Ch7 Dimensional Analysis And Similarity

Dimensional Analysis and Similarity

Dimensional Analysis Consider steady flow of an incompressible Newtonian fluid through long, smooth-walled, horizontal, circular pipe. We are interested in the pressure drop per unit length that develops along the pipe as a result of friction This problem cannot generally be solved analytically without the use of experimental data Before running experiment decide on the factors, or variables, that effects the pressure drop per unit length, Δpℓ. Those are the pipe diameter, D, fluid density, ρ, fluid viscosity, μ, and mean velocity, V, This relationships can be expressed as pl  f  D,  ,  , V  The objective of experiment is to determine the nature of this function. During experiment change one of the variables, while holding all others constant, and measure the corresponding pressure drop. Experimental results can be represented graphically Another way is to use dimensionless parameters.

Dimensional Analysis  VD  Dpl     V 2    Number of variables reduced from five to two Number and complexity of experiments also reduced. Results of experiments are represented by a single curve Two groups in the above equation are dimensionless

Thus, results presented in the form of the graph are independent of the system of units used. This type of analysis is called dimensional analysis Dimensional analysis is a method for reducing the number and complexity of experimental variables which effect a given physical phenomenon. Dimensional analysis is based on the Buckingham pi theorem

Buckingham Pi Theorem How many dimensionless products are required to replace the original list of variables? Buckingham pi theorem states: If an equation involving k variables is dimensionally homogeneous, it can be reduced to a relationship among k – r independent dimensionless products, where r is the minimum number of reference dimensions required to describe the variables.

Dimensionless products are referred to as “pi terms” and denoted by letter Π

Pi theorem is based on the principle of dimensional homogeneity, which states: If an equation truly expresses a proper relationship between variables in a physical process, it will be dimensionally homogeneous; i.e. each of its additive terms will have the same dimensions.

Thus, dimensionally homogeneous equation involving k variables

u1  f  u2 , u3 , ..., uk 

can be written in dimensionless form

1     2 ,  3 , ...,  k  r 

Determination of Pi Terms Method of Repeating Variables Step 1.

List all the variables that are involved in the problem.

Step 2.

Express each of the variables in terms of basic dimensions.

Step 3.

Determine the required number of pi terms.

Step 4.

Select a number of repeating variables.

Step 5.

Form a pi term by multiplying one of the nonrepeating variables by the product

of repeating

variables each raised to an exponent that will make the combination

dimensionless. Step 6.

Repeat Step 5 for each of the remaining repeating variables.

Step 7.

Check all the resulting pi terms to make sure they are dimensionless.

Step 8.

Express the final form relationship among the pi terms and think about 1 as  a  2 ,  3 , ...,  k r

what it means.



where Π1 contains dependent variable in the numerator



Method of Repeating Variables 1. For the steady flow of an incompressible Newtonian fluid through long, smooth-walled, horizontal, circular pipe, the pressure drop per unit length along the pipe (k = 5) pl  f  D,  ,  , V 

2. Variables in terms of basic dimensions (r = 3) pl B FL3

 B FL4T 2

D BL

 B FL2T

V B LT 1

3. Apply pi theorem; k – r = 2. Two pi terms are required. 4. Select repeating variables: D, V, and ρ 5,6. Form pi terms 1  pl D V  a

b

c

 FL   L   LT   FL 3

a

1

b

similarly 2 

4

T

2

 DV 



c

0 0

BF L T

0

1 

pl D V 2

Method of Repeating Variables 7. Check dimensions





  L   LT 

3

FL pl D 1  B 2 V FL4T 2

1

2

B F 0 L0T 0

or alternatively p D  ML T   L   B V  ML   LT  2

1

2

l

2

3

1

2

B M 0 L0T 0

8. Express result of dimensional analysis as  DV   Dpl     V 2   

Example 7.1 A thin rectangular plate having a width w and a height h is located so that it is normal to a moving stream of fluid. Assume the drag that the fluid exerts on the plate is a function of w and h, the fluid viscosity and density, μ and ρ, respectively, and the velocity V of the fluid approaching the plate. Determine a suitable set of pi terms to study this problem experimentally.

Answer:  w Vw  D    ,  w2 V 2  h  

Selection of Variables •

Selection of variables is the most important and difficult step in applying dimensional analysis.

•

For convenience, term variable will be used to indicate any quantity involved, including dimensional and nondimensional constants.

•

There is no simple procedure whereby the variables can be easily identified. Generally, one must rely on a good understanding of the phenomenon involved and the governing physical laws.

•

If extraneous variables are included, then too many pi terms appear in the final solution, and it may be difficult, time consuming, and expensive to eliminate these experimentally.

•

If important variables are omitted, then an incorrect result will be obtained.

•

Most engineering problems involve simplifying assumptions. Simplicity and accuracy must be balanced.

Selection of Variables •

Variables can be classified into three general groups: –

Geometry. Geometric characteristics can be described by a series of lengths and angles.

–

Material Properties. Response of a system to applied external effects such as forces, pressures, and changes in temperature is dependent on the nature of materials involved in the system. Therefore, material properties that relate the external effects and the responses must be included as variables. For example, for Newtonian fluids the viscosity of the fluid is the property that relates the applied forces to the rates of deformation fo the fluid.

–

External effects. Variables that produce a change in the system

Selection of Variables. Summary •

Clearly define the problem. What is the main variable of interest (the dependent variable)?

•

Consider the basic laws that govern the phenomenon. Even a crude theory that describes the essential aspects of the system may be helpful.

•

Start the variable selection process by grouping the variables into three broad classes: geometry, material properties, and external effects.

