FUNDAMENTALSFUNDAMENTALS OFOF FLUIDFLUID MECHANICSMECHANICS ChapterChapter 77 DimensionalDimensional AnalysisAnalysis Modeling,Modeling, andand SimilitudeSimilitude JyhJyh--CherngCherng ShiehShieh Department of Bio-Industrial Mechatronics Engineering National Taiwan University 12/24/2007 1 MAINMAIN TOPICSTOPICS

™™DimensionalDimensional AnalysisAnalysis ™™BuckinghamBuckingham PiPi TheoremTheorem ™™DeterminationDetermination ofof PiPi TermsTerms ™™CommentsComments aboutabout DimensionalDimensional AnalysisAnalysis ™™CommonCommon DimensionlessDimensionless GroupsGroups inin FluidFluid MechanicsMechanics ™™CorrelationCorrelation ofof ExperimentalExperimental DataData ™™ModelingModeling andand SimilitudeSimilitude ™™TypicalTypical ModelModel StudiesStudies ™™SimilitudeSimilitude BasedBased onon GoverningGoverning DifferentialDifferential EquationEquation

2 DimensionalDimensional AnalysisAnalysis 1/41/4

™™AA typicaltypical fluidfluid mechanicsmechanics problemproblem inin whichwhich experimentationexperimentation isis required,required, considerconsider thethe steadysteady flowflow ofof anan incompressibleincompressible NewtonianNewtonian fluidfluid throughthrough aa long,long, smoothsmooth-- walled,walled, horizontal,horizontal, circularcircular pipe.pipe. ™™AnAn importantimportant characteristiccharacteristic ofof thisthis system,system, whichwhich wouldwould bebe interestinterest toto anan engineerengineer designingdesigning aa pipeline,pipeline, isis thethe pressurepressure dropdrop perper unitunit lengthlength thatthat developsdevelops alongalong thethe pipepipe asas aa resultresult ofof friction.friction.

3 DimensionalDimensional AnalysisAnalysis 2/42/4

™™TheThe firstfirst stepstep inin thethe planningplanning ofof anan experimentexperiment toto studystudy thisthis problemproblem wouldwould bebe toto decidedecide onon thethe factors,factors, oror variables,variables, thatthat willwill havehave anan effecteffect onon thethe pressurepressure drop.drop. ™™PressurePressure dropdrop perper unitunit lengthlength ∆p = f (D,ρ,µ, V) l 9 Pressure drop per unit length depends on FOUR variables: sphere size (D); speed (V); fluid density (ρ); fluid viscosity (m)

4 DimensionalDimensional AnalysisAnalysis 3/43/4

™™ToTo performperform thethe experimentsexperiments inin aa meaningfulmeaningful andand systematicsystematic manner,manner, itit wouldwould bebe necessarynecessary toto changechange onon ofof thethe variable,variable, suchsuch asas thethe velocity,velocity, whichwhich holdingholding allall otherother constant,constant, andand measuremeasure thethe correspondingcorresponding pressurepressure drop.drop. ™™DifficultyDifficulty toto determinedetermine thethe functionalfunctional relationshiprelationship betweenbetween thethe pressurepressure dropdrop andand thethe variousvarious factsfacts thatthat influenceinfluence it.it.

5 SeriesSeries ofof TestsTests

6 DimensionalDimensional AnalysisAnalysis 4/44/4

™™Fortunately,Fortunately, therethere isis aa muchmuch simplersimpler approachapproach toto thethe problemproblem thatthat willwill eliminateeliminate thethe difficultiesdifficulties describeddescribed above.above. DDCollectingCollecting thesethese variablesvariables intointo twotwo nondimensionalnondimensional combinationscombinations ofof thethe variablesvariables (called(called dimensionlessdimensionless productproduct oror dimensionlessdimensionless groups)groups) 9 Only one dependent and one D∆p ⎛ ρVD ⎞ independent variable l = φ⎜ ⎟ 2 ⎜ µ ⎟ 9 Easy to set up experiments to ρV ⎝ ⎠ determine dependency 9 Easy to present results (one graph) 7 PlotPlot ofof PressurePressure DropDrop DataData UsingUsing ……

3 D∆p L(F/ L ) 0 0 0 l = =& F L T ρV2 (FL−4T2 )(FT−1)2

−4 2 −1 ρVD (FL T )(LT )(L) 0 0 0 =& = F L T µ (FL−2T)

dimensionless product or dimensionless groups

8 BuckinghamBuckingham PiPi TheoremTheorem 1/51/5

™ A fundamental question we must answer is how many dimensionless products are required to replace the original list of variables ? ™ The answer to this question is supplied by the basic theorem of dimensional 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 Theorem Pi terms 9 BuckinghamBuckingham PiPi TheoremTheorem 2/52/5

™™GivenGiven aa physicalphysical problemproblem inin whichwhich thethe dependentdependent variablevariable isis aa functionfunction ofof kk--11 independentindependent variables.variables.

u1 = f (u 2 , u 3 ,....., u k ) DDMathematically,Mathematically, wewe cancan expressexpress thethe functionalfunctional relationshiprelationship inin thethe equivalentequivalent formform

g(u1, u 2 , u 3 ,....., u k ) = 0

Where g is an unspecified function, different from f. 10 BuckinghamBuckingham PiPi TheoremTheorem 3/53/5

™™TheThe BuckinghamBuckingham PiPi theoremtheorem stastatestes that:that: GivenGiven aa relationrelation amongamong kk variablesvariables ofof thethe formform g(u1, u 2 , u 3 ,....., u k ) = 0 ™™TheThe kk variablesvariables maymay bebe groupedgrouped intointo kk--rr independentindependent dimensionlessdimensionless products,products, oror ΠΠ terms,terms, expressibleexpressible inin functionalfunctional formform byby Π1 = φ(Π 2 , Π 3 ,,,, Π k−r ) rr ???? ΠΠ???? or φ(Π1, Π 2 , Π 3 ,,,, Π k−r ) = 0

11 BuckinghamBuckingham PiPi TheoremTheorem 4/54/5

™™TheThe numbernumber rr isis usually,usually, butbut notnot always,always, equalequal toto thethe minimumminimum numbernumber ofof independindependentent dimensionsdimensions requiredrequired toto specifyspecify thethe dimensionsdimensions ofof allall thethe parameters.parameters. UsuallyUsually thethe referencereference dimensionsdimensions requiredrequired toto describedescribe thethe variablesvariables willwill bebe thethe basicbasic dimensionsdimensions M,M, L,L, andand TT oror F,F, L,L, andand T.T. ™™TheThe theoremtheorem doesdoes notnot predicpredictt thethe functionalfunctional formform ofof φφ oror ϕϕ .. TheThe functionalfunctional relationrelation amongamong thethe independentindependent dimensionlessdimensionless productsproducts ΠΠ mustmust bebe determineddetermined experimentally.experimentally. ™™TheThe kk--rr dimensionlessdimensionless productsproducts ΠΠ termsterms obtainedobtained fromfrom thethe procedureprocedure areare independent.independent.

12 BuckinghamBuckingham PiPi TheoremTheorem 5/55/5

™™AA ΠΠ termterm isis notnot independentindependent ifif itit cancan bebe obtainedobtained fromfrom aa productproduct oror quotientquotient ofof thethe ototherher dimensionlessdimensionless productsproducts ofof thethe problem.problem. ForFor example,example, ifif

2Π Π 3 / 4 Π = 1 or Π = 1 5 6 2 Π 2 Π 3 Π 3

thenthen neitherneither ΠΠ5 nornor ΠΠ6 isis independentindependent ofof thethe otherother dimensionlessdimensionless productsproducts oror dimensionlessdimensionless groupsgroups..

13 DeterminationDetermination ofof PiPi TermsTerms 1/121/12

™™SeveralSeveral methodsmethods cancan bebe usedused toto formform thethe dimensionlessdimensionless products,products, oror pipi term,term, thatthat arisearise inin aa dimensionaldimensional analysis.analysis. ™™TheThe methodmethod wewe willwill describedescribe inin detaildetail isis calledcalled thethe METHODMETHOD ofof repeatingrepeating variables.variables. ™™RegardlessRegardless ofof thethe methodmethod toto bebe usedused toto determinedetermine thethe dimensionlessdimensionless products,products, oneone beginsbegins byby listinglisting allall dimensionaldimensional variablesvariables thatthat areare knownknown (or(or believed)believed) toto affectaffect thethe givengiven flowflow phenomenon.phenomenon. ™™EightEight stepssteps listedlisted belowbelow outlineoutline aa recommendedrecommended procedureprocedure forfor determiningdetermining thethe ΠΠ terms.terms.

14 DeterminationDetermination ofof PiPi TermsTerms 2/122/12

™™StepStep 11 ListList allall thethe variables.variables. 1 >>ListList allall thethe dimensionaldimensional variablesvariables involved.involved. >>KeepKeep thethe numbernumber ofof variablesvariables toto aa minimum,minimum, soso thatthat wewe cancan minimizeminimize thethe amountamount ofof laboratorylaboratory work.work. >>AllAll variablesvariables mustmust bebe independent.independent. ForFor example,example, ifif thethe crosscross--sectionalsectional areaarea ofof aa pipepipe isis anan importantimportant variable,variable, eithereither thethe areaarea oror thethe pipepipe diameterdiameter couldcould bebe used,used, butbut notnot both,both, sincesince theythey areare obviouslyobviously notnot independent.independent.

γ=ρ×g, that is, γ,ρ, and g are not independent.

