms, forming positively charged ions, in a process known as the ionization. Particles lose kinetic energy when causing dissociation and ionization. This lost energy is then not reflected in the temperature. Thus, when vibration, dissociation, and ionization occur, the internal energy and enthalpy increase more rapidly than the temperature. Then the simple linear relations just given for u and T are no longer correct. Stated another way, the temperature of a gas rises less rapidly with heat addition when it is at temperature levels where significant vibration, dissociation, or ionization take place. The phenomenon may be thought of as analogous to phase change, in which enthalpy increases with heat addition while temperature does not. As a result of these phenomena, high flame temperatures determined with ideal gas enthalpies may be overestimated. At temperatures exceeding 2000K (3600°R), flame temperature calculations based on the JANAF gas enthalpies may start to become inaccurate. At these temperature levels, dissociation (and at still higher temperatures, ionization) starts to influence the composition of the gases and hence their thermodynamic properties. The progression of dissociation and ionization with temperature is shown for air at sea level density in Figure 3.6. It is seen that, at this density, little dissociation of nitrogen occurs below about 6000 K. but that oxygen starts to dissociate significantly above 3000 K.. At lower densities, the onset of dissociation occurs at progressively lower temperatures. In Figure 3.6, the number of particles per initial atom of air may be obtained by multiplying the ordinates by 1.993. Usually, dissociation does not seriously influence combustion calculations when the oxidizer is air, but the high combustion temperatures resulting from use of pure oxygen may be significantly influenced by dissociation. The reader is referred to advanced thermodynamics, physical chemistry, and advanced engineering texts for methods of predicting the effects of dissociation and ionization.

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Bibliography and References 1. Anon, Steam, Its Generation and Use, 39th ed. New York,: Babcock and Wilcox, 1978. 2. Singer, Joseph G. (Ed.), Combustion / Fossil Power Systems. Windsor, Conn.: Combustion Engineering Inc., 1981. 3. Anon., Classification of Coals by Rank, Standard D-388. Philadelphia: American Society for Testing Materials, Section 5, Volume 05.05, 1983. 4. Chase, M. W. Jr., et.al., JANAF Thermochemical Tables, 3rd ed., J. Phys. Chem. Ref. Data 14, Supplement No. 1, 1985. 5. Van Wylen, Gordon J., and Sonntag, Richard E., Fundamentals of Classical Thermodynamics. New York: Wiley, 1986.

128 6. El-Wakil, M. M., Powerplant Technology. New York: McGraw-Hill, 1984. 7. Campbell, Ashley S., Thermodynamic Analysis of Combustion Engines. New York: Wiley, 1979. 8. Culp, Archie W., Principles of Energy Conversion. New York: McGraw-Hill, 1979. 9. Wood, Bernard D., Applications of Thermodynamics. Reading, Mass.: AddisonWesley, 1982. 10. Lefebvre, Arthur H., Gas Turbine Combustion, New York: McGraw-Hill, 1983. 11. Baumeister, Theodore, and Marks, Lionel S. (Eds.), Standard Handbook for Mechanical Engineers, 7th ed. New York: McGraw-Hill, 1967. 12. Sorenson, Harry A., Energy Conversion Systems. New York: Wiley, 1983. 13. Hill, Phillip G., Power Generation. Cambridge, Mass.: MIT Press, 1977. 14. Anon., Coal Quality Information Book, CS-5421, Interim Report. Palo Alto: Electric Power Research Institute, December 1987. 15. Moeckel, W. E., and Weston, Kenneth C., “Composition and Thermodynamic Properties of Air in Chemical Equilibrium,” NACA TN 4265, April 1958. EXERCISES 3.1 Determine the mass fractions of a mixture of six grams of carbon, three grams of sulfur, and one gram of sodium chloride. 3.2 Determine the mole fractions of a gas consisting of a mole of oxygen, eight moles of nitrogen, a mole of CO, and two moles of CO2. Determine also the mass fractions. What is the average molecular weight of the gas? 3.3 Write the balanced reaction equation for the complete combustion of sulfur in oxygen. What are the mass and mole fractions of oxygen in the reactants? 3.4 Write the balanced reaction equation for the complete combustion of carbon in oxygen. What are the mass and mole fractions of oxygen in the reactants? 3.5 Write the balanced reaction equation for the complete combustion of carbon in air. What are the mass and mole fractions of fuel, air, oxygen, and nitrogen in the reactants?

129 3.6 Write the balanced reaction equation for the complete combustion of ethane, C2H6, in air. What are the mass and mole fractions of fuel, air, oxygen, and nitrogen in the reactants? What are the mass and mole fractions of carbon dioxide and water vapor in the combustion products? 3.7 Write the balanced reaction equation for the complete combustion of propane, C3H8,, in air. What are the mass and mole fractions of fuel, air, oxygen, and nitrogen in the reactants? What are the mass and mole fractions of carbon dioxide and water vapor in the combustion products. What are the mass and mole air-fuel ratios? 3.8 Write the balanced reaction equation for the complete combustion of C8H18 in air. What are the mass and mole fractions of fuel, air, oxygen, and nitrogen in the reactants? What are the mass and mole fractions of carbon dioxide and water vapor in the combustion products? What are the mass and mole air-fuel ratios? 3.9 Gasoline, sometimes represented as C8H18, is burned in 25% excess air mass. What are the mass and mole stoichiometric and actual air-fuel ratios? Determine the mass and mole fractions of the combustion products. 3.10 Determine the lower and higher heating values of methane using the JANAF table of heats the formation. 3.11 Determine the as-fired stoichiometric and actual air-fuel ratios for Greene, Pennsylvania raw coal (Table 3.3) with 5% moisture and the mass and mole flue gas compositions for combustion with 20% excess air. 3.12 Compare the stoichiometric and actual air-fuel ratios and the mole flue gas composition for combustion with 20% excess air for the following raw and clean (process #1) coals (Table 3.3): (a) Freestone, Texas, big brown lignite; (b) Indiana, Pennsylvania, Freeport (upper); (c) British Columbia, Hat Creek (A zone); (d) Perry, Illinois no. 6; (e) Muhlenberg, Kentucky no. 9; (f) Nicholas, West Virginia, Kittanning; (g) Belmont, Ohio, Pittsburgh; (h) Big Horn, Montana, Robinson; (i) Greene, Pennsylvania, Sewickley; (j) Kanawha, West Virginia, Stockton-Lewiston; (k) Belmont, Ohio, Waynesburg. 3.13 If Union, Kentucky no. 11 raw coal has 10% moisture, as mined, determine the asmined proximate and ultimate analyses for this coal. 3.14 If Big Horn, Montana, Robinson raw coal has 15% moisture, as-mined, what are its as-mined proximate and ultimate analyses? 3.15 Determine the ultimate analyses of the raw coals listed in Exercise 3.12 (a-e) assuming 10% as-mined moisture.

130 3.16 Write the balanced chemical equation for the combustion of methane in stoichiometric air. Use the table of heats of formation to determine the heats of reaction of methane (in Btu/lbm and Btu/lb-mole), with products and reactants all at the standard state and product water as liquid. What are the values if the product water is vapor? Would the heat of reaction be different if the combustion were in pure oxygen? Compare your results with tabulated heating values for methane. 3.17 Solve Exercise 3.16 in SI units: kJ/gm-mole and kJ/kg. 3.18 Calculate the heating values, in Btu/lbm and Btu/lb-mole, for the complete combustion of hydrogen, with product water in liquid and in vapor phases. Compare with tabulated heating values. Repeat the calculations in SI units. 3.19 Determine the heat transferred when ethane is burned (a) in stoichiometric air, and (b) in 100% excess air. In both cases the reactants are in the standard state and products at 1000K. Use a heat of formation of –36,420 Btu/lb-mole. 3.20 What is the adiabatic flame temperature for the combustion of ethane in air, ignoring dissociation? Use a heat of formation of –36,420 Btu/lb-mole of ethane. 3.21 Compare the adiabatic flame temperatures for the stoichiometric combustion of hydrogen in air and in pure oxygen, ignoring dissociation. 3.22 Determine the adiabatic flame temperature for the stoichiometric combustion in air of Illinois no. 6 coal after the clean #2 process. Determine also its heat of combustion, and compare with the tabulated value. 3.23* Develop a spreadsheet that determines the air-fuel ratio for coal characterized by a dry ultimate analysis, such as given in Table 3.3. Apply it to several of the coals in the table, as assigned by your instructor. 3.24* Develop a spreadsheet that determines the air-fuel ratio for a coal characterized by a dry ultimate analysis, such as given in Table 3.3, and a given moisture content. Apply it to a coal in the table and several different moisture contents, as assigned by your instructor. 3.25* Develop a spreadsheet that determines the mass and mole, wet and dry flue gas compositions for a coal characterized by a dry ultimate analysis, such as given in Table 3.3, and a given percentage of excess air. Apply it to several coals in the table for 20% and 40% excess air. ______________________ *Exercise numbers with an asterisk involve computer usage.

131 3.26* Develop a spreadsheet that determines the average molecular weight, and the mass and mole, wet and dry flue gas compositions for a coal characterized by a dry ultimate analysis, such as given in Table 3.3, a given percentage of excess air, and a given moisture content. Apply it to a coal in the table for 10% moisture and 10% and 20% excess air. 3.27* Develop a spreadsheet that determines the theoretical air-fuel ratio for a gas characterized by any combination of the components of Table 3.4. Apply it to the gases in the table, as assigned by your instructor. 3.28* Develop a spreadsheet that determines the average molecular weight, and the mass and mole, wet and dry flue gas compositions for a gas characterized by any combination of the components of Table 3.4 and a given percentage of excess air. Apply it to several the gases in the table for 20% excess air. 3.29 What are the theoretical and actual air-fuel ratios and the wet and dry mole flue gas compositions for 20% excess air for the Oklahoma natural gas in Table 3.4? 3.30 What are the theoretical and actual air-fuel ratios and the wet and dry mole flue gas compositions for 15% excess air for the Ohio natural gas in Table 3.4? 3.31 An adiabatic gas turbine combustor burns methane at 77°F with air at 400°F. The combustion products emerge from the combustion chamber at 3200°F. What is the airfuel ratio? What is the equivalent external heat transferred per lbm of air to produce this temperature rise, assuming a mean heat capacity of 0.24 Btu/lbm-R? 3.32 An adiabatic gas turbine combustor burns methane at 25°C with air at 250°C. The combustion products emerge from the combustion chamber at 2000K. What is the airfuel ratio? What is the equivalent external heat transferred per kilogram of air to produce this temperature rise, assuming a mean heat capacity of 1.005 kJ/kg-K? 3.33 An adiabatic combustor burns methane with 400% excess air. Both air and methane are initially at 298K. What is the exit temperature? 3.34 An adiabatic combustor burns methane with 500% theoretical air. Both air and methane are initially at 77°F. What is the flame temperature? 3.35 An adiabatic combustor burns methane in 100% excess oxygen. Both fuel and oxidizer enter at the JANAF tables reference temperature. What is the flame temperature? 3.36 Calculate and tabulate the higher and lower heating values of methane, in kJ/kgmole and in kJ/kg.

132 3.37 Determine the theoretical air-fuel ratio and the air-fuel ratio for 20% excess air for Muhlenberg, Kentucky, raw coal with 10% moisture. 3.38 Determine the air-fuel ratio and wet and dry flue gas mole and mass fractions for dry Muhlenberg, Kentucky no. 9 raw coal with 20% excess air. 3.39 Solve Exercise 3.38 for the coal having 10% moisture in the as-fired condition. 3.40 Solve Exercise 3.38 for 40% excess air. 3.41* Set up a spreadsheet to solve Exercises 3.38–3.40 where it is necessary to change only one parameter for each of the latter cases. 3.42 Using the compound composition data of Table 3.4, calculate the ultimate (elemental) analysis, and compare with the tabular results for the following natural gases: (a) Pennsylvania, (b) Southern California, (c) Ohio, (d) Louisiana, (e) Oklahoma. 3.43* Develop an interactive computer program to solicit and receive ultimate analysis data for an arbitrary coal, arbitrary as-fired ash and moisture fractions, an excess air percentage, and output the appropriate air-fuel ratio and mass and mole, wet and dry flue gas compositions. 3.44* Develop an interactive computer program to solicit and receive ultimate analysis data for an arbitrary coal, arbitrary as-fired ash and moisture fractions, and Orsat CO, CO2, and O2 data. The program should determine the actual operating air-fuel ratio. 3.45* Apply the spreadsheet of Example 3.13 to determine the adiabatic flame temperature, heat of combustion, and wet flue gas composition for stoichiometric combustion of the liquid fuels in Table 3.6. 3.46* Develop a well-organized spreadsheet in which the user may enter a coal ultimate analysis for C, H, O, N, and S; dry flue gas mole compositions for CO, CO2, and 02; and as-fired moisture and ash mass fractions to determine theoretical and actual air-fuel ratios and percentage of excess air.

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CHAPTER FOUR ASPECTS of STEAM POWER PLANT DESIGN

4.1 Introduction After studying the fundamental thermodynamic cycles of steam power plants and considering the characteristics and thermochemistry of fuels, it is appropriate to consider the design of the systems and flow processes that are operative in steam plants and other large-scale power production facilities. This chapter will focus first on the processing of several fundamental streams that play a major role in power plant operation. Up to this point, a great deal of attention has been focused on the water path from the point of view of the thermodynamics of the steam cycle. Additional aspects of the water path related to plant design are considered here. Another fundamental flow in the power plant, the gas stream, includes the intake of combustion air, the introduction of fuel to the air stream, the combustion process, combustion gas cooling in the furnace heat exchange sections, and processing and delivery of the gas stream to the atmosphere through a chimney or stack. A third important stream involves the transportation and preparation of fuel up to the point that it becomes part of the combustion gas. A major non-physical aspect of power production is the economics of power plant design and operation. This is considered in conjunction with some preliminary design analyses of a prototype plant. Environmental considerations also play an important part in planning and design. The chapter concludes with back-of-the-envelope type calculations that define the magnitudes of the flows in a large plant and identify major design aspects of steam power plants. 4.2 The Water Path The Liquid-Water-to-Steam Path

Several pumps are employed in the feedwater path of a steam power plant to push the working fluid through its cycle by progressively elevating the pressure of the water from the condenser to above the turbine throttle pressure. These pumps are usually driven by electric motors powered by electricity generated in the plant or by steam turbines powered by steam extracted from the main power cycle.

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The power requirement of a pump is proportional to the liquid mass-flow rate and the pump work, as given by Equation 2.9, and inversely proportional to the pump efficiency: Power = mvsat p/pump

[ft-lbf /s | kW]

The pumps are required to overcome frictional pressure losses in water-flow and steam-flow passages, to provide for the pressure differences across turbines, and to elevate the liquid to its highest point in the steam generator. The pump power requirements are typically a small percentage of the gross power output of the plant. Thus condensate leaving the condenser passes through one or more pumps and feedwater heaters on its way to the steam generator. A typical shell-and-tube closed feedwater heater is shown in Figure 4.1. Normally, feedwater passes through the tubes while extracton steam enters at the top and condenses as it flows over the tubes to the bottom exit.

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After passing through the chain of feedwater heaters and pumps, the feedwater enters the steam generator through the economizer. An economizer is a combustiongas-to-feedwater tubular heat exchanger that shares the gas path in a steam generator, as seen in Figure 4.2. The economizer heats the feedwater by transferring to it some of the remaining energy from the cooled exhaust gas before the gas passes to the air heater, the pollution control equipment, and the stack.

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Steam Generators Figures 4.3 and 4.4 show two-drum steam generators in which the design vaporization pressure is below the critical pressure of water. Hot, subcooled liquid feedwater passes from the economizer and through the boiler tube walls to the drum loop located near the top of the steam generator. Liquid water circulates by free convection through the many boiler tubes between the drums until it is vaporized by the hot gas stream flowing over the tubes. A fixed liquid level is maintained in the upper steam drum, where the steam separates from the liquid and passes to the superheater. Solids settle in the bottom of the so-called “mud drum” below it. A steam drum mounted on a railroad flat car en route to a construction site is shown in Figure 4.5. The many stub tubes around the bottom and on top are to be

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connected to steam-generating loops and to steam-superheating pipes, respectively, as seen at the top of Figure 4.4. The large pipes on the bottom and the ends are for connection to downcomers, which supply recirculated liquid water to various heating circuits in the steam generator. Steam produced in the steam drum, at a boiling temperature corresponding to the vapor pressure in the drum, passes to superheater tube or plate heat-exchanger banks. The superheater tube banks are located in the gas path upstream of the drum loop, as seen in Figures 4.3 and 4.4, taking advantage of the highest gas temperatures there to superheat the steam to throttle temperature. The hottest gases are used to heat the

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hottest water-tube banks, to minimize the irreversibility associated with the heat transfer through the large temperature differences between the combustion gas and the steam or liquid water. The dry steam from the superheater then passes from the steam generator through the main steam line to the HP turbine. The progression of tube banks, with decreasing water temperatures exposed to successively cooler gas temperatures from the secondary superheater to the economizer to the air heater are also seen in the universal-pressure (supercritical pressure) steam generator in Figure 4.2. As the design throttle steam pressure increases toward the critical pressure of water, the density difference between liquid water and vapor decreases, finally vanishing at the critical point (3208.2 psia, 705.47°F.). As a consequence, in steamgenerator boiling loops, natural convection water circulation—which is driven by the density difference between liquid and steam—becomes impractical at pressures above about 2500 psia. Thus modern high-throttle-pressure power plants use circulating pumps to provide forced to circulation to augment or replace natural circulation of water in the steam generator. In single-drum steam generators, water flows downward from the steam drum through large pipes called downcomers located outside the furnace wall, then through circulating pumps to headers at the bottom of the steam generator. From the headers, water flows upward in vertical tubes forming the inside of the furnace walls. The water is heated by the furnace gases as it rises, and eventually boils and forms a two-phase flow that returns to the steam drum. There, vapor separates and passes to the superheater. Steam generators may utilize natural convection flow through downcomers and

139 vapor-laden upward flow through the tube walls alone or may combine natural convection with the use of booster pumps to provide adequate circulation for a wider range of operating loads. It is important to recognize that at the same time steam is being generated in the boiler, the tube walls are being cooled by the water. Adequate water circulation must be ensured to provide waterside heat transfer rates high enough to maintain tube wall temperatures below their limiting design values and thereby to avoid tube failure. A once-through supercritical steam generator, operates at a throttle pressure above the critical pressure of water as in the Riverside station discussed in Chapter 2. There are no drums and no water recirculation in a once-through steam generator. Water from the economizer passes to the bottom of the furnace, where it starts its upward flow through the furnace tube walls. Steam formed in the tubes flows upward to be collected in headers and mixed to provide a unifrom feed to the superheater. The feedwater passes directly from the liquid to the vapor phase as it is heated at a pressure above the saturation pressure. It may be compared to a flow of water pumped through a highly heated tube with a downstream valve. The state of the steam emerging from the tube depends on the valve setting, the heat addition rate, and the feedwater flow rate. In the same way, the steam conditions at the turbine throttle may be adjusted by changing the turbine throttle valve setting, the fuel firing rate, and the feedwater flow rate. If the flow rate is decreased by closing the throttle valve, it is necessary to decrease the fuel firing rate to maintain the same thermodynamic conditions at the throttle. On the other hand, if the rate of heat transfer is increased without changing the flow rate, the steam discharge temperature will increase. Other adjustments, such as increasing condenser cooling-water flow rate, may then be appropriate to avoid an increase in condenser temperature and pressure. Similarly, an increase in fuel flow rate must be accompanied by an increase in air flow rate to maintain a constant air-fuel ratio. In cycles with reheat, the reduced-pressure steam from the HP turbines passes through the cold reheat line to the reheater section in the steam generator, where the steam temperature is returned to approximately the original throttle temperature. The steam then returns to the next turbine through the hot reheat steam line, as Figure 2.13 indicates. After leaving the LP turbine, low-pressure steam then passes over the water-cooled tubes in the condenser and returns to the feedwater heating system as saturated liquid condensate. A condensate pump then raises the pressure of the liquid and transports it to the first low-pressure feedwater heater, where it begins another trip through the cycle. In order to avoid corrosion, scaling and the deposits of solids along the water path can result in losses of efficiency and unscheduled shutdowns, water of extreme purity is required in the steam cycle. Chemical and filtration processes are employed to ensure that high water quality is maintained, to avoid deterioration or clogging of water path components. An example of the potential deposits when proper water treatment is neglected is seen in Figure 4.6. The deaerator, an open feedwater heater mentioned in

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Chapter 2, provides for the removal of noncondensable gases, particularly oxygen, from the working fluid. The deaerator allows noncondensable gases to escape to the atmosphere through a vent condenser, while accompanying steam is retained by condensing it on cool surfaces and returning it to the feedwater heater stream by gravity flow. The turbine room at the Bull Run coal-burning power plant of the Tennessee Valley Authority (TVA) is shown in Figure 2.3. Electrical generators are seen in the left and right foreground. Behind them, high-pressure turbines on the left are seen joined to low-pressure turbines on the right by two large, vee-shaped crossover steam lines. The side-by-side condensers are seen on either side of the low pressure turbines, a departure from the usual practice of locating the condenser below the low-pressure turbines. Figure 4.7 shows the turbogenerator room at TVA’s Brown’s Ferry nuclear power plant with a turbine in the foreground. The Condenser Cooling-Water Loop The cooling loop, in which water passes through tubes in the condenser removing heat from the condensing steam, is an important water path in large steam plants. This cooling water, clearly separate from the working fluid, is usually discharged into a nearby body of water (a river, or a natural or man-made lake) or into the atmosphere. Figure 4.8 shows a typical wood-framed, induced-draft cooling tower used to dissipate heat from the condenser cooling water into the atmosphere. The tower is usually located a few hundred yards from a plant. Typically, the cooling water entering the tower is exposed to a flow of air created by upward-blowing fans at the bases of the funnels at the top of the towers. A fraction of the condenser cooling water, which passes over extensive aerating surfaces in the tower, evaporates and exits to the

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atmosphere, cooling the rest of the water. The remaining chilled water is then returned to the condenser by a cooling-water circulating pump. A continuing supply of liquid makeup water is required for these towers to compensate for vapor loss to the atmosphere. In areas where large structures associated with power plants are acceptable, large natural-draft cooling towers may be used. Figure 4.9 shows two large natural-draft hyperbolic cooling towers serving a large power plant. The height of these towers, which may reach over 500 feet, creates an upward draft due to the difference in density between the warm air in the tower and the cooler ambient air. Heat from the condenser cooling water warms the air, inducing an upward air flow through the heat transfer surfaces at the base. These towers offer long-term fan-power savings over mechanical draft towers. Under some conditions, these power savings may offset high construction costs of hyperbolic towers.

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143 4.3 The Fuel Path for a Coal-Burning Plant The supply and handling of fuel for a modern coal-burning power plant is a complex and expensive undertaking. In contrast to the relatively simple steady flow of fluid fuels in power plants that consume natural gas and fuel oil, solid-fuel-burning plants offer major and continuing challenges to engineers. The discussion here focuses on these operations and their challenges. Getting the Coal to the Plant The source of coal for a plant may be a surface mine or a deep underground mine. Power plants are sometimes located adjacent to mines, where conveyors may provide the only transportation required. This significantly reduces coal transportation costs which otherwise can be higher than the cost of the coal alone. Such plants are called mine-mouth plants. A mine-mouth plant may be an attractive option if its selection does not result in significant transmission costs to bring the electrical power to distant load centers where the utilities’ customers are located. Today the power plant and coal mines are likely to be a considerable distance apart, perhaps a thousand miles or more. The most widely used modern transportation link between mine and plant is the unit train, a railroad train of about a hundred cars dedicated to transporting a bulk product such as coal. Although slurry pipelines (a slurry is a fluid mixture of solid lumps and liquid, usually water, which can be pumped through pipes by continuous motion) sometimes offer attractive technical solutions to coal transport problems, economic and political forces frequently dictate against their use. Dedicated truck transport is an occasional short-haul solution, and barges are sometimes used for water transport. Here, we will focus on unit trains. Several unit trains may operate continuously to supply a single plant. Trains carrying low-sulfur coal from Wyoming, Montana, and the Dakotas supply coal to plants as distant as Michigan, Illinois, and Oklahoma. This strange situation, in which utilities located in states with large quantities of coal purchase coal from distant states, was a response to pollution control requirements. It was preferred to purchase lowsulfur coal from a distant state rather than pay a high price for sulfur removal equipment (perhaps 10% of the cost of the plant construction), some of which has a reputation for unreliability. In response to such choices, the Oklahoma legislature passed a law requiring utilities to burn at least 10% Oklahoma coal in their coal-burning steam generators. Such mixing of small quantities of high-sulfur coal with low-sulfur coal is an expedient to protect local businesses and to spread out resource utilization geographically. New plants, however, no longer have the option to choose low-sulfur coal or sulfur removal equipment. “Best available control technology,” BACT, has become the rule. The Environmental Protection Agency now requires new plants to have scrubbers (sulfur removal equipment) even if the plants use low-sulfur coal, and they are required to employ the currently most effective pollution control technology.

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Coal Unloading and Storage On arrival at the plant, the unit train passes through an unloading station. Some coal cars have doors on the bottom that open and dump their load to a conveyor below. Others have couplings between cars that allow the rotation of individual cars about their coupling-to-coupling axis, by a dumping machine, without detachment from the train, as seen in Figure 4.10. The figure shows a breaker in the dumping facility that reduces large coal chunks to a smaller, more uniform size for transport on a belt conveyor. The under-track conveyor at the unloading station then carries the newly arrived coal up and out to a bunker or to a stacker-reclaimer in the coal yard as seen in Figure 4.11. The stacker-reclaimer either feeds the coal through a crusher to the plant or adds it to the live storage pile. A permanent coal storage pile sufficient to supply the plant for several months is usually maintained. While the first-in-first-out approach common in handling perishable goods seems logical, a first-in-last-out storage system is usually used. A primary reason for this approach is the hazard and expense of coal pile fires, which can occur due to spontaneous combustion. Once a stable storage pile is achieved by packing and other treatment to restrict air acccess, it is usually not disturbed unless coal must be withdrawn by the stacker-reclaimer to satisfy unusual demands caused by labor strikes, extreme weather, rail accidents, or the like.

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146 A conveyor transports coal from the reclaimer to a crusher house, where hammer mills, ball crushers, or roller crushers break up large chunks to a more manageable size. Another conveyor may then carry the crushed coal to one of several bunkers or silos for temporary storage prior to firing. Some of these features may be seen in the photograph of the PSO Northeastern Station in Figure 2.1. The rate of feeding coal from the silos is controlled to maintain the desired steam generator energy-release rate. In a pulverized-coal plant, the coal is fed from the silos to pulverizers, where it is further reduced in size to a powdery form. Warm air drawn through an air preheater in the steam generator by the primary air fan flows through the pulverizer, where it picks up the fine coal particles and transports them pneumatically through piping to the steam generator burners. Several arrangements of silos, feeders, pulverizers, and pneumatic transport systems are seen in Figures 4.2 to 4.4. 4.4 The Gas Path Fans While natural or free convection may be used to provide combustion air to small boilers and heaters, modern power plants employ large fans or blowers to circulate air to the burners and to assist flue gas in escaping from the furnace. These fans are called forced-draft fans and induced-draft fans, respectively. A common arrangement of these fans is shown in Figure 4.4. Atmospheric air drawn into the steam generator by one or more forced-draft fans is heated as it passes through the cold gas side of an air heater on its way to the furnace. At the same time, combustion gases that have passed through the furnace heat transfer sections are cooled as they passed through the hot side of the air heater on their way to induced-draft fans and thence to the stack. In the case of pulverized-coal-burning plants, primary air fans, as seen in Figures 4.3 and 4.4, supply enough pre-heated air to pulverizers to transport coal pneumatically to the burners. Primary air usually pre-heated to 300-600°F to dry the coal as it passes through the pulverizer. With a forced-draft fan alone, the furnace pressure is above atmospheric pressure, causing large outward forces on the furnace walls and a tendency for leakage of combustion gas from the furnace. On the other hand, the use of an induced-draft fan alone would cause the furnace pressure to be below atmospheric pressure, producing large inward forces on the walls and possible air leakage into the furnace. The forces on the walls, which can be significant, can be minimized by keeping the furnace pressure near atmospheric by using balanced draft, that is, the use of both forced- and induced-draft fans, which produce a gas path pressure distribution such as shown by the heavy line in Figure 4.12. In this design the forced-draft fan raises the pressure to 15 in. of water gauge entering the steam generator, and the induced-draft fan depresses its inlet pressure about 21 in. of water below atmospheric. As a result, the furnace inside-wall pressure is less than an inch of water below atmospheric. This substantially reduces both the potential for furnace leakage and the forces on the furnace walls.

147

The power requirements for fans may be determined in much the same way that pump power requirements are determined. Fans are primarily gas-moving devices that produce small pressure rises. Pressure and density changes across fans are usually small fractions of the fan inlet values. This justifies the approximation that the fan process is incompressible. Fan power requirements then closely follow the pump power prediction method discussed earlier. Thus, for a forced-draft fan, power may be estimated by using PowerFD = Qair pair/FDfan

[ft-lbf/s | kW]

For the induced-draft fan PowerID = ,airQair pgas(1 + F/A)/,gasFDfan

[ft-lbf/s | kW]

where Qair is the volume flow rate of air entering the forced-draft fan. The second equation accounts for the additional fuel mass handled by the induced-draft fan and the

148

density of the gas leaving the furnace, assuming no leakage or diversion of air from that leaving the forced-draft fan. The drawing of a centrifugal forced-draft fan is shown in Figure 4.13; a photograph of a rotor and open housing is presented in Figure 4.14. In a centrifugal fan, air is spun by the rotor blades, producing tangential motion and pressure rise and leaving behind a vacuum for air to flow in along the axis of the fan. It the fan entrance is open to the atmosphere, its exhaust is pressurized; if its exhaust is atmospheric, its entrance pressure is below atmospheric. Fans typically do not produce large pressure rises but do produce large flows of gases. A diagram of an axial-flow fan designed for induced-draft use is shown in Figure 4.15. Figure 4.16 presents a photo of the same type of fan. Induced-draft fans must be able to withstand high-temperature service and erosion due to airborne particulates. Large electrostatic precipitators located upstream of the induced-draft fans remove most of the flyash by inducing a static charge on the flowing particles and collecting them on plates of opposite charge. The plates are periodically rapped mechanically to free ash deposits that drop to the bottom of the precipitator and are collected and removed. Figure 2.1 shows electrostatic precipitators, to the right of the steam generator. The large structure behind the fans, air heater, and ducting and below the stack in Figure 4.4 is an electrostatic precipitator. Air Preheaters The air leaving the forced-draft fan usually flows through an air preheater to a windbox around the furnace and then to the burners. A Ljungstrom rotary air preheater, used

149

150

in many large plants, is shown in Figures 4.17 and 4.18. The rotary air heater is a slowly rotating wheel with many axial-flow passages, having large surface area and heat capacity, through which air and flue gas pass in counterflow parallel to the wheel axis. When the wheel surfaces heated by the flue gas rotate to the air side, they are cooled by the air from the forced-draft fan. As result, the air temperature rises several hundred degrees before passing to the furnace windbox. Although the Ljungstrom rotary air heater is widely used in utility and industrial power plants, heat-pipe air and process heaters are now being considered and applied for use in power plants and industry. A heat pipe, shown in Figure 4.19, is a sealed tube in which energy is transported from one end to the other by a thermally driven vapor. The heat-pipe working fluid absorbs heat and vaporizes at the lower, hot end. After rising to the higher, cold end, the vapor condenses, releasing its heat of vaporization, which is carried away by conduction and convection through external fins to the combustion air. The liquid then returns to the hot end by gravity and/or by capillary action through wicking, to complete the cycle. The wicking may be spiral grooves around the inside of the tube that ensure that the entire inside surface is wetted for maximum heat transfer. The wicking in the cold section is particularly important, because it provides increased surface area that increases inside-gas heat transfer rates. The outside the tube is usually finned to provide adequate external heat transfer rates, both from the flue gas and to the incoming air.

151

152

As a heat transfer device, a well-designed heat pipe has an effective thermal conductance many times that of a copper rod. Note that the energy transfer inside the heat pipe is essentially isothermal, since the liquid and the vapor are in near equilibrium. Although heat pipes will operate in a horizontal orientation, their effectiveness is augmented by gravity by inclining them about 5°– 10° to assist in liquid return to the hot end. In power plant air heater applications (see Figures 4.19 and 4.20), finned heat pipes supported by a central partition between the incoming air stream and the flue gas

153 stream are free to expand outward. This reduces thermal expansion problems and virtually eliminates the possibility of leakage between the flows. Such heaters often have been installed in process plants and have been retrofitted in power plants originally built without air heaters, because of their ease of installation and compact size compared with stationary tubular air heaters. EXAMPLE 4.1

The heat-pipe air heater in Figure 4.20 has an air flow of 360,800 lbm/hr and a flue gas flow rate of 319,000 lbm/hr. The flue gas enters at 705°F and leaves at 241°F; the combustion air enters at 84°F. What is the rate of energy recovery from the flue gas, and what is the air temperature entering the windbox? Assume a flue gas heat capacity of 0.265 Btu/lbm-R. Solution

The rate of heat recovery from the flue gas is mcp (Tout – Tin) = 319,000(0.265)(705 – 241) = 39,224,240 Btu/hr The air temperature is then 84 + 39,224,240/[(0.24)(360,800)] = 537°F _____________________________________________________________________ An analysis of the gas flow through a steam generator must take into account the streamwise pressure rise through the fans and pressure losses due to friction and losses through flow restrictions and turns. These include losses due to flow through furnace tube and plate heat exchanger banks and other passages, such as in both the air and the gas passes through the air heater. Power Plant Burners Burner design depends on the choice of fuel and the steam generator design. Figure 4.21 shows a burner designed for forced-draft applications burning natural gas and oil. Oil and steam under pressure are mixed in the central feed rod to atomize the oil to a fine mist coming out of the oil tip. The cone at the oil tip stabilizes the flame in the surrounding air flow. The gas pilot next to the oil feed rod provides a continuous ignition source. A separate duct for the natural gas supply feeds gas to the two types of gas tip. Separate air registers control the flow of air to the gas and oil tips. Figure 4.22 shows an oil burner, with a water-cooled throat, installed in a furnace wall. Registers that control the flow of air from the windbox are also visible.

154

In the case of the pulverized-coal plant, primary air flows through the pulverizer and carries the fuel directly to the burners (Figure 4.23). Secondary (and sometimes tertiary) air helps to control the temperature of the control nozzle and of the furnace wall, and mixes with the combustion gases to provide for essentially complete combustion of the fuel. Features of a burner designed for pulverized-coal firing in planar furnace walls are shown in Figure 4.24. Note that the secondary air flow through the windbox registers helps to cool the nozzle through which the coal and the primary air flow. Another approach to burning pulverized coal uses corner burners in tangentially fired steam generators. A plan view of such a furnace is shown in Figure 4.25. The burners induce a circular motion, in the horizontal plane, on the upward-rising combustion gases, promoting vigorous mixing, which hastens completion of combustion in the furnace. For control purposes, the corner burners can be tilted in the vertical direction to adjust the furnace heat transfer distribution. A cyclone furnace type of burner installation, used for burning slagging coals (those that form liquid ash, or slag, at moderate temperatures) in steam generators, is shown in Figure 4.2. The cyclone furnace is a cylindrical furnace with very large

155

156

volumetric heat release rates that lies adjacent to and opens onto the main furnace. Details of a cyclone furnace are shown in Figures 4.26 and 4.27. The coal supplied to cyclone furnaces, which is crushed but not pulverized, is fed to the cyclone by a mechanical feeder. The coal and primary air entering the cyclone

157

move tangentially to the inside of the furnace cylinder. There the momentum of the coal carried by the swirl flows forces the coal pieces toward the cylindrical burner wall. The very high temperature and vigorous mixing produce a high rate of burning. As a result, combustion is virtually complete by the time the combustion gas flow enters the

158 main furnace. The cooled walls stimulate formation of a protective slag layer on the cylinder walls. Because the main furnace is only required for steam generation and to cool the combustion gases, and not to provide time for completion of combustion, the cyclone furnace steam generator can be significantly smaller than the pulverized-coal steam generator. A steady flow of slag drains from the cyclone furnace into a slag tank at the bottom of the main furnace. In both cyclone and pulverized-coal steam generators the combustion gases flow upward from the burners, transferring heat to the tube walls by radiation and convection. The cooling gases then flow through superheater, reheat, and boiler tube or plate sections. The combustion gas temperature drops as it passes through these steam generator sections in essentially a counterflow arrangement with the water flow. The combustion gases undergo their final cooling as they pass through the economizer and then the air preheater. From there they pass through an electrostatic precipitator for removal of airborne particles and through scrubbers for control of oxides of nitrogen and sulfur (NOx and SOx) and through an induced-draft fan before entering the stack. The serious degradation of the environment caused by oxides of sulfur and nitrogen in the flue gas of power plants and from other sources has led to widespread chemical processing of flue gases. Figure 4.28 shows a schematic diagram of the gas flow path for removal of NOx and SOx after particulate removal in the precipitator and passage through an induced-draft fan. In this scheme, gas-to-gas heat exchangers (GGH) provide the proper temperatures for the flue gas desulfurization (FGD) unit and the DENOX catalyst unit. This additional equipment increases the pressure drop through the system, sometimes necessitating an additional fan. With high smokestacks, the stack effect also influences the gas path and must be taken into account.The stack effect is the upward movement of exhaust gas produced by the density difference between the hot gases inside the stack and the surrounding cooler atmospheric air. Because of hot gas buoyancy, a smaller pressure gradient along the stack length is required to expel the combustion gases from the stack. This effect is opposed by the usual viscous friction pressure losses. The diameter and height of the stack control the relative influence of frictional forces in opposing the stack effect. Other considerations, such as cost and a possible need to disperse the stack gas above a particular height, also have a significant influence on these dimensions. It is obvious that the air heater of the steam generator should extract as much energy from the combustion gas as possible to maximize its regenerative heat transfer effect. This implies cooling the gas to a low tempeature. However, practical limits exist on the minimum combustion gas temperature, to avoid the condensation of water vapor in the presence of sulfur and nitrogen compounds in the gas and to meet the temperature requirements of the pollution control equipment. Condensation of water vapor in the presence of gaseous oxides of sulfur and nitrogen leads to the formation of acids that erode the materials on which the liquid condenses. The temperature at which the vapor condenses is called the acid dew point. Typical acid dew points for coal range to about 320°F. As a result, stack gas design temperatures may exceed that value, depending on the coal and flue gas treatment.

159

4.5 Introduction to Engineering Economics The success of any engineering undertaking depends on adequate financial planning to ensure that the proceeds of the activity will exceed the costs. The construction of a new power plant or the upgrading of an old one involves a major financial investment for any energy company. Financial planning therefore starts long before ground is broken, detailed design is begun, and orders are placed for equipment. Cost analysis and fiscal control activities continue throughout the construction project and the operating life of the plant. This section briefly introduces fundamentals of engineering economics, with a slant toward power plant cost analysis as well as issues of maintenance and equipment replacement.

160 The cost to construct a power plant, waterworks, dam, bridge, factory, or other major engineering work is called its capital cost. It is common to discuss the capital cost of building a power plant in terms of dollars per kilowatt of plant power output. A plant may cost $1100 per kilowatt of installed power generation capacity, for instance. In addition to the cost of building the plant, there are many additional expenditures required to sustain its operation. These are called operating costs. They may be occasional, or they may occur regularly and continue throughout the life of the plant. Often these costs are periodic, or are taken to be periodic for convenience of analysis. There are, for instance, annual fuel costs, salary expenses, and administrative and maintenance costs that are not associated with the initial cost of the plant but are the continuing costs of generating and selling power. Operating costs are sometimes related to the amount of electrical energy sold. Usually they are expressed in cents per kilowatt-hour of energy distributed to customers. Thus the expenses associated with power generation and other business endeavors may be thought of as two types: (1) initial costs usually associated with the purchase of land, building site preparation, construction, and the purchase of plant equipment; and (2) recurring operating costs of a periodic or cyclic nature. It is frequently desirable to express all costs on a common basis. The company and its investors may wish to know what annual sum of money is equivalent to both the capital and operating costs. The company may, for example, borrow money to finance the capital cost of the plant and then pay the resulting debt over the expected useful life of the plant, say, 30 or 40 years. On the other hand, they may wish to know what present sum would be required to ensure the payment of all future expenses of the enterprise. It is clear that $100 in hand today is not the same as $100 in hand ten years from now. One difference is that money can earn interest. One hundred dollars invested today at 8% annual compound interest will become $215.89 in ten years. Clearly, an important aspect of engineering economics is the time value of money. Compound Interest If Alice lends Betty $500, who agrees to pay $50 each year for five years for the use of the money, together with the original $500, then at the end of the fifth year, Alice will have earned $250 in simple interest and receive a total of $750 in return. The annual interest rate is i = Annual interest / Capital = 50 / 500 = 0.1 or (0.1)(100) = 10% rate of return. If, however, Betty keeps the interest instead of paying it to Alice annually, and eventually pays 10% on both the retained interest and the capital, the deal involves compound interest. The total sum to be returned to Alice after 5 years is computed as follows: At the end of the first year Alice has earned $50 in interest. The interest for the

161 next year should be paid on the original sum and on the $50 interest earned in the first year, or $550. The interest on this sum for the second year is 0.1 × $550 = $55. The following table shows the calculation of the annual debt for the five-year loan of $500 at 10% interest: At the End

The Accumulated Debt is:

This Sum

First year

$500 + 0.1 x $500 =

$550.00

Second year

$550 + 0.1 x $550 =

$605.00

Third year

$605 + 0.1 x $605 =

$665.50

Fourth year

$665.50 + 0.1 x $665.50 =

$732.05

Fifth year

$732.05 + 0.1 x $732.05 =

$805.26

of:

It is evident that the interest earned on the preceding interest accumulation causes the annual indebtedness to grow at a increasing rate. It can be shown that the future sum, S, is given by S = P (1 + i)n where P is the principal, the initial sum invested; i is the interest rate, and n is the number of investment periods, in this case the number of years. Here the factor multiplying the principal, S / P = (1 + i)n is called the compound amount factor, CAF. The difference between simple and compound interest may not be spectacular for short investment periods but it is very impressive for long periods of time such as the operating life of a power plant. For our example, the CAF is (1 + 0.1)5 = 1.6105, and S = 500(1.6105) = $805.26. Now, consider the following closely related problem. EXAMPLE 4.2

What sum is required now, at 8% interest compunded annually, to produce one million dollars in 25 years? Solution

The future sum is S = P (1 + i)n = 1,000,000 = P (1 + 0.08)25 Solving for P, the present sum is 1,000,000/(1.08)25 = $146,017.90. Thus, compound interest brings a return of almost over seven times the original investment here. The

162 same present sum invested at 8% simple interest for twenty-five years would produce a future sum of less than half a million dollars. _____________________________________________________________________ In the example, the inverse of the CAF was used to determine the present worth of a future sum. The inverse of the CAF is called the present-worth factor, ( PWF): PWF = P/ S = 1 / (1 + I)n Thus we see that the time value of money is related to the compound interest that can be earned, and that taking compounding into account can be important. To recklessly adapt an old adage, “A dollar in the hand is worth two (or more) in the future (if invested wisely).” Capital Recovery Another important aspect of compound interest is the relationship between a present sum of money and a regular series of uniform payments. Consider a series of five annual payments of R dollars each, when the interest rate is i. What is the present dollar equivalent of these payments? Applying the CAF as in the preceding example, with R as the future sum, the present sum associated with the first payment is R/(1 + i). The present sum associated with the second payment is R/(1 + i)2. Thus the present worth of the five payments is P = R [ (1 + i) – 1 + (1 + i) – 2 + (1 + i) – 3 + (1 + i) – 4 + (1 + i) – 5 ] It may be shown that this expression can be written as P = R [(1 + i)5 – 1]/[i(1 + i)5]. The factor multiplying the annual sum R is called the series present-worth factor, SPWF, which for n years is: SPWF = P/ R = [(1 + i)n – 1]/[i(1 + i)n]

(4.1)

Solving for R, we obtain an expression for the regular annual payment for n years needed to fund a present expenditure of P dollars at an interest rate i. The resulting factor is called the capital recovery factor, CRF, which is the reciprocal of the series present worth factor: CRF = R / P = i(1 + i) n / [(1 + i) n – 1]

(4.2)

163 EXAMPLE 4.3

What uniform annual payments are required for forty years at 12% interest to retire the debt associated with the purchase of a $500,000,000 power plant? Solution

Using equation (4.2), we get R = Pi (1 + i)/[(1 + i) n – 1] = 5×108(0.12)(1 + 0.12) 40/(1.1240 – 1) = $60,651,813 This sum may be regarded as part of the annual operating expense of the plant. It must be recovered annually by the returns from the sale of power. _____________________________________________________________________ 4.6 A Preliminary Design Analysis of a 500-MW Plant Consider the design of a 500-megawatt steam power plant with a heat rate of 10,000 Btu/kW-hr and a water-cooled condenser with a 20°F cooling-water temperature rise produced by heat transfer from the condensing steam. The plant uses coal with a heating value of 10,000 Btu/lbm. Let us estimate the magnitude of some of the parameters that characterize the design of the plant. The reader should verify carefully each of the following calculations. A 500-megawatt plant operating at full load produces 500,000 kW and an annual electrical energy generation of 500,000 ( 365 ( 24 = 4.38 × 109 kW-hr With a heat rate of 10,000 Btu/kW-hr, this requires a heat addition rate of 500,000 ( 10,000 = 5 × 109 Btu/hr Coal with an assumed heating value of 10,000 Btu/lbm must therefore be supplied at a rate of 5 × 109 / 104 = 500,000 lbm/hr or 500,000 / 2000 = 250 tons/hr. A dedicated coal car carries about 100 tons. Hence the plant requires 250 /100 = 2.5 cars per hour of continuous operation. A coal unit train typically has about 100 cars. Then the plant needs 2.5 ( 24 / 100 = 0.6 unit trains per day, or a unit train roughly every two days. If coal costs $30 per ton, the annual cost of fuel will be 30 ( 250 ( 24 ( 365 = $65,700,000 The cost of fuel alone per kW-hr, based on 100% annual plant capacity, will be 65,700,000/(500,000 ( 365 ( 24) = $0.015/kW-hr ; 1.5 cents/kW-hr

164 The annual plant factor, or annual capacity factor, expressed as a decimal fraction , is the ratio of the actual annual generation to the annual generation at 100 % capacity. If the coal has 10% ash, the plant will produce 250 ( 0.1 = 25 tons of ash per hour. Under some circumstances the ash may be used in the production of cement or other paving materials. If it is not marketable, it is stabilized and stored in nearby ash ponds until it can be moved to a permanent disposal site. Similarly, if 2% of the coal is sulfur and half of it is removed from the combustion products, 2.5 tons per hour is produced for disposal. If the sulfur is of sufficient purity, it may be sold as an industrial chemical. With an air-fuel ratio of 14, an air flow rate of 14 ( 500,000 = 7,000,000 lbm/hr is required for combustion. This information is important in determining the size of the induced- and forced-draft fans, that of their driving motors or turbines, and of the plant’s gas path flow passages. The heat rate of 10,000 Btu/kW-hr corresponds to a thermal efficiency of 3413/10,000 = 0.3413 or 34.13%. If we approximate the heat of vaporization of water as 1000 Btu/lbm, the throttle steam flow rate, with no superheat, would be about 10,000 ( 500,000 / 1000 = 5,000,000 lbm/hr This determines the required capacity of the feedwater pumps and is important in sizing the passages for the water path. The above thermal efficiency implies that about 65% of the energy of the fuel is rejected into the environment, mostly through the condenser and the exiting stack-gas energy. As an upper limit, assume that all of the heat is rejected in the condenser. Thus (1 – 0.3413)(5 × 109) = 3.29 × 109 Btu/hr must be rejected to condenser cooling water. With 20° water temperature rise in the condenser, this rate of cooling requires a cooling-water flow rate to the condenser of 3.29 × 109/(1.0 × 20) = 1.65 × 108 lbm / hr assuming a water heat capacity of 1.0 Btu/lbm-R. This gives information relevant to the design sizing of cooling-water lines, cooling towers, and water pump capacities. These back-of-the-envelope calculations should not be regarded as precise, but they are reasonable estimates of the magnitudes of important power plant parameters. Such estimates are useful in establishing a conceptual framework of the relationships among design factors and of the magnitude of the design problem. EXAMPLE 4.4

Relating to the above rough design of a 500-MW plant, and assuming the capital cost information of Example 4.3, determine the capital cost per kW of generation capacity and estimate the minimum cost of generation for the plant if it is predicted to have an

165 annual plant factor of 80% and maintenance and administrative costs of $0.007 /kW-hr. Solution

The unit cost of the power plant is $500,000,000/(500 (1000) = $1000 per kW-hr of capacity. The capital cost part of the annual cost of power generation is (60,651,813 (100)/(365 ( 24 ( 0.8 (500,000) = 1.73 cents per kW-hr The cost of coal was determned to be 1.5 cents/kW-hr. The minimum cost of producing electricity is then 1.73 + 1.5 + 0.7 = 3.93 cents per kW-hr _____________________________________________________________________ Bibliography and References

1. Singer, Joseph G., (Ed.), Combustion /Fossil Power Systems. Windsor, Conn.: Combustion Engineering, Inc.,, 1981. 2. Anon., Steam, It’s Generation and Use. New York: Babcock and Wilcox, 1978. 3. Hensley, John C., Cooling Tower Fundamentals. Mission, Kan.: Marley Cooling Tower Co., 1985. 4. Li, Kam W., and Priddy, A. Paul, Power Plant Systems Design. New York: Wiley, 1985. EXERCISES

4.1 Derive an equation for the sum, S, resulting from P dollars invested at simple interest rate i for a period of n years. 4.2 For the power plant design discussed in Section 4.6, estimate the horsepower of a motor required to drive the fans used to overcome a steam generator gas-path pressure drop of 1 psia. Assume a fan efficiency of 80%. What is the fractional and percentage reduction in power plant output due to the fans? 4.3 Estimate the horsepower required by the feedwater pumps in the Section 4.6 design if the HP-turbine throttle pressure is 3200 psia. Assume a pump efficiency of 70%. What fractional and percentage reduction of the power plant output does this represent?

166 4.4 What are the annual savings in fuel costs in the Section 4.6 plant design if the plant heat rate can be reduced to 8500 Btu/kW-hr? 4.5 If the total capital cost of the Section 4.6 plant design is $600,000,000 and the annual administrative and maintenance costs are one cent per kW-hr, what is the minimum cost of electricity per kW-hr, assuming an annual interest rate of 9% and an expected plant lifetime of thirty-five years? 4.6 What is the present worth of a sequence of five annual payments of $4500, $6500, $3500, $7000, and $10,000 at an annual interest rate of 8%? 4.7 You have collected the following data on 1.5-MW steam turbines, as alternatives to the purchase of utility power, for a new process plant to operate at 60% plant factor: Turbine Number

Heat rate, Btu/kW-hr Installed cost Estimated annual maintenance cost

1

2

3

12,000

10,600

9,500

$124,000

$190,000

$245,000

$2,000

$1,800

$2,550

Coal (14,000 Btu/lbm) is the fuel to be used, at a cost of $26 per ton. Assuming an annual interest rate of 8%, compare the annual cost of of the turbines for thirty-year turbine lifetimes. Which turbine would you select? What other factors would you consider before making a decision? 4.8 Using the data of Exercise 4.7, compare the turbines on the basis of present worth of all costs. 4.9 For the power plant design discussed in Section 4.6, estimate the motor horsepower required to drive a 75% efficient fan that is used to overcome a steam generator gas-path pressure drop of 50 kPa. 4.10 Estimate the total power required by 65% efficient feedwater pumps operating in parallel in the Section 4.6 design for a throttle pressure of 20 MPa. 4.11 Work out a back-of-the-envelope analysis similar to that of Section 4.6 in SI units. 4.12* Develop an interactive computer program that implements a steam power plant system analysis of the type presented in Section 4.6 at one of the following levels, to be

167 assigned by your instructor. Level 1: User supplies parameters in response to screen prompts, and one or more output screens display the resulting input and output parameters. Level 2: Same as Level 1, but also provide a capability for the user to change the design by varying one input while holding all others constant. Level 3: Allow the user to select a dependent variable from a list of outputs, and a parameter to be varied and its range from another list. Display a graph of the variation of user-selected outputs over the range of the parameter. 4.13* Construct a spreadsheet that systematizes the computations for a steam power plant along the lines presented in this chapter. Set up a version of the spreadsheet that allows easy variation of input parameters. Use the spreadsheet to develop graphs that show the influence of plant heat rate on fuel costs and sulfur byproduct production. 4.14 An electric utility, expecting to increase its system capacity by 400 megawatts, must choose between a high-technology combined-cycle plant, at a cost of $1800 per kW of installed capacity, and an oil-burning steam plant, at $1150 per kW. The combined-cycle plant has a variable cost of 18 mills per kW-hr, while the oil-burning plant variable cost is 39 mills per kW-hr. For an annual plant factor of 0.6 and fixed charges of 15% of the capital cost, determine (a) the total annual cost of operation of each plant, and (b) the cost of electricity, in cents per kW-hr, for each plant. 4.15 Estimate the mass-flow rate of makeup water required by an evaporative cooling tower satisfying the cooling requirements of the example power plant of Section 4.6. 4.16 An 800-MW steam power plant operates at a heat rate of 8700 Btu/kW-hr. It has a 16°F rise in condenser cooling-water temperature. Neglecting energy losses, estimate the condenser cooling-water flow rate and the flow-rate of cooling-tower makeup water. Estimate the amount of pump power required to circulate the cooling water to the cooling tower. 4.17 Plot a curve of condenser cooling-water flow rate and makeup-water flow rate as a function of condenser cooling-water temperature rise for Riverside Station Unit #1. 4.18 The average temperature in an 800 ft. high power plant exhaust gas stack is 350°F and the ambient temperature is 60°F. Neglecting fluid friction and exhaust gas kinetic energy, estimate the pressure inside the base of the stack. 4.19 The average temperature in a 200 meter power plant exhaust gas stack is 150°C, and the ambient temperature is 20°C. Neglecting fluid friction and exhaust gas momentum, estimate the pressure, in kPa, at the inside of the base of the stack.

168 4.20 Estimate the heat transfer rates in the Riverside Station Unit #1 in the air preheater, an economizer that heats liquid water to saturation, the boiling surfaces, the reheater, and the superheater. Estimate the temperature drops in the combustion gas across each of these, assuming that they are arranged in the same order as just listed. 4.21 After completing Exercise 4.20, estimate the flow area of combustion gas through a crossflow economizer in the Riverside Station Unit #1, and define a suitable design. 4.22 After completing Exercise 4.20, estimate the flow area of combustion gas through a pendant superheater consisting of parallel U-tubes in cross flow, and define a suitable design. 4.23 The following is a list of ten air and gas path components of a steam power plant. Number the components so as to put them in order, starting with the air into the plant as number 1 and concluding with the flue gas out as number 10. _________

superheater

__________

boiler tubewall

_________

economizer

__________

windbox

_________

induced-draft fan

__________

air heater, gas side

_________

burner

__________

electrostatic precipitator

_________

forced-draft fan

__________

air heater, air side

4.24 Upon hiring on with Hot Stuff Engineering Company after graduation, you purchase a $30,000 automobile to establish an image as a prosperous engineer. You pay no money down, but 1% interest per month, compounded monthly, for four years. What are your monthly payments? What will your payments be if you are paying simple interest? 4.25 A forced-draft fan with an efficiency of 70% supplies 1,000,000 ft3 per minute of air to a furnace that produces a pressure drop of 0.7 psia. What is the fan power requirement, in horsepower and in kilowatts?

169

CHAPTER FIVE GAS TURBINES AND JET ENGINES

5.1 Introduction History records over a century and a half of interest in and work on the gas turbine. However, the history of the gas turbine as a viable energy conversion device began with Frank Whittle's patent award on the jet engine in 1930 and his static test of a jet engine in 1937. Shortly thereafter, in 1939, Hans von Ohain led a German demonstration of jet-engine-powered flight, and the Brown Boveri company introduced a 4-MW gas-turbine-driven electrical power system in Neuchatel, Switzerland. The success of the gas turbine in replacing the reciprocating engine as a power plant for high-speed aircraft is well known. The development of the gas turbine was less rapid as a stationary power plant in competition with steam for the generation of electricity and with the spark-ignition and diesel engines in transportation and stationary applications. Nevertheless, applications of gas turbines are now growing at a rapid pace as research and development produces performance and reliability increases and economic benefits. 5.2 An Ideal Simple-Cycle Gas Turbine The fundamental thermodynamic cycle on which gas turbine engines are based is called the Brayton Cycle or Joule cycle. A temperature-entropy diagram for this ideal cycle and its implementation as a closed-cycle gas turbine is shown in Figure 5.1. The cycle consists of an isentropic compression of the gas from state 1 to state 2; a constant pressure heat addition to state 3; an isentropic expansion to state 4, in which work is done; and an isobaric closure of the cycle back to state 1. As Figure 5.1(a) shows, a compressor is connected to a turbine by a rotating shaft. The shaft transmits the power necessary to drive the compressor and delivers the balance to a power-utilizing load, such as an electrical generator. The turbine is similar in concept and in many features to the steam turbines discusssed earlier, except that it is designed to extract power from a flowing hot gas rather than from water vapor. It is important to recognize at the outset that the term "gas turbine" has a dual usage: It designates both the entire engine and the device that drives the compressor and the load. It should be clear from the context which meaning is intended. The equivalent term “combustion turbine” is also used occasionally, with the same ambiguity.

170

Like the simple Rankine-cycle power plant, the gas turbine may be thought of as a device that operates between two pressure levels, as shown in Figure 5.1(b). The compressor raises the pressure and temperature of the incoming gas to the levels of p2 and T2. Expansion of the gas through the turbine back to the lower pressure at this point would be useless, because all the work produced in the expansion would be required to drive the compressor.

171 Instead, it is necessary to add heat and thus raise the temperature of the gas before expanding it in the turbine. This is achieved in the heater by heat transfer from an external source that raises the gas temperature to T3, the turbine inlet temperature. Expansion of the hot gas through the turbine then delivers work in excess of that needed to drive the compressor. The turbine work exceeds the compressor requirement because the enthalpy differences, and hence the temperature differences, along isentropes connecting lines of constant pressure increase with increasing entropy (and temperature), as the figure suggests. The difference between the turbine work, Wt, and the magnitude of the compressor work, |Wc|, is the net work of the cycle. The net work delivered at the output shaft may be used to drive an electric generator, to power a process compressor, turn an airplane propeller, or to provide mechanical power for some other useful activity. In the closed-cycle gas turbine, the heater is a furnace in which combustion gases or a nuclear source transfer heat to the working fluid through thermally conducting tubes. It is sometimes useful to distinguish between internal and external combustion engines by whether combustion occurs in the working fluid or in an area separate from the working fluid, but in thermal contact with it. The combustion-heated, closed-cycle gas turbine is an example, like the steam power plant, of an external combustion engine. This is in contrast to internal combustion engines, such as automotive engines, in which combustion takes place in the working fluid confined between a cylinder and a piston, and in open-cycle gas turbines. 5.3 Analysis of the Ideal Cycle The Air Standard cycle analysis is used here to review analytical techniques and to provide quantitative insights into the performance of an ideal-cycle engine. Air Standard cycle analysis treats the working fluid as a calorically perfect gas, that is, a perfect gas with constant specific heats evaluated at room temperature. In Air Standard cycle analysis the heat capacities used are those for air. A gas turbine cycle is usually defined in terms of the compressor inlet pressure and temperature, p1 and T1, the compressor pressure ratio, r = p2/p1, and the turbine inlet temperature, T3, where the subscripts correspond to states identified in Figure 5.1. Starting with the compressor, its exit pressure is determined as the product of p1 and the compressor pressure ratio. The compressor exit temperature may then be determined by the familiar relation for an isentropic process in an ideal gas, Equation (1.19): T2 = T1( p2 /p1)(k–1)/k

[R | K]

(5.1)

For the two isobaric processes, p2 = p3 and p4 = p1. Thus the turbine pressure ratio, p3/p4, is equal to the compressor pressure ratio, r = p2 /p1. With the turbine inlet temperature T3 known, the turbine discharge temperature can be determined from T4 = T3/( p2/p1)(k–1)/k

[R | K]

(5.2)

172 and the temperatures and pressures are then known at all the significant states. Next, taking a control volume around the compressor, we determine the shaft work required by the compressor, wc, assuming negligible heat losses, by applying the steady-flow energy equation: 0 = h2 – h1 + wc or wc = h1 – h2 = cp( T1 – T2)

[Btu/lbm | kJ/kg]

(5.3)

Similarly, for the turbine, the turbine work produced is wt = h3 - h4 = cp ( T3 – T4)

[Btu/lbm | kJ/kg]

(5.4)

The net work is then wn = wt + wc = cp ( T3 – T4 + T1 – T2)

[Btu/lbm | kJ/kg]

(5.5)

Now taking the control volume about the heater, we find that the heat addition per unit mass is qa = h3 – h2 = cp ( T3 – T2)

[Btu/lbm | kJ/kg]

(5.6)

The cycle thermal efficiency is the ratio of the net work to the heat supplied to the heater:

th = wn /qa

[dl]

(5.7)

which by substitution of Equations (5.1), (5.2), (5.5), and (5.6) may be simplified to

th = 1 – ( p2/p1) – (k–1)/k

[dl]

(5.8)

It is evident from Equation (5.8) that increasing the compressor pressure ratio increases thermal efficiency. Another parameter of great importance to the gas turbine is the work ratio, wt /|wc|. This parameter should be as large as possible, because a large amount of the power delivered by the turbine is required to drive the compressor, and because the engine net work depends on the excess of the turbine work over the compressor work. A little algebra will show that the work ratio wt /|wc| can be written as: wt /|wc| = (T3 /T1) / ( p2 /p1) (k–1)/k [dl]

(5.9)

173

Note that, for the ideal cycle, the thermal efficiency and the work ratio depend on only two independent parameters, the compressor pressure ratio and the ratio of the turbine and compressor inlet temperatures. It will be seen that these two design parameters are of utmost importance for all gas turbine engines. Equation (5.9) shows that the work ratio increases in direct proportion to the ratio T3 /T1 and inversely with a power of the pressure ratio. On the other hand, Equation (5.8) shows that thermal efficiency increases with increased pressure ratio. Thus, the desirability of high turbine inlet temperature and the necessity of a tradeoff involving pressure ratio is clear. Equation (5.9) also suggests that increases in the ratio T3 /T1 allow the compressor pressure ratio to be increased without reducing the work ratio. This is indicative of the historic trend by which advances in materials allow higher turbine inlet temperatures and therefore higher compressor pressure ratios. It was shown in Chapter 1 that the area of a reversible cycle plotted on a T–s diagram gives the net work of the cycle. With this in mind, it is interesting to consider a family of cycles in which the compressor inlet state, a, and turbine inlet temperatures are fixed, as shown in Figure 5.2. As the compressor pressure ratio pb/pa approaches 1, the cycle area and hence the net work approach 0, as suggested by the shaded cycle labeled with single primes. At the other extreme, as the compressor pressure ratio approaches its maximum value, the net work also approaches 0, as in the cycle denoted by double primes. For intermediate pressure ratios, the net work is large and positive, indicating that there is a unique value of compressor pressure ratio that maximizes the net work. Such information is of great significance in gas turbine design,

174

because it indicates the pressure ratio that yields the highest power output for given turbomachine inlet temperatures and mass flow rate. This is an important approach to the pressure ratio tradeoff mentioned earlier. It will be considered from an analytic viewpoint for a more realistic gas turbine model in a later section. Up to this point the discussion has focused on the closed-cycle gas turbine, an external combustion or nuclear-heated machine that operates with a circulating fixed mass of working fluid in a true cyclic process. In fact, the same Air Standard cycle analysis may be applied to the open-cycle gas turbine. The open cycle operates with atmospheric air that is pressurized by the compressor and then flows into a combustion chamber, where it oxidizes a hydrocarbon fuel to produce a hot gas that drives the turbine. The turbine delivers work as in the closed cycle, but the exiting combustion gases pass into the atmosphere, as they must in all combustion processes. A diagram of the cycle implementation is shown in Figure 5.3. Clearly, the opencycle gas turbine is an internal combustion engine, like the automotive engine. Note that the diagram is consistent with Figure 5.1 and all the preceding equations in this chapter. This is true because (1) the atmosphere serves as an almost infinite source and sink that may be thought of as closing the cycle, and (2) the energy released by combustion has the same effect as the addition of external heat in raising the temperature of the gas to the turbine inlet temperature. A cutaway of an open-cycle utility gas turbine is presented in Figure 5.4.

175

5.4 Realistic Simple-Cycle Gas Turbine Analysis The preceding analysis of the Air Standard cycle assumes perfect turbomachinery, an unachievable but meaningful ideal, and room-temperature heat capacities. Realistic quantitative performance information can be obtained by taking into account efficiencies of the compressor and the turbine, significant pressure losses, and more realistic thermal properties. Properties for Gas Turbine Analysis It is pointed out in reference 1 that accurate gas turbine analyses may be performed using constant heat capacities for both air and combustion gases. This appears to be a specialization of a method devised by Whittle (ref. 4). The following properties are therefore adopted for all gas turbine analyses in this book: Air: cp = 0.24 Btu/lbm-R or 1.004 kJ /kg-K k = 1.4 implies k /(k – 1) = 3.5 Combustion gas: cp,g = 0.2744 Btu/lbm-R

or

1.148 kJ /kg-K

kg = 1.333 implies kg/(kg – 1) = 4.0 The properties labeled as combustion gas above are actually high-temperature-air properties. Because of the high air-fuel ratio required by gas turbines, the

176 thermodynamic properties of gas turbine combustion gases usually differ little from those of high-temperature air. Thus the results given below apply equally well to closed-cycle machines using air as the working fluid and to open-cycle engines. Analysis of the Open Simple-Cycle Gas Turbine A simple-cycle gas turbine has one turbine driving one compressor and a power-consuming load. More complex configurations are discussed later. It is assumed that the compressor inlet state, the compressor pressure ratio, and the turbine inlet temperature are known, as before. The turbine inlet temperature is usually determined by the limitations of the high-temperature turbine blade material. Special metals or ceramics are usually selected for their ability to withstand both high stress at elevated temperature and erosion and corrosion caused by undesirable components of the fuel. As shown in Figure 5.3, air enters the compressor at a state defined by T1 and p1. The compressor exit pressure, p2, is given by p2 = rp1

[lbf /ft2 | kPa]

(5.10)

where r is the compressor pressure ratio. The ideal compressor discharge temperature, T2s is given by the isentropic relation T2s = T1 r (k–1)/k

[R | K]

(5.11)

The compressor isentropic efficiency, defined as the ratio of the compressor isentropic work to the actual compressor work with both starting at the same initial state and ending at the same pressure level, may be written as

c = ( h1 – h2s )/( h1 – h2 ) = ( T1 – T2s )/( T1 – T2 )

[dl]

(5.12)

Here the steady-flow energy equation has been applied to obtain expressions for the work for an irreversible adiabatic compressor in the denominator and for an isentropic compressor in the numerator. Solving Equation (5.12) for T2, we get as the actual compressor discharge temperature: T2 = T1 + ( T2s – T1 ) /  c

[R | K]

(5.13)

Equation (5.3) then gives the work needed by the compressor, wc: wc = cp ( T1 – T2 ) = cp ( T1 – T2s )/ c

[Btu /lbm | kJ/kg]

(5.14)

Note that the compressor work is negative, as required by the sign convention that defines work as positive if it is produced by the control volume. The compressor power requirement is, of course, then given by mawc [Btu/hr | kW], where ma is the

177 compressor mass flow rate [lbm / hr | kg / s]. After leaving the compressor at an elevated pressure and temperature, the air then enters the combustion chamber, where it completely oxidizes a liquid or a gaseous fuel injected under pressure. The combustion process raises the combustion gas temperature to the turbine inlet temperature T3. One of the goals of combustion chamber design is to minimize the pressure loss from the compressor to the turbine. Ideally, then, p3 = p2, as assumed by the Air Standard analysis. More realistically, a fixed value of the combustor fractional pressure loss, fpl, (perhaps about 0.05 or 5%) may be used to account for burner losses: fpl = (p2 – p3)/p2

[dl]

(5.15)

Then the turbine inlet pressure may be determined from p3 = (1 – fpl) p2

[lbf /ft2 | kPa]

(5.16)

Rather than deal with its complexities, we may view the combustion process simply as one in which heat released by exothermic chemical reaction raises the temperature of combustion gas (with hot-air properties) to the turbine inlet temperature. The rate of heat released by the combustion process may then be expressed as: Qa = ma(1 + f )cp,g(T3 – T2)

[Btu/hr | kW]

(5.17)

where f is the mass fuel-air ratio. The term ma(1 + f) is seen to be the sum of the air and fuel mass flow rates, which also equals the mass flow rate of combustion gas. For gas turbines it will be seen later that f is usually much less than the stoichiometric fuel-air ratio and is often neglected with respect to 1 in preliminary analyses. The turbine in the open-cycle engine operates between the pressure p3 and atmospheric pressure, p4 = p1, with an inlet temperature of T3. If the turbine were isentropic, the discharge temperature would be T4s = T3( p4 /p3 ) (kg– 1) / kg

[R | K]

(5.18)

From the steady-flow energy equation, the turbine work can be written as wt = cp,g (T3 – T4 ) = t cp,g( T3 – T4s)

[Btu/lbm | kJ/kg]

(5.19)

referenced to unit mass of combustion gas, and where t is the turbine isentropic efficiency. The turbine power output is then ma(1 + f)wt, where, as seen earlier, ma(1 + f) is the mass flow rate of combustion gas flowing through the turbine. The net work based on the mass of air processed and the net power output of the gas turbine, Pn, are then given by

178 wn = (1 + f )wt + wc

[Btu/lbm air| kJ/kg air]

(5.20)

[Btu/hr | kW]

(5.21)

and Pn = ma [(1 + f )wt + wc ]

and the thermal efficiency of the engine is

th = Pn /Qa

[dl]

(5.22)

EXAMPLE 5.1

A simple-cycle gas turbine has 86% and 89% compressor and turbine efficiencies, respectively, a compressor pressure ratio of 6, a 4% fractional pressure drop in the combustor, and a turbine inlet temperature of 1400°F. Ambient conditions are 60°F and one atmosphere. Determine the net work, thermal efficiency, and work ratio for the engine. Assume that the fuel-mass flow rate is negligible compared with the air flow rate. Solution

The notation for the solution is that of Figure 5.3. The solution details are given in Table 5.1 in a step-by-step spreadsheet format. Each line presents the parameter name, symbol, and units of measure; its value; and the right-hand side of its specific determining equation.

179 When an entire cycle is to be analyzed, it is best to start at the compressor with the inlet conditions and proceed to calculate successive data in the clockwise direction on the T-s diagram. The compressor isentropic and actual discharge temperatures and work are determined first using Equations (5.11), (5.13), and (5.14). The turbine pressure ratio is determined next, accounting for the combustor pressure loss, using Equation (5.16). The isentropic relation, Equation (5.18), gives the isentropic turbine exit temperature, and the turbine efficiency and Equation (5.19) yields the true turbine exit temperature and work. Once all the turbomachine inlet and exit temperatures are known, other cycle parameters are easily determined, such as the combustor heat transfer, net work, thermal efficiency, and work ratio. _____________________________________________________________________ An important observation may be made on the basis of this analysis regarding the magnitude of the compressor work with respect to the turbine work. Much of the turbine work is required to drive the compressor. Compare the work ratio of 1.66, for example, with the much higher values for the steam cycles of Chapter 2 (the Rankinecycle pumps have the same function there as the compressor here). Example 2.4 for the Rankine cycle with a 90% turbine efficiency has a work ratio of 77.2. Thus the gas turbine’s pressurization handicap relative to the Rankine cycle is substantial. The unimpressive value of the thermal efficiency of the example gas turbine, 25% (not typical of the current state of the art) compares with a Carnot efficiency for the same cycle temperature extremes of 72% The large amount of compressor work required clearly contributes to this weak performance. Nevertheless, current gas turbines are competitive with many other engines on an efficiency basis, and have advantages such as compactness and quick-start capability relative to Rankine cycle power plants. One approach to the improvement of thermal efficiency of the gas turbine will be addressed later in Section 5.5. First let’s look at what can be done about gas turbine work. Maximizing the Net Work of the Cycle Using Equations (5.14) and (5.19), we can rewrite the cycle net work as wn = t cp,g ( T3 – T4s) – cp( T2s – T1 )/c

[Btu/lbm | kJ/kg]

(5.23)

where the fuel-air ratio has been neglected with respect to 1. In the following, the combustor pressure losses and the distinction between hot-gas and air heat capacities will be neglected but the very important turbomachine efficiencies are retained. Nondimensionalizing the net work with the constant cpT1 we get: wn/cpT1 = t (T3 /T1)(cp,g/cp)(1 – r – (k–1)/k) + (1 – r (k–1)/k)/c

[dl]

(5.24)

By differentiating wn with respect to the compressor pressure ratio r and setting the

180

result equal to 0, we obtain an equation for r*, the value of r that maximizes the net work with fixed turbomachine efficiencies and with a constant ratio of the temperatures of the turbomachine inlets, T3 /T1. For constant gas properties throughout, the result is r* = (c t T3/T1)k/2(k–1)

[dl]

(5.25)

This relation gives a specific value for the compressor pressure ratio that defines an optimum cycle, in the sense of the discussion of Figure 5.2. There it was established qualitatively that a cycle with maximum net work exists for a given value of T3/T1. Equation (5.25) defines the condition for this maximum and generalizes it to include turbomachine inefficiency. The pressure ratio r* given by Equation (5.25) increases with increasing turbomachine efficiencies and with T3/T1. This is a clear indicator that increasing turbine inlet temperature favors designs with higher compressor pressure ratios. This information is important to the gas turbine designer but does not tell the whole design story. There are other important considerations; for example, (1) compressors and turbines become more expensive with increasing pressure ratios, and (2) the pressure ratio that maximizes thermal efficiency is different from that given by Equation (5.25). Figure 5.5 shows the influence of compressor pressure ratio on both efficiency and net work and the position of the value given by Equation (5.25). Thus, when all factors are taken into account, the final design pressure ratio is likely to be in the vicinity of, but not necessarily identical to, r*.

181 5.5 Regenerative Gas Turbines It was shown in Chapter 2 that the efficiency of the Rankine cycle could be improved by an internal transfer of heat that reduces the magnitude of external heat addition, a feature known as regeneration. It was also seen in Chapter 2 that this is accomplished conveniently in a steam power plant by using a heat exchanger known as a feedwater heater. Examination of Example 5.1 shows that a similar opportunity exists for the gas turbine cycle. The results show that the combustion process heats the incoming air from 924°R to 1860°R and that the gas turbine exhausts to the atmosphere at 1273°R. Thus a maximum temperature potential of 1273 – 924 = 349°F exists for heat transfer. As in the Rankine cycle, this potential for regeneration can be exploited by incorporation of a heat exchanger. Figure 5.6 shows a gas turbine with a counterflow heat exchanger that extracts heat from the turbine exhaust gas to preheat the compressor discharge air to Tc ahead of the combustor. As a result, the temperature rise in the combustor is reduced to T3 – Tc, a reduction reflected in a direct decrease in fuel consumed. Note that the compressor and turbine inlet and exit states can be the same as for a simple cycle. In this case the compressor, turbine, and net work as well as the work ratio are unchanged by incorporating a heat exchanger. The effectiveness of the heat exchanger, or regenerator, is a measure of how well it uses the available temperature potential to raise the temperature of the compressor discharge air. Specifically, it is the actual rate of heat transferred to the air divided by the maximum possible heat transfer rate that would exist if the heat exchanger had infinite heat transfer surface area. The actual heat transfer rate to the air is mcp(Tc – T2), and the maximum possible rate is mcp(T4 – T2). Thus the regenerator effectiveness can be written as

reg = ( Tc – T2 )/( T4 – T2 )

[dl]

(5.26)

and the combustor inlet temperature can be written as Tc = T2 + reg( T4 – T2 )

[R | K]

(5.27)

It is seen that the combustor inlet temperature varies from T2 to T4 as the regenerator effectiveness varies from 0 to 1. The regenerator effectiveness increases as its heat transfer area increases. Increased heat transfer area allows the cold fluid to absorb more heat from the hot fluid and therefore leave the exchanger with a higher Tc. On the other hand, increased heat transfer area implies increased pressure losses on both air and gas sides of the heat exchanger, which in turn reduces the turbine pressure ratio and therefore the turbine work. Thus, increased regenerator effectiveness implies a tradeoff, not only with pressure losses but with increased heat exchanger size and complexity and, therefore, increased cost.

182

The exhaust gas temperature at the exit of the heat exchanger may be determined by applying the steady-flow energy equation to the regenerator. Assuming that the heat exchanger is adiabatic and that the mass flow of fuel is negligible compared with the air flow, and noting that no shaft work is involved, we may write the steady-flow energy equation for two inlets and two exits as q = 0 = he + hc – h2 – h4 + w = cp,gTe + cpTc – cpT2 – cp,gT4 + 0 Thus the regenerator combustion-gas-side exit temperature is: Te = T4 – (cp/cp,g)( Tc – T2 )

[R | K]

(5.28)

While the regenerator effectiveness does not appear explicitly in Equation (5.28), the engine exhaust temperature is reduced in proportion to the air temperature rise in the regenerator, which is in turn proportional to the effectiveness. The dependence of the exhaust temperature on reg may be seen directly by eliminating Tc from Equation (5.28), using Equation (5.27) to obtain T4 – Te = reg (cp/cp,g)(T4 – T2)

[R | K]

(5.29)

183

The regenerator exhaust gas temperature reduction, T4 – Te, is seen to be jointly proportional to the effectiveness and to the maximum temperature potential, T4 – T2. The regenerator, like other heat exchangers, is designed to have minimal pressure losses on both air and gas sides. These may be taken into account by the fractional pressure drop approach discussed in connection with the combustor. EXAMPLE 5.2

Let’s say we are adding a heat exchanger with an effectiveness of 75% to the engine studied in Example 5.1. Assume that the same frictional pressure loss factor applies to both the heat exchanger air-side and combustor as a unit, and that gas-side pressure loss in the heat exchanger is negligible. Evaluate the performance of the modified engine. Solution

The solution in spreadsheet format, expressed in terms of the notation of Figures 5.6 and 5.7, is shown in Table 5.2. Examination of the spreadsheet and of the T-s diagram in Figure 5.7 shows that the entry and exit states of the turbomachines are not influenced by the addition of the heat exchanger, as expected. (There would have been a slight influence if a different pressure loss model had been assumed.) With the heat exchanger, it is seen that the combustor inlet temperature has increased about 262° and the exhaust temperature reduced 229°. The net work and

184

work ratio are clearly unchanged. Most importantly, however, the thermal efficiency has increased 10 percentage counts over the simple cycle case in Example 5.1. Such a gain must be traded off against the added volume, weight, and expense of the regenerator. The efficiency gain and the associated penalties may be acceptable in stationary power and ground and marine transportation applications, but are seldom feasible in aerospace applications. Each case, of course, must be judged on its own merits. _____________________________________________________________________ Figure 5.8 shows the influence of regenerator effectiveness and turbine inlet temperature on the performance of the gas turbine, all other conditions being the same as in the example. The values for reg = 0 correspond to a gas turbine without regenerator. The abscissa is arbitrarily truncated at reg = 0.8 because gas turbine heat exchanger effectivenesses usually do not exceed that value. The impressive influence of both design parameters is a strong motivator for research in heat exchangers and hightemperature materials. The use of regeneration in automotive gas turbines is virtually mandated because good fuel economy is so important.

