FUNDAMENTALS OF FLUID MECHANICSFLUID MECHANICS Chapter …cau.ac.kr/~jjang14/FME/Chap7.pdf · FLUID MECHANICSFLUID MECHANICS Chapter 7 Dimensional AnalysisChapter 7 Dimensional Analysis

FUNDAMENTALS OFFUNDAMENTALS OFFLUID MECHANICSFLUID MECHANICSFLUID MECHANICSFLUID MECHANICS

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

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MAIN TOPICSMAIN TOPICSMAIN TOPICSMAIN TOPICS

Di i l A l iDi i l A l iDimensional 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 DatappModeling and SimilitudeModeling and SimilitudeTypical Model StudiesTypical Model StudiesTypical Model StudiesTypical Model StudiesSimilitude Based on Governing Differential EquationSimilitude Based on Governing Differential Equation

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Dimensional AnalysisDimensional Analysis 1/41/4Dimensional Analysis Dimensional Analysis 1/41/4

A A typical fluid mechanics problemtypical fluid mechanics problem in which in which experimentation is required consider the experimentation is required consider the steady flow of an steady flow of an incompressible Newtonian fluid through a long, smoothincompressible Newtonian fluid through a long, smooth--walled, horizontal, circular pipewalled, horizontal, circular pipe. .

An important characteristic of this systemAn important characteristic of this system, which would , 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 lengthpressure drop per unit length that develops along the pipe that develops along the pipe as a result of friction.as a result of friction.

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The The first step in the planning of an experimentfirst step in the planning of an experiment to study to study this problem would be to this problem would be to decide the factors, or variablesdecide the factors, or variables, , that will have an effect on the that will have an effect on the pressure droppressure drop..

Pressure drop per unit lengthPressure drop per unit lengthp p gp p g

),,,( VDfp

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 (ρ); fluid viscositysphere size (D); speed (V); fluid density (ρ); fluid viscositysphere size (D); speed (V); fluid density (ρ); fluid viscosity sphere size (D); speed (V); fluid density (ρ); fluid viscosity (m)(m)

4


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

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Series of TestsSeries of TestsSeries of TestsSeries of Tests

6


Fortunately, there is a Fortunately, there is a much simpler approachmuch simpler approach to the to the problem that will eliminate the difficulties described problem that will eliminate the difficulties described above.above.

Collecting these variables into two nonCollecting these variables into two non--dimensional dimensional ggcombinations of the variablescombinations of the variables (called dimensionless (called dimensionless product or product or dimensionless groupsdimensionless groups))

Only one dependent and one Only one dependent and one independent variableindependent variable

VDpD Easy to set up experiments to Easy to set up experiments to

determine dependencydetermine dependency Easy to present results (one graph)Easy to present results (one graph)

V2

7

Easy to present results (one graph)Easy to present results (one graph)

Plot of Pressure Drop Data UsingPlot of Pressure Drop Data UsingPlot of Pressure Drop Data Using …Plot of Pressure Drop Data Using …000

3TLF)L/F(LpD

21242 TLF

)FT)(TFL(V

000124 )L)(LT)(TFL(VD

0002 TLF

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

dimensionless product or dimensionless product or dimensionless groupsdimensionless groups

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Buckingham Pi TheoremBuckingham Pi Theorem 1/51/5Buckingham Pi Theorem Buckingham Pi Theorem 1/51/5

A fundamental question we must answer is A fundamental question we must answer is how many how many dimensionless products are requireddimensionless products are required to replace the original list of to replace the original list of variables ?variables ?variables ?variables ?

The answer to this question is supplied by the The answer to this question is supplied by the basic theorem of basic theorem of dimensional analysisdimensional analysis that statesthat statesdimensional analysisdimensional analysis that statesthat states

If an equation involving k variables is dimensionally homogeneous it can be reduced to a relationship amonghomogeneous, it can be reduced to a relationship among k-r independent dimensionless products, where r is the minimum number of reference dimensions required tominimum number of reference dimensions required to describe the variables.

Pi termsPi terms9

Buckingham Pi TheoremBuckingham Pi TheoremPi termsPi terms


Given a physical problem in which the Given a physical problem in which the dependent variabledependent variableis a function of kis a function of k--1 independent variables1 independent variables..

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

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

0)u,.....,u,u,u(g k321 ), ,,,(g k321wwhere g is an unspecified function, different from f.here g is an unspecified function, different from f.

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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

ThTh k i blk i bl b d i tb d i t kk i d d ti d d t

0)u,.....,u,u,u(g k321 The The k variablesk variables may be grouped into may be grouped into kk--r independent r independent

dimensionless productsdimensionless products, or Π terms, expressible in , or Π terms, expressible in f ti l f bf ti l f bfunctional form byfunctional form by

),,,,,( rk321 ???? Π??Π??0),,,,,,(or rk321

r ?? r ?? Π??Π??

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ThTh b fb f i ll b l l hi ll b l l hThe The number of number of rr is usually, but not always, equal to the 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 parametersspecify the dimensions of all the parameters UsuallyUsually thethespecify the dimensions of all the parametersspecify the dimensions of all the parameters. . UsuallyUsually the the reference dimensionsreference dimensions required to describe the variables required to describe the variables will be the basic dimensionswill be the basic dimensions M L and T or F L and TM L and T or F L and Twill be the basic dimensions will be the basic dimensions M, L, and T or F, L, and TM, L, and T or F, L, and T..