15 DeterminationDetermination ofof PiPi TermsTerms 3/123/12

™™StepStep 11 ListList allall thethe variables.variables. 2 >>LetLet kk bebe thethe numbernumber ofof variables.variables. >>Example:Example: ForFor pressurepressure dropdrop perper unitunit length,length, k=5.k=5. (All(All variablesvariables areare ∆∆pp ,, D,D,µµ,,ρρ,, andand VV )) l ∆p = f (D,ρ,µ,V) l

16 DeterminationDetermination ofof PiPi TermsTerms 4/124/12

™™StepStep 22 ExpressExpress eacheach ofof thethe variablesvariables inin termsterms ofof basicbasic dimensions.dimensions. FindFind thethe numbernumber ofof referencereference dimensions.dimensions. >>SelectSelect aa setset ofof fundamentalfundamental (primary)(primary) dimensions.dimensions. >>ForFor example:example: MLT,MLT, oror FLT.FLT. >>Example:Example: ForFor pressurepressure dropdrop perper unitunit lengthlength ,, wewe choosechoose FLT.FLT. ∆p = FL−3 D = L ρ = FL−4T2 l & & & −2 −1 r=3r=3 µ =& FL T V =& LT 17 DeterminationDetermination ofof PiPi TermsTerms 5/125/12

™™StepStep 33 DetermineDetermine thethe requiredrequired numbernumber ofof pipi terms.terms. >>LetLet kk bebe thethe numbernumber ofof variablesvariables inin thethe problem.problem. >>LetLet rr bebe thethe numbernumber ofof referencereference dimensionsdimensions (primary(primary dimensions)dimensions) requiredrequired toto describedescribe thesethese variables.variables. >>TheThe numbernumber ofof pipi termsterms isis kk--rr >>Example:Example: ForFor pressurepressure dropdrop perper unitunit lengthlength k=5,k=5, rr == 3,3, thethe numbernumber ofof pipi termsterms isis kk--rr=5=5--3=2.3=2.

18 DeterminationDetermination ofof PiPi TermsTerms 6/126/12

™™StepStep 44 SelectSelect aa numbernumber ofof repeatingrepeating variables,variables, wherewhere thethe numbernumber requiredrequired isis equalequal toto thethe numbernumber ofof referencereference dimensions.dimensions. >>SelectSelect aa setset ofof rr dimensionaldimensional variablesvariables thatthat includesincludes allall thethe primaryprimary dimensionsdimensions (( repeatingrepeating variables).variables). >>TheseThese repeatingrepeating variablesvariables willwill allall bebe combinedcombined withwith eacheach ofof thethe remainingremaining parameters.parameters. NoNo repeatingrepeating variablesvariables shouldshould havehave dimensionsdimensions thatthat areare powerpower ofof thethe dimensionsdimensions ofof anotheranother repeatingrepeating variable.variable. >>Example:Example: ForFor pressurepressure dropdrop perper unitunit lengthlength (( rr == 3)3) selectselect ρρ ,, V,V, D.D. 19 DeterminationDetermination ofof PiPi TermsTerms 7/127/12

™™StepStep 55 FormForm aa pipi termterm byby multiplyingmultiplying oneone ofof thethe nonrepeatingnonrepeating variablesvariables byby thethe productproduct ofof thethe repeatingrepeating variables,variables, eacheach raisedraised toto anan exponentexponent thatthat willwill makemake thethe combinationcombination dimensionless.dimensionless. 1 >>SetSet upup dimensionaldimensional equations,equations, combiningcombining thethe variablesvariables selectedselected inin StepStep 44 withwith eacheach ofof thethe otherother variablesvariables ((nonrepeatingnonrepeating variables)variables) inin turn,turn, toto formform dimensionlessdimensionless groupsgroups oror dimensionlessdimensionless product.product. >>ThereThere willwill bebe kk –– rr equations.equations. >>Example:Example: ForFor pressurepressure dropdrop perper unitunit lengthlength

20 DeterminationDetermination ofof PiPi TermsTerms 8/128/12

™™StepStep 55 (Continued)(Continued) 2 Π = ∆p DaVbρc 1 l −3 a −1 b −4 2 c 0 0 0 (FL )(L) (LT ) (FL T ) =& F L T F :1+ c = 0 L : −3 + a + b − 4c = 0 ∆p D T : −b + 2c = 0 l Π1 = ⇒ a = 1,b = −2,c = −1 ρV2

21 DeterminationDetermination ofof PiPi TermsTerms 9/129/12

™™StepStep 66 RepeatRepeat StepStep 55 forfor eacheach ofof thethe remainingremaining nonrepeatingnonrepeating variables.variables. a b c Π2 = µD V ρ

−2 a −1 b −4 2 c 0 0 0 (FL T)(L) (LT ) (FL T ) =& F L T F :1+ c = 0 L : −2 + a + b − 4c = 0 µ T :1− b + 2c = 0 Π2 = ⇒ a = −1,b = −1,c = −1 DVρ

22 DeterminationDetermination ofof PiPi TermsTerms 10/1210/12

™™StepStep 77 CheckCheck allall thethe resultingresulting pipi termsterms toto makemake suresure theythey areare dimensionless.dimensionless. >>CheckCheck toto seesee thatthat eacheach groupgroup obtainedobtained isis dimensionless.dimensionless. >>Example:Example: ForFor pressurepressure dropdrop perper unitunit lengthlength .. ∆p D l 0 0 0 0 0 0 Π1 = =& F L T =& M L T ρV2

µ 0 0 0 0 0 0 Π2 = =& F L T =& M L T DVρ

23 DeterminationDetermination ofof PiPi TermsTerms 11/1211/12

™™StepStep 88 ExpressExpress thethe finalfinal formform asas aa relationshiprelationship amongamong thethe pipi terms,terms, andand thinkthink aboutabout whatwhat isis means.means. >>ExpressExpress thethe resultresult ofof thethe dimensionaldimensional analysis.analysis.

Π1 = φ(Π 2 , Π 3 ,,,, Π k−r ) >>Example:Example: ForFor pressurepressure dropdrop perper unitunit lengthlength .. ∆p D ⎛ µ ⎞ Dimensional analysis will not provide l = φ⎜ ⎟ the form of the function. The function 2 ⎜ ⎟ the form of the function. The function ρV ⎝ DVρ ⎠ can only be obtained from a suitable set of experiments.

24 DeterminationDetermination ofof PiPi TermsTerms 12/1212/12

™™TheThe pipi termsterms cancan bebe rearranged.rearranged. ForFor example,example, ΠΠ2,, couldcould bebe expressedexpressed asas ρVD Π = 2 µ ∆p D ⎛ ρVD ⎞ l = φ⎜ ⎟ ρV2 ⎝ µ ⎠

25 ExampleExample 7.17.1 MethodMethod ofof RepeatingRepeating VariablesVariables zz AA thinthin rectangularrectangular plateplate havinghaving aa widthwidth ww andand aa heightheight hh isis locatedlocated soso thatthat itit isis normalnormal toto aa movingmoving streamstream ofof fluid.fluid. AssumeAssume thatthat thethe drag,drag, D,D, thatthat thethe fluidfluid exertsexerts onon thethe plateplate isis aa functionfunction ofof ww andand h,h, thethe fluidfluid viscosity,viscosity, µµ ,and,and ρρ,, respectively,respectively, andand thethe velocity,velocity, V,V, ofof thethe fluidfluid approachingapproaching thethe plate.plate. DetermineDetermine aa suitablesuitable setset ofof pipi termsterms toto studystudy thisthis problemproblem experimentally.experimentally.

26 ExampleExample 7.17.1 SolutionSolution1/51/5

™™DragDrag forceforce onon aa PLATEPLATE D = f (w, h,ρ,µ, V) >>StepStep 1:List1:List allall thethe dimensionaldimensional variablesvariables involved.involved. D,w,h,D,w,h, ρρ,,μμ,V,V k=6k=6 dimensionaldimensional parameters.parameters. >>StepStep 2:Select2:Select primaryprimary dimensionsdimensions M,L,M,L, andand T.T. ExpressExpress eacheach ofof thethe variablesvariables inin termsterms ofof basicbasic dimensionsdimensions

−2 D =& MLT w =& L h =& L −1 −1 −3 −1 µ =& ML T ρ =& ML V =& LT

27 ExampleExample 7.17.1 SolutionSolution2/52/5

>>StepStep 3:3: DetermineDetermine thethe requiredrequired numbernumber ofof pipi terms.terms. kk--rr=6=6--3=33=3 >>StepStep 4:Select4:Select repeatingrepeating variablesvariables w,V,w,V,ρρ.. >>StepStep 5~6:combining5~6:combining thethe repeatingrepeating variablesvariables withwith eacheach ofof thethe otherother variablesvariables inin turn,turn, toto formform dimensionlessdimensionless groupsgroups oror dimensionlessdimensionless products.products.