185

Figure 5.9 shows the layout of a regenerative gas turbine serving a pipeline compressor station. Gas drawn from the pipeline may be used to provide the fuel for remotely located gas-turbine-powered compressor stations. (A later figure, Figure 5.12, shows details of the turbomachinery of this gas turbine.)

186 5.6 Two-Shaft Gas Turbines Problems in the design of turbomachinery for gas turbines and in poor part-load or off-design performance are sometimes avoided by employing a two-shaft gas turbine, in which the compressor is driven by one turbine and the load by a second turbine. Both shafts may be contained in a single structure, or the turbines may be separately packaged. Figure 5.10 shows the flow and T-s diagrams for such a configuration. The turbine that drives the compressor is called the compressor turbine. The compressor, combustor, compressor-turbine combination is called the gas generator, or gasifier, because its function is to provide hot, high-pressure gas to drive the second turbine, the power turbine. The compressor-turbine is sometimes also referred to as the gasifier turbine or gas-generator turbine. The analysis of the two-shaft gas turbine is similar to that of the single shaft machine, except in the determination of the turbine pressure ratios. The pressure rise produced by the compressor must be shared between the two turbines. The manner in which it is shared is determined by a power, or work, condition. The work condition expresses mathematically the fact that the work produced by the gasifier turbine is used to drive the compressor alone. As a result, the gas generator turbine pressure ratio, p3/p4, is just high enough to satisfy the compressor work requirement. Thus the compressor power (work) input is the same as the delivered gasgenerator turbine power(work) output: |wc| = mech (1 + f )wt

[Btu /lbm | kJ/kg]

(5.30)

where f is the fuel–air ratio and mech is the mechanical efficiency of transmission of power from the turbine to the compressor. The mechanical efficiency is usually close to unity in a well-designed gas turbine. For this reason, it was not included in earlier analyses. The gasifier turbine work may be written in terms of the turbine pressure ratio: wt = t cp,gT3( 1 – T4s /T3 ) = t cp,gT3[1 – 1/( p3/p4)(kg-1)/kg]

[Btu /lbm | kJ/kg]

(5.31)

With the compressor work determined, as before, by the compressor pressure ratio and the isentropic efficiency, the compressor-turbine pressure ratio, p3/p4, is obtained by combining Equations (5.30) and (5.31): p3/p4 = [ 1 – |wc| /mecht cp,g (1 + f )T3]–kg/(kg-1) = ( 1 – wf )–kg/(kg-1)

[dl]

(5.32)

where wf is the positive, dimensionless work factor, |wc| /mecht cp,g (1 + f )T3, used as

187

a convenient intermediate variable. The power turbine pressure ratio may then be determined from the identity p4/p5 = p4/p1 = ( p4/p3)( p3/p2)( p2/p1). This shows that the power turbine pressure ratio is the compressor pressure ratio divided by the gasifier turbine pressure ratio when there is no combustion chamber pressure loss ( p3 = p2). With the pressure ratios known, all the significant temperatures and performance parameters may be determined. EXAMPLE 5.3

Let’s consider a two-shaft gas turbine with a regenerative air heater. The compressor pressure ratio is 6, and the compressor and gas generator turbine inlet temperatures are 520°R and 1860°R, respectively. The compressor, gasifier turbine, and power turbine isentropic efficiencies are 0.86, 0.89, and 0.89, respectively. The regenerator effectiveness is 75%, and a 4% pressure loss is shared by the high-pressure air side of the regenerator and the combustor. Determine the pressure ratios of the two turbines, and the net work, thermal efficiency, and work ratio of the engine.

188

Solution

The solution in spreadsheet form shown in Table 5.3 follows the notation of Figure 5.10. The solution proceeds as before, until the calculation of the turbine pressure ratios. The available pressure ratio shared by the two turbines is p3/p5 = p3/p1 = (p2/p1)(p3/p2) = r ( 1 – fpl) = 5.76. The gasifier turbine pressure ratio is determined by the work-matching requirement of the compressor and its driving turbine, as expressed in Equation (5.32), using the dimensionless compressor work factor, wf. The resulting gas generator and power turbine pressure ratios are 2.61 and 2.2, respectively. Comparison shows that the design point performance of the two-shaft gas turbine studied here is not significantly different from that of the single-shaft machine considered an Example 5.2. While the performance of the two machines is found to be essentially the same, the single-shaft machine is sometimes preferred in applications

189

with fixed operating conditions where good part-load performance over a range of speeds is not important. On the other hand, the independence of the speeds of the gas generator and power turbine in the two-shaft engine allows acceptable performance over a wider range of operating conditions. _____________________________________________________________________ Let us examine further the characteristics of regenerative two-shaft gas turbines, starting with the spreadsheet reproduced in Table 5.3. By copying the value column of that spreadsheet to several columns to the right, a family of calculations with identical methodologies may be performed.. The spreadsheet /EDIT-FILL command may then be used to vary a parameter in a given row by creating a sequence of numbers with a specified starting value and interval. Such a parametric study of the influence of compressor pressure ratio on two-shaft regenerative gas turbine performance is shown in Table 5.4, where the pressure ratio is varied from 2 to 7. The fifth numeric column contains the values from Table 5.3. The data of Table 5.4 are included in the Example

190 5-4.wk3 spreadsheet that accompanies this text. Table 5.4 shows that, for the given turbine inlet temperature, the thermal efficiency maximum is at a pressure ratio between 4 and 5, while the net work maximum is at a pressure ratio of about 7. The work ratio is continually declining because the magnitude of the compressor work requirement grows faster with compressor pressure ratio than the turbine work does. Figure 5.11, plotted using the spreadsheet, compares the performance of the regenerative two-shaft gas turbine with a nonregenerative two-shaft engine (reg = 0 ). Net work for both machines has the same variation with pressure ratio. But notice the high efficiency attained with a low-compressor pressure ratio, a significant advantage attributable to regeneration. A cutaway view of a two-shaft regenerative gas turbine of the type used in pipeline compressor stations such as that shown in Figure 5.9 is seen in Figure 5.12. The figure shows that the compressor blade heights decrease in the direction of flow as the gas is compressed. The exhaust from the last of the sixteen compressor stages is reduced in velocity by a diffusing passage and then exits through the right window-like flange, which connects to a duct (not shown) leading to the regenerator. The heated air from the external regenerator reenters the machine combustor casing, where it flows around and into the combustor cans, cooling them. The air entering near the combustor fuel nozzles mixes with the fuel and burns locally in a near-stoichiometric mixture. As the mixture flows downstream, additional secondary air entering the combustor through slots in its sides mixes with, and reduces the temperature of, the combustion gas before it arrives at the turbine inlet. A cutaway view of an industrial two-shaft gas turbine with dual regenerators is presented in Figure 5.13. From the left, the air inlet and radial compressor and axial flow gasifier turbine and power turbine are seen on the axis of the machine, with the combustion chamber above and one of the rotary regenerators at the right. Due to the relatively low pressure ratios required by regenerative cycles, centrifugal compressors are normally used in regenerative machines because of their simplicity, good efficiency, compactness, and ruggedness. Performance data for the turbine of Figure 5.13 is graphed in Figure 5.14. The GT 404 gas turbine delivers about 360 brake horsepower at 2880-rpm output shaft speed. The torque-speed curve of Figure 5.14 shows an important characteristic of two-shaft gas turbines with respect to off-design point operation. Whereas the compressor pressure ratio and output torque of a single-shaft gas turbine drop as the shaft speed drops, the compressor speed and pressure ratio in a two-shaft machine is independent of the output speed. Thus, as the output shaft speed changes, the compressor may maintain its design speed and continue to develop high pressure and mass flow. Thus the torque at full stall of the output shaft of the GT404 is more than twice the full-load design torque. This high stall torque is superior to that of reciprocating engines and is important in starting and accelerating rotating equipment that has high initial turning resistance. This kind of engine may be used in truck, bus, and marine applications as well as in an industrial setting.

191

192

193

A unique patented feature of some of the Allison gas turbines, called “power transfer,” is the ability to link the duel shafts. A hydraulic clutch mechanism between the two turbine shafts acts to equalize their speeds. This tends to improve part-load fuel economy, and provides engine braking and overspeed protection for the power turbine. When the clutch mechanism is fully engaged, the shafts rotate together as a single-shaft machine. 5.7 Intercooling and Reheat Intercooling It has been pointed out that the work of compression extracts a high toll on the output of the gas turbine. The convergence of lines of constant pressure on a T-s diagram indicates that compression at low temperatures reduces compression work. The ideal compression process would occur isothermally at the lowest available temperature. Isothermal compression is difficult to execute in practice. The use of multistage compression with intercooling is a move in that direction. Consider replacing the isentropic single-stage compression from p1 to p2 = p2* in Figure 5.15 with two isentropic stages from p1 to pis and pi* to p2s. Separation of the compression processes with a heat exchanger that cools the air at Tis to a lower temperature Ti* acts to move the final compression process to the left on the T-s

194 diagram and reduces the discharge temperature following compression to T2s. A heat exchanger used to cool compressed gas between stages of compression is called an intercooler. The work required to compress from p1 to p2s = p2* = p2 in two stages is wc = cp [(T1 – Tis) + ( T1* – T2s)]

[Btu/lbm | kJ/kg]

Note that intercooling increases the net work of the reversible cycle by the area is–i*–2s–2*–is. The reduction in the work due to two-stage intercooled compression is also given by this area. Thus intercooling may be used to reduce the work of compression between two given pressures in any application. However, the favorable effect on compressor work reduction due to intercooling in the gas turbine application may be offset by the obvious increase in combustor heat addition, cp (T2* – T2s), and by increased cost of compression system. The next example considers the selection of the optimum pressure level for intercooling, pi = pis = pi*. EXAMPLE 5.4

Express the compressor work, for two-stage compression with intercooling back to the original inlet temperature, in terms of compressor efficiencies and pressure ratios. Develop relations for the compressor pressure ratios that minimize the total work of compression in terms of the overall pressure ratio. Solution

Taking Ti* = T1 as directed in the problem statement, and letting r = p2/p1, r1 = pis/p1 and r2 = p2s/pi* = r/r1 as in Figure 5.15, we get for the compression work, wc = cp [(T1 – Tis)/c1 + ( T1 – T2s)/c2 ] = cp T1[(1 – Tis /T1)/c1 + ( 1 – T2s / T1)/c2 ] = cp T1[(1 – r1(k – 1 )/k )/c1 + ( 1 – r2(k – 1 )/k )/c2 ]

[Btu /lbm | kJ/kg]

Eliminating r2, using r = r1r2, yields wc = cp T1[(1 – r1(k – 1 )/k )/c1 + ( 1 – (r/r1) (k – 1 )/k )/c2 ] Differentiating with respect to r1 for a fixed r and setting the result equal to zero, we obtain – r1– 1/k /c1 + (r (k – 1 )/k /c2) r1 – (2k – 1 )/k = 0 which simplifies to

195

r1opt = (c1/c2)k/(2k – 1) r1/2 Using this result we find also that r2opt = (c2/c1)k/(2k – 1) r1/2 Examination of these equations shows that, for compressors with equal efficiencies, both compressor stages have the same pressure ratio, which is given by the square root of the overall pressure ratio. For unequal compressor efficiencies, the compressor with the higher efficiency should have the higher pressure ratio. _____________________________________________________________________ Reheat Let us now consider an improvement at the high-temperature end of the cycle. Figure 5.16 shows the replacement of a single turbine by two turbines in series, each with appropriately lower pressure ratios, and separated by a reheater. The reheater may be a combustion chamber in which the excess oxygen in the combustion gas leaving the first turbine burns additional fuel, or it may be a heater in which external combustion provides the heat necessary to raise the temperature of the working fluid to Tm*. The high temperature at the low-pressure turbine inlet has the effect of increasing the area of the cycle by m–m*–4–4*m and hence of increasing the net work.

196 Like intercooling, the increase in net work is made possible by the spreading of the constant pressure lines on the T-s diagram as entropy increases. Thus the increase in turbine work is

wt = cp,g [(Tm* – T4) – ( Tm – T4*) ]

[Btu/lbm | kJ/kg]

(5.33)

Also as with intercooling, the favorable effect in increasing net work is offset by the reduction of cycle efficiency resulting from increased addition of external heat from the reheater: qrh = cp,g (Tm* – Tm)

[Btu/lbm | kJ/kg]

(5.34)

As with intercooling, the question arises as to how the intermediate pressure for reheat will be selected. An analysis similar to that of Example 5.4 shows the unsurprising result that the reheat pressure level should be selected so that both turbines have the same expansion ratio if they have the same efficiencies and the same inlet temperatures. Combining Intercooling, Reheat, and Regeneration Because of their unfavorable effects on thermal efficiency, intercooling and reheat alone or in combination are unlikely to be found in a gas turbine without another feature that has already been shown to have a favorable influence on gas turbine fuel economy: a regenerator. The recuperator or regenerator turns disadvantage into advantage in a cycle involving intercooling and/or reheat. Consider the cycle of Figure 5.17, which incorporates all three features. The increased turbine discharge temperature T4 produced by reheat and the decreased compressor exit temperature T2 due to intercooling both provide an enlarged temperature potential for regenerative heat transfer. Thus the heat transfer cp (Tc – T2) is accomplished by an internal transfer of heat from low pressure turbine exhaust gas. This also has the favorable effect of reducing the temperature of the gas discharged to the atmosphere. The requisite external heat addition for this engine is then qa = cp [(T3 – Tc) + ( Tm* – Tm)]

[Btu/lbm | kJ/kg]

(5.35)

Thus the combination of intercooling, reheat, and regeneration has the net effect of raising the average temperature of heat addition and lowering the average temperature of heat rejection, as prescribed by Carnot for an efficient heat engine.

197

The Ericsson cycle Increasing the number of intercoolers and reheaters without changing the overall pressure ratio may be seen to cause both the overall compression and the overall expansion to approach isothermal processes. The resulting reversible limiting cycle, consisting of two isotherms and two isobars, is called the Ericsson cycle. With perfect internal heat

198

transfer between isobaric processes, all external heat addition would be at the maximum temperature of the cycle and all heat rejected at the lowest temperature. Analysis of the limiting reversible cycle reveals, as one might expect, that its efficiency is that of the Carnot cycle. Plants with multistage compression, reheat, and regeneration can have high efficiencies; but complexity and high capital costs have resulted in few plants that actually incorporate all these features. 5.8 Gas Turbines in Aircraft –Jet engines Gas turbines are used in aircraft to produce shaft power and hot, high-pressure gas for jet propulsion. Turbine shaft power is used in turboprop aircraft and helicopters to drive propellers and rotors. A modern turboprop engine and an aircraft that uses it are shown in Figures 5.18 through 5.20. While its jet exhaust provides some thrust, the bulk of the propulsive thrust of the turboprop is provided by its propeller. The rear-

199

ward acceleration of a large air mass by the propeller is responsible for the good fuel economy of turboprop aircraft. Thus the turboprop is popular as a power plant for small business aircraft. At higher subsonic flight speeds, the conventional propeller loses efficiency and the turbojet becomes superior. Auxiliary power units, APUs, are compact gas turbines that provide mechanical power to generate electricity in transport aircraft while on the ground. The thermodynamic fundamentals of these shaft-power devices are the same those of stationary gas turbines, discussed earlier. Their design, however, places a premium on low weight and volume and conformance to other constraints associated with airborne equipment. Thus their configuration and appearance may differ substantially from those of other stationary gas turbines. The jet engine consists of a gas turbine that produces hot, high-pressure gas but has zero net shaft output. It is a gasifier. A nozzle converts the thermal energy of the hot, high-pressure gas produced by the turbine into a high-kinetic-energy exhaust stream. The high momentum and high exit pressure of the exhaust stream result in a forward thrust on the engine. Although the analysis of the jet engine is similar to that of the gas turbine, the configuration and design of jet engines differ significantly from those of most stationary gas turbines. The criteria of light weight and small volume, mentioned earlier, apply here as well. To this we can add the necessity of small frontal area to minimize the aerodynamic drag of the engine, the importance of admitting air into the

200

engine as efficiently (with as little stagnation pressure loss) as possible, and the efficient conversion of high-temperature turbine exit gas to a high-velocity nozzle exhaust. The resulting configuration is shown schematically in Figure 5.21. Up to now we have not been concerned with kinetic energy in the flows in gas turbines, because the flows at the stations of interest are usually designed to have low velocities. In the jet engine, however, high kinetic energy is present in the free stream ahead of the engine and in the nozzle exit flow. The analysis here will therefore be presented in terms of stagnation, or total, temperatures and pressures, where kinetic

201 energy is taken into account implicitly, as discussed in Section 1.7. The preceding analyses may be readily adapted to deal with the stagnation properties associated with compressible flow. In the following discussion, engine processes are first described and then analyzed. It should be recalled that if there are no losses, as in an isentropic flow, the stagnation pressure of a flow remains constant. All loss mechanisms, such as fluid friction, turbulence, and flow separation, decrease stagnation pressure. Only by doing work on the flow (with a compressor, for example) is it possible to increase stagnation pressure. In Figure 5.21, free-stream ambient air, denoted by subscript a, enters an engine inlet that is carefully designed to efficiently decelerate the air captured by its frontal area to a speed low enough to enter the compressor, at station 1, with minimal aerodynamic loss. There is stagnation pressure loss in the inlet, but efficient deceleration of the flow produces static and stagnation pressures at the compressor entrance well above the ambient free-stream static pressure. This conversion of relative kinetic energy of ambient air to increased pressure and temperature in the engine inlet is sometimes called ram effect. The compressor raises the stagnation pressure of the air further to its maximum value at station 2, using power delivered by the turbine. Fuel enters the combustion chamber and is burned with much excess air to produce the high turbine inlet temperature at station 3. We adopt here, for simplicity, the familiar idealization that no pressure losses occur in the combustion chamber. The hot gases then expand through the turbine and deliver just enough power to drive the compressor (the work condition again). The gases leave the turbine exit at station 4, still hot and at a stagnation pressure well above the ambient. These gases then expand through a nozzle that converts the excess pressure and thermal energy into a high-kinetic-energy jet at station 5. The forward thrust on the engine, according to Newton’s Second Law, is produced by the reaction to the internal forces that accelerate the internal flow rearward to a high jet velocity and the excess of the nozzle exit plane pressure over the upstream ambient pressure. Inlet Analysis Given the flight speed, Va , and the free-stream static temperature and pressure, Ta and pa, at a given altitude, the free-stream stagnation temperature and pressure are Toa = Ta + Va2 / 2cp

[R | K]

(5.36)

poa = pa( Toa/Ta ) k/(k – 1)

[lbf /ft2 | kPa]

(5.37)

and

Applying the steady-flow energy equation to the streamtube entering the inlet, we find

202 that the stagnation enthalpy hoa = ho1 for adiabatic flow. For subsonic flight and supersonic flight at Mach numbers near one, the heat capacity of the air is essentially constant. Thus constancy of the stagnation enthalpy implies constancy of the stagnation temperature. Hence, using Equation (5.36), To1 = Toa = Ta + Va2 / 2cp

[R | K]

(5.38)

The effects of friction, turbulence, and other irreversibilities in the inlet flow are represented by the inlet pressure recovery, PR, defined as PR = po1 / poa

[dl]

(5.39)

where an isentropic flow through the inlet has a pressure recovery of 1.0. Lower values indicate reduced inlet efficiency and greater losses. For subsonic flow, values on the order of 0.9 to 0.98 are typical. At supersonic speeds the pressure recovery decreases with increasing Mach number. Compressor Analysis With the stagnation conditions known at station 1 in Figure 5.21, the compressor pressure ratio, r = po2 /po1, now yields po2; and the isentropic relation, Equation (1.19), gives the isentropic temperaure, To2s: To2s = To1r(k – 1)/k

[R | K]

(5.40)

The actual compressor discharge stagnation temperature is then obtained from the definition of the compressor efficiency in terms of stagnation temperatures:

comp = ( To1 – To2s ) / ( To1 – To2 )

[dl]

(5.41)

Combustor and Turbine Analysis The turbine inlet temperature, T03, is usually assigned based on turbine blade material considerations. For preliminary analysis it may be assumed that there are negligible pressure losses in the combustion chamber, so that p03 = p02. As with the two-shaft gas turbine, the condition that the power absorbed by the compressor equal the power delivered by the turbine determines the turbine exit temperature, T04: cp(To2 – To1 ) = (1 + f )cp,gTo3 ( 1 – To4 / To3 )

[Btu | kJ]

(5.42)

where f is the engine fuel-air ratio, which often may be neglected with respect to 1 (as in our earlier studies) when high precision is not required.

203 The turbine efficiency equation then yields the isentropic discharge temperature T04s, and Equation (1.19) yields the turbine pressure ratio: To4s = To3 – ( To3 – To4) /turb

[R | K]

(5.43)

po3 /po4 = ( To3/To4s)kg/(kg – 1)

[dl]

(5.44)

Thus the stagnation pressure and temperature at station 4 are known. Note that the turbine pressure ratio is usually significantly lower than the compressor pressure ratio. Nozzle Analysis The flow is then accelerated to the jet velocity at station 5 by a convergent nozzle that contracts the flow area. A well-designed nozzle operating at its design condition has only small stagnation pressure losses. Hence the nozzle here is assumed to be loss-free and therefore isentropic. Under most flight conditions the exhaust nozzle is choked; that is, it is passing the maximum flow possible for its upstream conditions. A choked nozzle has the local flow velocity at its minimum area, or throat, equal to the local speed of sound. As a result, simple relations exist between the upstream stagnation conditions at station 4 and the choked conditions at the throat. Thus, for a choked isentropic nozzle, To4 = To5 = T5 + V5 2 / 2cp,g = T5 + kgRT5/2cp,g = T5 ( 1 + kgR/2cp,g) = T5( kg + 1 )/2

[R | K]

(5.45)

where kgR/cp,g = kg – 1. With kg = 1.333 for the combustion gas, this determines the exit temperature T5. Combining the isentropic relation with Equation (5.45) then gives the nozzle exit static pressure p5: p5 /po4 = (T5 /T04)kg/(kg – 1) = [2/(kg + 1)]kg/(kg – 1)

[dl]

(5.46)

EXAMPLE 5.5

The stagnation temperature and pressure leaving a turbine and entering a convergent nozzle are 970.2K. and 2.226 bar, respectively. What is the static pressure and temperature downstream if the nozzle is choked? If the free-stream ambient pressure is 0.54 bar, is the nozzle flow choked? Compare the existing nozzle pressure ratio with the critical pressure ratio.

204 Solution

If the nozzle is choked, then from Equations (5.45) and (5.46), T5 = 2To4/( kg + 1) = 2(970.2)/2.333 = 831.6 K and the static pressure at the nozzle throat is p5 = po4 [2/(kg + 1)]kg/(kg – 1) = 2.226( 2 / 2.333 )4 = 1.202 bar The fact that p5 > pa = 0.54 indicates that the nozzle is choked. The critical pressure ratio of the nozzle is po4 /p5 = [(kg + 1)/2]kg/(kg – 1) = (2.333/2)4 = 1.852 and the applied pressure ratio is 2.226/0.54 = 4.12. Thus the applied pressure ratio exceeds the critical pressure ratio.This also indicates that the nozzle is choked. _____________________________________________________________________ For the isentropic nozzle, the steady-flow energy equation gives 0 = h5 + V52/2 – ho4 or, with cp,g constant, V5 = [2cp,g(To4 – T5)]½

[ft/s | m/s]

(5.47)

Thus the jet velocity is determined from Equation (5.47), where T5 is obtained from Equation (5.45). The thrust of the engine is obtained by applying Newton’s Second Law to a control volume, as shown in Figure 5.22. If the mass flow rate through the engine is m, the rates of momentum flow into and out of the control volume are mVa and mV5, respectively. The net force exerted by the exit pressure is (p5 – pa)A5, where A5 is the nozzle exit area. Thus, applying Newton’s Second Law to the control volume, we can relate the force exerted by the engine on the gases flowing through, F, and the net exit pressure force to the rate of increase of flow momentum produced by the engine: m(V5 – Va) = F – (p5 – pa)A5

[lbf | kN]

F = m(V5 – Va) + ( p5 – pa)A5

[lbf | kN]

or (5.48)

Here, F is the engine force acting on the gas throughflow. The reaction to this force is the thrust on the engine acting in the direction of flight. Thus the magnitude of the

205

thrust is given by Equation (5.48). It is the sum of all the pressure force components acting on the inside the engine in the direction of flight. The exit area, A5, is related to the mass flow rate by m = A5,5V5

[lbm /s | kg /s]

(5.49)

where the density at station 5 is obtained from the perfect gas law using p5 and T5 from Equations (5.45) and (5.46). If A5 is known, the mass flow rate through the engine may be determined from Equation (5.49) and the thrust from Equation (5.48). Another type of nozzle used in high-performance engines and in rocket nozzles is a convergent-divergent nozzle, one in which the flow area first contracts and then increases. It differs from the convergent nozzle in that it can have supersonic flow at the exit. For such a fully expanded, convergent-divergent nozzle operating at its design condition, p5 = pa, and the engine thrust from Equation (5.48) reduces to m(V5 – Va). Jet Engine Performance It is seen that engine thrust is proportional to the mass flow rate through the engine and to the excess of the jet velocity over the flight velocity. The specific thrust of an engine is defined as the ratio of the engine thrust to its mass flow rate. From Equation (5.48)

206 the specific thrust is F/m = (V5 – Va) + ( p5 – pa)A5/m

[lbf -s/lbm | kN-s/kg]

(5.50)

Because the engine mass flow rate is proportional to its exit area, as seen in Equation (5.49), A5/m depends only on design nozzle exit conditions. As a consequence, F/m is independent of mass flow rate and depends only on flight velocity and altitude. Assigning an engine design thrust then determines the required engine-mass flow rate and nozzle exit area and thus the engine diameter. Thus the specific thrust, F/m, is an important engine design parameter for scaling engine size with required thrust at given flight conditions. Another important engine design parameter is the thrust specific fuel consumption, TSFC, the ratio of the mass rate of fuel consumption to the engine thrust TSFC = mf /F

[lbm / lbf-s | kg / kN-s ]

(5.51)

Low values of TSFC, of course, are favorable. The distance an aircraft can fly without refueling, called its range, is inversely proportional to the TSFC of its engines. The following example demonstrates the evaluation of these parameters. EXAMPLE 5.6

An aircraft flies at a speed of 250 m/s at an altitude of 5000 m. The engines operate at a compressor pressure ratio of 8, with a turbine inlet temperature of 1200K. The compressor and turbine efficiencies are 0.9 and 0.87, respectively, and there is a 4% pressure loss in the combustion chamber. The inlet total pressure recovery is 0.97, and the engine-mass flow rate is 100 kg/s. Use an engine mechanical efficiency of 0.99 and a fuel heating value of 43,000 kJ/kg. Assume that the engine has a convergent, isentropic, nozzle flow. Determine the nozzle exit area, the engine thrust, specific thrust, fuel flow rate, and thrust specific fuel consumption. Solution

The solution details are presented in Table 5.5 in spreadsheet form. At 5000 m altitude, the ambient static temperature and pressure are determined from standardatmosphere tables such as those given in Appendix H. The ambient stagnation pressure and temperature are then determined for the given flight speed. The stagnation temperature at the compressor entrance is the same as the free-stream value for an adiabatic inlet. The inlet pressure recovery determines the total pressure at the compressor entrance. Using the notation of Figures 5.21 and 5.22, and given the compressor pressure ratio and turbine inlet temperature, the stagnation conditions at the compressor, combustor, and turbine exits may be determined in the same way as the static conditions were determined earlier for stationary two-shaft gas turbines.

207

The calculated available nozzle pressure ratio po4/pa = 4.118 is then compared with the critical pressure ratio p04/pc = 1.852, which indicates that the convergent nozzle is choked; i.e., sonic velocity exists at the throat. The nozzle exit plane pressure must then be given by p04 divided by the critical pressure ratio. Equation (5.45) for a sonic condition then determines the nozzle exit plane temperature, and the exit plane density follows from the ideal gas law. Because the exit is choked, the exit plane temperature determines the exit velocity through the sonic velocity relation. The nozzle exit area may then be determined by using the given mass flow rate and

208 the exit velocity and density from Equation (5.49). The fuel-air ratio for the combustor is estimated from a simple application of the steady-flow energy equation to the combustor, which neglects the fuel sensible heat with respect to its heating value and assumes hot-air properties for the combustion products. The thrust, specific thrust, and TSFC are then found from Equations (5.48), (5.50), and (5.51). _____________________________________________________________________ The spreadsheet in Table 5.5 (available from Spreadsheet Examples as Example 5.6) is set up with conditional statements that treat the convergent nozzle for both the choked and subsonic exit conditions. The calculations of T5, V5, and p5 depend on whether the throat is choked or not. The format for the spreadsheet conditional statements is: Conditional test, Result if true, Result if not true The low value of fuel-air ratio, f = 0.0174, obtained in this example is typical of most gas turbines and jet engines. In comparison with the stoichiometric value of 0.068, it corresponds to an equivalence ratio of 0.256. Modern Jet Engines Full and cutaway views of a small modern jet engine used in business aircraft are shown in Figure 5.23. The engine is a turbofan engine, a type of gas turbine engine that is used in all large commercial aircraft and is gaining popularity in the business jet market. The fan referred to in the turbofan name is seen at the left in Figure 5.23(b). The incoming air splits after passing through the fan, with the flow through the outer annulus passing to a nozzle that expels it without heating. The inner-core flow leaving the fan passes through an axial flow compressor stage and a centrifugal compressor before entering the combustor, turbine stages, and exit nozzle. Thus the core flow provides the turbine power to drive the fan and the compressors. The thrust specific fuel consumption and the thrust/weight ratio of turbofans is superior to those of conventional jet engines, because a larger mass flow rate of air is processed and exits at high velocity. The lower average exit velocity of the turbofan engine (compared to turbojets) is secondary in importance to the increased total mass flow rate through the engine. A large turbofan engine designed to power Boeing 747 and 767, Airbus A310 and A300, and McDonnell-Douglas MD-11 aircraft is shown in Figure 5.24. In large turbofans and most large jet engines, axial compressors and turbines are used rather than centrifugal compressors. The axial compressors are capable of much higher pressure ratios, allow compression without turning the flow through a large angle, and have somewhat higher efficiencies than centrifugals. Large axial compressors have many axial stages and are capable of overall pressure ratios in excess of 30. Turbofans are studied in more detail in Chapter 9.

209

Afterburning The exhaust of a jet engine contains a large amount of unused oxygen because of the high air-fuel ratios necessary to limit the gas stagnation temperature to which the

210

turbine blading is exposed. This excess oxygen at the turbine exit makes it possible to burn additional fuel downstream and thereby to increase the nozzle exit temperature and jet velocity. By extending the interface between the turbine and the nozzle in a jet engine and by adding fuel spray bars to create a large combustion chamber called an afterburner, it is possible to dramatically increase the thrust of a jet engine. Much higher afterburner stagnation temperatures are allowed than those leaving the combustor, because (a) there is no highly stressed rotating machinery downstream of the afterburner, and (b) afterburner operating periods are usually limited to durations of a few minutes. Afterburners are used for thrust augmentation of jet aircraft to assist in takeoff and climb and to provide a brief high-speed-dash capability and increased maneuver thrust in military aircraft. However, the substantial fuel consumption penalty of afterburning restricts its use to brief periods of time when it is badly needed. The T-s diagram for a jet engine with afterburner seen in Figure 5.25 shows that afterburning is analogous to reheating in an open-cycle stationary gas turbine, as shown in Figure 5.16. The energy release in the afterburner at approximately constant stagnation pressure shifts the nozzle expansion process to the right on the T-s diagram. As a result, the nozzle enthalpy and temperature differences between T-s diagram

211

constant-pressure lines increase as the nozzle inlet stagnation temperature increases. This produces higher jet velocities. Using Equation (5.47) and the notation of Figure 5.16, we get for the ratio of fully expanded nozzle isentropic jet velocity with afterburning to that without: V/Vwo = [(Tom* – T4)/(Tom – T4*)]½ = (Tom* /Tom)½ [(1 – T4 /Tom*)/(1 – T4*/Tom)]½ = (Tom* /Tom)½ where the static-to-stagnation temperature ratios are eliminated using the corresponding equal isentropic pressure ratios. Thus, for example, the jet velocity ratio, and therefore the thrust ratio, for a reheat temperature ratio of 4 is about 2. The analysis of a jet engine with afterburner is illustrated in the following example. EXAMPLE 5.7

Consider the performance of the engine analyzed in Example 5.6 when an afterburner is added. Assume that heat addition in the afterburner raises the nozzle entrance stagnation temperature to 2000 K, with a 5% stagnation pressure loss in the afterburner. What is the increase in nozzle exit temperature, jet velocity, and thrust?

212

Solution

Table 5.6 is from an adaptation of the spreadsheet (Table 5.5) used for Example 5.6. The calculations are identical up to the turbine exit at station 4. Heat addition in the afterburner raises the stagnation temperature at the nozzle entrance, Tob, to 2000K, while the 5% pressure loss drops the stagnation pressure to 2.115 bar. Comparing the applied nozzle pressure ratio (pob/pa) with the critical pressure ratio (pob/pc) shows that

213 the nozzle remains choked. The remainder of the calculation follows Example 5.6 except that the fuel consumption associated with the afterburner temperature rise is taking into account in the fuel-air ratio, fuel flow rate, and specific fuel consumption. _____________________________________________________________________ Table 5.7 compares some of the performance parameters calculated in Examples 5.6 and 5.7. The table shows clearly the striking gain in thrust provided by the high nozzle exit temperature produced by afterburning and the accompanying high penalty in fuel consumption. Note, however, that even with afterburning the overall equivalence ratio 0.0448/0.068 = 0.659 is well below stoichiometric. Table 5.7

Examples 5.6 and 5.7 Compared

Parameter

Without Afterburner

With Afterburner

Nozzle entrance

970.2

2000.0

Nozzle exit

831.6

1714.3

564.1

809.9

54.7

87.99

Temperature, K

Nozzle exit velocity, m/s Thrust, kN Fuel-air Ratio

0.0174

0.0448

TSFC, kg/kN-s

0.032

0.051

Bibliography and References 1. Cohen, H., Rogers, G. F. C., and Saravanamuttoo, H. I. H., Gas Turbine Theory, 3rd Edition. New York: Longman Scientific and Technical, 1987. 2. Potter, J. H., “The Gas Turbine Cycle." ASME paper presented at the Gas Turbine Forum Dinner, ASME Annual Meeting, New York, N.Y., November 27, 1972. 3. Bathie, William W., Fundamentals of Gas Turbines. New York: Wiley, 1984. 4. Whittle, Sir Frank, Gas Turbine Aerothermodynamics. New York: Pergamon, 1981. 5. Wilson, David Gordon, The Design of High-Efficiency Turbomachinery and Gas Turbines. Cambridge, Mass.: MIT Press, 1984. 6. Chapman, Alan J., and Walker, William F., Introductory Gas Dynamics. New York: Holt, Rinehart and Winston, 1971.

214 7. Oates, Gordon C., Aerothermodynamics of Gas Turbine and Rocket Propulsion. Washington, D. C.: American Institute of Aerodynamics and Astronautics, 1988. 8. Bammert, K., and Deuster, G., “Layout and Present Status of the Closed-Cycle Helium Turbine Plant Oberhausen.” ASME paper 74-GT-132, 1974. 9. Bammert, Karl, “The Oberhausen Heat and Power Station with Helium Turbine.” Address on the Inauguration of the Helium Turbine Power Plant of EVO at Oberhausen-Sterkrade, Westdeutschen Allgemeinen Zeitung, WAZ, December 20, 1974. 10. Zenker, P., “The Oberhausen 50-MW Helium Turbine Plant,” Combustion, April 1976, pp. 21-25. 11. Weston, Kenneth C., “Gas Turbine Analysis and Design Using Interactive Computer Graphics.” Proceedings of Symposium on Applications of Computer Methods in Engineering, University of Southern California, August 23-26, 1977. 12. United States Standard Atmosphere, 1976, NOAA, NASA, and USAF, October 1976. 13. Weston, Kenneth C., “Turbofan Engine Analysis and Optimization Using Spreadsheets.” ASME Computers in Engineering Conference, Anaheim, California, July 30-August 3, 1989. 14. Kerrebrock, Jack K., Aircraft Engines and Gas Turbines. Cambridge, Mass.: MIT Press, 1983. 15. Dixon, S. L., Fluid Mechanics, Thermodynamics of Turbomachinery. New York: Pergamon Press, 1978. 16. Bammert, K., Rurik, J., and Griepentrog, H., “Highlights and Future Development of Closed-Cycle Gas Turbines.” ASME paper 74-GT-7, 1974. EXERCISES 5.1 For the Air Standard Brayton cycle, express the net work in terms of the compressor pressure ratio, r, and the turbine-to-compressor inlet temperature ratio, T3/T1. Nondimensionalize the net work with cpT1, and derive an expression for the pressure ratio that maximizes the net work for a given value of T3/T1. 5.2 For a Brayton Air Standard cycle, work out an expression for the maximum possible compressor pressure ratio for a given turbine-to-compressor inlet temperature

215 ratio. Draw and label the cycle on a T-s diagram. What is the magnitude of the net work for this cycle? Explain. 5.3 For a calorically perfect gas, write an expression for the temperature difference, T2 – T1, on an isentrope between two lines of constant pressure in terms of the initial temperature T1 and the pressure ratio p2/p1. Sketch a T-s diagram showing two different isentropes between the two pressure levels. Explain how your expression demonstrates that the work of an isentropic turbomachine operating between given pressure levels increases with temperature. 5.4 Derive an expression for the enthalpy difference, h2 – h1, along a calorically perfect gas isentrope spanning two fixed pressure levels, p2 and p1, in terms of the discharge temperature T2. Note that as T2 increases, the enthalpy difference also increases. 5.5 Derive Equation (5.9). 5.6 Derive Equations (5.24) and (5.25). 5.7 A simple-cycle stationary gas turbine has compressor and turbine efficiencies of 0.85 and 0.9, respectively, and a compressor pressure ratio of 20. Determine the work of the compressor and the turbine, the net work, the turbine exit temperature, and the thermal efficiency for 80°F ambient and 1900°F turbine inlet temperatures. 5.8 A simple-cycle stationary gas turbine has compressor and turbine efficiencies of 0.85 and 0.9, respectively, and a compressor pressure ratio of 20. Determine the work of the compressor and the turbine, the net work, the turbine exit temperature, and the thermal efficiency for 20°C ambient and 1200°C turbine inlet temperatures. 5.9 A regenerative-cycle stationary gas turbine has compressor and turbine isentropic efficiencies of 0.85 and 0.9, respectively, a regenerator effectiveness of 0.8, and a compressor pressure ratio of 5. Determine the work of the compressor and the turbine, the net work, the turbine and regenerator exit temperatures, and the thermal efficiency for 80°F ambient and 1900°F turbine inlet temperatures. Compare the efficiency of the cycle with the corresponding simple-cycle efficiency. 5.10 A regenerative-cycle stationary gas turbine has compressor and turbine isentropic efficiencies of 0.85 and 0.9, respectively, a regenerator effectiveness of 0.8, and a compressor pressure ratio of 5. Determine the work of the compressor and the turbine, the net work, the turbine and regenerator exit temperatures, and the thermal efficiency for 20°C ambient and 1200°C turbine inlet temperatures. Compare the efficiency of the cycle with the corresponding simple-cycle efficiency. 5.11 A two-shaft stationary gas turbine has isentropic efficiencies of 0.85, 0.88, and 0.9

216 for the compressor, gas generator turbine, and power turbine, respectively, and a compressor pressure ratio of 20. (a) Determine the compressor work and net work, the gas generator turbine exit temperature, and the thermal efficiency for 80°F ambient and 1900°F compressorturbine inlet temperatures. (b) Calculate and discuss the effects of adding reheat to 1900°F ahead of the power turbine. 5.12 A two-shaft stationary gas turbine has isentropic efficiencies of 0.85, 0.88, and 0.9 for the compressor, gas generator turbine, and power turbine, respectively, and a compressor pressure ratio of 20. (a) Determine the compressor work and net work, the gas generator turbine exit temperature, and the thermal efficiency for 20°C ambient and 1200°C compressorturbine inlet temperatures. (b) Calculate and discuss the effects of adding reheat to 1200°C ahead of the power turbine. 5.13 A two-shaft stationary gas turbine with an intercooler and reheater has efficiencies of 0.85, 0.88, and 0.9 for the compressor, gas-generator turbine, and power turbine, respectively, and a compressor pressure ratio of 5. (a) Determine the compressor work and net work, the gas generator turbine exit temperature, and the thermal efficiency for 80°F ambient and 1900°F turbine inlet temperatures. (b) Calculate and discuss the effect on thermal efficiency, exhaust temperature, and net work of adding a regenerator with an effectiveness of 75%. 5.14 A two-shaft stationary gas turbine with an intercooler and reheater has efficiencies of 0.85, 0.88, and 0.9 for the compressor, gas-generator turbine, and power turbine, respectively, and a compressor pressure ratio of 5. (a) Determine the compressor work and net work, the gas generator turbine exit temperature, and the thermal efficiency for 20°C ambient and 1200°C turbine inlet temperatures. (b) Calculate and discuss the effect on thermal efficiency, exhaust temperature, and net work of adding a regenerator with an effectiveness of 75%. 5.15 Consider a pulverized-coal-burning, single-shaft gas turbine in which the combustion chamber is downstream of the turbine to avoid turbine blade erosion and corrosion. The combustion gases leaving the burner heat the compressor discharge air through the intervening walls of a high temperature ceramic heat exchanger. (a) Sketch the flow and T-s diagrams for this gas turbine, showing the influence of pressure drops through the combustor and the heat exchanger. The ambient, turbine inlet, and combustor exhaust temperatures are 80°F, 1900°F, and 3000°F, respectively. The compressor pressure ratio is 5. Assume perfect turbomachinery.

217 (b) For zero pressure drops, determine the net work, the thermal efficiency, and the heat exchanger exhaust temperature. (c) If the coal has a heating value of 14,000 Btu/lbm, what is the coal consumption rate, in tons per hour, for a 50-MW plant? 5.16 Consider a pulverized-coal-burning, single-shaft gas turbine in which the combustion chamber is downstream of the turbine to avoid turbine blade erosion and corrosion. The combustion gases leaving the burner heat the compressor discharge air through the intervening walls of a high-temperature ceramic heat exchanger. (a) Sketch the flow and T-s diagrams for this gas turbine, showing the influence of pressure drops through the combustor and the heat exchanger. The ambient, turbine inlet, and combustor exhaust temperatures are 20°C, 1200°C, and 2000°C, respectively. The compressor pressure ratio is 5. Assume perfect turbomachinery. (b) For zero pressure drops, determine the net work, the thermal efficiency, and the heat exchanger exhaust temperature. (c) If the coal has a heating value of 25,000 kJ/kg, what is the coal consumption rate, in tons per hour, for a 50-MW plant? 5.17 A stationary gas turbine used to supply compressed air to a factory operates with zero external shaft load. Derive an equation for the fraction of the compressor inlet air that can be extracted ahead of the combustion chamber for process use in terms of the compressor pressure ratio, the ratio of turbine-to-compressor inlet temperatures, and the turbomachinery efficiencies. Plot the compressor mass extraction ratio as a function of compressor pressure ratio for temperature ratios of 3 and 5, perfect turbomachinery, and identical high- and low-temperature heat capacities. 5.18 A stationary gas turbine used to supply compressed air to a factory operates with zero external shaft load. Derive an equation for the fraction of the inlet air that can be extracted ahead of the combustion chamber for process use in terms of the compressor pressure ratio, the ratio of turbine-to-compressor inlet temperatures, and the turbomachinery efficiencies. What is the extraction mass flow for a machine that has a compressor pressure ratio of 10, turbomachine inlet temperatures of 1800°F and 80°F, turbomachine efficiencies of 90%, and a compressor inlet mass flow rate of 15 lbm/s? 5.19 A stationary gas turbine used to supply compressed air to a factory operates with zero external shaft load. Derive an equation for the fraction of the inlet air that can be extracted ahead of the combustion chamber for process use in terms of the compressor pressure ratio, the ratio of turbine-to-compressor inlet temperatures, and the turbomachinery efficiencies. What is the extraction mass flow for a machine that has a compressor pressure ratio of 8, turbomachine inlet temperatures of 1000°C and 25°C, turbomachine efficiencies of 90%, and a compressor inlet mass flow rate of 10 kg /s? 5.20 Do Exercise 5.18, but account for combustion chamber fractional pressure drops

218 of 3% and 5% of the burner inlet pressure. How does increased pressure loss influence process mass flow? 5.21 Do Exercise 5.19, but account for combustion chamber fractional pressure drops of 3% and 5% of the burner inlet pressure. How does increased pressure loss influence process mass flow? 5.22 A gas-turbine-driven car requires a maximum of 240 shaft horsepower. The engine is a two-shaft regenerative gas turbine with compressor, gas generator turbine, and power turbine efficiencies of 0.86, 0.9, and 0.87, respectively, and a regenerator effectiveness of 0.72. The compressor pressure ratio is 3.7, and the turbine and compressor inlet temperatures are 1800°F and 90°F, respectively. What air flow rate does the engine require? What is the automobile exhaust temperature? What are the engine fuel-air ratio and specific fuel consumption if the engine burns gaseous methane? 5.23 A gas-turbine-driven car requires a maximum of 150kW of shaft power. The engine is a two-shaft regenerative gas turbine with compressor, gas generator turbine, and power turbine efficiencies of 0.84, 0.87 and 0.9, respectively, and a regenerator effectiveness of 0.75. The compressor pressure ratio is 4.3, and the turbine and compressor inlet temperatures are 1250°C and 20°C, respectively. What air flow rate does the engine require? What is the automobile exhaust temperature? What are the engine fuel-air ratio and specific fuel consumption if the engine burns gaseous hydrogen? 5.24 A simple-cycle gas turbine is designed for a turbine inlet temperature of 1450°F, a compressor pressure ratio of 12, and compressor and turbine efficiencies of 84% and 88%, respectively. Ambient conditions are 85°F and 14.5 psia. (a) Draw and label a T-s diagram for this engine. (b) Determine the compressor, turbine, and net work for this cycle. (c) Determine the engine thermal efficiency. (d) Your supervisor has requested that you study the influence of replacing the compressor with dual compressors and an intercooler. Assume that the new compressors are identical to each other and have the same efficiencies and combined overall pressure ratio as the original compressor. Assume intercooling to 85°F with no intercooler pressure losses. Show clearly the T-s diagram for the modified system superimposed on your original diagram. Calculate the revised system net work and thermal efficiency. 5.25 Determine the air and kerosene flow rates for a 100-MW regenerative gas turbine with 1800K turbine inlet temperature, compressor pressure ratio of 5, and 1 atm. and 300K ambient conditions. The compressor and turbine efficiencies are 81% and 88%, respectively, and the heat exchanger effectiveness is 75%. Use a heating value for kerosene of 45,840 kJ/kg. What is the engine specific fuel consumption?

219 5.26 A regenerative gas turbine has compressor and turbine discharge temperatures of 350K and 700K, respectively. Draw and label a T-s diagram showing the relevant states. If the regenerator has an effectiveness of 70%, what are the combustor inlet temperature and engine exhaust temperature? 5.27 A gas turbine has a turbine inlet temperature of 1100K, a turbine pressure ratio of 6, and a turbine efficiency of 90%. What are the turbine exit temperature and the turbine work? 5.28 A two-shaft gas turbine with reheat has turbine inlet temperatures of 1500°F, a compressor pressure ratio of 16, and turbomachine efficiencies of 88% each. The compressor inlet conditions are 80°F and 1 atm. Assume that all heat capacities are 0.24 Btu/lbm-R and k = 1.4. (a) Draw T-s and flow diagrams. (b) Make a table of temperatures and pressures for all real states, in °R and atm. (c) What are the compressor work and power turbine work? (d) What is the power-turbine-to-compressor work ratio? (e) What is the cycle thermal efficiency? (f) Evaluate the recommendation to add a regenerator to the system. If a 4-count (0.04) increase in thermal efficiency can be achieved, the addition of the regenerator is considered economically feasible. Give your recommendation, supporting arguments, and substantiating quantitative data. 5.29 Consider a gas turbine with compressor and turbine inlet temperatures of 80°F and 1200°F, respectively. The turbine efficiency is 85%, and compressor pressure ratio is 8. (a) Draw coordinated T-s and plant diagrams. (b) What is the turbine work? (c) What is the minimum compressor efficiency required for the gas turbine to produce a net power output? (d) What is the thermal efficiency if the compressor efficiency is raised to 85%? 5.30 The first closed-cycle gas turbine power plant in the world using helium as a working fluid is a 50-MW plant located in Oberhausen, Germany (ref. 8). It was designed as an operating power plant and as a research facility to study aspects of component design and performance with helium as a working fluid. It has two compressors, with intercooling, connected directly to a high-pressure turbine. The highpressure turbine is in turn connected through a gearbox to a low-pressure turbine with no reheat. Helium is heated first by regenerator, followed by a specially designed heater that burns coke-oven gas. A water-cooled pre-cooler returns the helium to the low-pressure compressor inlet conditions. The high-pressure turbine mass flow rate is 84.4 kg/s, and the heater efficiency is 92.2%. The following design data are given in the reference:

220 Temperature, °C

Pressure, Bar

1. Low-pressure compressor inlet

25

10.5

2. Intercooler inlet

83

15.5

3. High-pressure compressor inlet

25

15.4

4. Regenerator inlet, high-pressure side

125

28.7

5. Heater inlet

417

28.2

6. High-pressure turbine inlet

750

27.0

7. Low-pressure turbine inlet

580

16.5

8. Regenerator inlet, low-pressure side

460

10.8

9. Precooler inlet

169

10.6

In the following, assume k = 1.67 and cp = 5.197 kJ/kg-K for helium. (a) Sketch and label flow and T-s diagrams for the plant. (b) What is the overall engine pressure ratio of the gas turbine? (c) Estimate the mechanical power output and plant thermal efficiency. Reference 8 gives 50-MW and 31.3% as net electrical output and efficiency, respectively. Evaluate your calculations with these data. 5.31 Determine the thrust for a turbojet engine flying at 200 m/s with a compressor inlet temperature of 27°C, a compressor pressure ratio of 11, a turbine inlet temperature of 1400K, and compressor and turbine efficiencies of 0.85 and 0.9, respectively. The engine mass flow rate is 20 kg/s. You may use k = 1.4 and cp = 1.005 kJ/kg-K throughout and neglect the differences between static and stagnation properties in the turbomachinery. Assume an ambient pressure of 0.2615 atmospheres. 5.32 The gas generator of a two-shaft gas turbine has a compressor pressure ratio of 5 and compressor and turbine inlet temperatures of 80°F and 2000°F, respectively, at sea level. All turbomachines have efficiencies of 90%, and the inlet air flow is 50 lbm/s. (a) What are the net work, pressure ratio, and horsepower of the power turbine and the cycle efficiency? (b) Suppose the power turbine is removed and the gas generator exhaust gas flows isentropically through a convergent-divergent propulsion nozzle that is fully expanded (exit pressure is ambient). What are the nozzle exhaust velocity and the static thrust? (c) Repeat part (b) for a choked conversion nozzle. 5.33 A gas turbine with reheat has two turbines with efficiencies e1 and e2. Derive relations for the turbine pressure ratios r1 and r2 that maximize the total turbine work for a given overall turbine pressure ratio, r, if both turbines have the same inlet temperature. How do the pressure ratios compare if the turbine efficiencies are equal?

221 If the efficiency of one turbine is 50% higher than the other, what is the optimum pressure ratio? 5.34* For a simple-cycle gas turbine, develop a multicolumn spreadsheet that tabulates and plots (a) the net work, nondimensionalized by using the product of the compressor constant-pressure heat capacity and the compressor inlet temperature, and (b) the thermal efficiency, both as a function of compressor pressure ratio (only one graph with two curves). Use 80°F and 2000°F as compressor and turbine inlet temperatures, respectively, and 0.85 and 0.90 as compressor and turbine efficiencies, respectively. Use appropriate constant heat capacities. Each of the input values should be entered in separate cells in each column so that the spreadsheet may be used for studies with other parametric values. Use your plot and table to determine the pressure ratio that yields the maximum net work. Compare with your theoretical expectation. 5.35* Solve Exercise 5.34 for a regenerative cycle. Use a nominal value of 0.8 for regenerator effectiveness. 5.36* Solve Exercise 5.34 for a two-shaft regenerative cycle. Use a nominal value of 0.8 for regenerator effectiveness and 0.88 for the power turbine efficiency. Account also for 3% pressure losses in both sides of the regenerator. 5.37 For the conditions of Exercise 5.31, but using more realistic properties in the engine hot sections, plot curves of thrust and specific fuel consumption as a function of gaseous hydrogen air-fuel ratio. Determine the maximum thrust corresponding to the stoichiometric limit for gaseous hydrogen fuel. 5.38* Use the spreadsheet corresponding to Table 5.4 to plot a graph showing the influence of turbine inlet temperature on net work and thermal efficiency for two-shaft gas turbines, with and without regeneration, for a compressor pressure ratio of 4. 5.39 Extend Example 5.6 by using hand calculations to evaluate the thrust and the thrust specific fuel consumption for the engine if it were fitted with a fully expanded (exit pressure equal to ambient pressure) convergent-divergent isentropic nozzle. 5.40* Modify the spreadsheet corresponding to Table 5.6 to evaluate the thrust and the thrust specific fuel consumption for the engine if it were fitted with a fully expanded (exit pressure equal to ambient pressure) convergent-divergent isentropic nozzle. 5.41 A gas turbine engine is being designed to provide work and a hot, high-velocity exhaust flow. The compressor will have a pressure ratio of 4 and an isentropic efficiency of 90% at the design point. The compressor and load are driven by separate ___________________ * Exercise numbers with asterisks involve computer usage.

222 turbines, but the overall expansion pressure ratio across the turbines will be 3 to 1, and the efficiency of each turbine will be 90%. The exhaust of the low-pressure turbine is expanded through a convergent nozzle to provide the high-velocity exhaust. At the design point the turbine inflow air temperature will be held to 1140°F, and the air flow rate will be 300,000 pounds mass per hour. Ambient conditions are 60°F and 14.7 psia. Calculate the brake power of the engine, in kW, and the temperatures at the entrance and the exit of nozzle. 5.42 A ramjet is a jet engine that flies at speeds high enough that the pressure rise produced by ram effect in the inlet makes a compressor and turbine unnecessary. At 50,000 feet and Mach 3, the inlet has a stagnation pressure recovery of 85%. Combustion raises the air temperature to 2500K. What is the thrust per unit mass flow rate of air and the exit velocity of the engine if the nozzle is (a) convergent, and (b) fully expanded (exit pressure equal to ambient pressure) convergent-divergent? 5.43 Using the First Law of Thermodynamics, derive an equation for the work of compression in a reversible steady flow in terms of volume and pressure. Use the equation to derive an expression for the reversible isothermal work of compression of a calorically perfect gas, with compressor pressure ratio as an independent variable. 5.44 Sketch a pressure-volume diagram comparing isothermal and isentropic compressions starting at the same state and having the same pressure ratio. Show for a thermally perfect gas that, at a given state, the isentrope has a steeper (negative) slope than an isotherm. Use your diagram to prove that the isothermal work of compression is less than the isentropic work. 5.45 A supersonic aircraft flies at Mach 2 at an altitude of 40,000 feet. Its engines have a compressor pressure ratio of 20 and a turbine inlet temperature of 2000°F. The inlets have total pressure recoveries of 89%, the compressors and turbines all have efficiencies of 90%, and the nozzles are convergent and isentropic. Determine the nozzle exit velocity and thrust, and estimate the thrust specific fuel consumption of each engine. The mass flow rate of air of each engine is 750 lbm/s. Use a fuel heating value of 18,533 Btu/lbm. 5.46 A supersonic aircraft flies at Mach 2 at an altitude of 13,000 m. Its engines have a compressor pressure ratio of 20 and a turbine inlet temperature of 1500K. The inlets have total pressure recoveries of 89%, the compressors and turbines all have efficiencies of 90%, and the nozzles are convergent and isentropic. Determine the nozzle exit velocity and thrust, and estimate the thrust specific fuel consumption of an engine for an engine air mass flow rate of 100 kg /s. Use a fuel heating value of 43,100 kJ/kg. 5.47 A supersonic aircraft flies at Mach 2 at an altitude of 40,000 feet. Its engines

223 have a compressor pressure ratio of 20 and a turbine inlet temperature of 2000°F. The inlets have total pressure recoveries of 89%, and the compressors and turbines all have efficiencies of 90%. The engines have afterburners that raise the temperature of the gas entering the nozzles to 3000°F. The nozzles are convergent and isentropic. Compare the design thrust and thrust specific fuel consumption with the afterburner on and off for an engine air mass flow rate of 100 lbm/s. 5.48* For a simple-cycle gas turbine, develop a multicolumn spreadsheet that tabulates and plots (a) the net work, nondimensionalized by using the compressor constantpressure heat capacity and the compressor inlet temperature, and (b) the thermal efficiency, as a function of compressor pressure ratio (only one graph with two curves). Use 30° C and 1500°C as compressor and turbine inlet temperatures, respectively, and 0.85 and 0.90 as compressor and turbine efficiencies, respectively. Assume appropriate constant heat capacities. Each of the input values should be entered in separate cells in each column, so that the spreadsheet may be used for studies with other parametric values. Use your plot and table to determine the pressure ratio that yields the maximum net work. Compare with your theoretical expectation. 5.49* Solve Exercise 5.48 for a regenerative cycle. Use a nominal value of 0.85 for regenerator effectiveness. 5.50 Methane is burned in an adiabatic gas turbine combustor. The fuel enters the combustor at the reference temperature for the JANAF tables and mixes with air compressed from 80°F through a pressure ratio of 16 with a compressor efficiency of 90.5%. Determine the equivalence ratio that limits the turbine inlet temperature to 2060°F by: (a) Using the JANAF tables. (b) Using an energy balance on the combustion chamber and the lower heating value for methane.

224 CHAPTER 6

RECIPROCATING INTERNAL COMBUSTION ENGINES

6.1 Introduction Perhaps the best-known engine in the world is the reciprocating internal combustion (IC) engine. Virtually every person who has driven an automobile or pushed a power lawnmower has used one. By far the most widely used IC engine is the spark-ignition gasoline engine, which takes us to school and work and on pleasure jaunts. Although others had made significant contributions, Niklaus Otto is generally credited with the invention of the engine and with the statement of its theoretical cycle. Another important engine is the reciprocating engine that made the name of Rudolf Diesel famous. The Diesel engine, the workhorse of the heavy truck industry, is widely used in industrial power and marine applications. It replaced the reciprocating steam engine in railroad locomotives about fifty years ago and remains dominant in that role today. The piston, cylinder, crank, and connecting rod provide the geometric basis of the reciprocating engine. While two-stroke-cycle engines are in use and of continuing interest, the discussion here will emphasize the more widely applied four-stroke-cycle engine. In this engine the piston undergoes two mechanical cycles for each thermodynamic cycle. The intake and compression processes occur in the first two strokes, and the power and exhaust processes in the last two. These processes are made possible by the crank-slider mechanism, discussed next. 6.2 The Crank-Slider Mechanism Common to most reciprocating engines is a linkage known as a crank-slider mechanism. Diagramed in Figure 6.1, this mechanism is one of several capable of producing the straight-line, backward-and-forward motion known as reciprocating. Fundamentally, the crank-slider converts rotational motion into linear motion, or vice-versa. With a piston as the slider moving inside a fixed cylinder, the mechanism provides the vital capability of a gas engine: the ability to compress and expand a gas. Before delving into this aspect of the engine, however, let us examine the crank-slider mechanism more closely.

225

It is evident from Figure 6.2 that, while the crank arm rotates through 180°, the piston moves from the position known as top-center (TC) to the other extreme, called bottom-center (BC). During this period the piston travels a distance, S, called the stroke, that is twice the length of the crank. For an angular velocity of the crank, T, the crank pin A has a tangential velocity component TS/2. It is evident that, at TC and at BC, the crank pin velocity component in the piston direction, and hence the piston velocity, is zero. At these points, corresponding to crank angle 2 = 0° and 180°, the piston reverses direction. Thus as 2 varies from 0° to 180°, the piston velocity accelerates from 0 to a maximum and then returns to 0. A similar behavior exists between 180° and 360°. The connecting rod is a two-force member; hence it is evident that there are both axial and lateral forces on the piston at crank angles other than 0° and 180°. These lateral forces are, of course, opposed by the cylinder walls. The resulting lateral force component normal to the cylinder wall gives rise to frictional forces between the piston rings and cylinder. It is evident that the normal force, and thus the frictional force, alternates from one side of the piston to the other during each cycle. Thus the piston motion presents a challenging lubrication problem for the control and reduction of both wear and energy loss. The position of the piston with respect to the crank centerline is given by x = (S/2)cos2 + LcosN

[ft | m]

(6.1)

where yA = (S/2)sin2 = LsinN can be used to eliminate N to obtain x/L = (S/2L)cos2 + [1– (S/2L)2 sin2 2 ]½

[dl]

(6.2)

Thus, while the axial component of the motion of the crank pin is simple harmonic, xA = (S/2)cos2, the motion of the piston and piston pin is more complex. It may be

226

seen from Equation (6.2), however, that as S/L becomes small, the piston motion approaches simple harmonic. This becomes physically evident when it is recognized that, in this limit, the connecting rod angle, N , approaches 0 and the piston motion approaches the axial motion of the crank pin. Equations (6.1) and (6.2) may be used to predict component velocities, accelerations, and forces in the engine. The volume swept by the piston as it passes from TC to BC is called the piston displacement, disp. Engine displacement, DISP, is then the product of the piston displacement and the number of cylinders, DISP = (n)(disp). The piston displacement is the product of the piston cross-sectional area and the stroke. The cylinder inside diameter (and, approximately, also the piston diameter) is called its bore. Cylinder bore, stroke, and number of cylinders are usually quoted in engine specifications along with or instead of engine displacement. It will be seen later that the power output of a reciprocating engine is proportional to its displacement. An engine of historical interest that also used the crank-slider mechanism is discussed in the next section. 6.3 The Lenoir Cycle An early form of the reciprocating internal combustion engine is credited to Etienne Lenoir. His engine, introduced in 1860, used a crank-slider-piston-cylinder arrangement

227

in which a combustible mixture confined between the piston and cylinder is ignited after TC. The resulting combustion gas pressure forces acting on the piston deliver work by way of the connecting rod to the rotating crank. When the piston is at BC, combustion gases are allowed to escape. The rotational momentum of the crank system drives the piston toward TC, expelling additional gases as it goes. A fresh combustible mixture is again admitted to the combustion chamber (cylinder) and the cycle is repeated. The theoretical Lenoir cycle, shown in Figure 6.3 on a pressure-volume diagram, consists of the intake of the working fluid (a combustible mixture) from state 0 to state 1, a constant-volume temperature and pressure rise from state 1 to state 2, approximating the combustion process, an isentropic expansion of the combustion gases to state 3, and a constant-pressure expulsion of residual gases back to state 0. Note that a portion of the piston displacement, from state 0 to state 1, is used to take in the combustible mixture and does not participate in the power stroke from state 2 to state 3. The engine has been called an explosion engine because the power delivered is due only to the extremely rapid combustion pressure rise or explosion of the mixture in the confined space of the cylinder. Hundreds of Lenoir engines were used in the nineteenth century, but the engine is quite inefficient by todays standards. In 1862, Beau de Rochas pointed out that the

228

efficiency of internal combustion could be markedly improved in reciprocating engines by compression of the air-fuel mixture prior to combustion. In 1876 Niklaus Otto (who is thought to have been unaware of Rochas’ suggestion) demonstrated an engine that incorporated this important feature, as described next. 6.4 The Otto Cycle The Otto cycle is the theoretical cycle commonly used to represent the processes in the spark ignition (SI) internal combustion engine. It is assumed that a fixed mass of working fluid is confined in the cylinder by a piston that moves from BC to TC and back, as shown in Figure 6.4. The cycle consists of isentropic compression of an air-fuel mixture from state 1 to state 2, constant-volume combustion to state 3, isentropic expansion of the combustion gases to state 4, and a constant-volume heat rejection back to state 1. The constant-volume heat rejection is a simple expedient to close the cycle. It obviates the need to represent the complex expansion and outflow of

229 combustion gases from the cylinder at the end of the cycle. Note that the Otto cycle is not concerned with the induction of the air-fuel mixture or with the expulsion of residual combustion gases. Thus only two mechanical strokes of the crank-slider are needed in the Otto cycle, even when it is used to represent an ideal four-stroke-cycle Otto engine. In this case the remaining strokes are used to execute the necessary intake and exhaust functions. Because it involves only two strokes, the Otto cycle may also represent a two-stroke-cycle engine. The two-stroke-cycle engine is in principle capable of as much work in one rotation of the crank as the four-stroke engine is in two. However, it is difficult to implement because of the necessity of making the intake and exhaust functions a part of those two strokes. It is therefore not as highly developed or widely used as the four-stroke-cycle engine. We will focus on the fourstroke-cycle here. The simplest analysis of the Otto cycle assumes calorically perfect air as the working fluid in what is called the Air Standard cycle analysis. Following the notation of Figure 6.4, the compression process can be represented by the isentropic relation for a calorically perfect gas, Equation (1.21), as p2/p1 = (V1/V2)k

[dl]

(6.3)

where the compression ratio, CR = V1/V2, is a fundamental parameter of all reciprocating engines. The diagram shows that the expansion ratio for the engine, V4 /V3, has the same value, V1/V2. The clearance volume, V2, is the volume enclosed between the cylinder head and the piston at TC. Thus the compression ratio may be expressed as the ratio of the sum of the clearance and displacement volumes to the clearance volume: CR = [V2 + (V1 – V2)]/V2 Thus, for a given displacement, the compression ratio may be increased by reducing the clearance volume. The efficiency of the cycle can be most easily determined by considering constantvolume-process heat transfers and the First Law cyclic integral relation, Equation (1.3). The heat transferred in the processes 2$3 and 4$1 are q2$3 = cv (T3 – T2)

[Btu/lbm | kj/kg]

(6.4)

[Btu/lbm | kJ/kg]

(6.5)

and q4$1 = cv (T1 – T4)

Both the expansion process, 3$4, and the compression process, 1$2, are assumed to be isentropic. Thus, by definition, they are both adiabatic. From the cyclic integral, the net work per unit mass is then: w = q2$3 + q4$1 = cv (T3 – T2 + T1 – T4)

[Btu/lbm | kJ/kg]

(6.6)

230 As before, the cycle thermal efficiency is the ratio of the net work to the external heat supplied:

0Otto = w/q2$3 = cv (T3 – T2 + T1 – T4) / [cv (T3 – T2)] = 1 + (T1 – T4) / (T3 – T2) = 1 – T1/T2 = 1 – 1 / CR k-1

[dl]

(6.7)

where Equation (1.20) has been used to eliminate the temperatures. Equation (6.7) shows that increasing compression ratio increases the cycle thermal efficiency. This is true for real engines as well as for the idealized Otto engine. The ways in which real spark ignition engine cycles deviate from the theoretical Otto cycle are discussed later. EXAMPLE 6.1

An Otto engine takes in an air-fuel mixture at 80°F and standard atmosphere presssure. It has a compression ratio of 8. Using Air Standard cycle analysis, a heating value of 20,425 Btu/lbm, and A/F = 15, determine: (a) The temperature and pressure at the end of compression, after combustion, and at the end of the power stroke. (b) The net work per pound of working fluid. (c) The thermal efficiency. Solution

We use the notation of Figure 6.4: (a) p2 = p1(V1/V2)k = 1(8)1.4 = 18.38 atm T2 = T1(V1/V2)k – 1 = (540)(8)0.4 = 1240.6°R T3 = T2 + qa /cv = T2 + (F/A)(HV)k/cp = 1240.6 + 1.4(20,425/15(0.24 = 9184°R p3 = p2T3 /T2 = 18.38(9184/1240.6) = 136.1 atm T4 = T3 /CRk–1 = 9184/ 80.4 = 3997.2°R p4 = p3 /CRk = 136.1/81.4 = 7.4 atm (b) The constant-volume heat addition is governed by the fuel-air ratio and the fuel heating value: qa = HV(F/A) = 20,425/15 = 1361.7 Btu/lbm of air

231 qr = cv (T1 – T4) = (0.24/1.4)( 540 – 3997.4) = – 592.7 Btu/lbm w = qa + qr = 1361.7 + ( – 592.7) = 769 Btu/lbm (c) The cycle termal efficiency may then be determined from the definition of the heat engine thermal efficiency or Equation (6.7):

0th = w/qa = 769/1361.7 = 0.565 0th = 1 – 1/80.4 = 0.565

_____________________________________________________________________ In view of the discussion of gas properties and dissociation in Chapter 3, the values of T3 and T4 in Example 6.1 are unrealistically high. Much of the energy released by the fuel would go into vibration and dissociation of the gas molecules rather than into the translational and rotational degrees of freedom represented by the temperature. As a result, significantly lower temperatures would be obtained. Thus, while the analysis is formally correct, the use of constant-low-temperature heat capacities in the Air Standard cycle makes it a poor model for predicting temperature extremes when high energy releases occur. Some improvement is achieved by using constant-hightemperature heat capacities, but the best results would be achieved by the use of real gas properties, as discussed in several of the references. 6.5 Combustion in a Reciprocating Engine The constant-volume heat transfer process at TC in the Otto cycle is an artifice to avoid the difficulties of modeling the complex processes that take place in the combustion chamber of the SI engine. These processes, in reality, take place over a crank angle span of 30° or more around TC. Let us consider aspects of these processes and their implementation in more detail. Normally, the mixture in the combustion chamber must have an air-fuel ratio in the neighborhood of the stoichiometric value for satisfactory combustion. A more or less homogeneous mixture may be produced outside the cylinder in a carburetor, by injection into the intake manifold, or by throttle-body injection into a header serving several intake manifolds. In the case of the carburetor, fuel is drawn into the engine from the carburetor by the low pressure created in a venturi through which the combustion air flows. As a result, increased air flow causes lower venturi pressure and hence increased fuel flow. The fuel system thus serves to provide an air-fuel mixture that remains close to the stoichiometric ratio for a range of air flow rates. Various devices designed into the carburetor further adjust the fuel flow for the special operating conditions encountered, such as idling and rapid acceleration. Maximum fuel economy is usually attained with excess air to ensure that all of the fuel is burned. A mixture with excess air is called a lean mixture. The carburetor

232 usually produces this condition in automobiles during normal constant-speed driving. On the other hand, maximum power is achieved with excess fuel to assure that all of the oxygen in the air in the combustion chamber is reacted. It is a matter of exploiting the full power-producing capability of the displacement volume. A mixture with excess fuel is called a rich mixture. The automotive carburetor produces a rich mixture during acceleration by supplying extra fuel to the air entering the intake manifold. The equivalence ratio is sometimes used to characterize the mixture ratio, whether rich or lean. The equivalence ratio, M, is defined as the ratio of the actual fuel-air ratio to the stoichiometric fuel-air ratio. Thus M > 1 represents a rich mixture and M < 1 represents a lean mixture. In terms of air-fuel ratio, M = (A/F)stoich /(A/F). Homogeneous air-fuel mixtures close to stoichiometric may ignite spontaneously (that is, without a spark or other local energy source) if the mixture temperature exceeds a temperature called the autoignition temperature. If the mixture is brought to and held at a temperature higher than the autoignition temperature, there is a period of delay before spontaneous ignition or autoignition This time interval is called the ignition delay, or ignition lag. The ignition delay depends on the characteristics of the fuel and the equivalence ratio and usually decreases with increasing temperature. In spark-ignition engines, compression ratios and therefore the temperatures at the end of compression are low enough that the air-fuel mixture is ignited by the spark plug before spontaneous ignition can occur. SI engines are designed so that a flame front will propagate smoothly from the spark plug into the unburned mixture until all of the mixture has been ignitied. However, as the flame front progresses, the temperature and pressure of the combustion gases behind it rise due to the release of the chemical energy of the fuel. As the front propagates, it compresses and heats the unburned mixture, sometimes termed the end-gas. Combustion is completed as planned when the front smoothly passes completely through the end-gas without autoignition. However, if the end-gas autoignites, a pinging or low-pitched sound called knock is heard. The avoidance of knock due to autoignition of the end-gas is a major constraint on the design compression ratio of an SI engine. If hot spots or thermally induced compression of the end-gas ignite it before the flame front does, there is a more rapid release of chemical energy from the end-gas than during normal combustion. Knock is sometimes thought of as an explosion of the end gas that creates an abrupt pulse and pressure waves that race back and forth across the cylinder at high speed, producing the familiar pinging or low-pitched sound associated with knock. Knock not only reduces engine performance but produces rapid wear and objectionable noise in the engine. Thus it is important for a SI engine fuel to have a high autoignition temperature. It is therefore important for SI engine fuel to have a high autoignition temperature. Thus the knock characteristics of commercially available fuels limit the maximum allowable design compression ratio for SI engines and hence limit their best efficiency. The octane number is a measure of a gasoline's ability to avoid knock. Additives such as tetraethyl lead have been used in the past to suppress engine knock. However, the accumulation of lead in the environment and its penetration into the food cycle has

233 resulted in the phaseout of lead additives. Instead refineries now use appropriate blends of hydrocarbons as a substitute for lead additives in unleaded fuels. The octane number of a fuel is measured in a special variable-compression-ratio engine called a CFR (Cooperative Fuels Research) engine. The octane rating of a fuel is determined by comparison of its knocking characteristics with those of different mixtures of isooctane, C8H18, and n-heptane, C7H16. One hundred percent isooctane is defined as having an octane number of 100 because it had the highest resistance to knock at the time the rating system was devised. On the other hand, n-heptane is assigned a value of 0 on the octane number scale because of its very poor knock resistance. If a gasoline tested in the CFR engine has the same knock threshold as a blend of 90% isooctane and 10% n-heptane, the fuel is assigned an octane rating of 90. In combustion chamber design, the designer attempts to balance many factors to achieve good performance. Design considerations include locating intake valves away from and exhaust valves near spark plugs, to keep end-gas in a relatively cool area of the combustion chamber and thereby suppress hot-surface-induced autoignition tendencies. Valves are, of course, designed as large as possible to reduce induction and exhaust flow restrictions. More than one intake and one exhaust valve per cylinder are now used in some engines to improve “engine breathing.” In some engines, four valves in a single cylinder are employed for this purpose. The valves are also designed to induce swirl and turbulence to promote mixing of fuel and air and to improve combustion stability and burning rate. Pollution and fuel economy considerations have in recent years profoundly influenced overall engine and combustion chamber design. Stratified-charge engines, for example, attempt to provide a locally rich combustion region to control peak temperatures and thus suppress NOx formation. The resulting combustion gases containing unburned fuel then mix with surrounding lean mixture to complete the combustion process, thus eliminating CO and unburned hydrocarbons from the exhaust. These processes occur at lower temperatures than in conventional combustion chamber designs and therefore prevent significant nitrogen reactions. 6.6 Representing Reciprocating Engine Perfomance In an earlier section, the theoretical work per unit mass of working fluid of the Otto engine was evaluated for a single cycle of the engine, using the cyclic integral of the First Law of Thermodynamics. The work done by pressure forces acting on a piston can also be evaluated as the integral of pdV. It is evident therefore that the work done during a single engine cycle is the area enclosed by the cycle process curves on the pressure-volume diagram. Thus, instead of using the cyclic integral or evaluating pdV for each process of the cycle, the work of a reciprocating engine can be found by drawing the theoretical process curves on the p–V diagram and graphically integrating them. Such a plot of pressure versus volume for any reciprocating engine, real or theoretical, is called an indicator diagram.

234 In the nineteenth and early twentieth centuries a mechanical device known as an engine indicator was used to produce indicator cards or diagrams to determine the work per cycle for slow-running steam and gas reciprocating engines. The indicator card was attached to a cylinder that rotated back and forth on its axis as the piston oscillated, thus generating a piston position (volume) coordinate. At the same time a pen driven by a pressure signal from the engine cylinder moved parallel to the cylinder axis, scribing the p-V diagram over and over on the card. The work of high speed engines is still evaluated from traces of pressure obtained with electronic sensors and displayed on electronic monitors and through digital techniques. The work done per cycle (from an indicator card, for instance) can be represented as an average pressure times a volume. Because the displacement volumes of engines are usually known, an engine performance parameter known as the mean effective pressure, MEP, is defined in terms of the piston displacement. The mean effective pressure is defined as the value of the pressure obtained by dividing the net work per cylinder per cycle at a given operating condition by the piston displacement volume: MEP = W/disp

[lbf/ft2 | kPa]

(6.8)

Thus the MEP is a measure of the effectiveness of a given displacement volume in producing net work. The power output of an engine with identical cylinders may be represented as the product of the work per cycle and the number of cycles executed per unit time by the engine. Thus if the engine has n cylinders, each executing N identical thermodynamic cycles per unit time, and delivering W work units per cylinder, with a piston displacement, disp, the power output is given by P = n(N(W = n(N (MEP ( disp

[ft-lbf /min | kW]

(6.9)

Expressed for the entire engine, the engine displacement is DISP = n(disp and the engine work is MEP (DISP. Hence the engine power is: P = N (MEP(DISP

[ft-lbf /min | kW]

(6.10)

where N, the number of thermodynamic cycles of a cylinder per unit time, is the number of crank-shaft revolutions per unit time for a two-stroke-cycle engine and one-half of the revolutions per unit time for a four-stroke-cycle engine. The factor of ½ for the four-stroke-cycle engine arises because one thermodynamic cycle is executed each time the crank rotates through two revolutions. EXAMPLE 6.2

What is the displacement of an engine that develops 60 horsepower at 2500 rpm in a four-stroke-cycle engine having an MEP of 120 psi?

235 Solution

From Equation (6.10), the displacement of the engine is DISP = P/(N (MEP) = (60)(33,000)(12)/[(2500/2)(120)] = 158.4 in3 Checking units: (HP)(ft-lbf/HP-min)(in/ft)/[(cycles/min)(lbf/in2)] = in3 _____________________________________________________________________ If the work is evaluated from an indicator diagram the work is called indicated work; the MEP is called the indicated mean effective pressure, IMEP; and the power is indicated power, IP. Note that the indicated work and power, being associated with the work done by the combustion chamber gases on the piston, do not account for frictional or mechanical losses in the engine, such as piston-cylinder friction or the drag of moving parts (like connecting rods) as they move through air or lubricating oil. Brake Performance Parameters Another way of evaluating engine performance is to attach the engine output shaft to a device known as a dynamometer, or brake. The dynamometer measures the torque, T, applied by the engine at a given rotational speed. The power is then calculated from the relation P = 2B(rpm (T

[ft-lbf /min | N-m/min]

(6.11)

A simple device called a prony brake, which was used in the past, demonstrates the concept for the measurement of the shaft torque of engines. Figure 6.5 shows the prony brake configuration in which a stationary metal band wrapped around the rotating flywheel of the engine resists the torque transmitted to it by friction. The product of the force measured by a spring scale, w, and the moment arm, d , gives the resisting torque. The power dissipated is then given by 2B(rpm)w (d. Modern devices such as water brakes and electrical dynamometers long ago replaced the prony brake. The water brake is like a centrifugal water pump with no outflow, mounted on low-friction bearings, and driven by the test engine. As with the prony brake, the force required to resist turning of the brake (pump) housing provides the torque data. This, together with speed measurement, yields the power output from Equation (6.11). The power dissipated appears as increased temperature of the water in the brake and heat transfer from the brake. Cool water is circulated slowly through the brake to maintain a steady operating condition. The torque measured in this way is called the brake torque, BT, and the resulting power is called the brake power, BP. To summarize: while indicated parameters relate to gas forces in the cylinder, brake parameters deal with output shaft forces. Thus the brake power differs from the indicated power in that it accounts for the effect of all of the energy losses in the engine. The difference between the two is referred to as the friction power, FP. Thus FP = IP – BP.