ThThisis theorem does theorem does not predict not predict the particular the particular functional functional formform of of or or . The . The functional relationfunctional relation among the among the ggindependent dimensionless products Π independent dimensionless products Π must be determined must be determined experimentallyexperimentally..

The kThe k--r dimensionless r dimensionless Π Π terms obtained from the terms obtained from the procedure are independent.procedure are independent.

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A Π term is not independent if it can be obtained from a A Π 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

4/3

23

4/31

632

15 or2

then neither Πthen neither Π55 nor Πnor Π66 is independent of the otheris independent of the other

332

then neither Πthen neither Π55 nor Πnor Π66 is independent of the other is independent of the other dimensionless products or dimensionless products or dimensionless groupsdimensionless groups..

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

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 pp g y gg y g((dimensionaldimensional)) variables that are known (or believed) to variables that are known (or believed) to affect the given flow phenomenonaffect the given flow phenomenon..

Eight steps listedEight steps listed below outline a recommended below outline a recommended procedure for determining the Π terms.procedure for determining the Π terms.procedure for determining the Π terms.procedure for determining the Π terms.

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

List all the dimensional variables involved.List all the dimensional variables involved.Keep the number of variables to a minimumKeep the number of variables to a minimum, so that we , so that we

can minimize the amount of laboratory workcan minimize the amount of laboratory workcan minimize the amount of laboratory work.can minimize the amount of laboratory work.All variables must be independentAll variables must be independent. For example, if the . For example, if the

crosscross sectional area of a pipe is an important variablesectional area of a pipe is an important variablecrosscross--sectional area of a pipe is an important variable, sectional area of a pipe is an important variable, either the area or the either the area or the pipe diameter could be used, but pipe diameter could be used, but not bothnot both sincesince they are obviously not independentthey are obviously not independentnot both, not both, since since they are obviously not independentthey are obviously not independent..

Ex. Ex. γγ==ρρ××g, that is, g, that is, γγ,,ρρ, and g are not independent., and g are not independent.15


Step 1 List all the variables. Step 1 List all the variables. 22Let k be Let k be the number of variablesthe number of variables..Example: For pressure drop per unit length, k=5. (All Example: For pressure drop per unit length, k=5. (All

variables arevariables are pp DD andand V )V )variables are variables are pp,, D,D,,,, and , and V )V )

)V,,,D(fp )V,,,D(fp

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Step 2 Step 2 Express each of the variables in terms of Express each of the variables in terms of basic basic dimensionsdimensions and and find find the number of reference the number of reference dimensionsdimensions..Select a set of fundamental (primary) dimensionsSelect a set of fundamental (primary) dimensionsSelect 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

r=3r=3

17

LTVTFL


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.ppLet r be the number of reference dimensions (primary Let r be the number of reference dimensions (primary

dimensions) required to describe these variablesdimensions) required to describe these variablesdimensions) required to describe these variables.dimensions) required to describe these variables.The number of pi terms is kThe number of pi terms is k--rr

l d i l h k 3l d i l h k 3Example: 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 termsthe number of pi terms is is kk--r=5r=5--3=23=2..

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l b f i i bll b f i i blStep 4 Step 4 Select a number of repeating variablesSelect 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 Select a set of r dimensional variablesr dimensional variables that that includes all includes all

the primary dimensions the primary dimensions (repeating (repeating variables).variables).These repeating variables will all be combined with These repeating variables will all be combined with

each of the remaining parameters. each of the remaining parameters. 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.

12

243 TFLLDFLp

19

12 LTVTFL


i bi b l i l i f hl i l i f hStep 5 Form a pi term by Step 5 Form a pi term by multiplying one of the multiplying one of the nonrepeating variablesnonrepeating variables by by the product of the the product of the repeating variablesrepeating variables, each raised to an exponent that , each raised to an exponent that will make the combination dimensionless. will make the combination dimensionless. 11Set up dimensional equations, Set up dimensional equations, combining the variables combining the variables

selected in Step 4selected in Step 4 with with each of the other variableseach of the other variables(nonrepeating variables) in turn, to form dimensionless (nonrepeating 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

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Step 5 (Continued) Step 5 (Continued) 22

cbaVD cba1 VDp

000c24b1a3

0c1:FTLF)TFL()LT()L)(FL( 000c24b1a3

0c2b:T0c4ba3:L

Dp

1c,2b,1a0c2b:T

21V

Dp

21


Step 6 Repeat Step 5 for each of the remaining Step 6 Repeat Step 5 for each of the remaining nonnon--repeating variables.repeating variables.p gp g

cba2 VD

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

0c4ba2:L0c1:F

1c1b1a0c2b1:T

DV2

22

1c,1b,1a


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

000000Dp 00000021 TLMTLF

VDp

0000002 TLMTLF

DV

23

DV


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.g p ,g p ,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 .

),,,,,( rk321

p p p p gp p p p g

Dp Dimensional analysis will not provide Dimensional analysis will not provide the form of the function The functionthe form of the function The function

DVV2 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 experimentsof experiments..