28 ExampleExample 7.17.1 SolutionSolution3/53/5

a b c −2 a −1 b −3 c 0 0 0 Π1 = Dw V ρ = (MLT )(L) (LT ) (ML ) = M L T M :1 + c = 0 L :1 + a + b − 3c = 0 D Π = T : −2 − b = 0 1 2 2 >> a = −2, b = −2,c = −1 w V ρ

a b c a −1 b −3 c 0 0 0 Π 2 = hw V ρ = L(L) (LT ) (ML ) = M L T M : c = 0 L :1 + a + b − 3c = 0 h Π = T : b = 0 2 w >> a = −1, b = 0,c = 0

29 ExampleExample 7.17.1 SolutionSolution4/54/5

a b c −1 −1 a −1 b −3 c 0 0 0 Π3 = µw V ρ = (ML T )(L) (LT ) (ML ) = M L T M :1+ c = 0 L : −1+ a + b − 3c = 0 µ T : −1− b = 0 Π3 = >> a = −1,b = −1,c = −1 wVρ

30 ExampleExample 7.17.1 SolutionSolution5/55/5

>>StepStep 7:7: CheckCheck allall thethe resultingresulting pipi termsterms toto makemake suresure theythey areare dimensionless.dimensionless. >>StepStep 8:8: ExpressExpress thethe finalfinal foformrm asas aa relationshiprelationship amongamong thethe pipi terms.terms. TheThe functionalfunctional relationshiprelationship isis

Π1 = φ(Π 2 , Π 3 ), or D ⎛ h µ ⎞ = φ⎜ , ⎟ w 2V 2ρ ⎝ w wVρ ⎠ 31 SelectionSelection ofof VariablesVariables 1/41/4

™™OneOne ofof thethe mostmost important,important, andand difficult,difficult, stepssteps inin applyingapplying dimensionaldimensional analysisanalysis toto anyany givengiven problemproblem isis thethe selectionselection ofof thethe variablesvariables thatthat areare involved.involved. ™™ThereThere isis nono simplesimple procedureprocedure wherebywhereby thethe variablevariable cancan bebe easilyeasily identified.identified. Generally,Generally, oneone mustmust relyrely onon aa goodgood understandingunderstanding ofof thethe phenomenonphenomenon involvedinvolved andand thethe governinggoverning physicalphysical laws.laws. ™™IfIf extraneousextraneous variablesvariables areare included,included, thenthen tootoo manymany pipi termsterms appearappear inin thethe finalfinal solution,solution, andand itit maymay bebe difficult,difficult, timetime consuming,consuming, andand expensiveexpensive toto eliminateeliminate thesethese experimentally.experimentally.

32 SelectionSelection ofof VariablesVariables 2/42/4

™™IfIf importantimportant variablesvariables areare omitteomitted,d, thenthen anan incorrectincorrect resultresult willwill bebe obtained;obtained; andand again,again, thisthis maymay proveprove toto bebe costlycostly andand difficultdifficult toto ascertain.ascertain. ™™MostMost engineeringengineering problemsproblems involveinvolve certaincertain simplifyingsimplifying assumptionsassumptions thatthat havehave anan influenceinfluence onon thethe variablesvariables toto bebe considered.considered. ™™UsuallyUsually wewe wishwish toto keepkeep ththee problemsproblems asas simplesimple asas possible,possible, perhapsperhaps eveneven ifif somesome accuracyaccuracy isis sacrificedsacrificed

33 SelectionSelection ofof VariablesVariables 3/43/4

™™AA suitablesuitable balancebalance betweenbetween simplicitysimplicity andand accuracyaccuracy isis anan desirabledesirable goal.~~~~~goal.~~~~~ ™™VariablesVariables cancan bebe classifiedclassified intointo threethree generalgeneral group:group: BBGeometry:Geometry: lengthslengths andand angles.angles. BBMaterialMaterial Properties:Properties: relaterelate thethe externalexternal effectseffects andand thethe responses.responses. BBExternalExternal Effects:Effects: produce,produce, oror tendtend toto produce,produce, aa changechange inin thethe system.system. SuchSuch asas force,force, pressure,pressure, velocity,velocity, oror gravity.gravity.

34 SelectionSelection ofof VariablesVariables 44--1/41/4

™™PointsPoints shouldshould bebe consideredconsidered inin thethe selectionselection ofof variables:variables: BBClearlyClearly definedefine thethe problem.problem. WhatWhat’’ss thethe mainmain variablevariable ofof interest?interest? BBConsiderConsider thethe basicbasic lawslaws thatthat governgovern thethe phenomenon.phenomenon. BBStartStart thethe variablevariable selectionselection processprocess byby groupinggrouping thethe variablesvariables intointo threethree broadbroad classes.classes.

ContinuedContinued

35 SelectionSelection ofof VariablesVariables 44--2/42/4

™™PointsPoints shouldshould bebe consideredconsidered inin thethe selectionselection ofof variables:variables: BBConsiderConsider otherother variablesvariables thatthat maymay notnot fallfall intointo oneone thethe threethree categories.categories. ForFor example,example, timetime andand timetime dependentdependent variables.variables. BBBeBe suresure toto includeinclude allall quantitiesquantities thatthat maymay bebe heldheld constantconstant (e.g.,(e.g., g).g). BBMakeMake suresure thatthat allall variablesvariables areare independent.independent. LookLook forfor relationshipsrelationships amongamong subsetssubsets ofof thethe variables.variables.

36 DeterminationDetermination ofof ReferenceReference DimensionDimension 1/3

™™WhenWhen toto determinedetermine thethe numbernumber ofof pipi terms,terms, itit isis importantimportant toto knowknow howhow manymany referencereference dimensionsdimensions areare requiredrequired toto describedescribe thethe variables.variables. ™™InIn fluidfluid mechanics,mechanics, thethe requiredrequired numbernumber ofof referencereference dimensionsdimensions isis three,three, butbut inin sosomeme problemsproblems onlyonly oneone oror twotwo areare required.required. ™™InIn somesome problems,problems, wewe occasiooccasionallynally findfind thethe numbernumber ofof referencereference dimensionsdimensions neededneeded toto describedescribe allall variablesvariables isis smallersmaller thanthan thethe numbernumber ofof basicbasic dimensionsdimensions.. IllustratedIllustrated inin ExampleExample 7.2.7.2.

37 ExampleExample 7.27.2 DeterminationDetermination ofof PiPi TermsTerms zz AnAn open,open, cylindricalcylindrical tanktank havinghaving aa diameterdiameter DD isis supportedsupported aroundaround itsits bottombottom circumferencecircumference andand isis filledfilled toto aa depthdepth hh withwith aa liquidliquid havinghaving aa specificspecific weightweight γγ.. TheThe verticalvertical deflection,deflection, δδ ,, ofof thethe centercenter ofof thethe bottombottom isis aa functionfunction ofof D,D, h,h, d,d, γγ,, andand E,E, wherewhere dd isis thethe thicknessthickness ofof thethe bottombottom andand EE isis thethe modulusmodulus ofof elasticityelasticity ofof thethe bottombottom material.material. PerformPerform aa dimensionaldimensional analysisanalysis ofof thisthis problem.problem.

38 ExampleExample 7.27.2 SolutionSolution1/31/3

™™TheThe verticalvertical deflectiondeflection δ = f ()D,h,d,γ , E δ = L D = L ForFor F,L,T.F,L,T. PiPi terms=6terms=6--2=42=4 h = L ForFor M,L,TM,L,T PiPi terms=6terms=6--3=33=3 d = L γ = FL−3,γ = ML−2T−2 E = FL−2 ,E = ML−1T−2

39 ExampleExample 7.27.2 SolutionSolution2/32/3

™™ForFor F,L,TF,L,T system,system, PiPi terms=6terms=6--2=42=4 D and γ are selected as repeating variables

a1 b1 a2 b2 Π1 = δD γ Π2 = hD γ

a3 b3 a4 b4 Π3 = dD γ Π4 = ED γ

δ h d E δ ⎛ h d E ⎞ Π1 = ,Π2 = ,Π3 = ,Π4 = ⇒ = Φ⎜ , , ⎟ D D D Dγ D ⎝ D D Dγ ⎠

40 ExampleExample 7.27.2 SolutionSolution3/33/3

™™ForFor M,L,TM,L,T system,system, PiPi terms=6terms=6--3=33=3 ?? AA closercloser looklook atat thethe dimensionsdimensions ofof thethe variablesvariables listedlisted revealreveal thatthat onlyonly twotwo referencereference dimensions,dimensions, LL andand MTMT-2 areare required.required.

41 DeterminationDetermination ofof ReferenceReference DimensionDimension 2/3

EXAMPLEEXAMPLE ∆h = f (D,γ,σ)

MLTMLT SYSTEMSYSTEM FLTFLT SYSTEMSYSTEM ∆h D γ σ ∆h D γ σ M M F F L L L L L2T2 T2 L3 L

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

42 DeterminationDetermination ofof ReferenceReference DimensionDimension 3/3

SetSet DimensionalDimensional MatrixMatrix MLTMLT SYSTEMSYSTEM FLTFLT SYSTEMSYSTEM ∆h D γ σ ∆h D γ σ M 0 0 1 1 F 0 0 1 1 L 1 1 − 2 0 L 1 1 − 3 −1 T 0 0 − 2 − 2 T 0 0 0 0 ∆h Π = Rank=2Rank=2 PiPi term=4term=4--2=22=2 1 D σ ∆h ⎛ σ ⎞ Π2 = ⇒ = Φ⎜ ⎟ D2γ D ⎜ D2γ ⎟ ⎝ ⎠ 43 UniquenessUniqueness ofof PiPi TermsTerms 1/41/4

™™TheThe PiPi termsterms obtainedobtained dependdepend onon thethe somewhatsomewhat arbitraryarbitrary selectionselection ofof repeatingrepeating variables.variables. ForFor example,example, inin thethe problemproblem ofof studyingstudying thethe pressurepressure dropdrop inin aa pipe.pipe. ∆p = f (D,ρ,µ,V) l Selecting D,V, and ρ as ∆p D ⎛ ρVD ⎞ l repeating variables: = Φ1⎜ ⎟ ρV2 ⎝ µ ⎠

Selecting D,V, and µ as ∆p D2 ⎛ ρVD ⎞ l repeating variables: = Φ2 ⎜ ⎟ Vµ ⎝ µ ⎠ 44 UniquenessUniqueness ofof PiPi TermsTerms 2/42/4

∆p D ⎛ ρVD ⎞ l = Φ1⎜ ⎟ BothBoth areare correct,correct, andand bothboth wouldwould ρV2 ⎝ µ ⎠ leadlead toto thethe samesame finalfinal equationequation forfor thethe pressurepressure drop.drop. ThereThere isis notnot aa ∆p D2 ⎛ ρVD ⎞ l uniqueunique setset ofof pipi termsterms whichwhich = Φ2 ⎜ ⎟ Vµ ⎝ µ ⎠ arisesarises fromfrom aa dimensionaldimensional analysisanalysis.. TheThe functionsfunctions ΦΦ1 andand ΦΦ2 areare willwill bebe differentdifferent becausebecause thethe dependentdependent pipi termsterms areare differentdifferent forfor thethe twotwo relationships.relationships.