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Friction power varies with engine speed and is difficult to measure directly. An engine is sometimes driven without fuel by a motor-dynamometer to evaluate friction power. An alternative to using friction power to relate brake and indicated power is through the engine mechanical efficiency, 0m:

0m = BP/IP

[dl]

(6.12)

Because of friction, the brake power of an engine is always less than the indicated power; hence the engine mechanical efficiency must be less than 1. Clearly, mechanical efficiencies as close to 1 as possible are desired. The engine indicated power can also be expressed in terms of torque, through Equation (6.11). Thus an indicated torque, IT, can be defined. Similarly, a brake mean effective pressure, BMEP, may be defined that, when multiplied by the engine displacement and speed, yields the brake power, analogous to Equation (6.10). Table 6.1 summarizes these and other performance parameters and relations. The thermal efficiency, as for other engines, is a measure of the fuel economy of a reciprocating engine. It tells the amount of power output that can be achieved for a given rate of heat release from the fuel. The rate of energy release is, in turn, the product of the rate of fuel flow and the fuel heating value. Thus, for a given thermal efficiency, power output can be increased by employing a high fuel flow rate and/or selecting a fuel with a high heat of combustion. If the thermal efficiency is evaluated using the brake power, it is called the brake thermal efficiency, BTE. If the evaluation uses the indicated power, it is called the indicated thermal efficiency, ITE.

237 It is common practice in the reciprocating engine field to report engine fuel economy in terms of a parameter called the specific fuel consumption, SFC, analogous to the thrust specific fuel consumption used to describe jet engine performance. The specific fuel consumption is defined as the ratio of the fuel-mass flow rate to the power output. Typical units are pounds per horsepower-hour or kilograms per kilowatt-hour. Obviously, good fuel economy is indicated by low values of SFC. The SFC is called brake specific fuel consumption, BSFC, if it is defined using brake power or indicated specific fuel consumption, ISFC, when based on indicated power. The SFC for a reciprocating engine is analogous to the heat rate for a steam power plant in that both are measures of the rate of energy supplied per unit of power output, and in that low values of both are desirable. Volumetric Efficiency The theoretical energy released during the combustion process is the product of the mass of fuel contained in the combustion chamber and its heating value if the fuel is completely reacted. The more air that can be packed into the combustion chamber, the Table 6.1

Engine Performance Parameters Indicated

Brake

Friction

Mean effective pressure

IMEP

BMEP

Power

IP

BP

Torque

IT

BT

FT = IT – BT m = BT / IT

Thermal efficiency

ITE

BTE

m = BTE / ITE

Specific fuel consumption

ISFC

BSFC

FMEP = IMEP – BMEP m = BMEP / IMEP FP = IP – BP

m = BHP / IHP

m = ISFC / BSFC

more fuel that can be burned with it. Thus a measure of the efficiency of the induction system is of great importance. The volumetric efficiency, 0v, is the ratio of the actual mass of mixture in the combustion chamber to the mass of mixture that the displacement volume could hold if the mixture were at ambient (free-air) density. Thus the average mass-flow rate of air through a cylinder is 0v (disp) DaN. Pressure losses across intake and exhaust valves, combustion-chamber clearance volume, the influence of hot cylinder walls on mixture density, valve timing, and gas inertia effects all influence the volumetric efficiency. EXAMPLE 6.3

A six-cylinder, four-stroke-cycle SI engine operates at 3000 rpm with an indicated mean effective pressure of five atmospheres using octane fuel with an equivalence ratio

238 of 0.9. The brake torque at this condition is 250 lbf–ft., and the volumetric efficiency is 85%. Each cylinder has a five inch bore and 6 inch stroke. Ambient conditions are 14.7 psia and 40°F. What is the indicated horsepower, brake horsepower, and friction horsepower; the mechanical efficiency; the fuel flow rate; and the BSFC? Solution

The six cylinders have a total displacement of DISP = 6B×52×6/4 = 706.86 in3 Then the indicated horsepower is IP = MEP×DISP×N /[12×33,000]

[lbf /in2][in3][cycles/min]/[in/ft][ft-lbf /HP-min]

= (5)(14.7)(706.86)(3000/2)/[12×33,000] = 196.8 horsepower The brake horsepower, from Equation (6.11), is: BP = 2B × 3000 × 250 / 33,000 = 142.8 horsepower Then the friction power is the difference between the indicated and brake power: FP = 196.8 – 142.8 = 54 horsepower and the mechanical efficiency is

0m = 142.8/196.8 = 0.726 The ambient density is

Da = 14.7 × 144/ [53.3 × 500] = 0.0794 lbm /ft3 and the mass flow rate of air to the engine is ma = 0.85×0.0794×706.86×(3000/2)/1728 = 41.4 lbm /min For octane the stoichiometric reaction equation is C8H18 + 12.5O2 + (12.5×3.76)N2 $ 8CO2 + 9H2O + (12.5×3.76)N2 The fuel-air ratio is then F/A = 0.9×[(8×12) + (18×1)]/[12.5(32 + 3.76×28)] = 0.0598 lbm-fuel /lbm-air

239 The fuel flow rate is mf = ma (F/A) = 41.4 × 0.0598 = 2.474 lbm /min The brake specific fuel consumption is BSFC = 60 mf /BHP = 60×2.474/142.8 = 1.04 lbm /BHP-hr ____________________________________________________________________ 6.7 Spark-Ignition Engine Performance A typical indicator diagram showing intake and exhaust processes, valve actuation, and spark timing for a four-stroke-cycle SI engine is shown in Figure 6.6. It is assumed that an appropriate air-fuel mixture is supplied from a carburetor through an intake manifold to an intake valve, IV, and that the combustion gas is discharged through an exhaust valve, EV, into an exhaust manifold. The induction of the air-fuel mixture starts with the opening of the intake valve at point A just before TC. As the piston sweeps to the right, the mixture is drawn into the cylinder through the IV. The pressure in the cylinder is somewhat below that in the intake manifold due to the pressure losses across the intake valve. In order to use the momentum of the mixture inflow through the valve at the end of the intake stroke to improve the volumetric efficiency, intake valve closure is delayed to shortly after BC at point B. Power supplied from inertia of a flywheel (and the other rotating masses in the engine) drives the piston to the left, compressing and raising the temperature of the trapped mixture. The combustion process in a properly operating SI engine is progressive in that the reaction starts at the spark plug and progresses into the unburned mixture at a finite speed. Thus the combustion process takes time and cannot be executed instantaneously as implied by the theoretical cycle. In order for the process to take place as near to TC as possible, the spark plug is fired at point S. The number of degrees of crank rotation before TC at which the spark occurs is called the ignition advance. Advances of 10° to 30° are common, depending on speed and load. The spark advance may be controlled by devices that sense engine speed and intake manifold pressure. Microprocessors are now used to control spark advance and other functions, based on almost instantaneous engine performance measurements. Recalling the slider-crank analysis, we observ that the piston velocity at top center is momentarily zero as the piston changes direction. Therefore no work can be done at this point, regardless of the magnitude of the pressure force. Thus, to maximize the work output, it is desired to have the maximum cylinder pressure occur at about 20° after TC. Adjustment of the spark advance (in degrees before TC) allows some control of the combustion process and the timing of peak pressure. For a fixed combustion duration, the combustion crank-angle interval must increase with engine speed. As a consequence, the ignition advance must increase with increasing engine speed to

240

maintain optimum timing of the peak pressure. Following combustion, the piston continues toward bottom center as the high pressure gases expand and do work on the piston during the power stroke. As the piston approaches BC, the gases do little work on the piston as its velocity again approaches zero. As a result, not much work is lost by early opening of the exhaust valve before BC (at point E) to start the blowdown portion of the exhaust process. It is expedient to sacrifice a little work during the end of the power stroke in order to reduce the work needed to overcome an otherwise-high exhaust stroke cylinder pressure. Inertia of the gas in the cylinder and resistance to flow through the exhaust valve opening slow the drop of gas pressure in the cylinder after the valve opens. Thus the gases at point E are at a pressure above the exhaust manifold pressure and, during blowdown, rush out through the EV at high speed. Following blowdown, gases remaining in the cylinder are then expelled as the piston returns to TC. They remain above exhaust manifold pressure until reaching TC because of the flow resistance of the exhaust valve. The EV closes shortly after TC at point C, terminating the exhaust process. The period of overlap at TC between the intake valve opening at point A and exhaust valve closing at point C in Figure 6.6 allows more time for the intake and exhaust processes at high engine speeds, when about 10 milliseconds may be available for these processes. At low engine speed and at idling there may be some mixture loss through the exhaust valve and discharge into the intake manifold during this valve overlap period. The combined exhaust and induction processes are seen to form a “pumping loop” that traverses the p-V diagram in a counterclockwise direction and therefore

241

represents work input rather than work production. The higher the exhaust stroke pressure and the lower the intake stroke pressure, the greater the area of the pumping loop and hence the greater the work that must be supplied by the power loop (clockwise) to compensate. Great attention is therefore paid to valve design and other engine characteristics that influence the exhaust and induction processes. Volumetric efficiency is a major parameter that indicates the degree of success of these efforts. Performance Characteristics A given ideal Otto-cycle engine produces a certain amount of work per cycle. For such a cycle, MEP = W/disp is a constant. Equating the power equations (6.9) and (6.11) shows that the average torque is proportional to MEP and independent of engine engine speed. Therefore power output for the ideal engine is directly proportional to the number of cycles executed per unit time, or to engine speed. Thus an Otto engine has ideal torque and power characteristics, as shown by the solid lines in Figure 6.7. The characteristics of real engines (represented by the dashed lines) tend to be similar in nature to the ideal characteristics but suffer from speed-sensitive effects, particularly at low or high speeds. Torque and power characteristics for a 3.1 liter V6 engine (ref. 9) are shown by the solid lines in Figure 6.8. Note the flatness of the torque-speed curve and the expected peaking of the power curve at higher speed than the torque curve. Rather than present graphical characteristics such as this in their

242

brochures, automobile manufacturers usually present only values for the maximum power and torque and the speeds at which they occur. Engine characteristics such as those shown in the figure are invaluable to application engineers seeking a suitable engine for use in a product. 6.8 The Compression-Ignition or Diesel Cycle The ideal Diesel cycle differs from the Otto cycle in that combustion is at constant pressure rather than constant volume. The ideal cycle, shown in Figure 6.9, is commonly implemented in a reciprocating engine in which air is compressed without fuel from state 1 to state 2. With a typically high compression ratio, state 2 is at a temperature high enough that fuel will ignite spontaneously when sprayed directly into the air in the combustion chamber from a high-pressure fuel injection system. By controlling the fuel injection rate and thus the rate of chemical energy release in relation to the rate of expansion of the combustion gases after state 2, a constant-

243

pressure process or other energy release pattern may be achieved as in Figure 6.9. For example, if the energy release rate is high, then pressure may rise, as from 2 to 3', and if low may fall to 3''. Thus constant-pressure combustion made possible by controlling the rate of fuel injection into the cyclinder implies the use of a precision fuel injection system. Instead of injecting fuel into the high-temperature compressed air, the cycle might be executed by compression of an air-fuel mixture, with ignition occurring either spontaneously or at a hot spot in the cylinder near the end of the compression process. Inconsistency and unpredictability of the start of combustion in this approach, due to variations in fuel and operating conditions, and to lack of control of the rate of heat release with the possibility of severe knock, makes the operation of such an engine unreliable, at the least, and also limits the maximum compression ratio. The Diesel engine therefore usually employs fuel injection into compressed air rather than carbureted mixture formation. In the Air Standard cycle analysis of the Diesel cycle, the heat addition process is at constant pressure: q2$3 = cp(T3 – T2)

[Btu/lbm | kJ/kg]

(6.13)

and, as with the Otto cycle, the closing process is at constant volume: q4$1 = cv(T1 – T4)

[Btu/lbm | kJ/kg]

(6.14)

244 The net work and thermal efficiency are then: w = q2$3 + q4$1 = cp(T3 – T2) + cv(T1 – T4) = cvT1[k(T3/T1 – T2/T1) + 1 – T4/T1]

[Btu/lbm | kJ/kg]

(6.15)

0Diesel = w/q2$3 = 1 + q4-$1/q2$3 = 1 + (cv/cp)(T1 – T4)/(T3 – T2) = 1 – (1/k)(T1/T2)(T4/T1 – 1)/(T3/T2 – 1)

[dl]

(6.16)

The expressions for the net work and cycle efficiency may be expressed in terms two parameters, the compression ratio, CR = V1/V2 (as defined earlier in treating the Otto cycle) and the cutoff ratio, COR = V3/V2. The temperature ratios in Equations (6.15) and (6.16) may be replaced by these parameters using, for the constant-pressure process, COR = V3/V2 = T3/T2 and by expanding the following identity: T4 /T1 = (T4/T3)(T3/T2)(T2 /T1) = (V3 /V4)k-1(V3/V2)(V1/V2)k-1 = [(V3/V4)(V1/V2)]k-1COR = (COR)k-1COR = CORk where the product of the volume ratios was simplified by recognizing that V4 = V1. Thus the nondimensionalized net work and Diesel-cycle thermal efficiency are given by w /cvT1 = kCRk-1(COR – 1) + (1 – CORk)

[dl]

(6.17)

and

0Diesel = 1 – (1/k)[(CORk – 1)/(COR – 1)]/CRk-1

[dl]

(6.18)

where the cutoff ratio, COR, is the ratio of the volume at the end of combustion, V3, to that at the start of combustion, V2. Thus the cutoff ratio may be thought of as a measure of the duration of fuel injection, with higher cutoff ratios corresponding to longer combustion durations.

245 Diesel-cycle net work increases with both compression ratio and cutoff ratio. This is readily seen graphically from Figure 6.9 in terms of p-V diagram area. As with the Otto cycle, increasing compression ratio increases the Diesel-cycle thermal efficiency. Increasing cutoff ratio, however, decreases thermal efficiency. This may be rationalized by observing from the p-V diagram that much of the additional heat supplied when injection is continued is rejected at increasingly higher temperatures. Another view is that heat added late in the expansion process can produce work only over the remaining part of the stroke and thus adds less to net work than to heat rejection. EXAMPLE 6.4

A Diesel engine has a compression ratio of 20 and a peak temperature of 3000K. Using an Air Standard cycle analysis, estimate the work per unit mass of air, the thermal efficiency, the combustion pressure, and the cutoff ratio. Solution

Assuming an ambient temperature and pressure of 300K and 1 atmosphere, the temperature at the end of the compression stroke is T2 = (300)(20)1.4 – 1 = 994.3K and the combustion pressure is p2 = (1)(20)1.4 = 66.3 atm Then the cutoff ratio is V3/V2 = T3/T2 = 3000/994.3 = 3.02 The expansion ratio is calculated as follows: V4 /V3 = (V1/V2)/(V3 /V2) = 20/3.02 = 6.62 T4 = T3 (V3 /V4)1.4 – 1 = 3000/6.620.4 = 1409K w = 1.005(3000 – 994.3) + (1.005/1.4)(300 – 1409) = 1219.6 kJ/kg qa = 1.005(3000 – 994.3) = 2015.7 kJ/kg

0th = w/qa = 1219.6/2015.6 = 0.605, or 60.5% _____________________________________________________________________

246

6.9 Comparing Otto-Cycle and Diesel-Cycle Efficiencies A reasonable question at this point is: Which cycle is more efficient, the Otto cycle or the Diesel cycle? Figure 6.10 assists in examining this question. In general notation, the cycle efficiency may be written as

0th = wnet /qin = wnet /(wnet + |qout|) = 1 /(1 + |qout| /wnet)

[dl]

(6.19)

Comparing the Otto cycle 1–2–3–4 and the Diesel cycle with the same compression ratio 1–2–3'–4, we see that both have the same heat rejection but that the Otto cycle has the higher net work. Equation (6.19) then shows that, for the same compression ratio, the Otto cycle has the higher efficiency. It has been observed that Diesel-cycle efficiency decreases with increasing cutoff ratio for a given compression ratio. Let us examine the limit of the Diesel-cycle efficiency for constant CR as COR approaches its minimum value, 1. We may write Equation (6.18) as

0Diesel = 1 – 1 /(kCRk-1) f (COR) where f(COR) = (CORk – 1)/(COR – 1). Applying L'Hospital's rule, with primes

247 designating differentiation with respect to COR, to the limit of f(COR) as COR $1, yields lim f(COR) = lim (CORk – 1)'/ Lim (COR– 1)' = lim kCORk – 1 = k

COR$1

COR$1

COR$1

and lim0

COR$1 Diesel

= 1 – 1 /CRk – 1 = 0Otto

Thus the limit of the Diesel-cycle efficiency as COR approaches 1 is the Otto cycle efficiency. Hence Equation (6.18) shows that the efficiency of the Diesel cycle must be less than or equal to the Otto-cycle efficiency if both engines have the same compression ratio, the same conclusion we reached by examination of the p-V diagram. Suppose, however, that the compression ratios are not the same. Compare the Otto cycle 1–2'–3'–4 with the Diesel cycle 1–2–3'–4 having the same maximum temperature in Figure 6.10. The Otto cycle has a smaller area, and therefore less work, than the Diesel cycle, but the same heat rejection. Equation (6.19) demonstrates that the Otto cycle has a lower thermal efficiency than the Diesel cycle with the same maximum temperature. The conclusion that must be drawn from the above comparisons is quite clear. As in most comparative engineering studies, the result depends on the ground rules which were adopted at the start of the study. The Otto cycle is more efficient if the compression ratio is the same or greater than that of the competing Diesel cycle. But knock in spark-ignition (Otto) engines limits their compression ratios to about 12, while Diesel-engine compression ratios may exceed 20. Thus, with these higher compression ratios, the Air Standard Diesel-cycle efficiency can exceed that of the Otto cycle. In practice, Diesel engines tend to have higher efficiencies than SI engines because of higher compression ratios. 6.10 Diesel-Engine Performance In 1897, five years after Rudolph Diesel's first patents and twenty-one years after Otto's introduction of the spark-ignition engine, Diesel's compression-ignition engine was proven to develop 13.1 kilowatts of power with an unprecedented brake thermal efficiency of 26.2% (ref. 7). At that time, most steam engines operated at thermal efficiencies below 10 %; and the best gas engines did not perform much better than the steam machines. Diesel claimed (and was widely believed) to have developed his engine from the principles expounded by Carnot. He had developed "the rational engine." Whether his claims were exaggerated or not, Diesel's acclaim was well deserved. He had developed an engine that operated at unprecedented temperatures and pressures, had proven his concept of ignition of fuel by injection into the compressed high-temperature air, and had overcome the formidable problems of injecting a variety of fuels in appropriate

248 amounts with the precise timing required for satisfactory combustion. His is a fascinating story of a brilliant and dedicated engineer (refs. 7, 8). In the Diesel engine, the high air temperatures and pressures prior to combustion are attributable to the compression of air alone rather than an air-fuel mixture. Compression of air alone eliminates the possibility of autiognition during compression and makes high compression ratios possible. However, because of the high pressures and temperatures, Diesel engines must be designed to be structurally more rugged. Therefore, they tend to be heavier than SI engines with the same brake power. The energy release process in the Diesel engine is controlled by the rate of injection of fuel. After a brief ignition lag, the first fuel injected into the combustion chamber autoignites and the resulting high gas temperature sustains the combustion of the remainder of the fuel stream as it enters the combustion chamber. Thus it is evident that the favorable fuel characteristic of high autoignition temperature for an SI engine is an unfavorable characteristic for a Diesel engine. In the Diesel engine, a low autoignition temperature and a short ignition delay are desirable. Knock is possible in the Diesel engine, but it is due to an entirely different cause than knock in a spark-ignition engine. If fuel is ignited and burns as rapidly as it is injected, then smooth, knock-free combustion occurs. If, on the other hand, fuel accumulates in the cylinder before ignition due to a long ignition lag, an explosion or detonation occurs, producing a loud Diesel knock. The cetane number is the parameter that identifies the ignition lag characteristic of a fuel. The cetane number, like the octane number, is determined by testing in a CFR engine. The ignition lag of the test fuel is compared with that of a mixture of n-cetane, C16H34, and heptamethylnonane, HMN (ref. 10). Cetane, which has good ignition qualities, is assigned a value of 100; and HMN, which has poor knock behavior, a value of 15. The cetane number is then given by the sum of the percentage of n-cetane and 0.15 times the percentage of HMN in the knock-comparison mixture. A cetane number of 40 is the minimum allowed for a Diesel fuel. 6.11 Superchargers and Turbochargers The importance of the volumetric efficiency, representing the efficiency of induction of the air-fuel mixture into the reciprocating-engine cylinders, was discussed earlier. Clearly, the more mixture mass in the displacement volume, the more chemical energy can be released and the more power will be delivered from that volume. During the Second World War, the mechanical supercharger was sometimes used with SI aircraft engines to increase the power and operational ceiling of American airplanes. Today supercharging is used with both Diesel engines and SI engines. The supercharger is a compressor that supplies air to the cylinder at high pressure so that the gas density in the cylinder at the start of compression is well above the free-air density. The piston exhaust gases are allowed to expand freely to the atmosphere through the exhaust manifold and tailpipe. The supercharger is usually driven by a belt or gear train from the engine crank shaft.

249

Figure 6.11 shows a modification of the theoretical Otto cycle to accommodate mechanical supercharging. The supercharger supplies air to the engine cyclinders at pressure p7 in the intake process 7 $ 1. The processes 4 $ 5 $ 6 purge most of the combustion gas from the cylinder. The most striking change in the cycle is that the induction-exhaust loop is now traversed counterclockwise, indicating that the cylinder is delivering net work during these processes as well as during the compressionexpansion loop. It should be remembered, however, that part of the cycle indicated power must be used to drive the external supercharger. The turbosupercharger or turbocharger, for short, is a supercharger driven by a turbine using the exhaust gas of the reciprocating engine, as shown schematically in Figure 6.12. A cutaway view of a turbocharger is shown in Figure 6.13(a). Figure 6.13(b) presents a diagram for the turbocharger. Compact turbochargers commonly increase the brake power of an engine by 30% or more, as shown in Figure 6.8, where the performance of an engine with and without turbocharging is compared. There, a substantial increase in peak torque and flattening of the torque-speed curve due to turbocharging is evident. For a supercharged engine, the brake power, BP, is the indicated power (as in Figure 6.11) less the engine friction power and the supercharger shaft power: BP = DISP ( IMEP ( N – Pm – FP

[ft-lbf /min | kJ/s]

(6.15)

250

where Pm is the supercharger-shaft mechanical power supplied by the engine (0 for a turbocharger). The IMEP includes the positive work contribution of the exhaust loop. The exhaust back pressure of the reciprocating engine is higher with a turbocharger than for a naturally aspirated or mechanically supercharged engine because of the drop in exhaust gas pressure through the turbine. The engine brake power increases primarily because of a higher IMEP due to the added mass of fuel and air in the cylinder during combustion. Intercooling between the compressor and the intake manifold may be used to further increase the cylinder charge density. Turbocharging may increase engine efficiency, but its primary benefit is a substantial increase in brake power. In a turbocharged engine, a wastegate may be required to bypass engine exhaust gas around the turbine at high engine speeds. This becomes necessary when the compressor raises the intake manifold pressure to excessively high levels, causing engine knock or threatening component damage. Thirty to forty percent of the exhaust flow may be bypassed around the turbine at maximum speed and load (ref. 1).

251

252 6.12 The Automobile Engine and Air Pollution Since the Second World War, concern for environmental pollution has grown from acceptance of the status quo to recognition and militance of national and international scope. Among other sources, causes of the well-known Los Angeles smog problem were identified as hydrocarbons (HC) and oxides of nitrogen (NOx) in exhaust emissions from motor vehicle reciprocating engines. As a result, national and California automobile air pollution limits for automobiles have been established and toughened. Prior to the Clean Air Act of 1990, the U.S. federal exhaust-gas emissions standards limited unburned hydrocarbons, carbon monoxide, and oxides of nitrogen to 0.41, 3.4, and 1.0 g/mile, respectively. According to reference 12, today it takes 25 autos to emit as much CO and unburned hydrocarbons and 4 to emit as much NOx as a single car in 1960. The reference anticipated that, led by existing California law and other factors, future engine designs should be targeted toward satisfying a tailpipe standard of 0.25, 3.4, 0.4 g/mile. Indeed, the 1990 Clean Air Act (refs. 15,16) specified these limits for the first 50,000 miles or five years of operation for all passenger cars manufactured after 1995. In addition to the regulations on gaseous emissions, the Clean Air Act of 1990 adopted the California standard for particulate matter of 0.08 g/mile for passenger cars. The standards on particulates are particularly difficult for the Diesel engine, because of its of soot-producing tendency. The automobile air pollution problem arises in part because the reactions in the exhaust system are not in chemical equilibrium as the gas temperature drops. Oxides of nitrogen, once formed in the cylinder at high temperature, do not return to equilibrium concentrations of nitrogen and oxygen in the cooling exhaust products. Likewise, CO formed with rich mixtures or by dissociation of CO2 in the cylinder at high temperature does not respond rapidly to an infusion of air as its temperature drops in the exhaust system. Their concentrations may be thought of as constant or frozen. Unburned hydrocarbons are produced not only by rich combustion but also by unburned mixture lurking in crevices (such as between piston and cylinder above the top piston ring), by lubricating oil on cylinder walls and the cylinder head that absorbs and desorbs hydrocarbons before and after combustion, and by transient operating conditions. Starting in 1963, positive crankcase ventilation was used in all new cars to duct fuel-rich crankcase gas previously vented to the atmosphere back into the engine intake system. Later in the ‘60s, various fixes were adopted to comply with regulation of tailpipe unburned hydrocarbons and CO, including lowering compression ratios. In 1973, NOx became federally regulated, and exhaust gas recirculation (EGR) was employed to reduce NOx formation through reduced combustion temperatures. At the same time, HC and CO standards were reduced further, leading to the use of the oxidizing catalytic converter. Introduction of air pumped into the tailpipe provided additional oxygen to assist in completion of the oxidation reactions. In 1981, a reducing catalytic converter came into use to reduce NOx further. This device does not perform well in an oxidizing atmosphere. As a result, two-stage catalytic converters were applied, with the first stage reducing NOx in a near-stoichiometric mixture and the

253 second oxidizing the combustibles remaining in the exhaust with the help of air introduced between the stages. This fresh air does not the increase NOx significantly, because of the relatively low temperature of the exhaust. The three-way catalytic converter using several exotic metal catalysts to reduce all three of the gaseous pollutants was also introduced. The use of catalytic converters to deal with all three pollutants brought about significant simultaneous reductions in the three major gaseous pollutants from automobiles. This allowed fuel-economy-reducing modifications that had been introduced earlier to satisfy emission reduction demands to be eliminated or relaxed, leading to further improvements in fuel economy. Catalytic converters, however, require precise control of exhaust gas oxygen to near-stoichiometric mixtures. The on-board computer has made possible control of mixture ratio and spark timing in response to censor outputs of intake manifold pressure, exhaust gas oxygen, engine speed, air flow, and incipient knock. The oxygen, or lambda, censor located in the exhaust pipe upstream of the three-way converter or between the two-stage converters is very sensitive to transition from rich to lean exhaust and allows close computer control of the mixture ratio to ensure proper operation of the catalytic converter. Computer control of carburetors or fuel injection as well as other engine functions has allowed simultaneous improvement in fuel economy and emissions in recent years. Thus, while emissions have been drastically reduced since 1974, according to reference 11 the EPA composite fuel economy of the average U.S. passenger car has nearly doubled; although this improvement has not come from the engine alone. Despite the hard-won gains in emissions control and fuel economy, further progress may be expected. EXAMPLE 6.5

The 1990 NOx emissions standard is 0.4 grams per mile. For an automobile burning stoichiometric octane with a fuel mileage of 30 mpg, what is the maximum tailpipe concentration of NOx in parts per million? Assume that NOx is represented by NO2 and that the fuel density is 692 kilograms per cubic meter. Solution

For the stoichiometric combustion of octane, C8H18, the air-fuel ratio is 15.05 and the molecular weight of combustion products is 28.6. The consumption of octane is mf = (692)(1000)(3.79×10-3)/ 30 = 87.4 g/mile [Note: (kg/m3)(g/kg)(m3/gal)/(mile/gal) = g/mile.] The concentration of NOx is the ratio of the number of moles of NOx to moles of combustion gas products: mole Nox /mole cg = (mNOx /mf)(mf / mcg)(Mcg /MNOx) = (0.4/87.4)(28.6/46)/ (15.05 + 1) = 0.0001773

254 or 177.3 parts per million (ppm). _____________________________________________________________________ Bibliography and References 1. Heywood, John B., Internal Combustion Engine Fundamentals. New York: McGraw-Hill, 1988. 2. Ferguson, Colin R., Internal Combustion Engines. New York: Wiley, 1986. 3. Adler, U., et al., Automotive Handbook, 2nd ed. Warrendale, Pa.: Society of Automotive Engineers., 1986. 4. Lichty, Lester C., Internal Combustion Engines. New York: McGraw Hill, 1951. 5. Crouse, William H., Automotive Engine Design. New York: McGraw-Hill, 1970. 6. Obert, Edward, Internal Combustion Engines, Analysis and Practice. Scranton, Pa.: International Textbook Co., 1944. 7. Grosser, Morton, Diesel: The Man and the Engine. New York: Atheneum, 1978. 8. Nitske, W. Robert, and Wilson, Charles Morrow, Rudolph Diesel: Pioneer of the Age of Power. Norman, Okla.: University of Oklahoma Press, 1965. 9. Demmler, Albert W. Jr., et al., “1989 Technical Highlights of Big-three U.S. Manufacturers,” Automotive Engineering. Vol. 96, No. 10, October 1988, p. 81. 10. Anon., “Ignition Quality of Diesel Fuels by the Cetane Method,” ASTM D 613-84, 1985 Annual Book of ASTM Standards, Section 5. 11. Amann, Charles A., “The Automotive Spark Ignition Engine-A Historical Perspective,” American Society of Mechanical Engineers, ICE-Vol. 8, Book No. 100294, 1989. 12. Amann, Charles A., “The Automotive Spark-Ignition Engine-A Future Perspective,” Society of Automotive Engineers Paper 891666, 1989. 13. Amann, Charles A., “The Passenger Car and the Greenhouse Effect,” Society of Automotive Engineers Paper, 1990. 14. Taylor, Charles Fayette, The Internal Combustion Engine in Theory and Practice, 2nd ed., revised. Cambridge, Mass.: MIT Press, 1985.

255 15. Public Law 101-549, “An Act to Amend the Clean Air Act to Provide for Attainment and Maintenance of Health, Protection, National Air Quality Standards, and Other Purposes,” November 15, 1990. 16. Anon., “Provisions–Clean Air Amendments,” Congressional Quarterly, November 24, 1990.

EXERCISES 6.1 Plot dimensionless piston position against crank angle for S/2L = 0.5, 0.4, 0.3, and 0.2. 6.2* Obtain expressions for the piston velocity and acceleration as a function of the crank angle, constant angular velocity, and S/2L ratio. Use a spreadsheet to calculate and plot velocity and acceleration against crank angle for S/2L = 0.5, 0.4, 0.3, and 0.2. 6.3 Determine the equation for the piston motion for a scotch yoke mechanism in terms of crank angle. Obtain an equation for the piston velocity for a crank that turns with a given angular velocity, T. 6.4 Derive an equation for the Otto-engine net work by integration of pdV for the Air Standard cycle. Compare with Equation (6.6).

6.5* Use a spreadsheet to calculate and plot cycle efficiency as a function of compression ratio for the Diesel cycle for cutoff ratios of 1, 2, and 3. Indentify the Otto-cycle efficiency on the plot. Explain and show graphically from the plot how a Diesel engine can be more efficient than an Otto engine. 6.6 A single-cylinder Air Standard Otto engine has a compression ratio of 8.5 and a peak temperature of 3500°F at ambient conditions of 80°F and one atmosphere. Determine the cycle efficiency, maximum cylinder pressure, and mean effective pressure. 6.7 A six-cylinder engine with a compression ratio of 11 runs at 2800 rpm at 80°F and 14.7 psia. Each cylinder has a bore and stroke of three inches and a volumetric efficiency of 0.82. Assume an Air Standard, four-stroke Otto cycle _______________________ * Exercise numbers with an asterisk indicate that they involve computer usage.

256 with stoichiometric octane as fuel. Assume that the energy release from the fuel is equally divided between internal energy increase in cylinder gases and cylinder wall heat loss. What are the cylinder mean effective pressures and the engine horsepower and specific fuel consumption? 6.8 A single-cylinder four-stroke-cycle spark-ignition engine has a BSFC of 0.4 kg/kW-hr and a volumetric efficiency of 78% at a speed of 45 rps. The bore is 6 cm and the stroke is 8.5 cm. What is the fuel flow rate, fuel-air ratio, and brake torque if the brake power output is 6 kW with ambient conditions of 100 kPa and 22°C? 6.9 A single-cylinder four-stroke-cycle spark-ignition engine operating at 3500 rpm has a brake mean effective pressure of 1800 kPa and a displacement of 400 cm3. Atmospheric conditions are 101kPa and 27°C. (a) If the stroke is 6 cm, what is the bore? (b) What is the brake power? (c) If the mass air-fuel ratio is 16 and the fuel flow rate is 0.00065 kg/s, what is the volumetric efficiency? (d) Compare your results with the performance of a two-cylinder engine with the same overall geometric characteristics. 6.10 A four-cylinder four-stroke-cycle spark-ignition engine operating at 3500 rpm has a brake mean effective pressure of 80 psi and a displacement of 400 cm3. Atmospheric conditions are one bar and 80°F. (a) If the stroke is 3 inches, what is the bore? (b) What is the brake power? (c) If the mass air-fuel ratio is 16 and the fuel flow rate is 0.00065 kg/s, what is the volumetric efficiency? 6.11 A four-cylinder four-stroke-cycle spark-ignition engine with 200 cm3 displacement and operating in air at 27°C and 110 kPa has a friction power of 27 kW and a brake power output of 136 kW at 3600 rpm. (a) What is the mechanical efficiency? (b) If it has a volumetric efficiency of 74% and burns liquid methanol with 15% excess air, what is the brake specific fuel consumption? 6.12 An eight-cylinder four-stroke-cycle engine has a bore of three inches and a stroke of 4 inches. At a shaft speed of 3000 rpm, the brake horsepower is 325 and the mechanical efficiency is 88%. Fuel with a heating value of 19,000 Btu/lbm is supplied at a rate of 80 lbm/hr. What are the engine displacement, BMEP, brake

257 torque, and indicated specific fuel consumption in lbm/HP-hr? 6.13 An eight-cylinder four-stroke-cycle engine has a bore of 10 cm and a stroke of 12 cm. At a shaft speed of 53 rps, the brake power is 300kW and the mechanical efficiency is 85%. Fuel with a heating value of 40,000 kJ/kg is supplied at a rate of 40 kg/hr. What are the engine displacement, BMEP, brake torque, and indicated specific fuel consumption in kg/kW-hr? 6.14 Consider a naturally aspirated eight-cylinder four-stroke-cycle Diesel engine with a compression ratio of 20 and cutoff ratio of 2.5. Air is inducted into the cylinder from the atmosphere at 14.5 psia and 80°F. Assume an Air Standard cycle. (a) Determine the temperatures and pressures immediately before and after combustion. (b) What is the heat added in the combustion process, in Btu/lbm? (c) What is the net work, in Btu/lbm, and the thermal efficiency? (d) If the volumetric efficiency is 85%, the engine displacement is 300 in3, and the engine speed is 2000 rpm, what is the mass flow rate of air through the engine in lbm/min? (e) What is the engine horsepower? (f) Assuming that losses through the valves cause a 20-psi pressure differential between the pressures during the exhaust and intake strokes, estimate the actual and fractional losses, in horsepower, due to these processes. Sketch the appropriate p-V diagram. 6.15 A two-cylinder four-stroke-cycle engine produces 30 brake horsepower at a brake thermal efficiency of 20% at 2600 rpm. The fuel is methane burning in air with an equivalence ratio of 0.8 and a heating value of 21,560 Btu/lbm. Ambient conditions are 520°R and 14.7 psia. The engine mechanical efficiency is 88%, and the volumetric efficiency is 92%. What are the fuel flow rate, the displacement volume per cylinder, and the brake specific fuel consumption? What is the bore if the bore and stroke or equal? 6.16 Sketch carefully a single p–V diagram showing Otto and Diesel cycles having the same minimum and maximum temperatures. Shade the area representing the difference in net work between the cycles. Repeat for cycles having the same compression ratio. Discuss the implications of these sketches. 6.17 An eight-cylinder reciprocating engine has a 3-in. bore and a 4-in. stroke and runs at 1000 cycles per minute. If the brake horsepower is 120 and the mechanical efficiency is 80%, estimate the indicated mean effective pressure.

258 6.18 Consider a naturally aspirated eight-cylinder two-stroke-cycle Diesel engine with a compression ratio of 20 and a cutoff ratio of 2.5. Air is inducted into the cylinder at 1 atm and 23°C. Assume an Air Standard cycle. (a) Determine the temperatures and pressures immediately before and after combustion. (b) What is the heat added in the combustion process, in kJ/kg? (c) What is the net work, in kJ/kg, and the thermal efficiency? (d) If the volumetric efficiency is 85%, the engine displacement is 2500 cc, and the engine speed is 2200 rpm, what is the mass flow rate of air through the engine, in lbm per minute? (e) What is the engine horsepower? (f) If losses through the valves cause a 120-kPa pressure differential between the pressures during the exhaust and intake strokes, estimate the actual and fractional losses, in horsepower, due to these processes. Sketch the appropriate p–V diagram. 6.19 A twelve-cylinder four-stroke-cycle Diesel engine has a 4-in. bore, a 4.5-in. stroke, and a compression ratio of 20. The mechanical efficiency is 89%, the cutoff ratios is 2, and the engine speed is 1200 rpm. The air entering the cylinder is at 14.5 psia and 60° F. Assuming Air Standard cycle performance, determine the cycle temperatures, indicated power, IMEP, and engine brake horsepower. 6.20 A hypothetical engine cycle consists of an isentropic compression, a constantpressure heat addition, and a constant-volume blowdown, consecutively. (a) Draw and label a p–V diagram for the cycle. (b) Use the cyclic integral of the First Law to derive an equation for the cycle net work in terms of the temperature. (c) Use the definition of mechanical work to derive an equation for the cycle net work also. Show that your equation is equivalent to the result obtained in part (b) using the cyclic integral. (d) Express an equation for the cycle thermal efficiency in terms of cycle temperature ratios and k. (e) If T1 = 60°F and the volume ratio is 10, determine the other cycle temperatures; and compare the cycle efficiency with the efficiency the Otto cycle having the same compression ratio. 6.21 A slightly more complex model of a reciprocating engine cycle than those discussed combines constant-pressure and constant-volume heat additions in a single Air Standard cycle. (a) Sketch and label a p–V diagram for this cycle that consists of the following

259 consecutive processes: isentropic compression, constant-volume heat addition, constant-pressure heat addition, isentropic expansion, and constant-volume blowdown. (b)The engine may be characterized by three parameters: the compression ratio; the Diesel engine cutoff ratio; and a third parameter, the ratio of the pressure after to that before the constant-volume combustion. Define these parameters in terms of the symbols in your sketch and derive an equation for the thermal efficiency of the cycle. (c) Show how varying the parameters appropriately reduces your efficiency equation to the equations for the Otto and Diesel cycles. 6.22 As a plant engineer you must recommend whether electric power for a plant expansion (2.5 MW continuous generation requirement) will be purchased from a public utility or generated using a fully attended Diesel-engine-driven generator or an automatic remotely controlled gas turbine generator set. The price of electricity is 4.8 cents per kW-hr, and the price of natural gas is 60 cents per thousand cubic feet. Both Diesel engine and gas turbine are to be natural-gas fired. The gas turbine has a heat rate of 11,500 Btu/kW-hr, and the Diesel engine 13,200 Btu/kW-hr. The initial costs of Diesel engine and gas turbine are $750,000 and $850,000, respectively. Control equipment for the gas turbine costs an additional $150,000. The engines and control equipment are estimated to have a useful life of 20 years. The annual wages and benefits for a Diesel engine operating engineer working eight-hour daily shifts is $36,000. Assume a 10% per annum interest-rate. Evaluate the alternatives for a natural gas heating value of 1000 Btu/ft3, and present a table of their costs, in cents per kW-hr. Discuss your recommendation. 6.23 Evaluate the alternatives in Exercise 6.22 based on the present-worth method. 6.24 In terms of the notation of Figure 6.3, what are the piston displacement, compression ratio, and expansion ratio for the Lenoir cycle? 6.25 What are the fuel and air flow rates and brake specific fuel consumption for an eight-cylinder engine with a 3.75-in. bore and 3.5-in. stroke delivering 212 horsepower at 3600 rpm with a brake thermal efficiency of 25%? The fuel is C8H18, and the equivalence ratio is 1.2. What is the power per cubic inch of engine displacement? 6.26* Construct a spreadsheet to perform an Air Standard cycle analysis for a Diesel engine with a compression ratio of 20 and a range of peak temperatures from 1000K to 3000K, in 500° increments. Use it to tabulate and plot both the net work per unit mass of air and the thermal efficiency against the cutoff ratio.

260 6.27 Determine the maximum tailpipe concentrations of the three federally regulated gaseous pollutants based on the existing standards for an automobile that achieves 28 mile/gal of iso-octane. Assume that the engine mixture equivalence ratio is 0.9, that NOx is represented by NO2 and unburned hydrocarbons by monatomic carbon, and that the fuel density is 700 kg/m3. 6.28 A single-cylinder Air Standard Otto engine has a compression ratio of 9.0 and a peak temperature of 3000°F at 80°F and one atmosphere ambient conditions. Determine the net work, cycle efficiency, maximum cylinder pressure, and mean effective pressure. 6.29 A six-cylinder engine with a compression ratio of 11 runs at 3200 rpm and 80°F and 14.7 psia. Each cylinder has a bore of 3 inches, a stroke of 3.25 inches, and a volumetric efficiency of 0.85. Assume an Air Standard four-stroke Otto cycle with stoichiometric octane as fuel. Assume that the energy release from the fuel is equally divided between internal energy increase in cylinder gases and cylinder wall heat loss. What are the cylinder mean effective pressures and the engine horsepower and specific fuel consumption? Assume a heating value of 20,600 Btu/lbm. 6.30 An eight-cylinder four-stroke-cycle compression-ignition engine operates with a fuel-air ratio of 0.03 at 2400 rpm. It has a turbocharger and intercooler, as diagrammed nearby, with compressor pressure ratio of 1.7 and intercooler exit temperature of 320K. The engine bore and stroke are 10 cm and 12 cm, respectively. The compressor efficiency, turbine efficiency, and volumetric efficiency are 70%, 75%, and 87%, respectively. The entrance temperature of the turbine gases is 1000K. What are the compressor power, the turbine pressure ratio, and the engine power, in kW and in horsepower? Assume that the engine is constructed of ceramic components that minimize engine heat losses so that they may be neglected—and ideal “adiabatic” engine.

261

261 CHAPTER 7 THE WANKEL ROTARY ENGINE

7.1 A Different Approach to the Spark-Ignition Engine The reciprocating internal combustion engine has served mankind for over a century, and will continue to do so for the foreseeable future. The Wankel rotary engine, a much more recent development, is said to have been conceived in its present form in 1954 (ref. 2). An implementation of the rotary engine used in the 1990 Mazda RX-7 automobile and its turbocharger are shown in Figures 7.1(a) and 7.1(b). As of 1987, over 1.5 million Wankel engines had been used in Mazda automobiles (ref. 6). The rotary engine has a host of advantages that make it a formidable contender for some of the tasks currently performed by reciprocating engines. The piston in a fourstroke-cycle reciprocating engine must momentarily come to rest four times per cycle as its direction of motion changes. In contrast, the moving parts in a rotary engine are in continuous unidirectional motion. Higher operating speeds, ease of balancing, and absence of vibration are a few of the benefits. The high operating speeds allow the engine to produce twice as much power as a reciprocating engine of the same weight. It has significantly fewer parts and occupies less volume than a reciprocating engine of comparable power. With all these advantages, why are there so few Wankel engines in service? Part of the answer lies in the reciprocating engine’s remarkable success in so many applications and its continuing improvement with research. Why change a good thing? Manufacturing techniques for reciprocating engines are well known and established, whereas production of rotary engines requires significantly different tooling. It must be admitted, however, that the rotary engine has some drawbacks. A major problem of the Wankel automobile engine is that it does not quite measure up to the fuel economy of some automotive reciprocating SI engines. It is the judgment of some authorities that it does not offer as great a potential for improvement in fuel economy and emissions reduction as reciprocating and gas turbine engines. However, although the rotary engine may never dominate the automotive industry, it is likely to find applications where low weight and volume are critical, such as in sports cars, general aviation, and motorcycles.

262

While the rotary engine may not enjoy the great success of reciprocating engines, it is worthy of study as an unusual and analytically interesting implementation of the familiar Otto cycle. Even the present success of this latter-day Otto engine should serve as an inspiration to those who search for novel ways of doing things. This chapter is a tribute to Felix Wankel and those who are helping to develop this remarkable engine. 7.2 Rotary Engine Operation Figure 7.2 shows a cross-section of a rotary engine. The stationary housing encloses a moving triangular rotor that rotates with its apexes in constant contact with the housing inner surface. Air and combustion gases are transported in the spaces between the rotor and the housing. The rotor rides on an eccentric that is an integral part of a shaft, as shown in the dual rotor crank shaft of Figure 7.3(a). The housing and rotor of a rotary engine designed for aircraft application are shown in Figure 7.3(b). The operation of the Wankel engine as an Otto-cycle engine may be understood by following in Figure 7.4 the events associated with the counterclockwise movement of a gas volume isolated between the housing and one of the rotor flanks. The air-fuel mixture may be supplied, by a conventional carburetor, through the intake port labeled I in Figure 7.4(a). As the shaft and rotor turn, the intake port is covered, trapping a fixed mass of air and fuel (assuming no leakage). This is analogous to the gas mass captured within the cylinder-piston volume by closure of a reciprocating engine intake valve. As the rotor continues to turn, the captured (crosshatched) volume contained

263

between the rotor and housing decreases, compressing the air-fuel mixture [part (b)]. When it reaches the minor diameter, the active mixture volume is a minimum corresponding to the volume at top center in the reciprocating engine. One or more spark plugs, as indicated at the top of each housing, initiate combustion, causing rapid rises in pressure and temperature [part (c)]. The hot, high-pressure combustion gas [part (d)] transmits a force to the eccentric through the rotor. Note that, during the

264

265 power phase, the line of action of the force F provides a counterclockwise torque acting about the shaft axis. As the rotation proceeds, the expanding gases drive the rotor until the exhaust port is exposed, releasing them [part (e)]. The exhaust process continues as the intake port opens to begin a new cycle. This port overlap is apparent in the lower volume shown in part (b). In summary, each flank of the rotor is seen to undergo the same intake, compression, ignition, power, and exhaust processes as in a four-stroke-cycle reciprocating Otto engine. All three flanks of the rotor execute the same processes at equally spaced intervals during one rotor rotation. Hence three power pulses are delivered per rotation of the rotor. Because there are three shaft rotations per rotor rotation, the Wankel engine has one power pulse per shaft rotation. Thus it has twice as many power pulses as a singlecylinder four-stroke-cycle reciprocating engine operating at the same speed, a clear advantage in smoothness of operation. This feature of one power pulse per shaft rotation causes many people to compare the Wankel engine with the two-stroke-cycle reciprocating engine. 7.3 Rotary Engine Geometry The major elements of the rotary engine—the housing and the rotor— are shown in cross-section in Figure 7.2. The housing inner surface has a mathematical form known as a trochoid or epitrochoid. A single-rotor engine housing may be thought of as two parallel planes separated by a cylinder of epitrochoidal cross-section. Following the notation of Figure 7.5, the parametric form of the epitrochoid is given by x = e cos 3 + Rcos

[ft | m]

(7.1a)

y = e sin 3 + R sin 

[ft | m]

(7.1b)

where e is the eccentricity and R is the rotor center-to-tip distance. For given values of e and R, Equations (7.1) give the x and y coordinates defining the housing shape when  is varied from 0 to 360 degrees. The rotor shape may be thought of as an equilateral triangle, as shown in Figures 7.2 and 7.4 (flank rounding and other refinements are discussed later in the chapter). Because the rotor moves inside the housing in such a way that its three apexes are in constant contact with the housing periphery, the positions of the tips are also given by equations of the form of Equations (7.1): x = e cos 3 + R cos( + 2n*/3)

[ft | m]

(7.2a)

y = e sin 3 + R sin( + 2n*)

[ft | m]

(7.2b)

where n = 0, 1, or 2, the three values identifying the positions of the three rotor tips, each separated by 120°. Because R represents the rotor center-to-tip distance, the

266

motion of the center of the rotor can be obtained from Equations (7.2) by setting R = 0. The equations and Figure 7.5 indicate that the path of the rotor center is a circle of radius e. Note that Equations (7.1) and (7.2) can be nondimensionalized by dividing through by R. This yields a single geometric parameter governing the equations, e/R, known as the eccentricity ratio. It will be seen that this parameter is critical to successful performance of the rotary engine. The power from the engine is delivered to an external load by a cylindrical shaft. The shaft axis coincides with the axis of the housing, as seen in Figure 7.2. A second circular cylinder, the eccentric, is rigidly attached to the shaft and is offset from the shaft axis by a distance, e, the eccentricity. The rotor slides on the eccentric. Note that the axes of the rotor and the eccentric coincide. Gas forces exerted on the rotor are transmitted to the eccentric to provide the driving torque to the engine shaft and to the external load. The motion of the rotor may now be understood in terms of the notation of Figure 7.5. The line labeled e rotates with the shaft and eccentric through an angle 3, while the line labeled R is fixed to the rotor and turns with it through an angle  about the moving eccentric center. Thus the entire engine motion is related to the motion of these two lines. Clearly, the rotor (and thus line R) rotates at one-third of the speed of the shaft, and there are three shaft rotations for each rotor revolution.

267 EXAMPLE 7.1

Derive expressions for the major (largest) and minor (smallest) diameters of an epitrochoid in terms the notation of Figure 7.5. Solution

The major diameter is defined by adding the lengths of the lines representing the eccentricity and the rotor radius when they are horizontal and colinear or by using Equation 7.1(a). Thus the major diameter at y = 0 corresponds to  = 0° and 180°, for which x = e + R and x = – e – R, respectively. The distance between these x coordinates is the length of the major diameter 2(e + R). The minor diameter is similarly defined along x = 0, but with e and R lines oppositely directed. The two cases correspond to  = 90° and 270°. For  = 90°, the e line is directed downward and the R line upward in Figure 7.5. This yields y = R – e and, by symmetry, the minor diameter is 2(R – e). Hence Major diameter = 2(R + e) Minor diameter = 2(R – e) _____________________________________________________________________

7.4 A Simple Model for a Rotary Engine Additional important features of the rotary engine can be easily studied by considering an engine with an equilateral triangular rotor. Figure 7.6 shows the rotor in the position where a rotor flank defines the minimum volume. We will call this position top center, TC, by analogy to the reciprocating engine. The rotor housing clearance parameter, d, is the difference between the housing minor radius, R – e, and the distance from the housing axis to mid-flank, e + R cos 60 = e + R/2: d = (R – e) – (e + R/2) = R/2 – 2e

[ft | m]

(7.3)

Setting the clearance to zero establishes an upper limiting value for the eccentricity ratio: (e/R)crit = 1/4. Study of Equations (7.1), at the other extreme, shows that, for e/R = 0, the epitrochoid degenerates to a circle. In this case the rotor would spin with no eccentricity and thus produce no compression and no torque. Thus, for the flat-flanked rotor, it is clear that usable values of e/R lie between 0 and 0.25. Now let's examine some other fundamental parameters of the flat-flanked engine model. Consider the maximum mixture volume shown in Figure 7.7. For a given rotor width w, the maximum volume can be determined by calculating the area between the housing and the flank of the rotor. Using Equations (7.1), the differential area 2y dx can be written as:

268

269 dAmax = 2y dx = 2(e sin3 + R sin) d(e cos3 + R cos)

[ft2 | m2]

(7.4)

Dividing by R2 and differentiating on the right-hand side, we obtain an equation for the dimensionless area in terms of the eccentricity ratio and the angle : Amax/R2 = – 2

[(e/R)sin3 + sin][3(e/R)sin3 + sin]d [dl]

(7.5)

In order for the differential area to sweep over the maximum trapped volume in Figure 7.7, the limits on the angle  must vary from 0° to 60°. Thus integration of Equation (7.5) with these limits and using standard integrals yields Amax/R2 = * [(e/R)2 + 1/3] – 31/2/4[1 – 6(e/R)]

[dl]

(7.6)

Similarly, using Figure 7.6 and the differential volume shown there, the nondimensionalized minimum area can be written as: Amin/R2 = * [(e/R)2 + 1/3] – 31/2/4 [1 + 6(e/R)]

[dl]

(7.7)

These maximum and minimum volumes (area-rotor width products) are analogous to the volumes trapped between the piston and cylinder at BC and TC in the four-stroke reciprocating engine. In that engine the difference between the volumes at BC and TC is the displacement volume, and their ratio is the compression ratio. A little thought should convince the reader that the analogy holds quantitatively for the displacement and compression ratio of the rotary engine. Therefore, subtracting Equation (7.7) from Equation (7.6) gives the displacement for a rotor width w for one flank of the flat-flanked engine as disp = 3 (31/2 wR2(e/R)

[ft3 | m3]

(7.8)

[dl]

(7.9)

and forming their ratio yields the compression ratio as Amax/R2 * [(e/R)2 + 1/3] – 31/2/4[1 – 6(e/R)] CR = --------- = ------------------------------------------Amin/R2 * [(e/R)2 + 1/3] – 31/2/4 [1 + 6(e/R)]

Thus the displacement increases with increases in rotor width, the square of the rotor radius, and with the eccentricity ratio, whereas the compression ratio is independent of size but increases with increase in eccentricity ratio.

270 EXAMPLE 7.2

What are the displacement and the compression ratio for a flat-flanked rotary engine with a rotor radius of 10 cm, an eccentricity of 1.5 cm, and a rotor width of 2.5 cm? Solution

For this engine, e/R = 1.5/10 = 0.15. Equation (7.8) then yields the displacement: 3(3)0.5(0.15)(10)2(2.5) = 194.9 cm3 or (194.9)(0.0610) = 11.89 in.3 Equation (7.9) can be written as CR = (a + b)/(a – b) where a = (3.14159)[(0.15)2 + 1/3] – 31/2/4 = 0.6849, and b = 3( 31/2(0.15)/2 = 0.3897. Then CR = (0.6849 + 0.3897)/(0.6849 - 0.3897) = 3.64 _____________________________________________________________________ The very low compression ratio of Example 7.2 would yield a poor Otto-cycle thermal efficiency. The compression ratio could be increased by increasing e/R, but it would still be low for most applications. It is therefore important to consider the favorable influence of flank rounding on rotary engine performance. 7.5 The Circular-Arc-Flank Model While the triangular rotor model represents a possible engine and is useful as a learning tool, such an engine would perform poorly compared with one having a rotor with rounded flanks. A more realistic model is one in which the triangular rotor is augmented with circular-arc flanks, as shown in Figure 7.8. The radius of curvature, r, of a flank could vary from infinity, corresponding to a flat flank, to a value for which the arc touches the minor axis of the epitrochoid. Note that the center of curvature of an arc terminated by two flank apexes depends on the value of r. It can also be seen from Figure 7.8 that r is related to the angle, , subtended by the flank arc by or

r sin( /2) = R sin(*/3) = 31/2R/2

[ft | m]

r/R = 31/2/[2sin( /2)]

[dl]

(7.10)

Thus either the included angle, , or the radius of curvature, r, may be used to define the degree of flank rounding for a given rotor radius R.

271

Clearance with Flank Rounding

The additional area obtained by capping a side of a triangle with a circular arc is called a segment. The segment height, h, shown in Figure 7.8, is the difference between r and the projection of r on the axis of symmetry: h/R = (r/R)[1 – cos( /2)]

[dl]

(7.11)

Substitution of Equation (7.10) in Equation (7.11) yields h/R = 31/2 [1 – cos( /2)] / [2sin( /2)]

[dl]

(7.12)

It is evident from the figure that the clearance for the rotor with circular arc flanks is the difference between the clearance of the flat-flanked rotor and h. Thus, using Equation (7.3), the clearance is given by d/R = 1/2 – 2(e/R) – 31/2 [1 – cos( /2)] / [2sin( /2)]

[dl]

In a real engine, of course, the clearance must be non-negative.

(7.13)

272 Added Volume per Flank Due to Rounding The segment area is the difference between the pie-shaped area of the sector subtended by its included angle, , and the enclosed triangular area. The sector area, or volume per unit rotor width, is the fraction of the area of a circle of radius, r, subtended by the angle ; i.e., * r2 ( /2* ) = r2 /2. Thus using Equation (7.10), the dimensionless segment volume is As /R2 = (Asec – Atri ) /R2 = (r/R)2( – sin ) /2 = (3/8)( – sin

) /sin2 /2

[dl]

(7.14)

Displacement and Compression Ratio It was pointed out earlier that the displacement of the flat-flanked engine is the difference between the maximum and minimum capture volumes, and is given by Equation (7.8). This is true also for the engine with rounded flanks. The additional volume added to the rotor by flank rounding subtracts from both of the flat-flanked maximum and minimum capture volumes, leaving the difference unchanged. Thus the displacement of one flank of a rounded-flank engine is disp = 3 (31/2 wR2 (e/R)

[ft3 | m3]

(7.15)

Likewise, the ratio of the maximum and minimum capture volumes given by Equations (7.6) and (7.7), corrected for the segment volume from Equation (7.14) provides a relation for the rounded-flank engine compression ratio:

* [(e/R)2 + 1/3] – 31/2/4 [1 – 6(e/R)] – As /R2

CR = ----------------------------------------------------* [(e/R)2 + 1/3] – 31/2/4 [1 + 6(e/R)] – As /R2

[dl]

(7.16)

The added rotor volume due to rounding subtracts from the flat-flanked capture volumes and therefore reduces the denominator of Equation (7.16) more than the numerator. As a result, the compression ratio is greater for rounded-flank than for flatflanked engines. Rotary engines usually have the maximum rounding possible consistent with adequate engineering clearances. Effect of the Recess Volume Equation (7.16) accounts for flank rounding but not for the recess usually found in rotor flanks. The additional capture volume associated with the recess is seen in Figure 7.9. Its influence on the displacement and compression ratio may be reasoned in the

273

same way as with the segment volume. The recess increases both minimum and maximum mixture volumes by the same amount. It therefore has no effect on displacement and it decreases the compression ratio. Figure 7.10 shows the influence of flank rounding and recession on clearance and compression ratio. While flank recession reduces the compression ratio for given values of and e/R, it improves the shape of the long, narrow combustion pocket forming the minimum capture volume. Rotary engines usually have more than one spark plug, to help overcome the combustion problems associated with this elongated shape. EXAMPLE 7.3

Rework Example 7.2, taking into account a flank-arc included angle of 0.65 radians. What is the flank clearance for this engine? Solution

Because flank rounding does not influence it, the displacement is still 194.9 cm3. Equation (7.16) rewritten using the notation of Example 7.2 becomes

274 CR = (a + b – As /R2) / (a – b – As /R2) where As /R2 = (3/8)[0.65 – sin(0.65)]/sin2(0.65/2) = 0.1648. Then the compression ratio is CR = [0.6849 + 0.3897 – 0.1648] / [0.6849 – 0.3897 – 0.1648] = 6.98 This represents a significant improvement over the value of 3.64 for the flat-flanked rotor. The flank clearance is given by Equation (7.13): d = 10{0.5 – 2(0.15) – 31/2[1 – cos(0.65/2)]} / [2sin(0.65/2)] = 0.58 cm. _____________________________________________________________________ We have already noted that the displacement volume associated with one flank of the rotary engine produces one power stroke during each rotor revolution and during three shaft rotations. Because there are three flanks per rotor, a rotor executes one complete thermodynamic cycle per shaft rotation. Thus the power produced by a single rotor is determined by the displacement volume of a single flank and the rotational speed: (disp [cm3/Rev])(MEP [kN/cm2])(N [Rev/min]) Power = ------------------------------------------------------(60 [sec/min])(100 [cm/m])

[kW]

(disp [in3/Rev])(MEP [lb/in2])(N [Rev/min]) Power = --------------------------------------------------(12 [in/ft])(33,000 [ft-lb/HP-min])

[HP]

or

7.6 Design and Performance of the Wankel Engine It is evident from Figure 7.4 that, in the Wankel engine, the opening and closing of the intake and exhaust ports by the motion of the rotor apexes serves a function equivalent to that of mechanical valves in reciprocating engines. This simple operation in the Wankel engine eliminates the need for many of the moving parts required by the reciprocating engine, such as cams, camshafts, tappets, valves, and lifters. There are, in fact, many more parts in a reciprocating engine than in a comparable rotary engine. However, sealing at the apexes and sides of the rotor is critical for efficient operation of the rotary engine. Significant pressure differences between the three active mixture volumes of a rotor in different phases of the Otto cycle require efficient seals

275

analogous to piston rings in the reciprocating engine. These are needed to avoid leakage between adjacent volumes, which causes a loss of compression and power. Seal friction has been estimated to account for about 25% of rotary engine friction. Spring loaded, self-lubricating apex seals, as shown in Figure 7.11, allow for sliding with low friction over the treated-chrome-alloy-plated housing inner surface. The figure shows improvements in apex seal design (ref. 6). The three-piece seal design, with two leaf springs rather than one, decreases seal mass through reduced thickness, and offers a configuration that promotes area contact rather than line contact between seal and rotor. Side seals are also important to maintain pressure integrity of each flank mixture pocket. Reductions in the thickness of both apex and side seals have decreased friction with the housing by reducing the seal area producing the friction-

276 causing normal force on the housing. Oil seals, also on the rotor sides, are used to control oil consumption. Though the peripheral intake port shown in Figures 7.4 and 7.9 provides better performance under heavy loads than a single side port, its associated intake-exhaust port overlap may allow excessive flow of exhaust gas into the fresh mixture, causing unreliable combustion in low-speed operation. Consequently, one or more side intake ports, in addition to or instead of a peripheral intake port, are sometimes used. Side ports, of course, are also opened and closed by rotor motion. In addition to reducing intake-exhaust overlap at light loads, side intake ports also induce combustionenhancing swirl in the air-fuel mixture. It is evident that the moving combustion volume at the time of ignition has a long and narrow flame propagation path. Rounded rotor flanks are usually recessed to provide a wider flame front path between the two lobes of the active volume. In highspeed operation, the brief time for combustion may dictate additional design features. Multiple spark plugs, swirl induced by side intake ports and multiple ports, the "squish" produced by the the relative motion of the walls of the active volume, fuel injection, and stratified-charge design all can contribute to improvement of the combustion process. It may be noted in Figures 7.3 and 7.9 that an internal ring gear is attached to the rotor. This gear meshes with a stationary gear attached to the engine housing. The function of this gearset is to position the rotor as the shaft turns—not to transmit torque. Engine torque, as indicated earlier, is transmitted by direct contact of surface forces between the rotor and the eccentric. Stratified-Charge Rotary Engine Reference 7 discusses the design and performance of stratified-charge rotary engines developed for commercial aviation propulsion and APU (auxiliary power unit) application as well as for marine, industrial, and military requirements. Figure 7.12 shows a direct fuel injection configuration that has performed well under a wide range of speed, load, and environmental conditions and with a variety of liquid fuels. The reference reports a lack of octane and cetane sensitivities, so that diesel, gasoline, and jet fuel can all be used with this configuration. As air in the rotor recess passes below, the spark plug ignites a locally rich pilot stream that in turn ignites the fuel from the main injector. The net fuel-air ratio is lean, resulting in improved fuel economy over normal carburetion. Figure 7.13 presents data for full-load brake horsepower and specific fuel consumption obtained with Jet-A fuel for the twin-rotor 2034R engine. The maximum takeoff power at 5800 rpm was 430 horsepower, with a brake specific fuel consumption (BSFC) of 0.44 lbm/BHP-hr. Throughout a range of loads and altitude conditions the engine operates with a fuel-air ratio between 0.035 and 0.037, well below the stoichiometric value. The reference reports a best thermal efficiency of 35.8% (BSFC = 0.387 lbm/BHP-hr) at 3500 rpm and 225-horsepower output.

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278 Closure Continued engineering research on the rotary engine has resulted in performance improvements through improved seals, lean-burn combustion, fuel injection, integral electronic control, improved intake design, weight reduction, and turbocharging. Despite vehicle weight increases, the Mazda RX-7 with a two-rotor 80-in.3 -displacement engine improved 9.4% in fuel consumption and 8% in power output between 1984 and 1987 (ref. 6). During this time period, the addition to the engine of a turbocharger with intercooling increased its power output by 35%. Reference 8 reports that the Mazda RX-Evolv, a year-2000 concept car, has a naturally-aspirated rotary engine called “RENESIS.” The two-rotor, side intake and exhaust engine is reported to have reduced emissions and improved fuel economy and to have attained 280 horsepower at 9000 rpm and 226 N-m torque at 8000 rpm. EXAMPLE 7.4

If the BMEP of the 11.89-in3-diplacement engine in Example 7.2 is 150 psi at 4000 rpm, what is the brake horsepower? Solution

The brake horsepower is BHP = (150)(4000)(11.89)/(12)(33000) = 18 horsepower or BHP = (18)(0.746) =13.44kW _____________________________________________________________________ Bibliography and References 1. Cole, David E. "The Wankel Engine," Scientific American, Vol. 227, No. 2, (August 1972): 14–23. 2. Ansdale, R. F., The Wankel R C Engine. South Brunswick, N.J.: A. S. Barnes, 1969. 3. Yamamoto, Kenichi, Rotary Engine, Tokyo: Sankaido Co., 1981. 4. Weston, Kenneth C. "Computer Simulation of a Wankel Rotary Engine—Analysis and Graphics." Proceedings of the Conference of the Society for Computer Simulation, July 1986, pp. 213-216. 5. Weston, Kenneth C., "Computerized Instruction in the Design of the Wankel Rotary Engine." ASEE Annual Conference Proceedings, June 1988.

279 6. Fujimoto, Y. et al., "Present and Prospective Technologies of Rotary Engine." Society of Automotive Engineers Paper 870446, 1987. 7. Mount, Robert E., and LaBouff, Gary A., “Advanced Stratified-Charge Rotary Engine Design.” Society of Automotive Engineers Paper 890324 (also in SAE SP-768, Rotary Engine Design; Analysis and Developments), 1989. 8. Jost, Kevin, (ed.), “Global Concepts–Mazda RX–Evolv,” Automotive Engineering International, Vol 8, No.8, (August 2000), p 59. EXERCISES

7.1 Using graph paper, plot, on a single sheet, epitrochoids for e/R = 0, 0.15, 0.2, 0.25, 0.3, and 0.4. On a separate sheet, draw the epitrochiod and the triangular rotor for three rotor positions (separated by 30°) for the case of e/R = 0.15. 7.2 Verify the results of Example 7.1 by specializing Equations (7.1) for the appropriate values of . 7.3 Derive Equation (7.5) using a differential area given by (x – xf) dy, where xf is the constant x-coordinate of the flank. 7.4 Following the approach in the derivation of Equations (7.3)–(7.6), and using the notation of Figure 7.6, derive Equation (7.7). 7.5 Derive Equation (7.7) using a differential area (y – yf ) dx, where yf is the constant y-coordinate of the flank. 7.6 Derive Equation (7.8). 7.7 Derive Equation (7.9). 7.8 Show that the radius of curvature for a circular-arc flank that touches the epitrochoid at its midpoint is given by r/R = 1 – e/R + 3(e/R)/(1 – 4e/R) 7.9 Use Equation (7.13) to derive an expression for the limiting value of e/R as a function of the flank included angle. Plot the limiting value of e/R as a function of the included angle. 7.10 Solve Example 7.3, accounting for a rotor flank recess of 3% of R2w.

280 7.11 If combustion takes place in an engine rotor rotation interval of 40° in an engine operating at 8000 rpm, how much time is available for the combustion process? 7.12* Develop a single-column spreadsheet that determines the compression ratio, clearance ratio, and nondimensional displacement for a given value of e/R and flank-rounding angle. Use a copy command to replicate the column, forming a table of alternative design characteristics, for a reasonable range of rounding angles. 7.13* Use the spreadsheet graphics option to develop plots of compression ratio and clearance ratio, as seen in Figure 7.10. 7.14 A snowmobile single-rotor Wankel engine developed 6.65 brake horsepower at 5500 rpm with a displacement volume of 108 cc and a compression ratio of 8.5. The fuel consumption was 2.77 lbm/hr. Determine the BMEP (in psi), the BSFC (in lbm/hp-min), the brake torque (in lbf-ft), and the brake thermal efficiency, assuming a fuel heating value of 18,900 Btu/lbm. 7.15 A rotary engine mounted on a dynamometer develops 23 lbf-ft of torque at 5000 rpm. When driven by a motor-generator at the same speed, a torque of 7 lbf-ft is required. Determine the brake and indicated horsepower and the engine mechanical efficiency. What additional information is needed to determine the indicated mean effective pressure? 7.16 A rotary engine has an eccentricity of 2 in. and an equilateral triangular rotor with a tip radius of 10 in. (a) Determine the major and minor diameters of the epitrochoidal housing. (b) Sketch the housing and its axes of symmetry and the rotor when it is in the nominal spark-plug firing position. (c) For the configuaration of part (b), determine the minimum rotor clearance. (d) Write equations for the relations between the shaft speed (rpm), the spark plug firing rate (FR), and the rotor speed (RS). Identify any new symbols used. 7.17 A rotary engine has an eccentricity of 3 cm and an equilateral triangular rotor with a tip radius of 13 cm. (a) Determine the major and minor diameters of the epitrochoidal housing. (b) Sketch the housing and its axes of symmetry and the rotor when it is in the nominal spark-plug firing position. (c) For the configuaration of part (b), determine the minimum rotor clearance.

281 (d) Write equations for the relations between the shaft speed (rpm), the spark plug firing rate (FR), and the rotor speed (RS). Identify any new symbols used. 7.18 A rotary engine with a flat-flanked rotor has a ratio of maximum to minimum housing inside diameters of 1.4. What is the engine compression ratio? 7.19 The major diameter of the epitrochoidal housing of a flat-flanked single-rotor industrial rotary engine is 39.6 inches. The engine turns at 1000 rpm while delivering 500 brake horsepower at a BMEP of 79.2 psi. If the eccentricity ratio is 0.14, what are the minor diameter, the rotor thickness, and the rotor displacement, [in.3]? 7.20 A flat-flanked dual-rotor industrial rotary engine has a 60-cm minor diameter.The engine delivers 800kW from an IMEP of 700kPa at a shaft speed of 20 rps. The mechanical efficiency is 89%, and the eccentricity ratio is 0.16. Determine the major diameter and the thickness and displacement of each rotor. 7.21 A Wankel rotary engine has an eccentricity of 2.2 in. and a major diameter of 28 in. It has a compression ratio of 9.5 and a 600 in.3 displacement. Determine the rotor width and the rotor sector included angle if the rotor flanks are circular and have no indentations. 7.22 A Wankel rotary engine has an eccentricity of 2.5 cm and a major diameter of 32 cm. The engine compression ratio is 9.0, and the displacement is 540 cm3. What are the eccentricity ratio, the rotor width, and the rotor sector included angle if the rotor flanks are circular and have no indentations?

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CHAPTER 8 REFRIGERATION AND AIR CONDITIONING

8.1 Introduction Up to this point we have considered fossil-fueled heat engines that are currently in use. These devices have provided society's answers to the thermodynamic question: How can the chemical energy of fossil fuels be converted into mechanical work and motive power? Let us now turn our attention to the another great thermodynamic question: How can thermal energy be transferred from cold to warmer regions? The well-known Clausius statement of the Second Law of Thermodynamics asserts: It is impossible to construct a device that, operating in a cycle, has no effect other than the transfer of heat from a cooler to a hotter body. Thus the Clausius statement tells us that energy (heat) will not flow from cold to hot regions without outside assistance. The devices that provide this help are called refrigeration units and heat pumps. Both types of devices satisfy the Clausius requirement of external action through the application of mechanical power or natural transfers of heat (more on this later). The distinction between refrigerator and heat pump is one of purpose more than technique. The refrigeration unit transfers energy (heat) from cold to hot regions for the purpose of cooling the cold region while the heat pump does the same thing with the intent of heating the hot region. The following will focus on refrigeration and make the distinction between refrigeration and heat pumps only when it is essential to the discussion. It should be pointed out that any heat engine cycle, when reversed, becomes a refrigeration cycle, becuase the cyclic integral of the heat transfer, and thus the net work, becomes negative. This implies heat rejection at higher than the lowest cycle temperature. The vapor compression refrigeration system is the mainstay of the refrigeration and air conditioning industry. Absorption refrigeration provides an alternative to the vapor compression approach, particularly in applications where a heat source is economical and readily available. This chapter considers both of these system types in turn, and closes with a discussion of moist air behavior and its influence on air conditioning system design. Numerous other specialized refrigeration systems exist. As just stated, in principle any heat engine cycle, when reversed, becomes a refrigeration cycle. Such cycles are usually discussed in thermodynamics texts. Other systems that are occasionally used in special applications include thermoelectric coolers, which employ electrical work, and Hilsch or vortex tubes, which employ compressed gas as an energy source.

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8.2 Vapor Compression Refrigeration Consider an insulated cold region of temperature TL as shown in Figure 8.1. Heat leakage from the surroundings to the system tends to increase the system’s temperature. In order to keep the cold region at temperature TL, the conservation of energy requires the removal of an amount of heat equal to the energy inflow. This is done by a cold region heat exchanger that has an even colder liquid flowing through it to carry away the heat. If the fluid is a saturated liquid, it will evaporate and absorb energy from the cold region in its heat of vaporization. Such a heat exchanger is called an evaporator. Thus the basic problem of refrigeration may be reduced to one

284 of providing a mechanism to supply cool saturated liquid or a mixture of liquid and vapor, the refrigerant, to an evaporator. Vapor compression refrigeration, as the name suggests, employs a compression process to raise the pressure of a refrigerant vapor flowing from an evaporator at pressure p1 to p2, as shown in Figure 8.2. The refrigerant then flows through a heat exchanger called a condenser at the high pressure, p2 = p3, through a throttling device, and back to the low pressure, p1, in the evaporator. The pressures p2 = p3 and p4 = p1 correspond to refrigerant saturation temperatures, T3 and T1 = T4, respectively. These temperatures allow natural heat exchange with adjacent hot and cold regions from high temperature to low. That is, T1 is less than TL; so that heat, QL, will flow from the cold region into the evaporator to vaporize the working fluid. Similarly, the temperature T3 allows heat, QH, to be transferred from the working fluid in the condenser to the hot region at TH. This is indicated by the arrows of Figure 8.2. Thus the resulting device is one in which heat is transferred from a low temperature, TL, to a high temperature, TH, using a compressor that receives work from the surroundings, therein satisfying the Clausius statement. The throttling device, as shown in Figure 8.2, restrains the flow of refrigerant from the condenser to the evaporator. Its primary purpose is to provide the flow resistance necessary to maintain the pressure difference between the two heat exchangers. It also serves to control the rate of flow from condenser to evaporator. The throttling device may be a thermostatic expansion valve (TEV) controlled by evaporator exit temperature or a long, fine-bore pipe called a capillary tube. For an adiabatic throttling device, the First Law of Thermodynamics requires that h3 = h4 for the irreversible process, because Q and W are zero and kinetic energy change is negligible. Thus saturated liquid at T3 flashes to a mixture of liquid and vapor at the evaporator inlet at the enthalpy h4 = h3 and pressure p4 = p1. Also the evaporator entrance has the quality x4 and temperature T4 = T1. Heat from the cold source at TL > T4 boils the mixture in the evaporator to a saturated or slightly superheated vapor that passes to the suction side of the compressor. The compressor in small and medium-sized refrigeration units is usually a reciprocating or other positive-displacement type, but centrifugal compressors often are used in large systems designed for commercial and industrial service. It may be noted from the T-s diagram in Figure 8.2 that the vapor compression cycle is a reversed Rankine cycle, except that the pressure drop occurs through a throttling device rather than a turbine. In principle, a turbine or expansion device of some kind could be used to simultaneously lower the refrigerant pressure and produce work to reduce the net work required to operate the compressor. This is very unlikely because of the difficulty of deriving work from a mixture of liquid and vapor and because of the low cost and simplicity of refrigeration throttling devices. An exploded view of a through-the-wall type room air conditioner commonly used in motels and businesses is shown in Figure 8.3. A fan coil unit on the space side is the evaporator. A thermally insulating barrier separates a hermetically sealed, electric-motor-driven positive displacement compressor unit and a finned-tube heat exchanger condenser from the room on the outdoor side.

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286

Figure 8.4 shows a packaged air-conditioning unit designed for in-space use or for a nearby space with short duct runs. Units sometimes are designed to operate with either one or two compressors, coils, and fans to better accommodate varying cooling demands. A unit with watercooled condensers such as that shown requires an external heat sink, usually provided by a nearby ground-level or rooftop evaporative cooling tower. Rather than being combined in a single enclosure, refrigeration units frequently are installed as split systems. Figure 8.5 shows an uncovered rooftop condensing unit that contains a compressor and air cooled condenser. Such units are, of course, covered to resist the outdoor environment over many years. Cooled refrigerant is piped in a closed circuit to remote air distribution units that contain cooling coils (evaporators) and throttling devices. Figure 8.6 shows a skid-mounted air-cooled condensing unit also designed to function with remote evaporators in applications such as walk-in coolers. Refrigerants Refrigerants are specially selected substances that have certain important characteristics including good refrigeration performance, low flammability and toxicity, compatibility with compressor lubricating oils and metals, and good heat transfer characteristics. They are usually identified by a number that relates to their molecular composition.

