taiwan shieh fluid07

1

FUNDAMENTALS OFFUNDAMENTALS OFFLUID MECHANICSFLUID MECHANICS

Chapter 7 Dimensional AnalysisChapter 7 Dimensional AnalysisModeling, and SimilitudeModeling, and Similitude

JyhJyh--CherngCherng ShiehShiehDepartment of BioDepartment of Bio--Industrial Industrial MechatronicsMechatronics Engineering Engineering

National Taiwan UniversityNational Taiwan University12/24/200712/24/2007

2

MAIN TOPICSMAIN TOPICS

Dimensional AnalysisDimensional AnalysisBuckingham Pi TheoremBuckingham Pi TheoremDetermination of Pi TermsDetermination of Pi TermsComments about Dimensional AnalysisComments about Dimensional AnalysisCommon Dimensionless Groups in Fluid MechanicsCommon Dimensionless Groups in Fluid MechanicsCorrelation of Experimental DataCorrelation of Experimental DataModeling and SimilitudeModeling and SimilitudeTypical Model StudiesTypical Model StudiesSimilitude Based on Governing Differential EquationSimilitude Based on Governing Differential Equation

3

Dimensional Analysis Dimensional Analysis 1/41/4

A typical fluid mechanics problem in which A typical fluid mechanics problem in which experimentation is required, consider the steady flow of an experimentation is required, consider the steady flow of an incompressible Newtonian fluid through a long, smoothincompressible Newtonian fluid through a long, smooth--walled, horizontal, circular pipe. walled, horizontal, circular pipe. An important characteristic of this system, which would An important characteristic of this system, which would be interest to an engineer designing a pipeline, is the be interest to an engineer designing a pipeline, is the pressure drop per unit length that develops along the pipe pressure drop per unit length that develops along the pipe as a result of friction.as a result of friction.

4


The first step in the planning of an experiment to study The first step in the planning of an experiment to study this problem would be to decide on the factors, or this problem would be to decide on the factors, or variables, that will have an effect on the pressure drop.variables, that will have an effect on the pressure drop.Pressure drop per unit lengthPressure drop per unit length

)V,,,D(fp µρ=∆ l

Pressure drop per unit length depends on FOUR variables:Pressure drop per unit length depends on FOUR variables:sphere size (D); speed (V); fluid density (sphere size (D); speed (V); fluid density (ρρ); fluid viscosity ); fluid viscosity (m)(m)

5


To perform the experiments in a meaningful and To perform the experiments in a meaningful and systematic manner, it would be necessary to change on of systematic manner, it would be necessary to change on of the variable, such as the velocity, which holding all other the variable, such as the velocity, which holding all other constant, and measure the corresponding pressure drop.constant, and measure the corresponding pressure drop.Difficulty to determine the functional relationship between Difficulty to determine the functional relationship between the pressure drop and the various facts that influence it.the pressure drop and the various facts that influence it.

6

Series of TestsSeries of Tests

7


Fortunately, there is a much simpler approach to the Fortunately, there is a much simpler approach to the problem that will eliminate the difficulties described problem that will eliminate the difficulties described above.above.Collecting these variables into two Collecting these variables into two nondimensionalnondimensionalcombinations of the variables (called dimensionless combinations of the variables (called dimensionless product or dimensionless groups)product or dimensionless groups)

Only one dependent and one Only one dependent and one independent variableindependent variableEasy to set up experiments to Easy to set up experiments to determine dependencydetermine dependencyEasy to present results (one graph)Easy to present results (one graph)

⎟⎟⎠

⎞⎜⎜⎝

⎛µ

ρφ=

ρ

∆ VDV

pD2l

8

Plot of Pressure Drop Data Using Plot of Pressure Drop Data Using ……

0002124

3

2 TLF)FT)(TFL(

)L/F(LV

pD==

ρ

∆−−

&l

0002

124TLF

)TFL()L)(LT)(TFL(VD==

µρ

−

−−&

dimensionless product or dimensionless product or dimensionless groupsdimensionless groups

9

Buckingham Pi Theorem Buckingham Pi Theorem 1/51/5

A fundamental question we must answer is how many A fundamental question we must answer is how many dimensionless products are required to replace the original listdimensionless products are required to replace the original list of of variables ?variables ?The answer to this question is supplied by the basic theorem of The answer to this question is supplied by the basic theorem of dimensional analysis that statesdimensional analysis that 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.

Buckingham Pi TheoremBuckingham Pi Theorem Pi termsPi terms

10


Given a physical problem in which the dependent variable Given a physical problem in which the dependent variable is a function of kis a function of k--1 independent variables.1 independent variables.

Mathematically, we can express the functional relationship Mathematically, we can express the functional relationship in the equivalent formin the equivalent form

)u,.....,u,u(fu k321 =

0)u,.....,u,u,u(g k321 =

Where g is an unspecified function, different from f.Where g is an unspecified function, different from f.

11


The Buckingham Pi theorem states that: Given a relation The Buckingham Pi theorem states that: Given a relation among k variables of the formamong k variables of the form

The k variables may be grouped into kThe k variables may be grouped into k--r independent r independent dimensionless products, or dimensionless products, or ΠΠ terms, expressible in terms, expressible in functional form byfunctional form by

0)u,.....,u,u,u(g k321 =

0),,,,,,(or

),,,,,(

rk321

rk321

=ΠΠΠΠφ

ΠΠΠφ=Π

−

−r ?? r ?? ΠΠ????

12


The number r is usually, but not always, equal to the The number r is usually, but not always, equal to the minimum number of independent dimensions required to minimum number of independent dimensions required to specify the dimensions of all the parameters. Usually the specify the dimensions of all the parameters. Usually the reference dimensions required to describe the variables reference dimensions required to describe the variables will be the basic dimensions M, L, and T or F, L, and T.will be the basic dimensions M, L, and T or F, L, and T.The theorem does not predict the functional form of The theorem does not predict the functional form of φφ or or ϕϕ . The functional relation among the independent . The functional relation among the independent dimensionless products dimensionless products ΠΠ must be determined must be determined experimentally.experimentally.The kThe k--r dimensionless products r dimensionless products ΠΠ terms obtained terms obtained from the procedure are independent.from the procedure are independent.

13


A A ΠΠ term is not independent if it can be obtained from a term is not independent if it can be obtained from a product or quotient of the other dimensionless products of product or quotient of the other dimensionless products of the problem. For example, ifthe problem. For example, if

then neither then neither ΠΠ55 nor nor ΠΠ66 is independent of the other is independent of the other dimensionless products or dimensionless products or dimensionless groupsdimensionless groups..

23

4/31

632

15 or2

Π

Π=Π

ΠΠΠ

=Π

14

Determination of Pi Terms Determination of Pi Terms 1/121/12

Several methods can be used to form the dimensionless Several methods can be used to form the dimensionless products, or pi term, that arise in a dimensional analysis.products, or pi term, that arise in a dimensional analysis.The method we will describe in detail is called the The method we will describe in detail is called the METHOD of repeating variables.METHOD of repeating variables.Regardless of the method to be used to determine the Regardless of the method to be used to determine the dimensionless products, dimensionless products, one begins by listingone begins by listing all all dimensional variablesdimensional variables that are known (or believed) to that are known (or believed) to affect the given flow phenomenon.affect the given flow phenomenon.Eight steps listedEight steps listed below outline a recommended below outline a recommended procedure for determining the procedure for determining the ΠΠ terms.terms.

15


Step 1 List all the variables.Step 1 List all the variables. 11

List all the dimensional variables involved.List all the dimensional variables involved.Keep the number of variables to a minimum, so that we Keep the number of variables to a minimum, so that we can minimize the amount of laboratory work.can minimize the amount of laboratory work.All variables must be independent. For example, if the All variables must be independent. For example, if the crosscross--sectional area of a pipe is an important variable, sectional area of a pipe is an important variable, either the area or the pipe diameter could be used, but either the area or the pipe diameter could be used, but not both, since they are obviously not independent.not both, since they are obviously not independent.

γγ==ρρ××g, that is, g, that is, γγ,,ρρ, and g are not independent., and g are not independent.

16


Step 1 List all the variables. Step 1 List all the variables. 22

Let k be the number of variables.Let k be the number of variables.Example: For pressure drop per unit length, k=5. (All Example: For pressure drop per unit length, k=5. (All variables are variables are ∆∆ppll,, D,D,µµ,,ρρ, and , and V )V )


17


Step 2 Express each of the variables in terms of Step 2 Express each of the variables in terms of basic dimensions. Find the number of reference basic dimensions. Find the number of reference dimensions.dimensions.

Select a set of fundamental (primary) dimensions.Select a set of fundamental (primary) dimensions.For example: MLT, or FLT.For example: MLT, or FLT.Example: For pressure drop per unit length , we choose Example: For pressure drop per unit length , we choose FLT.FLT.

12

243

LTVTFL

TFLLDFLp−−

−−

==µ

=ρ==∆

&&

&&&lr=3r=3

18


Step 3 Determine the required number of pi terms.Step 3 Determine the required number of pi terms.Let k be the number of variables in the problem.Let k be the number of variables in the problem.Let r be the number of reference dimensions (primary Let r be the number of reference dimensions (primary dimensions) required to describe these variables.dimensions) required to describe these variables.The number of pi terms is The number of pi terms is kk--rrExample: For pressure drop per unit length k=5, r = 3, Example: For pressure drop per unit length k=5, r = 3, the number of pi terms is the number of pi terms is kk--rr=5=5--3=2.3=2.