45 UniquenessUniqueness ofof PiPi TermsTerms 3/43/4

EXAMPLEEXAMPLE Π1 = Φ(Π2,Π3 )

Form a new pi term ' a b Π2 = Π2Π3 ' ' Π1 = Φ1(Π2,Π3)= Φ2 (Π2,Π2 )

All are correct

46 UniquenessUniqueness ofof PiPi TermsTerms 4/44/4

∆p D ⎛ ρVD ⎞ Selecting D,V, and ρ as l = Φ ⎜ ⎟ 2 1⎜ ⎟ repeating variables: ρV ⎝ µ ⎠ ∆p D ρVD ∆p D2 l × = l ρV2 µ Vµ

∆p D2 ⎛ ρVD ⎞ l = Φ2 ⎜ ⎟ Vµ ⎝ µ ⎠

47 CommonCommon DimensionlessDimensionless GroupsGroups 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.

48 FroudeFroude NumberNumber 1/21/2

V V 2 ρV 2 L2 Fr = >> Fr 2 = = gL gL ρgL3 ™ In honor of William Froude (1810~1879), a British civil engineer, mathematician, and naval architect who pioneered the use of towing tanks for the study of ship design. ™ Froude number is the ratio of the forces due to the acceleration of a fluid particles (inertial force) to the force due to gravity (gravity forces). ™ Froude number is significant for flows with free surface effects. ™ Froude number less than unity indicate subcritical flow and values greater than unity indicate supercritical flow.

49 FroudeFroude NumberNumber 2/22/2

2 * 2 * dV dV V dV V * dVs s s * s F = V m a s = = Vs = Vs * I s * dt ds l ds l ds

* Vs * s Vs = s = V l

2 * 2 FI V * dVs V V Fr = = Vs * ≡ ≡ FG gl ds gl gl

50 ReynoldsReynolds NumberNumber 1/21/2 ρV V Re = l = l µ υ ™ In honor of Osborne Reynolds (1842~1912), the British engineer who first demonstrated that this combination of variables could be used as a criterion to distinguish between laminar and turbulent flow. ™ The Reynolds number is a measure of the ration of the inertia forces to viscous forces. ™ If the Reynolds number is small (Re<<1), thi sis an indication that the viscous forces ar dominant in the problem, and it may be possible to neglect the inertial effects; that is, the density of the fluid will no be an important variable.

51 ReynoldsReynolds NumberNumber 2/22/2

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

52 EulerEuler numbernumber p ∆p Eu = ≡ ρV 2 ρV 2 ™ In honor of Leonhard Euler (1707~1783), a famous Swiss mathematician who pioneered work on the relationship between pressure and flow. ™ Euler’s number is the ratio of pressure force to inertia forces. It is often called the pressure coefficient, Cp.

53 CavitationCavitation NumberNumber p − p Ca = r v 1 ρV 2 2 ™ For problems in which cavitation is of concern, the dimensionless 1 2 group ( p r − p v ) / 2 ρ V is commonly used, where pv is the vapor pressure and pr is some reference pressure. ™ The cavitation number is used in the study of cavitation phenomena. ™ The smaller the cavitation number, the more likely cavitation is to occur.

54 CauchyCauchy NumberNumber && MachMach NumberNumber 1/21/2

ρV 2 V V ρ ρV 2 Ca = Ma = = = V Ma 2 = = Ca E υ c E υ E υ E υ ρ ™ The Cauchy number is named in honor of Augustin Louis de Cauchy (1789~1857), a French engineer, mathematician, and hydrodynamicist. ™ The Mach number is named in honor of Ernst Mach (1838~1916), an Austrian physicist and philosopher. ™ Either number may be used in problems in which fluid compressibility is important.

55 CauchyCauchy NumberNumber && MachMach NumberNumber 2/22/2

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

56 StrouhalStrouhal NumberNumber 1/21/2 ω St = l V ™ In honor of Vincenz Strouhal (1850~1922), who used this parameter in his study of “singing wires.” The most dramatic evidence of this phenomenon occurred in 1940 with the collapse of the Tacoma Narrow bridges. The shedding frequency of the vortices coincided with the natural frequency of the bridge, thereby setting up a resonant condition that eventually led to the collapse of the bridge. ™ This parameter is important in unsteady, oscillating flow problems in which the frequency of the oscillation is ω .

57 StrouhalStrouhal NumberNumber 2/22/2

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

For example, in a certain Reynolds number range, a periodic flow will develop downstream from a cylinder placed in a moving stream due to

a regular patterns of vortices that are shed from the body. 58 WeberWeber NumberNumber 1/21/2 ρV 2 We = l σ ™ Named after Moritz Weber (1871~1951), a German professor of naval mechanics who was instrumental in formalizing the general use of common dimensionless groups as a basis for similitude studies. ™ Weber number is important in problem in which there is an interface between two fluids. In this situation the surface tension may play an important role in the phenomenon of interest.

59 WeberWeber NumberNumber 2/22/2

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

60 CorrelationCorrelation ofof ExperimentalExperimental DataData

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

61 ProblemsProblems withwith OneOne PiPi TermTerm

™™TheThe functionalfunctional relationshiprelationship forfor oneone PiPi term.term.

Π1 = C where C is a constant. The value of the constant must still be determined by experiment.

62 ExampleExample 7.37.3 FlowFlow withwith OnlyOnly OneOne PiPi TermTerm z 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, d, the particle velocity, V, and the fluid viscosity, μ. Determine, with the aid the dimensional analysis, how the drag depends on the particle velocity.

63 ExampleExample 7.37.3 SolutionSolution

™ ™TheThe dragdrag −2 D =& F d =& L µ =& FL T D = f (d, V,µ) −3 −1 ρ =& ML V =& LT D Π = = C D = CµVd 1 µVd D ∝ V For a given particle and fluids, the drag varies directly with the velocity

64 ProblemsProblems withwith TwoTwo oror MoreMore PiPi TermTerm 1/2

™™ProblemsProblems withwith twotwo pipi termsterms

Π1 = Φ(Π 2 ) An empirical relationship is the functional relationship valid over the range of Π2. among the variables can the be determined by varying

Π2 and measuring the corresponding value of Π1.

The empirical equation Dangerous to relating Π2 and Π1 by using extrapolate beyond a standard curve-fitting valid range technique. 65 ProblemsProblems withwith TwoTwo oror MoreMore PiPi TermTerm 2/2

™™ProblemsProblems withwith threethree pipi terms.terms. To determine a suitable empirical equation Π1 = Φ()Π2 ,Π3 relating the three pi terms.

To show data correlations on simple graphs.

Families curve of curves

66 ModelingModeling andand SimilitudeSimilitude

ToTo developdevelop thethe proceduresprocedures forfor designingdesigning modelsmodels soso thatthat thethe modelmodel andand prototypeprototype willwill behavebehave inin aa similarsimilar fashionfashion…………..

67 ModelModel vs.vs. PrototypePrototype 1/21/2

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

68 ModelModel vs.vs. PrototypePrototype 2/22/2

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

69 SimilaritySimilarity ofof ModelModel andand PrototypePrototype

™ What conditions must be met to ensure the similarity of model and prototype? ~ Geometric Similarity >Model and prototype have same shape. >Linear dimensions on model and prototype correspond within constant scale factor. ~ Kinematic Similarity >Velocities at corresponding points on model and prototype differ only by a constant scale factor. ~ Dynamic Similarity >Forces on model and prototype differ only by a constant scale factor.

70 TheoryTheory ofof ModelsModels 1/51/5

™™TheThe theorytheory ofof modelsmodels cancan bebe readilyreadily developeddeveloped byby usingusing thethe principlesprinciples ofof dimensionaldimensional analysis.analysis. ™™ForFor givengiven problemproblem whichwhich cancan bebe describeddescribed inin termsterms ofof aa setset ofof pipi termsterms asas This relationship can be formulated Π1 = φ(Π2 ,Π3,,,Πn ) with a knowledge of the general nature of the physical phenomenon and the variables involved.

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

71 TheoryTheory ofof ModelsModels 2/52/5

™™AA similarsimilar relationshiprelationship cancan bebe writtenwritten forfor aa modelmodel ofof thisthis prototype;prototype; thatthat is,is, Π1m = φ(Π 2m , Π 3m ,,,, Π nm ) wherewhere thethe formform ofof thethe functionfunction willwill bebe thethe samesame asas longlong asas thethe samesame phenomenonphenomenon isis involvedinvolved inin bothboth thethe prototypeprototype andand thethe model.model.

TheThe prototypeprototype andand thethe modelmodel mustmust havehave thethe samesame phenomenon.phenomenon. 72 TheoryTheory ofof ModelsModels 3/53/5

™™ModelModel designdesign (the(the modelmodel isis designeddesigned andand operated)operated) conditions,conditions, alsoalso calledcalled similaritysimilarity requirementsrequirements oror modelingmodeling laws.laws.