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pp


The pi terms can be rearranged. For example, ΠThe pi terms can be rearranged. For example, Π22, could be , could be expressed asexpressed as

VD

2

VDDp

Vp

2

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Example 7.1 Method of Repeating Example 7.1 Method of Repeating VariablesVariables

A hi l l h i id h d h i h h iA hi l l h i id h d h i h h iA 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 theAssume that the dragdrag D that the fluid exerts on the plateD that the fluid exerts on the plateAssume that the Assume that the dragdrag, D, that the fluid exerts on the plate , D, that the fluid exerts on the plate is a is a function offunction of w and h, the fluid viscosity, w and h, the fluid viscosity, µµ ,and ,and ρρ, , respectively and the velocity V of the fluid approachingrespectively and the velocity V of the fluid approachingrespectively, 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.p p yp p y

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Example 7 1Example 7 1 SolutionSolution1/51/5Example 7.1 Example 7.1 SolutionSolution1/51/5

Drag force on a PLATEDrag force on a PLATE

)V,,,h,w(fD Step 1:List all the dimensional variables involved. Step 1:List all the dimensional variables involved.

D hD h VV k 6k 6 di i l tdi i l t

)V,,,h,w(fD

D,w,h, D,w,h, ρ,μ,Vρ,μ,V k=6k=6 dimensional parameters.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

2 LhLMLTD

1311

2

LTVMLTML

LhLwMLTD

27

V


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

Step 4:Select repeating variables w,V,Step 4:Select repeating variables w,V,..Step 5~6:combining the repeating variables with eachStep 5~6:combining the repeating variables with eachStep 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 productsgroups or dimensionless productsgroups or dimensionless products.groups or dimensionless products.

2 LhLMLTD

1311

2

LTVMLTML

LhLwMLTD

28

V


000c3b1a2cba

03b10c1:M

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

D

12b20b2:T

0c3ba1:L

221

VwD

1c,2b,2a Vw

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

0c3ba1:L0c:M

)()()(2

h

0c0b1a0b:T

0c3ba1:L

wh

2

29

0c,0b,1a


0003b111b

0c1:MTLM)ML()LT()L)(TML(Vw 000c3b1a11cba

3

0b1:T0c3ba1:L

31c,1b,1a

0b1:T

wV3

30


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.ppThe functional relationship isThe functional relationship is

or),,( 321

wV,

wh

VwD

2231

wVwVw

Selection of VariablesSelection of Variables 1/41/4Selection of Variables Selection of Variables 1/41/4

O f h i d diffi l i l iO f h i d diffi l i l iOne 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 dimensional analysis to any given problem is the selection selection of the variablesof the variables that are involvedthat are involvedof the variablesof the variables that are involved.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 Generallyeasily identified Generally one must rely on a goodone must rely on a goodeasily identified. Generally, easily identified. Generally, one must rely on a good one must rely on a good understanding of the phenomenon involved and the understanding of the phenomenon involved and the governing physical lawsgoverning physical lawsgoverning physical lawsgoverning 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 difficultterms appear in the final solution and it may be difficultterms 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.

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p yp y


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 g g p p y gg g p p y gassumptions 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..possible, perhaps even if some accuracy is sacrificedpossible, perhaps even if some accuracy is sacrificed..

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A A suitable suitable balancebalance between between simplicity and accuracysimplicity and accuracy is an is an desirable goaldesirable goal.~~~~~.~~~~~ EnricoEnrico FermiFermi

Variables can be classified into three general group:Variables can be classified into three general group:GeometryGeometry: lengths and angles: lengths and anglesGeometryGeometry: lengths and angles.: lengths and angles.Material PropertiesMaterial Properties: relate the external effects and the : relate the external effects and the

responsesresponsesresponses.responses.External EffectsExternal Effects (Input & Output)(Input & Output): produce, or tend to : produce, or tend to

d h i h S h fd h i h S h fproduce, a change in the system. Such as force, produce, a change in the system. Such as force, pressure, gravity,pressure, gravity,(<(<-- Input)Input) or velocityor velocity (<(<-- Output)Output)..

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Selection of VariablesSelection of Variables 44--1/41/4Selection 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. What’s the main variable Clearly define the problem. What’s the main variable y py p

of interest?of interest?Consider the basic laws that govern the phenomenon.Consider the basic laws that govern the phenomenon.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 classesvariables into three broad classesvariables into three broad classes.variables into three broad classes.

ContinuedContinued

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

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).constant (e.g., g).constant (e.g., g).

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

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Determination of Reference DimensionDetermination of Reference Dimension 1/31/3Determination 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, In fluid mechanics, the required number of reference the required number of reference qqdimensions is three, but dimensions is three, but in some problems only one or twoin some problems only one or twoare requiredare 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 isneeded to describe all variables isreference 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.

37

a p e 7. .a p e 7. .

Example 7 2 Determination of Pi TermsExample 7 2 Determination of Pi TermsExample 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 . The . The vertical deflectionvertical deflection, , , of the center of the bottom is a , of the center of the bottom is a functionfunction of D, h, d, of D, h, d, , 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.