19


Step 4 Select a number of repeating variables, Step 4 Select a number of repeating variables, where the number required is equal to the number where the number required is equal to the number of reference dimensions.of reference dimensions.

Select a set of r dimensional variables that includes all Select a set of r dimensional variables that includes all the primary dimensions ( repeating variables).the primary dimensions ( repeating variables).These repeating variables will all be combined with These repeating variables will all be combined with each of the remaining parameters. No repeating each of the remaining parameters. No repeating variables should have dimensions that are power of the variables should have dimensions that are power of the dimensions of another repeating variable.dimensions of another repeating variable.Example: For pressure drop per unit length ( r = 3) Example: For pressure drop per unit length ( r = 3) select select ρρ , V, D., V, D.

20


Step 5 Form a pi term by multiplying one of the Step 5 Form a pi term by multiplying one of the nonrepeatingnonrepeating variables by the product of the variables by the product of the repeating variables, each raised to an exponent that repeating variables, each raised to an exponent that will make the combination dimensionless. will make the combination dimensionless. 11

Set up dimensional equations, combining the variables Set up dimensional equations, combining the variables selected in Step 4 with each of the other variables selected in Step 4 with each of the other variables ((nonrepeatingnonrepeating variables) in turn, to form dimensionless variables) in turn, to form dimensionless groups or dimensionless product.groups or dimensionless product.There will be k There will be k –– r equations.r equations.Example: For pressure drop per unit lengthExample: For pressure drop per unit length

21


Step 5 (Continued) Step 5 (Continued) 22

cba1 VDp ρ∆=Π l

1c,2b,1a0c2b:T

0c4ba3:L0c1:F

TLF)TFL()LT()L)(FL( 000c24b1a3

−=−==⇒=+−

=−++−=+

=−−− &

21V

Dpρ

∆=Π l

22


Step 6 Repeat Step 5 for each of the remaining Step 6 Repeat Step 5 for each of the remaining nonrepeatingnonrepeating variables.variables.

cba2 VD ρµ=Π

1c,1b,1a0c2b1:T

0c4ba2:L0c1:F

TLF)TFL()LT()L)(TFL( 000c24b1a2

−=−=−=⇒=+−

=−++−=+

=−−− &

ρµ

=ΠDV2

23


Step 7 Check all the resulting pi terms to make Step 7 Check all the resulting pi terms to make sure they are dimensionless.sure they are dimensionless.

Check to see that each group obtained is dimensionless.Check to see that each group obtained is dimensionless.Example: For pressure drop per unit length .Example: For pressure drop per unit length .

0000002

00000021

TLMTLFDV

TLMTLFV

Dp

==ρ

µ=Π

==ρ

∆=Π

&&

&&l

24


Step 8 Express the final form as a relationship Step 8 Express the final form as a relationship among the pi terms, and think about what is means.among the pi terms, and think about what is means.

Express the result of the dimensional analysis.Express the result of the dimensional analysis.

Example: For pressure drop per unit length .Example: For pressure drop per unit length .

⎟⎟⎠

⎞⎜⎜⎝

⎛ρ

µφ=

ρ

∆DVV

Dp2l

),,,,,( rk321 −ΠΠΠφ=Π

Dimensional analysis will not provide Dimensional analysis will not provide the form of the function. The function the form of the function. The function can only be obtained from a suitable set can only be obtained from a suitable set of experiments.of experiments.

25


The pi terms can be rearranged. For example, The pi terms can be rearranged. For example, ΠΠ22, could , could be expressed asbe expressed as

⎟⎟⎠

⎞⎜⎜⎝

⎛µ

ρφ=

ρ

∆ VDV

Dp2l

µρ

=ΠVD

2

26

Example 7.1 Method of Repeating Example 7.1 Method of Repeating VariablesVariables

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

27

Example 7.1 Example 7.1 SolutionSolution1/51/5

Drag force on a PLATEDrag force on a PLATE

Step 1:List all the dimensional variables involved. Step 1:List all the dimensional variables involved. D,w,h, D,w,h, ρρ,,μμ,V,V k=6 dimensional parameters.k=6 dimensional parameters.Step 2:Select primary dimensions M,L, and T. Express Step 2:Select primary dimensions M,L, and T. Express each of the variables in terms of basic dimensionseach of the variables in terms of basic dimensions

)V,,,h,w(fD µρ=

1311

2

LTVMLTML

LhLwMLTD−−−−

−

==ρ=µ

===

&&&

&&&

28


Step 3: Determine the required number of pi terms. Step 3: Determine the required number of pi terms. kk--rr=6=6--3=33=3

Step 4:Select repeating variables w,V,Step 4:Select repeating variables w,V,ρρ..Step 5~6:combining the repeating variables with each Step 5~6:combining the repeating variables with each of the other variables in turn, to form dimensionless of the other variables in turn, to form dimensionless groups or dimensionless products.groups or dimensionless products.

29


1c,2b,2a0b2:T

0c3ba1:L0c1:M

TLM)ML()LT()L)(MLT(VDw 000c3b1a2cba1

−=−=−=>>=−−

=−++=+

==ρ=Π −−−

ρ=Π 221

VwD

0c,0b,1a0b:T

0c3ba1:L0c:M

TLM)ML()LT()L(LVhw 000c3b1acba2

==−=>>=

=−++=

==ρ=Π −−

wh

2 =Π

30


1c,1b,1a0b1:T

0c3ba1:L0c1:M

TLM)ML()LT()L)(TML(Vw 000c3b1a11cba3

−=−=−=>>=−−

=−++−=+

==ρµ=Π −−−−

ρµ

=ΠwV3

31


Step 7: Check all the resulting pi terms to make sure Step 7: Check all the resulting pi terms to make sure they are dimensionless. they are dimensionless. Step 8: Express the final form as a relationship among Step 8: Express the final form as a relationship among the pi terms.the pi terms.The functional relationship isThe functional relationship is

⎟⎟⎠

⎞⎜⎜⎝

⎛ρ

µφ=

ρ

ΠΠφ=Π

wV,

wh

VwD

or),,(

22

321

32

Selection of Variables Selection of Variables 1/41/4

One of the most important, and difficult, steps in applying One of the most important, and difficult, steps in applying dimensional analysis to any given problem is the selection dimensional analysis to any given problem is the selection of the variables that are involved.of the variables that are involved.There is no simple procedure whereby the variable can be There is no simple procedure whereby the variable can be easily identified. Generally, one must rely on a good easily identified. Generally, one must rely on a good understanding of the phenomenon involved and the understanding of the phenomenon involved and the governing physical laws. governing physical laws. If extraneous variables are included, then too many pi If extraneous variables are included, then too many pi terms appear in the final solution, and it may be difficult, terms appear in the final solution, and it may be difficult, time consuming, and expensive to eliminate these time consuming, and expensive to eliminate these experimentally. experimentally.

33


If important variables are omitted, then an incorrect result If important variables are omitted, then an incorrect result will be obtained; and again, this may prove to be costly will be obtained; and again, this may prove to be costly and difficult to ascertain. and difficult to ascertain. Most engineering problems involve certain simplifying Most engineering problems involve certain simplifying assumptions that have an influence on the variables to be assumptions that have an influence on the variables to be considered.considered.Usually we wish to keep the problems as simple as Usually we wish to keep the problems as simple as possible, perhaps even if some accuracy is sacrificedpossible, perhaps even if some accuracy is sacrificed

34


A suitable balance between simplicity and accuracy is an A suitable balance between simplicity and accuracy is an desirable goal.~~~~~desirable goal.~~~~~Variables can be classified into three general group:Variables can be classified into three general group:

Geometry: lengths and angles.Geometry: lengths and angles.Material Properties: relate the external effects and the Material Properties: relate the external effects and the responses.responses.External Effects: produce, or tend to produce, a change External Effects: produce, or tend to produce, a change in the system. Such as force, pressure, velocity, or in the system. Such as force, pressure, velocity, or gravity.gravity.

35

Selection of Variables Selection of Variables 44--1/41/4

Points should be considered in the selection of variables:Points should be considered in the selection of variables:Clearly define the problem. WhatClearly define the problem. What’’s the main variable s the main variable of interest?of interest?Consider the basic laws that govern the phenomenon.Consider the basic laws that govern the phenomenon.Start the variable selection process by grouping the Start the variable selection process by grouping the variables into three broad classes.variables into three broad classes.

ContinuedContinued

36

Selection of Variables Selection of Variables 44--2/42/4

Points should be considered in the selection of variables:Points should be considered in the selection of variables:Consider other variables that may not fall into one the Consider other variables that may not fall into one the three categories. For example, time and time dependent three categories. For example, time and time dependent variables.variables.Be sure to include all quantities that may be held Be sure to include all quantities that may be held constant (e.g., g).constant (e.g., g).Make sure that all variables are independent. Look for Make sure that all variables are independent. Look for relationships among subsets of the variables.relationships among subsets of the variables.