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The ASHRAE Handbook (ref. 1) identifies a large number of refrigerants by number, as shown in Table 8.1. Inorganic refrigerants are designated by 700, plus their molecular weight. For hydrocarbon and halocarbon refrigerants, the number scheme XYZ works as follows: (1) Z, on the right is the number of fluorine atoms; (2) Y is the number of hydrogen atoms plus one; and (3) the leftmost digit, X, is one less than the number of carbon atoms in the compound. Two important examples are refrigerants R-12 and R-22. R-12, dichlorodifluoromethane, has two fluorine, one carbon, and two chlorine atoms in a methane-type structure. Thus the halogens,

288

289

290

291

chlorine and fluorine, replace hydrogen atoms in the CH4 molecular structure as shown in Figure 8.7. R-22, monochlorodifluoromethane, has a similar structure to R-12, except for a single hydrogen atom replacing a chlorine atom. Charts of the thermodynamic properties of these refrigerants are given in Appendix F. The commonly used chlorofluorocarbon (CFC) refrigerants are a cause of great concern, because their accumulation in the upper atmosphere creates a “hole” in the ozone layer that normally shields the earth from solar ultraviolet radiation (refs. 8 and 9). In 1987, more than 35 countries, including the United States, signed the Montreal Protocol on Substances that Deplete the Ozone Layer. The Montreal Protocol called for a freeze in 1989 and reductions in the 1990s on the production levels of R-11, R-12, R-113, R-114, and R-115. The halocarbon refrigerants, some of which are also widely used as aerosol propellants, foams, and solvents, are now categorized as chlorofluorocarbons (CFCs), hydrochlorofluorocarbons (HCFCs), or hydrofluorocarbons (HFCs). The HFCs, lacking chlorine, are no threat to the ozone layer but are not in common usage as refrigerants. CFCs, which contain more chlorine than do HCFCs, are the most serious offenders, are very stable, and do not break down rapidly in the lower atmosphere. The Clean Air Act of 1990 (ref. 15) mandated termination of production in the United States of all CFCs such as R-12 by the year 2000. Government data indicate that, because of the structural difference between them, R-12 has twenty times the ozone-depletion potential in the upper atmosphere of R-22. Nevertheless, R-22 and other HCFCs are also scheduled by the law for phaseout of production by the year 2030. Thus, the search for alternate refrigerants to replace those used in existing applications (worth hundreds of billions of dollars) has assumed enormous importance. It is a difficult, expensive, and continuing task to which industry is vigorously applying its efforts. Charts of thermodynamic properties for two of the newer refrigerants, R-123 and R134a, are given in Appendix F.

292 Vapor-Compression Cycle Analysis A vapor-compression cycle was shown in Figure 8.2, The work required by the refrigeration compressor, assuming adiabatic compression, is given by the First Law of Thermodynamics: w = h1 – h2

[Btu/lbm | kJ/kg]

(8.1)

where the usual thermodynamic sign convention has been employed. The enthalpies h1 and h2 usually are related to the temperatures and pressures of the cycle through the use of charts of refrigerant thermodynamic properties such as those given in Appendix F. In the ideal vapor compression cycle, the compressor suction state 1 is assumed to be a saturated vapor. The state is determined when the evaporator temperature or pressure is given. For the ideal cycle, for which compression is isentropic, and for cycles for which the compression is determined using a compressor efficiency, state 2 may be defined from state 1 and the condensing temperature or pressure by using the chart of refrigerant thermodynamic properties. Assuming no heat exchanger pressure losses, the evaporator and condenser heat transfers are easily determined per unit mass of refrigerant by application of the First Law of Thermodynamics: qL = h1 – h4

[Btu/lbm | kJ/kg]

(8.2)

qH = h3 – h2

[Btu/lbm | kJ/kg]

(8.3)

The evaporator heat transferred, qL, is commonly referred to as the refrigeration effect, RE. The product of the refrigerant mass flow rate and RE, the rate of cooling produced by the unit, is called the refrigeration capacity [Btu/hr | kW]. Applying the First Law to the refrigerant in the system as a whole, we find that the work and heat-transfer terms are related by [Btu/lbm | kJ/kg] (8.4) qL + qH = w where qH and w are negative for both refrigerators and heat pumps. Hence qL + |w| = |qH|

[Btu/lbm | kJ/kg]

(8.5)

Equation (8.5) is written here with absolute values to show that the sum of the compressor work and the heat from the low-temperature source is the energy transferred by the condenser to the high-temperature region. This may be seen graphically by addition of the enthalpy increments representing Equations (8.1) to (8.3) in the pressure-enthalpy diagram shown in Figure 8.8. The p–h diagram is applied often in refrigeration work because of its ease of use in dealing with enthalpy differences and constant-pressure processes. EXAMPLE 8.1

Derive Equation (8.1) by using Equations (8.2) to (8.4).

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Solution

From Equation (8.4), w = qL + qH = (h1 – h4) + (h3 – h2) But for the adiabatic throttling valve, h3 = h4. Hence, w = h1 – h2 as in Equation (8.1). _____________________________________________________________________________ System Performance As mentioned in Chapter 1, a measure of the efficiency of a refrigerator or heat pump is the coefficient of performance, COP. The COP for a refrigerator is defined as the ratio of the useful

294 effect or desired energy transfer accomplished by the evaporator (RE) to the energy cost to achieve the effect (the compressor work), in equivalent units: COPr = qL/|w|

[dl]

(8.6)

Like a refrigerator, a heat pump may be a vapor compression system; but unlike the refrigerator it takes energy from a cold source and transfers it to a hot region for the purpose of raising the temperature of the region or compensating for heat losses from the region. Some commercial heat pumps are systems that combine both cooling and heating actions in a single package. These systems are discussed more fully later. For a heat pump, the useful effect or desired energy transfer is the passage of energy from the condenser to the high-temperature region at TH. Thus the COP for a heat pump is COPhp = |qH| / |w|

[dl]

(8.7)

It is evident from Equations (8.6) and (8.7) that high values of COP are desired and may be achieved by minimizing the compressor work input for given values of heat transfer. It may be particularly advantageous to apply a vapor compression system in situations and applications in which simultaneous heating and cooling functions are required–to cool a computer room with the evaporator while heating rooms on the cold side of the building with the condenser, for example. EXAMPLE 8.2

Show that the COP for a heat pump exceeds 1. Solution

Combining Equations (8.7) and (8.5) yields COPhp = |qH| / |w| = (qL + |w|) / |w| = 1 + qL / |w| = 1 + COPr > 1 Thus, because qL > 0 and COPr >0 , the heat pump COP must always exceed 1. This is also evident directly from the definition Equation (8.7) because the First Law [Equation (8.5)] requires that |qH| > |w|. Note the relationship of refrigerator and heat pump COPs for the same cycle. _____________________________________________________________________________ The T-s diagram in Figure 8.2 shows that condenser and evaporator pressures of both refrigerators and heat pumps are determined largely by the source and sink (cold-region and hotregion) temperatures, TL and TH, respectively. Thus, the greater the difference between the hotregion and cold-region temperatures (the application temperature difference), the higher the compressor pressure ratio and the higher the compressor work. For low application temperature differences, |w| can be significantly less than qL; and thus COPr may also exceed 1. Conversely, large application temperature differences lead to low coefficients of performance.

295 Prior to the twentieth century, the cooling of both people and perishable goods was achieved by using ice formed in the winter by nature from liquid water. Ice from lakes and ponds in northern climates was cut, hauled to ice storage houses, and delivered to customers as needed. As late as the 1930s and ’40s, the ice vendor would make a daily delivery of ice so that perishable goods could be kept fresh in insulated ice storage boxes, the predecessors of the modern refrigerator. By then, ice was being made by freezing water via large refrigeration units with ammonia compressors; and comfort air cooling was being installed in the best theaters and wealthiest homes in the United States. It is not surprising that a measure of rate of cooling, or cooling capacity, is often related to the cooling capability of a ton of ice. A ton of refrigeration is defined as the rate of cooling produced by melting a ton of ice in 24 hours. Based on this definition, and on the fact that the latent heat of fusion of water is 144 Btu/lbm, it may be deduced that a ton of refrigeration is a rate of cooling of 200 Btu/min, or 12,000 Btu/hour (see Exercise 8.3). Thus a 3-ton air conditioner is one that nominally removes 36,000 Btu/hr from the cooled space. When shopping for an air conditioner, one encounters labels that give the product's energy efficiency ratio (EER), that is, the ratio of the cooling capacity of the unit, measured in Btu/hr, to the power required to operate it, in watts. It is evident that the EER is proportional to the COP, neglecting fan power. Applying the conversion factor between Btus and watt-hours one sees that EER = 3.413(COP. EXAMPLE 8.3

For an ideal vapor compression refrigeration system operating with refrigerant 22 at an evaporator temperature of 0°F and condensing at 100°F, find the following: the compressor suction and discharge pressures, enthalpies, and specific volumes; the condenser discharge pressure and enthalpy; the refrigeration COP; and the refrigerant mass flow rate and power requirement for a 10-ton refrigeration unit. Solution

Following the notation of Figures 8.2 and 8.8, from the chart (Appendix F) for refrigerant 22 at T1 = 0°F, the other properties at state 1 are p1 = 38 psia, h1 = 104 Btu/lbm, v1 = 1.4 ft3/lbm, and s1 = 0.229 Btu/lbm-R. The saturated-liquid condenser discharge properties at T3 = 100°F are p3 = 210 psia and h3 = 39 Btu/lbm. The compressor discharge-state properties at s2 = s1 and p2 = p3 = 210 psia are h2 = 123 Btu/lbm, T2 = 155°F, and v2 = 0.31 ft3/lbm. The evaporator inlet enthalpy is the same as that at condenser discharge, h4 = h3 = 39 Btu/lbm. The refrigeration effect and the compressor work are then RE = h1 – h4 = 104 – 39 = 65 Btu/lbm w = h2 – h1 = 123 - 104 = 19 Btu/lbm

296 Thus COPr = RE /w = 65/19 = 3.42. The rate of cooling, or cooling capacity, for a 10-ton unit is 10×200 = 2000 Btu/min. The refrigerant mass flow rate is the capacity divided by the refrigeration effect = 2000/65 = 30.8 lbm/min. The power required by the compressor is the product of the mass flow rate and the compressor work = 30.8×19 = 585.2 Btu/min, or 585.2×60/ 3.413 = 10,290 W, or 10.29 kW. The ideal EER may then be calculated from the capacity and power as 2000×60/10,290 = 11.7 Btu/Watt-hr, or from the COP as 3.413×3.42 = 11.7 Btu/Watt-hr. _____________________________________________________________________________ Compressors While most small- and medium-capacity refrigeration systems use hermetically sealed, electricmotor-driven compressor units or open (externally powered) reciprocating compressors, centrifugal compressors are frequently found in large units for cooling buildings and for industrial applications. The reciprocating compressor has much in common geometrically with a simple two-stroke reciprocating engine with intake and exhaust valves. As in that case, the compressor clearance volume Vc is the volume at top center, and the piston sweeps out the displacement volume, as indicated in Figure 8.9. The processes 1-2-3-4-1 on the idealized pressure-volume diagram represent the following:

+ 1–2 Both valves are closed. Compression of the maximum cylinder volume V1 = Vc + Vd of refrigerant vapor through the pressure ratio p2/p1 to a volume V2.

+ 2–3 Exhaust valve is open. Discharge of refrigerant through the exhaust valve at

condenser pressure p3 until only the clearance volume V3 = Vc remains when the piston is at top center.

+ 3–4 Both valves are closed. Expansion of the clearance gas with both valves closed from V3 to V4. Note that the inlet valve cannot open until the cylinder pressure drops to p4 = p1 without discharging refrigerant back into the evaporator.

+ 4–1 Intake valve is open. Refrigerant is drawn from the evaporator into the cylinder at

constant pressure p1 through an intake valve by the motion of piston. Refrigerant in the amount V1 – V4 is processed per cycle.

Assuming polytropic compression and expansion processes with the same exponent k: V4 = V3( p3/p4)1/ k = Vc( p2/p1)1/ k

297

Then the volume of refrigerant vapor processed per cycle (or per shaft revolution) V1 – V4 = Vd + Vc – Vc ( p2/p1)1/ k = Vd – Vc [( p2/p1)1/ k – 1] is less than the displacement volume and depends on the compressor pressure ratio. Neglecting the difference between the refrigerant density leaving the evaporator and that in the compressor cylinder just before compression, we may write the compressor volumetric efficiency as the ratio of V1 – V4 to the displacement:

v = (V1 – V4)/Vd = 1 – (Vc /Vd) [( p2/p1)1/ k – 1] Examination of the compressor processes for different pressure ratios, as in Figure 8.9 ( p2' /p1, for example), shows that the refrigerant volume processed per cycle, and thus the volumetric

298 efficiency, decreases with increasing pressure ratio. It is also evident that the clearance volume must be kept small to attain high volumetric efficiency. It is clear that, for a given positivedisplacement compressor, the volumetric efficiency limits the usable pressure ratio and thus the difference between the condensing and evaporating temperatures. EXAMPLE 8.4

What is the volumetric efficiency of the reciprocating compressor used with the refrigeration system of Example 8.3 if the clearance volume is 3% of the displacement volume and the polytropic exponent is 1.16? Solution

The compressor pressure ratio is p2/p1 = 210/38 = 5.526. From the preceding equation, the volumetric efficiency is then

v = 1 – (0.03) [(5.526)1/1.16 – 1] = 0.899

_____________________________________________________________________________ The centrifugal compressor used in refrigeration is similar in principle to that introduced earlier in connection with the turbocharger. It is fundamentally different than the positivedisplacement reciprocating compressor in that it operates on a fluid dynamic principal in which kinetic energy imparted to the refrigerant by an impeller is converted to pressure analogous to that indicated by the Bernoulli equation. The high-speed rotor accelerates and thus increases the kinetic energy of the refrigerant, which is then converted to pressure in diffuser passages. Because the efficiency of small centrifugal compressors is limited, centrifugals are used primarily in large-scale applications with cooling loads in tens or hundreds of tons of refrigeration. The scroll compressor, shown in Figure 8.10, though based on a century-old concept, is a recent development in rotary compressors. In it, an electric-motor-driven orbiting scroll meshes with a stationary scroll. Six crescent-shaped pockets formed between the involute scrolls trap, transport, and compress low-pressure refrigerant from the outside of the scrolls to the discharge passage at the scroll center. This process is more easily visualized by following the shaded pocket of refrigerant in Figure 8.11. Some of the advantages attributed to the scroll compressor are: continuous and smooth compression with 100% volumetric efficiency, no valves, physical separation of suction and discharge lines (which avoids the temperature cycling of reciprocating compressors), and higher compression efficiency than the best reciprocating units (ref. 10). Seasonal energy efficiency ratios (EERs averaged over a cooling season) in excess of 10 are readily attained with this technology. An animation of a scroll compressor may be seen on the internet at www.copeland-corp.com. Suction and Subcooling Considerations Let's examine two items of concern with respect to some vapor compression systems. In systems with reciprocating compressors, there is a danger that, due to changing cooling loads, that the liquid refrigerant in the evaporator may not be completely vaporized, causing slugs of liquid to

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enter the compressor. Because liquids are essentially incompressible, positive-displacement compressors with fixed clearance volumes can be damaged when such "slugging" occurs. The use of a thermostatic expansion valve (TEV) that responds to change in the degree of superheat in the suction line provides one solution to this problem. A bulb filled with refrigerant attached to the suction line, when heated by superheated vapor, transmits an increasing pressure signal to a diaphragm in the TEV, which adjusts the valve flow area and in turn changes the mass

300

flow rate of refrigerant. This control is usually set to maintain a minimum of about five degrees of superheat to avoid liquid slugs entering the compressor inlet. A second concern is the possibility of entry of vapor into the throttling valve if the refrigerant at the condenser exit is not completely condensed. Because vapor occupies much more space than liquid, the throttling valve will not function properly if vapor can enter from the condenser. One approach to dealing with this is to locate a liquid receiver downstream of the condenser to assure the availability of liquid to the expansion device. Both of the above concerns may be dealt with simultaneously by the addition of a suction-line heat exchanger that superheats the evaporator discharge about five degrees, ensuring that only vapor enters the compressor. The heat exchanger that provides suction superheat from state 6 to state 1 in Figure 8.12 may be set up to receive heat from the subcooling of the condenser discharge from state 3 to state 4. This ensures the absence both of vapor entering the throttling valve and of liquid slugs entering the compressor. Note that the subcooling also tends to increase the refrigeration effect over that of the ideal cycle by decreasing the enthalpy entering the

301

evaporator. This effect may be accomplished by running the compressor suction line next to the condenser discharge, allowing some thermal contact. Combining Heating and Cooling in a Single System It is possible to combine both heating and cooling functions in a single system by providing heat exchangers that can operate as both evaporator and condenser and a control system that can reroute the flow of refrigerant when switching functions is required. Figure 8-13 presents a

302 schematic diagram for such a system, commonly called a heat pump (context usually determines whether the term “heat pump” refers to a device that heats only or that combines heating and cooling functions). The key component in a commercial heat pump is a reversing valve. With the valve shown in the figure, rotation through an angle of 90° reroutes the flow of refrigerant from the indoor coil to the outdoor coil, and vice versa. As a result of this change, the indoor coil that served as a condenser in the winter becomes an evaporator in the summer. The outdoor coil changes accordingly. Separate throttling devices may be used to accommodate differing load conditions in winter and summer. One-way check valves ensure that refrigerant flow is through the appropriate throttling device during each season. 8.3 Absorption Refrigeration Example 8.3 shows that vapor compression refrigeration requires a significant supply of work from an electric motor or other source of mechanical power. Absorption refrigeration is an alternate approach to cooling that is largely thermally driven and requires little external work. This form of refrigeration is growing in importance as energy conservation considerations demand closer scrutiny of the disposition of heat rejection from thermal processes. Absorption refrigeration provides a constructive means of utilizing waste heat or heat from inexpensive sources at a temperature of a few hundred degrees, as well as directly from fossil fuels. The eventual abolition of the use of CFCs may also boost absorption refrigeration technology. This system relies on the fact that certain refrigerant vapors may be dissolved in liquids called absorbents. For instance, water vapor is a refrigerant that tends to dissolve in liquid lithium bromide, an absorbent. Just as when they condense, vapors release heat when they go into solution. This heat must be removed from the system in order to maintain a constant temperature. Thus, cooling causes vapor to be absorbed in absorbents, just as cooling causes vapor to condense. On the other hand, heating tends to drive vapor out of solution just as it turns liquid to vapor. This solution phenomenon and the fact that pumping liquid requires a relatively small amount of work compared with that required to compress a gas are the secrets of absorption refrigeration. Consider the schematic diagram in Figure 8.14, which shows a basic absorption refrigeration unit. The condenser / throttling valve / evaporator subsystem is essentially the same as in the vapor compression system diagram of Figure 8.2. The major difference is the replacement of the compressor with a different form of pressurization system. This system consists primarily of an absorber at the pressure of the evaporator, a vapor generator at the pressure of the condenser, and a solution pump. A second throttling valve maintains the pressure difference between the absorber and the generator. The system operates as follows: Refrigerant vapor from the evaporator flows into the absorber, where it mixes with the absorbent. The mixture is cooled by heat transfer QA to air or water at the temperature of the environment, causing the vapor to go into solution. The refrigerant-absorbent solution flows to the solution pump where it is pressurized to the pressure level of the generator and condenser. Heat from an energy source, QG, then drives the vapor from the cold liquid solution. The vapor flows into the condenser while the heated liquid in the

303

304

generator passes back to the absorber through an absorbent throttling valve, thus completing the absorbent loop. As in vapor compression systems, the condensed vapor from the condenser passes through the refrigerant throttling valve into the evaporator as a chilled refrigerant liquid-vapor mixture, where its vaporization creates the desired refrigeration effect. The evaporator vapor then flows back to the absorber to again go into solution and repeat the process. It should be noted that the only work required is a relatively small amount needed to operate the solution pump. Thus the major energy requirement of an absorption refrigeration system is a transfer of heat QG from a source of moderate temperature, such as an engine combustion-gas exhaust, steam from a low-pressure boiler, or perhaps a heat transfer fluid from solar collectors. The coefficient of performance of an absorption system is therefore defined as the ratio of the refrigeration capacity to the rate of heat addition in the generator. This definition is fundamentally different

305 from that of the COP of a vapor compression system, because the energy source is in the form of heat rather than work. Therefore, it should not be surprising that absorption system COPs are much lower than those for vapor compression systems. 8.4 Comfort Air Conditioning Professionals define air conditioning to include heating, humidifying, dehumidifying, circulating, cleaning, and those other operations involving air (in addition to cooling) that provide thermal comfort to people and effectiveness to processes. Studies of what constitutes thermal comfort to individuals have identified air temperature, humidity, ventilation, air movement, and air cleanliness as important control parameters. The systems needed to control all of these parameters are discussed in many references, including refs. 1 through 5. This chapter is concerned primarily with temperature and humidity control. Space cooling and space heating are the primary functions that rely on vapor compression and absorption systems. In specifying systems to perform these functions, it is necessary to define the winter heating loads and summer cooling loads [Btu/hr | kW] for the space to be heated or cooled. Most HVAC (heating, ventilating, and air conditioning) professionals now have computer programs that assist in defining these loads. The determination of cooling loads involves delimiting the space to be conditioned and defining the appropriate indoor and outdoor design conditions. For example, for some applications the indoor design temperature and humidity might be specified as 74°F and 50% relative humidity based on thermal comfort data and experience. For a given locale, the ASHRAE Handbook gives outdoor temperatures that are not exceeded during a given fraction of the summer season. For example, the Handbook indicates that the temperature in Tulsa, Oklahoma is usually not greater than 98°F during 97.5% of the four-month period from June to September. This, therefore, might be selected as a suitable summer outdoor design temperature. A thermal cooling load analysis is performed with these design conditions and data on the structure and contents of the enclosure of the conditioned space for times when the load is expected to be greatest. The analysis involves heat transfer through walls, roofs, doors, and windows; and internal heat generation due to electrical devices, gas burning appliances, human occupants, and other sources. At such times, solar radiation is usually a strong influence. In most applications, additions of water vapor to air in the space must be considered also. The results of these calculations yield the design rates at which heat and moisture must be removed from the space. The air conditioning evaporator cooling capacity and the fan air circulation rate are then sized to provide for these near-extreme rates. Cooling and heating loads at a given locale may be expected to vary approximately linearly with the deviation of the ambient temperature from a break-even temperature of about 65°F. At this outdoor temperature, no mechanical heating or cooling is required for thermal comfort. The break-even ambient temperature is usually below the indoor design temperature because of the necessity of a temperature difference for the rejection of heat through the walls required to balance the energy generation from sources in the space. Figure 8.15 shows typical performance behavior of vapor compression heating and cooling equipment superimposed on corresponding load curves. The equipment is necessarily oversized for mild temperatures. In the case of the heat

306

pump, there is an ambient temperature at which the equipment capacity matches the heating load, the balance point. At lower temperatures the heat pump is unable to maintain the design indoor temperature, a heating deficit. This situation is usually handled by the use of electrical resistance heaters to supplement the heat pump output. The electric resistance heaters usually are switched on in banks as the ambient temperature decreases further below the break-even point. While the heat pump itself may be quite efficient, pure resistance heating with COP = 1.0 results in degradation in overall efficiency and economy while the resistance heaters are in operation. Operationally, for the frequent periods when capacity exceeds load, a thermostat is usually used to cycle the system on and off to provide for indoor comfort in a limited range of temperature around the indoor design value. More sophisticated systems resulting from advances in motor control, compressors, and fans may allow the system to operate at part load continuously, to avoid the discomfort and inefficiency of temperature cycling associated with simple thermostat operation. The dual-fuel heat pump, a new approach to dealing with the weaknesses of the heat pump at low temperatures, has been proposed and studied by the Electric Power Research Institute, an equipment manufacturer, and several electric utilities (ref. 14). Instead of expensive and energywasteful resistance heating making up the heating deficit when ambient temperatures require operation below the heat pump balance point, a gas furnace built into the heat pump package comes online and economically satisfies the need for additional heat. The combined electric heat pump and gas unit operate together, unless the ambient temperature drops to still lower

307 temperature levels at which the cost of gas operation falls below the cost of electricity. Below this break-even point electric heat pump component shuts down and gas heat is used exclusively. The break-even point may be adjusted when energy prices change. According to the reference, in regions where gas prices are high and electricity low, dual-fuel heat pump users will benefit from substantial reductions in operating costs and the utilities from increased off-peak sales in the winter months. However, the purchase price of the dual-fuel heat pump unit is expected to be higher than combined gas heating and electric cooling systems now used. This creates a marketing problem, because consumers often are more impressed by low purchase price than by low life-cycle costs resulting from low operating costs. 8.5 Moist Air and Air Conditioning System Design Psychrometrics Although the mass fraction of water vapor in ambient air usually is numerically small, water vapor can have a significant influence on an individual's thermal comfort and on the design of air conditioning equipment. Most people can recall instances of discomfort associated with both excessively dry or moist air. Numerous industrial processes also require moisture control to maintain high standards of quality and productivity. The engineering fundamentals of dealing with moist air and the application of those fundamentals are treated here. The study of the behavior of moist air is called psychrometrics. We will consider moist air as a two-component mixture of dry air and water vapor, and treat both components and the mixture as ideal gases in the context of the mixture theory discussed in Chapter 3. Humidity Ratio Properties of moist air are commonly referred to the mass of dry air mda, apart from the water vapor, contained in the air. For example, the humidity ratio is defined as the ratio of the mass of water vapor to the mass of dry air present in the moist air: W = mwv/mda

[dl]

(8.8)

Together, the humidity ratio and the ambient temperature and pressure define the thermodynamic state of moist air, and therefore may be used to define other properties of moist air. Using the ideal gas law, the humidity ratio may be written as W = ( pwvV/RwvT)/( pdaV/RdaT) = ( pwv/pda)(Rda/Rwv) = (pwv/pda)(Mwv/Mda) = 0.622( pwv/pda)

[dl]

(8.9)

308 where we have used the facts that the water vapor and dry air occupy the same space and therefore have the same volume, and that both components are in thermal equilibrium and therefore are at the same temperature. The ratio of gas constants in Equation (8.9), expressed in terms of the Universal Gas Constant, reduces to the inverse ratio of molecular weights Mwv /Mda = 18/28.9 = 0.622. Thus the humidity ratio is equivalent to the ratio of partial pressures of water vapor and dry air. Going one step further, consider the total air pressure P, the pressure that we measure with a barometer, as the sum of partial pressures of dry air and water vapor according to Dalton's Law: P = pda + pwv

[lbf/ft2 | kPa]

(8.10)

The humidity ratio then may be written as: W = 0.622pwv/(P – pwv)

[dl]

(8.11)

Thus, for a given ambient pressure, the humidity ratio is determined by the existing partial pressure of water vapor. It is evident that because pwv << P, the humidity ratio varies approximately linearly with the partial pressure of water vapor in the air. Humidity ratio is sometimes expressed in grains of water vapor per pound of dry air where there are 7000 grains per pound. Saturated Air Moist air is said to be saturated when its humidity ratio is a maximum for the existing temperature and total pressure. The humidity ratio for saturated air may be determined using Equation (8.11) with the saturation pressure of water vapor obtained from saturated steam tables at the known temperature. EXAMPLE 8.5

Determine the partial pressures of water vapor and dry air and the humidity ratio in saturated moist air for an ambient total pressure and temperature of 14.7 psia and 80°F, respectively. Solution

From the saturated-steam tables at 80°F, the partial pressure of water vapor is 0.50683 psia. The partial pressure of the dry air component is then 14.7 – 0.50683 = 14.2 psia, and the humidity ratio is W = 0.622(0.50683)/(14.7 – 0.50683) = 0.0222 Thus, at 80°F, the moist air may have no more than 0.0222 lb (or 155.4 grains) of moisture per pound of dry air. _____________________________________________________________________________

309 While air at a given temperature and pressure can have no more water vapor than indicated by the humidity ratio for saturated air, the air can be drier. This is, of course, indicated by lower values of humidity ratio and also by relative humidity, as discussed next. Relative Humidity and Dew Point An alternative parameter for indicating the moisture content of air is the relative humidity, M. The relative humidity is defined as the ratio of the mole fraction of water vapor in unsaturated air to the mole fraction of water vapor in air that is saturated at the same temperature and total pressure:

M = xwv /xwv,sat

[dl]

(8.12)

It has been shown in Chapter 3 that the mole fraction of a component of an ideal gas is the ratio of the partial pressure of that component to the total pressure of the gas. Thus the relative humidity may be written as

M = (pwv /P)/(pwv /P)sat = pwv /pwv,sat

[dl]

(8.13)

where the partial pressures are evaluated at the same temperature and total pressure. The relative humidity of the air may be related to the p-V diagram for its water component in Figure 8-16. The diagram shows two isotherms where T1 = T2 > T3. Considering the top isotherm, the relative humidity of state 1 is the ratio of the pressure of water vapor at state 1 to that at the saturation vapor state at the same temperature, state 2. The figure shows that the maximum pressure of water vapor possible at a given temperature corresponds to the saturation pressure of water at that temperature. Thus increasing amounts of water vapor in air at a given ambient temperature are indicated by higher relative humidity, higher humidity ratio, higher water vapor partial pressure, and by vapor states closer to the saturated-vapor line for water. Thus the term "saturated air" implies that the water vapor in the air is a saturated vapor. These two uses of the term "saturated" should be distinguished to avoid confusion. When a given mass of moist air cools slowly, the mass of water vapor in the air remains constant until the air becomes saturated. Thus the humidity ratio and the partial pressure of the water vapor are also fixed. This process is depicted by the line of constant pressure from 1 to 3 to 4 in Figure 8.16. Note that, once state 4 is reached, no further cooling can occur without condensation of water vapor. This condition, at which liquid water first appears, is known as the dew point. The formation of dew is a familiar early morning occurrence in which water vapor condenses, leaving drops of liquid on grass and other surfaces that are cooler than the surrounding air. The temperature at which water vapor begins to condense is therefore called the dew point temperature. It is evident from Figure 8.16 that as the air temperature at state 1 drops, approaching the dew point temperature, the relative humidity increases until it finally reaches 100% at state 4. Thus the dew point temperature for any moist air state may be obtained from the saturated-steam tables, if the partial pressure of the water vapor is known.

310

EXAMPLE 8.6

In ambient moist air, the temperature and pressure are 70°F and 14.7 psia, respectively, and the relative humidity is 50%. What are the partial pressures of dry air and water vapor and the humidity ratio and dew point? Solution

At 70°F, the steam tables give the saturation pressure of water as 0.36292 psia. This is the partial pressure of water vapor in saturated air. From Equation (8.13), the partial pressure of water vapor in the ambient air is then (0.5)(0.36292) = 0.181, and the corresponding partial pressure of dry air is 14.7 – 0.181 = 14.52 psia. Equation (8.11) then gives the humidity ratio: W = 0.622(0.181)/14.52 = 0.0078 lb of water vapor per pound of dry air.

311 The tables for saturated steam give the dew point temperature of about 51°F for the vapor partial pressure of 0.181 psia. _____________________________________________________________________________ Enthalpy of Moist Air Since any extensive property of a mixture of ideal gases is the sum of the corresponding properties of the components, the enthalpy of moist air is the sum of the enthalpies of dry air and water vapor. Hence H = Hda + Hwv = mdahda + mwvhwv

[Btu | kJ]

where the mass of the mixture is mda + mwv. In order to define an enthalpy per unit mass for the mixture, it must be decided which mass to select as a reference for the specific enthalpy of the mixture. As with the humidity ratio, the usual choice in psychrometrics is the mass of dry air. Thus the specific enthalpy of moist air is given by h = H /mda: h = hda + (mwv/mda)hwv = hda + Whwv

[Btu/lbm | kJ/kg]

(8.14)

Because the temperature range of most air conditioning operations is limited, it is usually satisfactory to use constant heat capacities in dealing with enthalpy calculations. Selecting the reference temperatures for enthalpy as 0° on both scales, the enthalpies of the dry air, expressed in terms of Fahrenheit and Celsius temperatures, may be written as hda = cpdaT [F] = 0.24T [F]

[Btu/lbm]

hda = cpdaT [C] = 1.004T [C]

[kJ/kg]

and

The enthalpy of water vapor is the sum of the sensible and latent enthalpies: hwv = hs + hfg = cpwvT + hfg = 0.446T [F] + 1061.2

[Btu/lbm]

(8.15a)

= 1.867 T [C] + 2501.3

[kJ/kg]

(8.15b)

EXAMPLE 8.7

Compare the enthalpy of water vapor from Equation (8.15a) with that from the steam tables for 32°, 100°, 150°, and 200°F.

312 Solution

For 32°F, the enthalpy is hwv = 0.446(32) + 1061.2 = 1075.47 Btu/lbm This compares with 1075.5 Btu/lbm from the steam tables. The comparison for other temperatures is as follows: Temperature, °F

Enthalpy, Btu/lbm – Equation (8.15a)

Enthalpy, Btu/lbm – Steam Tables

32

1075.47

1075.5

100

1105.8

1105.1

150

1128.1

1126.1

200

1150.4

1146

Thus the approximation is quite accurate in and beyond the air conditioning range, but it decreases in accuracy as temperature increases. _____________________________________________________________________________ Combining the above relations, we may then write the enthalpies of moist air as h = cpdaT [F] + W{cpwvT [F] + hfg} = 0.24T [F] + W{0.446T [F] + 1061.2}

[Btu/lbm]

(8.16a)

[kJ/kg]

(8.16b)

h = cpdaT [C] + W{cpwvT [C] + hfg} = 1.004T [C] + W{1.867T [C] + 2501.3} Adiabatic Saturation Consider a steady-flow process in which air flows through a liquid water spray in a duct where the air entrains moisture until it becomes saturated. If the duct is insulated so that the flow is adiabatic, and the water spray and the moist air leaving the duct are at the same temperature, the process is referred to as an adiabatic saturation process. It is shown shortly that measurement of the temperature and pressure of the air entering the duct and the final moist air temperature, called the adiabatic saturation temperature, determines the state of the moist air entering the duct. Thus two temperature measurements and a pressure measurement are sufficient to determine the state of moist air.

313

Figure 8.17 shows a schematic of an adiabatic saturation device in which liquid water at the adiabatic saturation temperature T2 is evaporated at a rate mw into a moist air stream flowing at a rate of mda(1 + W) through the duct. Thus conservation of mass applied separately to both dry air and water components yields mda2 = mda1 = mda

[lbm/s | kg/s]

mdaW1 + mw = mdaW2

[lbm/s | kg/s]

Combining these equations, we find that the ratio of the mass flow rate of entrained water or makeup water to the mass of dry air is mw/mda = (W2 – W1)

[dl]

The steady flow form of the First Law yields mda 1(hda 1 + W1hwv 1) + mw hf = mda 2(hda 2 + W2hwv 2) where hf is the enthalpy of the saturated liquid at T2. Substituting Equation (8.17) into this

(8.17)

314 equation yields (hda1 + W1hwv1) + (W2 – W1)hf = (hda 2 + W2hwv 2) or W1(hwv 1 – hf) = cpda(T2 – T1) + W2(hwv 2 – hf)

[Btu/lbm | kJ/kg]

W1 ={cpda(T2 – T1) + W2(hwv 2 – hf)}/(hwv 1 – hf)

[dl]

or (8.18)

Given the initial temperature T1, hwv 1 is determined from Equation (8.15). If T2 and P1 = P2 are also measured, the saturation partial pressure, the humidity ratio W2, the vapor enthalpy hwv 2, and the liquid-water enthalpy hf = hf(T2) may also be determined using the steam tables and Equations (8.11) and (8.15). Equation (8.18) may then be solved for the upstream (ambient) humidity ratio. The following example demonstrates the use of the adiabatic saturation concept in the determination of the ambient moist air state. EXAMPLE 8.8

Ambient air at 14.7 psia and 70°F drops to 56°F in passing through an adiabatic saturator. What is the ambient humidity ratio and the relative humidity? Solution

Saturated air at 56°F yields a partial pressure of 0.22183 psia, a humidity ratio W2 = 0.622(0.22183)/(14.7 – 0.22183) = 0.00953 a water-vapor enthalpy hwv 2 = 1086 Btu/lbm, and makeup-water enthalpy hf = 24.059 Btu/lbm. At the inlet where T1 = 70°F, the water-vapor enthalpy hwv 1 = 1092.1 Btu/lbm. Thus, from Equation (8.18) W1 = [0.24(56 –70) + 0.00953(1086 – 24.059)] /(1092.1 – 24.059) = 0.00633 lb of water vapor per pound of dry air. Solving Equation (8.11) for the partial pressure of water vapor gives pwv 1 = (W1)(P)/(0.622 + W1) = (0.00633)(14.7)/(0.622 + 0.00633) = 0.1481 psia Because the saturation partial pressure of water vapor at 70°F is pwv 1, sat = 0.36292 psia, the relative humidity at state 1 is

M = 0.1481/0.36292 = 0.408, or 40.8%. _____________________________________________________________________________

315

We may summarize the adiabatic saturation process as one in which liquid water evaporates into the air flow, increasing the mass of water vapor in the air and therefore increasing the humidity ratio and, according to Equation (8.11), increasing the partial pressure of the vapor also. Because the flow is adiabatic, the heat of vaporization for the water must come from the flow of moist air, decreasing the air temperature. The increase in vapor partial pressure implies that the air dew point temperature increases as the flow passes through the saturator, ending at the adiabatic saturation temperature. In order to distinguish among the various temperatures dealt with in psychrometrics, the temperature of moist ambient air, as measured by a mercury-in-glass thermometer, a thermocouple, or an equivalent device, is known as the dry bulb temperature. The sling psychrometer, a more convenient and inexpensive device for determining moist air states than the adiabatic saturator, uses two mercury-in-glass thermometers rigidly attached to a frame with a pivoted handle as shown in Figure (8.18). The bulb of one thermometer, called the wet bulb, is wrapped in moistened wicking, and the other is left bare to measure the dry-bulb temperature. In operation, the user twirls the psychrometer for a minute or two until the readings of the thermometers stabilize. Evaporation of water from the wicking cools the wet bulb in proportion to the dryness of the ambient air. The temperature difference between the bulbs is commonly called the wet-bulb depression. Tables are sometimes provided to give relative humidity in terms of wet-bulb depression, but a psychrometric chart may be conveniently used to identify the moist air properties, as discussed in the next section. The wet-bulb temperature, as an approximation to the adiabatic saturation temperature, together with the dry-bulb temperature and barometric pressure, determines the state of the moist ambient air. While the wet-bulb process differs fundamentally from the adiabatic saturation process, in practice it provides a reliable approximation to the wet-bulb temperature. Thus the adiabatic saturation temperature is sometimes referred to as the thermodynamic wet-bulb temperature. Devices that give direct readings of humidity are called hygrometers. These devices are usually more complicated and expensive and less reliable than the psychrometer. They

316

commonly make use of properties of materials that are influenced by the presence of water vapor, such as elongation and shrinkage. Human hair, paper, and nylon are examples of such materials. More information on hygrometers is available in reference 2. The Psychrometric Chart The properties and processes described above are conveniently related on a graph commonly known as a psychrometric chart. Although close examination of the skeleton chart of Figure 8.19 and the detailed chart in the Appendix shows that it is not quite a rectangular coordinate plot, for most purposes it may be thought of as such, with humidity ratio as ordinate and dry bulb temperature as abscissa. A psychrometric chart is constructed for a specified value of barometric pressure. Thus charts for different barometric pressures should be selected for analyses for high altitudes, such as at Denver’s, where atmospheric pressure is low, and those near sea level, such as Houston’s, where it is normally high.

317 While most psychrometric charts are based on the precise methods of reference 7, it is instructive to consider how a chart could be constructed based on the ideal gas theory discussed earlier and on the steam tables. The curve labeled "saturation" on the psychrometric chart corresponds to states of air containing saturated water vapor at each dry-bulb temperature. For each value of dry-bulb temperature, the saturation partial pressure obtained from the steam tables determines the saturation humidity ratio for a specified atmospheric pressure [Equation (8.11)]. Thus the saturation curve relates the dew point temperature (measured on the abscissa) to the humidity ratio and to the corresponding partial pressure of water vapor. The saturation line also defines the conditions for 100% relative humidity. For a given dry-bulb temperature, Equation (8.13) shows that the vapor partial pressure varies linearly with M: pwv = M(pwv, sat For a given value of M, the humidity ratio can then be established using Equation (8.11). Hence the value of M and the dry-bulb temperature determine the value of the ordinate W. Thus a line of constant relative humidity M lies between the horizontal axis (pwv = 0 or W = 0) and the chart saturation curve at a vertical position in proportion of the magnitude of M. Lines of constant volume of moist air per unit mass of dry air may be approximated using the ideal gas law applied to the dry air component: v = RdaT/pda = RdaT/(P – pwv)

[ft3/lbm | m3/kg]

(8.19)

where Rda is the gas constant for dry air. Because dry air and water vapor occupy the same volume independently, the specific volume of moist air based on unit mass of dry air is the same as that for dry air. For zero humidity ratio, the air is dry and pwv = 0. Then the dry-bulb temperature and ambient pressure alone determine the specific volume for each point on the horizontal axis of the chart. As the humidity ratio and hence the vapor partial pressure increase at constant ambient pressure, Equation (8.19) shows that the dry-bulb temperature must decrease to hold v constant. This establishes the negative slope of constant-specificvolume lines on the psychrometric chart. It is also easily seen from Equation (8.19) that, for dry air along the horizontal axis, lines of higher specific volume are located at higher dry-bulb temperatures. The same conclusion follows for any constant value of humidity ratio. Similar arguments may be used, together with Equation (8.14), to explain the slope and variation of lines of constant enthalpy with dry-bulb temperature on the psychrometric chart. Processes on the Psychrometric Chart Consider point A on the psychrometric chart in Figure 8.19. The processes of heating and cooling to points B and C, respectively, are called sensible heating and sensible cooling. The term “sensible” implies change in dry-bulb temperature without change in moisture content. These processes occur at constant humidity ratio and thus at constant dew point temperature, Tdp.

318

Next consider processes in which either moistening or drying of air takes place. An example of a humidification, or air-moistening, process already considered is the adiabatic saturation process. It is represented by movement along a line of constant wet-bulb temperature toward the saturation curve, such as the line AD on the figure. A pure humidification process is one in which the humidity ratio increases at constant dry-bulb temperature. The line AE in Figure 8.19 represents this process. The process requires the addition of water vapor to the air, with a consequent energy increase corresponding to the heat of vaporization of water, (WE – WA)hfg. The enthalpy increase between points A and E indicated by the chart is equal to this quantity because no sensible heating has occurred. The process is said to involve latent heat as opposed to sensible heat. Thus an isothermal change in humidity ratio represents a latent heat load associated with the moist air mass and the equipment dealing with it. The reverse process EA may be thought of as pure dehumidification. The modifier "pure" is meant to suggest that there are other, more common and easily executed, humidification and dehumidification processes in which dry bulb temperature change is involved. Examples are air washer processes like the adiabatic saturator and desiccant drying (to be discussed later). Let us now consider the steady-flow, adiabatic mixing of two moist air streams at states i and o to form a single stream at state m as shown in Figure 8.20. Conservation of mass of dry air and water vapor yields mi + mo = mm

[lbm/s | kg/s]

(8.20)

319 and miWi + moWo = mmWm Dividing through by the mixture mass flow rate, the latter equation becomes Wm = (mi /mm)Wi + (mo /mm)Wo

[dl]

(8.21)

Thus the mixture humidity ratio is the mass-weighted average of the humidity ratios of the two streams. Applying conservation of energy to the adiabatic mixing process, we obtain 0 = mmhm – mihi – moho which may be written as hm = (mi/mm)hi + (mo/mm)ho

[Btu/lbm | kJ/kg]

(8.22)

This equation is of the same form as Equation (8.21). Thus, given the flow rates and properties of the two streams, the mass flow of the mixture is the sum of the masses of the two inflowing streams and the humidity ratio and enthalpy of the mixture may be obtained by calculating the weighted means of the humidity ratios and enthalpies from Equations (8.21) and (8.22). A graphical solution for the mixture state can be easily obtained on the psychrometric chart. For fixed states i and o, Equations (8.21) to (8.22) depend on a single parameter, mi/mm, or mo/mm = mo/(mo + mi) = 1 – mi /mm Rearranging Equations (8.21) and (8.22) they become (Wm – Wi)/(Wo – Wi) = mo/mm

[dl]

(8.23a)

(hm – hi)/(ho – hi) = mo/mm

[dl]

(8.23b)

Eliminating the mass ratio between these equations defines a linear relation between the mixture humidity ratio Wm and enthalpy hm: Wm = Wi + (hm – hi)(Wo – Wi)/(ho – hi)

[dl]

as shown in Figure 8.20. The explicit linear dependence of Wm and hm on mo/mm is shown by rearranging Equations (8.23a and b) into Wm = Wi + (Wo – Wi)(mo/mm)

[dl]

(8.23c)

hm = hi + (ho – hi)(mo/mm)

[Btu/lbm | kJ/kg]

(8.23d)

320 When the flow rates and properties of states i and o are known, the latter two equations define the mixture state. These equations are the basis for a graphical method to determine the mixture state, as suggested earlier. A straight line connecting states i and o is drawn on the psychrometric chart, and the mass fraction mo/mm in Equations (8.23) is laid out as the fraction of the distance between i and o measured from state i. The resulting point defines the properties of the mixture state. EXAMPLE 8.9

Ten lbm/sec of outdoor air at 100°F dry-bulb and 60% relative humidity at sea level is mixed with 20 lbm/sec of indoor air at 75°F dry-bulb and 40% relative humidity. Determine the moist air properties of the mixture. Solution

From the psychrometric chart for the outdoor air state: ho = 51.8 Btu/lbm and Wo = 0.0252 lbm water vapor per lbm dry air. For indoor air: hi = 26 Btu/lbm and Wi = 0.0074 lbm of water vapor per lbm dry air. From Equations (8.20) and (8.21), the humidity ratio is then: Wm = (10/30)(0.0252) + (20/30)(.0074) = 0.01333 lbm water vapor per lbm dry air, and the enthalpy is hm = (10/30)(51.8) + (20/30)(26) = 34.6 Btu/lbm Other mixture properties from the psychrometric chart are then 83.4°F dry-bulb, 65°F dew point, and 70.8°F wet-bulb temperatures, 54% relative humidity, and 13.97 ft3/lbm specific volume. __________________________________________________________________________ Psychrometric Analysis of Space Cooling An important application of psychrometrics is in the analysis and design of air conditioning systems. In comfort cooling, a control volume (see Figure 8.21), usually referred to as a space or zone, is identified as a room or group of rooms that have common thermal conditions and are to be put under the control of a single thermostat or control. Those conditions are usually defined by the atmospheric pressure, dry-bulb temperature, a humidity parameter such as humidity ratio, and specified sensible and latent cooling loads, qs and qL. The design sensible and latent loads are defined based on the characteristics of the boundaries of the zone and on its external and desired internal environments, as discussed earlier. The sensible heat factor, SHF = qs /qt = qs /(qs + qL) [dl] (8.24) defines the fraction of the total load qt due to the sensible load. The sensible heat factor, SHF,

321

together with the sensible load, is frequently used in defining the conditions for cooling a zone. Consider the conditioned space in Figure 8.21 as having specified latent and sensible loads. It is desired to hold the dry-bulb temperature of the indoor space at TDBi and relative humidity, Mi (the wet-bulb temperature TWBi or humidity ratio Wi could be used instead). The design outdoor environment is represented by design dry-bulb and wet-bulb temperatures, TDBe and TWBe, respectively, as shown on the psychrometric chart in Figure 8.22. A system is to be designed to supply cool, dry air to offset the sensible and latent loads imposed on the zone. The system is represented by a frictionless adiabatic ducting system that supplies conditioned air, the supply air (state su on the chart), to replace an equal mass of room air at the space conditions (state i). Part of the air leaving the zone is exhausted to the environment; and the balance, the return air, is mixed with fresh outdoor ventilation air (state e). The mixture of return air and ventilation air (state m) is cooled and dehumidified by a cooling coil and circulated back to the space by a supply-air fan. It is desired to determine suitable values for supply-air temperature, coil cooling rate, and fan volume flow rate. Assuming that air leaves the space at temperature TDBi, the steady-flow form of the First Law applied to the space yields qt = qs + qL = msu(hi – hsu)

[Btu/hr | kW]

The enthalpy difference in this equation can be written as hi – hsu = hi – h3 + h3 – hsu

[Btu/lbm | kJ/kg]

(8.25)

322

This decomposition is seen graphically in the psychrometric chart in terms of the triangle i $su $3. Thus qt = msu(hi – h3 + h3 – hsu)

[Btu/hr | kW]

(8.26)

Because the sensible heat, reflecting the supply-air dry-bulb temperature rise, may be written as qs = msu(h3 – hsu) [Btu/hr | kW] (8.27a) it is clear from Equation (8.26) that the latent heat is given by qL = msu(hi – h3)

[Btu/hr | kW]

(8.27b)

Equations (8.27) show that both sensible and latent loads are proportional to the supply-air mass flow rate. Graphically, Equations (8.27) indicate that the line segments su $3 and 3 $i represent sensible and latent zone heat loads, respectively. The sensible heat factor may be expressed as

323 SHF = qs /qt = [(hi – h3)/(h3 – hsu) + 1]-1

[dl]

(8.28)

It is seen from Equation (8.28) and Figure 8.22 that the space sensible heat factor depends on the slope of the room condition line. The ASHRAE psychrometric chart includes a "protractor" that relates the condition line slope to the sensible heat factor. Thus the design indoor state and the cooling loads determine the room condition line. Selection of the supply-air dry-bulb temperature together with the room condition line determines the supply-air state. The supply-air temperature must be chosen within the capabilities of available refrigeration cooling coils. Note that the coil cooling load qc = msu(hm – hsu)

[Btu/hr | kW]

exceeds the space-cooling load whenever warm ventilation air is provided. The fan mass flow rate requirement may be evaluated from Equation (8.25) or (8.27) when the supply-air state is known. The fan volume flow rate is then msuvm [ft3/hr | m3/s]. However, it is usually quoted in ft3/min or CFM as CFM = msuvm/60

[ft3/min]

where the mass flow rate is in lbm/hr of dry air. There is a significant tradeoff here which may be considered with the help of the psychrometric chart. Higher supply-air dry-bulb temperatures allow higher evaporator coil temperatures, which improve the system cooling COP. However, the resulting lower supplyair to room-air temperature difference [according to Equation (8.25)] implies a larger air flow rate and thus a larger fan and larger ducts. A moisture mass balance on the cooling coil gives the mass rate of condensation of water from the supply air: mw = msu(Wm – Wsu)

[lbm/hr | kg/s]

It should be recognized that this is usually greater than the rate of removal of water vapor from the zone because of the higher vapor content of the ventilation air. EXAMPLE 8.10

A single zone is to be maintained at 70°F dry-bulb and 50 % relative humidity by a refrigeration coil in a forced-air system. The total cooling load on the zone is 55,000 Btu/hr and the latent load is 11,000 Btu/hr. Outdoor air at 98°F dry-bulb and 80°F wet-bulb is provided at a mass rate that is 25% of the supply-air rate. Air leaving the coil is at 50°F drybulb. (a) Show the states of the indoor and outdoor air on a psychrometric chart and identify their enthalpy, wet-bulb temperature, humidity ratio, specific volume, and relative humidity. (b) Show the mixture line and state on a psychrometric chart. (c) Show the space and coil condition lines on a psychrometric chart. What are the

324 enthalpy, wet-bulb temperature, humidity ratio, specific volume, and relative humidity of the supply air? (d) What are the mass and volume flow rates of the outdoor and supply air? (e) What is the total coil cooling load, in Btu/hr and in tons? Solution

(a) From the psychrometric chart, the zone and outdoor properties are: TDBi = 70°F TDBe = 98°F Mi = 50% TWBe = 80°F We = 0.018 Wi = 0.0078 he = 43.4 Btu/lbmda hi = 25.3 Btu/lbmda vi = 13.51 ft3/lbmda ve = 14.47 ft3/lbmda TWBi = 58°F Me = 47% (b) For the properties listed above and for mo/mm = 0.25 in Equation (8.23), the mixture enthalpy and humidity ratio are hm = 25.3 + 0.25(43.4 – 25.3) = 29.83 Btu/lbmda Wm = 0.0078 + 0.25(0.018 – 0.0078) = 0.01035 lbm/lbda It is seen in Figure 8.22 that these properties verify that the mixture state lies on a straight line between the outdoor and indoor states. Thus the mixture line and either hm or Wm could have been used to identify state m. The mixture dry-bulb temperature read from the chart is 77°F and the specific volume is 13.7 ft3/lbmda. (c) The sensible heat factor is SHF = 44,000/55,000 = 0.8. A line parallel to the room condition line is obtained using the chart protractor and the space sensible heat factor. Using two triangles, a parallel line through state i gives the space (room) condition line shown in Figure 8.22. From the intersection of the space condition line and the 50°F dry-bulb temperature line, the supply-air properties are: hsu = 19.2 Btu/lbmda Wsu = 0.0066 vsu = 12.95 ft3/lbmda Msu = 86% The coil condition line is drawn to connect states m and su. (d) The enthalpy of the supply air rises as it mixes with room air. The rise rate must equal the total space load in order for the space temperature to remain unchanged. Thus msu = (qs + qL)/(hi – hsu) = 55,000/(25.3 – 19.2) = 9016.4 lbmda/hr Using the vsu, the supply air volume flow rate is 9016.4(12.95)/60 = 1946 cfm. Similarly, the

325 on-coil volume flow rate is 9016.4(13.7)/60 = 2059 cfm. The outdoor-air mass and volume flow rates are then, respectively, 9016.4/4 = 2254.1 lbmda/hr and 2254.1(14.45)/60 = 543 cfm. Note that, because of the different specific volumes, the outdoor-air volume flow rate is not one-fourth of the supply-air rate; and the supply-air and on-coil volume flow rates are not equal. (e) The mixed ventilation and return air are cooled and dehumidified by heat transfer to the refrigeration coil. The coil load is therefore qcoil = msu(hm – hsu) = 9016.4(29.83 – 19.2) = 95,844 Btu/hr, or 95,844/12,000 = 8 tons The coil load is greater than the space-cooling load because of the cooling and dehumidification of the outdoor ventilation air. __________________________________________________________________________ Dehumidification Without Cooling Coils It was shown in Example 8.10 that dehumidification usually accompanies cooling in typical space-cooling systems using refrigerant coils. In industrial processes and other applications where precise humidity control is important, special equipment often is used. Substances known as desiccants, which absorb water vapor, may be employed in such dehumidification applications, alone or in combination with refrigerant coil cooling and dehumidification. Examples of solid desiccants are silica gel, lithium chloride, and activated alumina. You may have observed that cartons which contain merchandise that must be kept dry are often shipped with small packets of a desiccant such as silica gel. The difference in water vapor pressure between neighboring regions is a driving force for water vapor transport by diffusion from the high-vapor-pressure region to the lower-pressure region. A desiccant absorbs water vapor from air adjacent to its surface, lowering the local vapor pressure and thereby attracting an inflow of additional vapor from the surroundings. As a desiccant absorbs water vapor, however, the water vapor pressure at its surface gradually increases. The material will continue to accept water vapor until its vapor pressure equals or exceeds that of the surrounding air. Thus the material must be reactivated to continue the drying process. Since vapor pressure increases with temperature, a desiccant can be reactivated by heating it with hot air. At high temperatures the desiccant vapor pressure exceeds that of the surrounding air, and water vapor then passes from the desiccant to the surrounding air. Commercial dehumidifiers may use liquid or solid desiccants. Figure 8.23(a) shows a schematic of a compact desiccant-wheel dehumidification system that has a wide range of industrial applications and that allows the drying of process air to a very low dew point. In the dehumidifier, the process air passes through a duct leading to a sector of a rotating, fiberglass-composite honeycomb wheel having many small passages impregnated with lithium chloride dessicant. The water-laden sector rotates slowly into another duct, sealed off from

326

the process air, through which a separate air flow passes at a temperature from 250°– 300° F. This hot air reactivates the desiccant in preparation for its next exposure to the process air. Though heat (not work) is required to reactivate the desiccant, the process requires less energy than other dehumidification processes. A desiccant-wheel storage room installation is shown in Figure 8.23(b). As in the figure, heated outdoor air is usually provided for reactivation of the desiccant. The reader is referred to references 11 and 12 for further information on dessicant dehumidification.

327 Bibliography and References 1.

Parsons, Robert A. (Ed.), ASHRAE Handbook of Fundamentals. Atlanta: American Society of Heating, Refrigerating, and Air-Conditioning Engineers, 1989, p. 16.3.

2.

McQuiston, Faye C. and Parker, Jerald D., Heating Ventilating, and Air Conditioning, 3rd ed. Wiley, New York, N.Y., 1988.

3.

Stoecker, W. F. and Jones, J. W., Refrigeration and Air Conditioning, 2nd ed. McGraw-Hill, New York, N.Y., 1982.

4.

Threlkeld, James L., Thermal Environmental Engineering, 2nd ed. Prentice-Hall, Inc., Engelwood Cliffs, N.J., 1970.

5.

Jennings, Burgess H., The Thermal Environment – Conditioning and Control, Harper and Row, 1978.

6.

Stoecker, W.F., Design of Thermal Systems, 3rd ed. McGraw-Hill, New York, N.Y., 1989.

7.

Goff, John A. and Gratch, S., “Thermodynamic Properties of Moist Air," ASHRAE Handbook of Fundamentals, Atlanta: American Society of Heating, Refrigerating, and Air-Conditioning Engineers, 1972.

8.

Derra, Skip, "CFCs No Easy Solutions," R&D Magazine, May 1990, pp. 56-66.

9.

Moore, Taylor, "CFCs: The Challenge of Doing Without," EPRI Journal, Vol. 14, No. 6, September 1989.

10. Purvis, Ed, "Applications of Scroll Compressors in Heat Pumps," Energy Technology XVI, Government Institutes, Inc., February 1989, pp. 410-420. 11. Parsons, Robert A. (Ed.), “Sorbents and Desiccants,” ASHRAE Handbook of Fundamentals. Atlanta: American Society of Heating, Refrigerating, and AirConditioning Engineers, 1989, Ch 19. 12. Parsons, Robert A. (Ed.), “Sorption Dehumidification and Pressure Drying Equipment,” ASHRAE Handbook of Equipment. Atlanta: American Society of Heating, Refrigerating, and Air-Conditioning Engineers, 1988, Ch 7. 13. Althouse, Andrew D., et al., Modern Refrigeration and Air-Conditioning. South Holland, Ill.: Goodheart-Wilcox Co., 1975.

328 14. York, Thomas, “Heat Pumps: Developing the Dual Fuel Option,” EPRI Journal, December 1990, pp. 22-27. 15. Public Law 101-549, “ An Act to Amend the Clean Air Act to Provide for Attainment and Maintenance of Health Protection, National Air Quality Standards, and Other Purposes,” November 15, 1990. 16. Anon., Refrigeration and Ice Making. Alymer, Ontario: Ontario Arenas Association Inc., March 1988. EXERCISES 8.1 Derive a relationship between the coefficient of performances for the heat pump and refrigerator using the ideal cycles of Figure 8.1. 8.2 Carefully sketch a flow diagram and T-s diagram for a refrigeration cycle that uses a single heat exchanger for suction line heating and condenser discharge subcooling. Explain how the addition of this heat exchanger influences the cycle refrigeration effect? 8.3 Derive the conversion factors in Btu/hr and in kJ/hr, for a ton of refrigeration, based on the latent heat of fusion of ice of 144 Btu/lb. 8.4 Derive the following convenient relation for the horsepower required to produce a cooling capacity of one ton of refrigeration: horsepower/ton = 4.72/COP. 8.5 When shopping for an air conditioner, one encounters labels that give the product's EER. What is the coefficient of performance for an air conditioner with an EER of 9.3? 8.6 Determine the capacity and COP of a vapor compression chiller operating with a reciprocating compressor between 35°C condensing and –10°C evaporating with refrigerant 22. The compressor has a volumetric efficiency of 70% and operates at 1800 rpm with a displacement volume of 750 cubic centimeters. 8.7 Compare the refrigeration effect, compressor work, and COPr for refrigerants 12 and 22 for a refrigeration system that has a 40°F evaporator and condenses at 140°F. 8.8 Compare the refrigeration effect, COPr, compressor work, and EER for an R-12 refrigeration unit with 120°F condensing and 50°, 30°, and 10°F evaporators. Tabulate the results. 8.9 Compare the refrigeration effect, COPr, compressor work, and EER for an R-22 refrigeration unit with 120°F condensing and 50°, 30°, and 10°F evaporators. Tabulate the results.

329 8.10 Compare the refrigeration effect, COPr, compressor work, and EER for an R-22 refrigeration unit with 100°, 120°, and 140°F condensing and 50°F evaporator. Tabulate the results. 8.11 Compare the refrigeration effect, COPr, compressor work, and EER for an R-12 refrigeration unit with 100°, 120°, and 140°F condensing and 50°F evaporator. Tabulate the results. 8.12 Compare the results of Exercise 8.10 with those for a similar system, accounting for a compressor with a 75% isentropic efficiency. 8.13 Compare the results of Exercise 8.11 with those for a similar system, accounting for a compressor with a 75% isentropic efficiency. 8.14 Consider a two-stage refrigeration system that operates with a 40°F evaporator and condenses at 140°F. Draw labeled T-s and flow diagrams for the cycle that consists of two ideal vapor compression cycles connected by a heat exchanger that serves as the condenser for the low-pressure cycle and the evaporator for the high-pressure cycle. The evaporating and condensing temperature in the heat exchanger is 80°F. For R-12, determine the ratio of refrigerant flows for the two cycles and compare the refrigeration effect, total compressor work, COPr, and EER with that for the single-stage system. 8.15 Consider a two-stage refrigeration system that operates with a 40° F evaporator and condenses at 140°F. Draw labeled T–s and flow diagrams for the cycle that consists of two ideal vapor compression cycles connected by a heat exchanger that serves as the condenser for the low-pressure cycle and the evaporator for the high-pressure cycle. The evaporating and condensing temperature in the heat exchanger is 80°F. For R-22, determine the ratio of refrigerant flows for the two cycles; and compare the refrigeration effect and the total compressor work per unit mass in the evaporator, the COPr, and the EER with that for the single-stage system. 8.16 A Diesel engine develops 500 horsepower at an efficiency of 38%. Almost all of the heat loss from the engine is transferred through the lubricating oil, through the cooling water, and through an exhaust gas heat exchanger to the generator of an absorption refrigeration system with a COP of 0.8. Derive an equation for the capacity of the evaporator in terms of the engine efficiency and power output and the COP of the refrigeration system. What is the capacity of the evaporator of the refrigeration system in Btu/min and in tons? 8.17 The volumetric efficiency, mvs /(N(disp), of a reciprocating compressor may be crudely modeled as: Vol. eff. = 1 + C – Cr(1/n)

330 where C is the ratio of the clearance volume to the piston displacement disp, vs is the compressor suction specific volume, r is the compressor pressure ratio, and n is the polytropic exponent. Taking C = 0.15 and n = 1.2, plot capacity and compressor power versus evaporator temperature for a condensing temperature of 140°F for R-12. The compressor has a 3 in3 displacement and operates at 1200 rpm. 8.18 To attempt to compare COPs for vapor compression and absorption refrigeration systems, consider a vapor compression system (with COPr) driven by a heat engine (with efficiency 0E). The combined system is one that accepts heat from a hightemperature source and uses it to cool a low-temperature region. Work out an equation for the COP for the combined system COPa,and compare it with the COPr for the corresponding vapor compression system. What is COPa, when COPr = 6 and the engine efficincy is 25%? 8.19 Determine the COP of an ideal vapor-compression-refrigeration system using R-22 with a 40°F evaporator and a condensing temperature of 110°F. Next, assume that the work for the system is provided by a Carnot engine operating between 250°F and 100°F reservoirs. Diagram the system. Considering the combined system as a model of an absorption system, compare its absorption COP with the COP of the vapor compression system alone. Discuss the conditions under which higher-absorption COPs could be attained. 8.20 Sketch and discuss thoroughly the slope and variation of lines of constant enthalpy on the psychrometric chart. 8.21 Compare the sea level relative humidity, enthalpy, specific volume, wet-bulb temperature, and dew point using (a) steam tables and ideal gas mixture relations and (b) the psychrometric chart for the following conditions: 1. 60°F dry-bulb, W = 0.01. 2. 90°F dry-bulb, W = 0.02. 3. 90°F dry-bulb, W = 0.01. 8.22 Compare the sea level humidity ratio, enthalpy, specific volume, wet bulb temperature and dew point using (a) steam tables and ideal gas relations and (b) the psychrometric chart for the following conditions: 1. 60°F dry-bulb, M = 50% 2. 90°F dry-bulb, W = 80% 3. 90°F dry-bulb, W = 40% 8.23 Compare the sea level relative humidity, humidity ratio, enthalpy, specific volume, and dew point using (a) steam tables and ideal gas relations, and (b) the psychrometric chart for the following conditions: 1. 60°F dry-bulb, 55°F wet-bulb.

331 2. 90°F dry-bulb, 55°F wet-bulb. 3. 90°F dry-bulb, 80°F wet-bulb. 8.24 Two hundered lbm/hr of moist air at 65°F dry-bulb and 50°F wet-bulb temperatures mixes with 40 lbm/hr of air at 100°F and 85°F wet-bulb. What is the mixture dry-bulb temperature, humidity ratio, relative humidity, and dew point? What are these parameters if the second stream flow rate is increased to 80 lbm/hr? 8.25 One hundred lbm/hr of moist air at 70°F dry-bulb and 50% relative humidity mixes with 35 lbm/hr of air at 90°F wet-bulb and 70% relative humidity. What are the mixture drybulb and wet-bulb temperatures, humidity ratio, relative humidity, enthalpy, specific volume, and dew point? 8.26 Determine the sensible, latent, and total heat transfer for moist air flowing over a cooling coil. The on-coil wet-bulb and dry-bulb temperatures are 85°F and 105°F, respectively, and the coil-leaving conditions are 60°F dry-bulb and 60% relative humidity. How much water is condensed per unit mass of dry air? 8.27 Determine the sensible, latent, and total heat transfer for moist air flowing over a cooling coil. The on-coil wet-bulb and dry-bulb temperatures are 75°F and 95°F, respectvely, and the coil-leaving conditions are 50°F dry-bulb and 85% relative humidity. How much water is condensed per unit mass of dry air? 8.28 Air at sea level has a dry-bulb temperature of 80°F and an adiabatic saturation temperature of 60°F. Evaluate the humidity ratio, relative humidity, and dew point temperature using (a) equations and steam tables and (b) the psychrometric chart method. 8.29

Air at sea level has a dry-bulb temperature of 85°F and an adiabatic saturation temperature of 55°F. Evaluate the humidity ratio, relative humidity, and dew point temperature using (a) equations and steam tables and (b) the psychrometric chart method.

8.30 A zone is to be maintained at 72°F dry-bulb and 50% relative humidity. The zone sensible and latent loads are 150,000 Btu/hr and 50,000 Btu/hr, respectively. The supply-air temperature is to be 48°F. What is the supply-air mass flow rate and volume flow rate? 8.31 Rework the parts of Example 8.10 that are changed by reducing ventilation air to 10% of supply air. Compare the coil cooling load and volume flow rate with the given results. 8.32 Rework the parts of Example 8.10 that are changed by increasing the supply-air dry-

332 bulb temperature to 55°F. Compare the coil mass and volume flow rates and cooling load with the given results. 8.33 A room is to be maintained at 74°F dry-bulb and 62°F wet-bulb by a vapor compression air conditioner. The total cooling load on the zone is 24,000 Btu/hr, and the latent load is 4,000 Btu/hr. Outdoor air is provided for ventilation at 100°F dry-bulb and 60% relative humidity. The ventilation mass rate is 15% of the supply-air rate. Air leaving the coil is at 52°F dry-bulb. (a) Show the significant states and processes on the psychrometric chart, and construct a table of wet- and dry-bulb and dew point temperatures, enthalpy, humidity ratio, specific volume and relative humidity for the space, outdoor, mixture, and supply- air conditions. (b) What are the mass and volume flow rates of return air, ventilation air, and supply air? (c) What is the coil cooling load in Btu/hr and in tons? 8.34 Resistance heaters are switched on in 5-kW increments to satisfy the demand for heat from the electric heat pump system characterized by Figure 8.15. At what temperatures should the first and second 5-kW banks be turned on? 8.35 Indicate whether the following are true or false: (a) An isotherm on a steam p-V diagram is a line of constant relative humidity for a given total moist air pressure. (b) The saturated vapor line on a steam p-V diagram is a line of constant relative humidity for a given total moist air pressure. (c) Increasing partial pressure of water vapor for fixed moist air pressure always increases relative humidity. (d) Increasing partial pressure of water vapor for a fixed moist air pressure and fixed dry-bulb temperature always increases relative humidity. (e) Decreasing dry-bulb temperature for fixed moist air pressure and fixed water vapor pressure always increases relative humidity.

333 CHAPTER 9 ADVANCED SYSTEMS 9.1 Introduction For decades the response to the ever-growing need for electric generation capacity was to build a new steam power plant, one not very different from the last. Today the energy conversion engineer is faced with a variety of issues and emerging technologies and a changing social and technological climate in which a diversity of approaches is likely to be accepted. This chapter intends to indentify some of these concerns and opportunities. No claim of completeness is made. No chapter, or book for that matter, could thoroughly cover this domain. The reader is referred to the bibliography at the end of the chapter as a starting point for continued study. A few characteristics of importance in new power initiatives are: low capital and operating costs, ability to operate with a variety of fuels and with high tolerance to fuel variability, short construction time, low emission of pollutants, marketable or at least inert and easily disposable waste products, and high efficiency, maintainability, financeability, and reliability. Increasingly, the new initiative may take the form of repowering the the old plant so as to increase efficiency, meet pollution standards, and minimize the financial impact of meeting new power demands. The improvement of the efficiency of power plants using conventional cycles is usually evolutionary in nature, by virtue of high temperature limitations and advances in materials. Hence, only gradual improvements in efficiency can be expected. On the other hand, significant improvements in efficiency can sometimes be obtained by combining conventional cycles in appropriate ways. Such power plants are referred to as combined-cycle plants. This chapter will examine the characteristics of several combined-cycle plants. It is evident from the study of the Rankine and Brayton cycles, and in fact all heat engines, that the rejection of large amounts of thermal energy to the surroundings accompanies the production of useful power. This heat rejection can be reduced by improving the thermal efficiency of the cycle but cannot be eliminated. If this energy is not to be wasted, it is logical to seek applications where both power and rejected heat may be utilized. Power plants that produce mechanical or electrical power and utilize “waste heat” for industrial processes are called cogeneration plants. Several examples of cogeneration are considered in this chapter. District heating and other possible applications of waste heat are also discussed. Another key problem facing the energy conversion engineer is the anticipated scarcity, in a few decades, of fuels such as natural gas and oil, relative to the vast resources of coal available in the United States and elsewhere. Perhaps future power plants should utilize this coal and nuclear energy to save the natural gas and

334 petroleum for industrial feedstocks and other more critical future needs. On the other hand, serious problems exist with respect to utilization of these resources. Nuclear power, an important alternative, replete with problems, is considered in the next chapter. Much of the readily available coal has unacceptably high sulfur, which significantly degrades the environment when released from power plant stacks in untreated combustion products. The well-known problem of acid rain has been attributed to emissions from coal-burning power plants. Thus the search for technology to utilize medium- and high-sulfur coal and to reduce levels of pollutant emissions of all types is an important area for research and development. In this chapter advanced technologies that may contribute solutions to these and other crucial problems are considered. Some recent U.S. Department of Energy efforts in these areas may be found in Reference 66. 9.2 Combined-Cycle Power One of the unfavorable characteristics of the gas turbine is that the exhaust gas issuing from the turbine is at high temperature, thereby wasting much energy and creating a local hazard. One solution to this problem was considered in Chapter 5: the addition of an exhaust gas heat exchanger to preheat the combustion air. The resulting regenerative cycle was found to be much more efficient than the corresponding simple cycle and to produce a lower exhaust gas temperature. An alternative approach to dealing with the high gas turbine exhaust temperature is to provide a separate bottoming cycle to convert some of the energy of the turbine exhaust into additional power. Let’s consider the use of a Rankine cycle that uses gas turbine exhaust as its energy source. It is clear that, if the Rankine cycle does not interfere with the operation of the gas turbine, the combined cycle will produce additional power and will have a higher efficiency than the gas turbine alone. Even if more heat is required for the Rankine bottoming cycle to produce additional work, the overall combined efficiency will increase if the additional work is large enough and supplemental heat small enough. A combined gas turbine–Rankine cycle can be implemented in several ways. One method makes use of the fact that the exhaust of gas turbines usually has a high residual oxygen content because of the high air-fuel ratio required to limit the turbine inlet temperature when burning conventional fossil fuels. This hot, oxygen-rich exhaust gas can be used instead of air as the oxidizer in a steam generator as shown in Figure 9.1. For a moderate expenditure of additional natural gas in the furnace the resulting combustion products can provide heat for a hightemperature steam cycle with conventional steam plant technology. The Horseshoe Lake combined-cycle plant was designed in this way to yield additional power and high efficiency when operating in the combined mode, and to operate with the gas turbine alone or with the steam turbine alone by direct-firing

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of the steam generator with fuel and air. The plant, about ten miles east of Oklahoma City, was the first its kind, first producing power in 1963. It was designed for a net output of 200 MW and a net heat rate of 9350 Btu/kW-hr. The plant cross-section and photos of the turbines are seen in Figures 9.2–9.4. Note the evaporative cooler to reduce the temperature and to increase the density of the air entering the gas turbine compressor. This reduces compressor work and increases inlet mass flow rate. Twenty years of successful operating experience with the Horseshoe Lake plant and information on similar German plants is documented in reference 46. An open-cycle gas turbine also may be linked to a steam cycle through what may be considered a gas turbine exhaust heat exchanger containing an economizer, a boiler, and perhaps a superheater. This device, called a heat-recovery steam generator (HRSG), may be used to create and superheat steam as in a conventional steam cycle. A flow diagram for such a cycle is shown in Figure 9.5, together with a T-s diagram for the steam bottoming cycle. Gas turbine exhaust gas cools as it superheats, boils, and then warms liquid water in counterflow as it passes through the HRSG.

336

337 Combined-Cycle Analysis Assuming a constant heat capacity for the gas turbine combustion gas, we may compare gas temperatures with adjacent HRSG local water temperatures on the Ts diagram, Figure 9.5, by applying the steady-flow First Law of Thermodynamics to appropriate sections of the HRSG: ms(h12 – h11) = mgCpg(T6 – T7) = msA1

[Btu/hr | kJ/hr]

(9.1)

ms(h13 – h12) = mgCpg(T5 – T6) = msA2

[Btu/hr | kJ/hr]

(9.2)

ms(h8 – h13) = mgCpg(T4 – T5) = msA3

[Btu/hr | kJ/hr]

(9.3)

where ms and mg are the mass flow rates for the steam and gas turbine cycles, respectively, and the Ai (i = 1, 2, 3) are the areas on the T-s diagram representing heat transfer per unit mass of steam. By adding the three equations, we obtain the same equation as would result from application of the steady-flow First Law of Thermodynamics to the entire HRSG: ms(h8 – h11) = mgCpg(T4 – T7) = ms ( A1 + A2 + A3)

[Btu/hr | kJ/hr]

(9.4)

Thus the enthalpy rise of the steam in the HRSG is controlled by the ratio of mass flow rates and the hot-gas temperature drop. Expressing gas temperature in the HRSG in terms of steam enthalpy allows us to condense these equations into T = T7 + (ms /mgCpg)(h – h11)

[R | K]

(9.5)

It is evident that the gas temperature is linearly proportional to the water enthalpy on a T-h diagram, as shown in Figure 9.6. The abscissa may be viewed as the cumulative heat transfer per unit mass of water, which is in turn proportional to the exhaust gas heat transfer. The temperature difference T6 – T12, known as the pinchpoint temperature difference, is at a critical location in the heat recovery steam generator, because it occurs at the point of minimum temperature difference between the two fluids. It should exceed some minimum design value (about 30°F) for all operating conditions of the system to make effective use of all of the HRSG heat transfer surface. Smaller temperature differences would substantially increase the heat transfer surface area needed, while significantly larger values would necessitate reducing the boiling temperature and would adversely affect the combined-cycle thermal efficiency.

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EXAMPLE 9.1

Let us examine several possible Rankine bottoming cycles for the gas turbine considered earlier in Example 5.1. There a gas turbine with a compressor pressure ratio of 6 and a turbine inlet temperature of 1860°R was analyzed. The turbine exhaust temperature was found to be 1273°R, or 813°F. With 813°F as the HRSG gas entrance temperature, select the steam turbine throttle temperature as 700°F and consider Rankine cycles with a range of boiling temperatures, a condensing temperature of 100°F, and a pinchpoint temperature difference of at least 30°F. Determine a cycle with a satisfactory boiling temperature, and compare it with

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other cycles in the family. Take the HRSG gas exit temperature as 350°F to avoid condensation. Solution

The spreadsheet in Table 9.1 follows the notation of Figures 9.5 and 9.6. It tabulates three combined-cycle cases in which the boiling temperatures, T12 = T13, were selected as 300°, 400°, and 500°F. These temperatures, together with the throttle and condensing temperatures and the turbine efficiency, determine the Rankine-cycle thermodynamic conditions (neglecting pump work). Equation (9.5) relates the HRSG enthalpy changes to the corresponding gas temperatures. Thus with a mass ratio Rmass given by (h8 – h11)/(T4 – T7), the pinchpoint temperature T6 for Case 2 is T6 = T7 + (ms /mgCpg)(h12 – h11) = T7 + (h12 – h11)/Rmass = 350 + (375.1 – 68)/2.82 = 350 + 108.9 = 459°F

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and the pinchpoint temperature difference is T6 – T12 = 459 – 400 = 59°F Again, by Equation (9.5), T5 = 350 + (1201 – 68)/2.82 = 350 + 401.8 = 752°F With temperatures and enthalpies determined, work, heat, and efficiency values may be determined as usual, observing carefully the distinction between steam-mass and gas-mass references. It is seen that Case 2 has a satisfactory pinchpoint temperature difference and a combined-cycle efficiency of 39%, which is significantly greater than that of the gas turbine cycle (25%) and Rankine-cycle (29%) operating separately. The temperature distributions for this case are shown in Figure 9.7. __________________________________________________________________

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More than one gas turbine could be used in conjunction with a single steam turbine by using a larger HRSG and multiple gas turbine exhaust ducts. Alternatively, each gas turbine may have its own HRSG, with all HRSGs thermally coupled to the steam turbine through feedwater and steam line headers. An HRSG may be designed to burn additional fuel at its inlet using oxygen remaining in the gas turbine combustion gas to raise the temperature the HRSG gas and provide higher heat transfer to the water. Such a design is called a fired HRSG, and its use is usually referred to as supplemental firing. Figure 9.8 shows large heat-recovery steam generators built for a nominal 450MW combined-cycle plant in Texas utilizing three 100-MW gas turbines and one 140-MW steam turbine. The Comanche power station, located near Lawton, Oklahoma, is an example of a combined-cycle facility that employs HRSGs. Two Westinghouse simplecycle natural-gas-fired gas turbines, as diagramed in Figure 9.9, drive 85-MW electrical generators and exhaust into separate HRSGs designed for supplemental firing. Steam at 1200 psia and 950°F produced by the HRSGs is supplied to one nonreheat, single-extraction (not shown) steam turbine that drives a 120-MW electrical generator. The Comanche unit, first operated in 1974 (ref. 10) was upgraded in 1986 (ref. 11), resulting in a measured plant heat rate of 8508 Btu/kW-hr (40.1% thermal efficiency), with a gas turbine inlet temperature of 1993°F, an HRSG gas inlet

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temperature of 1200°F, and supplemental firing. In 1986, the plan was indentified by the Utility Data Institute as the most efficient steam-electric generating station in the United States, with an average net plant heat rate of 8821 Btu/kW-hr (thermal efficiency of 38.7%). High-temperature combined-cycle plants are now achieving thermal efficiencies exceeeding 50%. A combined-cycle heat rate for a United Technologies Turbo Power FT8 gas turbine is said to operate at a gross plant heat rate of 6815 Btu/kW-hr (50.1%) based on lower heating value (ref. 51). Reference 50 indicates that the Pegus Unit 12 combined-cycle cogeneration plant in the Netherlands produces 223.3 MW at maximum electrical output, with a net electrical yield of 51.74% based on lower heating value. According to reference 67, “Both GE and Siemens Westinghouse turbines will be able to break the 60 per cent efficiency barrier in combined-cycle operation, and a 3 to 6 per cent reduction in CO2 emissions should be possible because of the increased efficiency.”