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The vertical deflectionThe vertical deflection

EdfDf E,,d,f,Df

L

F F L T Pi t 6F F L T Pi t 6 2 42 4 LhLD

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 Ld

Lh

212

223

TMLEFLE

TML,FL

39

TMLE,FLE


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

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

2211 ba2

ba1 hDD

4433 ba4

ba3

21

EDdD

DE,

Dd,

Dh

DDE,

Dd,

Dh,

D 4321

40

DDDDDDDD


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 listedA closer look at the dimensions of the variables listedA closer look at the dimensions of the variables listed A closer look at the dimensions of the variables listed reveal that reveal that ““only two reference dimensions,only two reference dimensions,”” L and L and MTMT--22 are requiredare required..MTMT are requiredare required..

41

Determination of Reference DimensionDetermination of Reference Dimension 2/32/3Determination of Reference Dimension Determination of Reference Dimension 2/32/3

DfhEXAMPLEEXAMPLE ,,DfhEXAMPLEEXAMPLE

Dh Dh

MLT SYSTEMMLT SYSTEM FLT SYSTEMFLT SYSTEM

222 TM

TLMLL

Dh

LF

LFLL 3

222 TTL LL

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

42

Determination of Reference Dimension Determination of Reference Dimension 3/33/3

(option)(option)S Di i l M iS Di i l M iSet Dimensional MatrixSet Dimensional Matrix

MLT SYSTEMMLT SYSTEM FLT SYSTEMFLT SYSTEM

1100MDh

1100FDh

2200T0211L1100M

0000T1311L

1100F

2200T 0000T

h1

D1

222h

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

43

222

DDD

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

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

)V,,,D(fp

VDDp

)V,,,D(fp

Selecting D,V, and as i i bl

Vp

12repeating variables:

VD

VDp

2

2

Selecting D,V, and as repeating variables:

44

V


Both are correct, and both would Both are correct, and both would

VDV

Dp12

,,

lead to the same final equation for lead to the same final equation for the pressure drop. the pressure drop. There is There is not a not a

V2

2p pp p

unique set of pi termsunique set of pi terms which which arises from a dimensional analysisarises from a dimensional analysis. .

VD

VDp

2

2

yyThe functions The functions ΦΦ11 and and ΦΦ22 are will are will be different because the dependent be different because the dependent

pppi terms are different for the two pi terms are different for the two relationships.relationships.`̀

45

pp


EXAMPLEEXAMPLE 321 ,EXAMPLEEXAMPLE

ba' Form a new pi termForm a new pi term

''

b3

a22 Form a new pi termForm a new pi term

2223211 ,,

All are correct

46


VDV

Dp12

Selecting D,V, and as repeating variables:

DpVDDp 2

V2 repeating variables:

VDV

Dp12

V

DpVDV

Dp2

V

VD

VDp

2

2

VDDp 2

V

V

p2

47

Common Dimensionless GroupsCommon Dimensionless Groups 1/21/2Common Dimensionless GroupsCommon Dimensionless Groups 1/21/2

A list of variables that commonly arise in fluid ymechanical problems.

Possible to provide a physical interpretation to the dimensionless groups

hi h b h l f l iwhich can be helpful in assessing their influence in a particular applicationin a particular application.

48

Reynolds NumberReynolds Number 1/21/2Reynolds Number Reynolds Number 1/21/2

VVR

In honor of Osborne Re nolds (1842 1912) the British engineerIn honor of Osborne Re nolds (1842 1912) the British engineer

VVRe

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 be who first demonstrated that this combination of variables could be used as a criterion to distinguish between laminar and turbulent flow.used as a criterion to distinguish between laminar and turbulent flow.gg

The Reynolds number is a measure of the ration of the The Reynolds number is a measure of the ration of the inertia forces inertia forces to viscous forcesto viscous forces..

If the If the Reynolds number is smallReynolds number is small (Re<<1), this(Re<<1), this is an indication that is an indication that the the viscous forces arviscous forces aree dominant dominant in the problem, and it may be in the problem, and it may be possible to neglect the inertial effects; that is, the density of the fluidpossible to neglect the inertial effects; that is, the density of the fluidpossible to neglect the inertial effects; that is, the density of the fluid possible to neglect the inertial effects; that is, the density of the fluid will no be an important variable.will no be an important variable.

49

Reynolds NumberReynolds Number 2/22/2Reynolds Number Reynolds Number 2/22/2

Fl i h ll R ld bFl i h ll R ld b l f dl f d Flows with very small Reynolds numbersFlows with very small Reynolds numbers are commonly referred to are commonly referred to as “creeping flows”.as “creeping flows”.

For large Reynolds number flowFor large Reynolds number flow thethe viscous effects are smallviscous effects are small For large Reynolds number flowFor large Reynolds number flow, the , the viscous effects are small viscous effects are small relative to inertial effectsrelative to inertial effects and for these cases and for these cases it may be possible to it may be possible to neglect the effect of viscosityneglect the effect of viscosity and consider the problem as one and consider the problem as one i l i “ i ” fl idi l i “ i ” fl idinvolving a “nonviscous” fluid.involving a “nonviscous” fluid.