37

Determination of Reference Dimension Determination of Reference Dimension 1/31/3

When to determine the number of pi terms, it is important When to determine the number of pi terms, it is important to know how many reference dimensions are required to to know how many reference dimensions are required to describe the variables.describe the variables.In fluid mechanics, the required number of reference In fluid mechanics, the required number of reference dimensions is three, but in some problems only one or two dimensions is three, but in some problems only one or two are required.are required.In some problems, we occasionally find the number of In some problems, we occasionally find the number of reference dimensionsreference dimensions needed to describe all variables is needed to describe all variables is smaller than the number of basic dimensionssmaller than the number of basic dimensions. Illustrated in . Illustrated in Example 7.2.Example 7.2.

38

Example 7.2 Determination of Pi TermsExample 7.2 Determination of Pi Terms

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

39


The vertical deflectionThe vertical deflection

( )EdhDf ,,,, γδ =

For F,L,T. Pi terms=6For F,L,T. Pi terms=6--2=42=4For M,L,T Pi terms=6For M,L,T Pi terms=6--3=33=3

212

223

TMLE,FLE

TML,FL

LdLhLD

L

−−−

−−−

==

=γ=γ

====δ

40


For F,L,T system, Pi terms=6For F,L,T system, Pi terms=6--2=42=4

⎟⎟⎠

⎞⎜⎜⎝

⎛γ

Φ=δ

⇒γ

=Π=Π=Πδ

=ΠDE,

Dd,

Dh

DDE,

Dd,

Dh,

D 4321

D and D and γγ are selected as repeating variablesare selected as repeating variables

4433

2211

ba4

ba3

ba2

ba1

EDdD

hDD

γ=Πγ=Π

γ=Πγδ=Π

41


For M,L,T system, Pi terms=6For M,L,T system, Pi terms=6--3=3 ?3=3 ?

A closer look at the dimensions of the variables listed A closer look at the dimensions of the variables listed reveal that only two reference dimensions, reveal that only two reference dimensions, L and MTL and MT--22

are required.are required.

42


( )σγ=∆ ,,Dfh

222 TM

TLMLL

Dh σγ∆

LF

LFLL

Dh

3

σγ∆

EXAMPLEEXAMPLE

MLT SYSTEMMLT SYSTEM FLT SYSTEMFLT SYSTEM

Pi term=4-3=1 Pi term=4-2=2

43


2200T0211L1100M

Dh

−−−

σγ∆

0000T1311L

1100FDh

−−

σγ∆

Dh

1∆

=Π

⎟⎟⎠

⎞⎜⎜⎝

⎛

γ

σΦ=

∆⇒

γ

σ=Π 222

DDh

D

Set Dimensional MatrixSet Dimensional MatrixMLT SYSTEMMLT SYSTEM FLT SYSTEMFLT SYSTEM

Rank=2 Pi term=4Rank=2 Pi term=4--2=22=2

44

Uniqueness of Pi Terms Uniqueness of Pi Terms 1/41/4

The Pi terms obtained depend on the somewhat arbitrary The Pi terms obtained depend on the somewhat arbitrary selection of repeating variables. For example, in the selection of repeating variables. For example, in the problem of studying the pressure drop in a pipe.problem of studying the pressure drop in a pipe.

⎟⎟⎠

⎞⎜⎜⎝

⎛µ

ρΦ=

ρ

∆ VDV

Dp12

l

⎟⎟⎠

⎞⎜⎜⎝

⎛µ

ρΦ=

µ∆ VD

VDp

2

2l


Selecting D,V, and ρ as repeating variables:

Selecting D,V, and µ as repeating variables:

45


Both are correct, and both would Both are correct, and both would lead to the same final equation for lead to the same final equation for the pressure drop. the pressure drop. There is not a There is not a unique set of pi termsunique set of pi terms which which arises from a dimensional analysisarises from a dimensional analysis. . The functions The functions ΦΦ11 and and ΦΦ22 are will are will be different because the dependent be different because the dependent pi terms are different for the two pi terms are different for the two relationships.relationships.

⎟⎟⎠

⎞⎜⎜⎝

⎛µ

ρΦ=

ρ

∆ VDV

Dp12

l

⎟⎟⎠

⎞⎜⎜⎝

⎛µ

ρΦ=

µ∆ VD

VDp

2

2l

46


( )321 ,ΠΠΦ=Π

( ) ( )'2223

'211 ,, ΠΠΦ=ΠΠΦ=Π

EXAMPLEEXAMPLE

b3

a2

'2 ΠΠ=ΠForm a new pi termForm a new pi term

All are correct

47


µ∆

=µ

ρ×

ρ

∆V

DpVDV

Dp 2

2ll

⎟⎟⎠

⎞⎜⎜⎝

⎛µ

ρΦ=

µ∆ VD

VDp

2

2l

⎟⎟⎠

⎞⎜⎜⎝

⎛µ

ρΦ=

ρ

∆ VDV

Dp12

l Selecting D,V, and ρ as repeating variables:

48

Common Dimensionless GroupsCommon Dimensionless Groups 1/21/2

A list of variables that commonly arise in fluid mechanical problems.Possible to provide a physical interpretation to the dimensionless groups which can be helpful in assessing their influence in a particular application.

49

Froude Number Froude Number 1/21/2

In honor of William Froude (1810~1879), a British civil engineerIn honor of William Froude (1810~1879), a British civil engineer, , mathematician, and naval architect who pioneered the use of towimathematician, and naval architect who pioneered the use of towing ng tanks for the study of ship design.tanks for the study of ship design.Froude number is the ratio of the forces due to the acceleratioFroude number is the ratio of the forces due to the acceleration of a n of a fluid particles (inertial force) to the force due to gravity (grfluid particles (inertial force) to the force due to gravity (gravity avity forces).forces).Froude number is significant for flows with free surface effectsFroude number is significant for flows with free surface effects..Froude number less than unity indicate subcritical flow and valuFroude number less than unity indicate subcritical flow and values es greater than unity indicate supercritical flow.greater than unity indicate supercritical flow.

3

2222

gLLV

gLVFr

gLVFr

ρ

ρ==>>=

50

Froude Number Froude Number 2/22/2

*

*s*

s

2s

ss

s dsdVVV

dsdVV

dtdVa

l===

l

ssVVV *s*

s ==

mdsdVVVF *

*s*

s

2

Il

=

lll gV

gV

dsdVV

gV

FFrF

2

*

*s*

s

2

G

I ≡≡==

51

Reynolds Number Reynolds Number 1/21/2

In honor of Osborne Reynolds (1842~1912), the British engineer In honor of Osborne Reynolds (1842~1912), the British engineer who first demonstrated that this combination of variables could who first demonstrated that this combination of variables could be be used as a criterion to distinguish between laminar and turbulentused as a criterion to distinguish between laminar and turbulent flow.flow.The Reynolds number is a measure of the ration of the inertia foThe Reynolds number is a measure of the ration of the inertia forces rces to viscous forces.to viscous forces.If the Reynolds number is small (Re<<1), If the Reynolds number is small (Re<<1), thithi sis an indication that sis an indication that the viscous forces the viscous forces arar dominant in the problem, and it may be dominant in the problem, and it may be possible to neglect the inertial effects; that is, the density opossible to neglect the inertial effects; that is, the density of the fluid f the fluid will no be an important variable.will no be an important variable.

υ=

µρ

=ll VVRe

52

Reynolds Number Reynolds Number 2/22/2

Flows with very small Reynolds numbers are commonly referred to Flows with very small Reynolds numbers are commonly referred to as as ““creeping flowscreeping flows””..For large Reynolds number flow, the viscous effects are small For large Reynolds number flow, the viscous effects are small relative to inertial effects and for these cases it may be possirelative to inertial effects and for these cases it may be possible to ble to neglect the effect of viscosity and consider the problem as one neglect the effect of viscosity and consider the problem as one involving a involving a ““nonviscousnonviscous”” fluid.fluid.Flows with Flows with ““largelarge”” Reynolds number generally are turbulent. Flows Reynolds number generally are turbulent. Flows in which the inertia forces are in which the inertia forces are ““smallsmall”” compared with the viscous compared with the viscous forces are characteristically laminar flowsforces are characteristically laminar flows..

53

Euler numberEuler number

In honor of Leonhard Euler (1707~1783), a famous Swiss In honor of Leonhard Euler (1707~1783), a famous Swiss mathematician who pioneered work on the relationship between mathematician who pioneered work on the relationship between pressure and flow.pressure and flow.EulerEuler’’s number is the ratio of pressure force to inertia forces. It iss number is the ratio of pressure force to inertia forces. It isoften called theoften called the pressure coefficient, Cp.pressure coefficient, Cp.

22 Vp

VpEu

ρ∆

≡ρ

=

54

CavitationCavitation NumberNumber

For problems in which For problems in which cavitationcavitation is of concern, the dimensionless is of concern, the dimensionless group is commonly used, where group is commonly used, where ppvv is the vapor is the vapor pressure and ppressure and prr is some reference pressure.is some reference pressure.The The cavitationcavitation number is used in the study of number is used in the study of cavitationcavitation phenomena.phenomena.The smaller the The smaller the cavitationcavitation number, the more likely number, the more likely cavitationcavitation is to is to occur.occur.