Π 2 = Π 2m Π 3 = Π 3m .....Π n = Π nm ™™TheThe formform ofof ΦΦ isis thethe samesame forfor modelmodel andand prototype,prototype, itit followsfollows thatthat This is the desired prediction equation and

indicates that the measured of Π1m obtained Π1 = Π1m with the model will be equal to the corresponding Π1 for the prototype as long as the other Π parameters are equal.

73 TheoryTheory ofof ModelsModels 4/54/5 –– SummarySummary11

™™TheThe prototypeprototype andand thethe modelmodel mustmust havehave thethe samesame phenomenon.phenomenon.

ForFor prototypeprototype Π1 = φ(Π2 ,Π3,,,Πn )

ForFor modelmodel Π1m = φ(Π 2m , Π 3m ,,,, Π nm )

74 TheoryTheory ofof ModelsModels 5/55/5 –– SummarySummary22

™™TheThe modelmodel isis designeddesigned andand operatedoperated underunder thethe followingfollowing conditionsconditions (called(called designdesign conditions,conditions, alsoalso calledcalled similaritysimilarity requirementsrequirements oror modelingmodeling laws)laws)

Π 2 = Π 2m Π 3 = Π 3m .....Π n = Π nm

™™TheThe measuredmeasured ofof ΠΠ1m obtainedobtained withwith thethe modelmodel willwill bebe equalequal toto thethe correspondingcorresponding ΠΠ1 forfor thethe prototype.prototype.

Π1 = Π1m CalledCalled predictionprediction equationequation

75 TheoryTheory ofof ModelsModels EXAMPLEEXAMPLE 11

™ Example: Considering the drag force on a sphere. F ⎛ ρVD ⎞ F = f (D, V,ρ,µ) = f1⎜ ⎟ ρV 2 D 2 ⎝ µ ⎠ BThe prototype and the model must have the same phenomenon.

Fm ⎛ ρ m Vm D m ⎞ F ⎛ ρVD ⎞ = f1⎜ ⎟ = f1⎜ ⎟ 2 2 ⎜ µ ⎟ 2 2 ⎜ ⎟ ρ m Vm D m ⎝ m ⎠ ρV D ⎝ µ ⎠prototype BDesign conditions. ⎛ ρVD ⎞ ⎛ ρVD ⎞ ⎜ ⎟ = ⎜ ⎟ ⎝ µ ⎠mod el ⎝ µ ⎠prototype

⎛ F ⎞ ⎛ F ⎞ BThen … ⎜ ⎟ = ⎜ ⎟ ⎜ ρV 2 D 2 ⎟ ⎜ ρV 2 D 2 ⎟ ⎝ ⎠mod el ⎝ ⎠prototype 76 TheoryTheory ofof ModelsModels EXAMPLEEXAMPLE 22

™ Example: Determining the drag force on a thin rectangular plate (w × h in size) D ⎛ w ρVw ⎞ = Φ⎜ , ⎟ D = f ()w, h,µ,ρ, V 2 2 ⎜ ⎟ w ρV ⎝ h µ ⎠ BThe prototype and the model must have the same phenomenon.

D ⎛ w ρ V w ⎞ D ⎛ w ρVw ⎞ m = Φ⎜ m , m m m ⎟ = Φ⎜ , ⎟ 2 2 ⎜ ⎟ 2 2 ⎜ h µ ⎟ w m ρ m Vm ⎝ h m µ m ⎠ w ρV ⎝ ⎠ prototype

w w ρ V w ρVw BDesign conditions. m = , m m m = h m h µ m µ

2 2 D D ⎛ w ⎞ ⎛ ρ ⎞⎛ V ⎞ BThen … = m >> D = ⎜ ⎟ ⎜ ⎟⎜ ⎟ D 2 2 2 2 ⎜ ⎟ ⎜ ⎟⎜ ⎟ m w ρV w m ρ m Vm ⎝ w m ⎠ ⎝ ρ m ⎠⎝ Vm ⎠ 77 ExampleExample 7.57.5 PredictionPrediction ofof PrototypePrototype PerformancePerformance fromfrom ModelModel DataData 1/2 z 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 past this type of bluff body, vortices may develop on the downwind side that are shed in a regular fashion at some definite frequency. Since these vortices can create harmful periodic forces acting on the structure, it is important to determine the shedding frequency. For the specific structure of interest, D=0.1m, H=0.3m, and a representative wind velocity 50km/hr. Standard air can be assumed. The shedding frequency is to be determined through the use of a small-scale model that is to be tested in a water tunnel. For the model Dm=20mm and the water temperature is 20℃.

78 ExampleExample 7.57.5 PredictionPrediction ofof PrototypePrototype PerformancePerformance fromfrom ModelModel DataData 2/2

Determine the model dimension, Hm, and the velocity at which the test should be performed. If the shedding frequency ω for the model is found to be 49.9Hz, what is the corresponding frequency for the prototype? ~ For air at standard condition µ = 1.79 ×10 −5 kg / m − s,ρ = 1.23kg / m 3 −3 3 ~ For water at 20℃, µ water = 1×10 kg / m − s,ρ water = 998kg / m

79 ExampleExample 7.57.5 SolutionSolution1/41/4

99 StepStep 1:List1:List allall thethe dimensionaldimensional variablesvariables involved.involved. ωω D,H,V,D,H,V,ρρ,,μμ k=6k=6 dimensionaldimensional variables.variables. 99 StepStep 2:Select2:Select primaryprimary dimensionsdimensions F,LF,L andand T.T. ListList thethe dimensionsdimensions ofof allall variablesvariables inin termsterms ofof primaryprimary dimensions.dimensions. r=3r=3 primaryprimary dimensionsdimensions −1 ω =& T D =& L H =& L −1 −4 2 −2 V =& LT ρ =& FL T µ =& ML T

80 ExampleExample 7.57.5 SolutionSolution2/42/4

99 StepStep 3:3: DetermineDetermine thethe requiredrequired numbernumber ofof pipi terms.terms. kk--rr=6=6--3=33=3 99 StepStep 4:Select4:Select repeatingrepeating variablesvariables D,V,D,V, μμ.. 99 StepStep 5~6:combining5~6:combining thethe repeatingrepeating variablesvariables withwith eacheach ofof thethe otherother variablesvariables inin turn,turn, toto formform dimensionlessdimensionless groups.groups. ωD H Π = ωD a1 V b1 µc1 = Π = HD a 2 V b2 µ c2 = 1 V 2 D ρVD Π = ρD a 3 V b3 µ c3 = 3 µ

81 ExampleExample 7.57.5 SolutionSolution3/43/4

ωD ⎛ H ρVD ⎞ 99 TheThe functionalfunctional relationshiprelationship isis = φ⎜ , ⎟ V ⎝ D µ ⎠

Strouhal number

99 TheThe prototypeprototype andand thethe modelmodel mustmust havehave thethe samesame phenomenon.phenomenon.

ωm D m ⎛ H m ρ m Vm D m ⎞ = φ⎜ , ⎟ Vm ⎝ D m µ m ⎠

82 ExampleExample 7.57.5 SolutionSolution4/44/4

99 TheThe DesignDesign conditions.conditions. H Hm ρVD ρmVmDm = = D Dm µ µm ThenThen…….. H µm ρ D H m = Dm = ... = 60mm Vm = V = ... = 13.9m / s D µ ρm Dm

V Dm ω = ωm = ... = 29.0Hz Vm D

83 ModelModel ScalesScales

™™TheThe ratioratio ofof aa modelmodel variablevariable toto thethe correspondingcorresponding prototypeprototype variablevariable isis calledcalled thethe scalescale forfor thatthat variable.variable.

l1 = l1m ⇒ λ = l1m = l 2m Length Scale l l 2 l 2m l1 l 2

Vm Velocity Scale λ = V V

ρm Density Scale λ = ρ ρ

µm λ = Viscosity Scale µ µ

84 ValidationValidation ofof ModelsModels DesignDesign

™™TheThe purposepurpose ofof modelmodel designdesign isis toto predictpredict thethe effectseffects ofof certaincertain proposedproposed changeschanges inin aa givengiven prototype,prototype, andand inin thisthis instanceinstance somesome actualactual prototypeprototype datadata maymay bebe available.available. ValidationValidation ofof modelmodel designdesign ?? ™™TheThe modelmodel cancan bebe designed,designed, constructed,constructed, andand tested,tested, andand thethe modelmodel predictionprediction cancan bebe comparedcompared withwith thesethese data.data. IfIf thethe agreementagreement isis satisfactorysatisfactory,, thenthen thethe modelmodel cancan bebe changedchanged inin thethe desireddesired manner,manner, andand thethe correspondingcorresponding effecteffect onon thethe prototypeprototype cancan bebe predictedpredicted withwith increasedincreased confidence.confidence.

85 DistortedDistorted ModelsModels

™™InIn manymany modelmodel studies,studies, toto achieveachieve dynamicdynamic similaritysimilarity requiresrequires duplicationduplication ofof severalseveral dimensionlessdimensionless groups.groups. ™™InIn somesome cases,cases, completecomplete dynamicdynamic similaritysimilarity betweenbetween modelmodel andand prototypeprototype maymay notnot bebe attainable.attainable. IfIf oneone oror moremore ofof thethe similaritysimilarity requirementsrequirements areare notnot met,met, forfor example,example, ifif Π 2 ≠ Π 2 m ,, ththenen itit followsfollows thatthat thethe predictionprediction equatequationion

Π1 = Π1m isis notnot true;true; thatthat is,is, Π1 ≠ Π1m ™™MODELSMODELS forfor whichwhich oneone oror moremore ofof thethe similaritysimilarity requirementsrequirements areare notnot satisfiedsatisfied areare calledcalled DISTORTEDDISTORTED MODELSMODELS..