Flows with “large” Reynolds number generally are Flows with “large” Reynolds number generally are turbulentturbulent. Flows . Flows in which the inertia forces are “small” compared with the viscousin which the inertia forces are “small” compared with the viscousin which the inertia forces are small compared with the viscous in which the inertia forces are small compared with the viscous forces are characteristically laminar flowsforces are characteristically laminar flows..

50

Correlation of Experimental DataCorrelation of Experimental DataCorrelation 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 groupsbetween the groupsbetween the groupsbetween the groups..

To determine this relationship, suitable experimental data must be To determine this relationship, suitable experimental data must be obtainedobtainedobtained.obtained.

The degree of difficulty depends on the number of pi terms.The degree of difficulty depends on the number of pi terms.

51

Problems with One Pi TermProblems with One Pi TermProblems with One Pi TermProblems with One Pi Term

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

C C1

where C is a constant. The value of the constant must still be d i d b idetermined by experiment.

52

Example 7 3 Flow with Only One Pi TermExample 7 3 Flow with Only One Pi TermExample 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 falls very Assume that the drag, D, acting on a spherical particle that falls very slowly through a viscous fluid is a function of the particle diameter, slowly through a viscous fluid is a function of the particle diameter, d the particle velocity V and the fluid viscosity μ Determined the particle velocity V and the fluid viscosity μ Determined, the particle velocity, V, and the fluid viscosity, μ. Determine, d, the particle velocity, V, and the fluid viscosity, μ. Determine, with the aid the dimensional analysis, how the drag depends on the with the aid the dimensional analysis, how the drag depends on the particle velocity. particle velocity. p yp y

53

Example 7 3Example 7 3 SolutionSolutionExample 7.3 Example 7.3 SolutionSolution

The drag The drag

)Vd(fD 13

2

LTVMLTFLLdFD

),V,d(fD 13 LTVML

D dVCDCdV

D1

VD 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

54

Problems with Two or More Pi TermProblems with Two or More Pi Term 1/21/2Problems 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 An empirical relationship isAn empirical relationship is)( 21

the functional relationship the functional relationship among the variables can theamong the variables can the

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

among the variables can the among the variables can the be determined by varying Πbe determined by varying Π22and measuring the and measuring the corresponding value of Πcorresponding value of Π11..

The empirical equation The empirical equation Dangerous toDangerous top qp qrelating Πrelating Π22 and Πand Π11 by using by using a standard curvea standard curve--fitting fitting

Dangerous to Dangerous to extrapolate beyond extrapolate beyond valid rangevalid range

55technique.technique.

Problems with Two or More Pi TermProblems with Two or More Pi Term 2/22/2Problems 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 , To determine a suitable empirical equation To determine a suitable empirical equation 321 , p qp qrelating the three pi terms.relating the three pi terms.

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

Families curve of curvesFamilies curve of curves56

Families curve of curvesFamilies curve of curves

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 fashion…….will behave in a similar fashion…….

57

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

M d l ? AM d l ? A d ld l ii t ti f h i l tt ti f h i l t th tth tModel ? A Model ? A modelmodel is a is a representation of a physical systemrepresentation of a physical system that may that may be used to be used to predict the behavior of the systempredict the behavior of the system in some desired respect. in some desired respect. Mathematical or computer models may also conform to this Mathematical or computer models may also conform to this d fi iti i t t ill b i h i l d ld fi iti i t t ill b i h i l d ldefinition, 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 be Prototype? The physical system for which the prediction are to be made. made.

Models that resemble the prototype but are generally of a different Models that resemble the prototype but are generally of a different sizesize, may involve different fluid, and often operate under different , may involve different fluid, and often operate under different conditionsconditionsconditions. conditions.

Usually a model is smaller than the prototypeUsually a model is smaller than the prototype. . Occasionally, if the prototype is very small, it may be advantageous Occasionally, if the prototype is very small, it may be advantageous y p yp y y gy p yp y y g

to have a model that is larger than the prototype so that it can to have a model that is larger than the prototype so that it can bbe e more easily studied. For example, large models have been used to more easily studied. For example, large models have been used to study the motion of red blood cells.study the motion of red blood cells.

58

yy

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

With the successful development of a valid modelWith the successful development of a valid model, it is possible to , it is possible to predict the behavior of the prototypepredict the behavior of the prototype under a certain set of under a certain set of conditionsconditionsconditions.conditions.

There is an inherent There is an inherent danger in the use of modelsdanger in the use of models in that predictions in that predictions can be made that are in error and the error not detected until thecan be made that are in error and the error not detected until thecan be made that are in error and the error not detected until the can be made that are in error and the error not detected until the prototype is found not to perform as predicted.prototype is found not to perform as predicted.

It is imperative thatIt is imperative that the model be properly designed and tested andthe model be properly designed and tested and It is imperative that It is imperative that the model be properly designed and tested and the model be properly designed and tested and that the results be interpreted correctlythat the results be interpreted correctly..

59

Similarity of Model and PrototypeSimilarity of Model and PrototypeSimilarity of Model and PrototypeSimilarity of Model and Prototype

Wh di i bWh di i b h i il i f d l dh i il i f d l dWhat conditions must be met to What conditions must be met to ensure the similarity of model and ensure the similarity of model and prototypeprototype??