2

vr

V21

ppCaρ

−=

221

vr V/)pp( ρ−

55

Cauchy Number & Mach Number Cauchy Number & Mach Number 1/21/2

The Cauchy number is named in honor of The Cauchy number is named in honor of AugustinAugustin Louis de Louis de Cauchy (1789~1857), a French engineer, mathematician, and Cauchy (1789~1857), a French engineer, mathematician, and hydrodynamicisthydrodynamicist..The Mach number is named in honor of Ernst Mach (1838~1916), The Mach number is named in honor of Ernst Mach (1838~1916), an Austrian physicist and philosopher.an Austrian physicist and philosopher.Either number may be used in problems in which fluid Either number may be used in problems in which fluid compressibility is important.compressibility is important.

CaEVMa

EV

EV

cVMa

EVaC

22

2

=ρ

=ρ

=

ρ

==ρ

=υυυυ

56

Cauchy Number & Mach Number Cauchy Number & Mach Number 2/22/2

Both numbers can be interpreted as representing an index of the Both numbers can be interpreted as representing an index of the ration of inertial force to compressibility force, where V is thration of inertial force to compressibility force, where V is the flow e flow speed and c is the local sonic speed.speed and c is the local sonic speed.Mach number is a key parameter that characterizes compressibilitMach number is a key parameter that characterizes compressibility y effects in a flow.effects in a flow.When the Mach number is relatively small (say, less than 0.3), tWhen the Mach number is relatively small (say, less than 0.3), the he inertial forces induced by the fluid motion are not sufficientlyinertial forces induced by the fluid motion are not sufficiently large large to cause a significant change in the fluid density, and in this to cause a significant change in the fluid density, and in this case the case the compressibility of the fluid can be neglected.compressibility of the fluid can be neglected.For truly incompressible flow, c=For truly incompressible flow, c=∞∞ so that M=0.so that M=0.

57

StrouhalStrouhal Number Number 1/21/2

In honor of In honor of VincenzVincenz StrouhalStrouhal (1850~1922), who used this parameter (1850~1922), who used this parameter in his study of in his study of ““singing wires.singing wires.”” The most dramatic evidence of this The most dramatic evidence of this phenomenon occurred in 1940 with the collapse of the Tacoma phenomenon occurred in 1940 with the collapse of the Tacoma Narrow bridges. The shedding frequency of the vortices coincidedNarrow bridges. The shedding frequency of the vortices coincidedwith the natural frequency of the bridge, thereby setting up a with the natural frequency of the bridge, thereby setting up a resonant condition that eventually led to the collapse of the brresonant condition that eventually led to the collapse of the bridge.idge.This parameter is important in unsteady, oscillating flow probleThis parameter is important in unsteady, oscillating flow problems ms in which the frequency of the oscillation is in which the frequency of the oscillation is ωω ..

VtS lω=

58

StrouhalStrouhal Number Number 2/22/2

This parameter represents a measure of This parameter represents a measure of the ration of inertial force due to the the ration of inertial force due to the unsteadiness of the flow (local unsteadiness of the flow (local acceleration) to the inertial forces due to acceleration) to the inertial forces due to change in velocity from point to point in change in velocity from point to point in the flow field (convective acceleration). the flow field (convective acceleration). This type of unsteady flow may develop This type of unsteady flow may develop when a fluid flows past a solid body (such when a fluid flows past a solid body (such as a wire or cable) placed in the moving as a wire or cable) placed in the moving stream.stream.

For example, in a certain Reynolds number range, a periodic flowFor example, in a certain Reynolds number range, a periodic flow will will develop downstream from a cylinder placed in a moving stream duedevelop downstream from a cylinder placed in a moving stream due to to a regular patterns of vortices that are shed from the body.a regular patterns of vortices that are shed from the body.

59

Weber Number Weber Number 1/21/2

Named after Moritz Weber (1871~1951), a German professor of Named after Moritz Weber (1871~1951), a German professor of naval mechanics who was instrumental in formalizing the general naval mechanics who was instrumental in formalizing the general use of common dimensionless groups as a basis for similitude use of common dimensionless groups as a basis for similitude studies.studies.Weber number is important in problem in which there is an interfWeber number is important in problem in which there is an interface ace between two fluids. In this situation the surface tension may plbetween two fluids. In this situation the surface tension may play an ay an important role in the phenomenon of interest.important role in the phenomenon of interest.

σρ

=l2VWe

60

Weber Number Weber Number 2/22/2

Weber number is the ratio of inertia forces to surface tension fWeber number is the ratio of inertia forces to surface tension forces.orces.Common examples of problems in which Weber number may be Common examples of problems in which Weber number may be important include the flow of thin film of liquid, or the formatimportant include the flow of thin film of liquid, or the formation of ion of droplets or bubbles.droplets or bubbles.The flow of water in a river is not affected significantly by The flow of water in a river is not affected significantly by surefacesurefacetension, since inertial and gravitational effects are dominant tension, since inertial and gravitational effects are dominant (We>>1).(We>>1).

61

Correlation of Experimental DataCorrelation of Experimental Data

Dimensional analysis only provides the dimensionless groups Dimensional analysis only provides the dimensionless groups describing the phenomenon, and not the specific relationship describing the phenomenon, and not the specific relationship between the groups.between the groups.To determine this relationship, suitable experimental data must To determine this relationship, suitable experimental data must be be obtained.obtained.The degree of difficulty depends on the number of pi terms.The degree of difficulty depends on the number of pi terms.

62

Problems with One Pi TermProblems with One Pi Term

The functional relationship for one Pi term.The functional relationship for one Pi term.

C1 =Π

where C is a constant. The value of the constant must still be determined by experiment.

63

Example 7.3 Flow with Only One Pi TermExample 7.3 Flow with Only One Pi Term

Assume that the drag, D, acting on a spherical particle that falAssume that the drag, D, acting on a spherical particle that falls very ls very slowly through a viscous fluid is a function of the particle diaslowly through a viscous fluid is a function of the particle diameter, meter, d, the particle velocity, V, and the fluid viscosity, d, the particle velocity, V, and the fluid viscosity, μμ. Determine, . Determine, with the aid the dimensional analysis, how the drag depends on twith the aid the dimensional analysis, how the drag depends on the he particle velocity. particle velocity.

64

Example 7.3 Example 7.3 SolutionSolution

The drag The drag

),V,d(fD µ= 13

2

LTVMLTFLLdFD

−−

−

==ρ

=µ==

&&

&&&

VD

dVCDCdV

D1

∝

µ==µ

=Π

For a given particle and fluids, the drag varies For a given particle and fluids, the drag varies directly with the velocitydirectly with the velocity

65

Problems with Two or More Pi Term Problems with Two or More Pi Term 1/21/2

Problems with two pi termsProblems with two pi terms)( 21 ΠΦ=Π

the functional relationship the functional relationship among the variables can the among the variables can the be determined by varying be determined by varying ΠΠ22 and measuring the and measuring the corresponding value of corresponding value of ΠΠ11..

The empirical equation The empirical equation relating relating ΠΠ22 and and ΠΠ11 by using by using a standard curvea standard curve--fitting fitting technique.technique.

An empirical relationship is An empirical relationship is valid over the range of valid over the range of ΠΠ22..

Dangerous to Dangerous to extrapolate beyond extrapolate beyond valid rangevalid range

66

Problems with Two or More Pi Term Problems with Two or More Pi Term 2/22/2

Problems with three pi terms.Problems with three pi terms.

( )321 ,ΠΠΦ=Π

Families curve of curvesFamilies curve of curves

To determine a suitable empirical equation To determine a suitable empirical equation relating the three pi terms.relating the three pi terms.

To show data correlations on simple graphs.To show data correlations on simple graphs.

67

Modeling and SimilitudeModeling and SimilitudeTo develop the procedures for designing To develop the procedures for designing models so that the model and prototype models so that the model and prototype will behave in a similar fashionwill behave in a similar fashion…………..

68

Model vs. Prototype Model vs. Prototype 1/21/2

Model ? A model is a representation of a physical system that maModel ? A model is a representation of a physical system that may y be used to predict the behavior of the system in some desired rebe used to predict the behavior of the system in some desired respect. spect. Mathematical or computer models may also conform to this Mathematical or computer models may also conform to this definition, our interest will be in physical model.definition, our interest will be in physical model.Prototype? The physical system for which the prediction are to bPrototype? The physical system for which the prediction are to be e made. made. Models that resemble the prototype but are generally of a differModels that resemble the prototype but are generally of a different ent size, may involve different fluid, and often operate under diffesize, may involve different fluid, and often operate under different rent conditions. conditions. Usually a model is smaller than the prototype. Usually a model is smaller than the prototype. Occasionally, if the prototype is very small, it may be advantagOccasionally, if the prototype is very small, it may be advantageous eous to have a model that is larger than the prototype so that it canto have a model that is larger than the prototype so that it can vevemore easily studied. For example, large models have been used tomore easily studied. For example, large models have been used tostudy the motion of red blood cells.study the motion of red blood cells.