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9.3 Integrated Gasification Combined-Cycle (IGCC) Power Plants The Cool Water IGCC Plant

One approach to the problem of clean coal utilization lies in the technology exemplified by the Cool Water integrated gasification combined cycle (IGCC) power plant located near Barstow, California. This plant, which went into operation in 1984, demonstrated the capability of producing power for the

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Southern California Edison System with very low levels of pollutants by using both low-sulfur and medium-sulfur coals. The Cool Water demonstration plant shown in Figure 9.10 (named after the ranch on which it is located) utilized a coal gasification technique known as the Texaco process. The medium-heating-value synthetic gas called syngas produced by the process has about one-third the heating value of natural gas, about 265 Btu per standard cubic foot on a dry basis. As it is produced, the syngas is first cooled, then treated to remove pollutants, and finally burned to drive turbine generators, as seen in Figure 9.11. The initial cooling in the syngas coolers produces saturated steam,which is later superheated by gas turbine exhaust gas in an HRSG, to power a steam turbine (path A in the diagram). The syngas emerging from the coolers (path B) is processed to remove particulates and sulfur and to control oxides of nitrogen, and is then burned in the gas turbine to produce additional power. The high-temperature gas turbine exhaust (path C) then passes through the heat-recovery steam generator, adding energy to the steam before it passes to the steam turbine. Thus the gasifier flows and the steam turbine and gas turbine flows interact, hence the name “integrated gasification combined cycle,” IGCC. The Texaco process requires oxygen of at least 95% purity to gasify the coal in the gasifier. The Cool Water plant thus has a small, independently owned oxygen plant (seen surrounding the single tower left of center in figure 9.10) that separates

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oxygen from air to feed the gasifier. The Cool Water plant produces oxygen of 99.5% purity in order to produce argon as a by-product (as mentioned earlier, the process otherwise requires only 95% purity). Nitrogen is also a by-product. Oxygen is produced continually for gasifier use, but oxygen is also stored so that the IGCC plant can continue to operate if the oxygen plant is shut down. A coal slurry, a mixture of nominally 60% coal and 40% water produced in a wet grinding process, is introduced with oxygen into the Texaco gasifier (Figure 9.11), where partial combustion of the coal takes place at about 600 psig and 2500°F. The gasifier yields a mixture of mainly carbon dioxide, carbon monoxide, and hydrogen gases, with sulfur primarily in the form of hydrogen sulfide. A relatively inert slag containing most of the mineral matter of the coal passes from the gasifier into a pool of water in the bottom of the radiant cooler. The slag is taken out periodically through a lockhopper process. The seal at the bottom of the radiant cooler is maintained by the water that is recycled. The syngas, after cooling in the radiant and convective coolers, passes through a carbon scrubber, where a water spray removes most of the particulates and further cools the gas. After additional cooling to ambient temperature, the gas flows to a sulfur-removal unit, where a solvent removes the hydrogen sulfide and therefore most of the sulfur from the stream. The relatively particle-free and sulfur-free gas is then saturated with moisture to control the formation of oxides of nitrogen (NOx) during combustion in the gas turbine. The water-quenching process suppresses NOx formation by reducing the gas combustion temperature, and it also increases the turbine power output by adding to the mass flow in the gas turbine combustor. Combustion gases from the gas turbine then pass through the heat

346 recovery steam generator, where they produce additional steam as they drop in temperature to about 400°F. The combustion products leaving the plant are remarkably pollutant free. Performance of the Cool Water Station is summarized in Table 9.2 below. Table 9.2 Cool Water IGCC Station Nominal Performance Gas turbine electric generator power

65 MW

Steam turbine electric generator power

+55 MW

Gross power

120 MW

Air-separation oxygen plant

–17 MW

Internal plant consumption

– 7 MW

Net power

96 MW

Design heat rate

11,510 Btu/kW-hr

Observed heat rate

10,950 Btu/kW-hr

Source: Reference 3. The operators of the Cool Water station have demonstrated values of critical pollutants well below current environmental limits for both permit and regulatory limits and New Source Performance Standards (NSPS), as indicated in Table 9.3. The Cool Water plant has operated successfully with Utah run-of-mine coal with 0.4% sulfur, Illinois #6 coal with 3.1% sulfur, and Pittsburg #8 coal with 2.9% sulfur. The sulfur removal process in the plant yields about 99.6% pure elemental sulfur, which can be sold for the production of sulfuric acid and fertilizers. The slag produced by the gasifier is considered nonhazardous and suitable for the production of road-making materials or cement. Table 9.3 Emissions from the Cool Water Station HRSG (lb/million Btu) High Sulfur Coal SO2

Low Sulfur Coal SO2

NOx

CO

0.16

0.033

0.13

0.07

0.01

Utah coal

–

0.018

0.07

0.004

0.001

Illinois #6

0.068

–

0.094

0.004

0.009

Pittsburgh #8

0.122

–

0.066

<0.002

0.009

Federal NSPS Source: Reference 3

0.6

0.24

0.6

–

0.03

Permit and regulatory limit

Particulate Matter

The net effect of the plant then is to generate power efficiently by utilizing widely available coals with sulfur content up to 3.5% in an environmentally sound

347 way while producing nonhazardous waste that offers the possibility of constructive use. It should be understood that the Cool Water plant was built as a demonstration plant to prove a technology on a large but not a full scale. Engineering studies based on the Cool Water operating experience (ref. 4) have indicated that a 360MW full-scale IGCC plant using Illinois #6 coal could be built with a net heat rate of 9000 Btu/kW-hr at a capital cost of $1530/kW and with operating and maintenance costs of 2.3 cents/kW-hr (or 5.2 cents/kW-hr, which includes all fixed charges) (ref. 3). These costs are competitive with those for conventional new coal-burning power plants using flue gas desulfurization. An important feature of the IGCC plant concept is the attractiveness of phased construction. A conventional coal-burning steam power plant must be constructed as a unit and takes a relatively long time to erect. On the other hand, one or more gas turbines of a planned IGCC plant may be quickly put into service using natural gas as a fuel. Additional gas and steam turbines may be added later to transform the plant into a combined-cycle plant, with coal gasifiers and an oxygen plant added still later at a third stage. Because the units may be paid for as they are built, phased construction offers significant financial benefits as well as ordely growth. Significant materials problems have been overcome in the Texaco and Cool Water plant technology. The inside of the gasifier requires refractory materials to withstand the severely corrosive high-temperature environment. Cool Water experience (ref . 5) indicates that 10,000–14,000 hours of refractory life is attainable. This implies that gasifier overhaul will be required at least every two years. Additional problems with cooler tube-wall materials, coal slurry pumps, and piping; and other severe material operating environments offer challenges for materials research to improve IGCC operation. Several of the references expand on the idea of cogeneration–polygeneration– by pointing out that, based on the Cool Water technology today, a single facility whose only major requirements are air, water, and coal can simultaneously produce electricity, steam, sulfur, inert slag for road construction, oxygen, nitrogen, hydrogen, carbon monoxide, carbon dioxide, argon, methanol, other chemical products, and even syngas. The Dow Gasification Process Another coal gasification process combined-cycle system (CGCC), developed by Dow Chemical Co., operates in Plaquemine, Louisiana. Like the IGCC system, the Dow process reacts a coal slurry with oxygen to produce syngas, from which most of the sulfur is removed for by-product use. The raw syngas from the gasifier produces steam in an HRSG, is cooled, passes through pollutant removal equipment, and is burned in gas turbines. Reference 47 indicates that “of the new coal-based technologies, the CGCC system has the highest efficiency and the lowest emission of environmental pollutants.” The 161-MW plant, built in the remarkable time of twenty-one months, was completed in 1987. The reference indicates for CGCC plants a net heat rate based on lower heating value of 8670

348 Btu/kW-hr (39.4%) and a total capital cost of $1201/kW, based on 1988 dollars. Additional information on combined-cycle coal gasification systems is given references 58, 59, and 65. Tampa Electric’s 250-MW IGCC Polk Power Station began commercial operation in 1996. A website (ref. 65) indicates that the station is 10-12% more efficient than conventional coal-fired plants.Three dimensional views may be seen at the website. 9.4 Combustion in Fluidized Beds The need to develop environmentally sound methods for utilizing a variety of coals, industrial and municipal wastes, and other solids as fuels has stimulated research in a variety of areas. Another prominent and promising technology applicable to these goals is fluidized bed combustion (FBC). Atmospheric fluidized bed combustors (AFBCs) operate near atmospheric pressure; pressurized fluidized combustors (PFBCS) are enclosed in a pressure vessel and operate at a pressure of about 12 atmospheres. As the name suggests, a principal unique feature of this technology is that combustion takes place in a bed of solid particles supported in vigorous, turbulent motion on an upwardly directed stream of air. The bed may consist of a combination of particles of fuel, sand, ash, waste materials, limestone, and other compounds, depending on the function and design of the fluidized bed combustor. The key point is that these materials mingle and react in continually changing orientations, providing ample opportunity for intimate contact of fuel and oxidizer and for removal of combustion products, while supported on the fluidizing air stream. The name bubbling bed is sometimes used to decribe this action. It has been found that horizontal water tubes located within the fluidized bed, in crossflow to the upward air stream, experience very high heat transfer coefficients in comparison with those in normal furnace convective and radiative environments. It is even more important that fluidized beds containing coal and limestone produce combustion gases with both low sulfur content and low concentrations of oxides of nitrogen. Limestone reacts with sulfur in the coal to produce particles of calcium sulfate that are removed as bed materials are renewed. More specifically, the limestone, CaCO3, reacts to form CO2 and lime, CaO: CaCO3 => CaO + CO2 and the CaO reacts with the sulfate vapors: H2SO4 + CaO => CaSO4 + H2O. The calcium sulfate forms as a solid that becomes a bed material, and the water vapor passes off as a component of the flue gas.

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Oxides of nitrogen (NOx) produced by the FBC are maintained at low values, because bed and gas temperatures are well below those at which NOx forms in conventional combustors. The bed temperature is determined primarily by the rates of air and fuel supplied. Normally, FBC bed temperatures are about 1550°F, compared with conventional furnace temperatures in the neighborhood of 3000°F. The reactivity of nitrogen is low at 1500°F, and chemical equilibrium calculations and laboratory observations show that the nitrogen in flue gas is almost entirely in its normal, diatomic form. The combustion of coal or other solids occurs largely at the interface between the solid and the surrounding oxidizing gases. The rate at which the solid burns depends on the rate at which oxygen is brought to the solid interface and on the rate at which combustion products are removed, as well as on the rate of chemical reaction at the interface itself. The vigorous relative motion of the bed particles and the intervening air flow provide an excellent mechanism for delivery of oxygen to and the transport of combustion products from the interface. Normally, FBC occurs with enough excess air, in the primary supporting air

350 stream and possibly in a secondary over-fire air flow, that combustion is virtually complete. Flyash and other airborne particles are removed by centrifugal separators and baghouse filters. Solids from the separators may be reinjected into the bed to further ensure almost complete burnup of carbon. The circulating fluidized bed (CFB) combustor (refs. 26 and 30) is one in which smaller solid bed materials are carried upward by the combustion air/gas stream. A return passage transports the unburnt and inert particles and part of the combustion gas back to the main furnace, allowing the remaining flue gas to pass to the heat-recovery area, as seen in Figure 9.12. The solid bed materials continue to burn as they circulate, thus maintaining an approximately uniform temperature of about 1550°F throughout the furnace. As a result, there is a long residence time for particles of the furnace to complete their reactions. A mechanical cyclone separator built into the furnace helps to separate the particles from the exiting flue gas. As a result, reinjection of the unburnt carbon makes possible very high combustion efficiencies. According to reference 26, CFB designs achieve higher combustion efficiency, reduced NOx emissions, minimum CO formation, and reduced limestone utilization in capturing SO2 when compared with bubbling fluidized bed combustors. Much continues to be learned about problems and opportunities inherent in fluidized bed combustion as more units come into use. In a December 1998 work (ref. 65), the U.S. Department of Energy (DOE) proposed a 379-MWe, pressurized circulating fluidized bed combustor combined-cycle plant with a net efficiency of 47%. 9.5 Energy Storage It has been observed that there is no existing means of storing electrical power on a large scale. As a consequence, power generation varies from instant to instant, to satisfy the immediate demands of consumers. Utility generation capacity must therefore be great enough to satisfy the peak demand, or the utility must purchase power at a premium from other utilities to make up its generation deficit. Demand varies from place to place, seasonally, daily, and hourly. For instance, the loads of utilities in the southern United States are usually greatest during hot summer days, when air conditioning and industrial demands coincide. As a result, southern utilities may have excess capacity at night and in the winter. It were possible to generate a full capacity during off-peak hours and store the energy in excess of demand, the utilities could operate with installed capacity below the demand peak and operate more units as base-load plants close to their high-efficiency design points.

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Pumped Storage One approach to off-peak energy storage utilizes a high reservoir into which water from a low elevation is pumped, using electricity generated during off-peak hours. Thus energy may be stored in the form of potential energy of the elevated water for later use during peak loads. There are about thirty such pumped-storage facilities in the United States. Figure 9.13 depicts the two phases of operation of a pumped-storage facility. These installations usually employ motor-generator sets driven by hydraulic turbines. Mechanically reversible pump-turbines generate electricity by using water from an elevated reservoir during periods of peak demand. When there is an excess of base-load power, the motor-generators can be reversed to drive the pump-turbines as pumps for filling the reservoir. It is clear that net energy is lost in the use of pumped storage. Its success relies on the availability of cheap power during off-peak hours and consistent demand for electricity, with its associated high price during peak hours. Pumped storage allows utilities to generate more electricity with their most efficient base-load plants instead of handling peaks with less efficient equipment. Figure 9.14 is a photograph of the Salina pumped-storage facility (ref. 34) of the Grand River Dam Authority, located about 50 miles from Tulsa, Oklahoma. The facility was designed for gradual expansion, in three steps, from the 130-MW configuration that went online in 1968, to the current 260-MW facility shown in

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the figure, to a 520-MW final modification. The eventual completion of the project will require the development of an additional upper reservoir as well as construction of six additional penstocks and the installation of their hydroelectric pump-turbine motor-generator sets. The present plant is capable of producing power at full-rated load for eight hours a day or at part load for twelve hours a day. The reservoir is 251 feet above the lower lake. The penstocks are 14 feet in diameter and 720 feet long. Reference 36 identifies an early pumped-storage facility named Rocky River at New Milford, Conn., which was in operation in 1928. The limited number of high head surface sites for such facilities, the high capital cost of building a dam, and the large land area impacted by these facilities make the future use of pumped storage questionable. Reference 37, however, predicted continued growth in pumped storage capacity. One possibility that would support such growth is to have the upper pool at ground level and to use underground mines for the lower pool (ref. 38). Another is the development of low head facilities.

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Compressed-Air Energy Storage A relatively new approach to energy storage is compressed-air energy storage (CAES), which employs underground caverns for storage of air under pressure (refs. 25 and 49). Air is compressed into the cavern using off-peak power and later is released to oxidize fuel in a gas turbine combustor to generate electrical power during hours of peak demand. The first CAES plant, a 290-MW unit, first operated at Huntorf, West Germany, in 1978. A second CAES plant located in McIntosh, Ala, with phased construction of two 110-MW units, first went on-line in May, 1991 (ref. 64). CAES plants are located in the vicinity of underground caverns. The caverns may be natural or may be created and/or enlarged by solution-mining of underground salt domes. In solution-mining, water is pumped into a salt formation, the water dissolves salt locally, enlarging the cavern, and the resulting brine is pumped to the surface, where the salt is driven from the solution and the water reused. Figure 9.15 shows a schematic of the McIntosh plant. During off-peak hours a motor-generator, powered by electricity produced elsewhere in the utility system and with turbine-expander clutch disengaged, drives the compressor set that packs air into the salt cavern. Later, air is allowed to escape from the cavern to oxidize fuel at the high cavern pressure, forming combustion gas to pass through the gas turbines (expanders) that drive the motor-generator in the generation mode with

354 the compressor clutch disengaged. The generator thus is able to return electricity to the utility system during periods of peak demand. The CAES plant is essentially a gas turbine in which the compression process is decoupled from the power delivery process. During daily cyclical operation, the mass of air supplied to the cavern at night is utilized during the day. Since the charging period may be different than the utilization period, the average mass rates of storage and delivery may differ. The McIntosh plant is designed for twenty-six hours of storage capacity and therefore does not operate on a strict daily cycle but rather on a weekly cycle that takes advantage of less expensive weekend power. The presence of the recuperator in the McIntosh plant, the first in use in a CAES facility, provides regeneration, which, as discussed in Chapter 5, substantially reduces gas turbine fuel consumption and thus improves plant efficiency. Regeneration may also be accomplished by absorption and storage of the heat of compression of the air in an aftercooling, high heat capacity heat exchanger such as a pebble bed recuperator. The stored energy is later returned to the air discharged from the cavern as it flows to the turbines. The success of a CAES plant, like that of pumped storage, depends on the availability of cheap, off-peak power. The objective is to drive the compressor train during these periods so that high-priced power may be produced later in periods of high demand. It should be noted that the compression process is not only decoupled mechanically from the power delivery process, but the compression process may take place over a longer period of time than the period of CAES power production. This allows the use of smaller compressors than would be needed for a conventional peaking gas turbine. According to reference 25, the Huntorf plant compresses for 4 hours, and the McIntosh plant for 1.7 hours, for each hour of power production. CAES may become a more viable option in the United States than surface pumped storage because of the existence of more potential sites and lower land surface area requirements for CAES. 9.6 District Heating and Cooling and Cogeneration Comfort heating and cooling in homes, businesses, and industry consumes large quantities of energy. Much of this low-temperature-energy use is accomplished by the direct or indirect burning of fossil fuels and high temperature. Electrical resistance heating, especially, and heat pumps may be included in this because the electricity they consume is produced largely from high-temperature sources. Many believe that these are inappropriate uses of fossil fuels, in a conservation sense, because of the unnecessary loss of the availability of high-temperature energy to do work. It is simply a reflection of the desirability of using high temperatures where needed and low-temperature sources for low-temperature functions. For instance, it is no revelation that some of the obligatory heat rejection by modern heat engines is at a high enough temperature to supply energy for comfort heating and cooling.

355 District heating, the provision of heat to a populated area by a nearby central heating plant, has been in use for one hundred years or more. More recently the addition of cooling to such plants has become widespread. Now, colleges and universities frequently have such a facility. Shopping malls and blocks of business districts in many cities also take advantage of the economic benefits provided by a central plant. This is usually accomplished through the use of boilers that produce hot water or steam for heating, and by vapor compression or steam-driven absorption refrigeration machines that produce chilled water for cooling. Water, steam, or brine is usually used to deliver the energy to the user. In municipal facilities, the steam, water, or brine is metered and circulated in insulated pipes under the streets to air-handling units and other point-of-use devices in the customers’ buildings, and thence to return pipes that bring the fluid back to the central plant or, in single-pipe one-way systems, to sewer mains. Until recently, most of the district heating facilities the United States did not find significant advantage in producing electrical power and using the waste heat for district heating, cooling, or industrial process energy. With Europe’s more limited energy resources, combined heat and power (CHP), or cogeneration, the synergistic generation of electric power and heat, found more extensive use there than in United States following the Second World War. In fact, a number of steam turbines and closed-cycle gas turbines burning a variety fuels were developed for simultaneous electric power generation and district heating or industrial cogeneration purposes (refs. 12 and 15–20). These activities are closely related to total energy systems, which seek to utilize natural gas for other purposes while generating electricity (ref. 17). In the United States this term has been used in the past for the promotion of natural-gas-burning systems that provided heating, cooling, and electricity for shopping malls, colleges, and similar customers. Some of the possible cogeneration steams include: 1. Steam turbine power with condenser heat rejection for low-temperature processes, facility heating, or district heating. 2. Steam turbine power with steam extraction or use of a back-pressure turbine for process or district heating use. 3. Steam turbine power with exhaust steam or steam extraction heat transfer to absorption refrigeration system generators for chilling processes or district cooling in summer. 4. Closed-cycle gas turbine power with coolers (intercoolers and pre-coolers) used for district heating. 5. Closed-cycle gas turbine power with coolers used with absorption refrigeration system for chilling processes or district cooling in summer.

356 6. Open-cycle gas turbine power with exhaust heat recovery for process use or district heating. 7. Diesel or gas reciprocating engine power with water-jacket cooling, oil cooler, and/or exhaust gas heat recovery used for process or district heating. A measure of the efficiency of energy utilization in a cogeneration plant is the energy utilization factor, EUF, which is the sum of the net work, w, and the useful heat produced, qu, divided by the energy supplied to achieve the combined heat and work, qin, (ref. 18): EUF = (w + qu )/qin

[dl]

(9.6)

Care should be taken when considering the EUF and the thermal efficiency of a plant. The EUF is not restricted by the Carnot efficiency and can therefore approach 100%. For instance, when there is no useful heat transfer, the EUF is the plant thermal efficiency. At the other extreme, when no power is produced the plant can utilize almost all the energy supplied by the fuel for useful heat; and therefore the EUF could approach 100%. Thus the EUF is a measure of the extent of productive use of the energy source, with no consideration for work–useful heat proportions. An EUF in excess of 80 percent is possible for a CHP plant. Other efficiencies for CHP plants may be defined. For instance, a weighting factor might be used in the numerator of the EUF to attempt to give appropriate weight to both heat and work. For example, since conventional power plants convert heat to work with about 33% efficiency, one might define a special EUF* as (w + 0.33qu )/qin for comparison with thermal efficiencies. While such factors may be useful in evaluating the design of a plant, they should be applied with care. Reference 18 discusses alternative definitions more fully. 9.7 Electricity Generation and Legislation Historically, the electric power industry in the United States developed by recognizing the economic advantages of scale of large central plants that used extensive power transmission and distribution systems, following the lead of the great governmental hydroelectric power projects. They also recognized the enhanced growth potential inherent in the society’s becoming “totally electric.” The productive disposition of condensor rejected heat had no place as a revenue producer in this scheme, for the important reasons that it would reduce plant efficiency and power output and could not reach across the miles as power transmission lines could. Individual consumers as well as industry happily accepted this approach to the electrification of America, as electricity prices dropped decade after decade. Natural-gas and fuel oil companies were also there to satisfy the vast needs for heat. More recently, energy-consuming industries came to recognize the possibilities of simultaneous power and heat generation to satisfy their energy requirements. At

357 the same time, a consciousness grew that there are limits to the world’s energy resources and that a more thoughtful stewardship of them would be prudent, when it is economically attractive. However, in the early 1970s the OPEC oil embargo brought these notions into clearer focus, and federal legislation in 1978 brought cogeneration to everyone’s attention with the enactment of PURPA, the Public Utility Regulatory Policies Act (ref. 14). PURPA, a part of the National Energy Act, was intended to bring competition into the electric power generation business and to increase the national effectiveness of energy utilization by making cogeneration more economically attractive. It required that utilities purchase power from qualified cogenerators at a reasonable price, created tax incentives for developers, and established the Federal Energy Regulatory Commission, FERC, to regulate and administer the activity. Prospective cogenerators must apply to the FERC for certification as a qualified facility, QF. The FERC, however, left the deliberation of what constitutes a fair price to state public utility commissions. As a result, industrial cogeneration thrives in some states and is virtually nonexistent and others. PURPA requires utilities to buy electricity from QFs at the marginal cost of new generation, that is, the cost of electricity generated by new power plant if it were built, the so-called avoided cost. The growth of cogeneration, state and national actions to change the regulatory structure of the electric utility industry, the difficulties of acquiring capital for large, long-term projects and other uncertainties are significantly changing the outlook and structure of the utility industry (ref. 21). Utilities now appear to be looking toward incremental and modular growth and the avoidance of long-term commitment, which is reflected in their reluctance or inability to undertake the construction of large, capital-intensive base-load plants. Legislative changes in the 1990's and beyond are bringing substantial deregulation to the industry, open access to power transmission systems, and the introduction of merchant plants created to offer electricity to the highest bidder in the new free-enterprise climate (ref. 66). 9.8 Steam-Injected Gas Turbines Water injection has been used for many years for brief augmentation of the thrust of jet engines. More recently, liquid water injection and steam injection have been used to control the formation of NOx in gas turbines. Injection of water or steam into the combustion chamber reduces the combustion temperature, which in turn suppresses the formation of NOx caused by high temperature. Power output is maintained or increased because the injection increases both the turbine mass flow and the energy extraction by the turbine. The latter is possible because the heat capacity of steam is almost twice that of normal combustion products. Thus the enthalpy change of steam for a given temperature drop is about double that of air or combustion gas. If water is injected as a liquid, additional energy must be extracted from the combustion gas to vaporize the water.

358

As a consequence of these considerations, an important cogeneration technology is emerging. The steam-injected gas turbine, STIGTM* or SIGT, uses the gas turbine exhaust flow through an HRSG to produce steam that is partially or entirely injected into the gas turbine combustion chamber and possibly the compressor and/or turbine, as indicated in Figure 9.16, resulting in augmented gas turbine power output. Reduced steam requirements for process use in an industrial cogeneration plant may be used to increase electrical power generation, which in turn may be sold to a local utility under PURPA if it is not needed for other on-site uses. References 13, 22, 23, and 31 show increased power output, increased thermal efficiency, and reduced NOx as benefits of steam injection. Table 9.4, for example, lists performance measurements of the effects of steam injection into two industrial gas turbines. The table shows that as the cogeneration process steam percentage is reduced and the steam flow to the gas turbine increases accordingly, substantial increases in power output and thermal efficiency are consistently attained. Thus the steam-injected gas turbine holds considerable promise for cogeneration and process applications. A cross-section of the General Electric LM 5000 gas turbine without steam injection is shown in Figure 9.17; a photo of the same is presented in Figure 9.18. The LM 5000 is a compact, high-performance, aeroderivative gas turbine intended for marine and industrial applications. It is derived from the CF6 family of highbypass-ratio turbofan engines but burns either distillates or natural gas fuels. The LM 5000 has a dual-rotor gas generator and a three-stage power turbine. The manufacturer quotes a power output of 46,200 shaft horsepower (34,451 kW) and a heat rate of 9160 Btu/kW-hr at a power turbine speed of 3600 rpm for the LM 5000 without STIG. ____________________ *STIG is a trademark of General Electric Co., U.S.A.

359 Table 9.4 Performance of Steam-Injected Gas Turbines with Unfired HRSG Cogeneration Steam Percentage

Power Output (kW)

Thermal Efficiency

Allison 501-KH 100%

3500

0.24

50%

4750

0.30

0%

6000

0.35

100%

33000

0.33 (0.36)

50%

40000

0.36 (0.38)

0%

47000

0.38 (0.42)

General Electric LM5000

Adapted from reference 22. Data in parentheses from references 56 and 57.

360

361

A schematic of the LM 5000 STIG configuration is shown in Figure 9.19; a photo of the same is presented in Figure 9.20. The steam manifolds and injection lines may be seen in both figures. Table 9.4 shows the performance of this configuration without supplemental firing of the HRSG. With full STIG and supplemental firing, the power output increases to 72,100 shaft horsepower (53,765 kW) and a heat rate of 7580 Btu/kW-hr (or a thermal efficiency of 45%) is attained.

362

Figure 9.21 shows the average steam generation capability of the engine exhaust with an unfired HRSG having a 30°F pinchpoint temperature difference. It is seen that, as expected, higher steam flow rates are obtained with reduced steam temperature and pressure for given HRSG exhaust and turbine exhaust temperatures. A study the economic advantages of steam-injected gas turbines for utility use relative to the combined-cycle power plant (ref. 33) has shown combined cycles to be superior in unit sizes above 50 MW. However, utilities are becoming increasingly interested in plants of 50 MW or less because of their small size and quick availability. Intercooled Steam-Injected Gas Turbines A modification of the steam-injected gas turbine is ISTIG, or, intercooled STIG, in which steam is injected into compressor bleed air for turbine cooling, together with steam injection into the combustor and into one or more turbine stages (ref. 22). The enhanced blade cooling allows increased turbine inlet temperature and further power and efficiency increases. The reference predicts efficiencies for ISTIG turbines better than for existing combined cycles and comparable to advanced combined cycles. Thus STIG and ISTIG show great promise for cogeneration applications and are likely to find their way into future power generation plans. Steam-Injected Gas Turbine Analysis The influence of steam injection into the combustor can be analyzed with a model that accounts for the major effects on the engine performance: the added mass and

363

364 heat capacity of steam in the turbine flow. The model assumes the following:

+Downstream steam injection has no effect on the compressor. +The properties of steam may be represented by constant values of the heat capacity Cps: 1.86 kJ/kg-K and 0.444 Btu / lbm -R.

+The heat capacity of the mixture of combustion gas and steam may be represented by a temperature-independent mass-weighted heat capacity given by Cpm = Cpg + (ms /ma )Cps

[Btu/lbm-R | kJ/kg-K]

where ms /ma is the steam-air mass ratio. Here, as before, the mass of fuel is neglected with respect to the mass of air.

+The isentropic exponent of the mixture, km , remains the isentropic exponent of combustion gas with no water injection, km = kg = 4/3.

+The fuel control system maintains the turbine inlet temperature at a constant value, irrespective of steam injection rate.

EXAMPLE 9.2

Consider the injection of steam in the combustion chamber of the single-shaft gas turbine studied in Example 5.1. Five pounds of steam are injected for every hundred pounds of compressor discharge air. The fuel flow rate is adjusted to maintain the turbine inlet temperature at a constant 1400°F. Compare the power output for a compressor flow rate of 100 lbm /s, the thermal efficiency, the work ratio, and the fuel-air ratio with like parameters for the machine with no injection. Assume that the steam injected is saturated at the combustor pressure level and is heated by the turbine exhaust from pressurized feedwater at 70°F in a heatrecovery steam generator as shown in Figure 9.22. Solution

Table 9.5 presents the spreadsheet solution as well as the no-injection reference solution, which is repeated there for convenience. The algorithm is based on the assumptions just enumerated and uses the HRSG analysis techniques discussed in section 9.2 in connection with the combined-cycle steam generator study. Steam conditions entering the combustor are obtained from the saturated-vapor tables, assuming constant pressure mixing at the known combustor pressure level; feed water conditions are for saturated liquid water at 70°F.

365

366

A 3-point increase in thermal efficiency and a substantial increase in power output due to steam injection are evident. The steam generator easily provides the required steam mass with a large pinchpoint temperature difference. This indicates that the turbine exhaust gas is capable of producing a still larger steam fraction with a smaller pinchpoint temperature difference while maintaining acceptable exhaust gas temperature with respect to the exhaust gas dew point. __________________________________________________________________ By expanding the spreadsheet, with mass rate as a parameter, we present the influences of increased steam mass rate on performance and on steam generator steam-air temperature differences in Figures 9.23 and 9.24, respectively. The figures indicate that steam-air mass ratios up to about 0.18 are possible with this configuration and that substantial performance benefits are the result.

367

368

9.9 Resource Recovery The disposal of municipal and industrial solid waste has become a national concern with significant energy conversion aspects. It has been estimated that 50% of sanitary landfills that are convenient or accessible to urban areas would be fully utilized in the 1990s and that satisfactory new sites will be progressively more difficult to find. This has led to the rapid growth of the resource recovery industry. Resource recovery deals with the environmentally sound disposal of municipal and industrial waste. It starts, at one extreme, with garbage landfills and incineration and extends to the ultimate conversion of components of waste to their useful constituent elements, with recovery of available energy in the process. Vigorous activity in this area is producing a variety of approaches to the problem. Many of the solutions focus on the development of a central waste disposal facility that receives and prepares waste for efficient landfill disposal. This section considers a modern facility that reduces the volume of about 1125 tons per day of solid waste by over 90% by burning, to produce steam for industrial use and to generate electricity when steam is not needed.

369

Figure 9.25 shows a modern facility designed to serve the waste disposal needs of almost 500,000 Tulsa-area residents. The plant is similar in many respects to the power plant steam generators discussed earlier. Major exceptions are the modifications required to deal with the unique problems of handling municipal solid waste (MSW). The right foreground of Figure 9.25 shows where waste haulers’ trucks are weighed on outside scales and enter the facility’s tipping floor. Large noncombustible materials and other refuse that cannot be burned in the facility, such as steel bars and tires, are left outside the plant by the hauler for separate disposal. The trucks dump the MSW in or near a large refuge storage pit adjacent to the tipping floor and leave from the opposite side of the tipping floor from which they entered. Haulers are held responsible for the waste that they deliver, although overhead crane operators who feed refuse to the three boilers also attempt to sort the refuse, to help achieve uniform heat release in the furnace and to keep large, noncombustible, hazardous, and otherwise inappropriate materials from entering the furnaces. Figure 9.26 shows how refuse handled by cranes is hoisted from the holding pit and lowered through near-vertical feed chutes (4) to inclined reciprocating-grate furnaces (5). There, rugged hydraulically operated rams meter the flow of refuse as they push it onto the grates. The combination of underfire air from a forceddraft fan (6) and vertically reciprocating grates keeps the refuse in continuous motion. Overfire air assures 98% burnup of combustible materials. Supply air for the forced-draft fans is drawn from the tipping-floor enclosure (1), to retain odors within the plant. At the end of the grate, a variable-speed ash discharge roller controls the rate of discharge and hence the depth of the ash bed at the end the grate. The ash falls

370 from the roller into a water-bath ash discharger that cools the residue. The water also seals the bottom of the furnace, which is slightly below atmospheric pressure. The cooling ash residue is pushed by a ram out of the water bath, up an inclined surface, and into a compartment where water is allowed to drain off and evaporate from the residue for fifteen minutes. The ash residue then falls to a conveyor, where it is transported to the ash house. There, salvageable metals are separated and recycled. The remaining ash is then trucked to a sanitary landfill. The plant achieves about a 90% reduction in refuse volume delivered to the landfill. In the furnaces, the combustion gas from the burning MSW on the grates heats the water in the welded-membrane water walls of the furnaces and the various steam generator tube banks (8-12). Each of the steam generators produces 88,500 lbm/hr of steam at 680 psia and 700°F. The entire steam production passes in a 12inch underground steam line to a refinery about a mile away, for process use. In figure 9.25, the refinery process plant utilizing the steam is located just beyond the tank farm in the upper left. A steam turbine and generator set rated at 16 MW is available for electrical generation as an alternative to refinery use of steam production. The electricity may be used on site or sold to the local electric utility. After leaving the economizer (12), combustion gas passes into an electrostatic precipitator (14), where most of the fly ash remaining from the numerous passes through the boiler is collected. The precipitators have automatic rapping systems that free the collected particles, allowing them to drop into flyash hoppers to be transported to the residue conveyor. Induced-draft fans (16) transport the cleaned combustion gas from the precipitators to the stacks. Tulsa’s Walter B. Hall Resource Recovery Facility, described here, is an environmentally sound example of an increasing number of facilities operating or under construction. These facilities typically are externally neat and are suited for operation in industrial and some commercial locations. Massive reductions in waste volumes are achieved in these facilities, with the possibility of generating steam for process use, district heating, and steam turbine generation of electricity. In resource recovery facilities of differing design, fluidized bed combustors might be employed, and the released refuse heat might instead be used in connection with closed-cycle gas turbines or other heat-driven devices. 9.10 Polytropic Efficiency To this point the performance of turbomachinery has been represented by isentropic efficiencies. In comparisons of turbomachines with differing pressure ratios, the use of the isentropic efficiency gives an undeserved advantage to some machines over others with different pressure ratios. Another approach to efficiency, called the small-stage efficiency or polytropic efficiency, is considered here as an alternative and, under certain circumstances, a more consistent way of representing the quality of turbine and compressor performance.

371

Axial compressors and turbines consist of alternating rows of stationary, or stator, blades and rotating, or rotor, blades, with the rotor rows firmly attached to a rotating shaft. In a turbine, the stationary blades act as nozzles to increase flow velocity, and the rotor blades downstream turn and decelerate (change the momentum of) the flow. The reaction to the momentum change is a force with a large component in the direction of blade rotation. This blade force delivers torque and power about the rotor axis. The combination of the stator row and a rotor row is called a stage. Instead of thinking in terms of efficiency for the entire machine, we focus our attention on the efficiency of a single turbine stage. The efficiency of a stage may be defined in terms analogous to the definition of the isentropic efficiency. Consider the T-s diagram of Figure 9.27, which shows the temperature drop of a calorically perfect gas in a stage of a multistage turbine. We assume here that all stages have identical pressure ratios. In the notation of the figure, the turbine isentropic efficiency is given by (T1 – T2)/(T1 – T2i), and by analogy the stage isentropic efficiency is *T/ *Ti. The expansion process in the turbine may be thought of as a stairstep sequence of expansions through individual small stages, each having its own efficiency. A few such steps are indicated in the figure. It should be observed that, in each successive stage, the isentropic temperature drop *Ti moves to the right on the diagram and therefore is larger than the corresponding drop between the same pressure levels on the expansion line from 1 to 2i; i.e., *Ti > *Ts. Thus the irreversibility of the expansions through earlier stages results in a sum of stage isentropic temperature drops greater than the overall isentropic temperature drop and hence greater work-producing capability for successive stages.

372 Let us now imagine the turbine as comprised of an infinite number of stages with infinitesimal pressure drops of equal efficiency. Thus the *Ts become dTs and the small-stage efficiency becomes

s = dT/dTi

[dl]

(9.7)

The isentropic relation for a calorically perfect gas, Equation (1.19), may be written in the form T/p (k – 1)/k = a constant. Using differentiation by parts yields dTi – [(k – 1)/k]Tp – 1dp = 0 which combined with Equation (9.7), gives dT/T = [s (k – 1)/k] p – 1dp

[dl]

(9.8a)

If it is assumed that the stage efficiency is constant, integration of Equation (9.8a) between the turbine inlet and exit states gives T1/T2 = (p1/p2)s (k– 1)/k

[dl]

(9.8b)

Thus the requirement that all stages have equal efficiency yields a temperature pressure relationship of the same form as Equation (1.19), except for the exponent. Such relationships, of the form T1/T2 = (p1/p2) (n – 1)/n are called polytropic. For example, the isentropic Equation (1.19) is polytropic with n = k. Because the use of constant small-stage efficiency yields a pressure-temperature relation of polytropic form, s is often called the polytropic efficiency. The exponent for the turbine expansion with constant stage efficiency is then given by (n – 1)/n = s(k –1)/k

[dl]

(9.9)

If the polytropic efficiency is unity, n becomes k, and Equations (9.8) become the usual isentropic relation. Values of s less than 1 reduce the turbine temperature ratio, T1/T2, for a given turbine pressure ratio, p1/p2, below the isentropic value in qualitatively the same way that decreasing turbine isentropic efficiency does. A similar statement applies to turbine work. The turbine isentropic efficiency, t, can be expressed in terms of the polytropic efficiency by substitution in the isentropic efficiency definition:

t = (T2/T1 – 1)/(T2i /T1 – 1) = (1/r s(k –1)/k – 1)/ (1/r (k –1)/k – 1)

[dl]

(9.10)

373

where r is the turbine pressure ratio, p1/p2. If we define a compressor polytropic efficiency as a small-stage isentropic efficiency, a similar analysis yields relations analogous to Equations (9.8) – (9.10): T2/T1 = ( p2/p1) (k– 1)/(ks)

[dl]

(9.11)

(n – 1)/n = (k –1)/(ks)

[dl]

(9.12)

c = (r(k –1)/k – 1)/ (r (k –1)/(ks) – 1) [dl]

(9.13)

where c and r are the compressor isentropic efficiency and pressure ratio, respectively. Figure 9.28 shows isentropic efficiencies for both compressors and turbines as functions of pressure ratio for two values of polytropic efficiencies. Applying L’Hospital’s rule to Equations (9.10) and (9.13), we can show that the isentropic efficiency of both turbines and compressors approaches the polytropic efficiency in the limit as the pressure ratio approaches 1. This fact is evident in the figure, and it is apparent that as pressure ratio increases, isentropic efficiencies increase for turbines and decrease for compressors for fixed values of the polytropic efficiencies.

374 The polytropic efficiencies may be regarded as measures of the internal quality of turbomachines, that is, superior internal blade passage design. For example, for two turbines having the same internal aerothermodynamic quality, Figure 9.28 indicates that a machine with a high pressure ratio will have a higher overall isentropic efficiency than one with a low pressure ratio. As another example, the compressor curves imply that, in a comparison of two compressors having the same isentropic efficiency, the one with the higher pressure ratio has a superior aerothermodynamic quality. This suggests that parametric studies involving varying compressor pressure ratio should use a constant value of polytropic efficiency rather than constant isentropic efficiency to represent comparable compressor quality. 9.11 Turbofan Engines The turbofan engine, ducted fan, or fanjet, discussed briefly in Chapter 5, is the dominant gas turbine engine in commercial aircraft and is extensively employed in military aircraft is well. Its primary feature is a large fan that accelerates a large mass of unheated air in an annular duct surrounding the central core engine, as in Figures 9.29 and 9.30, which show, respectively, a cutaway diagram and a photograph of the General Electric CF6-80C2 high-bypass-ratio engine. The large fan diameter produces a large jet exhaust consisting of a cylindrical wake of hot combustion gas surrounded by an annular flow of slower-moving warm air. The bypass ratio, B, is the ratio of the mass flow rate through the outer cooler duct, mc, to the flow rate of the hot core engine, mh: B = mc/mh

[dl]

(9.14)

Bypass ratios range from 0 for the pure turbojet engine studied in Chapter 5 to values in the neighborhood of 10. The bypass ratio is a design parameter that is primarily determined by the mission of the aircraft. High-bypass-ratio engines are desirable for long-range commercial aircraft because of their excellent fuel economy. The CF6-80C2 engine has a bypass ratio of 5.05 and a total airflow of 1769 lbm/s (802 kg/s). The bypass air may have its own nozzle, separate from the core engine as in the CF6 engine, or the core and bypass flows may be mixed in a specially designed nozzle. The mixing nozzle helps to reduce jet noise by transferring momentum from the fast-moving core gas to the slower-moving bypass air, thereby reducing the wake shear noise source. The mixing process, however, involves a thrust-loss penalty.

375

376

Figure 9.31 presents some of the nomenclature and notation that will be used in discussing the turbofan engine. For the configuration shown in the figure, the fan pressurizes the compressor inlet air as well as delivering the power needed to accelerate the bypass air through its nozzle downstream. While other fan configurations are possible, this frequently used arrangement is the only one analyzed here. In addition to the bypass ratio, a second important design parameter is the fan pressure ratio, the ratio of the stagnation pressure downstream to that upstream of the fan: FPR = po2/po1

[dl]

(9.15)

The fan pressure ratio, together with the bypass ratio, determines the power transferred between the hot core engine and the bypass flow. For a given core engine configuration, higher bypass ratios and fan pressure ratios cause more power to be extracted from the turbine and passed to the bypass air by the fan. This produces higher bypass duct thrust. However, as more power is extracted from the core flow, the core nozzle velocity and core engine thrust are reduced. The determination of the design values of these parameters therefore involves a complex tradeoff with numerous other design factors. The combination of a compressor and a turbine joined by a shaft is sometimes referred to as a spool. High-pressure-ratio engines are frequently arranged in a twin-spool or even a three-spool configuration. In a twin-spool engine, a highpressure turbine drives a high-pressure compressor with a hollow shaft, and a lowpressure turbine delivers power to a low-pressure compressor and/or a fan by means of a shaft that turns inside of the high-pressure-spool shaft. The turbofan configuration in Figure 9.31 has the low-pressure turbine driving both the fan and

377 the low-pressure compressor, as does the PW 4000 fanjet shown in Figure 5.24. In another arrangement, the low pressure turbine powers the fan only, and the entire compressor is driven by the high-pressure turbine. Regardless of engine layout, the objective of the fanjet is to accelerate a large mass of air and thereby increase the propulsive efficiency of the engine. Let us now consider the analysis of the turbofan configuration shown in Figure 9.31. The following parameters are assumed to be specified:

+ Ambient conditions at a specified altitude and flight speed or Mach number, pa, ta, ca, or Ma

+ Inlet stagnation-pressure recovery, IPR + High-pressure-turbine inlet stagnation temperature, To5 + Fan pressure ratio, FPR = po2/po1 + Bypass ratio, B + Low-pressure-compressor pressure ratio, LPCPR = po3/po2 + High-pressure-compressor pressure ratio, HPCPR = po4/po3 + Low- and high-pressure compressor efficiencies, LPC, HPC + Low- and high-pressure turbine efficiencies, LPT, HPT + Fan efficiency, F (defined analogous to compressor efficiency) The inlet is assumed adiabatic, the nozzles are assumed to be isentropic, and all mechanical efficencies are taken to be unity. The calculation procedure parallels that of the turbojet analysis of Chapter 5. Free-stream stagnation conditions are determined from the ambient conditions and flight speed or Mach number. For an adiabatic inlet, the stagnation temperature at the fan face, To1, is the same as the free-stream value, Toa, and the fan face stagnation pressure, po1, is given by the product of the inlet stagnation-pressure recovery and the free-stream stagnation pressure: po1 = IPR(poa. The conditions immediately downstream of the fan, assumed to be the same for both hot and cold paths, are given by po2 = po1(FPR

[lbf /in2 | kPa]

(9.16)

To2 = To1 + To1[FPR(k – 1)/k – 1]/F

[R | K]

(9.17)

and

378 The pressurized flow in the bypass duct then accelerates from po2 and To2 through the convergent nozzle to a high velocity c9 with static pressure p9, contributing m9(c9 – ca) + A9( p9 – pa) to the engine thrust. Here, p9 = pa if the nozzle pressure ratio, po2/p9, is less than or equal to the critical pressure ratio; otherwise it is equal to the critical pressure ratio. The stagnation conditions downstream of both low- and high-pressure compressors are determined in the same way as for the fan. For instance, for the low-pressure compressor, po3 = po2(LPCPR

[lbf /in2 | kPa]

(9.18)

To3 = To2 + To2[LPCPR(k – 1)/k – 1]/LPC

[R | K]

(9.19)

po4 = po3(HPCPR

[lbf /in2 | kPa]

(9.20)

To4 = To3 + To3[HPCPR(k – 1)/k – 1]/HPC

[R | K]

(9.21)

and

Similarly, for the high-pressure compressor,

and

Neglecting combustor pressure losses, the high-pressure-turbine inlet pressure is po5 = po4 and the known turbine inlet temperature is To5. Recognizing that, apart from assumed-small frictional losses, the power delivered by the high-pressure turbine is delivered to the high-pressure compressor; the steady-flow form of the First Law of Thermodynamics applied to the high-pressure spool yields the stagnation temperature upstream of the low-pressure turbine: m8Cpa(To4 – To3 ) = m8Cpg(To5 – To6) To6 = To5 – (Cpa /Cpg)(To4 – To3 )

[R | K]

(9.22)

The turbine isentropic efficiency definition then gives the isentropic discharge temperature, To6s, which in turn yields the high-pressure-turbine pressure ratio. A similar procedure for the low-pressure turbine spool results in the energy rate balance: (m9 + m8)Cpa(To2 – To1) + m8Cpa(To3 – To2) = m8Cpg(To6 – To7)

[Btu/hr | kW]

which, after dividing by m8Cpa and using the bypass ratio equation, yields (B + 1)(To2 – To1) + (To3 – To2) = (Cpg / Cpa)(To6 – To7)

[R | K]

379 This equation may be solved for the nozzle entrance stagnation temperature: To7 = To6 – (Cpa /Cpg)[(B + 1)(To2 – To1) + (To3 – To2 )]

[R | K]

(9.23)

The turbine pressure ratio follows from the application of the isentropic turbine efficiency definition and the isentropic temperature-pressure relation, as in the earlier high-pressure-turbine analysis. Alternatively, polytropic efficiencies may be used in the solutions. This is done for the fan and compressors by replacing Equations (9.16) to (9.21) by their polytropic equivalents: po2 = po1(FPR

[lbf /in2 | kPa]

(9.24)

To2 = To1 + To1[FPR(n – 1)/n – 1]

[R | K]

(9.25)

po3 = po2(LPCPR

[lbf /in2 | kPa]

(9.26)

To3 = To2 + To2[LPCPR(n – 1)/n – 1]

[R | K]

(9.27)

po4 = po3(HPCPR

[lbf /in2 | kPa]

(9.28)

To4 = To3 + To3[HPCPR(n – 1)/n – 1]

[R | K]

(9.29)

where (n – 1)/n is given by Equation (9.12) for the appropriate compressor poltytropic efficiency. Equations (9.22) and (9.23) are still applicable for the analysis of the turbines. However, polytropic equations, using exponents given by Equation (9.9): n /(n – 1) = k /(k – 1) s where  s is the appropriate turbine polytropic efficiency, are required to determine the turbine pressure ratios, namely po6/po5 = (To6/To5)n/(n – 1)

[dl]

(9.30)

po7/po6 = (To7/To6)n/(n – 1)

[dl]

(9.31)

and

Once po7 and To7 are known, the core nozzle may be treated in the same way as the turbojet nozzle in Chapter 5. The engine thrust is then the sum of the thrusts produced by the core and bypass flows.

380 EXAMPLE 9.3

A turbofan engine has a fan and a low-pressure compressor driven by the lowpressure turbine, as shown in Figure 9.31. It operates at a flight velocity of 200 m/s at 12,000 meters altitude, where the ambient temperature and pressure are 216.65K and 0.1933 bar, respectively. The overall pressure ratio the engine is 19, and the fan and low-pressure-compressor pressure ratios are 1.65 in 2.5, respectively. The fan, compressors, and turbines have 90% polytropic efficiencies. Assume an isentropic inlet and separate isentropic convergent nozzles, a fuel heating value of 43,000 kJ/kg, and a combustor total pressure loss of 6.5%. Determine the core engine and bypass duct exit velocities, the engine thrust specific fuel consumption, thrust, and specific thrust for an engine with an inlet air flow of 100 kg/s and a bypass ratio of 3.0. Determine whether the nozzles are choked. Solution

Table 9.6 tabulates, in spreadsheet format, the design data for the example and systematically calculates the engine parameters. After computing the fan and LPturbine exit total pressures, the nozzles may be checked for choking at the throats. The applied pressure ratios are compared with the critical pressure ratios for cold bypass air (k = 1.4) and for hot gas (k = 4/3), respectively. Branching in the computation required by the presence or absence of choking is easily handled by the @IF function of popular spreadsheets, which allows a conditional selection between specified alternatives, as discussed previously in connection with Example 5.6. TABLE 9.6

Spreadsheet Solution to Example 9.3

AEROTHERMODYNAMIC ANALYSIS OF A FANJET Ca Ta Pa Fuel HV Cpa Cpa/Cpg OPR FPR LPCPR HPCPR B mair cpeta (n-1)/n l comp

200 m/s 216.65 K 0.1933 Bar 43000 kJ/kg 1.005 kJ/kg 0.8754 19.0000 1.6500 2.5000 4.6100 3.0000 100 kg/s 0.9 0.3175

Flight velocity Ambient temperature Ambient pressure Heating value Air heat capacity Air/gas heat capacity ratio Overall pressure ratio Fan pressure ratio Low pressure compressor pressure ratio HPCPR = OPR/[(FPR)(LPCPR)] Bypass Ratio Inlet total mass flow rate Compressor and fan polytropic efficiencies (n-1)/n = (k-1)/[k+cpeta] (for fan & comp.)

381 tpeta n/(n-1) l turbine (dp/p) l combust To5 To5 To1=Toa Poa = Po1 Po2 Po3 Po4 Po5 To2 To3 To4 To6 To7 Po6 Po7 Po7/Pc8 Po7/Pa Po2/Pc9 Po2/Pa T8

0.9 4.4400 0.065 1027 C 1300 K 236.550 K 0.263 Bar 0.434 Bar 1.085 Bar 4.995 Bar 4.671 Bar 277.31 K 370.93 K 602.39 K 1097.38 K 872.68 K 2.199 Bar 0.795 Bar 1.8530 4.1100 1.8930 2.2440 748.02 K

C8

535.02 m/s

P8 Rho8 A8/m8 T9

0.429 Bar 0.200 kg/m^3 0.0094 m^2-s/kg 231.09 K

C9

304.72 m/s

P9 rho9 A9/m9 Specific Thrust, ST

0.229 Bar 0.346 kg/m^3 0.009 m^2-s/kg 242.9 N-s/kg

Thrust f/a mfuel TSFC TSFC' ST | core m

24294.4 N 0.0186 0.466 kg/s 0.000019 kg/N-s 0.069 kg/N-hr 971.8 N-s/kg

Turbine polytropic efficiency Turbine polytropic factor, k/[(k-1)+tpeta] Combustor fractional pressure drop High pressure turbine inlet temperature To5 (K) = To5 (C) + 273 Toa=Ta+Ca^2/(2+Cpa*1000) Po1 = Poa = Pa+(Toa/Ta)^3.5 Po2=FPR+Po1 Po3=(LPCPR)Po2 Po4=Po3(HPCPR) Po5=Po4+[1 - (dp/p)] | combust To2=To1+FPR^[(n-1)/n] I comp To3=To2+(LPCPR)^[(n-1)/n] | comp To4=To3+HPCPR^[(n-1)/n] | comp To6=To5-(Cpa/Cpg)+(To4-To3) To7=To6-(Cpa/Cpg)[(B+1)(To2-To1)+(To3-To2)] Po6=Po5+(To6/To5)^[n/(n-1)] | turbine Po7=Po6+(To7/To6)^[n/{n-1)] | turbine Crit. press. ratio: Po7/Pc8=[(4/3+1)/2]^4 Core flow is choked Crit. Press. ratio: Po2/Pc9=[(7/5+1)/2]^3.5 Bypass flow is choked IF (Po7/Pc8) < (Po7/Pa) THEN T8=2+To7/(4/3+1) ELSE T8=To7/(Po7/Pa)^0.25 IF (Po7/Pc8)<(Po7/Pa) THEN C8=(287+T8+4/3)^0.5 ELSE C8 = [(To7-T8)+2000+Cpg)]^0.5 IF Pc8>Pa THEN P8=Po7/(Po7/Pc8) ELSE P8=Pa Rho8=100+P8/(0.287*T8) A8/m8 = 1 / (C8+Rho8) IF Pc9>Pa THEN T9=2To2/(7/5+1) ELSE T9=To2/(Po2/Pa)^(1/3.5) IF Pc9>Pa THEN C9=(287+T8+7/5)^0.5 ELSE C=[(To2-T9)+2000+Cpa]^0.5 IF Pc9>Pa THEN P9=Po2/(Po2/Pc9) ELSE P9=Pa Rho9=100+P9/(0.287+T9) A9/m9=1 / (C9+Rho9) ST = C8+C9+B-Ca+(B+1)+[(P8-Pa)+(A8/m8) +(P9-Pa)+(A9/m9)+B]+10^5/(B+1) Total engine thrust= mair(specific thrust) f/a=Cpg(To5-To4)/HV Fuel flow rate=mair+(f/a)/(B+1) TSFC=(f/a)/[spec. thrust)(B+1)] TSFC'=3600+TSFC ST based on m8 = (B+1)+ST

The specific thrust is presented in two forms, one based on the total mass flow rate to the engine and the other based on the mass flow rate to the core engine. The former, ST, allows easy comparison with engines of comparable frontal area, while the latter, ST | core m, is useful in showing the thrust increase due to adding

382 a fan to a given core engine. Both definitions reduce to the common turbojet definition for B = 0, but deviate progressively with increasing bypass ratio. __________________________________________________________________ It is of interest to compare the performance of a turbofan and twin-spool turbojet in flight at the same altitude and velocity based on our computations. This is easily accomplished with the spreadsheet of Table 9.6 by setting FPR = 1.0 and B = 0 for the turbojet engine. Table 9.7 shows some of the parameters resulting from this comparison. Table 9.7

Turbofan–Turbojet Comparison Turbofan

Turbojet

Core jet velocity, m/s

535.02

566.88

Bypass jet velocity, m/s

304.72

---------

Thrust, N

24,294.4

67,745

Specific thrust based on total mass, N-s/kg

242.9

677.5

Specific thrust based on core mass, N-s/kg

971.8

677.5

Fuel mass flow rate, kg/s

0.466

1.862

TSFC, kg/N-hr

0.069

0.099

It is evident that the extraction of power to drive the fan reduces the core jet velocity. The lower jet velocities substantially reduce the turbofan engine thrust for the same size inlet and total engine mass flow rate. Thus the high-bypass-ratio turbofan is less likely to be used for military applications requiring high speed and therefore high thrust per unit of frontal area. This is reflected in the specific thrust based on total engine mass flow. On the other hand, the specific thrust based on core mass flow shows that a significant increase in thrust can be achieved by adding a fan to an existing turbojet design. The major advantage of the fanjet is shown in the fuel flow rate and TSFC comparisons, where the superior fuel economy of the fanjet appears. The success of the turbofan or ducted-fan engine has made it clear that further advances in jet engine fuel economy are possible with higher bypass ratios. The large-diameter cowlings necessary for drastic increases in turbofan bypass ratio, however, appear impractical. Still, advances in propeller technology now make flight at high subsonic Mach numbers possible with gas-turbine-driven unducted fans.

383

Figures 9.32 and 9.33 show a UDF®* engine featuring contrarotating fans installed in an engine nacelle. A twin-spool gas generator upstream of the fans supplies combustion gas to the cantilevered fan section. In this design the forward fan blades are coupled directly to the outer fan-turbine blades, as shown in Figure 9.34, and rotate clockwise. The free-wheeling aft fan section, which includes the _________ * UDF® is a registered trademark of General Electric Co., U.S.A..

384 inner fan-turbine blades, rear fairing, and tailcone assembly, is cantilevered at the rear of the engine and rotates counterclockwise. With unducted-fan engines, very high bypass ratios are possible. The UDF engine, with an overall pressure ratio of approximately 42 at top of climb, has demonstrated substantial SFC reductions over modern turbo fans, approximately 20% lower than the best of the current turbofan engines. In-cabin noise, originally expected to be a problem with unducted fans, was very low–equal to or less than that of today’s turbofan-powered aircraft (ref. 55). Despite the attractiveness of the unducted fan concept, it appears that in the next decade large transport aircraft will use ducted fans that incorporate more modest increases in bypass ratio together with increases in overall pressure ratio for improved performance. Reference 63, for example, anticipated the family of 75,000 to 95,000 pounds-thrust turbofans coming into use around 1995. They were expected to have a bypass ratio of 9 and overall pressure ratio about 45 to give a 9% improvement in SFC over then-existing engines. The reference anticipated that the engines would have a low speed, low pressure ratio fan for low noise. The fans were expected to be made of composite materials to save about 25% weight over competing metal fans. Bibliography and References 1. Curran, Paul F., "Clean Power at Cool Water," Mechanical Engineering, Vol. 109, No. 8 (August 1987): 68-71. 2. Grover, Ralph W., et al.; "Preliminary Environmental Monitoring Results for The Cool Water Coal Gasification Program." Proceedings of the EPRI Sixth Annual Conference on Coal Gasification, Palo Alto, Calif., October 15–16, 1986. 3. Watts, Donald H., Dinkel, Paul W., and McDaniel, John E., "Cool Water IGCC Performance to Date and its Future in the Electric Utility Industry," U.S. Utility Symposium, Palm Springs, Calif., October 1987. 4. Gluckman, Michael J., et al., "A Reconciliation of the Cool Water IGCC Plant Performance and Cost Data with Equivalent Projections for a Mature 360-MW Commercial IGCC Facility." Cool Water Gasification Program, October 1987. 5. Fahrion, M. E., "Materials Development Experience at Cool Water Coal Gasification Plant." Cool Water Coal Gasification Program, October 1987. 6. Anon., "Boilers and Auxiliary Equipment," Power, June 1988: 91–229. 7. Ehrlich, Shelton, "Fluidized-Combustion Boiler Technology–1988 Update." Missouri Valley Electric Association Engineering Conference, Kansas City, Mo., March 1988. 8. Ehrlich, Shelton, "Fluidized Combustion: Is It Achieving Its Promise?" Electric Power Research Institute, Palo Alto, Calif.

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387 41. Brunner, Calvin R., Handbook of Hazardous Waste Incineration. Blue Ridge Summit, Pa: Tab Books, 1989. 42. Spiewak, Scott A., Cogeneration and Small Power Production Manual, 3rd ed. Lilburn, Ga.: Fairmont Press, 1991. 43. Weston, Kenneth C., “Turbofan Engine Analysis and Optimization Using Spreadsheets,” Proceedings of the ASME Computers in Engineering Conference, August 1, 1989. 44. Cohen, H., Rogers, G. F. C., and Saravanamuttoo, H. I. H., Gas Turbine Theory, 3rd ed. New York: Longman, 1987. 45. Wilson, David Gordon, The Design of High-Efficiency Turbomachinery and Gas Turbines. Cambridge, Mass.: MIT Press, 1984. 46. Chenoweth, C. V., Schreiber, H., and Dolbec, A. C., “20 Years of Operation at Horseshoe Lake Station–Oklahoma Gas and Electric Company,” Procedings of the American Power Conference, Vol. 96, 1984. 47. Roll, Mark W., “The Dow Syngas Project,” Turbomachinery International, Vol. 32, No. 1, January-February 1991: 28-32. 48. Larson, E. D., and Williams, R. H., “Biomass-Gasifier Steam-Injected Gas Turbine Cogeneration,” ASME Journal of Engineering for Gas Turbines and Power, Vol 112, April 1990: 157-163. 49. Zaugg, P., and Stys, Z. S., “Air-Storage Power Plants with Special Consideration of USA Conditions,” Brown Boveri Review, Vol. 67 (December 1990): 723-733. 50. Kehlhofer, Rolf, Combined-Cycle Gas and Steam Turbine Power Plants. Lilburn, Ga: Fairview Press, 1991. 51. United Technologies Turbo Power and Marine Systems, Inc. FT8 Gas Turbine Brochure. 52. Goldstein, Gina, “Rechanneling the Waste Stream,” Mechanical Engineering, Vol. 111 (August 1989): 44-50. 53. Butler, Arthur J., and Hufnagel, Karl R., “Skagit County Resource Recovery Facility Design of a 2500-kW Waste-to-Energy Plant,” ASME ‘89–JPGC/Pwr-12, October 1989. 54. Douglas, John, “Beyond Steam,” EPRI Journal, December 1990: 4-11. 55., G.E. Aircraft Engines source, personal communication. 56. Riley, Shaun T., (manager of LM 5000 and Industrial programs, G. E. Marine and Industrial Engines & Service Div.), personal communication.

388 57. Riley, Shaun T., “STIG Increases Output from Aeroderivative Gas Turbines,” Modern Power Systems, June 1990: 41-47. 58. Stewart, Norman, “Shell Coal Gasification,” EPRI Journal, June 1989: 38-41. 59. Sundstrum, David G., “Status of the Dow Syngas Project: Three Years of Experience at the World’s Largest Single Train Coal Gasification Plant,” Proceedings of the American Power Conference, Vol. 52, 1990, pp. 512-515. 60. Anon., “Gasifier Demo Heralds New Era for Gas Turbines,” Power, October 1989: S-24–S-28. 61. Anon., “Cogen System Balances Steam Cycle Demand,” Power, October 1989: S-42–S-44. 62. Ullman, Ronald R., and Duffy, Thomas J., “Integrated Material and Energy Recovery from Municipal Solid Waste,” Proceedings of the American Power Conference, Vol. 52, 1990, pp. 570-577. 63. Smith Jeffrey J., “ The GE90 ... When is Big’ Big Enough?” ASME Turbo Expo–Land, Sea, and Air, Orlando, Fla., June 6, 1991. 64. Lamarre, Leslie, “Alabama Cooperative Generates Power from Air,” EPRI Journal, December 1991: 12-19. 65. United States Department of Energy website, www.fe.doe.gov. 66. Schroeter, Jeffrey W., “The Merchant Revolution,” Mechanical Engineering Power, May, 2000. 67. Lane, Jon, “More Power, Fewer Players,” Mechanical Engineering Power, May, 2000. EXERCISES

9.1

A 10-MW gas turbine operating at the conditions in Example 5.1 exhausts through a process heat exchanger, with the combustion gas leaving at 200°F. What is the rate of heat transfer to the process? If the heat transfer is to liquid water, what flow rate of water can be increased in temperature from 50°F to 160°F in the heat exchanger? How many homes with an average heating demand of 60,000 Btu/hr could be serviced by the gas turbine in a district heating application where there is a 20% energy loss in the system distributing heat to the customers? What is the unweighted system energy utilization factor?

9.2

Design the combustion gas heat exchanger required in Exercise 9.1. Indicate the type selected, the required surface area, the geometric configuration, and overall size.

389 9.3

The oxygen remaining in the exhaust gas of a 5-MW gas turbine supplies the oxidizer for a process heater that burns methane completely. The gas turbine operates at the conditions of Example 5.1. What is the nominal heat transfer rating if the heater is designed for an exit temperature of 300°F? What is the rated of fuel consumption and the unweighted energy utilization factor of this combined heat and power system? What would the maximum heater output be if the gas turbine were shut down and atmospheric air were supplied to the heater with the same supplemental fuel firing rate?

9.4

Compare the unweighted energy utilization factors for the simple-cycle and regenerative gas turbines of Examples 5.1 and 5.2, assuming their exhausts serve unfired process heat exchangers with 250°F exit gas temperatures. Compare the energy utilization factors, assuming the useful heat in each is weighted by the cycle’s thermal efficiency.

9.5

The exhaust of a 2-MW gas turbine, operating at the conditions of Example 5.1, transfers heat to an absorption chiller with a COP of 0.78. The exhaust gas leaves the chiller at a temperature of 220°F. How many tons of refrigeration can be produced by the chiller? What is the unweighted EUF based on shaft power and chiller rate of cooling?

9.6

A 20-MW simple-cycle gas turbine operates with compressor inlet conditions of 101 kPa and 15°C, a turbine inlet temperature of 1200°C, and a compressor pressure ratio of 12. The compressor and turbine isentropic efficiencies are 84% and 88%, respectively. The turbine exhaust flows through a process heat exchanger and exits to the atmosphere at 110°C. Determine the gas turbine cycle state properties, the thermal efficiency, and the work ratio, accounting for an 80-kPa pressure drop on the process heat exchanger gas turbine exhaust side. What is the rate of heat transfer to the process through the heat exchanger? If the heat transfer is to liquid water, what flow rate of water can be boiled at atmospheric pressure if water enters the heat exchanger at 30°C? How many homes with an average heating requirement of 20 kW can be heated in a district heating application where there is a 15% loss in the distribution of heat to customers? What is the unweighted system energy utilization factor?

9.7

Design the combustion gas heat exchanger required in Exercise 9.6. Indicate the type selected, the required surface area, the geometric configuration, overall size, and the estimated pressure drops. Can you improve significantly on the assumed heat exchanger pressure drop used in Example 9.6?

9.8

The exhaust of a 5-MW gas turbine supplies the oxidizer for a heater that burns methane completely. The gas turbine operates the 101 kPa and 15°C compressor inlet conditions and 1060K turbine inlet temperature. The engine has compressor and turbine isentropic efficiencies of 86% and 88%, respectively, and a compressor pressure ratio of 10. What is the nominal heat transfer rating of the heater? What is the total rate of fuel consumption and the unweighted energy utilization factor? Combustion gases leave the heater at 200°C.

390 9.9

The exhaust of a 2-MW gas turbine operating at 960°C turbine inlet temperature with a compressor pressure ratio of 9 transfers heat without loss to operate an absorption chiller with a COP of 0.83. The exhaust gas leaves the chiller at a temperature of 120°C. The compressor inlet conditions are 105 kPa and 25°C, and the compressor and turbine isentropic efficiencies are 82% and 86%, respectively. How many tons of refrigeration can be produced by the chiller?

9.10*Develop a spreadsheet like that in Table 9.1, and investigate the influence on the combined-cycle efficiency of varyingeach of the following: (a) HRSG exit temperature. (b) Steam turbine throttle temperature. (c) Gas turbine inlet temperature. (d) Compressor pressure ratio. 9.11 A closed-cycle gas turbine using air as a working fluid has an intercooler, a reheater, and a recuperator. The overall pressure ratio is 6. The low-pressure compressor has a pressure ratio of 2. Both compressors are driven by the high-pressure turbine. All turbomachines are 85% efficient. Compressor inlet temperatures are 80°F, and turbine inlet temperatures are 1500°F. Regenerator effectiveness is 75%. (a) Draw T-s and flow diagrams, and label both compatibly. (b) Identify actual temperatures, in degrees Rankine, at all stations of significance. (c) If the low-pressure compressor inlet pressure is 6 atm, what are the intercooler and reheater gas pressures? (d) Calculate the total compressor work, and compare with the work for a single compressor with 85% efficiency without intercooling. (e) What is the gas turbine net work? (f) What is the total external heat addition? (g) How much heat is available from the precooler for district heating ? (h)What is the plant thermal efficiency? (i) What is the plant energy utilization factor if 80% of the precooler heat rejection is used for district heating? (j) If the turbines deliver 100MW of power, what is the air flow rate and (k) the rate of consumption of coal, in tons per hour (12,000 Btu/lbm heating value and 90% combustion efficiency)? ______________________ * Exercise numbers with an asterisk involve computer usage. 9.12 A closed-cycle gas turbine using air as a working fluid has an intercooler, a reheater, and a recuperator. The overall pressure ratio is 6. The low-pressure compressor has a pressure ratio of 2. Both compressors are driven by the high-pressure turbine. All turbomachines are 85% efficient. Compressor inlet temperatures are 15°C, and turbine inlet temperatures are 1000°C. Regenerator effectivenesss is 75%. (a) Draw T-s and flow diagrams, and label both compatibly. (b) Identify actual temperatures, in degrees Kelvin, at all stations of significance.

391 (c) If the low-pressure compressor inlet pressure is 6 atm, what are the intercooler and reheater gas pressures? (d) Calculate the total compressor work, and compare with the work for a single compressor with 85% efficiency without intercooling. (e) What is the gas turbine net work? (f) What is the total external heat addition? (g) How much heat is available from the precooler for district heating? (h) What is the plant thermal efficiency? (i) What is the plant energy utilization factor if 80% of the precooler heat rejection is used for district heating? ( j) If the turbine delivers 100 MW of power, what is the air flow rate and (k) the rate of consumption of coal in kilograms per second (25,000 kJ/kg heating value and 90% combustion efficiency)? 9.13 Starting with the selected combined-cycle design for Example 9.1, consider a modification to the heat-recovery steam generator that allows saturated liquid water at state 12 to be extracted for industrial process use. Assume that the water is returned to the condenser as a saturated liquid at the condensing temperature. Assume that process use dictates the mass flow rate extracted, with the balance of the water going as steam to the steam turbine. Calculate and tabulate the heat supplied per unit of gas turbine mass flow as a function of the process liquid mass fraction of the total water flow entering the steam generator. Also calculate and plot the energy utilization factor as a function of the process mass fraction. If the gas turbine power output is 6 MW, what is the maximum saturated-liquid-water process heat transfer rate? 9.14* Starting with the selected combined-cycle design for Example 9.1, consider superheated steam at state 8 to be extracted for industrial process use. Assume that the water is returned to the condenser as a saturated liquid at the condensing temperature. Assume that process use dictates the mass flow rate extracted, with the balance of the steam created going to the steam turbine. Calculate and tabulate the process heat supplied per unit of gas turbine mass flow as a function of the process steam mass fraction of the water flow through the steam generator. Also calculate and plot the energy utilization factor as a function of the process mass fraction. If the gas turbine power output is 6 MW, what is the maximum superheatedsteam process heat transfer rate? 9.15 Extend the analysis of Table 9.1 to include supplemental firing of the HRSG, to provide an inlet gas temperature of 1500°F and a steam turbine throttle temperature of 1000°F. Determine the influence of boiling temperature on the pinchpoint temperature difference, and on the net work per pound of gas turbine flow; and compare the combined-cycle thermal efficiency with the efficiencies of the individual cycles. Discuss the results of the analysis. 9.16 Extend the analysis of Table 9.1, to consider a modification of the heatrecovery steam generator that allows saturated water vapor at state 13 to be extracted for industrial process use. Assume that the water is returned to the condenser as a saturated liquid at the condensing temperature. Assume that process use of the steam has priority, with the balance of the steam going to

392 the steam turbine. Calculate and tabulate the process heat supplied per unit of gas turbine mass flow as a function of the process water mass fraction of the water flow through the steam generator. Also calculate and plot the energy utilization factor as a function of the process mass fraction. It the total power output of the combined cycle is 6 MW, what is the maximum process heat transfer rate? 9.17 Extend Exercise 9.15 to consider a modification of the heat-recovery steam generator that allows superheated steam at state 8 to be extracted for industrial process use. Assume that the water is returned to the condenser as a saturated liquid at the condensing temperature. Assume that process use of the steam has priority, with the balance of the steam going to the steam turbine. Calculate and tabulate the process heat supplied per unit of gas turbine mass flow as a function of the process steam mass fraction of the water flow through the steam generator for a boiling temperature of 600°F. Also calculate and plot the energy utilization factor as a function of process mass fraction. If the total power output of the combined cycle is 20 MW with no process heat, what is the maximum process heat transfer rate? 9.18 Design an open-cycle regenerative gas turbine cogeneration system in which a fraction of the turbine exhaust gas can bypass the heat exchanger for process use. Prepare a report stating your design criteria, defining your analysis methodology, and presenting performance data for the nominal design condition that you selected. 9.19 Perform the design required in Exercise 9.18, and include consideration of performance for a range of off-design process heat transfer requirements that are lower than the design value. 9.20 A 20-MW electric motor in a simple compressed-air-storage plant drives a compressor with a pressure ratio of 10 and an efficiency of 85% for six hours nightly. What power output can be obtained with a turbine inlet temperature of 1600°F if the plant operates for four hours during the day at constant power output? The turbine efficiency is 90%. What is the air-fuel ratio if methane is the fuel used? What is the net generation efficiency, considering only the fuel consumption of the turbine? Assume a daily cycle, that cavern pressure changes are negligible, and that the heat of compression is dissipated before generation begins. 9.21* An ideal steam turbine operates with 1000°F, 2000-psia throttle, and 1-psia condenser and produces 15 MW without extraction. When steam is extracted for process use at 500 psia, after use it is condensed to a saturated liquid at that pressure and throttled to the condenser. Tabulate and plot the process heating rate and the EUF as a function of extraction mass fraction. 9.22* A 25 MW steam turbine operates in 1000°F and 2000 psia with an efficiency of 87%. Eighty percent of the condenser heat transfer is used for an industrial process. Tabulate and plot the process heating rate and the EUF as a function of condenser pressure between 1 psia and 2 atmospheres.

393 9.23* Develop the equations and an algorithm for the analysis of the steaminjected gas turbine with an unfired steam generator producing superheated steam at the combustion chamber pressure and using methane as a fuel . Use the JANAF tables for thermodynamic properties of steam in the gas turbine. State clearly the assumptions made. Write a computer programming implementing the algorithm. 9.24 Consider a compressor operating at a pressure ratio of 20 and a polytropic efficiency of 86% that compresses ambient air at 101 kPa and 15°C into a cavern. Assume that heat losses from the cavern maintain the air at 15°C and constant pressure during the filling period from midnight to 6 am daily. The compressor is driven by a 20 MW electric motor. What is the daily mass addition to the cavern? From 2 pm to 6 pm daily the same mass of air that was added to the cavern during the night is heated to 1200K and allowed to escape to the atmosphere at a constant flow rate through a turbine expander with a 90% isentropic efficiency. What is the expander power output? What is the daytime energy output? What is the fractional fuel consumption reduction if a regenerator with 80% effectiveness is added to the system? 9.25 Resolve Example 9.2 for superheated steam injection at 400°, 500°, 600°, and 700°F and the combustor pressure level. Write a brief report on your findings on the influence of temperature of injected steam on STIG performance. 9.26* Use the STIG spreadsheet shown in Table 9.5 to verify the performance calculations of Figures 9.23 and 9.24. 9.27* Investigate the influence of compressor pressure ratio variation on STIG performance for the model of Example 9.2, and prepare a memo reporting your results. 9.28* Evaluate the separate influences of steam heat capacity and added mass on the thermal efficiency, power output, and work ratio for the model of Example 9.2. 9.29 Consider a two-shaft gas turbine to be modified for steam injection. The compressor pressure ratio is 9.3, and the turbine inlet temperature is 982°C. The isentropic efficiencies of the compressor and turbines are 83% and 90%, respectively. The gas generator mechanical efficiency is 99%, and the power turbine drives an electrical generator that has a 93% efficiency. Accounting for a 4% pressure loss in the combustor and a fuel heating value of 43,000 kJ/ kg, compare the electrical power output, specific fuel consumption, thermal efficiency, and fuel-air ratios for 0.0 and 0.05 steam-air ratios. Briefly described your selection of steam system design conditions. 9.30 It has been decided that the heat-recovery steam generator for a steaminjected gas turbine must be retubed to continue running it in the steam injection mode. The expected cost of retubing is $230,000. Steam injection produces an additional 4000 MW-hr per year, adding two cents per kW-hr to

394 revenue. What is the break-even operating time to recover the cost of this maintenance operation? 9.31 Compare the isentropic efficiencies of two compressors, each having a polytropic efficiency of 87% and pressure ratios of 6 and 18. 9.32 What is the polytropic efficiency of a turbine, with a pressure ratio 30, that has the same isentropic efficiency as a turbine having a polytropic efficiency of 92% at a pressure ratio of 12? 9.33* Compare the performance of a twin-spool turbofan engine with separate fan and compressor to that of a turbojet engine, both with an overall pressure ratio of 30 and a turbine inlet temperature of 1300K, and designed for an altitude of 15,000m and a flight speed of 275 m/s. The bypass ratio is 6 and the fan pressure ratio is 1.6. Assume turbine, fan, and compressor polytropic efficiencies of 90% and a combustor pressure loss of 3% of the compressor exit total pressure. Compare specific thrusts for engines built from the same core engine. Compare, also, specific fuel consumption and jet velocities. 9.34* Build a multicase spreadsheet for the conditions of Exercise 9.33, and use it to plot specific thrust and TSFC as a function of fan pressure ratio. Use the spreadsheet to explore further the influence of varying bypass ratio for a value of fan pressure ratio suggested by your plot. Write a memo discussing briefly the results of your study. 9.35 Develop an analysis for a turbofan engine, sometimes called an aftfan engine, in which the fan is located at the same axial station and directly attached to a low pressure turbine dedicated to driving the fan. The highpressure turbine drives the compressor, and is located upstream of the fan. The compressor and fan each have their own separate inlets. 9.36 Express the work of a compressor in terms of its inlet temperature and pressure ratio using (a) the isentropic efficiency, and (b) the polytropic efficiency. Equate the relations, and solve for the compressor isentropic efficiency. Compare your result with Equation (9.13). 9.37 Express the work of a turbine in terms of its inlet temperature and pressure ratio using (a) the isentropic efficiency, and (b) the polytropic efficiency. Equate the relations, and solve for the turbine isentropic efficiency. Compare your result with Equation (9.10).