86 DistortedDistorted ModelsModels EXAMPLEEXAMPLE--11 1/31/3

™™DetermineDetermine thethe dragdrag forceforce onon aa surfacesurface ship,ship, completecomplete dynamicdynamic similaritysimilarity requiresrequires thatthat bothboth ReynoldsReynolds andand FroudeFroude numbersnumbers bebe duplicatedduplicated betweenbetween modelmodel andand prototype.prototype.

V V Fr = m = Fr = p m 1/ 2 p 1/ 2 Froude numbers (gl m ) (gl p ) V Vm l m pl p Re m = = Re p = Reynolds numbers υm υp 1/ 2 V ⎛ ⎞ ™™ToTo matchmatch FroudeFroude numbersnumbers m = ⎜ l m ⎟ ⎜ ⎟ betweenbetween modelmodel andand prototypeprototype Vp ⎝ l p ⎠

87 DistortedDistorted ModelsModels EXAMPLEEXAMPLE--11 2/32/3

™™ToTo matchmatch ReynoldsReynolds numbersnumbers betweenbetween modelmodel andand prototypeprototype 1/ 2 3 / 2 υ V υ ⎛ ⎞ ⎛ ⎞ m = m l m m = ⎜ l m ⎟ l m = ⎜ l m ⎟ ⎜ ⎟ ⎜ ⎟ υp Vp l p υp ⎝ l p ⎠ l p ⎝ l p ⎠

IfIf llm// llp equalsequals 1/100(a1/100(a typicaltypical lengthlength scalescale forfor shipship modelmodel tests)tests) ,, thenthen υυm//υυp mustmust bebe 1/1000.1/1000. >>>>>> TheThe kinematickinematic viscosityviscosity ratioratio requiredrequired toto duplicateduplicate ReynoldsReynolds numbersnumbers cannotcannot bebe attained.attained.

88 DistortedDistorted ModelsModels EXAMPLEEXAMPLE--11 3/33/3

™™ItIt isis impossibleimpossible inin practicepractice forfor thisthis model/prototypemodel/prototype scalescale ofof 1/1001/100 toto satisfysatisfy bothboth ththee FroudeFroude numbernumber andand ReynoldsReynolds numbernumber criteriacriteria;; atat bestbest wewe willwill bebe ableable toto satisfysatisfy onlyonly oneone ofof them.them. ™™IfIf waterwater isis thethe onlyonly practicalpractical liquidliquid forfor mostmost modelmodel testtest ofof freefree--surfacesurface flows,flows, aa fullfull--scalescale testtest isis requiredrequired toto obtainobtain completecomplete dynamicdynamic similaritysimilarity..

89 DistortedDistorted ModelsModels EXAMPLEEXAMPLE--22 1/21/2

™™InIn thethe studystudy ofof openopen channelchannel oror freefree--surfacesurface flows.flows. TypicallyTypically inin thesethese problemsproblems bothboth thethe ReynoldsReynolds numbernumber andand FroudeFroude numbernumber areare involved.involved. V V Fr = m = Fr = p m 1/ 2 p 1/ 2 Froude numbers (g m l m ) (g pl p )

ρ V ρ m Vm l m p pl p Re m = = Re p = Reynolds numbers µ m µ p 1/ 2 ™™ToTo matchmatch FroudeFroude numbersnumbers V ⎛ ⎞ m = ⎜ l m ⎟ = λ ⎜ ⎟ l betweenbetween modelmodel andand prototypeprototype Vp ⎝ l p ⎠

90 DistortedDistorted ModelsModels EXAMPLEEXAMPLE--22 2/22/2

™™ToTo matchmatch ReynoldsReynolds numbersnumbers betweenbetween modelmodel andand prototypeprototype V µ m / µ p l p m = l m = V µ ρ p l p V ρ / ρ m = m p l p p m l m Vp µ p ρ m l m µ m / µ p υ λ 3 / 2 = = m l ρ p / ρ m υp

IfIf llm// ll p equalsequals 1/100(a1/100(a typicaltypical lengthlength scalescale forfor shipship modelmodel tests)tests) ,, thenthen υυm//υυp mustmust bebe 1/1000.1/1000. >>>The>>>The kinematickinematic viscosityviscosity ratioratio requiredrequired toto duplicateduplicate ReynoldsReynolds numbersnumbers cannotcannot bebe attained.attained.

91 TypicalTypical ModelModel StudiesStudies

™™FlowFlow throughthrough closedclosed conduits.conduits. ™™FlowFlow aroundaround immersedimmersed bodies.bodies. ™™FlowFlow withwith aa freefree surface.surface.

92 FlowFlow ThroughThrough ClosedClosed ConduitsConduits 1/51/5

™™ThisThis typetype ofof flowflow includesincludes pipepipe flowflow andand flowflow throughthrough valves,valves, fittings,fittings, andand meteringmetering devices.devices. ™™TheThe conduitsconduits areare oftenoften circular,circular, theythey couldcould havehave otherother shapesshapes asas wellwell andand maymay containcontain expansionsexpansions oror contractions.contractions. ™™SinceSince therethere areare nono fluidfluid interfacesinterfaces oror freefree surface,surface, thethe dominantdominant forcesforces areare inertialinertial anandd viscousviscous forcesforces soso thatthat thethe ReynoldsReynolds numbernumber isis anan importimportantant similaritysimilarity parameter.parameter.

93 FlowFlow ThroughThrough ClosedClosed ConduitsConduits 2/52/5

™™ForFor lowlow MachMach numbersnumbers (Ma<0.3),(Ma<0.3), compressibilitycompressibility effectseffects areare usuallyusually negligiblenegligible forfor bothboth thethe flowflow ofof liquidsliquids oror gases.gases. ™™ForFor flowflow inin closedclosed conduitsconduits atat lowlow MachMach numbers,numbers, andand dependentdependent pipi term,term, suchsuch asas pressurepressure drop,drop, cancan bebe expressedexpressed asas ⎛ li ε ρVl ⎞ DependentDependent pipi term=term= φ⎜ , , ⎟ ⎝ l l µ ⎠

Where l is some particular length of the system and li represents a series of length terms, ε/ l is the relative roughness of the surface, and ρVl/µ is the Reynolds number.

94 FlowFlow ThroughThrough ClosedClosed ConduitsConduits 3/53/5

™™ToTo meetmeet thethe requirementrequirement ofof geometricgeometric similaritysimilarity ε ε ε lim = li m = ⇒ lim = l m = m = λ l l m l l m l li l ε

™™ToTo meetmeet thethe requirementrequirement ofof ReynoldsReynolds numbernumber

ρ V ρV V µ ρ µm l m ml m = l ⇒ m = m l = = 1 µ µm µ V µ ρm l m l m

V m = l ⇒ V = V / λ If the same fluid is used, then m l V l m

95 FlowFlow ThroughThrough ClosedClosed ConduitsConduits 4/54/5

™™TheThe fluidfluid velocityvelocity inin thethe modelmodel willwill bebe largerlarger thanthan thatthat inin thethe prototypeprototype forfor anyany lengthlength scalescale lessless thanthan 1.1. SinceSince lengthlength scalesscales areare typicallytypically mumuchch lessless thanthan unity.unity. ™™ReynoldsReynolds numbernumber similaritysimilarity maymay bebe difficultdifficult toto achieveachieve becausebecause ofof thethe largelarge modelmodel velocitiesvelocities required.required. V m = l ⇒ V = V / λ m l Vm > V V l m λ < 1 l

96 FlowFlow ThroughThrough ClosedClosed ConduitsConduits 5/55/5

™™WithWith thesethese similaritysimilarity requirementsrequirements satisfied,satisfied, itit followsfollows thatthat thethe dependentdependent pipi termterm willwill bebe equalequal inin modelmodel andand prototype.prototype. ForFor example,example, ∆p DependentDependent pipi termterm Π1 = ρV2 ™™TheThe prototypeprototype pressurepressure dropdrop 2 ρ ⎛ V ⎞ ⎜ ⎟ ∆p = ⎜ ⎟ ∆pm In general ∆p ≠ ∆pm ρm ⎝ Vm ⎠

97 ExampleExample 7.67.6 ReynoldsReynolds NumberNumber SimilaritySimilarity

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

98 ExampleExample 7.67.6 SolutionSolution

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

VmDm VD Vm D Rem = Re = ⇒ = νm ν V Dm Same fluid used in the model and prototype Q V A D m = m m = ... = m Q VA D

(3/12ft) Q = (30ft 3 / s) = 3.75cfs m (2ft)

99 FlowFlow AroundAround ImmersedImmersed BodiesBodies 1/71/7

™™ThisThis typetype ofof flowflow includesincludes flowflow aroundaround aircraft,aircraft, automobiles,automobiles, golfgolf balls,balls, andand building.building. ™™ForFor thesethese problems,problems, geometricgeometric andand ReynoldsReynolds numbernumber similaritysimilarity isis required.required. ™™SinceSince therethere areare nono fluidfluid interfaces,interfaces, surfacesurface tensiontension isis notnot important.important. Also,Also, gravitygravity willwill notnot affectaffect thethe flowflow pattern,pattern, soso thethe FroudeFroude numbernumber needneed notnot toto bebe considered.considered. ™™ForFor incompressibleincompressible flow,flow, thethe MachMach numbernumber cancan bebe omitted.omitted.