•

Consider other variables that may not fall into one of the above categories. For example, time will be an important variable if any of the variables are time dependent.

•

Be sure to include all quantities that enter the problem even though some of them may be held constant (e.g., the acceleration of gravity, g). For a dimensional analysis it is the dimensions of the quantities that are important – not specific values!

•

Make sure that all variables are independent. Look for relationships among subsets of the variables.

Example 7.2 An open, cylindrical tank having a diameter D is supported around its bottom circumference and is filled to a depth h with a liquid having a specific weight γ. The vertical deflection, δ, of the center of the bottom is a function of D, h, d, γ, and E, where d is the thickness of the bottom and E is the modulus of elasticity of the bottom material. Perform a dimensional analysis of this problem.

Answer:  h d E    , ,  D  D D D 

Comments:

Number of reference dimensions can differ from number of basic dimensions Required number of pi terms cannot be affected by using MLT instead of FLT system, or vice versa.

Determination of Pi Terms by Inspection Pi terms can be formed by inspection by making use of the fact that each pi term must be dimensionless For example, consider pressure drop per unit length along the a smooth pipe. As before, determine variables:

pl  f  D,  ,  , V 

Dimensions of variables:

pl B FL3

D BL

 B FL4T 2

 B FL2T

Number of pi terms is 5 -3 =2

Determine pi terms by inspection. Π1 must contain dependent variable.

1 

pl D V 2

Form the second pi term, Π2, by selecting variable that was not used in Π1

2 

 DV 

V B LT 1

Correlation of Experimental Data Dimensional analysis greatly facilitate the efficient handling, interpretation, and correlation of experimental data. If only one pi term is involved in a problem, it must be equal to a constant, that is

1  C

For example, assume that the drag acting on a spherical particle that falls very slowly through a viscous fluid is a function of the particle diameter, d, particle velocity, V, and the fluid viscosity, μ. Determine, with the aid of dimensional analysis, how the drag depends on the particle velocity.

Since

D  f  d ,V ,  

then

D C Vd

or

D  C Vd

Thus, for a given particle and fluid, the drag varies directly with the velocity. Approximate solution to this problem was obtained from theory. Stokes law gives drag as

D  3Vd

Correlation of Experimental Data For problems involving only two pi terms, results of an experiment can be conveniently presented in a simple graph

In addition to presenting data graphically, it is desirable to obtain an empirical equation relating Π1 and Π2 by using a standard curve-fitting technique. For example, Blasius equation 1 4

 VD  Dpl  0.1582   V 2   

is widely used for predicting the pressure drop in smooth pipes in the range 4x103
Correlation of Experimental Data As the number of pi terms increases, in becomes more difficult to display results in graphical form and to determine specific empirical equation that describe the phenomenon. For problems involving three pi terms, it is still possible to show data correlations on simple graphs by plotting families of curves

For problems involving more than two or three pi terms, it is often necessary to use a model to predict specific characteristics.

End of Lecture

END OF CHAPTER

Supplementary slides

Dimensional Analysis Nature of the function is difficult to obtain by using this approach Moreover, experiments (c) and (d) are difficult to carry out

Pressure drop as a function of several different factors

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Dimensional Analysis Qualitative description of physical quantities can be given in terms of basic dimensions such as mass, M, length, L, and time, T. Alternatively, force, F, L, and T can be used, since

F B MLT 2

Thus, dimensions of the two groups in the referenced equation are

 F 3   L 

L

Dpl B 2 V FL4T 2





LT 1



2

B F 0 L0T 0

4 2 1 VD  FL T   LT  L 0 0 0 B B F LT 2   FL T 

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Determination of Pi Terms Method of repeating variables Step 1.

List all the variables that are involved in the problem (determine k).

S  S0  V0t 

1 2 gt 2

Dimensional variables are S and t. All have dimension and can be nondimensionlalized. Dimensional constants are S0, V0 and g, are used to nondimensionlalize variables Pure constants have no dimensions Variables can be classified into three groups: geometry, material properties, and external effects. Keep the number of variables to minimum. All variables must be independent.

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Determination of Pi Terms Method of repeating variables Step 2.

Express each of the variables in terms of basic dimensions (determine r).

Ether set M, L, T or F, L, T can be used

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Determination of Pi Terms Method of repeating variables Step 3.

Determine the required number of pi terms (determine k – r).

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Determination of Pi Terms Method of repeating variables Step 4.

Select a number of repeating variables, where the number require is equal to

the number of

reference dimensions

All of the required reference dimensions must be included within the group of repeating variables, and each repeating variable must be dimensionally independent of the others (i.e. dimensions of one repeating variable cannot be reproduced by some combination of product of powers of the remaining repeating variables)

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Determination of Pi Terms Method of repeating variables Step 5.

Form a pi term by multiplying one of the nonrepeating variables by the product

of repeating

variables each raised to an exponent that will make the combination

dimensionless.

Each pi term will be of the form

ui u1a u2b u3c

where ui is one of the nonrepeating variables; u1, u2 and u3 are the repeating variables; and the exponents ai, bi, and ci are determined so that the combination is dimensionless.

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Determination of Pi Terms Method of repeating variables Step 7.

Check all the resulting pi terms to make sure they are dimensionless.

Substitute dimensions of variables into pi terms to confirm that they are dimensional. Good way to do this is to express the variables in term of M, L, T if the basic dimensions F, L, Twere used, or vice versa.

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Determination of Pi Terms Method of repeating variables Step 8.

Express the final form as a relationship among the pi terms and think about

what it means.

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