69

Model vs. Prototype Model vs. Prototype 2/22/2

With the successful development of a valid model, it is possibleWith the successful development of a valid model, it is possible to to predict the behavior of the prototype under a certain set of predict the behavior of the prototype under a certain set of conditions.conditions.There is an inherent danger in the use of models in that predictThere is an inherent danger in the use of models in that predictions ions can be made that are in error and the error not detected until tcan be made that are in error and the error not detected until the he prototype is found not to perform as predicted.prototype is found not to perform as predicted.It is imperative that the model be properly designed and tested It is imperative that the model be properly designed and tested and and that the results be interpreted correctly.that the results be interpreted correctly.

70

Similarity of Model and PrototypeSimilarity of Model and Prototype

What conditions must be met to ensure the similarity of model anWhat conditions must be met to ensure the similarity of model and d prototype?prototype?Geometric SimilarityGeometric Similarity

Model and prototype have same shape.Model and prototype have same shape.Linear dimensions on model and prototype correspond within Linear dimensions on model and prototype correspond within constant scale factor.constant scale factor.

Kinematic SimilarityKinematic SimilarityVelocities at corresponding points on model and prototype differVelocities at corresponding points on model and prototype differonly by a constant scale factor.only by a constant scale factor.

Dynamic SimilarityDynamic SimilarityForces on model and prototype differ only by a constant scale Forces on model and prototype differ only by a constant scale factor.factor.

71

Theory of Models Theory of Models 1/51/5

The theory of models can be readily developed by using The theory of models can be readily developed by using the principles of dimensional analysis.the principles of dimensional analysis.For given problem which can be described in terms of a For given problem which can be described in terms of a set of pi terms asset of pi terms as

),,,,( n321 ΠΠΠφ=ΠThis relationship can be formulated This relationship can be formulated with a knowledge of the general with a knowledge of the general nature of the physical phenomenon nature of the physical phenomenon and the variables involved. and the variables involved.

This equation applies to any system that is governed by the same variables.

72


A similar relationship can be written for a model of this A similar relationship can be written for a model of this prototype; that is,prototype; that is,

where the form of the function will be the same as long where the form of the function will be the same as long as the as the same phenomenonsame phenomenon is involved in both the is involved in both the prototype and the model.prototype and the model.

),,,,,( nmm3m2m1 ΠΠΠφ=Π

The prototype and the model must have The prototype and the model must have the same phenomenon.the same phenomenon.

73


Model design (the model is designed and operated) Model design (the model is designed and operated) conditions, also called similarity requirements or modeling conditions, also called similarity requirements or modeling laws.laws.

The The form of form of ΦΦ is the same for model and prototype, it is the same for model and prototype, it follows thatfollows that

nmnm33m22 ..... Π=ΠΠ=ΠΠ=Π

m11 Π=Π

This is the This is the desired prediction equationdesired prediction equation and and indicates that the measured of indicates that the measured of ΠΠ1m1m obtained obtained with the model will be equal to the with the model will be equal to the corresponding corresponding ΠΠ11 for the prototype as long as for the prototype as long as the other the other ΠΠ parameters are equal. parameters are equal.

74

Theory of Models Theory of Models 4/54/5 –– SummarySummary11

The prototype and the model must have the same The prototype and the model must have the same phenomenon. phenomenon.

),,,,( n321 ΠΠΠφ=ΠFor prototypeFor prototype

For modelFor model ),,,,,( nmm3m2m1 ΠΠΠφ=Π

75

Theory of Models Theory of Models 5/55/5 –– SummarySummary22

The model is designed and operated under the following The model is designed and operated under the following conditions conditions (called design conditions, also called similarity (called design conditions, also called similarity requirements or modeling laws)requirements or modeling laws)

The measured of The measured of ΠΠ1m1m obtained with the model will be obtained with the model will be equal to the corresponding equal to the corresponding ΠΠ11 for the prototype. for the prototype.

nmnm33m22 ..... Π=ΠΠ=ΠΠ=Π

m11 Π=Π Called prediction equationCalled prediction equation

76

Theory of Models Theory of Models EXAMPLE 1EXAMPLE 1

Example: Considering the drag force on a sphereExample: Considering the drag force on a sphere..

The prototype and the model must have the same phenomenon. The prototype and the model must have the same phenomenon.

Design conditions.Design conditions.

Then Then ……

⎟⎟⎠

⎞⎜⎜⎝

⎛µ

ρ=

ρ m

mmm12

m2

mm

m DVfDV

F

),,V,D(fF µρ=

prototype122

VDfDV

F⎟⎟⎠

⎞⎜⎜⎝

⎛µ

ρ=

ρ

prototypeelmod

VDVD⎟⎟⎠

⎞⎜⎜⎝

⎛µ

ρ=⎟⎟

⎠

⎞⎜⎜⎝

⎛µ

ρ

prototype22

elmod22 DV

FDV

F⎟⎟⎠

⎞⎜⎜⎝

⎛

ρ=⎟

⎟⎠

⎞⎜⎜⎝

⎛

ρ

⎟⎟⎠

⎞⎜⎜⎝

⎛µ

ρ=

ρ

VDfDV

F122

77

Theory of Models Theory of Models EXAMPLE 2EXAMPLE 2

Example: Determining the drag force on a thin rectangular plate Example: Determining the drag force on a thin rectangular plate ((ww×× hh in size)in size)

The prototype and the model must have the same phenomenon. The prototype and the model must have the same phenomenon.

Design conditions.Design conditions.

Then Then ……

( )V,,,h,wfD ρµ=

prototype22

Vw,hw

VwD

⎟⎟⎠

⎞⎜⎜⎝

⎛µ

ρΦ=

ρ⎟⎟⎠

⎞⎜⎜⎝

⎛µ

ρΦ=

ρ m

mmm

m

m2

mm2

m

m wV,hw

VwD

µρ

=µ

ρ=

VwwV,hw

hw

m

mmm

m

m

m

2

mm

2

m2

mm2

m

m22 D

VV

wwD

VwD

VwD

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛ρρ

⎟⎟⎠

⎞⎜⎜⎝

⎛=>>

ρ=

ρ

⎟⎟⎠

⎞⎜⎜⎝

⎛µ

ρΦ=

ρ

Vw,hw

VwD

22

78

Example 7.5 Prediction of Prototype Example 7.5 Prediction of Prototype Performance from Model Data Performance from Model Data 1/21/2

A long structural component of a bridge has the cross section A long structural component of a bridge has the cross section shown in Figure E7.5. It is known that when a steady wind blows shown in Figure E7.5. It is known that when a steady wind blows past this type of bluff body, vortices may develop on the downwipast this type of bluff body, vortices may develop on the downwind nd side that are shed in a regular fashion at some definite frequenside that are shed in a regular fashion at some definite frequency. cy. Since these vortices can create harmful periodic forces acting oSince these vortices can create harmful periodic forces acting on the n the structure, it is important to determine the shedding frequency. structure, it is important to determine the shedding frequency. For For the specific structure of interest, D=0.1m, H=0.3m, and a the specific structure of interest, D=0.1m, H=0.3m, and a representative wind velocity 50km/hr. Standard air can be assumerepresentative wind velocity 50km/hr. Standard air can be assumed. d. The shedding frequency is to be determined through the use of a The shedding frequency is to be determined through the use of a smallsmall--scale model that is to be tested in a water tunnel. For the scale model that is to be tested in a water tunnel. For the model Dm=20mm and the water temperature is 20model Dm=20mm and the water temperature is 20℃℃..

79

Example 7.5 Prediction of Prototype Example 7.5 Prediction of Prototype Performance from Model Data Performance from Model Data 2/22/2

Determine the model dimension, Determine the model dimension, HmHm, and the velocity at which the , and the velocity at which the test should be performed. If the shedding frequency test should be performed. If the shedding frequency ωω for the for the model is found to be 49.9Hz, what is the corresponding frequencymodel is found to be 49.9Hz, what is the corresponding frequencyfor the prototype? for the prototype? For air at standard condition For air at standard condition For water at 20For water at 20℃℃,,

35 m/kg23.1,sm/kg1079.1 =ρ−×=µ −

3water

3water m/kg998,sm/kg101 =ρ−×=µ −

80


Step 1:List all the dimensional variables involved. Step 1:List all the dimensional variables involved. ωωD,H,V,D,H,V,ρρ,,μμ k=6 dimensional variables.k=6 dimensional variables.Step 2:Select primary dimensions F,L and T. List the Step 2:Select primary dimensions F,L and T. List the dimensions of all variables in terms of primary dimensions. dimensions of all variables in terms of primary dimensions. r=3 primary dimensionsr=3 primary dimensions

TMLTFLLTV

LHLDT2241

1

−−−

−

=µ=ρ=

===ω

&&&

&&&

81


Step 3: Determine the required number of pi terms.Step 3: Determine the required number of pi terms.kk--rr=6=6--3=33=3

Step 4:Select repeating variables D,V, Step 4:Select repeating variables D,V, μμ..Step Step 5~6:combining the repeating variables with each of 5~6:combining the repeating variables with each of the other variables in turn, to form dimensionless groups.the other variables in turn, to form dimensionless groups.

µρ

=µρ=Π

=µ=Πω

=µω=Π

VDVD

DHVHD

VDVD

333

222111

cba3

cba2

cba1

82


The functional relationship isThe functional relationship is

The prototype and the model must have the same The prototype and the model must have the same phenomenon.phenomenon.