Geometric SimilarityGeometric Similarity Geometric SimilarityGeometric SimilarityModel and prototype have Model and prototype have same shapesame shape..Linear dimensions on model and prototype correspond withinLinear dimensions on model and prototype correspond withinLinear dimensions on model and prototype correspond within Linear dimensions on model and prototype correspond within

constant scale factorconstant scale factor.. Kinematic SimilarityKinematic Similarity

VelocitiesVelocities at corresponding points on model and prototype at corresponding points on model and prototype differ differ only by a constant scale factoronly by a constant scale factor..

D i Si il itD i Si il it Dynamic SimilarityDynamic SimilarityForcesForces on model and prototype on model and prototype differ only by a constant scale differ only by a constant scale

factorfactor60

factorfactor..

Theory of ModelsTheory of Models 1/51/5Theory of Models Theory of Models 1/51/5

Th h f d l b dil d l d b iTh h f d l b dil d l d b iThe 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.

F i bl hi h b d ib d i fF i bl hi h b d ib d i fFor 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

Thi l i hi b f l dThi l i hi b f l d),,,,( n321

This 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 phenomenonnature of the physical phenomenonnature 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.

61

g y


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,

hh th f f th f ti ill b thth f f th f ti ill b th ll),,,,,( nmm3m2m1

where where the form of the function will be the samethe form of the function will be the same as long 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.

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

62

the same phenomenon.the same phenomenon.


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

nmnm33m22 .....

The The form of Φform of Φ is the sameis the same for model and prototype, it for model and prototype, it f ll hf ll h

nmnm33m22

follows thatfollows that

This is the This is the desired prediction equationdesired prediction equation and and indicates that the measured of Πindicates that the measured of Π1m1m obtained obtained

m11 with the model will be equal to the with the model will be equal to the corresponding Πcorresponding Π11 for the prototype as long as for the prototype as long as the other Π parameters are equalthe other Π parameters are equal

63

the other Π parameters are equal. the other Π parameters are equal.

Theory of ModelsTheory of Models 4/54/5 SummarySummary11Theory 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 phenomenonphenomenon. .

)( 321 For prototypeFor prototype ),,,,( n321 For prototypeFor prototype

For modelFor model )( 321 For modelFor model ),,,,,( nmm3m2m1

64

Theory of ModelsTheory of Models 5/55/5 SummarySummary22Theory 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 (called design conditionsdesign conditions, also called similarity , also called similarity requirements or modeling laws)requirements or modeling laws)

nmnm33m22 .....

The measured of ΠThe measured of Π1m1m obtained with the model will be obtained with the model will be l t th di Πl t th di Π f th t tf th t t

nmnm33m22

equal to the corresponding Πequal to the corresponding Π11 for the prototype. for the prototype.

11 Called Called prediction equationprediction equationm11 p qp q

65

Theory of ModelsTheory of Models EXAMPLE 1EXAMPLE 1Theory of Models Theory of Models EXAMPLE 1EXAMPLE 1

E l C id i h d f hE l C id i h d f h Example: Considering the drag force on a sphereExample: Considering the drag force on a sphere..

),,V,D(fF

VDfF122

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

),,V,D(fF

VDF

DV

122

m

mmm12

m2

mm

m DVfDV

F

prototype122

VDfDV

F

Design conditions.Design conditions.prototypemodel

VDVD

Then …Then …

prototypemodel

FF

66prototype22

model22

DVDV

Theory of ModelsTheory of Models EXAMPLE 2EXAMPLE 2Theory of Models Theory of Models EXAMPLE 2EXAMPLE 2

E l D i i h d f hi l l (E l D i i h d f hi l l ( 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)

VhfD

VwwD

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

V,,,h,wfD

,hVw 22

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

prototype22

Vw,hw

VwD

m

mmm

m

m2

mm2

m

m wV,hw

VwD

Design conditions.Design conditions.

prototype mmmmm Vw

VwwV,hw

hw mmmm

Then …Then …

hh mm

m

22

22m

22 DVVwDDD

67

mmmm

2mm

2m

22 VwVwVw

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 downwindpast this type of bluff body vortices may develop on the downwindpast this type of bluff body, vortices may develop on the downwind past this type of bluff body, vortices may develop on the downwind side that are shed in a regular fashion at some definite frequency. side that are shed in a regular fashion at some definite frequency. Since these vortices can create harmful periodic forces acting on the Since these vortices can create harmful periodic forces acting on the p gp gstructure, it is important to determine the shedding frequency. For structure, it is important to determine the shedding frequency. For the specific structure of interest, D=0.2m, H=0.4m, and a the specific structure of interest, D=0.2m, H=0.4m, and a representative wind velocity 50km/hr. Standard air can be assumed. representative wind velocity 50km/hr. Standard air can be assumed. 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 thescale model that is to be tested in a water tunnel For thesmallsmall--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℃℃..

68

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 ω for the model test should be performed. If the shedding frequency ω for the model is found to be 49 9Hz what is theis found to be 49 9Hz what is the corresponding frequency for thecorresponding frequency for theis found to be 49.9Hz, what is the is found to be 49.9Hz, what is the corresponding frequency for the corresponding frequency for the prototypeprototype? ?