100 FlowFlow AroundAround ImmersedImmersed BodiesBodies 2/72/7

™™AA generalgeneral formulationformulation forfor thesethese problemsproblems isis

⎛ li ε ρVl ⎞ DependentDependent pipi term=term= φ⎜ , , ⎟ ⎝ l l µ ⎠ Where l is some characteristic length of the system and li represents other pertinent lengths, ε/ l is the relative roughness of the surface, and ρVl/µ 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 located in a long-test-section, meteorological

wind tunnel. 101 FlowFlow AroundAround ImmersedImmersed BodiesBodies 3/73/7

™™Frequently,Frequently, thethe dependentdependent variablevariable ofof interestinterest forfor thisthis typetype ofof problemproblem isis thethe drag,drag, D,D, developeddeveloped onon thethe body.body. The dependent pi terms would usually be expressed in the form of a drag coefficient D C = D 1 ρV2 2 2 l ™™ToTo meetmeet thethe requirementrequirement ofof geometricgeometric similaritysimilarity ε ε ε lim = li m = ⇒ lim = l m = m = λ l l m l l m l li l ε 102 FlowFlow AroundAround ImmersedImmersed BodiesBodies 4/74/7

™™ToTo meetmeet thethe requirementrequirement ofof ReynoldsReynolds numbernumber similaritysimilarity ρ V ρV V µ ρ υ m ml m = l ⇒ m = m l = m l µm µ V µ ρm l m υ l m 2 2 D D ρ ⎛ V ⎞ ⎛ ⎞ = m ⇒ D = ⎜ ⎟ ⎜ l ⎟ D 1 1 ⎜ ⎟ ⎜ ⎟ m ρV2 2 ρ V 2 2 ρm ⎝ Vm ⎠ ⎝ l m ⎠ 2 l 2 m m l m V m = l ⇒ V = V / λ TheThe samesame fluidfluid isis used,used, thenthen m l V l m

103 FlowFlow AroundAround ImmersedImmersed BodiesBodies 5/75/7

™™TheThe fluidfluid velocityvelocity inin thethe modelmodel willwill bebe largerlarger thanthan thatthat inin thethe prototypeprototype forfor anyany lengthlength scalescale lessless thanthan 1.1. SinceSince lengthlength scalesscales areare typicallytypically mumuchch lessless thanthan unity.unity. ™™ReynoldsReynolds numbernumber similaritysimilarity maymay bebe difficultdifficult toto achieveachieve becausebecause ofof thethe largelarge modelmodel velocitiesvelocities required.required.

104 FlowFlow AroundAround ImmersedImmersed BodiesBodies 6/76/7

™™HowHow toto reducereduce thethe fluidfluid velocityvelocity inin thethe modelmodel ??

AA differentdifferent fluidfluid isis usedused inin thethe modelmodel suchsuch thatthat υm / υ < 1 For example, the ratio of the kinematic viscosity of water to that of air is approximately 1/10, so that if the prototype fluid were air, test might be run on the model using water.

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

105 FlowFlow AroundAround ImmersedImmersed BodiesBodies 7/77/7

™™HowHow toto reducereduce thethe fluidfluid velocityvelocity inin thethe modelmodel ??

SameSame fluidfluid withwith differentdifferent density..density.. ρρm>>ρρ

AnAn alternativealternative wayway toto reducereduce VVm isis toto increaseincrease thethe airair pressurepressure inin thethe tunneltunnel soso thatthat

ρρm>>ρρ.. TheThe pressurizedpressurized tunnelstunnels areare obviouslyobviously complicatedcomplicated andand expensive.expensive.

106 ExampleExample 7.77.7 ModelModel DesignDesign ConditionsConditions andand PredictedPredicted PrototypePrototype PerformancePerformance

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

107 ExampleExample 7.77.7 SolutionSolution1/21/2

™The Reynolds numbers in model and prototype are the same. Thus, ρ V ρV m ml m = l µm µ

VVm=V,=V, llm// ll=1/10=1/10 ρ µ V µ m = m l = 10 m ρ µ Vm l m µ

TheThe samesame fluidfluid withwith ρρm==ρρ andand μμm==μμ cannotcannot bebe usedused ifif ReynoldsReynolds numbernumber similaritysimilarity isis toto bebe maintained.maintained. 108 ExampleExample 7.77.7 SolutionSolution2/22/2

WeWe assumeassume thatthat anan increasincreasee inin pressurepressure doesdoes notnot significantlysignificantly changechange thethe viscosityviscosity soso thatthat thethe requiredrequired increaseincrease inin densitydensity isis givengiven byby thethe relationshiprelationship ρ p ρ m = 10 For ideal gas m = m = 10 ρ p ρ D D = m D = ... 1 2 2 1 2 2 2 ρV l 2 ρmVm l m

109 FlowFlow AroundAround ImmersedImmersed BodiesBodies atat HighHigh ReynoldsReynolds NumberNumber 1/3 ™™Unfortunately,Unfortunately, inin manymany situasituationstions thethe flowflow characteristicscharacteristics areare notnot strongstrong influencedinfluenced byby thethe ReynoldsReynolds numbernumber overover thethe operatingoperating rangerange ofof interest.interest. ™™ConsiderConsider thethe variationvariation inin thethe dragdrag coefficientcoefficient withwith thethe ReynoldsReynolds numbernumber forfor aa smoothsmooth spheresphere ofof diameterdiameter dd placedplaced inin aa uniformuniform streamstream withwith approachapproach velocity,velocity, V.V. ™™AtAt highhigh ReynoldsReynolds numbersnumbers thethe dragdrag isis oftenoften essentiallyessentially independentindependent ofof thethe ReynoldsReynolds number.number.

110 FlowFlow AroundAround ImmersedImmersed BodiesBodies atat HighHigh ReynoldsReynolds NumberNumber 2/3

The effect of Reynolds number on the drag coefficient, CD for a 2 smooth sphere with CD = D/ ½ AρV , where A is the projected area of sphere, πd2/4. 111 FlowFlow AroundAround ImmersedImmersed BodiesBodies atat HighHigh ReynoldsReynolds NumberNumber 3/3 ™™ForFor problemsproblems involvinginvolving highhigh velocitiesvelocities inin whichwhich thethe MachMach numbernumber isis greatergreater thanthan aboutabout 0.3,0.3, thethe influenceinfluence ofof compressibility,compressibility, andand thereforetherefore thethe MachMach numbernumber (or(or CauchyCauchy number),number), becomesbecomes significant.significant. ™™InIn thisthis casecase completecomplete similaritysimilarity requiresrequires notnot onlyonly geometricgeometric andand ReynoldsReynolds numbernumber similaritysimilarity butbut alsoalso MachMach numbernumber similaritysimilarity soso thatthat V V c υ m = ⇒ = l m cm c cm υ m l

112 FlowFlow withwith aa FreeFree SurfaceSurface 1/81/8

™™ThisThis typetype ofof flowflow includesincludes flowflow inin canals,canals, rivers,rivers, spillways,spillways, andand stillingstilling basics,basics, asas wellwell asas flowflow aroundaround ship.ship. ™™ForFor thisthis classclass ofof problems,problems, gravitational,gravitational, inertialinertial forces,forces, andand surfacesurface tensiontension areare importantimportant and,and, therefore,therefore, thethe FroudeFroude numbernumber andand WeberWeber numbernumber becomebecome importantimportant similaritysimilarity parameters.parameters. ™™SinceSince therethere isis aa freefree surfacesurface withwith aa liquidliquid--airair interface,interface, forcesforces duedue toto surfacesurface tensiontension maymay bebe significant,significant, andand thethe WeberWeber numbernumber becomesbecomes anotheranother similaritysimilarity parameterparameter thatthat needsneeds toto bebe consideredconsidered alongalong withwith thethe ReynoldsReynolds number.number.

113 FlowFlow withwith aa FreeFree SurfaceSurface 2/82/8

™™AA generalgeneral formulationformulation forfor thesethese problemsproblems isis ⎛ ε ρV V ρV2 ⎞ Dependent pi term= φ⎜ li , , l , , l ⎟ ⎜ µ σ ⎟ ⎝ l l gl ⎠ Where l is some characteristic length of the system and li represents other pertinent lengths, ε/ l is the relative roughness of the surface, and ρVl/µ is the Reynolds number. ™™ToTo meetmeet thethe requirementrequirement ofof FroudeFroude numbernumber similaritysimilarity g = g Vm V m V = m = l m = λ l gml m gl V l

114 FlowFlow withwith aa FreeFree SurfaceSurface 3/83/8

™™ToTo meetmeet thethe requirementrequirement ofof ReynoldsReynolds numbernumber andand FroudeFroude numbernumber similaritysimilarity ρ V ρV V µ ρ υ m ml m = l ⇒ m = m l = m l µm µ V µ ρm l m υ l m υ m = (λ )3/ 2 υ l The working fluid for the prototype is normally either freshwater or seawater and the length scale is small. υ / υ = (λ )3/ 2 It is virtually impossible to satisfy m l , so models involving free-surface flows are usually distorted.