⎟⎟⎠

⎞⎜⎜⎝

⎛µ

ρφ=

ω VD,DH

VD

⎟⎟⎠

⎞⎜⎜⎝

⎛µ

ρφ=

ω

m

mmm

m

m

m

mm DV,DH

VD

StrouhalStrouhal numbernumber

83


The Design conditions.The Design conditions.

ThenThen……..m

m

DH

DH

=m

mmm DVVDµ

ρ=

µρ

mm60...DDHH mm === s/m9.13...V

DDV

mm

mm ==

ρρ

µµ

=

Hz0.29...D

DVV

mm

m==ω=ω

84

Model ScalesModel Scales

The ratio of a model variable to the corresponding The ratio of a model variable to the corresponding prototype variable is called the scale for that variable.prototype variable is called the scale for that variable.

2

m2

1

m1

m2

m1

2

1l

l

l

l

l

l

l

ll ==λ⇒=

ρρ

=λρm

VVm

V =λ

µµ

=λµm

Length ScaleLength Scale

Velocity ScaleVelocity Scale

Density ScaleDensity Scale

Viscosity ScaleViscosity Scale

85

Validation of Models DesignValidation of Models Design

The purpose of model design is to predict the effects of The purpose of model design is to predict the effects of certain proposed changes in a given prototype, and in this certain proposed changes in a given prototype, and in this instance some actual prototype data may be available.instance some actual prototype data may be available.Validation of model design ?Validation of model design ?The model can be designed, constructed, and tested, and The model can be designed, constructed, and tested, and the model prediction can be compared with these data. the model prediction can be compared with these data. If If the agreement is satisfactorythe agreement is satisfactory, then the model can be , then the model can be changed in the desired manner, and the corresponding changed in the desired manner, and the corresponding effect on the prototype can be predicted with increased effect on the prototype can be predicted with increased confidence.confidence.

86

Distorted ModelsDistorted Models

In many model studies, to achieve dynamic similarity In many model studies, to achieve dynamic similarity requires duplication of several dimensionless groups. requires duplication of several dimensionless groups. In some cases, complete dynamic similarity between In some cases, complete dynamic similarity between model and prototype may not be attainable. If one or more model and prototype may not be attainable. If one or more of the similarity requirements are not met, for example, of the similarity requirements are not met, for example, if , then it follows that the prediction equatif , then it follows that the prediction equation ion

is not true; that is, is not true; that is, MODELS for which one or more of the similarity MODELS for which one or more of the similarity requirements are not satisfied are called requirements are not satisfied are called DISTORTED DISTORTED MODELSMODELS. .

m22 Π≠Π

m11 Π=Π m11 Π≠Π

87

Distorted Models Distorted Models EXAMPLEEXAMPLE--1 1/31 1/3

Determine the drag force on a surface ship, complete Determine the drag force on a surface ship, complete dynamic similarity requires that both Reynolds and Froude dynamic similarity requires that both Reynolds and Froude numbers be duplicated between model and prototype.numbers be duplicated between model and prototype.

To match Froude numbers To match Froude numbers between model and prototypebetween model and prototype

2/1p

pp2/1

m

mm

)g(

VFr

)g(VFr

ll===

p

ppp

m

mmm

VReVRe

υ==

υ=

ll

Froude numbersFroude numbers

Reynolds numbersReynolds numbers2/1

p

m

p

mVV

⎟⎟⎠

⎞⎜⎜⎝

⎛=

l

l

88

Distorted Models Distorted Models EXAMPLEEXAMPLE--11 2/32/3

To match Reynolds numbers between model and To match Reynolds numbers between model and prototypeprototype

p

m

p

m

p

mVV

l

l=

υυ

2/3

p

m

p

m2/1

p

m

p

m⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟

⎟⎠

⎞⎜⎜⎝

⎛=

υυ

l

l

l

l

l

l

If If llmm/ / llp p equals 1/100(a typical length scale for ship equals 1/100(a typical length scale for ship model tests) , then model tests) , then υυmm//υυp p must be 1/1000.must be 1/1000.>>> The >>> The kinematickinematic viscosity ratio required to viscosity ratio required to duplicate Reynolds numbers cannot be attained.duplicate Reynolds numbers cannot be attained.

89


It is impossible in practice for this model/prototype scale It is impossible in practice for this model/prototype scale of 1/100 to satisfy both the Froude number and Reynolds of 1/100 to satisfy both the Froude number and Reynolds number criterianumber criteria; ; at best we will be able to satisfy only at best we will be able to satisfy only one of them.one of them.If water is the only practical liquid for most model test of If water is the only practical liquid for most model test of freefree--surface flows, a surface flows, a fullfull--scale test is required to obtain scale test is required to obtain complete dynamic similaritycomplete dynamic similarity..

90


In the study of open channel or freeIn the study of open channel or free--surface flows. surface flows. Typically in these problems both the Reynolds number Typically in these problems both the Reynolds number and Froude number are involved. and Froude number are involved.

To match Froude numbers To match Froude numbers between model and prototypebetween model and prototype

2/1pp

pp2/1

mm

mm

)g(

VFr

)g(VFr

ll=== Froude numbersFroude numbers

p

pppp

m

mmmm

VReVRe

µ

ρ==

µρ

=ll Reynolds numbersReynolds numbers

ll

lλ=⎟

⎟⎠

⎞⎜⎜⎝

⎛=

2/1

p

m

p

mVV

91


To match Reynolds numbers between model and To match Reynolds numbers between model and prototypeprototype

m

p

m

p

p

m

p

mVV

l

l

ρ

ρ

µµ

=

If If llmm/ / ll p p equals 1/100(a typical length scale for ship equals 1/100(a typical length scale for ship model tests) , then model tests) , then υυmm//υυp p must be 1/1000.must be 1/1000.>>>The >>>The kinematickinematic viscosity ratio required to viscosity ratio required to duplicate Reynolds numbers cannot be attained.duplicate Reynolds numbers cannot be attained.

p

m

mp

pm2/3

m

p

mp

pm

p

m

p

m

//

//

VV

υυ

=ρρ

µµ=λ

ρρ

µµ==

l

l

l

l

l

92

Typical Model StudiesTypical Model Studies

Flow through closed conduits.Flow through closed conduits.Flow around immersed bodies.Flow around immersed bodies.Flow with a free surface.Flow with a free surface.

93

Flow Through Closed Conduits Flow Through Closed Conduits 1/51/5

This type of flow includes pipe flow and flow through This type of flow includes pipe flow and flow through valves, fittings, and metering devices. valves, fittings, and metering devices. The conduits are often circular, they could have other The conduits are often circular, they could have other shapes as well and may contain expansions or contractions.shapes as well and may contain expansions or contractions.Since there are no fluid interfaces or free surface, the Since there are no fluid interfaces or free surface, the dominant forces are inertial and viscous forces so that the dominant forces are inertial and viscous forces so that the Reynolds number is an important similarity parameter.Reynolds number is an important similarity parameter.

94


For low Mach numbers (Ma<0.3), compressibility effects For low Mach numbers (Ma<0.3), compressibility effects are usually negligible for both the flow of liquids or gases.are usually negligible for both the flow of liquids or gases.For flow in closed conduits at low Mach numbers, and For flow in closed conduits at low Mach numbers, and dependent pi term, such as pressure drop, can be expressed dependent pi term, such as pressure drop, can be expressed asasDependent pi term=Dependent pi term= ⎟⎟

⎠

⎞⎜⎜⎝

⎛µ

ρεφ

l

ll

l V,,i

Where Where ll is some particular length of the system and is some particular length of the system and lliirepresents a series of length terms, represents a series of length terms, εε/ / ll is the relative is the relative roughness of the surface, and roughness of the surface, and ρρVVll//µµ is the Reynolds number.is the Reynolds number.

95


To meet the requirement of geometric similarityTo meet the requirement of geometric similarity

To meet the requirement of Reynolds numberTo meet the requirement of Reynolds number

ll

l

l

l

lll

l

l

lλ=

εε

==⇒ε

=ε

= mm

i

im

m

mi

m

im

mm

mm

m

mmmV

VVVl

lll

ρρ

µµ

=⇒µ

ρ=

µρ

If the same fluid is used, thenIf the same fluid is used, then ll

lλ=⇒= /VV

VV

mm

m

1m

m ==µµ

l

l

96


The fluid velocity in the model will be larger than that in The fluid velocity in the model will be larger than that in the prototype for any length scale less than 1. Since length the prototype for any length scale less than 1. Since length scales are typically much less than unity.scales are typically much less than unity.Reynolds number similarity may be difficult to achieve Reynolds number similarity may be difficult to achieve because of the large model velocities required.because of the large model velocities required.