395 C H A P T E R 10 Nuclear Power Plants*

10.1 Introduction Nuclear power is universally controversial. Many would say that it is also universally needed–as an alternative or supplement to power generated by fossil fuels. The combustion of fossil fuels produces carbon dioxide, now notorious for the threat of global warming. Nuclear power plants produce neither carbon dioxide nor oxides of sulfur and nitrogen, as does the burning of fossil fuels. Thus nuclear power reduces the global production of carbon dioxide and other pollutants, and helps to alleviate many of the pervasive problems of fossil fuel supply. Petroleum is least available in regions of widest use; natural gas is, for the time being, plentiful and sought after by all; and widely abundant coal has come to be regarded as the great Satan of air pollution. Water power is important, but it offers limited possibility for growth. Solar energy, while promising, is far from being a mainstay of the world’s energy supply. Thus sources other than fossil fuels and nuclear power offer little hope to become major suppliers during our lifetimes. Nuclear power, in stasis for many years, may make a comeback. Engineers have been quietly working on new and safer designs for nuclear power plants, and the political climate may be swinging slowly back in favor of nuclear power. According to references 31 and 34, there were 434 operating nuclear plants producing 17% (350,000 megawatts) of the world’s electricity in 1998. Regardless of one’s position towards it, nuclear power is a major factor in world power production. Knowledge of nuclear power is not American, French, Indian, or British; it is virtually universal. Nuclear power plants such as those shown in Figure 10.1 are producing power in many nations around the world. Blockage of the growth of nuclear power in the United States did not prevent the development and extensive use of nuclear-generated electricity in France or Canada. Should developing countries with minimal fossil resources not use nuclear power? Should a country that has seen the terror wrought by nuclear weapons be denied the benefits of electricity from the nucleus? Can attempts to halt the growth of nuclear power stop the proliferation of nuclear weapons? These and many other issues around nuclear power, which obviously extend far beyond the bounds of engineering, are much too broad to be pursued here. Nuclear power is controversial, but it is here; it is important and likely to remain so. *Thanks to Dr. Andrew A. Dykes for his valuable inputs and comments on this chapter.

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This chapter is concerned largely with nuclear fission power reactors. Most nuclear power plants use steam cycles that differ little from those of fossil fuel plants except for the source of heat for the steam generator and for steam supply conditions. Steam turbine cycles were considered in some detail in Chapter 2. Thus this chapter will focus primarily on (a) the characteristics of nuclear reactor steam supply systems and, to a lesser extent, on (b) the climate for future nuclear development. In preparation for this study let us first review relevant aspects of atomic and nuclear structure.

397 10.2 Review of Atomic Structure Atoms and Molecules Somewhat more than a hundred elements are known and are thought to be the building blocks of everything in the universe. The atom is the basic unit of structure for each element. An important connection between the microscopic world of the atom and the macroscopic world of experience is given by Avogadro's number. A gram-mole of any element has Avogadro's number (6.023 x 1023) of atoms. The atom may be considered as consisting of a positively charged nucleus at its center and one or more negative charges around the nucleus called electrons that make the atom electrically neutral. The electron is the fundamental unit of negative charge. It may be viewed as a particle which is much smaller than the nucleus and which orbits around the nucleus as a planet orbits the sun, or it may be viewed as a diffuse electron cloud around the nucleus. Still another concept is that of a particle whose location is not known but which is more likely to be in some places than others, its position being given by a probability distribution. We will not concern ourselves with the rationales for these views. Thus atoms consist of nuclei surrounded by electrons. The sizes of atoms are conveniently measured in Angstroms (10-8 cm). The nucleus typically is of the order of 10 –5 Angstroms. Thus the volume of the atom is largely due to the size of the outer electron’s orbit or to the atom’s electron cloud. Molecules are collections of atoms held together by electromagnetic forces between the nuclei and the electrons. Atoms and molecules can exist in a variety of energy states associated with their electron distributions. These microscopic states and their macroscopic influences are dealt with theoretically in the fields of quantum mechanics and statistical thermodynamics. Molecules and atoms can interact with each other to form different molecules in ways that are controlled by their electron structures. These interactions, called chemical reactions, have little to do with the nucleus. In Chapter 3, we considered aspects of these reactions that are relevant to combustion. The magnitudes of the energy associated with these chemical changes, while of great importance in thermal engineer-ing, are so small that they have no significant influence on the nuclei of the reacting atoms. Thus the nuclei may be thought of as merely going along for the ride when a chemical reaction occurs. An electrically neutral particle, however, can penetrate an atom's electromagnetic field and approach the nucleus, where it interacts via short range but powerful nuclear forces. Then the electrical forces holding the nucleus together may be overcome, resulting in changes in the nucleus. In these cases the interatomic forces are largely irrelevant and are overpowered by nuclear events. It is these changes that are the concern of this chapter.

398 The Nucleus The nucleus, for our purposes, may be thought of as being made up of integral numbers of protons and neutrons. The proton is a particle with a positive charge of the same magnitude as that of the electron, so that pairing a proton with an electron produces exact electrical neutrality. Thus protons account for the charge of the nucleus, and a like number of electrons ensures the electrical neutrality of the atom. Compared with the electron, the proton is a massive particle, having a mass which is about 1800 times the mass of the electron. Thus the hydrogen atom, which consists of one proton in the nucleus and a single electron in orbit around the nucleus, is electronically neutral and has a mass only slightly larger than that of the nucleus. Atoms larger than the hydrogen atom have more than one proton in the nucleus and have one or more neutrons as well. A neutron, as the name suggests, is an electrically neutral particle with a mass only slightly larger than that of the proton. As components of the nucleus, protons and neutrons are called nucleons, and are thought of interchangeably with respect to mass because their masses differ so little from each other. The number of protons in an atom of an element is called the atomic number of the element. Thus hydrogen has an atomic number of 1. The atomic number of a given element is unique to that element. Thus we could identify the elements by their atomic numbers rather than by their names if we wished. Elements are ordered in the periodic table in part by their atomic numbers. The mass number of an element is the number of nucleons in an atom of that element and is therefore the sum of the number of protons and neutrons in the nucleus. Atoms of a given element that have differing mass numbers are called isotopes of the element. A given isotope of an element is sometimes designated by a notation that includes the element's chemical symbol, its mass number, and its atomic number. For example, the most common isotope of uranium is denoted as 92U238, where 92 is the atomic number of the element uranium and 238 is the sum of the number of protons and neutrons in the isotope nucleus. The isotopes are also sometimes simply identified by their name or symbol and mass number, such as U-235 or Uranium-235. Other significant examples are the isotopes of hydrogen, deuterium, 1H2, and tritium, 1H3, which have, respectively, one and two neutrons accompanying the proton. These isotopes are sometimes written 1D2 and 1T3 to reflect their commonly used names. The form of water, H2O, in which the isotope deuterium replaces hydrogen is commonly called heavy water, D2O, because of the added mass of the extra neutron in each nucleus. It will be seen later that heavy water has characteristics that are advantageous in some nuclear processes. 10.3 Nuclear Reactions Just as chemical fuels undergo chemical reactions that release energy, nuclei may also participate in energy-releasing nuclear reactions. When this happens atoms of the

399 reacting elements are converted to atoms of other elements, the sort of transmutation sought by the alchemists of the past. Mass-Energy Equivalence Two nuclear processes are known to be capable of releasing energy on a scale large enough to influence the personal and business lives of humans. The reality of both processes has been amply demonstrated by the production and detonation of the atomic and hydrogen bombs. Both processes owe their energy release to the annihilation of matter consistent with the famous Einstein formula, E = mc2, which asserts the convertibility of mass to energy. Because the speed of light is so large, the equation shows that the annihilation of a small amount of mass yields a large quantity of energy. These energy releases are usually measured in MEV, millions of electron-volts. The electron-volt, EV, is defined as the energy required for an electron to pass through a potential difference of one volt. The reactions of individual nuclei typically produce particles with energies measured in MEV. On the other hand, as indicated earlier, the most energetic of chemical reactions releases much less energy, only a few EV per molecule. As a result of our encounter with Einstein's energy-mass relation we must adjust our philosophical position on the conservation laws of mass and energy and think instead in terms of conservation of mass-energy. Mass and energy may be thought of as different forms of the same thing. Any change in the mass of an isolated system must be accompanied by a corresponding change in system energy. This in no way influences the discussions in previous chapters, because changes in mass are entirely insignificant in chemical and other nonnuclear processes. Fission and Fusion The process known as nuclear fusion occurs in nature in the stars, including our own sun. Since the Second World War, scientists have been attempting to achieve the conditions for fusion in the laboratory. Because it can use heavy water from the sea as a fuel, controlled thermonuclear fusion offers the hope of vast quantities of power for many centuries in the future. Fusion occurs when light atoms interact to form a heavier atom in reactions such as D2 + 1D2 ; 2He3 + 0n1 + 3.2 MEV.

1

Here, two deuterium atoms collide to form helium-3 and a neutron while releasing 3.2 MEV of energy. Other fusion reactions exist that provide comparable amounts of energy. Using precise atomic masses measured with mass spectrometers, we can determine the differences of the masses of the reactants and products in this reaction. Application of the Einstein relation to the mass loss yields the same energy release (in this case 3.2 MEV) as is obtained by energy measurements. Thus the energy yield of

400

known nuclear reactions may be determined with only mass measurements. For over fifty years, researchers have pursued the goal of achieving controlled thermonuclear fusion on a scale suited for commercial power production. Since the reactants are two positively charged nuclei, they must have high kinetic energies to overcome their mutual repulsion. These high energies imply a gaseous state with enormously high temperature, a condition known as a plasma. Because solid materials cannot exist at plasma conditions and plasmas would be cooled by the presence of solids, magnetic confinement of plasmas has been one approach to achieving a thermonuclear plasma. Large experimental devices called Stellerators, Tokamaks, and mirror machines have been built to help solve the problems inherent in achieving largescale fusion reactions and in stably confining the associated thermonuclear plasma. While progress continues, controlled thermonuclear fusion remains, and will likely continue, in the research stage for many years. It will therefore not be considered further here. Whereas nuclear fusion annihilates mass by forming larger atoms from light atoms, fission is a process of breaking massive atoms into two large, more-or-less equal-sized atoms, with an accompanying mass loss and energy release. While controlled fusion remains elusive, nuclear fission has been producing electrical power on a commercial scale for decades. The remainder of this chapter therefore deals with the fundamentals and commercial use of nuclear fission.

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10.4 Fundamentals of Nuclear Fission Nuclear fission involves the breakup of certain massive elements resulting from collisions with neutrons. Uranium U-235, for instance, the isotope of uranium with 235 nucleons, forms an highly excited isotope with 236 nucleons upon capture of a neutron, as in Figure 10.2(a). These excited U-236 atoms are unstable and break up into a variety of pairs of large atoms shortly after they are created. The many such fissions occurring in a reactor may be expressed as an average reaction: 92

U235 + 0n1 ; 92U236; F1 + F2 + 2.47 0n1 + 203 MEV

where F1 and F2 represent fission fragment elements, as in Figure 10.2(b). On the average, 2.47 neutrons are released in the process, one of which must initiate another fission to sustain a chain reaction, as illustrated schematically in Figure 10.3. In addition alpha, beta, and gamma radiation is released. One of the many reactions that participate in the average reaction above creates xenon and strontium fission fragments and two neutrons: 92

U235 + 0n1 ; 92U236 ; 54Xe139 + 38Sr95 + 2 0n1 + energy

U-235 reactions exemplified by the Xe-Sr reaction create diverse fission fragments and small integral numbers of neutrons that average to 2.47 and release energies that average to 203 MEV. Over 80% of the 203 MEV of energy released by the average U-235 fission reaction is the kinetic energy of the fission fragments associated with their large mass and high velocity. Because of their high energy, the fission fragments become ionized and ionize nearby atoms as they tear their way through surrounding materials. The fission fragments, however, travel only a very short distance, for interactions cause them to lose much of their energy to the surrounding solid. Thus most of the fission

402 energy appears immediately as internal energy and therefore locally high temperature of the surrounding materials. It is, therefore, necessary that the fuel be cooled continuously to counter the fission heat generation to avoid temperature buildup and possible melting. The heated coolant is then provides the energy input for a thermodynamic (usually steam) cycle. As the fission fragments come to rest, their presence changes the character of the reactor materials. Non-fissionable material exists where fissionable and other atoms once resided. In time these materials may decay radioactively, forming still different substances and releasing additional heat. But, importantly, they absorb neutrons without offering the possibility of fission. Thus the fission fragments are said to poison the reactor. Fission Reactor Design Considerations While most of the fission heat generation is due to the fission fragments, reactor design focuses on the actions of neutrons. This is because neutron captures produce fissions, fissions produce more neutrons, and, as we will see later, fissile material may also be produced using neutrons. Thus neutrons are the currency of the reactor and may be used constructively or wastefully. The manner in which the neutrons in a reactor are used is called the neutron economy. The reactor designer pays close attention to all of the details of the neutron economy. One of these details is the distribution of the kinetic energy of the neutrons, that is, the fractions of the neutrons that lie in given energy ranges. Almost all commercial power reactors (as opposed to experimental reactors) are thermal reactors in which fission is caused by thermal neutrons. A thermal neutron is a neutron that is in thermal equilibrium with the surrounding atoms. This implies that they have energies on the order of 0.02 EV. Because they move relatively slowly, thermal neutrons are much more likely to cause fission of U-235 than are higher-energy neutrons; therefore, they are the choice for most power reactor designs. However, the neutrons created by the fission reaction are not thermal neutrons. The neutrons created by fission have kinetic energies that range from about 1 to 10 MEV. They are called fast neutrons because of this high kinetic energy. Thus, if a chain reaction is to be sustained in a thermal reactor, it is necessary for the fast neutrons to be slowed, or thermalized, to much lower energies so that they can cause fissions before they are absorbed in nonproductive captures in reactor material or before they escape from the reactor. They do this by colliding with certain other nuclei in the core, put there for that purpose. These nuclei are called moderators. A good moderator is a light element that, on collision with a neutron, is speeded up by the collision, and thereby extracts energy from the neutron without absorbing it. The moderator concept may be understood by considering what happens when a billiard cue ball hits another ball head-on. The cue ball stops, and the second ball carries away the kinetic energy. However, with balls of differing mass, if the second ball is heavy the cue ball bounces off without losing energy, whereas light balls are propelled

403 at high speed and extract significant energy from the cue ball. In neutron collisions with atoms, it is the lightest atoms that extract the most energy. Ordinary water, heavy water, graphite, and beryllium are all used as moderators because they are light and are poor absorbers of neutrons. Hydrogen and deuterium are the moderators in water. The oxygen in the water is not an effective moderator but has low absorption and therefore does not interfere significantly with the moderation process. It is evident that conservation of neutrons is a prime consideration in any reactor design. The number of neutrons per unit volume, or neutron density, in a reactor is an important design parameter. Neutrons diffuse about in the reactor as they are scattered and slowed by moderator atoms. Because they have no charge, they are uninfluenced by electromagnetic fields and therefore may travel further than charged particles. Four events can influence local neutron densities as they pass through the surrounding reactor core: 1. The neutrons can be captured by a fissionable atom and produce a fission. 2. They can be absorbed non-productively by fission products, structural materials, or nonfissioning fuel. 3. They can escape through the walls of the reactor. 4. They can be absorbed in nuclei that create more fuel. The last possibility will be considered in Sections 10.5 and 10.8. Events 2 and 3, absorption and escape, result in nonproductive waste of neutrons in the neutron economy. The necessity that enough events of type 1, rather than types 2 and 3, occur to sustain the chain reaction suggest several important considerations for reactor design: + The reactor should be large enough that only events of types 1, 2, and 4 take place and hence that escape of neutrons from the reactor is rare. Since reactor size is dictated primarily by the cooling requirements imposed by nuclear heat generation, this usually follows automatically from the design process. In addition, positioning moderating material such as water at the boundaries of the reactor as a reflector, to deflect escaping neutrons back into the reactor, may allow a more compact design, in some cases.

+ Materials to be used in the reactor design are selected so that type 2 events are

minimized. The gradual buildup of poisons must also be considered in designing for the change in reactor performance with time between refuelings.

+ The reactor should be designed to minimize the amount of structural materials in the active fuel region, to reduce the frequency of type 2 events.

+ Neutron-absorbing materials may be moved into and out of the reactor to change the average neutron density for reactor control purposes.

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A reactor can function over a range of neutron densities. When the neutron density and power levels are constant, the reactor is said to be critical. This implies that the number of neutrons producing fission at one instant is the same as at a later instant, a situation depicted in Figure 10.4. Since high neutron densities produce more fissions, each of which produces 203 MEV in the case of a U-235 thermal reactor, heat generation, and thus power output, increases with neutron density. Let's assume that 2.5 fast neutrons are produced per fission in a critical thermal reactor and track the activities of 100 fast neutrons created in an instant. In a critical reactor, about 40 of these fast neutrons must thermalize and undergo fission to produce another 100 fast neutrons. Figure 10.4 shows that 2 of the original 100 fast neutrons produce (fast) fissions and 5 more neutrons immediately so that there are 103 fast neutrons diffusing around. Of these, about 10 escape from the core or are absorbed in non-fissile materials while slowing down. Of the remaining 93 thermal neutrons, about

405 3 escape from the core and about 43 are absorbed in non-U-235 materials while diffusing around at thermal energies. The remaining 47 captures by U-235 result in 7 non-fission captures and 40 fissions. These 40 fissions, in turn, produce 100 fast neutrons for the next generation in a reactor operating at critical. This sequence representing a single generation in a critical reactor is repeated over and over by an enormous number of neutrons. By slight changes in the reactor configuration during operation (for instance, by addition or removal of a small amount of poisons), the neutron economy can be adjusted to increase or decrease the number of neutrons in the next generation and thus change the reactor operating condition. A parameter, k, called the multiplication factor, is defined as the ratio of the number of neutrons in one generation to the number in the preceding generation. Thus k = 1 for a critical reactor. If k is less than or greater than 1, the neutron density is decreasing or building and the reactor is said to be subcritical or supercritical, respectively. A reactor is designed to have a maximum value of k > 1 so that it may be brought up to a desired power level and maintained there over the duration of the fuel cycle. Once the reactor approaches the desired power level, control actions are taken to adjust the value of k to 1, bringing the reactor to critical and stabilizing its operation. The control action requires the introduction of additional neutron-absorbing material or the reduction of the amount of moderator in the reactor, to reduce the rate of buildup of the neutron density to zero. As fissionable material is depleted and poisons build up, control material is gradually withdrawn from the reactor to maintain critical operation. When the reactor can no longer maintain critical at its design power level with no control material present, it must be refueled. The presence of a nuclear heat source is the major difference between fossil fuel and nuclear plants. In that connection, there are several important factors that must be considered in the design of a nuclear power plant that are not factors in conventional power plant design and operation. First, the entire amount of fuel needed to operate a nuclear power plant for up to two years is loaded into the plant at one time. The rate at which power is generated must then be maintained by controlling the neutron chain reaction over the wide range of reactor operating and fuel depletion conditions that can arise between refueling operations. This calls for detailed planning and analysis, both before and throughout the operating cycle. Second, because the products of fission are highly radioactive and their rate of decay cannot be controlled, the heat from radioactive decay of fission products after shutdown amounts to as much as 7% of full power output. Consequently, provision must be made in the thermal design to remove this heat under all credible operating and accident conditions. (The reactor at Three Mile Island melted over an hour after the nuclear chain reaction was terminated, because the operators misinterpreted their instruments and turned off the emergency systems that were removing the decay heat.) The nuclear decay process is not self-limiting and has no maximum temperature, as with chemical reactions. If the heat generated is not removed from the reactor core, the core will melt and be destroyed.

406 Third, if radioactive materials from the reactor core find their way to the environment, they can be hazardous to nearby life. Although the proper cooling of the core will ensure that these materials will remain contained inside the fuel assemblies, a defense-in-depth approach to safety must be employed to make the possibility of a release extremely remote. Table 10.1

Nuclear Fuels

Fertile Fuels

Fuels Fissionable by Thermal Neutrons U-235

U-238

;

Pu-239

Th-232

;

U-233

10.5. Nuclear Fuels Uranium-235 is the only material that is both fissionable by thermal neutrons and found in nature in sufficient abundance for power production. Other fissile fuels are uranium233 and plutonium-239 (Table 10.1) which are created from thorium-232 and uranium238, respectively, by absorption of neutrons. Substances from which fissionable fuels are created, called fertile fuels, are transmuted into fissionable fuels in a reactor by extra neutrons not needed to sustain the fission chain reaction. Fertile fuel used in this way is said to have been converted. The resulting fissile materials may be processed to make new fuel elements when sufficient quantities have accumulated. Some of the converted material may be consumed directly by fissions during reactor operation. The composition of uranium ore is about 99.3% U-238 and 0.7% U-235. Because of the ore’s small percentage of U-235, it is difficult to design a water-cooled, thermal reactor that uses natural uranium. Therefore the power reactors in the United States and most other parts of the world are thermal reactors that employ uranium enriched to between 2% and 5% U-235. Such reactors use ordinary (light) water for both cooling and moderation and are therefore commonly called light-water reactors. Uranium enrichment is an expensive and difficult process because it involves separation of two isotopes of the same element, which rules out most chemical methods. Thus processes that rely on the small mass difference between U-235 and U-238 are usually used. The gaseous diffusion process involves conversion of a uranium compound processed from the ore to gaseous uranium hexafluoride, UF6. The gaseous UF6 flows in hundreds of stages of diffusion through porous walls that eventually produce separate UF6 gas flows containing enriched and depleted uranium. The enriched UF6 then is processed to UO2 powder which is sintered into hard ceramic fuel pellets such as those shown in Figure 10.5. The pellets are sealed in long cylindrical metal tubes for use in the reactor. Newer enrichment processes currently available or under development include: high speed centrifugal separation; which also relies on the uranium isotopic mass difference;

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a separation process that relies on differences in chemical reactivity between the isotopes; and laser enrichment, which relies on ionization of uranium by an intense light beam with subsequent chemical or physical separation of the ions (ref. 8). 10.6 Light-Water Power Reactors Power reactors active in the United States today are light-water reactors. They are designed so that the core is both moderated and cooled by highly purified water and therefore must use a fuel that fissions with thermal neutrons. Water has many advantages in thermal reactors. From a neutron point of view, H2O is an extremely efficient moderator. As we know from its extensive use in conventional power plants, water has excellent heat transfer characteristics, and the technologies of its use in steam power plants are well established. Water has disadvantages as well. To maintain its excellent moderation and heat transfer capabilities, it must remain a liquid. Thus water reactors are currently limited to producing hot liquid or steam with little superheat. Moreover, boiling temperatures suited to an efficient plant require very high pressures, as in fossil fuel plants. Thus water-cooled reactor cores must be encased in pressure vessels that operate with high temperatures nearby. In addition, they must endure, for the design life of the plant, the

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severe environment resulting from the fission reactions. Finally, and perhaps most importantly, should reactor pressure integrity be lost while the reactor is operating, the liquid water will flash to steam, losing much of its heat transfer advantages. All of these factors contribute significantly to the challenges that an engineer faces in the thermal and mechanical design of light-water reactors. There are two major types of light water reactors (Figures 10.6 and 10.7), which are differentiated primarily by the thermodynamic conditions of the water used to cool uranium fuel elements in the reactor vessel. The boiling water reactor (BWR) operates at a pressure that allows boiling of the coolant water adjacent to the fuel elements. The water in the pressurized water reactor (PWR) is at about the same temperature as in the BWR but is at a higher pressure, so that the reactor coolant remains a liquid throughout the reactor coolant loop. In addition to their use in utility power reactors, PWRs are used in American nuclear submarines. Boiling Water Reactors A schematic of the layout of the General Electric BWR/6 system is shown in Figure 10.8. There the turbines, condenser, pumps, and feedwater heaters studied in Chapter 2 appear in a familiar configuration. Water boils inside the reactor core, producing slightly radioactive steam that passes directly to the steam turbines. The radioactivity in the steam, however, has a half-life of only a few seconds. The carryover of radioactivity to the turbine-feedwater system is virtually nonexistent, and experience has shown that components outside the reactor vessel (turbine, condensate pump, etc.) may be serviced essentially as in a fossil-fueled system. Some other reactor designs, such as the pressurized water reactor to be considered later, have an additional separate water loop, as seen in Figure 10.7, that isolates the turbine steam loop from the reactor coolant to provide further assurance that the turbine-feedwater system components remain free of radioactivity.

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Fuel Assemblies Uranium appears in most boiling water reactors in the form of sintered cylindrical pellets of uranium dioxide (UO2) (Figure 10.5), about 0.4 inches in diameter and about 0.4 inches long. These pellets are stacked inside of long sealed zirconium alloy (zircaloy) tubes called fuel rods. Fuel rods, in turn, are mounted in an eight-by-eight array in a fuel bundle, as seen in Figure 10.9. The fuel bundle and the fuel channel that surrounds the fuel rods comprise a fuel assembly, as shown in Figure 10.10.

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The lower-tie-plate nose piece seen there, together with the fuel channel, directs the coolant water flow over the fuel rods inside of the fuel assembly. The fuel assemblies, mounted vertically in the core, are designed to minimize operating stresses on the fuel rods. For example, Figure 10.10 shows that the fuel rods are spring loaded so they are free to expand in the axial direction in response to changes in reactor operating temperatures. The Reactor Assembly The core of a BWR/ 6 1220-MWe reactor (MWe stands for electrical generator power output, as opposed to MWt for reactor thermal output) has 732 fuel assemblies and 177 control rod assemblies in an approximately 16-ft-diameter circular array, as shown in Figure 10.11. Cooling water receives heat from the fuel rods by forced-convection and

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two-phase nucleate boiling as it flows upward through the fuel assemblies. As the water cools the fuel assemblies, it also thermalizes the fast neutrons that diffuse through the core. Water that bypasses the fuel assemblies and flows upward on the outside of the fuel channels is confined by the cylindrical core shroud, as seen in Figure 10.11. This flow cools the channels and the neutron absorber control rods. The control rods are mounted in cruciform assemblies, as seen in Figure 10.12. The control rod assemblies fit vertically in a fuel module between 4 fuel assemblies in the core, as diagrammed in Figure 10.13. The control rod assemblies move vertically between the fuel bundles to change the effective multiplication factor to compensate for changes in reactor operating conditions due to buildup of poisons over time. The stainless-steel-clad control rods contain boron carbide (B4C), which absorbs neutrons and hence tends to

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terminate fission processes when in place in the core. The control rod assemblies are moved up and down through the core to change the rate of absorption of neutrons when significant changes in power level or adjustments to account for fuel burnup are required, or in the event of an emergency shutdown. The positions of the control rods are adjusted by hydraulic drives located below the core. The rods are fully inserted in the core during a shut down making the multiplication factor less than 1. Most of the control rods are fully out of the core during critical operation. When all of the rods are out of the reactor the multiplication factor slightly exceeds 1, which allows the neutron density and power level to increase. The bottom-entry fuel rod drives in the GE BWR are an unusual feature in reactors. Their location below the reactor simplifies the refueling process which is carried out from above in most reactor designs. The quality of the steam leaving the top of the core is approximately 11%, indicating that most of the water is still liquid and must be recirculated through the core for additional heating. The liquid water leaving the top of the core and the steam separators flows downward outside the core shroud, driven by the jet pumps located between the shroud and the reactor vessel wall, as shown in Figure 10.14. The jet pumps, in turn, are driven by recirculation pumps located outside the reactor vessel. The jet pumps induce the downward flow outside the shroud by momentum transfer to the slower-moving liquid.

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Vapor bubbling up through the fuel assemblies leaves the core and passes upward with the liquid. The flow is turned by stationary vanes in the steam separator, where the higher angular momentum of the liquid separates it from the vapor. The separated liquid flows to the outside of the reactor vessel where it is recirculated outside the core shroud. The steam passing through the separator is further dried in the steam dryer assembly before it leaves the reactor vessel as a slightly superheated or saturated vapor. The turbines are specially designed to operate with saturated vapor at the throttle and small amounts of liquid within. Liquid is separated from the wet steam leaving the HP turbine. Steam tapped from the HP-turbine-throttle steam line is used to reheat the HP exit steam before its entry to the LP turbines. The low HP-turbine-throttle conditions of 550°F and 1040 psia lead to a plant thermal efficiency of about 32%. In the boiling water reactor, control is primarily achieved by adjustment of the rate of recirculation through the reactor by the recirculation and jet pumps shown in Figures 10.11 and 10.14. Change in the rate of water recirculating through the core changes both the onset of boiling and the volume fraction of steam in the cooling channels, and thus the amount of moderator in the core at a given time. This allows significant adjustment of reactor power output without control rod movement. For example, increasing the recirculation and jet pump speeds sweeps bubbles away faster, increasing moderation and raising the power level until the additional boiling restores the proper void fraction for critical operation at a higher power level.

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As the coolant passes upward through the fuel assemblies, heat from the fuel rods produces vapor bubbles via nucleate-boiling heat transfer. Nucleate boiling is characterized by the local formation of bubbles of vapor that break away from the fuel rod surface, causing vigorous, agitated, fluid motion with resulting high heat transfer rates. This may be contrasted with film boiling, in which a stable vapor layer covers the tube surface, resulting in low heat transfer rates. Film boiling occurs at high surface-tobulk fluid temperature differences. It is crucial that a departure from nucleate boiling, DNB, be avoided, because reduced heat transfer coefficients and cooling rates produce drastic increases in tube temperatures, leading to fuel melting or zircaloy fuel rod burnout. Thus the ratio of maximum heat flux to the critical heat flux for DNB is a major thermal design parameter for the water reactor. The reactor fuel rod heatgeneration rate and flow-channel convective cooling are designed to maintain tube-to-fluid heat fluxes well below the unstable transition range between nucleate and film boiling. The maximum UO2 fuel temperature based on the maximum design fuel rod heat-generation rate of 13.4 kW/ft is approximately 3400°F, whereas the fuel melting temperature is about 5100°F. The boiling water reactor, like other reactors, has numerous active and passive safety systems. A thick pressure vessel, for instance, surrounds the reactor core. An even thicker concrete containment structure surrounds the pressure vessel to confine anything that may escape from it. An emergency core cooling system (ECCS) senses overheating of the core and supplies a flood of water to take away the heat generated by the fuel elements. These and other safety systems clearly reduce the danger of accidents but also increase the cost and complexity of plant operation.

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Pressurized Water Reactors The pressurized water reactor, PWR, is currently the predominant reactor type in the world. It is a light-water reactor that uses slightly enriched U-235 as fuel. Figure 10.15 shows the three reactor containment buildings of the Oconee PWR Nuclear Station. A major difference between BWRs and PWRs is that the pressure of the PWR coolant is above the saturation pressure (it is subcooled liquid) through the entire cooling loop so that there is no possibility of bulk boiling in the core. As shown in Figure 10.7, separate steam generators receive heat from the reactor liquid cooling loop, thus preventing radioactive material from entering the turbine power loop. Another difference is that control rods are at the top of the PWR and can drop by gravity into the reactor when necessary. Figure 10.16 shows a sectional view of the reactor building of the Oconee plant. The stairs and landings give some idea of the size of the equipment within the containment. Figure 10.17 gives a less cluttered view of the major components. Table 10.2 shows that, for PWRs, the turbine loop is at a lower pressure than the liquid in the reactor loop and therefore produces an outflow of steam to the turbine throttle with about 50 Fahrenheit degrees of superheat.

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We have seen that, in contrast to the BWR, which circulates the reactor coolant through the turbine, the PWR uses two loops: a primary loop that cools the reactor core, and a secondary loop, heated by the primary loop, that provides steam for the turbines, as diagrammed in Figure 10.18. The primary loop is contained completely within the reactor containment building (Figure 10.16) and is designed so that the water that cools the core is completely isolated from the environment. The secondary loop executes a steam turbine cycle similar to those of conventionally fueled power plants, with the exceptions that the steam is generated by heat transported from the reactor and that the maximum turbine inlet temperature and superheat are limited by the maximum temperature in the reactor core. The Babcock & Wilcox steam generator is a counterflow shell-and-tube heat exchanger, as seen in Figure 10.19. The primary water enters the top of the unit and flows downward through thousands of small-diameter tubes to provide a large heat transfer surface. Feedwater is piped into the bottom of the shell side of the steam generator and is first heated, then boiled by heat from the hot tubes containing the primary water flow. As the steam rises, it encounters the hotter portions of the primary tubing, reaching 50 Fahrenheit degrees of superheat at the top of the steam generator. An important thermal design parameter of the steam generator is its size. A larger unit increases the heat transfer area from the primary loop, but it also increases the capital cost of the plant, both in the cost of manufacturing the generator itself and in the larger size required of the primary containment. This must be balanced with the cost savings, lower heat exchanger effectiveness and consequent lower thermal efficiency achieved, if the heat transfer area is made smaller.

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The pressurizer (Figure 10.17) controls the pressure in the primary loop and serves as a surge tank to accommodate the reactor coolant water’s expansion and contraction with temperature changes. It is a large tank designed to contain saturated water and steam, each occupying about one-half of the tank volume. As shown in Figures 10.17 and 10.18, the pressurizer is connected to the primary loop at the outlet of the reactor through a single pipe. The temperature of the pressurizer contents are controlled with internal electric heaters and water sprays. The resulting saturation temperature establishes the operating pressure of the reactor. An important part of the PWR thermal design is sizing the pressurizer tank to permit the primary system to respond to all possible transients without bursting the coolant pipes. Recalling that water is an almost incompressible fluid that expands when heated, if the reactor power rises and heats the primary water to a higher temperature, the expanding water will flow into the pressurizer, compressing the steam bubble. Pressure sensors will detect the increased pressure and open spray valves at the top of the pressurizer to condense some of the steam, thus restoring a lower operating pressure. Conversely, if the primary loop temperature declines, heaters in the bottom

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of the pressurizer will turn on, creating more steam to fill the volume formerly occupied by the contracting water and thus increasing the reactor temperature and pressure. Although there is no bulk boiling in a PWR core, heat transfer to the coolant is by subcooled nucleate boiling. In subcooled nucleate boiling, steam bubbles are formed on the surface of the cladding. As the bubbles expand they become detached from the tubes and immediately collapse as they are swept into the coolant channel. The lateral motion in the coolant achieved by this action is an extremely effective mechanism to transport the energy from the cladding into the coolant stream. With subcoolednucleate-boiling heat transfer, the cladding surface temperature will stay within 10°F of the water's saturation temperature while providing the very high rates of heat transfer needed to remove the fission energy from the fuel rods. An extremely important design limitation is the critical heat flux at which steam bubbles form and grow so fast that they coalesce to form a vapor film over the clad. At this point the heat transfer undergoes a departure from nucleate boiling to film boiling, and the clad no longer touches liquid water. Since metal to vapor heat transfer is much more inefficient than nucleate boiling, the fuel rod temperature rises in order to transfer the accumulating heat that is created by fissions. Unfortunately, film boiling heat transfer coefficients are so low that the high surface temperatures needed to achieve a steady state are sufficient to melt the clad and severely damage the fuel rods. For this reason the Nuclear Regulatory Commission has established safety factors for the ratio between the critical heat flux and the maximum heat flux expected in a reactor under its most severe overpower transient conditions. In its application for an operating license, every nuclear power plant must be able to demonstrate through engineering analysis that it will maintain the required safety factor under all credible overpower and undercooling transients. The Babcock and Wilcox PWR uses U-235 enriched to about 3% in the form of UO2 fuel pellets (Figure 10.20) encased in a zircaloy-clad tube similar to the fuel rods described for the BWR. Table 10.2 shows that a 1300-MWe plant has 205 fuel assemblies with 54,120 fuel rods. Figure 10.21 shows fuel assemblies for an electric

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425 utility PWR and for the Nuclear Ship Savannah. Each fuel assembly for electric power generation is fitted for the inclusion of sixteen control rods, as seen attached to an activating spider in the left fuel assembly photo. Positioning of the fuel assemblies in the core and those assemblies containing control rods are shown in Figure 10.22. There are sixty-nine control rod assemblies in Oconee Unit 1, of which sixty-one are for control of power level; the remaining eight contain poisons in the lower part of the rods for shaping of the reactor power distribution. The control rod neutron-absorbing material, silver-indium-cadmium, is encased in stainless steel. 10.7 The CANDU Reactor The CANadian Deuterium Uranium, CANDU, reactor is a reactor of unique design that utilizes natural uranium as a fuel and heavy water as a moderator and coolant. These reactors produce a substantial saving due to the absence of fuel enrichment costs, but a large chemical plant is required to supply the quantities of heavy water required. The Pickering station near Toronto, an Ontario Hydro plant shown in Figure 10.23, uses eight CANDU reactors to generate 4800 MWe and has been generating power since 1971. One of the important and unique features of the CANDU reactor is that, whereas light-water reactors are shut down for refueling annually, CANDU reactors are refueled daily. The pressurized heavy-water-cooled fuel bundles are horizon-tally oriented in individual fuel channels inside the unpressurized "calandria," as diagrammed in Figure 10.24. Each bundle may be individually accessed, rearranged in

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the calandria, or replaced using special fuel handling equipment while the reactor is operating. Heavy water at atmospheric pressure in the calandria surrounds the fuel channels and moderates the reactor. Thus, in an emergency, the reactor can be shut down by draining the calandria to remove the moderator, thereby depriving the fuel of thermal neutrons. The pressurized heavy-water loop draws hot coolant from the fuel channels through headers to supply heat to steam generators as in a PWR. A lightwater loop passing through the steam generator in turn supplies steam to the turbine-feedwater loop. A turbine hall at the Pickering Station is shown in Figure 10.25. The largest cylindrical structure seen in Figure 10.23 is a vacuum building that connects with each of the reactor buildings. In the event of an emergency, escaping steam and radioactive materials would be drawn by vacuum into the structure. A coldwater spray would condense the steam to limit any pressure buildup. The thermal efficiency of a CANDU reactor plant is only about 29%, but the CANDU reactor uses a larger fraction of U-235 in uranium ore than other reactors and also makes better use of the U-238 to Pu-239 conversion process to extend fuel burnup.

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Moreover, statistics show that, among large reactors, CANDU reactors have outstanding reliability records, with annual capacity factors (the ratio of annual electrical energy output to maximum possible annual output) as high as 96% and cumulative capacity factors as high as 88% (ref. 14). 10.8 Fast Reactors Reactors may be designed to fission with fast neutrons, but these fast reactors must be more compact than thermal reactors so that the fast neutrons may produce fissions quickly before they are absorbed or moderated by surrounding materials. They are designed with structural materials that are poor absorbers and moderators of neutrons, such as stainless steel. The core of a fast reactor must contain a fissionable fuel of about 20% enrichment to compensate for the lowered probability of fissioning with high-energy neutrons. Because of their high fuel density, fast reactors have a high power density that poses a difficult cooling problem. One solution is the use of a liquid metal as coolant. Liquid metals such as sodium and potassium have excellent heat transfer characteristics and do not interfere significantly with neutron functions.

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The choice of fuel used for thermal and fast reactors depends on the fuel’s fission probability and the net number of neutrons produced per neutron absorbed. Most effective for thermal reactors is U-235, whereas Pu-239 is most suitable for fast reactors. In fissioning with fast neutrons, Pu-239 emits almost 20% more neutrons than does U-235. These additional neutrons are extremely important for making breeding practical, as will be discussed shortly. Because Pu-239 must be created from fertile U-238, a plutonium reactor can use fuel processed from fuel produced in a uranium thermal reactor or in another plutonium reactor. This would occur in the core of a plutonium fast reactor and in a blanket of U-238 surrounding the core, where additional plutonium is created using neutrons that escape from the core. Figure 10.26 diagrams two steps in a chain reaction in a fast reactor where a plutonium atom is created for each one consumed. Note that each fission must produce a minimum of two neutrons for this reaction to continue. As a practical matter, more than two neutrons are required for complete replacement because of non-productive neutron captures. A reactor that transmutes a fertile fuel to a fissionable fuel is called a converter. The conversion efficiency is the ratio of the number of new fissionable atoms produced to the number of atoms consumed in the fission. Thus the conversion efficiency of the reaction shown in Figure 10.26 is 1, because a new fissionable atom is created for each atom consumed. In this case there is no net fissionable fuel consumption. Fast Breeder Reactors The cover of a 1971 U. S. Atomic Energy Commission booklet (ref. 7), (see Figure 10.27) shows a notebook page that poses and answers the following question: Johnny had 3 truckloads of plutonium. He used 3 of them to light New York for 1 year. How much plutonium did Johnny have left? Answer: 4 truckloads. This neatly emphasizes the point that if unproductive neutrons in the process shown in Figure 10.26 were to

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transmute a U-238 atom to Pu-239 occasionally, the chain reaction would produce more fissionable fuel than it consumes. The reactor would have a conversion efficiency greater than 1. A reactor that does this is called a breeder reactor. Thus, the breeder reactor provides a means of increasing the world supply of fissionable fuel as it generates power. While it may appear that the breeder provides something for nothing, it actually only takes advantage of the possibility of conversion of the fertile fuels through more efficient use of fission neutrons. The breeder reactor is of great importance because it would allow the use of the vast store of U-238 in uranium ore that remains as a by-product of the U-235 enrichment process to provide fuel for current LWRs. This supply of U-238 has the potential to provide fuel for many years without further uranium mining. Fission of U-235 is currently the only natural large-scale source of neutrons. The continued use of low-conversion-efficiency reactors could preclude the eventual use of much of the energy resource of the U-238 in uranium ore. One possibility for the design of a breeder reactor is a liquid-metal fast breeder reactor, LMFBR, which has the characteristics described briefly in the preceding section. The development of such a reactor involves careful design of its neutron economy and the development of a system of fuel reprocessing and nuclear waste storage. These topics will be considered in the next section. In 1971, President Nixon set the development of a breeder reactor as a national goal and established a program for the development of an LMFBR pilot plant in Tennessee to be known as the Clinch River Breeder Reactor, CRBR. The vocal opposition of segments of the American public to all forms of nuclear power, coupled

430 with well-known nuclear accidents, bureaucratic and construction delays, and concern for diversion of nuclear fuels to foreign military or terrorist uses, brought the development of new nuclear plants in the United States, including the CRBR, to a standstill. While some other nations, notably Japan, France, and Russia, continue to support extensive nuclear establishments and research, the path to the development of breeder reactors has been slow. The French, in particular, have developed experience with their 233-MW Phénix reactor, in operation since 1974, and an advanced 1200MW LMFBR called the Super-Phénix, which went into operation in 1986. In 1994, the status of Super-Phénix was changed from production to a research facility; and in 1998 the French government directed that it be shut down, well before its planned 2015 shutdown (refs. 24 to 26). In Japan, delayed reporting of a minor sodium leak during commisioning of the Monjo breeder in 1995 produced a national turmoil that had serious repercussions for the Japanese breeder program. (Ref. 27). 10.9 Innovation in Reactor Design During the hiatus in new domestic nuclear plant orders in the United States since the late 1970s, the industry has been working to incorporate new ideas and recommendations, based on utility operating experience, into the design of plants that will be more capable of winning public, government, and industry acceptance. Popular opinion notwithstanding, the safety record of nuclear power plants in the United States has been a good one. No one has been killed in a nuclear accident in a U.S. power plant. Few other industries can approach that record. Nevertheless, serious and expensive accidents have taken place at home and abroad; and the threat of catastrophic accidents, while extremely remote, remains. Designs are being considered by industry, the U.S. Department of Energy (DOE), the Electric Power Research Institute (EPRI), and the Nuclear Regulatory Commission (NRC) that would provide built-in systems and safeguards that are passive, rather than active, that must be depended on to perform when a malfunction occurs (ref. 15). Providing for natural conduction/convection cooling of the core in case of an accident, rather than depending on pumps to force water through the core, is one of such features under consideration. The General Electric Company, with an international team, is developing a 600 MWe simplified / small boiling water reactor (SBWR) that incorporates this approach (ref. 28). Nuclear plant incidents sometimes result from failure of an operator to interpret instrument data and act thereon or from instrument failure. Redundant instrumentation is provided to avoid such problems, but operators do not always analyze the readings correctly or make the right decisions. If the plant is designed to minimize the necessity of such actions, the likelihood of accidents can be significantly reduced. Thus, much effort focuses on reducing the complexity of reactor control. Designs have been produced that appear to accomplish this to a significant degree. The NRC must, of course, conduct hearings and evaluate the adequacy of any design before a plant is built.

431 One of the problems of the power industry is the lack of standardization. Every American power plant is different from every other because of the number of reactor suppliers and power plant engineering firms and the diverse requirements of the different utilities that buy them. One of the obvious reasons for the reliability of the CANDU reactors is that they are planned and constructed by Ontario Hydro and Atomic Energy of Canada, largely for Ontario Hydro. American suppliers are also striving for standardization, but resolving the differences between differing requirements and standards is a formidable problem. An advanced boiling water reactor (ABWR) program with an NRC certification (ref. 29) addresses these issues by providing a standardized design that places pumps within the reactor envelope, eliminates much external piping, and incorporates other passive safety features. Reference 23 indicates that this system benefits from a new NRC licensing process in which safety issues are resolved, with full public participation, before construction begins. ABWR planning calls for a forty-eight-month construction schedule once a site has been approved. The long-term trend has been for utilities to seek larger and larger plants to take advantage of the economies of scale. When the annual rate of growth of the electrical power industry dropped drastically in the 1970s to 2–3% due to conservation and other factors, many of the utility requirements for large reactors became less critical or disappeared. Interest has appeared in smaller reactors that can be constructed economically and on a timely basis. One hope is that better quality control could be exercised, and better economics result if small reactors were built, entirely or in modular fashion, in a factory and shipped to and installed at the power plant site, rather than erected there. Some of theses issues are addressed in the ABWR and SBWR programs (ref. 23). 10.10. Nuclear Reprocessing and Waste Disposal Spent fuel is fuel that has resided in a reactor for a year or more, has been depleted of much of its fissionable material, and includes a buildup of radioactive fission products. When a reactor is shut down for refueling, spent fuel assemblies are removed and partially spent assemblies may be repositioned in the core to obtain further fuel utilization. Ideally, the spent fuel would be reprocessed to reclaim the unused and newly created fissionable materials for use in new fuel rods, and the high-level waste would be isolated to minimize the volume of highly toxic wastes. Even without breeder reactors, current spent fuel inventories can provide ample Pu-239 to support reprocessing. In the United States, by law, fuel assemblies are being stored indefinitely in reactor-fuel storage pools in nuclear power plants until legal decisions are made on whether reprocessing will take place and on the final disposition of nuclear waste. Unfortunately, these decisions are among the most politically difficult ones of our times.

432 10.11 Concluding Remarks Nuclear power engineering has become so hopelessly intertwined with politics that any speculation on the future of nuclear power must simultaneously consider both technical and political realities. Even the continuation and expansion of nuclear power generation in the United States is in question. A major concern for many years has been the availability of fissionable resources for an expanding nuclear power generation industry. It has been widely recognized that U-235, as the sole natural neutron resource, places limits on extensive nuclear development. For that reason, countries with nuclear generation capability have planned on developing breeder reactors to greatly extend that neutron resource. It is possible that the continuing long-term use of reactors that do not provide for efficient conversion of the fertile fuels may result in the waste of this enormous energy resource. While the American nuclear power industry languishes, foreign development of nuclear power continues in some areas. The eventual recovery of the American nuclear power industry may depend on the success of international nuclear development. The absence of CO2 production by nuclear plants, and its global warming implications, gives an added incentive for international perpetuation and development of the industry. Expansion of the nuclear power industry and development of a breeder reactor nuclear power economy inevitably entails the development of nuclear processing and hazardous waste disposal facilities and reactor fuel recycling. These activities arouse concerns about the possible diversion of nuclear materials for the production of weapons, (refs. 13 and 17–19). The tradeoffs among nuclear power, coal, and other energy conversion alternatives are also becoming more prominent as concerns over global warming and atmospheric pollution intensify. Well-thought-out and consistent government policies and regulation, as well as international cooperation, would give welcome direction to the power generation industry. Reference 16 observes that success in nuclear programs in other countries seems to correlate with limiting public intervention, and it questions whether the democratic institutions of the United States are consistent with the growth of the American nuclear power industry. Bibliography and References

1. Hulme, H. R., Nuclear Fusion. London: Wykeham Publications Ltd., 1969. 2. Semat, Henry, Introduction to Atomic and Nuclear Physics. New York : Holt, Rinehart & Winston, 1959. 3. Spitzer, Lyman, Physics of Fully Ionized Gases. New York: Interscience Publishers, 1956. 4. Glasstone, Samuel, Principles of Nuclear Reactor Engineering. New York: D. Van Nostrand, 1960.

433 5. Glasstone, S., and Sesonske, A., Nuclear Reactor Engineering. New York: Van Nostrand Reinhold, 1981. 6. Steam / Its Generation and Use, 39th ed. New York: Babcock and Wilcox, 1978. 7. Mitchell, Walter III, and Turner, Stanley E., Breeder Reactors. U. S. Atomic Energy Commission, 1971. 8. Nero, Anthony V. Jr., A Guidebook to Nuclear Reactors. Berkeley, Calif.: University of California Press, 1979. 9. Winterton, R. H. S., Thermal Design of Nuclear Reactors. New York: Pergamon Press, 1981. 10. Judd, A. M., Fast Breeder Reactors, An Engineering Introduction. New York: Pergamon Press, 1981. 11. Bromberg, Joan Lisa, Fusion. Cambridge, Mass.: MIT Press, 1982. 12. Robinson, Mark Aaron, 100 Grams of Uranium Equal 290 Tons of Coal. Kelso, Wash.: R & D Engineering, 1987. 13. Nuclear Proliferation and Safeguards–Summary. Congress of the United States, Office of Technology Assessment, OTA-E-148, March 1982. 14. Nuclear Power in an Age of Uncertainty. Congress of the United States, Office of Technology Assessment, OTA-E-216, February 1984. 15. Catron, Jack, "New Interest in Passive Reactor Designs." EPRI Journal, April/May 1989: 5–13. 16. Campbell, Jack, Collapse of an Industry-Nuclear Power and the Contradictions of U.S. Policy. Ithaca, New York: Cornell University Press, 1988. 17. Blocking the Spread of Nuclear Weapons. New York: Council on Foreign Relations, 1986. 18. Muller, Harald, A European Non-Proliferation Policy, New York: Clarendon Press, 1987. 19. Snyder, Jed C., and Wells, Samuel F. Jr., (Eds.) Limiting Nuclear Proliferation. Cambridge, Mass.: Ballinger, 1985.

434 20. Lovins, Amory, Soft Energy Paths. New York: Harper and Row, 1979. 21. Nealey, Stanley M., Nuclear Power Development–Prospects for the 1990s. Columbus, Ohio: Battelle Press, 1990. 22. “World List of Nuclear Power Plants.” Nuclear News, February 1991. 23. Wilkins, D. R., Quirk, J. F., and McCandless, A. H., “Status of Advanced Boiling Water Reactors.” American Nuclear Society, 7th Pacific Basin Nuclear Conference, March 1990. 24. Electricité de France, “Super Phénix Shutdown.” www.info-france-usa.org/america/embassy/nuclear/facts/superpr.htm (Oct. 30, 2000). 25. Electricité de France, “Permanent Shutdown.” www.info-france-usa.org/america/embassy/nuclear/facts/superpff.htm (Oct. 30, 2000). 26. Electricité de France, “Shutdown of the Superphénix breeder reactor.” www.info-france-usa.org/america/embassy/nuclear/facts/supersum.htm (Oct. 30, 2000). 27. Japan Nuclear Cycle Development Institute, “Monju Reactor Website.” www.jnc.go.jp/zmonju/mjweb/index.htm (October 30, 2000). 28. “The Simplified Boiling Water Reactor.” www.nuc.berkeley.edu/designs/sbwr/sbwr.html (December 19, 2000). 29. The Advanced Boiling Water Reactor.” www.nuc.berkeley.edu/designs/abwr/abwr.html (December 19, 2000). 30. Davis, Mary Bird, “Nuclear France: Materials and Sites–Creys-MalvilleSuperphénix.” www.francenuc.org/en_sites/rhone_crey_e.htm (Oct. 30, 2000). 31. Fetter, Steve, “Energy 2050,” Bulletin of the Atomic Scientists. July/August 2000. www.bullatomsci.org/issues/2000/ja00/ja00fetter.html (October 30, 2000). 32. Nuclear Regulatory Commission, “U.S. Commercial Nuclear Power Reactors.” www.nrc.gov/AEOD/pib/states.html (November 2, 2000). 33. Nuclear Regulatory Commission, “Nuclear Reactors.” www.nrc.gov/NRC/reactors.html (November 2, 2000).

435 34. Environment News Service, “World Total: 434 Nuclear Power Plants.” ens.lycos.com/ens/may99/1999L-05-06-06.html (November 4, 2000). EXERCISES

10.1 Consider the collision of a particle of mass m and velocity v with a stationary particle of mass M. Write energy and momentum equations for the collision. Derive an equation for the ratio of the final to the initial kinetic energy of the original moving particle in terms of the masses of both particles. Use the result to show why light atoms are used as moderators. 10.2 Evaluate the assertion of the title of reference 12 that 100 grams of uranium equal 290 tons of coal. Assume coal to be represented by pure carbon. Could you make the title of the book more accurate? If so how? 10.3 Estimate the core thermal power and thermal efficiency of a 1220-MWe boiling water reactor that has 46,376 fuel rods in a 150-in.-high core with a maximum fuel rod linear energy density of 13.4 kW / ft and a fuel rod peak-to-average power release of 2.2. If there are 748 fuel assemblies, what is the average number of fuel rods per assembly? Estimate the number of thermally inactive rods in a reactor with an eight-by-eight fuel assembly array. 10.4 Assuming the fuel temperature to be 295K, calculate the energy of a thermal neutron using 3kT/ 2 where k is the Boltzmann constant. 10.5 Study the literature and then discuss the details of the nuclear processes by which neutrons convert U-238 to fissionable fuel. Include nuclear reaction equations. 10.6 Study the literature, then discuss the details of the nuclear processes by which neutrons convert thorium to a fissionable fuel. Include nuclear reaction equations. 10.7 Sketch and label a PWR steam generator flow diagram and a T-s diagram for the two flows through the PWR steam generator, using the data from Table 10.2. 10.8 Determine the ratio of the reactor-loop flow rate to the steam flow rate from the data for the PWR given in Table 10.2. 10.9 Sketch a T-h diagram and determine the pinchpoint temperature difference for a PWR steam generator that has, respectively, 572.5°F and 630°F reactor inlet and outlet temperatures and steam generator inlet and outlet temperatures of 473°F and 603°F. 10.10 Discuss the changes in the neutron economy diagram (Figure 10.4) due to (a) insertion of control rods, (b) withdrawal of control rods, (c) increase and (d)

436 decrease in recirculation flow rate in the boiling water reactor. 10.11 List from BWRs, PWRs, CANDUs, and LMFBRs those power reactors that are (a) Thermal reactors (b) Fast reactors (c) Heavy-water reactors (d) Light-water reactors (e) Natural uranium reactors (f) Enriched uranium reactors (g) Plutonium reactors 10.12 Identify those power reactors that use moderators and those that do not. 10.13 Which power reactor (a) uses light water as coolant and has separate reactor coolant and steam loops, (b) uses heavy water and natural uranium, and (c) uses plutonium as fuel and liquid metal as coolant? 10.14 Based on data in Table 10.2 for the Oconee Unit 1 PWR, estimate the turbineloop thermodynamic conditions (temperature and pressure) and power delivered by a simple Rankine-cycle steam turbine with 85% efficiency. 10.15 Compare the reactor temperature rise shown in Table 10.2 for the Oconee Unit 1 with an analysis based on heat transfer data given in the table. Discuss your result. 10.16 Compare the reactor temperature rise shown in Table 10.2 for the 1300-MWe PWR with an analysis based on heat transfer data given in the table. Discuss your result. 10.17 Based on the data given in Table 10.2, determine the heat transfer rating for the Oconee Unit 1 PWR steam generator, and evaluate the average reactor-loop heat transfer loss rate and fraction. Discuss your result. 10.18 Based on data in Table 10.2, estimate the turbine-loop thermodynamic conditions (temperature and pressure) and power delivered by a simple Rankine-cycle steam turbine with 85% efficiency for the 1300-MW PWR. 10.19 Based on the data given in Table 10.2, determine the heat transfer rating for the 1300-MWe PWR steam generator, and evaluate the average reactor-loop heat transfer loss rate and fraction. Discuss your result. 10.20 Develop a thermal design for a 2000-MWe pressurized water reactor core with 0.4 in. diameter, 15-ft-long fuel rods having an average linear power output of 16 kW/ft. Assume an average film coefficient of 4500 Btu/hr-ft2-R. Prepare a report.

437

CHAPTER ELEVEN Energy System Alternatives Part 1. Electromagnetic Principles, Batteries, and Fuel Cells

11.1 Introduction So far we have considered energy conversion systems that are in widespread use. In this chapter we turn our attention to systems that, to date, have not had major impacts in power generation but appear to have the potential to profoundly influence society within the next few decades. Most of the preceding study has examined power and propulsion systems that were primarily the outgrowth of thermodynamics and fluid mechanics. We now consider systems among which are several of a fundamentally different character. To better understand these systems, we must invest a little more time in reviewing electromagnetic fundamentals. In earlier chapters we considered numerous systems that employ the energy conversion chain: Energy source Y Heat Y Mechanical energy Y Electricity In this chapter we are concerned with several conversion processes in which the heatto-mechanical energy transformation link is not essential: Energy source Y Electricity Such processes, in which the source energy is converted directly to electricity, are called direct energy conversion processes. While direct energy conversion techniques other than those discussed here exist, we will concentrate on a few that show potential for large-scale power production and that could reach commercialization in the next few decades. Specifically, we focus on fuel cells, solar photovoltaics, and magnetohydrodynamic systems. In addition, the application of batteries and hydrogen as energy storage media are considered. Let us first develop some principles needed in the analysis of these systems. It would be advantageous for students to relate the fundamental material reviewed in the following sections to chemistry and physics texts with which they are familiar.

438 11.2 Review of Electromagnetic Principles A few principles of electromagnetics, fundamental to alternative energy systems, are now reviewed. We will work almost exclusively with SI units in this chapter, except for a few problems framed in the English system. Charge A unit of electrical charge, q, is the coulomb [C]. The electron has a negative charge of 1.6×10 -19C and a mass of 9.1×10 -31kg. Even the smallest positively charged particle, the proton or hydrogen ion, has a mass more than a thousand times that of the electron. As a result, the electron is much more mobile than neighboring ions and is usually the dominant charge carrier. However, because electromagnetic theory is usually based on the concept of the positive charge, we will think of an electric current as a flow of positive charge, even though motion of electrons in the opposite direction usually transports the charge. Potential and Field A positively charged particle influenced by other charged particles may be thought of as being in a force field created by these particles. This field, known as an electric field, may be represented by nonintersecting, usually curved surfaces of constant potential (V) in three-dimensional space, as sketched in Figure 11.1. Consider a charged particle located between two potential surfaces of potential difference V2 – V1. A force is exerted on the particle in a direction perpendicular to the local potential surface. The larger the potential difference between the two surfaces, the larger the force. Also, the smaller the distance between surfaces of fixed potential difference, the larger the force. The direction of the force on positively charged particles is from high potential to low, and is oppositely directed for negative particles. It was indicated that the strength of the force acting on a charged particle between two potential surfaces depends on both the magnitude of V2 – V1 and how close together the surfaces are. An electric field vector is defined to account for both of these factors. In one dimension, the electric field intensity Ex is the negative of the rate of change of potential with respect to distance, – dV/dx, the limit of the local potential difference per unit distance in the direction normal to the potential surface. Thus Ex = – dV/dx . – lim (V2 – V1)/*x

[V/m]

(11.1)

For a three-dimensional potential field given by V = V(x, y, z), the electric field intensity is a vector, E (vectors are represented by boldface symbols here) which in terms of Cartesian unit vectors i, j, k is E = Exi + Ey j + Ez k

[V/m]

(11.2)

439

The electric field vector is related to the potential function V(x, y, z) by E = – LV (x, y, z) = – (iMV/Mx + jMV/My + kMV/Mz)

[V/m]

(11.3)

where the derivatives in Equation (11.3) are partial derivatives. For a one-dimensional potential, V = V(x), the derivatives with respect to y and z vanish; and the remaining partial derivative becomes the total derivative dV/dx, consistent with Equation (11.1). The electric field vector gives the magnitiude and direction of the force on a positive charge in three-dimensional space. The force on the charge due to the electric field is explicitly represented by the vector equation F = qE

[N]

(11.4)

Equation (11.4) indicates that a force of one newton is exerted on a charge of one coulomb in an electric field of one volt per meter. Hence one coulomb is one newton-meter per volt, or one joule per volt [1 C = 1 N-m/V = 1 J/V].

440 The work done to move a positive charge a small distance dl against the electric field (to a higher potential) is given by the scalar product of the force –F and the differential distance that the charge moves, dl = dxi + dyj + dzk. Using Equation (11.3), the work is w = –IF@dl = – qIE@dl = q (V 2 – V1)

[C-V]

(11.5)

Work is done on the positive charge when the differential distance (dx in Figure 11.1) has a component parallel to the force (and the electric field vector). Consistent with the observation made in connection with the units of Equation (11.4), Equation (11.5) shows that a coulomb-volt is the same as the usual measure of work, the newton-meter or joule. When the electric field is specified it becomes unnecessary to think in terms of the associated charge distribution. The electric field may give rise to or resist motions of positive charges or free electrons in the space occupied by the field, just as the original charge distribution would have. Current and Power A motion of charges through an electrically conducting medium is called a current. The current is the rate at which positive charge passes through a given surface. It is measured in coulombs per second [C/s] or amperes [A] I = dq/dt

[A]

(11.6)

A current flow produced by an electric field is in the direction of the field and thus is in the direction of decreasing potential, as shown in Figure 11.2. Correspondingly, electrons flow from low to high potential in a direction opposite to the conventional current. Power is the rate at which work is done. Using Equation (11.5) Power = dw/dt = Vdq/dt = VI

[W]

(11.7)

where we now use V for potential difference, as is common. Here it is seen that the unit of power, the watt [W], is the equivalent of the product of amperes and volts [A-V]. Current flows in a medium because of the forces exerted on the charges. For a given potential difference, the current is smaller or larger depending on the ease with which the charges pass through the medium. Thus the current is proportional to the applied potential difference and the conductance, G, of the conducting medium. The conductance is usually expressed in terms of its inverse, the resistance R = 1/G. Resistance is defined by Ohm's Law: R = V / I = 1/G

[5]

(11.8)

441

Ohm’s Law defines the unit of resistance, the ohm, as the ratio of a volt to an ampere 5 =V/A. Thus the unit of conductance is the inverse ohm or mho 5 –1. EXAMPLE 11.1

A 12-V battery produces a current of 7 A in a load resistor. Estimate the power delivered to the resistor by the electric field created by the terminals and the magnitude of the resistance. Explain the phenomena in terms of current and electron flow. Solution

The resistive load has a resistance of R = V/I = 12 / 7 = 1.71 5 The power output of the battery is IV =7 × 12= 84 W.

Current flows from the positive terminal through the load resistor to the negative terminal because of work done by the electric field applied by the battery terminals. In reality, work is done on electrons passing from the negative to the positive terminals. Chemical reactions within the battery provide the energy to move the electrons from the positive to the negative terminals within the battery. _____________________________________________________________________

442 The terms conductance and resistance refer to characteristics of a specific configuration of a conducting material. The terms electrical conductivity and electrical resistivity refer to intrinsic electrical properties of the material itself. We shall see that the latter properties often are used in dealing with currents in multidimensional fields. The current density J [A/m2] is defined as the current per unit cross-sectional area normal to the direction of the current. The electric field vector and the current density are related (Figure 11.2) in a generalized form of Ohm's Law by J = FE

[A/m2]

(11.9)

where F is the electrical conductivity of an isotropic conducting medium. In general, when a medium has conductivity varying in direction, the medium is called “anisotropic.” Here we need only consider the isotropic case, where the conductivity of the medium is the same in all directions. Note that Equation (11.9) indicates that F must have units of (A/m2)/(V/m) or (5-m)–1. For a cylindrically-shaped conductor, Equation (11.9) may be expressed in terms of the current and potential difference to show that the electrical conductivity is related to the resistance of the conductor by

F = J/E = (I/A)/(V/L) = L/(RA)

[(5-m)-1]

(11.10)

where L and A are the length and cross-sectional area of the conductor, respectively. Rewriting Equation (11.10) as R = L/(AF), we see that the resistance of a conductor is proportional to its length and inversely proportional to the electrical conductivity and cross-sectional area, as expected. The electrical resistivity D [5-m] is the inverse of the electrical conductivity. Thus the resistance of a cylindrical conductor may be written in terms of resistivity as R = LD/A. Magnetic Field We have seen that the electric field may be thought of as representing the effect of a charge distribution on a test charge. Now we turn our attention to a fundamentally different field. A stationary magnetic field exists between the north and south poles of a bar or horseshoe magnet (Figure 11.3). A magnet may be thought of as a collection of tiny magnetic elements, or domains, each having its own north and south poles, which are aligned to produce the magnetic field of the magnet. Magnetic field lines are always closed or terminate on magnetic poles, as opposed to electric field lines, which terminate on charges. It was observed by Oersted in 1819 that a small magnet deflects when placed in the neighborhood of a wire carrying an electric current. This is explained by the existence of circular magnetic field lines that surround the conductor, as shown in Figure 11.4. Thus not only magnets but electric currents and moving electric charges give rise to

443

magnetic fields. In view of the connection of currents and electric fields, Oersted’s discovery indicates that electric fields may produce magnetic fields. A magnetic field vector B, called the magnetic induction or magnetic flux density, is a measure of the strength of a magnetic field and indicates its direction at any point in space (Figures 11.3 and 11.4). Magnetic flux density is measured in webers per square meter [Wb/m2]. A weber is equivalent to a volt-second. The unit of weber per square meter is also called a tesla.

444 The magnetic field lines around a circular conductor of electricity are continuous concentric circles, as indicated in Figure 11.4. The direction of the magnetic flux or lines of force around a current carrying conductor is given by the direction of the fingers of the right hand when they are wrapped around it with the thumb pointing in the direction of the current flow. Twelve years after Oersted observed that electric fields (which produce charge motion, or electric currents) give rise to magnetic fields, Michael Faraday (1831) discovered that magnetic fields can create electric fields. Faraday observed that a current was induced in a stationary circuit by a changing magnetic field produced by a nearby moving magnet. That changing magnetic field could have been created by (1) a transient current in a nearby stationary circuit, (2) motion of a nearby circuit with a constant current, or (3) motion of a magnet in the vicinity. A current, and thus an electric field, is generated in any conductor that experiences a change in its local magnetic field, regardless of whether the observed change is due to the motion of a magnet or to motion of the conductor itself through a stationary magnetic field. The latter phenomenon is the basis of the operation of an electrical generator in which stationary magnets create a current in a rotating armature. The force exerted on a positive charge in a magnetic field is proportional to the vector product of the charge velocity and the magnetic field strength: F = qu×B

[N]

(11.11)

The vector product indicates that charges in a conductor moving with a velocity u through a magnetic field experience a force in a direction perpendicular to the plane defined by the vectors u and B. As indicated in Figure 11.5, the sense of the force is readily determined by rotating the four fingers of the right hand from the velocity vector to the B vector. The direction of the extended thumb during this operation gives the direction of the force. The force exerted on a coil of an armature of an electric motor is an example of this magnetic action. In general, both electric and magnetic fields occupy the same space, Thus it is not unusual to speak of a single electromagnetic field represented by the electric and magnetic field vectors. If the electric field intensity is ENand the magnetic flux density is B at a given location, the total electromagnetic force acting on a charged particle there, called the Lorentz force, is given by F = q(E' + u×B)

[N]

(11.12)

Thus the charge experiences a force that is the vector sum of the electric field vector and a vector perpendicular to both the charge velocity and magnetic flux vectors. In a coordinate system moving with the velocity u, the Lorentz force would be given by F = qE because the observer in this system sees no motion of the charge. Thus the electromagnetic field felt by the charge may be expressed as an electric field:

445

E = E' + u×B

[N/C]

(11.13)

Here E' is the field existing at the charge location in a stationary coordinate system and E is the effective electric field sensed by the charge. Thus in a moving coordinate system, the influence of the magnetic field appears as part of the electric field vector. While our purpose here has been to establish the concepts and relations necessary for an understanding of the energy conversion systems discussed in this chapter, we would be remiss to move on without a few closing comments. We have discussed the interactions of two invisible fields. These spatially varying fields may be thought of as existing in all of space. These are not merely mathematical abstractions but real phenomena. What better proofs of their reality than the electromagnetic waves that a portable radio picks out of space and converts to music to entertain us as we walk or jog, the warmth of solar radiation and the chill of its absence when passing through shade, and the interplanetary signals from space probes that provide amazing views of other planets? The study of the physics and mathematics of these fields and related phenomena culminated in the formulation of a set of four partial differential equations involving E and B that govern electromagnetics. These famous equations known as Maxwell's equations, were firmly established on a consistent basis by James Clerk Maxwell in

446

1860. They mark one of the great triumphs of science because they are capable of quantitatively defining details of countless phenomena in the areas of electricity, magnetism, and radiation. Their use in detailed research and development in energy conversion devices goes well beyond the brief introduction to electromagnetic theory given here. Interested readers may consult, among others, references 6, 26, and 64 for broader and deeper treatments of the subject. 11.3 Batteries and Fuel Cells Batteries and fuel cells, while structurally and functionally distinct, are based on similar electrochemical principles and technologies. For that reason we have chosen to introduce them together to emphasize their common scientific foundations. They are, however, easily distinguishable, because batteries are devices for the storage of electrical energy, while fuel cells are energy converters with no inherent energy storage capability. A battery contains a finite amount of chemicals that spontaneously react to produce a flow of electrons when a conducting path is connected to its terminals, as

447 shown in Figure 11.6. Fuel cells behave similarly, except that chemical reactants are continuously supplied from outside the cell and products are eliminated in continuous, steady streams. The chemical reactions in batteries and fuel cells are oxidation-reduction reactions. A common terminology exists for both types of cells. Both contain electrode pairs in contact with an intervening electrolyte (the charge-carrying medium), as in Figure 11.6. The electrolyte is a solution through which positively charged ions pass from the anode to the cathode. The negative electrode, called the anode, is where an oxidation reaction takes place, delivering electrons to the external circuit. The positive electrode, where reduction occurs, is called the cathode. Electrons flow through the external load from the negative to the positive electrode, while the conventional current flows in the opposite direction from high to low potential. At the cathode, electrons arriving from the external load are neutralized by reaction with positive ions from the electrolyte. The use of electrical storage cells dates back to the early 1800s. In 1800, Alessandro Volta gave the first description of an electrochemical cell in a letter to the British Royal Society. The invention of the telegraph in the 1830s provided strong motivation for the development and manufacture of this electrical energy source. Though the electrochemical cell provided electrical energy for the telegraph and other inventions, it would be half a century before central plant electrical power became available. In 1839, Sir William Grove demonstrated the fuel cell concept, but it was not until the 1950s that significant progress was made toward the development of useful fuel cells. Batteries designed for a single discharge cycle (flashlight batteries, for example) are called primary cells; those that can be discharged and recharged numerous times are called secondary cells. Fuel cells, on the other hand, utilize chemicals that flow into the cell to provide a more or less continuous supply of direct current. Before considering the exciting possibilities of fuel cells, we will take a brief look at batteries, with emphasis on secondary cells because of their current and future technical importance. Batteries Enormous numbers of batteries are used in a variety of applications, from the tiny button cells in wristwatches, to the cells in flashlights, to the starting, lighting, and ignition (SLI) batteries in automobiles, to the large batteries found in submarines for submersed operations, as well as in emergency power supplies, and remote relay stations. Batteries have been considered for some time for load-leveling power plants as alternatives to pumped hydroelectric and compressed-air storage. While batteries have been used as propulsion power sources for road vehicles for about a century (according to ref. 10 an electric car held the world land speed record of 66 mph in 1899), they have not been competitive with the fossil-fueled internal combustion engine. However,

448

a renewed interest in this application has developed as research in battery technology and the public sensitivity to pollution by internal combustion engine-propelled vehicles has become more advanced. It has been pointed out that batteries are basically devices for storing charge and automatically delivering an electrical current flow, on demand, for a limited time. A battery consists of one or more cells connected in series to provide a particular open circuit voltage or electromotive force (EMF). Each chemical cell typically has an EMF of about two volts, although the exact value is dependent on the specific cell reactants. A Simple Mathematical Model of a Battery. When current is drawn from a battery, the voltage across its terminals decreases below its EMF, largely due to the internal voltage drop associated with the battery's internal resistance. Thus a simple model of a battery involves a constant voltage source in series with a constant resistance. Following Figure 11.7, the terminal voltage for this model is given by the difference between the EMF and the internal resistive potential drop: V = E – IRi [V] (11.14) and the current is I = E /(Ri + Ro)

[A]

(11.15)

EXAMPLE 11.2

The voltage of a lead-acid battery with an EMF of 12.7 V is measured as 11.1 V when it delivers a current of 50 amperes to an external resistance. What are the internal and external resistances? What is the battery voltage and the current flow through a 5-5 resistor? Sketch the voltage–current characteristic.

449

Solution

Solving Equation (11.14) for Ri and Equation (11.15) for Ro, we obtain the internal and external resistances: Ri = (E – V)/I = (12.7 – 11.1)/50 = 0.032 5 Ro = E / I – Ri = 12.7/50 – 0.032 = 0.222 5 With the constant internal resistance established, the current flow through a 5-5 external resistor is then given by Equation (11.15): I = E/(Ri + Ro) = 12.7/(0.032 + 5) = 2.524 A Note that if Ro were infinite in Equation (11.15), the current would vanish and the cell voltage would be equal to the EMF, as in an open circuit. The cell voltage with the 5-5 resistor is given by equation (11.14): or

V = E – IRi = 12.7 – 2.524(0.032) = 12.62 V V = IRo = 2.524(5) = 12.62 V

By Equation (11.14), the voltage characteristic is a straight line (see Figure 11.8) with intercept E and slope –Ri. _____________________________________________________________________

450 While the linear model is useful and informative, it does not conform precisely to real battery characteristics. As in the model, real batteries are capable of providing current and power to a wide range of resistances with little voltage drop for short periods of time. However, the model fails for long periods of time and when large currents are drawn, because it does not account for the finite amount of stored chemical energy and reaction rate limitations encountered in real batteries. In these circumstances, the terminal voltage and power output drop drastically as cell chemicals are consumed. Secondary cell reactions are approximately internally reversible. Thus, by application of an external direct current source, internal battery reactions may be reversed over a period of time in a process called charging. In a conventional automotive application, electrical power for lighting, ignition, radio, and so on is supplied by an engine-driven alternator. Normally, the battery is required only for engine starting and to compensate for possible alternator output deficits when the engine runs slowly. A voltage regulator maintains a steady electrical system voltage and controls battery charging when excess alternator current is available and needed. According to reference 27, the regulator limits the charging voltage of a 12-V battery to about 14.4 V. Battery Performance Parameters. Important parameters characterizing battery performance are the storage capacity, cold-cranking amperes, reserve capacity, energy efficiency, and energy density. The storage capacity, a measure of the total electric charge of the battery, is usually quoted in ampere-hours rather than in coulombs ( 1 amp-hour = 3600 C). It is an indication of the capability of a battery to deliver a particular current value for a given duration. Thus a battery that can discharge at a rate of 5 A for 20 hours has a capacity of 100 A-hr. Capacity is also sometimes indicated as energy storage capacity, in watt-hours. Cold-cranking-amperage, CCA, for a battery composed of nominal 2-V cells is the highest current, in amperes, that the battery can deliver for 30 seconds at a temperature of 0°F and still maintain a voltage of 1.2 V per cell (ref. 27). The CCA is commonly used in automotive applications, where the problem of high engine resistance to starting in cold winter conditions is compounded by reduced battery performance. At 0°F the cranking resistance of an automobile engine may be increased more than a factor of two over its starting power requirement at 80°F, while the battery output at the lower temperature is reduced to 40% of its normal output. The reduction in battery output at low temperatures is due to the fact that chemical reaction rates decrease as temperature decreases. The reserve capacity rating is the time, in minutes, that an SLI battery will deliver 25 amperes at 80°F. This is an indication of how long the battery would continue to satisfy essential automotive operating requirements if the alternator were to fail. Other parameters evaluate batteries with respect to energy for traction or other deep-cycling applications where, in contrast to SLI applications, their state of charge becomes low regularly. The energy efficiency is the ratio of the energy delivered by a

451 fully charged battery to the recharging energy required to restore its original state of charge. For transportation purposes where weight and volume are critical, the energy density is an important parameter. It may be represented on a mass basis, in watt-hours per kilogram, or on a volume basis, in watt-hours per cubic meter. Analysis of a Cell. Consider a reversible cell with a terminal potential difference V discharging with a current I through a variable resistance. The flow of electrons through the load produces the instantaneous electrical power, IV. At low currents the cell voltage is close to the cell EMF and the external work done and power output are small. Suppose that chemical reaction in the battery consumes reactant at an electrode at a rate of nc moles per second. Electrons released in the reaction flow through the electrode to the external load at a rate proportional to the rate of reaction, jnc, where j is the number of moles of electrons released per mole of reactants. It will be seen later that, for a lead-acid battery, two moles of electrons are freed to flow through the external load for each mole of lead reacted. Hence, in this case j = 2. Thus jnc is the rate of flow of electrons from the cell, in moles of electrons per second. There are 6.023×1023 electrons per gram-mole of electrons, and each has a charge of 1.602×10-19 C. Thus the product F = (6.023×1023)(1.602×10-19) = 96,488 C/gram-mole is the charge transported by a gram-mole of electrons. The constant F is called Faraday's Constant, in honor of Michael Faraday, a great pioneer of electrochemistry. The electric current from a cell may then be related to the rate of reaction in the cell as I = jncF

[A]

(11.16)

and the instantaneous power delivered by the cell is: Power = jncFV

[W]

(11.17)

The cell electrode and electrolyte materials and cell design determine the maximum cell voltage. Equations (11.16) and (11.17) show that the nature and rate of chemical reaction control the cell current and maximum power output of a cell. Moreover, it is clear that the store of consumable battery reactants sets a limit on battery capacity. Lead-Acid Batteries. By far the most widely used secondary battery is the lead/sulfuric acid/lead oxide (Pb/H2SO4/PbO2), or lead-acid battery, by virtue of its widespread application in automomobiles for starting, lighting, and ignition and in

452

traction batteries used in rail and street vehicles, golf carts, and electrically powered industrial trucks. An automotive lead-acid battery is shown schematically in Figure 11.9. Today most lead-acid batteries are 12-V designs, assembled as six 2-V cells in series. Each cell consists of a porous sponge lead anode and a lead dioxide cathode (usually in the form of plates) separated by porous membranes in a sulfuric acid solution. The aqueous electrolyte contains positive hydrogen and negative sulfate ions resulting from dissociation of the sulfuric acid in solution: 2H2SO4 => 4H+ + 2(SO4)2 – At the anode, the lead releases two electrons to the external load as it is converted to lead sulfate:

453 Pb(s) + SO4 2 – (aq) => PbSO4 (s) + 2e– This shows that two moles of electrons are produced for each mole of sulfate ions and sponge lead reacted. Thus j = 2 for the lead-acid cell in Equations (11.16) and (11.17). Meanwhile at the cathode, lead dioxide is converted to lead sulfate in the reduction reaction involving hydrogen and sulfate ions from the electrolyte and electrons from the external circuit: PbO2 (s) + 4H+ (aq) + SO4 2 – (aq) +2e – <=> PbSO4 + 2H2O (l) The equations show that the sulfate radical (and thus sulfuric acid) is consumed at both electrodes, and that hydrogen ions combine with oxygen from the cathode material to form water that is retained in the cell. Thus the water formed as the battery discharges makes the sulfuric acid electrolyte more dilute. This is also seen in the net cell reaction, obtained as the sum of the electrode and electrolyte reactions: Pb(s) + PbO2 (s) + 2H2SO4 (aq) <=> 2PbSO4 (s) + 2H2O (l) It is clear that during battery discharge the lead and lead oxide electrodes are converted to lead sulfate. Furthermore, sulfuric acid is consumed and water is produced. The production of water and the consumption of sulfuric acid make the electrolyte more dilute and decrease its specific gravity. The specific gravity drops from about 1.265 when the battery is fully charged to about 1.12 when fully discharged (ref. 27). Thus the cell specific gravity is a measure of the state of discharge of the battery. A simple device called a hydrometer, which temporarily withdraws electrolyte from an open battery, gives a visual indication of the electrolyte specific gravity. The depth of the float in the hydrometer electrolyte indicates its specific gravity and thus the battery state of charge. When any of the reactants is depleted, the battery ceases to function. When the battery is charged by applying an external power source, the electron flow is reversed, the previously displayed reactions also reverse, and water is consumed to produce sulfate and hydrogen ions from the lead sulfate on the electrodes. EXAMPLE 11.3

Consider the battery model of Example 11.2 for the case of 50-A current. Determine the instantaneous power output, in watts, and the rate of consumption of sulfuric acid of the cell, in kilograms per hour. Solution Power = IV = (11.1) (50) = 555 W Two moles of electrons are produced for each mole of sulfuric acid consumed at the anode. From Equation (11.16),

454

nc = I/jF = 50/(2 @ 96,487) = 0.000259 gm-moles/s Because sulfuric acid is consumed at the cathode at the same rate as at the anode, the total rate of acid consumption is 0.000518 gm-moles /s, or 5.18 × 10 –7 kg-moles/s. The molecular weight of sulfuric acid is 98, so the mass rate of acid consumption is (5.18 × 10–7)(98) = 5.078 × 10–5 kg/s or (5.078 × 10 –5 )(3600) = 0.1828 kg/hr _____________________________________________________________________ Battery Applications. It is important to distinguish between SLI and traction batteries. The design of the SLI battery is governed primarily by the cranking requirements of the vehicle, usually a high current flow for a brief time interval. Once the engine is started, the alternator provides power for vehicle operating loads and to recharge the battery. During normal driving the alternator satisfies the power needs of lighting, ignition, and other electrical functions. Thus a SLI battery is usually close to fully charged and seldom deeply discharged.