115 FlowFlow withwith aa FreeFree SurfaceSurface 4/84/8

™™TheThe problemproblem isis furtherfurther complicatedcomplicated ifif anan attemptattempt isis mademade toto modelmodel surfacesurface tensiontension effecteffects,s, sincesince thisthis requiresrequires thethe equalityequality ofof WeberWeber numbers,numbers, whichwhich leadsleads toto thethe conditioncondition

ρ V 2 ρV2 σ /ρ V 2 m m l m = l ⇒ m m = m l m = ()λ 2 2 l σm σ σ /ρ V l For the kinematic surface tension σ/ρ. It is evident that the same fluid cannot be used in model and prototype if we are to have similitude with respect to surface tension effects for λ ≠ 1. l

116 FlowFlow withwith aa FreeFree SurfaceSurface 5/85/8

™™Fortunately,Fortunately, inin manymany problemsproblems involvinginvolving freefree--surfacesurface flows,flows, bothboth surfacesurface tensiontension andand viscousviscous effecteffect areare smallsmall andand consequentlyconsequently strictstrict adherenceadherence toto WeberWeber andand ReynoldsReynolds numbernumber similaritysimilarity isis notnot required.required. ™™Certainly,Certainly, surfacesurface tensiontension isis notnot importantimportant inin largelarge hydraulichydraulic structuresstructures andand rivers.rivers. ™™OurOur onlyonly concernconcern wouldwould bebe ifif inin aa modelmodel thethe depthsdepths werewere reducedreduced. toto thethe pointpoint wherewhere surfacesurface tensiontension becomesbecomes anan importantimportant factor.factor. HowHow toto overcomeovercome thisthis problemproblem ?? 117 FlowFlow withwith aa FreeFree SurfaceSurface 6/86/8

™™DifferentDifferent horizontalhorizontal andand verticalvertical lengthlength scales,scales, whichwhich introduceintroduce geometricgeometric distortion,distortion, areare oftenoften usedused toto eliminateeliminate surfacesurface tensiontension effectseffects inin thethe model.model. ™™AlthoughAlthough thisthis approachapproach eliminateseliminates surfacesurface tensiontension effectseffects inin thethe model,model, itit introducesintroduces geometricgeometric distortiondistortion thatthat mustmust bebe accountedaccounted forfor empirically,empirically, usuallyusually byby increasingincreasing thethe modelmodel surfacesurface roughness.roughness.

))VerificationVerification. testtest inin thethe modelmodel mustmust bebe made.made. ModelModel roughnessroughness cancan bebe adjustedadjusted toto givegive satisfactorysatisfactory agreementagreement betweenbetween modelmodel andand prototype.prototype.

118 FlowFlow withwith aa FreeFree SurfaceSurface 7/87/8

™™ForFor largelarge hydraulichydraulic structuresstructures,, suchsuch asas damdam spillways,spillways, thethe ReynoldsReynolds numbersnumbers areare largelarge soso thatthat viscousviscous forcesforces areare smallsmall inin comparisoncomparison toto thethe forcforcee duedue toto gravitygravity andand inertia.inertia. ™™InIn thisthis casecase ReynoldsReynolds numbernumber similaritysimilarity isis notnot maintainedmaintained andand modelsmodels areare designeddesigned onon thethe basisbasis ofof FroudeFroude numbernumber similarity.similarity.

119 FlowFlow withwith aa FreeFree SurfaceSurface 8/88/8

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

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. 120 ExampleExample 7.87.8 FroudeFroude NumberNumber SimilaritySimilarity

™ A certain spillway for a dam is 20 m wide and is designed to carry 125 m3/s at flood stage. A 1:15 model is constructed to study the flow characteristics through the spillway. Determine the required model width and flowrate. What operating time for the model corresponds to 1 24-hr period in the prototype? The effects of surface tension and viscosity are to be neglected.

121 ExampleExample 7.87.8 SolutionSolution1/21/2

™The width, wm, of the model of spillway is obtained from the length scale w 1 20m m = λ = w = = 1.33m w l 15 m 15 ™With the neglect of surface tension and viscosity, the dynamic similarity will be achieved if the Froude numbers are equal between model and prototype

Vm V gm = g V = m = l m = λ g g l ml m l V l 122 ExampleExample 7.87.8 SolutionSolution2/22/2

™The flowrate 2 Q V A ⎛ ⎞ m = m m = l m ⎜ l m ⎟ Q VA l ⎝ l ⎠ Q = (λ )5/ 2 Q = ... = 0.143m3 /s m l

V t t V = l m m = l m = λ l Vm l m t t Vm l

123 SimilitudeSimilitude BasedBased onon GoverningGoverning DifferentialDifferential EquationsEquations 1/5

™™ForFor aa steadysteady incompressibleincompressible twotwo--dimensionaldimensional flowflow ofof aa NewtonianNewtonian fluidfluid withwith constantconstant viscosity.viscosity. DDTheThe massmass conservationconservation equationequation isis ∂u ∂v + = 0 Has dimensions of 1/time. ∂x ∂y Has dimensions of 1/time. DDTheThe NavierNavier--StokesStokes equationsequations areare ⎛ ∂u ∂u ⎞ ∂p ⎛ ∂ 2u ∂ 2u ⎞ ρ⎜ u + v ⎟ = − + µ⎜ + ⎟ ⎜ ⎟ ⎜ 2 2 ⎟ Has dimensions of ⎝ ∂x ∂y ⎠ ∂x ⎝ ∂x ∂y ⎠ force/volume ⎛ ∂v ∂v ⎞ ∂p ⎛ ∂ 2 v ∂ 2 v ⎞ ρ⎜ u + v ⎟ = − − ρg + µ⎜ + ⎟ ⎜ ⎟ ⎜ 2 2 ⎟ ⎝ ∂x ∂y ⎠ ∂y ⎝ ∂x ∂y ⎠ 124 SimilitudeSimilitude BasedBased onon GoverningGoverning DifferentialDifferential EquationsEquations 2/5

™™HowHow toto nonnon--dimensionalizedimensionalize thesethese equationsequations ?? x y u v p t x* = y* = u* = v* = p* = t* = l l V V p0 τ ∂u V ∂u* ∂ 2u ∂ ⎛ ∂u ⎞ V ∂ 2u* = = = * 2 ⎜ ⎟ 2 *2 ∂x l ∂x ∂x ∂x ⎝ ∂x ⎠ l ∂x TheThe massmass conservationconservation equationequation ∂u∗ ∂v∗ + = 0 ∂x∗ ∂y∗

125 SimilitudeSimilitude BasedBased onon GoverningGoverning DifferentialDifferential EquationsEquations 3/5

TheThe NavierNavier--StokesStokes equationsequations Reynolds number

∗ ∗ ∗ ⎡ ⎤ ∗ ⎛ 2 ∗ 2 ∗ ⎞ ⎡ l ⎤ ∂u ∗ ∂u ∗ ∂u p0 ∂p ⎡ µ ⎤⎜ ∂ u ∂ u ⎟ + u + v = −⎢ ⎥ + ⎢ ⎥ 2 + 2 ⎢ τV ⎥ ∗ ∗ ∗ 2 ∗ ρV ⎜ ∗ ∗ ⎟ ⎣ ⎦ ∂t ∂x ∂y ⎣⎢ρV ⎦⎥ ∂x ⎣ l ⎦⎝ ∂x ∂y ⎠ ∗ ∗ ∗ ⎡ ⎤ ∗ ⎛ 2 ∗ 2 ∗ ⎞ ⎡ l ⎤ ∂v ∗ ∂v ∗ ∂v p0 ∂p ⎡ gl ⎤ ⎡ µ ⎤⎜ ∂ v ∂ v ⎟ + u + v = −⎢ ⎥ − + ⎢ ⎥ 2 + 2 ⎢ τV ⎥ ∗ ∗ ∗ 2 ∗ ⎢ 2 ⎥ ρV ⎜ ∗ ∗ ⎟ ⎣ ⎦ ∂t ∂x ∂y ⎣⎢ρV ⎦⎥ ∂y ⎣ V ⎦ ⎣ l ⎦⎝ ∂x ∂y ⎠

Strouhal number Euler number Euler number Reciprocal of the square of the Froude number

126 SimilitudeSimilitude BasedBased onon GoverningGoverning DifferentialDifferential EquationsEquations 4/5

™™FromFrom thesethese equationsequations itit followsfollows thatthat ifif twotwo systemssystems areare governedgoverned byby thesethese equations,equations, thenthen thethe solutionssolutions (in(in termsterms ofof u*,v*,p*,x*,y*,u*,v*,p*,x*,y*, andand t*)t*) willwill bebe thethe samesame ifif thethe fourfour parametersparameters areare equalequal forfor thethe twotwo systems.systems. ™™TheThe twotwo systemssystems willwill bebe dynamicallydynamically similar.similar. OfOf course,course, boundaryboundary andand initialinitial conditionsconditions expressedexpressed inin dimensionlessdimensionless formform mustmust alsoalso bebe equalequal forfor thethe twotwo systems,systems, andand thisthis willwill requirerequire completecomplete geometricgeometric similarity.similarity.

127 SimilitudeSimilitude BasedBased onon GoverningGoverning DifferentialDifferential EquationsEquations 5/5

™™TheseThese areare thethe samesame similaritysimilarity requirementsrequirements thatthat wouldwould bebe determineddetermined byby aa dimensionaldimensional analysisanalysis ifif thethe samesame variablesvariables werewere considered.considered. TheseThese variablesvariables appearappear naturallynaturally inin thethe equations.equations. ™™AllAll thethe commoncommon dimensionlessdimensionless groupsgroups thatthat wewe previouslypreviously developeddeveloped byby usingusing dimensionaldimensional analysisanalysis appearappear inin thethe governinggoverning equationsequations thatthat descdescriberibe fluidfluid motionmotion whenwhen thesethese equationsequations areare expressedexpressed inin termterm ofof dimensionlessdimensionless variables.variables. ™™TheThe useuse ofof governinggoverning equationsequations toto obtainobtain similaritysimilarity lawslaws providesprovides anan alternativealternative toto conventionalconventional dimensionaldimensional analysis.analysis.

128