VVm >ll

lλ=⇒= /VV

VV

mm

m

1<λl

97


With these similarity requirements satisfied, it follows that With these similarity requirements satisfied, it follows that the dependent pi term will be equal in model and the dependent pi term will be equal in model and prototype. For example,prototype. For example,

The prototype pressure dropThe prototype pressure drop21

Vp

ρ

∆=ΠDependent pi termDependent pi term

m

2

mmp

VVp ∆⎟⎟

⎠

⎞⎜⎜⎝

⎛ρρ

=∆ mpp ∆≠∆In general

98

Example 7.6 Reynolds Number SimilarityExample 7.6 Reynolds Number Similarity

Model test are to be performed to study the flow through a largeModel test are to be performed to study the flow through a largevalve having a 2valve having a 2--ftft--diameter inlet and carrying water at a diameter inlet and carrying water at a flowrateflowrate of of 30cfs. The working fluid in the model is water at the same 30cfs. The working fluid in the model is water at the same temperature as that in the prototype. Complete geometric similartemperature as that in the prototype. Complete geometric similarity ity exits between model and prototype, and the model inlet diameter exits between model and prototype, and the model inlet diameter is is 30 in. Determine the required 30 in. Determine the required flowrateflowrate in the model. in the model.

99

Example 7.6 Example 7.6 SolutionSolution

DD...

VAAV

QQ mmmm ===

To ensure dynamic similarity, the model tests should be run so that

m

m

m

mmm D

DVVVDDVReeR =⇒

ν=

ν=

Same fluid used in the model and prototypeSame fluid used in the model and prototype

cfs75.3)s/ft30((2ft)

(3/12ft)Q 3m ==

100

Flow Around Immersed Bodies Flow Around Immersed Bodies 1/71/7

This type of flow includes flow around aircraft, This type of flow includes flow around aircraft, automobiles, golf balls, and building.automobiles, golf balls, and building.For these problems, geometric and Reynolds number For these problems, geometric and Reynolds number similarity is required.similarity is required.Since there are no fluid interfaces, surface tension is not Since there are no fluid interfaces, surface tension is not important. Also, gravity will not affect the flow pattern, so important. Also, gravity will not affect the flow pattern, so the Froude number need not to be considered.the Froude number need not to be considered.For incompressible flow, the Mach number can be omitted.For incompressible flow, the Mach number can be omitted.

101


A general formulation for these problems isA general formulation for these problems is

⎟⎟⎠

⎞⎜⎜⎝

⎛µ

ρεφ

l

ll

l V,,i

Where Where ll is some characteristic length of the system and is some characteristic length of the system and lliirepresents other pertinent lengths, represents other pertinent lengths, εε/ / ll is the relative is the relative roughness of the surface, and roughness of the surface, and ρρVVll//µµ is the Reynolds number.is the Reynolds number.

Dependent pi term=Dependent pi term=

Model of the National Bank of Commerce, San Antonio, Texas, for measurement of peak, rms, and mean pressure distributions. The model is located in a long-test-section, meteorological wind tunnel.

102


Frequently, the dependent variable of interest for this type Frequently, the dependent variable of interest for this type of problem is the drag, D, developed on the body.of problem is the drag, D, developed on the body.

To meet the requirement of geometric similarityTo meet the requirement of geometric similarity

22D

V21

DClρ

=

The dependent pi terms would usually be expressed in the form of a drag coefficient

ll

l

l

l

lll

l

l

lλ=

εε

==⇒ε

=ε

= mm

i

im

m

mi

m

im

103


To meet the requirement of Reynolds number similarityTo meet the requirement of Reynolds number similarity

m

m

mm

mm

m

mmmV

VVVl

l

l

lll

υυ

=ρρ

µµ

=⇒µ

ρ=

µρ

m

2

m

2

mm2m

2mm

m22

DVVD

V21

D

V21

D⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛ρρ

=⇒ρ

=ρ l

l

ll

The same fluid is used, thenThe same fluid is used, then ll

lλ=⇒= /VV

VV

mm

m

104


The fluid velocity in the model will be larger than that in The fluid velocity in the model will be larger than that in the prototype for any length scale less than 1. Since length the prototype for any length scale less than 1. Since length scales are typically much less than unity.scales are typically much less than unity.Reynolds number similarity may be difficult to achieve Reynolds number similarity may be difficult to achieve because of the large model velocities required.because of the large model velocities required.

105


How to reduce the fluid velocity in the model ?How to reduce the fluid velocity in the model ?

A different fluid is used in the model such thatA different fluid is used in the model such that 1/m <υυ

For example, the ratio of the For example, the ratio of the kinematickinematic viscosity of water to that viscosity of water to that of air is approximately 1/10, so that if the prototype fluid werof air is approximately 1/10, so that if the prototype fluid were e air, test might be run on the model using water. air, test might be run on the model using water.

This would reduce the required model velocity, but This would reduce the required model velocity, but it still may be difficult to achieve the necessary it still may be difficult to achieve the necessary velocity in a suitable test facility, velocity in a suitable test facility, such as a water such as a water tunnel.tunnel.

VVm <

106


How to reduce the fluid velocity in the model ?How to reduce the fluid velocity in the model ?

Same fluid with different density.. Same fluid with different density.. ρρmm>>ρρ

An alternative way to reduce An alternative way to reduce VVmm is to is to increase the air pressure in the tunnel so that increase the air pressure in the tunnel so that ρρmm>>ρρ. The pressurized tunnels are obviously . The pressurized tunnels are obviously complicated and expensive.complicated and expensive.

107

Example 7.7 Model Design Conditions and Example 7.7 Model Design Conditions and Predicted Prototype PerformancePredicted Prototype Performance

The drag on an airplane cruising at 240 mph in standard air is tThe drag on an airplane cruising at 240 mph in standard air is to be o be determined from tests on a 1:10 scale model placed determined from tests on a 1:10 scale model placed inaina pressurized pressurized wind tunnel. To minimize compressibility effects, the airspeed iwind tunnel. To minimize compressibility effects, the airspeed in the n the wind tunnel is also to be 240 mph. Determine the required air wind tunnel is also to be 240 mph. Determine the required air pressure in the tunnel (assuming the same air temperature for mopressure in the tunnel (assuming the same air temperature for model del and prototype), and the drag on the prototype corresponding to aand prototype), and the drag on the prototype corresponding to ameasured force of 1 lb on the model. measured force of 1 lb on the model.

108


The Reynolds numbers in model and prototype are the same. Thus,

µρ

=µ

ρ ll VV

m

mmm

VVmm=V, =V, llmm// ll=1/10=1/10

µµ

=µµ

=ρρ m

mm

mm 10VV

l

l

The same fluid with The same fluid with ρρmm==ρρ and and μμmm==μμ cannot be cannot be used if Reynolds number similarity is to be used if Reynolds number similarity is to be maintained.maintained.

109


10m =ρρ

We assume that an increase in pressure does not We assume that an increase in pressure does not significantly change the viscosity so that the required significantly change the viscosity so that the required increase in density is given by the relationshipincrease in density is given by the relationship

10p

p mm =ρρ

=For ideal gas

...DVD

VD

2m

2mm2

1m

2221

=ρ

=ρ ll

110

Flow Around Immersed Bodies at High Flow Around Immersed Bodies at High Reynolds Number Reynolds Number 1/31/3

Unfortunately, in many situations the flow characteristics Unfortunately, in many situations the flow characteristics are not strong influenced by the Reynolds number over the are not strong influenced by the Reynolds number over the operating range of interest.operating range of interest.Consider the variation in the drag coefficient with the Consider the variation in the drag coefficient with the Reynolds number for a smooth sphere of diameter d Reynolds number for a smooth sphere of diameter d placed in a uniform stream with approach velocity, V.placed in a uniform stream with approach velocity, V.At high Reynolds numbers the drag is often essentially At high Reynolds numbers the drag is often essentially independent of the Reynolds number.independent of the Reynolds number.

111


The effect of Reynolds number on the drag coefficient, CD for a smooth sphere with CD = D/ ½ AρV2, where A is the projected area of sphere, πd2/4.

112


For problems involving high velocities in which the Mach For problems involving high velocities in which the Mach number is greater than about 0.3, the influence of number is greater than about 0.3, the influence of compressibility, and therefore the Mach number (or compressibility, and therefore the Mach number (or Cauchy number), becomes significant.Cauchy number), becomes significant.In this case complete similarity requires not only In this case complete similarity requires not only geometric and Reynolds number similarity but also Mach geometric and Reynolds number similarity but also Mach number similarity so thatnumber similarity so that

l

lm

mmm

m

cc

cV

cV

υυ

=⇒=

113

Flow with a Free Surface Flow with a Free Surface 1/81/8

This type of flow includes flow in canals, rivers, spillways, This type of flow includes flow in canals, rivers, spillways, and stilling basics, as well as flow around ship.and stilling basics, as well as flow around ship.For this class of problems, gravitational, inertial forces, For this class of problems, gravitational, inertial forces, and surface tension are important and, therefore, the and surface tension are important and, therefore, the Froude number and Weber number become important Froude number and Weber number become important similarity parameters.similarity parameters.Since there is a free surface with a liquidSince there is a free surface with a liquid--air interface, air interface, forces due to surface tension may be significant, and the forces due to surface tension may be significant, and the Weber number becomes another similarity parameter that Weber number becomes another similarity parameter that needs to be considered along with the Reynolds number.needs to be considered along with the Reynolds number.