455 The traction battery, on the other hand, usually has no energy converter available to continuously maintain the fully charged level. Thus the traction battery must be capable of operation from fully charged to almost fully discharged over a period of several hours. At least eight hours daily are normally required to return the battery to its fully charged level. This type of secondary cell is appropriate for vehicle propulsion and power plant peak-shaving applications. In the latter instance, off-peak base-load power is used to charge the battery. Later, during peak-demand periods, the battery returns a large fraction of the charging energy to the system. Such a battery is designed for thousands of charge and discharge cycles. According to reference 4, a 10-MW, 40 MW-hr lead-acid battery storage facility was built at the Southern California Edison Company at Chino. It was designed for load-leveling use for at least 2000 deep-discharge cycles and 80% energy efficiency. Six of the cells were tested at 50°C and achieved 2300 cycles while retaining 108% of rated capacity (ref. 33). The facility, shown in Figure 11.10, uses strings of 1032 deep-discharge, 2-V lead-acid cells connected in series to provide a nominal bus voltage of 2000 V DC. A total of 8256 2-V cells, each weighing 580 pounds, are used to store 40 MW-hr for a 4-hr daily discharge cycle. A power conditioning system acts as both an inverter to provide AC output to the utility and as a rectifier to convert utility AC to DC for battery charging. The system is designed to utilize the responsiveness of battery storage to deal with wide load swings. The Chino facility is said to be able to swing from 10-MW discharge to 10-MW charge in about 16 milliseconds. According to references 4 and 11, lead-acid batteries can now deliver energy densities of more than 40 W-h/kg. This is less than the expected performance of several other battery types. Nickel-iron batteries with an energy density of 50.4 W-hr/kg, for instance, were among 10 types endurance-tested in an electric vehicle. A van with nickel-iron batteries logged over 44,000 miles, as compared to fewer than 27,000 miles for the best performance by a lead-acid-battery-equipped van. Nickel-iron batteries, while noted for ruggedness and long life, suffer from a low energy efficiency of about 60% (ref. 34). Space limitations allow the present discussion to only hint at the more than a century of research and the sophistication of the science and engineering of batteries. Fuel Cells Whereas batteries store energy in the form of chemical energy and subsequently transform it to electricity, fuel cells are capable of continuous transformation of chemical energy into electrical energy without intermediate thermal-to-mechanical conversion. Fuel cells may be thought of as chemical reactors similar to storage batteries, but with external supplies of fuel and oxidizer that react for indefinitely long time periods without substantial change in cell materials. Thus fuel cells do not run down or require recharging. As in a storage battery, reactions take place at electrodes, giving rise to a

456

flow of electrons through an external circuit as depicted in Figure 11.11. An electrolyte, between the electrodes, again provides a medium for ion transfer. This allows electrons to flow through the external load while ion transport through the electrolyte maintains overall electrical neutrality of the cell. An important element of fuel cell design is that, like large batteries, they are built from a large number of identical unit cells. Each has an open-circuit voltage on the order of one volt, depending on the oxidation-reduction reactions taking place. The fuel cells are usually built in sandwich-style assemblies called stacks. The schematic in Figure 11.12 shows crossflows of fuel and oxidant through a portion of a stack.

457

Electrically conducting bipolar separator plates serve as direct current transmission paths between successive cells. This modular type of construction allows research and development of individual cells and engineering of fuel cell systems to proceed in parallel. A photograph of a 32-kW phosphoric acid fuel cell stack is shown in Figure 11.13(a). A conceptual design of a nominal 100-kW stack using three of the 32-kW stacks appears in Figure 11.13(b). Phosphoric acid fuel cells ideally operate with hydrogen as fuel and oxygen as oxidizer. To be economically practical in most applications, these fuel cells will probably require the use of a hydrocarbon fuel as a source of hydrogen and air as an oxidizer. Consider the basic chemical reactions in a hydrogen-oxygen fuel cell as shown in Figure 11.11. Hydrogen molecules supplied at the anode are oxidized (lose electrons), each forming two hydrogen ions that drift through an electrolyte to the cathode. Electrons liberated from the hydrogen at the anode pass through an external circuit to the cathode, where they combine in a reduction reaction with the H+ ions from the electrolyte and an external supply of oxygen to form water. The electrode and overall cell reaction equations are:

458 H2 Y 2H+ + 2e–

at the anode

2H+ + 2e– + 0.5O2 Y H2O

at the cathode

H2 + 0.5O2 Y H2O

overall cell reaction

Thus the sole reaction product of a hydrogen–oxygen fuel cell is water, an ideal product from a pollution standpoint. In manned-space-vehicle applications the water is used for drinking and/or sanitary purposes. In addition to electrons, heat is also a reaction product. This heat must be continuously removed, as it is generated, in order to keep the cell reaction isothermal. In the fuel cell, as in the battery, the reactions at the electrodes are surface phenomena. They occur at a liquid-solid or gas-solid interface and therefore proceed at a rate proportional to the exposed area of the solid. For this reason porous electrode materials are used, frequently porous carbon impregnated or coated with a catalyst to speed the reactions. Thus, because of microscopic pores, the electrodes effectively have a surface area many times their visible area. Any phenomenon that prevents the gas from entering the pores or deactivates the catalyst must be avoided if the cell is to function effectively over long time periods. Because of the reaction rate–area relation, fuel cell current and power output increase with increased cell area. The power density [W/m2] therefore is an important parameter in comparing fuel cell designs, and the power output of a fuel cell can be scaled up by increasing its surface area. The electrolyte acts as a medium for ion transport between electrodes. The rate of passage of positive charge through the electrolyte must match the rate of electron arrival at the opposite electrode to satisfy the physical requirement of electrical neutrality of the discharge fluids. Impediments to the rate of ion transport through the electrolyte can limit current flow and hence power output. Thus care must be taken in design to minimize the length of ion travel path and other factors that retard ion transport. Fuel Cell Thermodynamic Analysis and Efficiencies. The theoretical maximum work of an isothermal fuel cell (or other isothermal reversible control volume) is the difference in Gibbs’ function (or free energy) of the reactants (r) and the products (p). Because the enthalpy, entropy, and temperature are all thermodynamic properties, the Gibbs’ function, g = h – Ts, is also a thermodynamic property. From the First Law, the work leaving the control volume is w = hr – hp + q

[J/g-mole]

(11.18)

For a reversible isothermal process, q = T(sp – sr), and the maximum work can be written as wmax = hr – hp + T(sp – sr) = gr – gp

[J/g-mole]

(11.19)

459 Thus the maximum work output of which a fuel cell is capable is given by the decrease in the Gibbs function. But the maximum energy available from the fuel in an adiabatic steady-flow process is the difference in inlet and exit enthalpies, hr – hp. Defining the fuel cell thermal efficiency as wmax /(hr – hp), we obtain

0th = (gr – gp)/(hr – hp) = 1 – T(sr – sp)/(hr – hp)

[dl]

(11.20)

The enthalpies and Gibbs function may be obtained from the JANAF tables for the appropriate substances (ref. 77). EXAMPLE 11.4

Evaluate the open-circuit voltage, maximum work, and thermal efficiency for a direct hydrogen-oxygen fuel cell at the standard reference conditions for the JANAF tables. Consider the two cases when the product water is in the liquid and vapor phases. Solution

The maximum work of the fuel cell is given by equation (11.19). The thermodynamic properties needed are obtained from the formation properties of the JANAF thermodynamic tables: hp(l) = hf [H2O(l)] = – 285,830 kJ/kg-mole hp(g) = hf [H2O(g)] = – 241,826 kJ/kg-mole hr = hf [H2] + 0.5hf [O2] = 0.0 kJ/kg-mole gp (l) = gf [H2O(l)] = – 237,141 kJ/kg-mole gp(g) = gf [H2O(g)] = – 228,582 kj/kg-mole gr = gf [H2] + 0.5gf [O2] = 0.0 kJ/kg-mole Thus the maximum work for liquid-water product is: wmax(l) = gr – gp (l) = 0 – (– 237,141) = 237,141 kJ/kg-mole and for water-vapor product is wmax(g) = gr – gp (g) = 0 – (– 228,582) = 228,582 kJ/kg-mole of H2 consumed or water produced in the reaction. The theoretical ideal open-circuit voltage of the cell may also be determined from the Gibbs function using, for liquid-water product, EMF(l) = wmax(l)/( jF) = 237,141×103/(2×9.6487×107) = 1.229 V

460 For water-vapor product: EMF(g) = wmax(g)/( jF) = 228,582x103/(2×9.6487×107) = 1.18 V By Equation (11.20), the thermal efficiency with liquid-water product is

0th(l) = 237,141/285,830 = 0.8297 and with water-vapor product is

0th(g) = 228,582/241,826 = 0.945 Regardless of the phase of the product water, a large fraction of the available enthalpy of the reactant flow is converted to work in the ideal isothermal hydrogen-oxygen fuel cell. Because the fuel cell operates as a steady-flow isothermal process and not in a cycle, the Carnot efficiency limit does not apply. _____________________________________________________________________ Fuel cells, like other energy conversion systems do not function in exact conformance with simplistic models or without inefficiencies. Open-circuit potentials of 0.7–0.9 volts are typical, and losses or inefficiencies in fuel cells, often called polarizations, are reflected in a cell voltage drop when the fuel cell is under load. Three major types of polarization are: • Ohmic Polarization: The internal resistance to the motion of electrons through electrodes and of ions through the electrolyte. •

Concentration Polarization: Mass transport effects relating to diffusion of gases through porous electrodes and to the solution and dissolution of reactants and products.

•

Activation Polarization: Related to the activation energy barriers for the various steps in the oxidation-reduction reactions at the electrodes.

The net effect of these and other polarizations is a decline in terminal voltage with increasing current drawn by the load. This voltage drop is reflected in a peak in the power-current characteristic. The high thermal efficiencies predicted in Example 11.4 are based on operation of the fuel cell as an isothermal, steady-flow process. Although heat must be rejected, the transformation of chemical energy directly into electron flow does not rely on heat rejection in a cyclic process to a sink at low temperature as in a heat engine, which is why the Carnot efficiency limit does not apply. However, inefficiency associated with the various polarizations and incomplete fuel utilization reduce overall cell conversion efficiencies to well below the thermal efficiencies predicted in the example.

461 Fuel cell conversion efficiency, 0fc, is defined as the electrical energy output per unit mass (or mole) of fuel to the corresponding heating value of the fuel consumed. Defining the voltage efficiency, 0V, as the ratio of the terminal voltage to the theoretical EMF, V/EMF, and the current efficiency, 0I , as the ratio of the cell electrical current to the theoretical charge flow associated with the fuel consumption, we find that the conversion efficiency is related to the thermal efficiency by

0fc = (V/EMF)[I /( jFnc)]0th = 0V 0I 0th Thus the voltage efficiency accounts for the various cell polarizations previously described. Fuel cell conversion efficiencies in the range of 40–60% have been demonstrated and are possible in large-scale power plants. EXAMPLE 11.5

The DC output voltage of a hydrogen-oxygen fuel cell with liquid-water product is 0.8 V. Assuming a current efficiency of 1, what is the fuel cell conversion efficiency? Solution

From Example 11.4, the thermal efficiency is 0.8297 and the EMF is 1.229 V. The fuel cell conversion efficiency is then

0fc = 0V 0I 0th = (0.8/1.229)@[email protected] = 0.54 _____________________________________________________________________ If heat rejected in high-temperature fuel cells is used and not wasted, overall energy efficiencies as high as 80% may be attainable. In cells that operate at high temperatures, there is an opportunity to use the thermal energy of the cell reaction products for cogeneration purposes or in a combined cycle, with a steam turbine or a gas turbine using the waste heat. It is evident that the power output of fuel cells depends on the rate at which the reactants are consumed in the cell, just as the output of other steady-flow energy conversion systems depend on fuel and air supply rates. In low-temperature fuel cells, expensive catalysts (such as platinum) are required, to increase the rates of chemical reaction at the electrodes so that the fuel cells will be small enough to achieve reasonable power densities (per unit mass and volume). Because chemical reaction rates increase with temperature, high-temperature fuel cells can achieve satisfactory reaction rates without catalysts or with reduced quantities. This advantage of high temperature may be offset by increased corrosion rates and other adverse effects of high temperature that reduce the fuel cell’s active life. For utility applications, long, active cell life (about 40,000 hours) is considered an

462

important design characteristic. It has been suggested, however, that the fuel cell stack may be a relatively small part of the cost of generation in large combined-cycle fuel cell power plants designed for long life, so that it may be acceptable to replace fuel cell stacks after shorter operating periods than other plant components, as is done with fuel rods in nuclear reactors. While space vehicle fuel cells successfully operate on pure hydrogen and oxygen, stationary and vehicle power plants are more likely to be viable if they consume hydrocarbon fuels and air. Thus the goals of major fuel cell research and development programs focus on systems that incorporate the use of natural gas or other hydrogen-rich fuels. A utility fuel cell system appears schematically as in Figure 11.14. A fuel processor upstream of the fuel cell stack is used to prepare the incoming fuel. Its primary functions are conversion of the incoming fuel to a hydrogen-rich gas and removal of impurities such as sulphur. For natural gas, syngas, or other gaseous fuels,

463 the hydrocarbons may be converted by a steam-reforming process such as that for methane: CH4 + H2O Y CO + 3H2 to produce a mixture of hydrogen and carbon monoxide. The carbon monoxide in the products can then be converted to CO2 and produce more hydrogen in the so-called shift reaction: CO + H2O Y CO2 + H2 The direct current generated by a fuel cell for utility use must be converted to alternating current by a power conditioner, as the figure suggests. High-efficiency solid, state inverters are used for this function. Following Figure 11.14, the overall fuel cell system efficiency is then the product of a power conditioner efficiency (AC power output / DC power output), a fuel processing efficiency (enthalpy change of hydrogen / heating value of supply fuel), and the ratio of the sum of the fuel cell DC output and bottoming cycle power to the hydrogen enthalpy change. Fuel Cell Types. Table 11.1 indicates the major fuel cell types, their convenient abbreviations, and their nominal operating temperatures. Fuel cells are usually classified according to their electrolytes, as indicated by the types in the table. While the alkaline and polymer electrolyte fuel cells were successfully demonstrated in the NASA Gemini (1-kW PEFC), Apollo (1.5-kW AFC), and Space Shuttle Orbiter (7-kW AFC) programs, these were applications with short design operating lives that used expensive catalysts to attain satisfactory reaction rates and thus may be appropriate only to space and military applications and certain limited transportation uses. Table 11.1

Fuel Cells of Current Technical Interest

Type

Abbreviation

Operating Temperature

Polymer electrolyte fuel cell

PEFC

80°C

Alkaline fuel cell

AFC

100°C

Phosphoric acid fuel cell

PAFC

200°C

Molten carbonate fuel cell

MCFC

650°C

Solid oxide fuel cell

SOFC

1000°C

464

The last three fuel cell types listed in the table (PAFC, MCFC, and SOFC) are widely believed to be most likely to achieve commercialization for stationary power production early in the 21st century. Figure 11.15 summarizes the reactants and flow paths of these three types. The PAFC is most likely to be commercialized first, but significant progress is being made in the advanced high-temperature technologies represented by the MCFC and the SOFC. Large-scale phosphoric acid fuel cell power plants have been explored based on fuel cell stacks developed by International Fuel Cells, Inc. (IFC). A basic 10-ft2, 700kW unit is shown in Figure 11.16, together with its pressure vessel. According to reference 36, The Tokyo Electric Power Company 11-MW Goi Station had twenty 700-kW fuel cell assemblies with an expected heat rate of 8300 Btu/kW-hr using liquefied natural gas as fuel. More recently, the IFC website (ref. 81) states that the PC 25TM (a 200-kW unit) is one of five units comprising the “largest commercial fuel cell system” in the United States, which became operational in the year 2000 for the U.S. Postal Service in Anchorage, Alaska. The PC 25TM operates with hydrogen, natural gas, propane, butane, naptha, or waste gases as its energy source. The website states that IFC delivered its 200th PC25 in the year 2000, with 3.5 million total operating hours. The PC 25 originally went into production in 1991.

465

Reference 82 states that the ONSI Corporation (a subsidiary of IFC) PC 25 fuel cell “delivers 35 to 40 percent electrical efficiency and up to 80 percent total efficiency with heat recovery.” A 1997 DOE report (ref. 87) explores the potential for a PC 25 fuel cell cogeneration system retrofit in a large office building. The first molten carbonate fuel cell power plant, a 100-kW plant operated by Pacific Gas and Electric, despite being conservatively designed for simplicity and reliability, was expected to have a heat rate of about 7000 Btu/kW-hr without a bottoming cycle. This pilot plant, located at San Ramon, Calif., uses natural gas as fuel and aimed for a cell lifetime of 25,000 hours (refs. 15 and 36). Reference 78 indicates that verification and durability testing began in June 1991 leading to a series of 2-MW demonstration plants starting in 1994, with first commercial units scheduled for 1997. The Santa Clara Fuel Cell Demonstration project, fueled by natural gas, started construction in 1994, produced first power in March 1996, and concluded the demonstration in April 1997. Reference 84 indicates that it was “the largest and most efficient fuel cell power plant ever operated in the U.S...” Reference 85 announces the successful completion of the grid-connected operation of a 250-kW, Fuel Cell Energy, Inc., demonstration plant as of June 2000. The plant logged more than 11,800 hours and generated more than 1.8 million kilowatt-hours of electricity. The plant is to be refitted to demonstrate a very-high-efficiency combinedcycle operation with a gas turbine. Perhaps the most promising fuel cell type is the solid oxide fuel cell. The use of a solid electrolyte, as shown in Figure 11.17, allows a radical departure from the plate or sandwich type of design. The individual cells are of a closed-end cylindrical design with

466

fuel manifolded to flow over the outside electrode and air on the inside communicating with the inner electrode through a porous cylindrical support structure. The clever fuel, air, and exhaust manifolding and electrical connections of this design are apparent in Figures 11.17 and 11.18. The high-temperature design allows operation without catalysts and provides an excellent opportunity to use a bottoming cycle. Beyond high efficiency, there are several additional major characteristics of fuel cells that make them particularly attractive as energy conversion systems. First, fuel cell operation has been shown to occur with very low levels of environmental pollution (ref. 15). It has been projected that commercial fuel cells may attain pollution levels that are factors of ten below those of new conventional coal-burning power plants using the best available pollution control equipment (ref. 16). A second important characteristic is that, because most fuel cells operate with a hydrogen-rich fuel or pure hydrogen, the fuel can be obtained from a number of sources, such as petroleum, natural gas, naptha, methanol, and syngas made from coal. Thus a commercial stationary power plant will have a chemical processor upstream of the fuel cell stack as in Figure 11.14. The reforming and shift reactions likely to be used will produce carbon dioxide as a product. This somewhat tarnishes the environmental

467

beauty of the hydrogen-oxygen fuel cell concept. The chemical process to obtain hydrogen from some of the fuels may in some cases take place within the fuel cell stack. Electrolysis of water may also be used to produce hydrogen and oxygen for fuel cell use. This provides an opportunity for energy storage by using electricity generated by base-loaded nuclear or conventional fossil fuel plants operating at off-peak hours. The hydrogen and, when appropriate, oxygen could then be reacted in fuel cells to generate electricity during periods of peak electrical demand at the central plant site or at other locations. It has been pointed out that the modular nature of fuel cells and of fuel cell stacks is an important characteristic of this technology. Because of this modularity, a given cell design could lead to a range of power plant designs such as large utility-scale plants, smaller environmentally acceptable neighborhood plants, and industrial facilities providing combined heat and power. It has been demonstrated that scaling based on fuel cell modularity does not necessarily adversely influence stack efficiency. Their compact construction suggests that stacks and other power plant components could be factory-built and assembled in plug-in style at plant sites. The modular nature of the fuel cell offers real possibilities for distributed power production, contrary to the historic long-term trend in electrical power production.

468 Because of the expected short manufacturing time for fuel cells, their apparent environmental acceptability, and their ability to function well in a variety of sizes, they offer utilities alternatives to the long construction times and the financial risks of large fossil fuel and nuclear plants that have required as much as ten years to bring into service. Thus significant advantages may be expected to accrue from the ability to construct small, high-efficiency, environmentally acceptable plants in areas where demand is growing. While its expected high efficiency commends its use as a utility base-load plants with constant power output, the fuel cell also offers the potential for serving varying loads without substantial losses in performance. Thus fuel cells may prove useful in load-following applications as well as in base-load application. Another potential application for fuel cells is in the large vehicle transportation sector—buses, trucks, locomotives, and the like. A likely entry for the fuel cell may be in public transportation vehicles in large cities with significant pollution problems. Reference 18 reports on studies of alternatives to battery-powered trucks in Paris, where almost 200 electric garbage trucks are currently in use. It was found that fuelcell-powered trucks and, to a lesser degree, hybrid fuel cell/flywheel trucks (the flywheel stores extra energy for occasional vehicle accelerations while the fuel cell handles the base-load propulsion and flywheel reenergizing) would have significantly lower operating costs per ton of garbage per day than the existing battery-powered trucks. Studies also showed that diesel-powered buses were only about 10% to 20% cheaper to operate in Europe than fuel-cell-powered buses. A simulation study, reported in reference 67, considered hypothetical fuel-cellpowered bus designs operating on three urban bus routes. A bus design specification, resulting from the study, combined two 30-kW phosporic acid fuel cells with a 38-kWhr battery to power a 50-kW electric traction motor. The motor transmits power to a rear wheel-axle assembly through a conventional transmission. The specification was for a 23 passenger bus for use at speeds up to 45 mph. In April 2000, Daimler-Chrysler announced (ref. 88) an offering of 20-30 buses employing 250-kW fuel cells operating on hydrogen fuel for delivery in 2002. In these vehicles, hydrogen would be stored in eight tanks mounted on the roof over the forward axle. These studies suggest that fuel cell technology is maturing. In many cases the economics of these systems is not yet competitive with that of existing systems. However, further advances may soon make fuel cells competitive for some transportation and stationary applications. Reference 68 gives further evidence that the quest for commercial fuel cell technology is truly international in scope.

469 Bibliography and References

1. Angrist, Stanley W., Direct Energy Conversion, 3rd ed. Newton, Mass.: Allyn & Bacon, 1976. 2. Spitzer, Lyman, Physics of Fully Ionized Gases, New York: Wiley-Interscience, 1956. 3. Appleby, A. J., "Advanced Fuel Cells and Their Future Market," Ann. Rev. Energy, 13 (1988): 267–316. 4. McLarnon, Frank R., and Cairns, Elton J., "Energy Storage," Ann. Rev. Energy, 14 (1989): 241–271. 5. Mellde, Rolf W., "Advanced Automobile Engines For Fuel Economy, Low Emissions, and Multifuel Capability," Ann. Rev. Energy, 14 (1989): 425–444. 6. Jackson, John David, Classical Electrodynamics. New York: Wiley, 1962. 7. Fraas, Arthur P., Engineering Evaluation of Energy Systems. New York: McGraw-Hill, 1982. 8. Vincent, Colin A., et al., Modern Batteries. Baltimore: Edward Arnold, 1984. 9. Sperling, Daniel and DeLuchi, Mark A., “Transportation Energy Futures,” Ann. Rev. Energy, 14 ( 1989): 375–424. 10. Fickett, A. P., " Fuel Cell Power Plants," Scientific American, December 1978: 70. 11. Society of Automotive Engineers, Automotive Handbook, 4th ed., Robert Bosch GmbH, 1996. 12. Mayfield, Manville J., Beyma, Edmund F., and Nelkin, Gary A., "Update on U.S. Department of Energy’s Phosphoric Acid Fuel Cell Program." Proceedings of the Sixteenth Energy Technology Conference, February 28–March 2, 1989, Government Institutes, Inc., pp. 184–196. 13. Myles, K. M., and Krumpelt, M., "Status of Molten Carbonate Fuel Cell Technology." Proceedings of the Sixteenth Energy Technology Conference, February 28– March 2, 1989, Government Institutes, Inc., pp. 197-204. 14. Bates, J. Lambert, "Solid Oxide Fuel Cells: A Materials Challenge." Proceedings of the Sixteenth Energy Technology Conference, February 28–March 2, 1989, Government Institutes, Inc., pp. 205–219.

470 15. Gillis, Edward, "Fuel Cells," EPRI Journal, September 1989: 34–36. 16. Harder, Edwin L., Fundamentals of Energy Production. New York: Wiley, 1982. 17. Sissine, F., "Fuel Cells for Electric Power Production: Future Potential, Federal Role and Policy Options," in Fuel Cells:Trends In Research and Applications, Ed. A. J. Appleby. Washington: Hemisphere, 1987. 18. Appleby, A. J., "Phosphoric Acid Fuel Cells," in Fuel Cells:Trends In Research and Applications, Ed. A. J. Appleby. Washington: Hemisphere, 1987. 19. Appleby, A. J. (Ed.), Fuel Cells:Trends In Research and Applications, Washington: Hemisphere, 1987. 20. Ketelaar, J. A. A., "Molten Carbonate Fuel Cells," Fuel Cells:Trends In Research and Applications, Ed. A. J. Appleby. Washington: Hemisphere, 1987. 21. Kinoshita, K., McLarnon, R. R., and Cairns, E.J., Fuel Cells: A Handbook., U.S. Department of Energy METC-88/6096, 1988. 22. Berry, D.A., and Mayfield, M.J., Fuel Cells, Technology Status Report, U.S. Department of Energy METC-89/0266, 1988. 23. Anon., Fuel Cell Systems Program Plan, FY 1990, U.S.Department of Energy, FE-0106P, October 1989. 24. Huber, W. J., Proceedings of the First Annual Fuel Cells Contractors Review Meeting, U.S.Department of Energy METC-89/6105, May 1989. 25. Anon., 1988 Fuel Cell Seminar. Washington, D.C.: Courtesy Associates, 1988. 26. Feynman, Richard P., Leighton, Robert B. and Sands, Matthew, The Feynman Lectures on Physics, Vol. 2. Reading, Mass.: Addison-Wesley, 1964. 27. Anon., Battery Service Manual. Chicago: Battery Council International, 1987. 28. Kotz, John C., and Purcell, Keith F., Chemistry and Chemical Reactivity. Philadelphia: Saunders, 1987. 29. Goodman, Frank, "Power Electronics for Renewables," EPRI Journal, January/February 1988: 44–47.

471 30. Schaefer, John, "Photovoltaic Operating Experience," EPRI Journal, March 1988: 40–42. 31. Peterson, Terry, "Amorphous Silicon Thin Film Solar Cells," EPRI Journal, June 1988: 41–44. 32. Dostalek, Frank, "High Concentration Photovoltaics," EPRI Journal, March 1989: 46–49. 33. Morris, Douglas, "The Chino Battery Facility," EPRI Journal, March 1988: 46–50. 34. Purcell, Gary, and Driggans, Rick, "Electric Vehicle Testing at TVA," EPRI Journal, March 1988: 44–46. 35. Anon., The Storage Battery. Horsham, Pa.: Exide Corp., 1980. 36. Makanski, Jason, "Fuel Cells Extend Boundaries of Process/Power Integration," Power, May 1990: 82–86. 37. Chapman, Alan J., Heat Transfer, New York: Macmillan, 1968. 38. Swanson, Theodore D., "Large Scale Photovoltaic System Design Considerations," The Handbook of Photovoltaic Applications. Atlanta, Ga.: Fairmont Press, 1986, pp. 19–32. 39. Van Overstraeten, R.J., and Mertens, R.P., Physics, Technology, and Use of Photovoltaics. Boston: Adam Hilger, 1986. 40. McDaniels, David K., The Sun: Our Future Energy Source, 2nd ed. New York: Wiley, 1984. 41. Feynman, Richard P., Leighton, Robert B. and Sands, Matthew, The Feynman Lectures on Physics, Vol. 3, Reading, Mass.: Addison-Wesley, 1965. 42. Takahashi, K. and Konagai, M., Amorphous Silicon Solar Cells. New York: Wiley, 1986. 43. Duffie, John A., and Beckman, William A., Solar Engineering of Thermal Processes. New York: Wiley, 1980. 44. Rosa, Richard J., Magnetohydrodynamic Energy Conversion, rev. ed. Washington: Hemisphere, 1987.

472 45. Gregory, Derek P., “The Hydrogen Economy,” Scientific American, Vol. 228, No.1, (January 1973) 13–21. 46. Gregory, D. P. and Pangborn, J.B., "Hydrogen Energy", Ann. Rev. Energy, 1 (1976): 279–310. 47. Cogineni, M. Rao, Andrus, H.E. Jr., and Jones, T.J., "Advanced Energy Systems." Proceedings of the American Power Conference, Vol. 50, 1988, pp. 282–287. 48. Bajura, R. A., and Halow, J.S., " Looking Beyond the Demonstration Plants: Longer-Term, Coal-Based Technology Options." Proceedings of the American Power Conference, Vol. 50, 1988, pp. 49–56. 49. DeMeo, Edgar, et al., "Thin Films: Expanding the Solar Marketplace", EPRI Journal, March 1989: 4–15. 50. Wood, Bernard W., Applications of Thermodynamics, 2nd ed. Reading, Mass.: Addison-Wesley, 1982. 51. Womack, G. J., MHD Power Generation: Engineering Aspects. London: Chapman and Hall, 1969. 52. Soo, S. L., Direct Energy Conversion. Engelwood Cliffs, N.J.: Prentice Hall, 1968. 53. Kettani, M. Ali, Direct Energy Conversion. Reading, Mass.: Addison Wesley, 1970. 54. Sutton, George W., Direct Energy Conversion. New York: McGraw-Hill, 1966. 55. Pitts, Donald R., and Sissom, Leighton E., Heat Transfer, Shaum's Outline Series. New York: McGraw-Hill, 1977. 56. Anon., The Astronomical Almanac. Washington, D.C.: U. S. Government Printing Office, 1990. 57. Parsons, Robert A. (Ed.), The ASHRAE Handbook, HVAC Systems and Applications Volume. Atlanta, Ga.: American Society of Heating, Refrigerating, and Air Conditioning Engineers, 1987. 58. DeMeo, Edgar, "Getting Down to Business with Thin Films," EPRI Journal, March 1989: 4–15.

473 59. Taylor, Roger W., Cummings, John E., and Swanson, Richard M., "High Efficiency Photovoltaic Device Development: An Example of the R&D Process" Proceedings of the American Power Conference, Vol. 47, 1985, pp. 255–259. 60. Bockris, J. O. M., Energy The Solar-Hydrogen Alternative. New York: Wiley, 1975. 61. Williams, L. O., Hydrogen Power, Elmsford, N. Y.: Pergamon Press, 1980. 62. Sze, S. M., Semiconductor Devices Physics and Technology. New York: Wiley, 1985. 63. Moore, Taylor, et al., "Opening the Door for Utility Photovoltaics," EPRI Journal, January–February 1987: 5–15. 64. Griffiths, David, J., Introduction to Electrodynamics, 2nd ed. Engelwood Cliffs, N. J.: Prentice-Hall, 1989. 65. Anon., “Solar Electric Generating Stations (SEGS),” IEEE Power Engineering Review, August, 1989: 4–8. 66. Anon., “Promise of Solar Energy Being Fulfilled in California,” Power, October 1989: s32—s36. 67. Romano, Samuel, “Fuel Cells for Transportation,” Mechanical Engineering, August 1989: 74–77. 68. Hirschenhofer, J. H., “International Developments in Fuel Cells,” Mechanical Engineering, August 1989: 78–83. 69. Carlson, D. E., “Photovoltaic Technologies for Commercial Power Generation,” Annual Review of Energy, Vol. 15, 1990, pp. 85–98. 70. MacDonald, Gordon J., “The Future of Methane as an Energy Resource,” Annual Review of Energy, Vol. 15, 1990, pp. 53–83. 71. Moore, Taylor, et al., “On-Site Utility Applications for Photovoltaics,” EPRI Journal, March 1991: 26–37. 72. Moore, Taylor, et al., “Thin Films: Expanding the Solar Marketplace,” EPRI Journal, March 1989: 4–15.

474 73. Smock, Robert W., “Second Generation Fuel Cell Technology Moves Toward Demos,” Power Engineering, June 1990: 10. 74. Sapre, Alex R., “Properties, Performance and Emissions of Medium-Concentration Methanol-Gasoline Blends in a Single-Cylinder, Spark-Ignition Engine,” SAE Paper 881679, October 1988. 75. Douglas, John, “Beyond Steam: Breaking Through Performance Limits,” EPRI Journal, December 1990: 5–11. 76. Graff, Eric, Texas Instruments Co. Personal Communication, May 24, 1991. 77. Chase, M. W. Jr., et al., JANAF Thermodynamic Tables, 3rd ed., J. Phys. Chem. Ref. Data 14, Supplement No. 1, 1985. 78. Douglas, John, et al., “Fuel Cells for Urban Power,” EPRI Journal, September 1991: 5–11. 79. Howell, John R., Bannerot, Richard B., and Vliet, Gary C., Solar-Thermal Energy Systems. New York: McGraw-Hill, 1982. 80. Levine, Jules D., et al., “Basic Properties of the Spheral Solar Cell.” 22nd Photovoltaic Specialists Conference, Las Vegas, Nev., 1991. 81. International Fuel Cells: Clean, Reliable Fuel Cell Energy, www.internationalfuelcells.com/index_fl1.shtml, (November 14, 2000) 82. Rulseh, Ted, “Fuel Cells: From Promise to Performance,” Grid, Spring/Summer 2000: 15. 83. Santa Clara Demonstration Project, www.ttcorp.com/fccg/scdpnew1.htm, (November 14, 2000) 84. Fuel Cell Energy, Carbonate Fuel Cell Manufacturer, www.ttcorp.com/fccg/erc_abt.htm, (November 15, 2000) 85. Welcome to Fuel Cell Energy, Inc., www.fuelcellenergy.com/, What’s New?, (November 15, 2000) 86. Library of Fuel Cell Related Publications, http://216.51.18.233/biblio.html, (November 14, 2000)

475 87. Archer, David H., and Wimer, John G., “ A Phosphoric Acid Fuel Cell Cogeneration System Retrofit to a Large Office Building,” Department of Energy FETC-97/1044, April 1997., www.fetc.doe.gov/netltv/index.html, (November 15, 2000) 88. DaimlerChrysler Offers First Commercial Fuel Cell Buses to Transit Agencies, www.hfcletter.com/letter/may00/feature.html, (November 15, 2000)

EXERCISES

11.1* Derive an expression for the battery power output, in terms of E and Ri /Ro, for the linear model discussed in connection with Example 11.2. Nondimension-alize the power by dividing by E2/Ro. Use a spreadsheet to tabulate and plot the dimensionless power as a function of Ri /Ro. 11.2* Derive an expression for the linear-battery-model power output nondimensionalized by E2/Ri in terms of the internal-to-external resistance ratio. Use a spreadsheet to tabulate the dimensionless power function, and plot it. Is there a condition that produces an extreme value of the dimensionless power? If so, use calculus methods to derive the condition. 11.3 An automobile storage battery with an open-circuit voltage of 12.8V is rated at 260 A-hr. The internal resistance of the battery is 0.2 5. Estimate the maximum duration of current flow and its value through an external resistance of 1.8 5. 11.4 A battery electrical storage plant is to be designed for 20-MW peak power delivery for a duration of four hours. The plant uses 600 A-hr batteries operating at 400 volts DC. Estimate the minimum number of batteries and the current in each during peak operation. 11.5 A hydrogen-oxygen fuel cell operates with a voltage of 0.7v with water-vapor product. Calculate the work per kg-mole of hydrogen, in kJ and in kW-hr, and determine the cell efficiency. 11.6 A neighborhood fuel cell power plant is to be designed for an electrical power output of 2000 kW with liquid-water product. Estimate the flow rates of hydrogen and oxygen during peak power production, assuming that an 80% efficient power conditioner is used to convert DC to AC power and that the fuel cell efficiency is 55%. What is the plant heat rate? ___________________ * Exercise numbers with an asterisk involve computer usage.

476 11.7 A hydrogen-oxygen fuel cell has liquid water as product when it operates at 0.82 volts. What is the electrical energy output, in kJ / kg-mole of hydrogen, and the cell efficiency? 11.8 Prepare a typed three-page, double-spaced memorandum outlining the design of the power train of a hydrogen-oxygen fuel-cell-powered automobile, giving the design criteria, system description, and quantitative preliminary design data on the power and fuel supply systems. 11.9 A hydrogen-oxygen fuel cell stack produces 50 kW of DC power at an efficiency of 60%, with water-vapor as product. What is the hydrogen mass flow rate, in g/s, and the cell voltage? 11.10 A fuel cell power plant is to be designed for an electrical power output of 200MW with liquid-water product. Estimate the flow rates of hydrogen and oxygen during peak power production with an 85% efficient power conditioner and fuel cell efficiency of 55%. What is the plant heat rate?

477 CHAPTER ELEVEN Energy System Alternatives Part 2. MHD, Solar Energy, a Hydrogen Economy and Concluding Remarks

11.4 Magnetohydrodynamic Energy Conversion Magnetohydrodynamic energy conversion, popularly known as MHD, is another form of direct energy conversion in which electricity is produced from fossil fuels without first producing mechanical energy. The process involves the use of a powerful magnetic field to create an electric field normal to the flow of an electrically conducting fluid through a channel, as suggested by Equation (11.13) and depicted in Figure (11.19). The flow velocity u is parallel to the channel axis, taken in the ydirection. The drift of electrons induced by this lateral electric field produces an electric current, represented by the current density vector J. Electrodes in opposite side walls of the MHD flow channel provide an interface to an external circuit. Electrons pass from the fluid at one wall to an electrode, to an external load, to the electrode on the opposite wall, and then back to the fluid, completing a circuit. Thus the MHD channel flow is a direct current source that can be applied directly to an external load or can be linked with a power-conditioning inverter to produce alternating current. MHD effects can be produced with electrons in metallic liquids such as mercury and sodium or in hot gases containing ions and free electrons. In both cases, the electrons are highly mobile and move readily among the atoms and ions while local net charge neutrality is maintained. That is, while electrons may move with ease, any small volume of the fluid contains the same total positive charges on the ions and negative electron charges, because any charge imbalance would produce large electrostatic forces to restore the balance. Though liquid metal MHD has been demonstrated experimentally, most theoretical and experimental work and power plant development and application studies have focussed on high-temperature ionized gas as the working fluid. Unfortunately, most common gases do not ionize significantly at temperatures attainable with fossil fuel chemical reactions. This makes it necessary to seed the hot gas with small amounts of vapor of readily ionizable materials such as the alkali metals. The resulting ions and electrons make the hot gas sufficiently electrically conducting that it may be influenced by the applied magnetic field. The ionization potentials are measures of the energy needed to free valence electrons from an atom. Materials such as cesium and potassium have ionization potentials low enough that they ionize at the temperatures attainable with combustion

478

reactions in air. Recovery and reuse of seed materials from the MHD channel exhaust are usually considered necessary from both economic and pollution standpoints. Ionized Gases in Electromagnetic Fields Before analyzing the MHD channel, we will consider briefly the behavior of electrons in an ionized gas in the presence of electromagnetic fields. In a gas at or near equilibrium, atoms, ions, and electrons are in random motion. At any given spatial position their velocities are distributed about a mean velocity that increases with increase in the local temperature. Consider just one of the free electrons moving, without collision, in a plane normal to a uniform magnetic field, as in Figure 11.20. The electron experiences a constant force qceB normal to its path according to Equation (11.11). Here, q is the charge of the electron and ce the magnitude of its velocity. Because the force is normal to its path, the electron travels with constant velocity on a circular path around magnetic lines of force. By Newton’s Second Law, the force on the electron is F = mece2/r = qceB

[N]

(11.21)

It follows that the angular frequency of the electron about a line of force ce/r, called its cyclotron frequency, is

T = ce/r = qB/me

[s –1]

(11.22)

The electron cyclotron frequency is independent of electron velocity and is dependent

479

only on the magnetic field strength and electron properties. Although the cyclotron motions of electrons exist in gases when strong magnetic fields are present, the circular paths of the electrons may be disrupted by collisions with other particles. The likelihood of collisions between particles depends on their effective sizes: larger particles will collide more frequently. The probability of collision is taken as proportional to the collision cross-section Q of the particle, which may be thought of as its area. The frequency of collision of electrons Tc is given by the product of the electron number density, ne [electrons/m3], the collision cross-section, Q [m2], and the velocity, ce [m/s]:

Tc = neQce = 1/J

[collisions per s]

(11.23)

Here, the mean time between collisions, J [s], is the inverse of the collision frequency. The ratio of the cyclotron frequency to the collision frequency T/Tc , is called the Hall parameter. It indicates the relative importance of the magnetic field and collisions in controlling electron motion in the ionized gas. The Hall parameter is related to the magnetic field intensity by

T/Tc = TJ = qBJ /me = qB/nemeQce

[dl]

(11.24)

It is proportional to the number of cyclotron loops made per collision. A Hall parameter large compared with one indicates magnetic-field-dominated motion of

480 electrons, while a small value implies that collisions quickly break up ordered motions produced by the magnetic field. At least three velocities are of importance in a conducting gas in a MHD channel. First, the velocity of the gas stream is given by u (assumed constant for the present case for an appropriately designed channel). Secondly, the velocities of individual electrons ce, as just introduced, are distributed about an average value that increases with the local temperature. In the absence of electromagnetic fields, the average value of ce over all electrons is the flow velocity u; i.e., on the average the electrons move with the gas flow. When fields are present, however, there may be an average motion of electrons relative to the gas. It is the ease of this motion that determines the conductivity of the gas. The third velocity, the relative velocity of an electron we, is defined as the vector difference of its absolute velocity and the mean fluid velocity: we = ce – u

[m/s]

The drift velocity we, is the magnitude of the average of the relative velocities of the electrons. In the absence of fields, the average of ce is u, and thus the drift velocity is zero. When an electric field is present, however, the transport of negative charge by electrons represents a current flow in the gas. Another important parameter, the electron mobility :, is a measure of the response of electrons to an electric field. It is defined as the ratio of the magnitude of the electron drift velocity we to the local electric field intensity:

: = we /E

[m2/V-s]

(11.25)

If it is assumed that an electron loses all of its drift velocity on collision, the acceleration of the electron may be approximated by the ratio of the drift velocity to the mean time between collisions. Because the force due to the electric field is given by qE, Newton’s Second Law allows the drift velocity to be expressed as we = qEJ/me

[m /s]

(11.26)

The electron mobility can then be written as

: = qJ /me

[m2/V-s]

(11.27)

Using equation (11.24), the product :B becomes the Hall parameter:

:B = qJ B/me = TJ = T/Tc

[dl]

(11.28)

Thus the Hall parameter is large for gases of high electron mobility in strong magnetic fields. It will be seen that this can have a significant effect on MHD channel design. Assuming electrons as the dominant charge carriers, the current density can also

481 be related to the electron mobility through the drift velocity: J = neqwe = neq2JE/me = :neqE

[A /m2]

(11.29)

The electron conductivity of a stationary gas is then given by:

F = J/E = :neq

[(S-m)-1]

(11.30)

Thus high electron mobility and electron number density are essential to achieve the high conductivity needed in an MHD generator. Analysis of a Segmented Electrode MHD Generator Consider the one dimensional flow of a gas in an MHD channel coupled with a simple three-dimensional model of the electromagnetic phenomena. Rather than the continuous electrode configuration shown in Figure 11.19, we examine a refined configuration, shown in Figure 11.21. Here the electrodes, set in opposite electrically insulated channel walls, are segmented in the streamwise direction. This eliminates a return path along the wall for axial electrical currents in the flow. By the same notation as in Figure 11.19, a seeded, ionized gas flows through the segmented-electrode channel in the y-direction with a constant velocity u. A uniform magnetic field in the z-direction exists throughout the gas in the channel. A force given by qu×B, and thus an equivalent electric field u×B is imposed on the flow in the channel. Therefore, positive ions tend to drift in the positive x-direction and electrons drift in the negative x-direction toward the right electrodes. Because their mobility is much greater than that of the relatively massive ions, the electrons are the primary charge carriers. The electrons are collected at the right electrodes and flow through the external circuits returning to the channel at the left electrodes, as shown in Figure 11.21. When the channel is under an electrical load, the current density vector in the xdirection induces a force on the fluid in the negative y-direction. Thus the x-component of J interacts with the magnetic field to produce the axial electric field component Ey = – JB that opposes the flow velocity u. In order to maintain a constant velocity in the duct, a streamwise pressure gradient, dp/dy, must balance the force due to this axial electric field and the viscous forces. Thus, ignoring viscous resistance, the axial force on the gas per unit volume is Fy = – |J×B| = – JB = dp/dy

[N/m3]

(11.31)

where the negative sign indicates that the magnetic force is directed upstream. As a

482

result dp/dy < 0, indicating that the flow pressure drops as y increases. The resulting net pressure force in the positive y-direction balances the magnetic and viscous forces and maintains the flow velocity constant. A compressor is therefore required upstream of the channel to pressurize the flow, to support the field-induced streamwise pressure gradient, and thus to maintain the steady flow in the channel. With segmented electrodes there is no axial current in the channel (Jy = 0), and thus the current density component Jx = J is proportional to the net electric field in the x- direction: J = F (uB – Ex)

[A/m2]

(11.32)

The combined electrical resistance of the MHD channel flow and the external load governs the available potential at the MHD electrodes. If the external circuit is open, J = 0; hence, Equation (11.32) indicates that Ex|open = uB. With a finite external resistance, current flows and the electrode potential is reduced below the open-circuit value. Thus, under load, the channel voltage drops to a fraction K of the open-circuit voltage. Hence, we may write Ex = KuB, where K is called the channel load factor and where 0 # K # 1. The current density then becomes

483 J = FuB(1 – K)

[A/m2]

(11.33)

The electrical power delivered to the load per unit volume of channel is then given by Power|out = J@E = Fu2B2K(1 – K)

[W/m3]

(11.34)

Returning now to consideration of the streamwise electrical fluid interaction, we write the steady-flow form of the First Law for an adiabatic control volume, including the work done against the body force, as m(h1 + u12/2) = m(h2 + u22/2) + mw

[J/s]

(11.35)

where m is the channel mass flow rate. Work is positive here because it is done by the fluid in the channel to produce the electrical current flow to the external load. For constant velocity in the channel, this becomes h2 = h1 – w = h1 – (Power|out)(Volume)/m = h1 – Fu2B2K(1 – K)/D

[kJ/kg]

(11.36)

where D is the gas density. Thus the work delivered to the load reduces the thermal energy of the flow. We have seen that a compressor is required to pressurize the flow in the channel and that heating of the flow provides a high entrance enthalpy and work output. From Equations (11.31) and (11.33), the electrical retarding force on the flow is Fy = – FuB2 (1 – K)

[N/m3]

(11.37)

and the fluid power to push the gas through the channel per unit volume is Power|in = |Fyu| = Fu2B2 (1 – K)

[W/m3]

(11.38)

The Ohmic or I2R loss is given, using Equation (11.33), by J2/F = [FuB(1 – K)]2/F = F (uB)2(1 – 2K + K2) = Fu2B2(1 – K) – Fu2B2K(1 – K)

[W/m3]

(11.39)

Comparing equations (11.34) and (11.38) with equation (11.39), we see that the ohmic loss is the difference between the power required to push the flow through the channel and the useful power through the load. The efficiency of the channel is defined as the ratio of the Power|out to Power|in. By Equations (11.34) and (11.38), the MHD channel efficiency is

484

0 = Power|out /Power|in = K

[dl]

(11.40)

Thus the electrical efficiency of the segmented electrode MHD channel is equal to the channel load factor. Examination of Equation (11.34) shows that the power output vanishes when K = 0 and when K = 1. Thus there must be an intermediate value of K that maximizes the power output. By differentiation of Equation (11.34) with respect to K, the usual methods of calculus indicate that the power is maximized when K = 0.5. Thus operation at this value implies that 50% of the flow energy input to the channel is converted to electricity, and the remainder is dissipated in the flow channel. This energy is not lost from the flow but is an irreversibility that is reflected in a loss in ability of the flow to do work. EXAMPLE 11.6

A 10-m3 MHD generator with segmented electrodes has a short-circuit current density of 12,000 amperes per square meter. The gas conductivity is 20 (Ohm-m)-1. If flow and magnetic field conditions are unchanged when the load factor is 0.6, what is the output power? What is the actual current density in the channel? If the magnetic field is doubled in strength, by what factor would you expect the output power to change? Solution

For a short-circuit condition, the load factor is zero, and Equation (11.33) yields Jsc = FuB = 12000 A/m2 Then uB = Jsc/F = 12000/20 = 600 V/m. The power output is then given by Equation (11.34): Power = Fu2B2K(1 – K)V = 20(600)2(0.6)(0.4)(10) = 17,280,000 W = 17.28 MW The channel current density is then given by J = FuB(1 – K) = 20(600)(0.4) = 4800 A/m2. If the magnetic field strength is doubled, Equation (11.34) shows that the power output is increased by a factor of four, assuming there is no change in the flow or load conditions. ____________________________________________________________________

485

Other MHD Configurations The preceding discussion of the segmented electrode MHD channel provides a brief insight into the fundamentals of MHD power. Other electrode configurations include the Faraday continuous electrode generator and the so-called Hall generator. In the former case, with continuous electrodes, the electric field in the y-direction leads to a streamwise electrical current in a circuit that is completed in the continuous conducting electrodes. The resulting current along the channel axis, called the Hall current, contributes nothing to external power delivered in the Faraday generator. It is shown in references 1 and 44 that the power output for the continuous electrode generator is influenced by the Hall parameter and is given by Equation (11.34) with the right-hand side divided by the factor 1 + (:B)2. Thus high values of the Hall parameter drastically reduce the channel power output relative to the segmented electrode generator. Rather than attempting to circumvent the Hall current, the Hall generator seeks to employ it by using the electrode configuration shown in Figure 11.22. In the Hall configuration, the electrodes are designed to short circuit the cross-channel electric current and pass the axial Hall current through the load. The Hall generator power output also depends on the Hall parameter, but in a more complex way than in the continuous electrode Faraday configuration. The reader is referred to references 1 and 44 for further detail. MHD channels, like gas turbines, may be operated on open or closed cycles. Normally, combustion-driven MHD systems use an open-cycle, such as that shown in Figure 11.23, where pressurized combustion takes place upstream of the MHD channel. As with the gas turbine, the temperature of the channel exhaust is high, making it advantageous to employ regenerative heat exchange and/or a bottoming cycle such as the steam turbine in Figure 11.23. For MHD channels with steam bottoming cycles, overall efficiencies in the range of 50–60% are predicted.

486

If a DC source, rather that a load, is applied to the electrodes of a properly contoured Faraday channel, the energy in the flow may be increased and the flow accelerated. The electric field in this case exceeds the uB term in Equation (11.32), reversing the direction of the current density and thus the direction of the J × B force. The MHD channel may then be used for propulsion through expulsion of mass in the same way as in a jet engine or rocket.

487

11.5 Solar Energy The preceding pages have dealt with the conversion of energy from fossil and nuclear fuels, all finite resources. While the need may not be immediate, the day is approaching when the Earth’s fossil and nuclear resources will no longer satisfy man’s need to grow and prosper. We must therefore consider more seriously the effective utilization of the readily available and, for practical purposes, eternal energy source provided by the sun. Ancient peoples learned to use the sun for survival and comfort. Not very long ago a clothesline and wind formed the mechanism of a solar clothes dryer. Powered by the sun, the atmospheric engine that produces weather change provides the wind to fill sails and turn windmills to produce small amounts of power at a few select locations. Passive solar architecture has recently attracted widespread interest, and solar collectors mounted on roofs for water heating and space heating have become commonplace in many areas. Figure 11.24 shows fixed and tracking solar arrays that produce electricity to supplement or replace grid power at a remote location. A major problem connected with using the sun for power generation is the same characteristic that allows the survival of human life on earth: the low intensity levels of solar radiation. It is well known that the two planets closest to the sun have average temperatures higher than the Earth’s and that the outer planets are colder because of the inverse square law of radiative transfer of the sun’s radiation field. Higher solar fluxes at Earth’s surface would clearly make the sun a more easily engineered radiation source. Let us estimate the intensity of the solar radiation arriving at Earth.

488

Consider the geometry of the Earth–sun system and the spherically symmetric nature of the sun's radiation field presented in Figure 11.25. Assuming that the sun radiates energy uniformly in all directions at a rate of E kW, the radiant flux density M crossing any sphere concentric with the sun is given by dividing E by the area of the sphere. Thus the radiant flux density, or irradiance, at a distance r from the sun center is

M = E/(4B r 2)

[kW/m2]

(11.41)

In order to evaluate M, it is necessary to calculate the rate of energy emitted by the sun per unit area of its surface, Ms. While the temperature at the center of the sun is much hotter, solar radiation approximates blackbody radiation at a temperature of 5762 K. Using the Stefan-Boltzmann constant, F = 5.66961×10–8 W/m2-°K4, the blackbody radiation law (discussed later in more detail) gives the rate of energy emission per unit area of the sun’s surface as

Ms = FT4 = 5.66961 × 10-8 (5762)4 (10–3)= 62,495 kW/m2 Applying conservation of solar energy to concentric spheres makes it clear that energy emitted uniformly from a sphere at the sun’s radius rs will be distributed over a wider area when passing through a sphere of larger radius. Thus, eliminating E for two concentric spheres from Equation (11.41) shows that radiant flux density scales inversely as the area ratio or square of the radius ratio of the spheres:

M = Ms( rs /r ) 2

[kW/m2]

(11.42)

This and Equation (11.41) are both mathematical expressions of the inverse square law for solar radiation.

489 As in Figure 11.25, taking the sun’s diameter to be 865,400 miles and the Earth–sun distance as 93,000,000 miles, the solar radiant flux density, or solar constant of radiation, reaching Earth’s orbit is

Me = Ms( rs /re ) 2 = 62,495[(865,400/2)/93,000,000]2 = 1.353 kW/m2 Thus the value of the solar constant is the rate at which solar radiation crosses a unit area normal to the sun’s rays at the distance of Earth from the sun. A more precise value for the solar constant is Me = 1.357 kW/m2, or 430.2 Btu/hr-ft2 (ref. 39). While solar radiation may be viewed as a divergent spherical radiation field on the scale of the solar system, a calculation of the angle of divergence at the earth's distance indicates that its rays are essentially parallel on the scale of almost all human activities. Thus it is common to assume a constant radiation flux density with parallel rays in analyzing terrestrial solar radiation problems. Some of the solar radiation arriving at the earth’s surface is scattered and reflected in the atmosphere. Thus solar radiation consists of direct (parallel) and diffuse components. Moreover, absorption, reflection, and scattering in the atmosphere reduces the maximum direct radiation flux density arriving at sea level to about Mdir = 1 kW/m2. When the sky is clear, the direct component dominates. But the direct component may essentially vanish on overcast days, leaving only a diffuse component. The calculation of the solar performance of space vehicles is usually simplified by the absence of a diffuse component. However, for near-Earth satellites, radiation emitted from Earth and solar radiation reflected from Earth could be significant. Matter interacts with solar radiation in three basic ways: it can absorb the radiation, transmit it, or reflect it. These actions are represented by characteristics called, respectively, absorptance A, transmittance T, and reflectance R. Each is expressed as a fraction of the total incident radiation. Thus the radiation energy reflected from a surface with reflectance R and a given area S is RSM. Consider a layer of material with surface area S normal to a radiation field with radiant flux M. Conservation of energy requires that the rate of energy incident on the surface equal the sum of the rates of energy reflected, transmitted, and absorbed. Thus MS = RSM + TSM + ASM. Thus the sum of the reflectance, the transmittance, and the absorptance must equal 1: R+T+A=1

[dl]

(11.43)

Let us apply this statement of energy conservation to the Earth’s atmosphere: On a clear day, the atmospheric reflectance is small, hence any radiation not absorbed is transmitted to the surface (T + A . 1). On a cloudy day, reflection from and absorption by clouds are significant, and the transmitted radiation becomes a small fraction of the incoming solar irradiance (R + A . 1); only the diffuse component scattered by the clouds remains.

490 Spectral Characteristics Though it is usually convenient to deal with total radiation quantities, sometimes it is necessary to consider wavelength-dependent, or spectral, effects. The material radiative properties just introduced above are sometimes frequency or wavelength dependent. Thus, while solar radiation in space approximates a blackbody spectral distribution, radiation penetrating to Earth’s surface deviates significantly from the blackbody distribution because of scattering and strong absorption bands due to atmospheric water vapor, carbon dioxide, dust, and other substances that remove significant amounts of energy from the incident radiation at certain wave lengths. The eye itself is a highly wavelength-selective radiation sensor, being limited to observing radiation only in the range of about 0.4 to 0.7 microns. A micron is 10– 6 meters. Consider the Planck equation for the blackbody spectral distribution of radiation Fb(8), where wavelength 8 is in microns [: ] (refs. 37, 43 and 55): Fb(8) = 2Bhc28--5/(ehc/(k&T) – 1)

[Btu/(ft2-hr-:) | W/(m2-:)]

The Planck equation may be written in terms of a single parameter 8T, where wavelength is in microns and temperature is in appropriate absolute units: Fb(8T)/T5 = 2Bhc2 (8T) --5/(ehc/(k&T) – 1) = 3.742×108(8T) –5/(e14381/(&T) – 1)

[W/(m2-:-°K5)]

(11.44a)

= 1.187x108( 8T)--5/(e25896/(&T) – 1)

[Btu/(ft2-hr-:-°R5)]

(11.44b)

Figure 11.26 shows the Planck blackbody radiation distributions for two different temperatures as a function of wave length, one approximating solar radiation at 10,000 °R (5555.5°K) and the other much closer to, but above, the normal terrestrial temperature of 1000°R (555.5°K). At first glance the radiation levels appear comparable; but they actually they are vastly different from each other (by a factor of 105), because the Planck function is divided by FT5, to allow display on the same scale. The figure makes clear that in the spectral range to which the eye is sensitive, the visible range, little radiation emitted by relatively cool terrestrial objects is visible; and almost all of what we see is reflected solar radiation or other high-temperature-source radiation (such as from high-temperature lamp filaments or from flames). The vast difference in the wave lengths of the peaks shows why it is important to consider spectral effects in dealing with solar radiation.

491

The Wien displacement law, a simple relation derivable from the Planck equation, given in both English and SI units,

8maxT = 5215.6

[:-°R]

(11.45a)

= 2897.6

[:-°K]

(11.45b)

shows that the wavelength of the peak of the blackbody radiation distribution, 8max , is inversely proportional to the temperature, as may be verified numerically for the two peaks in Figure 11.26 ( 8max = 5.215: and 0.5215:). Integration of Fb over all wavelengths from zero to infinity, using Equation (11.44) yields the Stefan-Boltzmann law for the blackbody flux Mb: "

"

Mb = 1 Fb d8 = T41 [Fb(8T)/T5]d(8T) = F T 4 0

[Btu/hr-ft2|W/m2]

(11.46)

0

which was used earlier to estimate the radiative flux from the surface of the sun and the terrestrial solar constant. It is useful to sum the contributions to blackbody emission from zero to a given wavelength by an integration of Fb, as was done for the entire spectral range in Equation (11.46). Letting " = 8T, we define a function E(8T) as a dimensionless fraction of the blackbody irradiance from " = 0 to " = 8T:

492

&T

E(8T) = T4/Mb I [Fb(")/T5]d"

[dl]

0

Figure 11.27 shows a log-log graph of the Planck distribution and E(8T). A table of the latter function, calculated by numerical integration using a spreadsheet, is provided in Appendix I. The dimensional blackbody flux between any two wave lengths for a given temperature may be determined from a difference of sigma functions as:

M(8T)2 – M(8T)1 = (&T)2

= T 4I[Fb(")/T5]d" = [E(8T)2 - E(8T)1]FT4 (&T)1

[Btu/hr-ft2|W/m2]

EXAMPLE 11.7

Consider a window, normal to the sun’s rays, covering a solar collector mounted on a satellite in near-Earth orbit. The window has a transmittance of 0.9 in the range of wavelengths from 0.2–2.0 : and 0 at other wavelengths. What is the fraction of incident solar radiation transmitted by the window, and what is the rate of useful energy transfer to a circulating fluid in a 4 × 4-m flat plate collector if the collector absorbs 85% of the transmitted radiation?

493 Solution

The only radiation transmitted is between 0.2 and 2.0 :. Because the effective temperature of the solar radiation is 10,000° R, the values of 8T and E for these two wavelengths are and

8T = (0.2)(10000) = 2000 :-°R and

E(2000) = 0.0013

8T = (2.0)(10000) = 20,000 :-°R and

E(20,000) = 0.934

using the G table in Appendix I. The transmitted fraction of the blackbody incident radiant energy is the product of the fraction of radiation in the transmitted spectral range and the window transmittance: (0.9)(0.934 – 0.0013) = 0.839 The radiation that heats the circulating fluid is then the product of the transmitted spectral fraction, the absorptance, the area, and the near-earth solar constant: (0.839)(0.85)(42)(1.357) = 15.5 kW. ____________________________________________________________________ Example 11.7 dealt with a constant transmittance over a single spectral interval. More complex radiation characteristics may be treated in a similar way by summing the contributions of more than one spectral band. Earth-Sun Geometry and Solar Collectors Radiative transfer from the sun to a plane surface depends on the orientation of the surface to the sun’s rays. If parallel rays irradiate a surface, the total energy rate per unit area is given by the product of the irradiation M and the cosine of the angle between the surface normal and the sun’s rays: Mcos2. The following discusses ways in which the cosine function can be evaluated. Consider the component of radiation falling on a terrestrial horizontal surface with unit normal n, when the sun’s rays are in the direction of a second unit vector s. Following the notation of Figure 11.28, the two unit vectors n and s can be expressed in terms of the orthogonal unit vectors i, j, and k: n = nxi + nyj + nzk

and

s = sxi + syj + szk

where nx, ny, nz, and sx, sy and sz are the direction cosines of the two vectors. Note that the direction cosines must satisfy the following conditions:

494

n @ n = nx2 + ny2 + nz2 = 1

and

s @ s = sx2 + sy2 + sz2 = 1

for n and s each to be of unit magnitude. The fraction of the maximum solar radiation falling on the unit area, then, is given by the scalar product: cos 2 = n @ s = nxsx + nysy + nzsz

(11.47a)

Figure 11.29 shows these unit vectors set at the center of the earth. The unit vector n points to the zenith of the observer and is normal to the horizontal surface at P. Here the x-axis is taken through the intersection of the meridional plane containing n and the equatorial plane. The z-axis points north and the y axis is selected to complete a right-handed coordinate system. The angle L is the latitude of n. In this system the position of the sun is determined by two convenient angles: H and *. Examination of Figure 11.29 shows that the components of s on the coordinate axes are sx = cos * cos H sy = cos * sin H and sz = sin * Similarly, nx = cos L, ny = 0, and nz = sin L. Thus the cosine of the sun-surface angle given by n @ s using Equation (11.47a) is cos 2 = cos * cos H cos L + sin * sin L

(11.47)

495

Thus the cosine of the angle of the sun’s rays falling on a horizontal surface depends on the latitude of the surface and the angles H and * of the sun. Now consider the determination of H and *. First we define the ecliptic as the plane of earth’s annual motion around the sun, as shown in Figure 11.30. Because the earth’s axis always points in approximately the same direction in space and is inclined at 23.5° to the normal to the ecliptic, the apparent position of the sun as viewed from Earth ranges above and below the equator by 23.5° annually. This movement is designated by the solar declination angle * in Figure 11.29, measured from the equatorial plane along a meridian of longitude. It can be seen from Figure 11.30 that * varies from + 23.5° degrees on June 21 to – 23.5° on December 21. The hour angle H of the sun in Figure 11.29 is the angle measured in the plane of the equator between the observer’s meridional plane and the meridional plane of the sun. Earth rotates once or 360° about its axis in 24 hours; hence, the hour angle depends on time, as its name implies. If solar time is measured from solar noon (when the sun is at its highest point in the sky at the observer’s meridian), the hour angle may be computed from the solar time using the factor 360/24 = 15° per hour. The hour angle is positive when the sun is east of the observer and negative when it is west. Thus at 3:00 P.M. solar time, the sun is 3 hours past solar noon and, therefore, has an hour angle of 3×(–15) = – 45° (west of the observer).

496

Care should be taken to distinguish between solar time and time defined by law called civil time. In the continental United States, for instance, there are four time zones defined by law in which standard time decreases by one hour for each 15° of longitude from the east to the west coast. This allows a uniform time within a zone while approximating solar time. Exceptions have been made to accomodate irregular state boundaries. Thus standard time deviates from solar time by several minutes, depending on the longitude of the observer. Moreover, by law, daylight saving time differs from standard time. In the spring and fall, clocks are advanced and set back an hour, respectively. Under daylight saving time, the civil time in the summer would be 1 pm at any location where it is noon standard time. Thus solar time can deviate from daylight saving time by more than an hour at some locations. EXAMPLE 11.8

What is the incident radiative energy rate falling on a 7 × 10-m horizontal roof on a clear day at 40° north latitude, at a solar time of 10 A.M. on December 21? Solution

Figure 11.30 shows that on December 21 the solar declination angle * = – 23.5°. At 10 A.M. sun time, the hour angle is 2 hr × 15°/hr = 30°. By Equation (11.47), the angle of the sun relative to the roof vertical is cos 2 = cos * cos H cos L + sin * sin L = cos(–23.5°) cos(30°) cos(40°) + sin(–23.5°) sin(40°) = 0.352. Taking the solar constant at Earth’s surface as 1.0 kW / m2, the incident energy on the roof is (0.352)(70)(1.0) = 24.64 kW

497 Note that the solar angle 2 is independent of the sign of the hour angle. Thus cos2 is a symmetric function of H, and the same irradiance occurs at 10 A.M. and 2 P.M. solar time. ____________________________________________________________________ EXAMPLE 11.9

What is the irradiance on an east-facing vertical wall in kW / m2, under the conditions of Example 11.8? Solution

A unit vector on an east-facing wall at the observer location in Figure 11.29 is directed along the y-axis and is therefore simply j. The cosine of the angle between the east-facing-wall normal and the solar direction vector s is j @ s = sy = cos * sin H = cos(–23.5°) sin(30°) = 0.4585 The solar irradiance is then (1.0)(0.4585) = 0.4585 kW/m2. Here, the hour angle dependency is antisymmetric. It would give a negative result for 2 P.M. solar time (H = – 30°). Thus, after solar noon, when j @ s < 0, the negative sign indicates that the sun is behind the east-facing wall and that the wall of the building is in the shade. ____________________________________________________________________ Solar collectors can be fixed in position or they can track the sun. Fixed collectors for year-round use in the northern hemisphere are normally oriented facing the south with a tilt angle ß that is approximately equal to the collector latitude. For a southfacing collector tilted at an angle ß with respect to the horizontal, the collector normal makes an angle of L – ß with the equatorial plane in Figure 11.29. The angle of the collector normal, n, with the sun direction may then be determined using n = i cos (L – ß) + k sin (L – ß) Then cos 2 is cos 2 = n @ s = cos(L – ß) cos* cosH + sin(L – ß)sin* Note that this equation may also be obtained by replacing the latitude in Equation (11.47) with L – ß. For sun-tracking solar collectors, two degrees of angular freedom are required for perfect tracking. Typically, such collectors would pivot about horizontal and vertical axes dictated by the local solar azimuth and elevation. Their angular motion may be preprogrammed using astronomical data such as those from reference 56, or, it could be controlled by a sun-seeking control system.

498

Solar Thermal Electric Generating Stations Large thermal applications of solar energy to produce electrical power are located at two sites in the Mohave desert of California (refs. 65 and 66). These facilities, (Figure 11.31) called solar electric generating stations–-SEGS, consist of 14-MW, 30-MW, and 80-MW units employing line-focusing parallel-trough solar collectors to provide heat for reheat steam turbine systems. The SEGS plants in the Mojave Desert make up the world's largest parabolic trough facility. In the year 2000, there were nine plants, which provide a combined capacity of 354 MW (ref. 89). Deployment of additional plants is not expected to occur until at least 2002. When completed, the 12 SEGS units will have a total electrical generating capacity of about 600 megawatts. Reflected solar radiation from the mirrored-glass sun-tracking horizontal-axis parabolic collectors is focused on an evacuated tubular-heat-collection element through which a heat transfer fluid flows. After leaving the collector, the hot fluid heats water in a steam generator before returning to loop through the collectors, as seen in Figure 11.32. The resulting super heated steam is used in Rankine-cycle reheat steam turbine generators. Cooling towers reject heat from the condensers to the surrounding desert air. Supplemental heat is provided, when required, by boilers burning natural gas, as shown in Figure 11.32, or, in the newer designs, by gas-fired heaters that heat the collector heat transfer fluid. This supplemental heat is required to ensure that full SEGS plant design output is obtained during the periods of peak demand for electricity in Southern California–from noon to 6 pm between June and September.

499

The large land area required by SEGS plants is evident from Figure 11.31. References 65 and 66 give insight into some of the technical issues faced in the development of SEGS systems; issues that include periodic mirror washing, breakage of the heat collection elements, wind loading, and other operational considerations. Updated information on the SEGS system and other solar-thermal programs is given in references 89 to 92. Solar Photovoltaics Photovoltaic cells have gained wide use in recent years as small scale power sources for watches, calculators and other such devices. Less evident to the public is the widespread research and development effort on electric power production using solar photovoltaics. Over two billion dollars in corporate funds have been invested in photovoltaics (ref. 58). The major goal is to provide cheap and efficient direct conversion of solar photons into electricity, using light-sensitive devices with few or

500

no moving parts. Success could provide a sorely-needed long-term energy source that could benefit most of the peoples of the world. Photovoltaics rely on specially prepared materials called semiconductors, which produce useful current flows when illuminated with solar or artificial radiation. They are typically thin layers of materials such as crystalline silicon, with electrical leads mounted on opposite faces, as shown in Figure 11.33. The silicon is usually mounted on a substrate that provides both structural support and cooling to conduct away heat created by nonproductive radiation absorption. The exposed silicon is usually covered with a transparent, antireflecting protective coating to reduce energy loss via surface reflection. Crystalline semiconductors are near perfect geometric lattices of atoms that were usually produced by a crystal growth method known as the Czochralski process (CZ), which involves slowly turning a seed crystal of pure silicon as it is withdrawn from a bath of high-purity molten silicon. The CZ process forms an ingot that may be 15 cm in diameter and over one meter long. Thin wafers of silicon are then sliced from the ingot in a slow, delicate process. Other methods, such as casting of square ingots, are in use also. The research and development of more cost-efficient techniques for the large-scale production of semiconductors has a continuing high priority. The extensive use of semiconductors in the computer and electronics industries provides substantial motivation for continued research in semiconductor production methods.

501 After slicing, silicon crystals are "doped" with occasional special impurity atoms, perhaps a few foreign atoms per million silicon atoms in the lattice. This is typically achieved by heating the wafers and exposing their surfaces to vapors of the doping material, called a dopant. This process allows the dopant to diffuse to a controlled depth in the crystals. Semiconducting crystals are electrical insulators at low temperature and weak conductors of electricity at room temperature. The presence of impurities provides charge carriers that substantially enhance the material electrical conductivity when it is exposed to radiation. The following considers how photon absorption in the crystals produces an electrical potential and direct current. Electrons in semiconductors may exist in a band of bound energy states called a valence band. This is analogous to the bound energy levels of individual atoms. As with individual atoms, the absorption of a photon in a semiconductor may change the energy state of an electron. If the photon energy equals or exceeds a threshold level, the electron may be excited into a conduction band, where it is free to move throughout the material, obviously enhancing the electrical conductivity of the material. Consider silicon again as an example of semiconductor behavior. Silicon atoms have four electrons in their outer shell. If impurity atoms in a silicon crystalline lattice have five electrons in their outer shell (atoms such as arsenic or phosphorus) the silicon crystal has an overabundance of electrons (and positive charge) and is therefore known as an n-type (negative) semiconductor. The extra electrons associated with impurity atoms are loosely bound just below the conduction band. Small amounts of energy from the thermal motions in the lattice or from absorption of photons allow these electrons from the impurity atoms (called donor atoms) to move freely in the crystal. The result is a neutral material containing mobile negative-charge carriers. Now consider another type of crystalline silicon semiconductor, the p-type, which has positive charge carriers. If, in the silicon crystal,, the impurity atoms have three electrons instead of five in their outer shells (as in boron or gallium), the impurity atoms can easily snatch electrons from nearby silicon atoms leaving positively charged silicon sites called holes and fixed negative dopant atoms. These impurity sites are called acceptors because they draw electrons from silicon atoms, thus creating positive holes in the lattice. This type of semiconductor is called a p-type, because the holes are positive charge carriers. When an electron abandons one silicon site for another, a hole disappears at its destination and new hole is created at its point of origin. Thus a p-type semiconductor has holes that may move through the material as positive-charge carriers. When an external electric field is applied to an n-type semiconductor, electrons will move toward the region of high potential. However, the electrons are restrained by the positive fixed donor sites, resulting in a charge distribution in balance with the applied field. In a p-type semiconductor, its positive holes would move toward low potential until a stable charge distribution with the negative acceptor sites is established. Thus, because of the fixed charges at donor and acceptor sites, the

502 charge carriers will move just far enough to set up charge distributions that are in balance with the external field. Consider layers of p-type and n-type semiconductors brought together to form an electrically neutral slab with a common interface called a p-n junction. Because of concentration gradients of free electrons and holes across the junction, some electrons from the n-type will cross into the p-type, while holes move a short distance into the n-type. The diffusion of holes and electrons rapidly creates and is balanced by an electric field in a thin space-charge region less than a micron thick, as indicated in Figure 11.33. As a result, the p-type material is at a higher potential than the n-type. Thus the two materials on either side of the junction are each at different potentials without the benefit of an external electric field, thanks to the electric field at the p-n junction. Now let’s consider radiation incident on the p-n semiconductor. Photons have an energy h<, where h is Planck’s constant and < is the frequency of the radiation. When a p-type material near a p-n junction is irradiated by an energetic photon, the photon may be absorbed, raising an electron into the conduction band and leaving a hole behind. Thus a single photon creates an electron-hole pair. Conduction band electrons readily move across the junction into the n-type material to a region of lower potential energy, while the holes tend to move in the opposite direction and thus remain in the p-material. Likewise, electron-hole pairs created by irradiation of n-type material cause holes to move into the p-material, leaving the electrons behind. Thus the photons provide the energy for a pumping process that separates and selectively drives charges across the interface. These actions increase the positive charge in the p-type and the negative charge in the n-type materials, opposing the electric field produced by diffusion due to charge concentration gradients. This process continues until the electric fields balance each other. When an external load is connected to the irradiated cell, electrons flow from the n-type material through the external circuit while holes move in the opposite direction. The net current flow through the load is the sum of both the electron flow and the hole flow through the semiconductor layers. Since the number of charge carriers driven through the load depends on the number of photon excitations, the total cell current and power output depend on the junction surface area and on the intensity of solar irradiance. A silicon solar cell may be 10 cm by 10 cm, and have an open circuit voltage of about 0.5 V. At 10% efficiency it would deliver a power output of 1 W when receiving peak solar irradiance on Earth of 1.0 kW/m2. Power = (1.0)(10/100)2(0.1)(1000) = 1.0 W The band gap, or energy separation, between conduction and valence bands is 1.1 eV for silicon p-n junctions. This is the energy necessary to move an electron into the conduction band of the semiconductor.

503

Thus, to excite electrons into the conduction band, photons must have a frequency equal to or greater than E/h, the photon energy (equal to band gap energy) divided by Planck’s constant

< = (1.1 eV)(1.602×10–19 J/eV)/(6.626×10–34 J-s) = 2.66×1014 s –1 and a wave length no greater than c/<, the speed of light divided by the photon frequency:

8 = (2.9979×108 m/s)(10 6 :/m)/(2.66×1014 s–1) = 1.127 : Photons with lower frequencies and longer wave lengths, and therefore energies less than the band-gap energy, cannot raise electrons to the conduction band and may be transmitted through the material without effect. On the other hand, one photon can excite only one electron into the conduction band, and any excess photon energy beyond the band gap value only heats the semiconductor. Thus photon wave lengths greater and less than the bandgap value result in inefficiency in converting incident radiation to electricity. The p-n junction acts like a combination of a current source and a diode in parallel as shown in Figure 11.34. For a given level of cell irradiation, a cell electrical characteristic may be represented by I = Isc – Io(e qV/kT – 1)

[A]

(11.48)

where k is the Boltzmann constant, q is the electronic charge, and Io is called the dark current. The short-circuit current Isc (corresponding to a cell potential difference =0) is proportional to the rate of incident irradiance. The dark current Io is related to the cell open-circuit potential difference Voc by setting I = 0 in Equation (11.48). Thus:

504

Thus Io /Isc = 1 /(e qVoc/kT – 1). By combining these two equations, we can write the cell characteristic in nondimensional form as I /Isc = 1 – (Io /Isc)(e qV/kT – 1) = 1 – (e qV /kT – 1)/(e qVoc /kT – 1)

[dl]

(11.49)

The cell power output is the product of the current and cell potential difference: P = IV = [Isc – Io (eqV/kT – 1)]V

[W]

Nondimensionalizing the power by dividing it by IscVoc gives P/IscVoc = [1 – (e qV /kT – 1)/(e qVoc /kT – 1)](V/Voc)

[dl]

(11.50)

The nondimensional current and power characteristics are shown in Figure (11.35). It may be seen from the figure that peak cell power increases linearly with output voltage for small values of voltage, reaches a maximum at between 80% and 90% of the open circuit voltage, and drops rapidly thereafter. The maximum power is about 80% of IscVoc. For given values of Isc and Voc, the maximum power occurs at particular values of current and voltage (indicated by the circle on the current characteristic), and thus at a single load resistance. For other resistive loads the power output is reduced.