For air at standard conditionFor air at standard condition 35 m/kg23.1,sm/kg1079.1 For air at standard condition For air at standard condition For water at 20For water at 20℃℃,,

m/kg23.1,sm/kg1079.1 3

water3

water m/kg998,sm/kg101

69


Step 1:List all the dimensional variables involved. Step 1:List all the dimensional variables involved. ωωD,H,V,D,H,V,ρ,μρ,μ 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. p yp yr=3 primary dimensionsr=3 primary dimensions

1

TMLTFLLTV

LHLDT2241

1

TMLTFLLTV

70


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

Step 4:Select repeating variables D,V, Step 4:Select repeating variables D,V, μμ.. StepStep 5~6:combining the repeating variables with each of5~6:combining the repeating variables with each of 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.HD

DHVHD

VDVD 222111 cba

2cba

1

VDVD 333 cba

3

71


VDHD

The functional relationship isThe functional relationship is

VD,DH

VD

Strouhal numberStrouhal number

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

mmmmmm DVHD

m

mmm

m

m

m

mm ,DV

72


The Design conditions.The Design conditions.mHH

mmm DVVD

ThenThenmDD

m

Then….Then….

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

DDV

mm

mm

D Dmm

DV m Hz0.29...D

DVV

mm

m

73

Model ScalesModel ScalesModel 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

1

Length ScaleLength Scale

VVm

V Velocity ScaleVelocity Scale

m

V

Density ScaleDensity Scale

mViscosity ScaleViscosity Scale

74

Viscosity ScaleViscosity Scale

Distorted ModelsDistorted ModelsDistorted ModelsDistorted Models

I d l di hi d i i il iI d l di hi d i i il iIn many model studies, to achieve dynamic similarity In many model studies, to achieve dynamic similarity requires duplication of several dimensionless groupsrequires duplication of several dimensionless groups. .

I l d i i il i bI l d i i il i bIn 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 exampleof the similarity requirements are not met for exampleof the similarity requirements are not met, for example, of the similarity requirements are not met, for example, if , then it follows that the prediction equation if , then it follows that the prediction equation

is not true; that isis not true; that ism22

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

i i fi d ll di i fi d ll d

m11 m11

requirements are not satisfied are called requirements are not satisfied are called DISTORTED MODELSDISTORTED MODELS. .

75

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

D i h d f f hi lD i h d f f hi lDetermine 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 prototypenumbers be duplicated between model and prototypenumbers be duplicated between model and prototype.numbers be duplicated between model and prototype.

pm VFVF F d bF d b2/1

p

pp2/1

m

mm

)g(Fr

)g(Fr

ppVV

Froude numbersFroude numbers

TT h F d bh F d bp

ppp

m

mmm

VReVRe

Reynolds numbersReynolds numbers2/1

V To To match Froude numbersmatch Froude numbersbetween model and prototypebetween model and prototype p

m

p

mVV

76

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

To To match Reynolds numbersmatch Reynolds numbers between model and prototypebetween model and prototype

2/32/1

p

m

p

m

p

mVV

2/32/1

VV

p

m

p

m

p

m

p

m

2/1

p

m

p

mVV

ppp pppp

IfIf // l 1/100( t i l l th l f hil 1/100( t i l l th l f hi

pp

If If mm/ / p p equals 1/100(a typical length scale for ship equals 1/100(a typical length scale for ship model tests) , then υmodel tests) , then υmm/υ/υp p must be 1/1000.must be 1/1000.>>> The >>> The kinematic viscosity ratiokinematic viscosity ratio required to required to duplicate Reynolds numbers duplicate Reynolds numbers cannot be attainedcannot be attained..

77

Distorted ModelsDistorted Models EXAMPLEEXAMPLE--11 3/33/3Distorted Models Distorted Models EXAMPLEEXAMPLE 11 3/33/3

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 requiredscale test is required to obtain to obtain complete dynamic similaritycomplete dynamic similarity..

78

Typical Model StudiesTypical Model StudiesTypical 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.

79

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

This type of flow includes This type of flow includes flow through flow through pipes, pipes, valves, valves, fittings, and metering devicesfittings, and metering devices. .

The conduits are The conduits are often circularoften circular, they could have other , they could have other shapes as well and may contain expansions or contractions.shapes as well and may contain expansions or contractions.p y pp y p

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 thedominant forces are inertial and viscous forces so that thedominant 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 parameterReynolds number is an important similarity parameter..

80


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 p p p p pp p p p pasasDependent pi term=Dependent pi term=

V,,iDependent pi termDependent pi term

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

81


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

mmimmiim

imm

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

VVV 1m

mm

mm

m

mmmV

VVV

1

m

m

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

/VV

VV

mm

m

82


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 y y yy y ybecause of the large model velocities requiredbecause of the large model velocities required..

V VVm

/VV

VV

mm

m

83


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,

pD d t i tD d t i t

The prototype pressure dropThe prototype pressure drop21

Vp

Dependent pi termDependent pi term

The prototype pressure dropThe prototype pressure drop2

V

mmm

pV

p

mpp In general

84

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

This type of flow includes This type of flow includes flow around aircraft, flow around aircraft, automobiles, golf balls, and buildingautomobiles, golf balls, and building..