114


A general formulation for these problems isA general formulation for these problems is

To meet the requirement of Froude number similarityTo meet the requirement of Froude number similarity

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

σρ

µρε

φl

l

l

ll

l 2i V,

gV,V,,Dependent pi term=Dependent pi term=

Where Where ll is some characteristic length of the system and is some characteristic length of the system and lliirepresents other pertinent lengths, represents other pertinent lengths, εε/ / ll is the relative is the relative roughness of the surface, and roughness of the surface, and ρρVVll//µµ is the Reynolds number.is the Reynolds number.

ll gV

gV

mm

m =ggm =

ll

lλ== mm

VV

115


To meet the requirement of To meet the requirement of Reynolds number and Froude Reynolds number and Froude numbernumber similaritysimilarity

m

m

mm

mm

m

mmmV

VVVl

l

l

lll

υυ

=ρρ

µµ

=⇒µ

ρ=

µρ

The working fluid for the prototype is normally either freshwateThe working fluid for the prototype is normally either freshwater r or seawater and the length scale is small. or seawater and the length scale is small. It is virtually impossible to satisfy It is virtually impossible to satisfy , so models , so models involving freeinvolving free--surface flows are usually distorted.surface flows are usually distorted.

( ) 2/3m / lλ=υυ

2/3m )( lλ=υυ

116


The problem is further complicated if an attempt is made The problem is further complicated if an attempt is made to model surface tension effects, since this requires the to model surface tension effects, since this requires the equality of Weber numbers, which leads to the conditionequality of Weber numbers, which leads to the condition

( )22m

2mmm

2

m

m2

mm

VV

//VV

ll

lllλ==

ρσρσ

⇒σ

ρ=

σρ

For the For the kinematickinematic surface tension surface tension σσ//ρρ. It is evident that the s. It is evident that the same ame fluid cannot be used in model and prototype if we are to have fluid cannot be used in model and prototype if we are to have similitude with respect to surface tension effects for similitude with respect to surface tension effects for ..1≠λl

117


Fortunately, in many problems involving freeFortunately, in many problems involving free--surface surface flows, both surface tension and viscous effect are small flows, both surface tension and viscous effect are small and consequently strict adherence to Weber and Reynolds and consequently strict adherence to Weber and Reynolds number similarity is not required.number similarity is not required.Certainly, surface tension is not important in large Certainly, surface tension is not important in large hydraulic structures and rivers.hydraulic structures and rivers.Our only concern would be if in a model the depths were Our only concern would be if in a model the depths were reduced to the point where surface tension becomes an reduced to the point where surface tension becomes an important factor.important factor.

..

How to overcome this problem ?How to overcome this problem ?

118


Different horizontal and vertical length scales, which Different horizontal and vertical length scales, which introduce geometric distortion, are often used to eliminate introduce geometric distortion, are often used to eliminate surface tension effects in the model.surface tension effects in the model.Although this approach eliminates surface tension effects Although this approach eliminates surface tension effects in the model, it introduces geometric distortion that must in the model, it introduces geometric distortion that must be accounted for empirically, usually by increasing the be accounted for empirically, usually by increasing the model surface roughness. model surface roughness.

Verification test in the model must be made. Model Verification test in the model must be made. Model roughness can be adjusted to give satisfactory roughness can be adjusted to give satisfactory agreement between model and prototype.agreement between model and prototype.

..

119


For large hydraulic structures, such as dam spillways, the For large hydraulic structures, such as dam spillways, the Reynolds numbers are large so that viscous forces are Reynolds numbers are large so that viscous forces are small in comparison to the force due to gravity and inertia.small in comparison to the force due to gravity and inertia.In this case Reynolds number similarity is not maintained In this case Reynolds number similarity is not maintained and models are designed on the basis of Froude number and models are designed on the basis of Froude number similarity. similarity.

120


A scale hydraulic model (1:197) of the Guri Dam in Venezuela which is used to simulate the characteristics of the flow over and below the spillway and the erosion below the spillway.

The model Reynolds The model Reynolds numbers are large so that numbers are large so that they are not required to they are not required to equal to those of the equal to those of the prototype.prototype.

121

Example 7.8 Froude Number SimilarityExample 7.8 Froude Number Similarity

A certain spillway for a dam is 20 m wide and is designed to carA certain spillway for a dam is 20 m wide and is designed to carry ry 125 m125 m33/s at flood stage. A 1:15 model is constructed to study the /s at flood stage. A 1:15 model is constructed to study the flow characteristics through the spillway. Determine the requireflow characteristics through the spillway. Determine the required d model width and model width and flowrateflowrate. What operating time for the model . What operating time for the model corresponds to 1 24corresponds to 1 24--hr period in the prototype? The effects of hr period in the prototype? The effects of surface tension and viscosity are to be neglected.surface tension and viscosity are to be neglected.

122


The width, wm, of the model of spillway is obtained from the length scale

With the neglect of surface tension and viscosity, the dynamic similarity will be achieved if the Froude numbers are equal between model and prototype

m33.115

m20w151

ww

mm ===λ= l

ll gV

gV

mm

m = ggm =l

l

lλ== mm

VV

123


The flowrate

s/m143.0...Q)(Q

VAAV

QQ

32/5m

2mmmmm

==λ=

⎟⎠⎞

⎜⎝⎛==

l

l

l

l

l

ll

l

l

lλ=== m

m

mm

mm VV

tt

tt

VV

124

Similitude Based on Governing Differential Similitude Based on Governing Differential Equations Equations 1/51/5

For a steady incompressible twoFor a steady incompressible two--dimensional flow of a dimensional flow of a Newtonian fluid with constant viscosity.Newtonian fluid with constant viscosity.The mass conservation equation isThe mass conservation equation is

The The NavierNavier--Stokes equations areStokes equations are

Has dimensions of 1/time.Has dimensions of 1/time.0yv

xu

=∂∂

+∂∂

⎟⎟⎠

⎞⎜⎜⎝

⎛

∂

∂+

∂

∂µ+ρ−

∂∂

−=⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+∂∂

ρ

⎟⎟⎠

⎞⎜⎜⎝

⎛

∂

∂+

∂

∂µ+

∂∂

−=⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+∂∂

ρ

2

2

2

2

2

2

2

2

yv

xvg

yp

yvv

xvu

yu

xu

xp

yuv

xuu Has dimensions of Has dimensions of

force/volumeforce/volume

125


How to nonHow to non--dimensionalizedimensionalize these equations ?these equations ?

τ======

ttppp

Vvv

Vuuyyxx *

0

*****ll

2*

*2

22

2

*

*

xuV

xu

xxu

xuV

xu

∂

∂=⎟

⎠⎞

⎜⎝⎛∂∂

∂∂

=∂

∂

∂

∂=

∂∂

ll

0yv

xu

=∂

∂+

∂

∂∗

∗

∗

∗The mass conservation equationThe mass conservation equation

126


⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

∂

∂+

∂

∂⎥⎦

⎤⎢⎣

⎡ρµ

+⎥⎦

⎤⎢⎣

⎡−∂

∂

⎥⎥⎦

⎤

⎢⎢⎣

⎡

ρ−=

∂

∂+

∂

∂+

∂

∂⎥⎦⎤

⎢⎣⎡τ

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

∂

∂+

∂

∂⎥⎦

⎤⎢⎣

⎡ρµ

+∂

∂

⎥⎥⎦

⎤

⎢⎢⎣

⎡

ρ−=

∂

∂+

∂

∂+

∂

∂⎥⎦⎤

⎢⎣⎡τ

∗

∗

∗

∗

∗

∗

∗

∗∗

∗

∗∗

∗

∗

∗

∗

∗

∗

∗

∗

∗

∗∗

∗

∗∗

∗

∗

22

22

y

v

x

vVV

gyp

Vp

yvv

xvu

tv

V

y

u

x

uVx

pVp

yuv

xuu

tu

V

22

220

22

20

l

ll

l

l

StrouhalStrouhal numbernumber Euler numberEuler number

The The NavierNavier--Stokes equationsStokes equations Reynolds numberReynolds number

Reciprocal of the square Reciprocal of the square of the of the FroudeFroude numbernumber

127


From these equations it follows that if two systems are From these equations it follows that if two systems are governed by these equations, then the solutions (in terms governed by these equations, then the solutions (in terms of u*,v*,p*,x*,y*, and t*) will be the same if the of u*,v*,p*,x*,y*, and t*) will be the same if the four four parametersparameters are equal for the two systems.are equal for the two systems.The two systems will be dynamically similar. Of course, The two systems will be dynamically similar. Of course, boundary and initial conditionsboundary and initial conditions expressed in expressed in dimensionless form must also be equal for the two systems, dimensionless form must also be equal for the two systems, and this will require complete geometric similarity.and this will require complete geometric similarity.

128


These are the same similarity requirements that would be These are the same similarity requirements that would be determined by a dimensional analysis if the same variables determined by a dimensional analysis if the same variables were considered. These variables appear naturally in the were considered. These variables appear naturally in the equations.equations.All the common dimensionless groups that we previously All the common dimensionless groups that we previously developed by using dimensional analysis appear in the developed by using dimensional analysis appear in the governing equations that describe fluid motion when these governing equations that describe fluid motion when these equations are expressed in term of dimensionless variables.equations are expressed in term of dimensionless variables.The use of governing equations to obtain similarity laws The use of governing equations to obtain similarity laws provides an alternative to conventional dimensional provides an alternative to conventional dimensional analysis.analysis.

taiwan shieh fluid07

Documents