505 EXAMPLE 11.10

For a photocell with an open-circuit voltage of 0.6 V at 350K, evaluate qVoc/kT and the nondimensional current and power at V/ Voc = 0.8. Propose a simplification of the model given earlier for large values of the voltage parameter. Solution

For T = 350K and Voc = 0.6 V, the dimensionless voltage parameter is qVoc/kT = 1.602×10–19×0.6 / (1.38×10–23×350) = 19.9 Using Equation (11.49), the current ratio is then I /Isc = 1 – (e 0.8×19.9 – 1)/(e 19.9 – 1) = 0.9813 and the power ratio is P/IscVoc = (I/Isc)(V/Voc) = 0.9813×0.8 = 0.785 It is clear that for large values of the voltage parameter, 1 may be neglected with respect to the exponentials; hence we may write the following excellent approximations: I /Isc = 1 – e (qVoc/kT)(V/Voc – 1) and P/IscVoc = [1 – e (qVoc/kT)(V/Voc – 1)](V/Voc) The reader may verify numerically the accuracy of these approximations. ____________________________________________________________________ The cell efficiency is the ratio of the maximum cell power output to the rate of incident radiant energy normal to the cell, Pm /Einc, where Pm is the product of the cell current and the voltage at maximum power point. For a given irradiance, doubling the cell efficiency implies doubling the cell current and power output. Let us consider a few of the types of energy losses that contribute to low cell efficiency in converting energy from the radiation field into electricity. It has been seen that in silicon, the minimum photon energy required to create an electron-hole pair is 1.1 ev and that (1) photons with energies less than 1.1 ev are not absorbed by the semiconductor; and (2) the excess energy of photons with greater than the required energy is converted to heat, since a photon can create only one electron-hole pair. For silicon cells, these losses exceed 50% of the incident energy (ref. 11.59). Thus the nature of the solar spectrum and the excitation energy of

506

particular junction materials influences the cells’ efficiency. Analyses of these losses indicate that the silicon band-gap energy is lower than optimum for the solar specral distribution. References 39 and 42 indicate that cadmium telluride, CdTe, and gallium arsenide, GaAs, have near-optimum band-gap energies around 1.4 eV for the solar application. Another loss that contributes to inefficiency is reflection at the cell surface. Antireflection coatings are currently used to reduce this loss to a few percent of the incident energy. When these and other losses are accounted for, a maximum conversion efficiency of single-crystal silicon cells in sunlight is about 25%. Values of 23.2% and 22.3% have been obtained in two laboratories when high efficiency was the objective (ref. 69). The best efficiency of commercial silicon cells is about 15%. Solar Cell Design Other semiconducting materials that have received less attention than silicon have been found to use the solar spectrum more efficiently, as indicated above, and thus offer potential for higher conversion efficiency. Research is also in progress on cells made of thin layers of several junction materials that each work best in different parts of the spectrum. Thus, if a top layer converts high-frequency photons efficiently and allows lower frequency photons to pass to a lower layer to be converted by a semiconductor with a lower band-gap energy, a larger part of the spectrum may be used efficiently. For instance, as Figure 11.36 shows for a two junction cell,

507

ultraviolet and visible wavelength photons may be absorbed near the top junction and infrared photons near the bottom. Regardless of the individual cell characteristics, in order to achieve power outputs of significance for utility and industrial applications it is necessary to string the cells together in arrays. Identical cells connected in series all have the same current; and their combined voltage is the sum of the individual cell voltages at the given current flow, as shown in Figure 11.37. For example, using a typical value for the opencircuit voltage and the voltage condition for maximum cell power, in order to get a voltage of about 14 V to charge a 12-V battery it is necessary to bring together about 14 V/[(0.5Voc/cell) (.8V/Voc)] = 35 cells in series. For 10-cm by 10-cm cells, such a string would have a maximum power output of about (35 cells)(1 watt /cell) = 35 W and an area of 35×10×10 = 3500 cm2 or 0.35 m2. In order to achieve high current flow, it is necessary to connect cells or strings of cells in parallel. In this case the currents are additive for a given cell potential, as seen in Figure 11.38. While high voltages are desirable to reduce ohmic losses and reduce wire size, according to reference 38 limitations on insulations and safety concerns will limit the voltage levels achieved by solar arrays. Since the National Electric Code does not provide for equipment ratings above 250 V DC, higher voltages may imply increased costs for solar power systems. In the event of cell damage, loss of one cell in a parallel string of cells means the loss of only the power of that cell, whereas loss of a single cell in a series string

508

causes a large loss in power because of the large increase in series resistance of the string. Shading from the sun of a single cell in series has a similar effect, because the reduction in incident radiation cuts the current carrying capability of the cell and hence the entire string. Circuitry is usually provided to allow current to bypass a defective cell. EXAMPLE 11.11

A near-Earth satellite requires a maximum of 10 kW of electrical power. Estimate the required collector area and the number of 10-cm by 10-cm cells, each having a conversion efficiency of 14%. Solution

Using a solar constant of 1.357 kW/m2, the required area is determined by dividing the design power requirement by the amount of the incident radiation that is converted to electricity by the cells: 10/[(0.14)(1.357)] = 52.64 m2 The number of 10 by 10 cells is then 52.64/(10/100)2 = 5264 cells ____________________________________________________________________ The direct conversion of solar radiation to electricity is very attractive from the point of view of the long-term reliability of the source and environmental

509

acceptability. There are, however, several fairly obvious adverse characteristics of solar conversion. First, terrestrial applications suffer from the limited daily availability of sunshine. The frequent coincidence of the daily peak irradiance and utility demand peaks is advantageous but does not compensate for the wide daily variability of the solar source. The occurrence of cloudy days, sometimes several successive ones, and the inevitable daily sunset indicate the need for energy storage if solar energy is to provide a reliable primary source of electricity. For remote, stand-alone locations this implies the use of batteries or other storage devices to satisfy 24 hr/day demand. Figure 11.39 shows a remote photovoltaic-powered telecommunication site located on a mountain top. The system comprises the cell array, the storage batteries, the telecommunications load, and the electronics needed to control battery charging and load matching. The system shown is said to have recovered its cost in a year and to produce annual savings of $60,000. In industrialized countries, such as the United States, where utility grids penetrate to most parts of the country, linkages with utilities seem to offer the best hope for widespread photovoltaic use, because the existing utility grids can provide the needed backup. Both decentralized and central-station solar conversion appear to be possibilities for utility involvement. Moreover, the Public Utilities Regulatory Policy Act (PURPA) offers other possibilities by encouraging third-party as well as utility and customer ownership arrangements.

510

The high cost and low efficiency of solar cells are major deterrents to their widespread application for power generation. Although the costs of solar cells have been declining and cell efficiencies increasing over the years, solar cells still are economically suited for only a limited number of applications such as those where cost is not a major factor (e.g., military and aerospace applications) or where the alternatives are expensive (e.g., energy sources for remote installations). Nevertheless the annual production of photovoltaics has been increasing and the resulting production experience will help develop production techniques that, together with increased volume, should further drive down their prices. It is likely that the demand for photovoltaic power generation would increase dramatically if solar cells become available in a price range below $1/W, depending on the price of natural gas, oil and coal. Concentrating Solar Cell Systems One approach to reducing the cost and increasing the efficiency of photovoltaics is the development of a concentrating solar cell system that focuses radiation on the cell. The use of a Fresnel lens or other type of concentrator increases the incident energy on a given cell area and, consequently, the cell power output. A photograph of such a concentrating array is shown in Figure 11.40. Each of the 14 elements in each

511 parquet concentrates energy from a 5-inch-square Fresnel lens onto a 1/4-inch silicon cell. Unlike nonconcentrating cells, the concentrating units only function under clear sky conditions and derive no benefit from diffuse radiation. Cells with high concentration ratios must be cooled because of high solar intensities on the cells. Some of the details of the EPRI-Stanford cell system adapted from reference 63 are given in Figure 11.41. According to that reference and reference 49, efficiencies as high as 28% have been reached in experimental crystalline silicon concentrator cells. Reference 59 suggests that the required high efficiencies for commercialization of concentrating cells are being achieved today and that attention can be focussed on production and cost reduction efforts for this type.

512

Other Approaches Major reasons for the high cost of solar arrays is the complexity and labor intensiveness of production of high-quality crystalline silicon. Noncrystalline silicon, or amorphous silicon, may also be used for photovoltaic cells, but with a practical conversion efficiency limit of about 14%. The best commercially available amorphous cells have an efficiency of about 6%. Crystalline silicon currently has most of the market but amorphous silicon and other non-crystalline semiconductors are receiving considerable research attention. Whereas crystalline cells are cut from carefully grown crystals, amorphous cells may be created by vacuum deposition of a thin film of silicon on a substrate such as glass. While not without problems, manufacturing techniques for amorphous materials appear to offer greater potential for low-cost mass production than those for crystalline cells. One of the current research problems, however, is that amorphous silicon loses some of its effectiveness rapidly upon exposure to light. This has not stopped its use in some applications but is a serious concern. An exciting approach to photovoltaic power technology was announced in 1991 by Texas Instruments and Southern California Edison. The concept, referred to as Spheral SolarTM* technology, involves the use of 17,000 tiny silicon balls per 100 cm2 supported on flexible aluminum foil sheets, as seen in Figure 11.42. The spheres are metallurgical grade p-type silicon with surfaces doped with an n-type material so that each sphere is an individual solar cell. The spheres are set in perforations in a foil

513

layer that forms the contact with the n-layer. Both sides of the spheres are coated with transparent ethylene vinyl acetate. The rear tips of the spheres are etched, exposing p-type material where another foil layer forms the contact with the p-layer. Several features claimed for the spheral technology offer hope for a breakthrough in the production of solar electricity:

C The silicon used is inexpensive, low purity, metallurgical grade. C The technology uses low-cost production methods that provide an almost 100% yield of manufactured laminates.

C Since each sphere is a solar cell, the failure of an individual sphere has a negligible influence on system performance.

C The flexible nature of the laminate makes it adaptable to a wide range of applications (Figure 11.43).

Module efficiencies of about 8–10% are expected (ref. 76). Reference 80 indicates that the highest efficiency attained during R&D testing was 11.5%. A line for pilot production prototype modules was established in 1991, with production commercial planned for 1994. More recent information is available in reference 96. The application and sales of photovoltaics have been growing continually in the last 20 years. Though photovoltaics continue to offer great hope for large-scale power production, there remains a diversity of opinion regarding the avenues and rates of commercialization of photovoltaics in power production applications (ref. 49).

514 11.6 A Hydrogen Economy

Coal, natural gas, petroleum, and uranium are all primary energy sources, mainly for heat. On the other hand, electricity is a convenient form of energy, or work, derived from a primary source and may therefore may be called a secondary energy source. Electricity plays an important role in the world energy system because it is a convenient and inexpensive source of both heat and work. It is used as readily for lighting and cooling as for heating and mechanical power delivery. It is readily transmitted over moderate distances and is relatively safe and non-polluting. Its pervasive influences on society within the world’s energy system are such that we have what is sometimes referred to as “an electrical economy.” However, electricity has failed to gain a foothold in most modes of transportation, primarily because of the lack of a compact and flexible means of storing it. Some progress in this area may be anticipated because of renewed interest in electric cars, trucks and other road vehicles. A primary dissatisfaction with electricity as a means of energy application lies in the facts that an electrical economy requires the consumption of about three units of primary energy for each unit of secondary energy produced and that most of the primary energy conversion processes have adverse environmental consequences. Much interest has been expressed in supplementing or replacing electricity with another secondary energy source, with hydrogen being the leading contender. Methanol, ethanol, liquefied natural gas (LNG), and compressed natural gas (CNG) are other possibilities. A number of advantages may be attributed to employing hydrogen as a secondary energy source:

C Hydrogen may be produced by chemical processes from fossil fuels, by

hydrolysis of water with oxygen as a by-product, or by thermal decomposition. According to reference 16, electrolysis cells produce hydrogen at efficiencies between 60 and 69%; 75 to 80% was feasible with 1973 technology, and 85% is expected in a well-developed hydrogen economy.

C Hydrogen can be used as a fuel to produce power with almost no direct

environmental pollution, because the principal product is water vapor. Combustion of hydrogen produces no carbon products and therefore no “greenhouse gases.” Oxides of nitrogen would remain as pollutants but should be more easily controllable because of the absence of carbon and sulfur-containing pollutants.

C Internal combustion engines can use hydrogen with minor modifications, sometimes with improved efficiency.

C Fuel cells employing hydrogen can be used to produce electricity at high

efficiencies not bounded by the Carnot limit and with little environmental impact.

515

C Hydrogen could be produced by electrolysis at large base-loaded electrical

power plants during periods of low demand for electricity, thus providing an opportunity for electrical energy storage when used with fuel cells or other efficient energy converters.

C Hydrogen could be transported in pipelines similar to natural gas pipelines. In

some cases, existing pipelines could be used with relatively little modification. Liquid hydrogen has been routinely shipped in special rail cars and trucks for many years.

C Hydrogen can be stored in the same ways that natural gas and helium are stored now, underground or in liquefied form in special storage tanks.

C Hydrogen has a very high heating value on a mass basis but a low volumetric

value because of its high specific volume at standard pressure and temperature. The numerous attractive features of hydrogen as a fuel are to a certain extent offset by some serious concerns, which are the objects of extensive engineering research: First, the ignition energy of hydrogen is low and the flammability limits are wide, so hydrogen is readily ignited. While this is advantageous in combustion systems, great care must be taken in storing and handling hydrogen. Though an atypical event, the Hindenburg disaster perpetually reminds us of these concerns. On the other hand, because hydrogen is lighter than air and diffuses more rapidly than other gases, it is more readily diluted and disperses quickly, characteristics that tend to reduce fire and explosion hazards. The hazards associated with using hydrogen gas should not be minimized, but to put things in perspective, we should remember that society has enthusiastically adopted an automobile that commonly carries ten to twenty gallons of gasoline, a very hazardous fuel. Large storage tanks of gasoline are maintained and freely accessed by untrained users in the midst of residential areas. Trucks loaded with gasoline routinely service and pass through these same areas. Second, hydrogen’s high specific volume makes its storage a problem, especially in transportation applications, where space for fuel storage is costly. In addition to storage as a compressed gas, it may be liquefied and stored at cryogenic temperatures, or it may be stored in metal hydrides. In the latter cases, the additional equipment space, expense and complexity must be considered in evaluating feasibility and design. Finally, while gasoline and natural gas remain readily available and at a modest price, hydrogen will have little opportunity to become established as an important secondary fuel. For ground transportation, as a replacement fuel for gasoline, hydrogen must compete with relatively inexpensive methanol and ethanol (ref. 9). Heavily subsidized ethanol is currently available in the United States in a 10% blend with gasoline and there is considerable interest in compressed natural gas and richer methanol mixtures for ground vehicles (ref. 74)

516 EXAMPLE 11.12

Using reasonable estimates of unit efficiencies for generation, transport, and utilization, compare the overall energy efficiencies of nuclear-electric and nuclear-hydrogen economies in providing mechanical work. Solution

For the nuclear-electric economy, mechanical energy is produced in the sequence Fission 6Electricity 6Transmission 6Electric motor which, using unit efficiencies in the order given, would have an efficiency of about (0.32)(0.9)(0.9) = 0.259 For the nuclear-powered hydrogen economy the appropriate chain is Fission 6Electricity 6Electrolysis 6Hydrogen 6Transmission 6Fuel cell 6Motor Here the overall efficiency is estimated to be (0.32)(0.85)(0.9)(0.6)(0.9) = 0.132 Note that neither case includes the energy costs of mining and processing nuclear fuel. Nevertheless, this example suggests that, despite hydrogen’s many advantages, its use as a secondary fuel may lead to inefficient resource utilization. This conclusion is discussed more fully in reference 16. _____________________________________________________________________ One can readily appreciate the beauty of the concept of a solar-energy-driven hydrogen-powered cycle in which hydrogen produced by electrolysis of seawater using solar photovoltaic energy, reacts and delivers power and heat in a high-efficiency and non-polluting fuel cell at the point of use, with product water ascending into the atmosphere, where it eventually returns to the sea as rain. The problems in developing and adopting such a cycle for commercial use are great, but the future needs of Earth are so serious that such concepts at least provide directions in which research should proceed. While the hydrogen economy may not displace the electric economy, when the politics and economics are right, society will find ways to use hydrogen as a supplement to electricity as a secondary fuel. 11.7 Concluding Remarks

It is clear that the seriousness of Earth's energy and environmental problems and the growing diversity of propulsion and power production options ensures stimulating

517 careers for engineers far into the future. Attractive, mid-term alternatives exist to the conventional, state-of-the-art, coal-burning power plant with advanced scrubber technology. Despite the risks involved, it seems clear that numerous other power options will find commercial use as dictated by local and regional conditions and available technologies and resources. A recurring question in all of the developments we have discussed exists: Once the feasibility of a new technology has been established, how can it be rapidly brought to a commercial stage in the absence of massive government support for development and demonstration plants? Under existing economic and regulatory structures in the United States, power utilities cannot make massive investments in technologies that involve significant risks or costs that are not competitive with existing methods. The prices and risks of new technologies will remain high, however, until a significant number of units have been ordered that can absorb research, development, and production costs. In many cases it is unlikely that timely U.S. investment capital will be available for demonstration plants to provide visible confidence-building utility experience. While U.S. governmental support exists, it appears that much of the costs and risks, together with the successes, will be borne internationally. An important long-term question remains: How can humankind make the transition from a resource-depleting, highly polluting society to some semblance of a steady-state, environmentally acceptable, energy economy based on renewable energy sources? This is not a question for the energy conversion engineer alone. Indeed, it involves political, moral, and social issues that transcend the boundaries of science and engineering. Can a smooth transition be achieved that avoids cataclysmic change and minimizes painful societal dislocations? Perhaps the internationalization of the leadership in technology will help bring suitable global responses to these formidable challenges. Regardless of the future course, it must be a responsibility of engineers to keep the long-term needs of Earth and its inhabitants in mind as they make technical decisions to satisfy present needs and desires and continue the quest for technical progress.

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522 62. Sze, S. M., Semiconductor Devices Physics and Technology. New York: Wiley, 1985. 63. Moore, Taylor, et al., "Opening the Door for Utility Photovoltaics," EPRI Journal, January–February 1987: 5–15. 64. Griffiths, David, J., Introduction to Electrodynamics, 2nd ed. Engelwood Cliffs, N. J.: Prentice-Hall, 1989. 65. Anon., “Solar Electric Generating Stations (SEGS),” IEEE Power Engineering Review, August, 1989: 4–8. 66. Anon., “Promise of Solar Energy Being Fulfilled in California,” Power, October 1989: s32—s36. 67. Romano, Samuel, “Fuel Cells for Transportation,” Mechanical Engineering, August 1989: 74–77. 68. Hirschenhofer, J. H., “International Developments in Fuel Cells,” Mechanical Engineering, August 1989: 78–83. 69. Carlson, D. E., “Photovoltaic Techniques for Commercial Power Generation,” Annual Review of Energy, Vol. 15, 1990, pp. 85–98. 70. MacDonald, Gordon J., “The Future of Methane as an Energy Resource,” Annual Review of Energy, Vol. 15, 1990, pp. 53–83. 71. Moore, Taylor, et al., “On-Site Utility Applications for Photovoltaics,” EPRI Journal, March 1991: 26–37. 72. Moore, Taylor, et al., “Thin Films: Expanding the Solar Marketplace,” EPRI Journal, March 1989: 4–15. 73. Smock, Robert W., “Second Generation Fuel Cell Technology Moves Toward Demos,” Power Engineering, June 1990: 10. 74. Sapre, Alex R., “Properties, Performance and Emissions of Medium-Concentration Methanol-Gasoline Blends in a Single-Cylinder, Spark-Ignition Engine,” SAE Paper 881679, October 1988. 75. Douglas, John, “Beyond Steam: Breaking Through Performance Limits,” EPRI Journal, December 1990: 5–11.

523 76. Graff, Eric, Texas Instruments Co. Personal Communication, May 24, 1991. 77. Chase, M. W. Jr., et al., JANAF Thermochemical Tables, 3rd ed., J. Phys. Chem. Ref. Data 14, Supplement No. 1, 1985. 78. Douglas, John, et al., “Fuel Cells for Urban Power,” EPRI Journal, September 1991: 5–11. 79. Howell, John R., Bannerot, Richard B., and Vliet, Gary C., “Solar-Thermal Energy Systems.” New York: McGraw-Hill, 1982. 80. Levine, Jules D., et al., “Basic Properties of the Spheral Solar Cell.” 22nd Photovoltaic Specialists Conference, Las Vegas, Nev., 1991. 81. “International Fuel Cells: Clean, Reliable Fuel Cell Energy,” www.internationalfuelcells.com/index_fl1.shtml (November 14, 2000). 82. Rulseh, Ted, “Fuel Cells: From Promise to Performance,” Grid, Spring/Summer 2000: 15. 83. “Santa Clara Demonstration Project,” www.ttcorp.com/fccg/scdpnew1.htm (November 14, 2000). 84. “Fuel Cell Energy, Carbonate Fuel Cell Manufacturer,” www.ttcorp.com/fccg/erc_abt.htm (November 15, 2000). 85. “Welcome to Fuel Cell Energy, Inc.,” www.fuelcellenergy.com/homeframe.html (December 9, 2000). 86. “Library of Fuel Cell Related Publications,” http://216.51.18.233/biblio.html (November 14, 2000) 87. Archer, David H. and Wimer, John G., “ A Phosphoric Acid Fuel Cell Cogeneration System Retrofit to a Large Office Building,” Department of Energy FETC-97/1044, April 1997., www.fetc.doe.gov/netltv/index.html (November 15, 2000). 88. “DaimlerChrysler Offers First Commercial Fuel Cell Buses to Transit Agencies,” www.hfcletter.com/letter/may00/feature.html (November 15, 2000). 89. “TroughNet Projects - Projects Deployed,” www.eren.doe.gov/troughnet/deployed.html (November 23, 2000).

524 90. “Solar Fact Sheets: Solar Thermal Electricity, Solar Energy Industries Association,” www.seia.org/sf/sfsolthe.htm (November 23, 2000). 91. “Solar Thermal Electric Power Plant,” www.magnet.consortia.org.il/ConSolar/stepp.html (November 23, 2000). 92. “About FPL Energy,” www.fplenergy.com/aboutfpl/solar-1.htm (November 23, 2000). 93. “US Department of Energy PV Home Page,” www.eren.doe.gov/pv/pvmenu.cgi?site=pv&idx=0&body=video.html (November 24, 2000). 94. “Solar Information Center - Photovoltaic generation systems - Kyocera Solar, Inc.,” www.kyocerasolar.com/info/solarenergy.html (November 23, 2000). 95. “Photovoltaics Program,” http://www.sandia.gov/pv/ ( November 24, 2000). 96. “TI Seeks Buyer For Spheral Solar Technology,” www.ti.com/corp/docs/press/company/1995/510no.shtml (November 25, 2000).

EXERCISES

11.1* Derive an expression for the battery power output, in terms of E and Ri /Ro, for the linear model discussed in connection with Example 11.2. Nondimensionalize the power by dividing by E2/Ro. Use a spreadsheet to tabulate and plot the dimensionless power as a function of Ri /Ro. 11.2* Derive an expression for the linear-battery-model power output nondimensionalized by E2/Ri in terms of the internal-to-external resistance ratio. Use a spreadsheet to tabulate the dimensionless power function, and plot it. Is there a condition that produces an extreme value of the dimensionless power? If so, use calculus methods to derive the condition. 11.3

An automobile storage battery with and open-circuit voltage of 12.8V is rated at 260 A-hr. The internal resistance of the battery is 0.2 S. Estimate the maximum duration of current flow and its value through an external resistance of 1.8 S. _________________________ *Exercise numbers with an asterisk involve computer usage.

525 11.4

A battery electrical storage plant is to be designed for 20-MW peak power delivery for a duration of four hours. The plant uses 600 A-hr batteries operating at 400 volts DC. Estimate the minimum number of batteries and the current in each during peak operation.

11.5

A hydrogen-oxygen fuel cell operates with a voltage of 0.7 volts with watervapor product. Calculate the work per kg-mole of hydrogen, in kJ and in kWhr, and determine the cell efficiency.

11.6

A neighborhood fuel cell power plant is to be designed for an electrical power output of 2000 kW with liquid-water product. Estimate the flow rates of hydrogen and oxygen during peak power production, assuming that an 80% efficient power conditioner is used to convert DC to AC power and the fuel cell efficiency is 55%. What is the plant heat rate?

11.7

A hydrogen-oxygen fuel cell has liquid water as product when it operates at 0.82 volts. What is the electrical energy output in kJ / kg-mole of hydrogen, and the cell efficiency?

11.8

Prepare a typed three-page, double-spaced memorandum outlining the design of the power train of a hydrogen-oxygen fuel-cell-powered automobile, giving the design criteria, system description, and quantitative preliminary design data on the power and fuel supply system.

11.9

A hydrogen-oxygen fuel cell stack produces 50 kW of DC power at an efficiency of 60%, with water vapor as product. What is the hydrogen mass flow rate, in g/s, and the cell voltage?

11.10 A fuel cell power plant is to be designed for an electrical power output of 200 MW with liquid-water product. Estimate the flow rates of hydrogen and oxygen during peak power production with an 85% efficient power conditioner and fuel cell efficiency of 55%. What is the plant heat rate? 11.11 Determine the fraction of the solar spectrum that lies in the visible wavelengths between 0.4 and 0.7 :. What is the ratio of the energy in this visible range to that in a range of the same width between 2.0 and 2.3 :? 11.12 It is desired to have a window in a south-facing wall shaded by a 1-m overhang at noon on June 21 and fully exposed to the sun at noon on December 21. Determine the maximum vertical size of the window for a house at latitude 35° north.

526 11.13 A skylight in a horizontal roof is 10ft. above the floor of the room below. Determine the distance that the image of the skylight on the floor at noon moves between December 21 and June 21 at latitude 30° north. Sketch and label a plan view of the two positions. 11.14 Derive an equation for the duration of daylight as a function of latitude for December 21 and June 21. Use a spreadsheet to create a table and a plot of the hours of daylight as a function of latitude for December 21 and June 21. 11.15 Perform an analysis analogous to Example 11.12, comparing the energy efficiency of using utility central solar photovoltaic electricity with (1) an electric heat pump for space heating and (2) with solar hydrogen production for space heating using a furnace. 11.16 An MHD generator uses helium seeded with cesium at 2200K to give an electrical conductivity of 10 (S-m)–1. The gas travels at 1000 m/s in a magnetic field of 3 Webers/m2. What MHD generator volume is needed to produce 50MW of output power with a load factor is 0.6? 11.17 The electrical conductivity of C2H4 burned in oxygen at 3000K with a small amount of potassium seed is 60 (S-m)–1. An MHD generator operates with a gas velocity of 1000 m/s, the magnetic field intensity is 5 tesla, the channel is 1 meter square in cross-section, and the load factor is 0.5. What is the opencircuit voltage, the load potential difference, and the short-circuit and operating current density? If the electrode area is 50% of the wall area and the channel is 10 m long, what is the current and the power output? 11.18 Prepare a preliminary design of a solar photovoltaic system to provide 1.0 kW of stand-alone, 24 hour / day power to a travel trailer. Write a report giving details, including a schematic of the overall system, a cost estimate, the cell array design, and a discussion of the assumptions on which the design is based. 11.19 Estimate the electrical power requirement in kW, of a 1400-ft2 floor area (three bedroom home) with three occupants. Using your home power estimate, predict the power requirement for a city of 300,000 people. Use these results to estimate the area of silicon solar cells required to satisfy the community power requirements. Write a short narrative discussing your assumptions and analysis. 11.20 Estimate the efficiency of a parquet of an EPRI-Stanford concentrating solar cell.

527 11.21 According to reference 65, a 400-MW expansion of the SEGS plants in southern California is expected to cost $1.4 billion. What is the expected capital cost of the generation facilities, in dollars per kW of capacity? Estimate the minimum cost of electricity per kW-h generated by the new facilities if they are operated at full capacity for six hours per day for 30 years. 11.22 Evaluate the efficiency of a solar-photovoltaic-powered hydrogen economy for comparison with the nuclear-powered version given in Example 11.12. Consider the cases where the energy source is (a) photovoltaic electricity, and (b) the sun. 11.23 Assuming 8% efficiency for a spheral solar laminate, what is the maximum electrical output of a one square meter sheet? Estimate the peak and the annual electrical loads for three-person family home. Determine the required area for a fixed spheral solar installation. Indicate clearly the assumptions made. 11.24 Assuming 10% efficiency for a spheral solar laminate, what is the maximum electrical output of a one square foot sheet? Estimate the peak and the annual electrical loads for three-person family home. Determine the required area for a fixed spheral solar installation. Indicate clearly the assumptions made.

Table A.1 Physical Constants __________________________________________________ Name

Value

Units of measure

2.9979x108 m/s 2 6.673x10 –11 N-m2 / kg 6.022x1023 atoms/g-mole 1.38x10 –23 J / °K 0.082057 L-atm/g-mole -°K 8.314 kJ /kg-mole - °K 1545.3 ft-lbf /lb-mole - °R 1.986 Btu /lb-mole - °R 1.986 cal / g-mole - °K 0.730 atm-ft3 /lb-mole - °R 10.73 psia - ft3 /lb-mole - °R Volume of ideal gas, at STP 22.41 m3/kg-mole –27 Unified atomic mass unit, amu 1.660531x10 kg Planck's constant 6.626196x10–34 J-s Electron charge 1.602192×10–19 C Electron rest mass 9.109558x10–31 kg 5.485930x10–4 amu Proton rest mass 1.672614x10–27 kg 1.00727661 amu Neutron rest mass 1.674920x10–27 kg 1.00866520 amu Electron-charge-to-mass ratio 1.758803x1011 C / kg Stefan-Boltzmann constant 5.66961x10–8 W / m2 - °K4 Faraday's constant 9.6487x107 C / kg-mole ________________________________________________________ Speed of light in a vacuum Gravitational constant Avogadro's number Boltzmann constant Universal gas constant

Table A.2 Conversion Factors ________________________________________________________________________________ Length 1 ft 1 in 1 mile 1 km 1m

1 cm

Mass 1 Ibm 1 slug 1 kg 1 ton: metric short

= 0.3084 m = 2.540 cm = 5280 ft = 1.6093 km = 1000 m = 100 cm = 1.0936 yd = 3.2808 ft = 0.3937 in.

= 0.45359237 kg = 7000 grain = 32.174 Ibm = 14.5939 kg = 1000 g = 2.2046 Ibm = 1000 kg = 2000 Ibm

Pressure 1 kPa = I000 N / m2 = 20.886 Ibf /ft2 1 atm = 760 torr = 1.01325 x 105 N/m2 = 14.696 Ibf /in2 = 29.92 in. Hg 1 torr = 1 mm Hg = 1.933 × 10–2 psi 1 mm Hg = 0.01934 Ibf /in2 1 bar = 0.9869 atm = 105 N /m2 = 106 dyne/cm2 1 in. Hg = 0.491 Ibf /in2 = 0.0334 atm = 33,864 dyne/cm2 2 1 dyne/cm = 10 –1 N/m2

Volume I liter = 10 –3 m3 = 1000 cm3 = 0.035313 ft3 = 0.26417 gal = 61.025 in3 1 gal = 3.7854 liter = 0.13368 ft3 I ft3 = 28.317 liter = 7.4805 gal 3 1 in = 16.387 cm3

Force I Ibf = 4.4482 N = 444,800 dyne = 32.174 Ibm-ft/s2 1 N = 1 kg-m/s2 = 0.22481 Ibf 1 dyne= 1 g-cm/s2 = 10 –5 N Time 1h I min 1 ms 1 :s

= 3600 s = 60 s = 10 – 3s = 10 –6 s

Energy 1J = 1 kg-m2/s2 = 9.478×10 – 4 Btu = 107 erg 1 erg = 1 dyne-cm 1cal = 4.186 J 1 Btu = 252.16 cal = 1.05504 kJ = 778.16 ft-lbf 1 ft-lbf = 1.3558 J 1 ev = 1.602 x 10 –19 J

Table A.2 (continued) ________________________________________________________________________________ Temperature 1 °K = 1.8 °R °K = °C + 273.15 °C = (°F – 32)/1.8 °R = °F + 459.67 Electromagnetic units 1 ampere = 1 C/s = 1 W/v 1 volt = 1 J/C 1 ohm = 1 v/A 1 farad = 1 A-s/v 1 henry = 1 v-s/A 1 weber = 1 v-s 1 tesla = 1 Wb/m2 = 1 N/A-m = 104 gauss Density 1 Ibm/ft3 = 0.01602 g/cm3 = 16.02 kg / m3

Power 1W

1 kW

1 hp

= 1 J/s = 1 kg-m2/s3 = 860.42 cal /hr = 1000 W = 3412.2 Btu/h = 1.341 hp = 737.56 Ibf -ft/s = 2544.5 Btu/h = 745.7 W = 550 ft-lbf /s = 33,000 ft-lbf /min

Specific Energy 1 Btu/Ibm = 2.3259 kJ/kg Energy Flux 1 Btu/ft2-s = 11.356 kW/m2

APPENDIX B Properties of Steam (English units) Table B.1 Saturated Steam: Temperature Table Table B.2 Saturated Steam: Pressure Table Figure B.1 Mollier Diagram for Steam (two parts) Figure B.2 Temperature–Entropy Diagram for Steam Figure B.3 Pressure–Enthalpy Diagram for Steam

This page is intentionally blank.

APPENDIX C Properties of Steam (SI units) Table C.1 Saturated Steam (Temperature) From NBS/NRC Steam Tables by L. Haar, et.al., New York: Hemisphere Publishing, 1984; with permission

APPENDIX C Properties of Steam (SI units) Figure C.1 Isentropic Exponent for Steam Figure C.2 Mollier Diagram for Steam Figure C.3 Pressure–Enthalpy Diagram for Steam Figure C.4 Temperature–Entropy Diagram for Steam From NBS/NRC Steam Tables by L. Haar, et.al., New York: Hemisphere Publishing, 1984; with permission

AP P E N D I X D Enthalpy of Selected Substances Table D.1 Table D.2 Table D.3 Table D.4

Enthalpy (Btu/lb-mole) Enthalpy (Btu/Ibm) Enthalpy (kJ/kg) Enthalpy (kJ/g-mole)

Reproduced with Permission from: Chase, M.W. Jr., Davies, C.A., Downey, J.R., Frurip D.J., McDonald, R.A., JANAF Thermochernical Tables, Third Edition, Part 1 (Al-Co) and Part H (Cr-Zr), J. Phys. Chem. Ref. Data 14, Supplement No. 1, 1985. The primary data in the JANAF Tables is presented on a gram-mole basis in S1 units. He the primary enthalpy data is presented in units of kJ/gram-mole. Tables in other sets of uni are derived from this primary table. For instance enthalpy in kJ/kg is obtained by dividin the primary values by the appropriate molecular masses and multiplying by 1000. Mass-base values in the English system are obtained from the SI mass-based values by multiplying by th conversion factor 0.43021 Btu-kg/kJ-lbm. Mole-based values in the English system are obtaine directly from the primary values by multiplying by (0.43021)(1000 JANAF Table temperatures are given in degrees Kelvin. Rankine temperatures listed in t present tables are a factor of 1.8 times the Kelvin values

Table D.1 Enthalpy (Btu/lb-mole TEMPERATURE

ENTHALPY (Btu/lb-mole)

R K O2 N2 C CO CO2 H2 H2O SO2 _______________ _____________________________________________________________________ 0 0 -3736 -3730 -452 -3730 -4028 -3643 -4261 -4540 180 100 -2486 -2481 -426 -2482 -2777 -2352 -2846 -3105 360 200 -1234 -1229 -286 -1230 -1469 -1193 -1412 -1607 536.7

298.15

0

0

0

0

0

0

0

0

540 720 900 1080 1260 1440 1620 1800 1980 2160 2340 2520 2700 2880 3060 3240 3420 3600 3780 3960 4140 4320 4500 4680 4860 5040 5220 5400

300 400 500 600 700 800 900 1000 1100 1200 1300 1400 1500 1600 1700 1800 1900 2000 2100 2200 2300 2400 2500 2600 2700 2800 2900 3000

23 1301 2617 3977 5377 6812 8278 9767 11277 12803 14345 15899 17466 19044 20632 22230 23839 25458 27086 28725 30373 32030 33698 35374 37059 38753 40455 42166

23 1278 2543 3826 5135 6473 7840 9234 10652 12093 13553 15030 16522 18028 19544 21071 22607 24151 25702 27259 28822 30390 31963 33540 35122 36707 38295 39887

7 447 1017 1696 2459 3286 4161 5074 6018 6987 7976 8982 10004 11038 12085 13142 14208 15283 16367 17458 18557 19662 20774 21893 23018 24148 25285 26427

23 1280 2552 3847 5172 6529 7916 9331 10770 12231 13710 15205 16714 18234 19766 21307 22860 24412 25974 27542 29118 30684 32259 33846 35436 37030 38627 40226

30 1722 3573 5553 7638 9811 12059 14368 16728 19133 21574 24047 26546 29069 31612 34172 36748 39338 41940 44553 47177 49809 52450 55098 57754 60416 63084 65759

23 1273 2530 3791 5055 6325 7604 8897 10204 11528 12871 14232 15612 17011 18428 19862 21313 22780 24263 25759 27270 28793 30329 31877 33436 35006 36586 38177

27 1485 2979 4518 6106 7745 9438 11185 12988 14845 16753 18711 20715 22762 24848 26971 29128 31315 33531 35773 38040 40328 42637 44966 47312 49674 52051 54443

32 1828 3768 5827 7980 10205 12486 14811 17171 19559 21970 24401 26847 29307 31780 34263 36755 39257 41765 44281 46803 49332 51866 54405 56950 59499 62053 64611

Table D.1 continued. Enthalpy (Btu/lb-mole TEMPERATURE

ENTHALPY (Btu/lb-mole)

R K O2 N2 C CO CO2 H2 H2O SO2 _______________ _____________________________________________________________________ 5580 3100 43885 41481 27574 41829 68438 39777 56848 67173 5760 3200 45612 43079 28727 43434 71123 41387 59264 69740 5940 3300 47346 44678 29885 45041 73812 43006 61693 72310 6120 3400 49088 46281 31049 46651 76507 44634 64133 74885 6300 3500 50836 47885 32218 48263 79206 46271 66583 77462 6480 3600 52591 49492 33392 49877 81909 47917 69042 80044 6660 3700 54352 51101 34570 51493 84616 49571 71510 82630 6840 3800 56120 52712 35754 53111 87328 51233 73988 85219 7020 3900 57893 54324 36943 54731 90043 52904 76473 87811 7200 4000 59672 55939 38136 56353 92763 54583 78966 90407 7380 4100 61457 57555 39335 57976 95486 56269 81466 93005 7560 4200 63247 59173 40538 59601 98213 57964 83973 95608 7740 4300 65043 60793 41746 61228 100943 59667 86487 98214 7920 4400 66843 62414 42959 62856 103677 61377 89007 100822 8100 4500 68650 64037 44165 64486 106414 63095 91533 103434 8280 4600 70461 65661 45398 66118 109155 64820 94065 106049 8460 4700 72277 67287 46625 67751 111899 66553 96603 108668 8640 4800 74099 68914 47857 69385 114646 68292 99145 111289 8820 4900 75927 70543 49093 71021 117397 70039 101693 113913 9000 5000 77760 72173 50333 72658 120150 71792 104246 116541 9180 5100 79599 73805 51579 74297 122907 73552 106803 119171 9360 5200 81444 75438 52829 75937 125668 75317 109366 121805 9540 5300 83295 77073 54083 77578 128433 77089 111934 124441 9720 5400 85153 78709 55342 79222 131202 78867 114507 127081 9900 5500 87017 80347 56606 80861 133974 80649 117085 129723 10080 5600 88889 81986 57874 82504 136751 82437 119668 132368 10260 5700 90769 83626 59146 84150 139532 84230 122256 135016 10440 5800 92657 85268 60423 85799 142316 86026 124849 137667 10620 5900 94552 86912 61705 87448 145105 87827 127447 140321 10800 6000 96457 88558 62991 89099 147897 89630 130050 142979

Table D.2. Enthalpy (Btu/lbm) TEMPERATURE

ENTHALPY (Btu/lbm)

R K O2 N2 C CO CO2 H2 H2O SO2 _______________ ______________________________________________________________________ 0 0 -116.7 -133.2 -37.6 -133.2 -91.5 -1806.8 -236.5 -252.0 180 100 -77.7 -88.6 -35.5 -88.6 -63.1 -1166.9 -158.0 -172.3 360 200 -38.6 -43.9 -23.8 -43.9 -33.4 -592.0 -78.4 -89.2 536.7

298.15

0.0

0.0

0.0

0.0

0.0

0.0

0.0

0.0

540 720 900 1080 1260 1440 1620 1800 1980 2160 2340 2520 2700 2880 3060 3240 3420 3600 3780 3960 4140 4320 4500 4680 4860 5040 5220 5400

300 400 500 600 700 800 900 1000 1100 1200 1300 1400 1500 1600 1700 1800 1900 2000 2100 2200 2300 2400 2500 2600 2700 2800 2900 3000

0.7 40.7 81.8 124.3 168.0 212.9 258.7 305.2 352.4 400.1 448.3 496.9 545.8 595.1 644.8 694.7 745.0 795.6 846.5 897.6 949.2 1001.0 1053.0 1105.4 1158.1 1211.0 1264.2 1317.7

0.8 45.6 90.8 136.6 183.3 231.1 279.9 329.7 380.3 431.7 483.9 536.6 589.9 643.6 697.8 752.3 807.1 862.2 917.6 973.2 1029.0 1085.0 1141.1 1197.4 1253.9 1310.5 1367.2 1424.0

0.6 37.2 84.7 141.2 204.7 273.5 346.4 422.5 501.1 581.7 664.0 747.8 832.9 919.0 1006.1 1094.1 1182.9 1272.4 1362.7 1453.5 1545.0 1637.0 1729.6 1822.7 1916.4 2010.5 2105.1 2200.2

0.8 45.7 91.1 137.3 184.7 233.1 282.6 333.1 384.5 436.7 489.5 542.8 596.7 651.0 705.7 760.7 816.1 871.5 927.3 983.3 1039.5 1095.5 1151.7 1208.3 1265.1 1322.0 1379.0 1436.1

0.7 39.1 81.2 126.2 173.6 222.9 274.0 326.5 380.1 434.7 490.2 546.4 603.2 660.5 718.3 776.5 835.0 893.8 953.0 1012.3 1072.0 1131.8 1191.8 1251.9 1312.3 1372.8 1433.4 1494.2

11.3 631.4 1255.2 1880.2 2507.2 3137.4 3772.0 4413.1 5061.6 5718.4 6384.4 7059.6 7744.2 8438.0 9140.9 9852.4 10571.9 11299.6 12035.0 12777.4 13526.6 14282.3 15044.1 15811.9 16585.3 17364.0 18148.0 18936.9

1.5 82.4 165.4 250.8 338.9 429.9 523.9 620.9 720.9 824.0 929.9 1038.6 1149.8 1263.4 1379.2 1497.1 1616.8 1738.2 1861.2 1985.6 2111.4 2238.5 2366.6 2495.9 2626.1 2757.2 2889.2 3021.9

1.8 101.5 209.1 323.4 442.9 566.4 693.0 822.1 953.1 1085.6 1219.5 1354.4 1490.2 1626.7 1764.0 1901.8 2040.2 2179.0 2318.2 2457.9 2597.9 2738.2 2878.9 3019.8 3161.1 3302.6 3444.3 3586.3

Table D.2 continued. Enthalpy (Btu/lbm TEMPERATURE

ENTHALPY (Btu/lbm)

R K O2 N2 C CO CO2 H2 H2O SO2 _______________ ______________________________________________________________________ 5580 3100 1371.4 1480.9 2295.8 1493.3 1555.1 19730.8 3155.4 3728.5 5760 3200 1425.4 1538.0 2391.7 1550.6 1616.1 20529.3 3289.5 3871.0 5940 3300 1479.6 1595.1 2488.2 1608.0 1677.2 21332.5 3424.4 4013.7 6120 3400 1534.0 1652.3 2585.1 1665.5 1738.4 22140.0 3559.8 4156.6 6300 3500 1588.6 1709.6 2682.4 1723.0 1799.7 22952.0 3695.8 4299.6 6480 3600 1643.5 1766.9 2780.1 1780.7 1861.1 23768.2 3832.3 4442.9 6660 3700 1698.5 1824.4 2878.2 1838.4 1922.7 24588.6 3969.3 4586.5 6840 3800 1753.7 1881.9 2976.8 1896.1 1984.3 25413.3 4106.8 4730.2 7020 3900 1809.2 1939.5 3075.8 1953.9 2046.0 26242.0 4244.7 4874.0 7200 4000 1864.8 1997.1 3175.1 2011.9 2107.8 27074.6 4383.1 5018.1 7380 4100 1920.5 2054.8 3274.9 2069.8 2169.6 27911.4 4521.9 5162.4 7560 4200 1976.5 2112.6 3375.1 2127.8 2231.6 28751.9 4661.0 5306.8 7740 4300 2032.6 2170.4 3475.7 2185.9 2293.6 29596.6 4800.6 5451.5 7920 4400 2088.9 2228.3 3576.6 2244.0 2355.8 30444.8 4940.4 5596.3 8100 4500 2145.3 2286.2 3677.0 2302.2 2418.0 31296.9 5080.7 5741.2 8280 4600 2201.9 2344.2 3779.7 2360.5 2480.2 32152.6 5221.2 5886.4 8460 4700 2258.7 2402.2 3881.9 2418.8 2542.6 33012.2 5362.1 6031.7 8640 4800 2315.6 2460.3 3984.4 2477.1 2605.0 33875.0 5503.2 6177.2 8820 4900 2372.7 2518.5 4087.3 2535.5 2667.5 34741.4 5644.6 6322.9 9000 5000 2430.0 2576.7 4190.6 2594.0 2730.1 35611.0 5786.3 6468.7 9180 5100 2487.5 2635.0 4294.3 2652.5 2792.7 36484.0 5928.2 6614.7 9360 5200 2545.1 2693.3 4398.3 2711.0 2855.4 37359.8 6070.5 6760.9 9540 5300 2603.0 2751.6 4502.8 2769.6 2918.3 38238.8 6213.0 6907.3 9720 5400 2661.0 2810.0 4607.6 2828.3 2981.2 39120.5 6355.8 7053.8 9900 5500 2719.3 2868.5 4712.8 2886.8 3044.2 40004.6 6498.9 7200.4 10080 5600 2777.8 2927.0 4818.4 2945.5 3107.3 40891.5 6642.3 7347.3 10260 5700 2836.5 2985.6 4924.3 3004.3 3170.5 41780.5 6786.0 7494.2 10440 5800 2895.5 3044.2 5030.7 3063.1 3233.7 42671.7 6929.9 7641.4 10620 5900 2954.8 3102.9 5137.3 3122.0 3297.1 43564.7 7074.1 7788.7 10800 6000 3014.3 3161.7 5244.4 3180.9 3360.5 44459.5 7218.6 7936.2

Table D.3. Enthalpy (kJ/kg) TEMPERATURE

ENTHALPY (kJ/kg)

R K O2 N2 C CO CO2 H2 H2O SO2 _______________ ______________________________________________________________________ 0 0 -271.3 -309.5 -87.5 -309.6 -212.8 -4199.9 -549.7 -585.7 180 100 -180.6 -205.9 -82.5 -206.0 -146.7 -2712.3 -367.2 -400.6 360 200 -89.6 -102.0 -55.4 -102.0 -77.6 -1376.0 -182.2 -207.4 536.7

298.15

0.0

0.0

0.0

0.0

0.0

0.0

0.0

0.0

540 720 900 1080 1260 1440 1620 1800 1980 2160 2340 2520 2700 2880 3060 3240 3420 3600 3780 3960 4140 4320 4500 4680 4860 5040 5220 5400

300 400 500 600 700 800 900 1000 1100 1200 1300 1400 1500 1600 1700 1800 1900 2000 2100 2200 2300 2400 2500 2600 2700 2800 2900 3000

1.7 94.5 190.1 288.9 390.6 494.8 601.3 709.5 819.1 930.0 1042.0 1154.9 1268.7 1383.3 1498.7 1614.8 1731.7 1849.2 1967.5 2086.5 2206.3 2326.7 2447.8 2569.5 2691.9 2815.0 2938.6 3062.9

1.9 106.1 211.0 317.5 426.2 537.2 650.6 766.3 884.0 1003.5 1124.7 1247.3 1371.1 1496.0 1621.9 1748.6 1876.0 2004.2 2132.9 2262.1 2391.8 2522.0 2652.5 2783.4 2914.6 3046.2 3178.0 3310.1

1.3 86.5 196.9 328.3 475.9 635.8 805.3 982.0 1164.7 1352.1 1543.5 1738.3 1936.0 2136.2 2338.7 2543.3 2749.6 2957.7 3167.4 3378.7 3591.2 3805.2 4020.4 4236.9 4454.5 4673.3 4893.3 5114.2

1.9 106.2 211.7 319.2 429.2 541.8 656.9 774.4 893.8 1015.0 1137.7 1261.8 1387.0 1513.2 1640.3 1768.1 1897.0 2025.8 2155.5 2285.6 2416.4 2546.3 2677.0 2808.7 2940.7 3072.9 3205.5 3338.2

1.6 91.0 188.7 293.3 403.4 518.2 636.9 758.9 883.5 1010.5 1139.5 1270.1 1402.1 1535.3 1669.6 1804.8 1940.9 2077.7 2215.1 2353.1 2491.7 2630.7 2770.2 2910.1 3050.4 3190.9 3331.9 3473.1

26.3 1467.8 2917.7 4370.5 5827.9 7292.7 8767.9 10257.9 11765.4 13292.2 14840.3 16409.7 18001.0 19613.6 21247.5 22901.3 24573.9 26265.4 27974.7 29700.4 31442.0 33198.4 34969.2 36754.0 38551.6 40361.6 42184.0 44017.9

3.4 191.6 384.4 582.9 787.7 999.2 1217.7 1443.2 1675.8 1915.3 2161.5 2414.1 2672.7 2936.7 3205.9 3479.9 3758.1 4040.3 4326.2 4615.5 4907.9 5203.2 5501.1 5801.5 6104.2 6409.0 6715.7 7024.3

4.1 235.9 486.1 751.8 1029.5 1316.7 1611.0 1911.0 2215.5 2523.5 2834.6 3148.2 3463.8 3781.3 4100.2 4420.6 4742.2 5064.9 5388.6 5713.2 6038.6 6364.8 6691.8 7019.4 7347.7 7676.6 8006.1 8336.1

Table D.3 continued. TEMPERATURE

Enthalpy (kJ/kg) ENTHALPY (kJ/kg)

R K O2 N2 C CO CO2 H2 H2O SO2 _______________ ______________________________________________________________________ 5580 3100 3187.8 3442.4 5336.4 3471.2 3614.7 45863.1 7334.5 8666.7 5760 3200 3313.2 3574.9 5559.5 3604.4 3756.4 47719.2 7646.4 8997.9 5940 3300 3439.2 3707.7 5783.6 3737.8 3898.5 49586.3 7959.8 9329.5 6120 3400 3565.7 3840.7 6008.8 3871.3 4040.8 51463.3 8274.5 9661.7 6300 3500 3692.7 3973.8 6235.0 4005.1 4183.3 53350.7 8590.6 9994.3 6480 3600 3820.2 4107.1 6462.2 4139.1 4326.1 55248.0 8907.9 10327.4 6660 3700 3948.1 4240.7 6690.3 4273.2 4469.1 57154.8 9226.4 10661.0 6840 3800 4076.5 4374.3 6919.4 4407.4 4612.3 59071.9 9546.0 10995.0 7020 3900 4205.3 4508.2 7149.4 4541.8 4755.8 60998.0 9866.6 11329.4 7200 4000 4334.5 4642.2 7380.4 4676.4 4899.4 62933.5 10188.3 11664.4 7380 4100 4464.2 4776.3 7612.4 4811.1 5043.2 64878.5 10510.8 11999.7 7560 4200 4594.2 4910.6 7845.2 4946.0 5187.2 66832.3 10834.3 12335.4 7740 4300 4724.6 5044.9 8079.0 5081.0 5331.4 68795.6 11158.6 12671.6 7920 4400 4855.4 5179.5 8313.7 5216.1 5475.8 70767.4 11483.8 13008.2 8100 4500 4986.6 5314.2 8547.0 5351.4 5620.4 72748.0 11809.7 13345.2 8280 4600 5118.2 5448.9 8785.8 5486.8 5765.2 74737.1 12136.4 13682.6 8460 4700 5250.2 5583.9 9023.2 5622.3 5910.1 76735.1 12463.8 14020.4 8640 4800 5382.5 5718.9 9261.5 5757.9 6055.2 78740.6 12791.9 14358.6 8820 4900 5515.3 5854.1 9500.8 5893.7 6200.5 80754.5 13120.6 14697.2 9000 5000 5648.4 5989.4 9740.8 6029.5 6345.9 82775.8 13449.9 15036.2 9180 5100 5782.0 6124.8 9981.8 6165.5 6491.5 84805.1 13779.9 15375.6 9360 5200 5916.0 6260.3 10223.7 6301.6 6637.3 86840.8 14110.5 15715.4 9540 5300 6050.4 6396.0 10466.5 6437.8 6783.3 88883.9 14441.8 16055.6 9720 5400 6185.4 6531.8 10710.2 6574.2 6929.6 90933.5 14773.8 16396.1 9900 5500 6320.8 6667.7 10954.7 6710.3 7076.0 92988.6 15106.4 16737.0 10080 5600 6456.8 6803.7 11200.1 6846.6 7222.7 95050.1 15439.7 17078.3 10260 5700 6593.3 6939.8 11446.3 6983.2 7369.6 97116.6 15773.6 17420.0 10440 5800 6730.5 7076.1 11693.5 7120.0 7516.6 99188.0 16108.1 17762.0 10620 5900 6868.2 7212.5 11941.5 7256.9 7663.9 101264 16443.4 18104.4 10800 6000 7006.6 7349.1 12190.4 7393.9 7811.4 103344 16779.3 18447.3

Table D.4. Enthalpy (kJ/g-mole) TEMPERATURE

ENTHALPY (kJ/g-mole)

R K O2 N2 C CO CO2 H2 H2O SO2 _______________ ______________________________________________________________________ 0 0 -8.683 -8.670 -1.051 -8.671 -9.364 -8.467 -9.904 -10.552 180 100 -5.779 -5.768 -0.991 -5.769 -6.456 -5.468 -6.615 -7.217 360 200 -2.868 -2.857 -0.665 -2.858 -3.414 -2.774 -3.282 -3.736 536.7

298.15

0.000

0.000

0.000

0.000

0.000

0.000

0.000

0.000

540 720 900 1080 1260 1440 1620 1800 1980 2160 2340 2520 2700 2880 3060 3240 3420 3600 3780 3960 4140 4320 4500 4680 4860 5040 5220 5400

300 400 500 600 700 800 900 1000 1100 1200 1300 1400 1500 1600 1700 1800 1900 2000 2100 2200 2300 2400 2500 2600 2700 2800 2900 3000

0.054 3.025 6.084 9.244 12.499 15.835 19.241 22.703 26.212 29.761 33.344 36.957 40.599 44.266 47.958 51.673 55.413 59.175 62.961 66.769 70.600 74.453 78.328 82.224 86.141 90.079 94.036 98.013

0.054 2.971 5.911 8.894 11.937 15.046 18.223 21.463 24.760 28.109 31.503 34.936 38.405 41.904 45.429 48.978 52.548 56.137 59.742 63.361 66.995 70.640 74.296 77.963 81.639 85.323 89.015 92.715

0.016 1.039 2.365 3.943 5.716 7.637 9.672 11.795 13.989 16.240 18.539 20.879 23.253 25.658 28.090 30.547 33.026 35.525 38.044 40.581 43.134 45.704 48.289 50.889 53.503 56.131 58.773 61.427

0.054 2.976 5.931 8.942 12.023 15.177 18.401 21.690 25.035 28.430 31.868 35.343 38.850 42.385 45.945 49.526 53.136 56.744 60.376 64.021 67.683 71.324 74.985 78.673 82.369 86.074 89.786 93.504

0.069 4.003 8.305 12.907 17.754 22.806 28.030 33.397 38.884 44.473 50.148 55.896 61.705 67.569 73.480 79.431 85.419 91.439 97.488 103.562 109.660 115.779 121.917 128.073 134.246 140.433 146.636 152.852

0.053 2.959 5.882 8.811 11.749 14.702 17.676 20.680 23.719 26.797 29.918 33.082 36.290 39.541 42.835 46.169 49.541 52.951 56.397 59.876 63.387 66.928 70.498 74.096 77.720 81.369 85.043 88.740

0.062 3.452 6.925 10.501 14.192 18.002 21.938 26.000 30.191 34.506 38.942 43.493 48.151 52.908 57.758 62.693 67.706 72.790 77.941 83.153 88.421 93.741 99.108 104.520 109.973 115.464 120.990 126.549

0.074 4.250 8.758 13.544 18.548 23.721 29.023 34.428 39.914 45.464 51.069 56.718 62.404 68.123 73.870 79.642 85.436 91.250 97.081 102.929 108.792 114.669 120.559 126.462 132.376 138.302 144.238 150.184

Table D.4 continued. Enthalpy (kJ/g-mole TEMPERATURE

ENTHALPY (kJ/g-mole)

R K O2 N2 C CO CO2 H2 H2O SO2 _______________ ______________________________________________________________________ 5580 3100 102.009 96.421 64.095 97.229 159.081 92.460 132.139 156.140 5760 3200 106.023 100.134 66.775 100.960 165.321 96.202 137.757 162.106 5940 3300 110.054 103.852 69.467 104.696 171.573 99.966 143.403 168.081 6120 3400 114.102 107.577 72.172 108.438 177.836 103.750 149.073 174.065 6300 3500 118.165 111.306 74.889 112.185 184.109 107.555 154.768 180.057 6480 3600 122.245 115.041 77.617 115.937 190.393 111.380 160.485 186.058 6660 3700 126.339 118.781 80.357 119.693 196.686 115.224 166.222 192.068 6840 3800 130.447 122.525 83.109 123.454 202.989 119.089 171.980 198.086 7020 3900 134.569 126.274 85.872 127.219 209.301 122.972 177.757 204.111 7200 4000 138.705 130.027 88.646 130.989 215.622 126.874 183.552 210.145 7380 4100 142.854 133.784 91.432 134.762 221.951 130.795 189.363 216.186 7560 4200 147.015 137.545 94.229 138.540 228.290 134.734 195.191 222.235 7740 4300 151.188 141.309 97.037 142.321 234.637 138.692 201.034 228.292 7920 4400 155.374 145.078 99.856 146.106 240.991 142.667 206.892 234.356 8100 4500 159.572 148.850 102.658 149.895 247.354 146.660 212.764 240.427 8280 4600 163.783 152.625 105.526 153.687 253.725 150.670 218.650 246.506 8460 4700 168.005 156.405 108.378 157.483 260.103 154.698 224.548 252.592 8640 4800 172.240 160.187 111.240 161.282 266.489 158.741 230.458 258.685 8820 4900 176.488 163.973 114.114 165.084 272.882 162.801 236.380 264.785 9000 5000 180.749 167.763 116.997 168.890 279.283 166.876 242.313 270.893 9180 5100 185.023 171.556 119.892 172.699 285.691 170.967 248.258 277.007 9360 5200 189.311 175.352 122.797 176.511 292.109 175.071 254.215 283.128 9540 5300 193.614 179.152 125.713 180.326 298.535 179.190 260.184 289.257 9720 5400 197.933 182.955 128.640 184.146 304.971 183.322 266.164 295.392 9900 5500 202.267 186.761 131.577 187.957 311.416 187.465 272.157 301.534 10080 5600 206.618 190.571 134.524 191.775 317.870 191.621 278.161 307.683 10260 5700 210.987 194.384 137.482 195.603 324.334 195.787 284.177 313.838 10440 5800 215.375 198.201 140.451 199.434 330.806 199.963 290.204 320.000 10620 5900 219.782 202.023 143.429 203.268 337.288 204.148 296.244 326.169 10800 6000 224.210 205.848 146.419 207.106 343.779 208.341 302.295 332.346

APPENDIX E Properties of Selected Coals Table E.1 Analyses of Selected U.S. Coals, as mined Table E.2 Typical analyses of Coals of the World

APPENDIX F Thermodynamic properties of Refrigerants Figure F.1 Thermodynamic Properties of R-12 (English units) Figure F.2 Thermodynamic Properties of R-22 (English units) Figure F.3 Thermodynamic Properties of R-123 (English units) Figure F.4 Thermodynamic Properties of R-123a (English units) Figure F.5 Thermodynamic Properties of R-134a (SI units)

This page is intentionally blank.

APPENDIX H Properties of the 1976 U.S. Standard Atmosphere Table H.1 Properties of the 1976 U.S. Standard Atmosphere (SI units) Table H.2 Properties of the 1976 U.S. Standard Atmoshere (English units)

APPENDIX I Blackbody Spectral Distribution Function The Planck function divided by the fifth power of the temperature and the Stefan-Boltzmann constant is multiplied by 100,000 so that it plots nicely versus the wavelength-temperature product, Lambda-T, as in Figure 11.27. SIGMA was calculated by summing the product of the tabulated Planck function and the abscissa interval (100) to the desired value of lambda-T. The sum must then be divided by 1000 to compensate for the 100,000 factor. Note the distinction between the Stefan-Boltzmann constant (Sigma) and the cumulative radiation function (SIGMA). c1 = c2 =

Lambda-T micron-R

1000 1100 1200 1300 1400 1500 1600 1700 1800 1900 2000 2100 2200 2300 2400 2500 2600 2700 2800 2900 3000 3100 3200 3300 3400 3500 3600 3700 3800 3900

118700000 Btu-micron^4/ft^2-hr 25896 micron-degree R

Fb(Lambda-T)/Sigma-T^5 1/(micron-R) x100,000 3.926E-05 0.000256682 0.001181568 0.004164636 0.011928687 0.028995038 0.061770887 0.11819881 0.207027327 0.336873778 0.515283226 0.747952401 1.038219036 1.386845559 1.792070875 2.249871296 2.754359627 3.298254226 3.87336131 4.471028633 5.082543507 5.699460885 6.313857168 6.918512359 7.507027618 8.073887592 8.614477758 9.12506675 9.602762863 10.04545268

SIGMA dimensionless

3.926E-08 2.95942E-07 1.47751E-06 5.64215E-06 1.75708E-05 4.65659E-05 0.000108337 0.000226536 0.000433563 0.000770437 0.00128572 0.002033672 0.003071891 0.004458737 0.006250808 0.008500679 0.011255039 0.014553293 0.018426654 0.022897683 0.027980226 0.033679687 0.039993544 0.046912057 0.054419084 0.062492972 0.07110745 0.080232516 0.089835279 0.099880732

Lambda-T micron-R

4000 4100 4200 4300 4400 4500 4600 4700 4800 4900 5000 5100 5200 5300 5400 5500 5600 5700 5800 5900 6000 6100 6200 6300 6400 6500 6600 6700 6800 6900 7000 7100 7200 7300 7400 7500 7600 7700 7800 7900 8000 8100 8200 8300 8400 8500 8600 8700 8800 8900

Fb(Lambda-T)/Sigma-T^5 1/(micron-R) x100,000 10.45172852 10.82081001 11.152464 11.44692597 11.70482499 11.9271139 12.11500549 12.26991521 12.39341038 12.48716597 12.55292639 12.59247315 12.6075976 12.60007853 12.5716639 12.52405626 12.45890147 12.37778011 12.28220131 12.17359857 12.05332729 11.92266363 11.78280457 11.6348688 11.47989844 11.31886126 11.15265333 10.98210197 10.80796898 10.63095392 10.4516975 10.27078501 10.08874964 9.906075829 9.723202485 9.540526075 9.358403632 9.177155612 8.997068608 8.818397933 8.641370049 8.466184858 8.293017853 8.12202213 7.95333027 7.787056092 7.623296286 7.462131926 7.303629879 7.147844105

SIGMA dimensionless

0.110332461 0.121153271 0.132305735 0.143752661 0.155457486 0.167384599 0.179499605 0.19176952 0.204162931 0.216650096 0.229203023 0.241795496 0.254403094 0.267003172 0.279574836 0.292098892 0.304557794 0.316935574 0.329217775 0.341391374 0.353444701 0.365367365 0.377150169 0.388785038 0.400264937 0.411583798 0.422736451 0.433718553 0.444526522 0.455157476 0.465609173 0.475879958 0.485968708 0.495874784 0.505597986 0.515138513 0.524496916 0.533674072 0.54267114 0.551489538 0.560130908 0.568597093 0.576890111 0.585012133 0.592965463 0.60075252 0.608375816 0.615837948 0.623141578 0.630289422

Lambda-T micron-R

9000 9100 9200 9300 9400 9500 9600 9700 9800 9900 10000 10100 10200 10300 10400 10500 10600 10700 10800 10900 11000 11100 11200 11300 11400 11500 11600 11700 11800 11900 12000 12100 12200 12300 12400 12500 12600 12700 12800 12900 13000 13100 13200 13300 13400 13500 13600 13700 13800 13900

Fb(Lambda-T)/Sigma-T^5 1/(micron-R) x100,000 6.994816861 6.844579811 6.697155054 6.552556067 6.410788573 6.271851341 6.135736916 6.002432297 5.871919548 5.744176362 5.619176579 5.496890652 5.377286082 5.260327805 5.145978545 5.034199146 4.924948859 4.818185608 4.713866235 4.611946716 4.512382358 4.415127973 4.320138045 4.227366866 4.136768666 4.048297729 3.961908497 3.877555654 3.795194213 3.714779583 3.636267632 3.559614744 3.484777862 3.41171453 3.340382931 3.270741911 3.202751009 3.136370475 3.071561285 3.008285156 2.946504554 2.886182704 2.82728359 2.769771958 2.713613318 2.658773939 2.60522085 2.552921826 2.501845393 2.451960812

SIGMA dimensionless

0.637284239 0.644128818 0.650825973 0.65737853 0.663789318 0.670061169 0.676196906 0.682199339 0.688071258 0.693815435 0.699434611 0.704931502 0.710308788 0.715569116 0.720715094 0.725749293 0.730674242 0.735492428 0.740206294 0.744818241 0.749330623 0.753745751 0.758065889 0.762293256 0.766430025 0.770478322 0.774440231 0.778317787 0.782112981 0.78582776 0.789464028 0.793023643 0.796508421 0.799920135 0.803260518 0.80653126 0.809734011 0.812870381 0.815941943 0.818950228 0.821896732 0.824782915 0.827610199 0.830379971 0.833093584 0.835752358 0.838357579 0.840910501 0.843412346 0.845864307

Lambda-T micron-R

14000 14100 14200 14300 14400 14500 14600 14700 14800 14900 15000 15100 15200 15300 15400 15500 15600 15700 15800 15900 16000 16100 16200 16300 16400 16500 16600 16700 16800 16900 17000 17100 17200 17300 17400 17500 17600 17700 17800 17900 18000 18100 18200 18300 18400 18500 18600 18700 18800 18900

Fb(Lambda-T)/Sigma-T^5 1/(micron-R) x100,000 2.403238073 2.355647887 2.309161674 2.263751554 2.219390335 2.1760515 2.133709199 2.09233823 2.051914034 2.012412676 1.973810837 1.936085795 1.89921542 1.863178154 1.827953001 1.793519515 1.759857785 1.726948422 1.694772551 1.663311791 1.632548251 1.60246451 1.573043612 1.544269048 1.51612475 1.488595076 1.4616648 1.435319102 1.409543555 1.384324117 1.359647121 1.335499261 1.311867588 1.288739494 1.266102709 1.243945289 1.222255604 1.201022337 1.180234467 1.159881267 1.139952293 1.120437377 1.10132662 1.082610382 1.06427928 1.046324175 1.028736168 1.011506593 0.994627013 0.978089209

SIGMA dimensionless

0.848267545 0.850623193 0.852932354 0.855196106 0.857415496 0.859591548 0.861725257 0.863817595 0.865869509 0.867881922 0.869855733 0.871791819 0.873691034 0.875554212 0.877382165 0.879175685 0.880935543 0.882662491 0.884357263 0.886020575 0.887653124 0.889255588 0.890828632 0.892372901 0.893889025 0.895377621 0.896839285 0.898274604 0.899684148 0.901068472 0.902428119 0.903763618 0.905075486 0.906364226 0.907630328 0.908874274 0.910096529 0.911297552 0.912477786 0.913637667 0.91477762 0.915898057 0.916999384 0.918081994 0.919146273 0.920192597 0.921221334 0.92223284 0.923227467 0.924205556

Lambda-T micron-R

19000 19100 19200 19300 19400 19500 19600 19700 19800 19900 20000 20100 20200 20300 20400 20500 20600 20700 20800 20900 21000 21100 21200 21300 21400 21500 21600 21700 21800 21900 22000 22100 22200 22300 22400 22500 22600 22700 22800 22900 23000 23100 23200 23300 23400 23500 23600 23700 23800 23900

Fb(Lambda-T)/Sigma-T^5 1/(micron-R) x100,000 0.961885177 0.946007123 0.930447451 0.915198768 0.900253868 0.885605733 0.871247525 0.857172583 0.843374417 0.829846703 0.816583279 0.80357814 0.790825435 0.778319459 0.766054655 0.754025605 0.742227027 0.730653775 0.719300828 0.708163296 0.697236407 0.686515509 0.675996069 0.665673662 0.655543976 0.645602804 0.635846044 0.626269693 0.616869849 0.607642702 0.598584538 0.589691733 0.580960749 0.572388137 0.563970528 0.555704637 0.547587256 0.539615256 0.531785581 0.524095249 0.516541348 0.509121036 0.501831539 0.494670147 0.487634214 0.480721158 0.473928455 0.467253643 0.460694314 0.454248117

SIGMA dimensionless

0.925167442 0.926113449 0.927043896 0.927959095 0.928859349 0.929744954 0.930616202 0.931473375 0.932316749 0.933146596 0.933963179 0.934766757 0.935557583 0.936335902 0.937101957 0.937855982 0.938598209 0.939328863 0.940048164 0.940756327 0.941453564 0.942140079 0.942816075 0.943481749 0.944137293 0.944782896 0.945418742 0.946045011 0.946661881 0.947269524 0.947868108 0.9484578 0.949038761 0.949611149 0.95017512 0.950730824 0.951278411 0.951818027 0.952349812 0.952873908 0.953390449 0.95389957 0.954401402 0.954896072 0.955383706 0.955864427 0.956338355 0.956805609 0.957266303 0.957720552

Lambda-T micron-R

24000 24100 24200 24300 24400 24500 24600 24700 24800 24900 25000 25100 25200 25300 25400 25500 25600 25700 25800 25900 26000 26100 26200 26300 26400 26500 26600 26700 26800 26900 27000 27100 27200 27300 27400 27500 27600 27700 27800 27900 28000 28100 28200 28300 28400 28500 28600 28700 28800 28900

Fb(Lambda-T)/Sigma-T^5 1/(micron-R) x100,000 0.447912759 0.441685995 0.435565634 0.429549536 0.42363561 0.417821812 0.412106146 0.406486658 0.400961444 0.395528638 0.390186419 0.384933006 0.379766658 0.374685673 0.369688389 0.364773178 0.359938449 0.355182648 0.350504254 0.345901779 0.341373769 0.336918802 0.332535485 0.328222458 0.323978389 0.319801976 0.315691944 0.311647046 0.307666063 0.3037478 0.299891089 0.296094787 0.292357775 0.288678958 0.285057264 0.281491643 0.277981069 0.274524537 0.27112106 0.267769677 0.264469442 0.261219432 0.258018743 0.254866487 0.251761797 0.248703823 0.245691732 0.242724709 0.239801956 0.236922689

SIGMA dimensionless

0.958168464 0.95861015 0.959045716 0.959475265 0.959898901 0.960316723 0.960728829 0.961135316 0.961536277 0.961931806 0.962321992 0.962706925 0.963086692 0.963461378 0.963831066 0.964195839 0.964555778 0.96491096 0.965261464 0.965607366 0.96594874 0.966285659 0.966618194 0.966946417 0.967270395 0.967590197 0.967905889 0.968217536 0.968525202 0.96882895 0.969128841 0.969424936 0.969717294 0.970005973 0.97029103 0.970572521 0.970850503 0.971125027 0.971396148 0.971663918 0.971928387 0.972189607 0.972447625 0.972702492 0.972954254 0.973202958 0.973448649 0.973691374 0.973931176 0.974168099

Lambda-T micron-R

29000 29100 29200 29300 29400 29500 29600 29700 29800 29900 30000 30100 30200 30300 30400 30500 30600 30700 30800 30900 31000 31100 31200 31300 31400 31500 31600 31700 31800 31900 32000 32100 32200 32300 32400 32500 32600 32700 32800 32900 33000 33100 33200 33300 33400 33500 33600 33700 33800 33900

Fb(Lambda-T)/Sigma-T^5 1/(micron-R) x100,000 0.234086142 0.231291563 0.228538218 0.225825385 0.223152357 0.220518442 0.217922962 0.215365251 0.212844659 0.210360545 0.207912285 0.205499264 0.20312088 0.200776545 0.198465679 0.196187716 0.1939421 0.191728285 0.189545739 0.187393935 0.185272361 0.183180512 0.181117894 0.179084023 0.177078421 0.175100623 0.173150172 0.171226618 0.16932952 0.167458447 0.165612974 0.163792686 0.161997174 0.160226038 0.158478884 0.156755326 0.155054987 0.153377494 0.151722482 0.150089594 0.148478478 0.146888789 0.145320188 0.143772343 0.142244927 0.14073762 0.139250107 0.137782079 0.136333232 0.134903269

SIGMA dimensionless

0.974402185 0.974633476 0.974862015 0.97508784 0.975310992 0.975531511 0.975749434 0.975964799 0.976177644 0.976388004 0.976595916 0.976801416 0.977004537 0.977205313 0.977403779 0.977599967 0.977793909 0.977985637 0.978175183 0.978362577 0.978547849 0.978731029 0.978912147 0.979091231 0.97926831 0.97944341 0.979616561 0.979787787 0.979957117 0.980124575 0.980290188 0.980453981 0.980615978 0.980776204 0.980934683 0.981091438 0.981246493 0.981399871 0.981551593 0.981701683 0.981850161 0.98199705 0.98214237 0.982286143 0.982428388 0.982569125 0.982708375 0.982846157 0.982982491 0.983117394

Lambda-T micron-R

34000 34100 34200 34300 34400 34500 34600 34700 34800 34900 35000 35100 35200 35300 35400 35500 35600 35700 35800 35900 36000 36100 36200 36300 36400 36500 36600 36700 36800 36900 37000 37100 37200 37300 37400 37500 37600 37700 37800 37900 38000 38100 38200 38300 38400 38500 38600 38700 38800 38900

Fb(Lambda-T)/Sigma-T^5 1/(micron-R) x100,000 0.133491896 0.132098827 0.130723778 0.129366474 0.128026641 0.126704012 0.125398324 0.124109319 0.122836744 0.121580348 0.120339889 0.119115124 0.117905819 0.11671174 0.115532659 0.114368353 0.1132186 0.112083185 0.110961895 0.10985452 0.108760855 0.107680698 0.10661385 0.105560116 0.104519305 0.103491227 0.102475696 0.101472532 0.100481554 0.099502586 0.098535454 0.097579989 0.096636023 0.095703391 0.09478193 0.093871482 0.092971889 0.092082998 0.091204657 0.090336717 0.08947903 0.088631454 0.087793845 0.086966064 0.086147973 0.085339438 0.084540326 0.083750505 0.082969847 0.082198226

SIGMA dimensionless

0.983250886 0.983382985 0.983513708 0.983643075 0.983771101 0.983897805 0.984023204 0.984147313 0.98427015 0.98439173 0.98451207 0.984631185 0.984749091 0.984865803 0.984981335 0.985095704 0.985208922 0.985321006 0.985431967 0.985541822 0.985650583 0.985758264 0.985864877 0.985970438 0.986074957 0.986178448 0.986280924 0.986382396 0.986482878 0.98658238 0.986680916 0.986778496 0.986875132 0.986970835 0.987065617 0.987159489 0.987252461 0.987344544 0.987435748 0.987526085 0.987615564 0.987704195 0.987791989 0.987878955 0.987965103 0.988050443 0.988134983 0.988218734 0.988301703 0.988383902

Lambda-T micron-R

39000 39100 39200 39300 39400 39500 39600 39700 39800 39900 40000 40100 40200 40300 40400 40500 40600 40700 40800 40900 41000 41100 41200 41300 41400 41500 41600 41700 41800 41900 42000 42100 42200 42300 42400 42500 42600 42700 42800 42900 43000 43100 43200 43300 43400 43500 43600 43700 43800 43900 44000

Fb(Lambda-T)/Sigma-T^5 1/(micron-R) x100,000 0.081435515 0.080681594 0.079936341 0.079199638 0.078471367 0.077751414 0.077039664 0.076336008 0.075640335 0.074952537 0.074272508 0.073600143 0.07293534 0.072277996 0.071628013 0.070985292 0.070349736 0.069721249 0.069099739 0.068485113 0.067877279 0.067276149 0.066681634 0.066093647 0.065512104 0.064936919 0.06436801 0.063805295 0.063248693 0.062698126 0.062153516 0.061614785 0.061081858 0.06055466 0.060033118 0.059517159 0.059006713 0.058501708 0.058002076 0.057507748 0.057018658 0.056534738 0.056055925 0.055582153 0.055113359 0.054649481 0.054190457 0.053736227 0.053286731 0.05284191 0.052401706

SIGMA dimensionless

0.988465337 0.988546019 0.988625955 0.988705155 0.988783626 0.988861378 0.988938417 0.989014753 0.989090394 0.989165346 0.989239619 0.989313219 0.989386154 0.989458432 0.98953006 0.989601045 0.989671395 0.989741116 0.989810216 0.989878701 0.989946578 0.990013855 0.990080536 0.99014663 0.990212142 0.990277079 0.990341447 0.990405252 0.990468501 0.990531199 0.990593353 0.990654967 0.990716049 0.990776604 0.990836637 0.990896154 0.990955161 0.991013663 0.991071665 0.991129172 0.991186191 0.991242726 0.991298782 0.991354364 0.991409477 0.991464127 0.991518317 0.991572053 0.99162534 0.991678182 0.991730584