For these problems, geometric and Reynolds number For these problems, geometric and Reynolds number similarity is required.similarity is required.y qy q

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, soimportant. Also, gravity will not affect the flow pattern, soimportant. 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 omittedFor incompressible flow the Mach number can be omittedFor incompressible flow, the Mach number can be omitted.For incompressible flow, the Mach number can be omitted.

85


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

Vid id i

V,,i

WhereWhere is some characteristic length of the system andis some characteristic length of the system and ii

Dependent pi term=Dependent pi term=

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

Model of the National Bank of Commerce, San Antonio, Texas, for measurement of peak, rms, and mean pressure distributions. The model is l d i l i l i l

86

located in a long-test-section, meteorological wind tunnel.


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

22D

V1DC

To meet the requirement of geometric similarityTo meet the requirement of geometric similarityV

2

mmimmiim

87

imm


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

mmmmmm VVV

m

m

mm

mm

m

mmmV

VVV

22

m

2

m

2

mm22m

22D

VVD

V1D

V1D

mmmmmmV2

V2

VThe same fluid is used, thenThe same fluid is used, then

/VV

VV

mm

m

88


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 y y yy y ybecause of the large model velocitiesbecause of the large model velocities required.required.

89


How toHow to reduce the fluid velocity inreduce the fluid velocity inHow to How to reduce the fluid velocity in reduce the fluid velocity in the modelthe model ?? m

mm

m

mmm

VVVV

A A different fluid is useddifferent fluid is used in the model such thatin the model such that 1/m

For example, the ratio of the kinematic viscosity of water to that For example, the ratio of the kinematic viscosity of water to that p , yp , yof air is approximately 1/10, so that of air is approximately 1/10, so that if the prototype fluid were if the prototype fluid were air, test might be run on the model using waterair, 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

VVm t st ay be d cu t to ac eve t e ecessa yt st ay be d cu t to ac eve t e ecessa y

velocity in a suitable test facility, velocity in a suitable test facility, such as a water such as a water tunnel.tunnel.

90


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.. mm>>

An alternative way to reduce VAn alternative way to reduce Vmm is to is to increase the air pressure in the tunnel so thatincrease the air pressure in the tunnel so thatincrease the air pressure in the tunnel so that increase the air pressure in the tunnel so that mm>>. The pressurized tunnels are obviously . The pressurized tunnels are obviously complicated and expensive.complicated and expensive.complicated and expensive.complicated and expensive.

91

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

For a For a steady incompressible twosteady incompressible two--dimensionaldimensional flow of a flow of a Newtonian fluid with constant viscosity.Newtonian fluid with constant viscosity.

The mass conservation equation isThe mass conservation equation is

Has dimensions of 1/timeHas dimensions of 1/time0vu

The NavierThe Navier--Stokes equations areStokes equations are

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

2

2

2

2

yu

xu

xp

yuv

xuu Has dimensions of Has dimensions of

2

2

2

2

yv

xvg

yp

yvv

xvu

yforce/volumeforce/volume

92

yxyyx


Dimensionli e ith characteristic (standard) q antitiesDimensionli e ith characteristic (standard) q antitiesDimensionlize with characteristic (standard) quantities Dimensionlize with characteristic (standard) quantities such that such that the quantities of dimensionless parameters are the quantities of dimensionless parameters are OO(1)(1) tpulx *****(1).(1).

How to nonHow to non--dimensionalize these equations ?dimensionalize these equations ?ccccc ttt

ppp

Vuullxx *****

cccccc ttt

ppp

Vvv

Vuuyyxx ******

2*

*2

22

2

*

*

xuV

xu

xxu

xuV

xu cc

xxxxxx cc

0vu

The mass conservation equationThe mass conservation equation

93yx


R ld bR ld b

22

The NavierThe Navier--Stokes equationsStokes equations Reynolds numberReynolds number

22

22

2 yu

xu

Vxp

Vp

yuv

xuu

tu

Vt ccc

c

cc

c

22

22

22 yv

xv

VVg

yp

Vp

yvv

xvu

tv

Vt ccc

c

c

c

cc

c

ycc

Strouhal numberStrouhal number Euler numberEuler number Reciprocal of the sq areReciprocal of the sq areReciprocal of the square Reciprocal of the square of the Froude numberof the Froude number

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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 of u*,v*,p*,x*,y*, and t*) will be the same if the the four four parametersparameters are equal for the two systemsare 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 systemsdimensionless form must also be equal for the two systems, , and this will require complete geometric similarity.and this will require complete geometric similarity.

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Th h i il i i h ld bTh h i il i i h ld bThese 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 thewere considered These variables appear naturally in thewere considered. These variables appear naturally in the were considered. These variables appear naturally in the equations.equations.

All the common dimensionless groups that we previouslyAll the common dimensionless groups that we previouslyAll 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 thesegoverning equations that describe fluid motion when thesegoverning 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 lawsThe use of governing equations to obtain similarity lawsThe 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 analysisanalysis..

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FUNDAMENTALS OF FLUID MECHANICSFLUID MECHANICS Chapter …cau.ac.kr/~jjang14/FME/Chap7.pdf · FLUID MECHANICSFLUID MECHANICS Chapter 7 Dimensional AnalysisChapter 7 Dimensional Analysis