1

FacultyofEngineeringDepartmentofMechanicalEngineering

MCG3143:BiofluidmechanicsLecturesnotes

Summer2015

MarianneFenech

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CHAPTER 1 ......................................................................................................................................................... 5 

CARDIOVASCULAR PHYSIOLOGY ......................................................................................................................... 5 

SUGGESTED EXERCISES FROM THE TEXTBOOK: ...................................................................................................................... 5 

CHAPTER 2 ......................................................................................................................................................... 6 

FUNDAMENTALS OF FLUID MECHANICS ............................................................................................................. 6 

1. INTRINSIC PROPERTIES OF FLUID ..................................................................................................................................... 6 2 CONSERVATION LAWS ................................................................................................................................................... 6 

2.1. Mathematical tools ........................................................................................................................................ 6 2.2 Mass Conservation ........................................................................................................................................ 13 2.3. Conservation of momentum ........................................................................................................................ 18 2.4 form of fluid motions equations .................................................................................................................... 25 2.5. Dimensional analysis .................................................................................................................................... 27 2.6 Energy conservations & Bioheat Equation of Mammalian Tissue ................................................................ 31 

3 EXERCISES ................................................................................................................................................................ 39 4. SOLUTIONS .............................................................................................................................................................. 43 5. REFERENCES ............................................................................................................................................................ 44 

CHAPTER 3 ........................................................................................................................................................ 45 

MATHEMATICAL SOLUTIONS FOR BIOFLUID PROBLEMS .................................................................................... 45 

1 HOW TO SOLVE A PROBLEM? ....................................................................................................................................... 46 2 BOUNDARY CONDITIONS ............................................................................................................................................. 46 3 MATHEMATICAL SOLUTIONS FOR BIOFLUID PROBLEMS ...................................................................................................... 46 

3.1 Shear stress on arterial endothelial cells ....................................................................................................... 46 3.2 NS in a pipe ................................................................................................................................................... 49 3.2 1 Validity of the Hagen‐Poisseuille relationship in the cardiovascular system ............................................. 51 3.3 Pulsatile flow ................................................................................................................................................. 52 3.3.1 Effect of pulsatility ..................................................................................................................................... 52 3.3.2 Wormersley solution .................................................................................................................................. 53 

4. EXERCISES ............................................................................................................................................................... 59 5. SOLUTIONS .............................................................................................................................................................. 65 6. ASSIGNEMENT ......................................................................................................................................................... 65 7. REFERENCES ............................................................................................................................................................ 67 

CHAPTER 4 ........................................................................................................................................................ 68 

COMPUTATIONAL FLUID DYNAMIC (CFD) AND MEASUREMENT TECHNIQUES IN BIOMEDICAL ........................... 68 

1. COMPUTATIONAL FLUID DYNAMICS .............................................................................................................................. 68 2. FLOW MEASUREMENT IN THE CARDIOVASCULAR SYSTEM .................................................................................................. 68 3. EXERCICES ............................................................................................................................................................... 68 4. PROJECT : MEDICAL DEVICE DESIGN USING CFD ............................................................................................................. 69 

CHAPTER 5 ........................................................................................................................................................ 70 

FLOW OVER IMMERSED BODY (INCOMPRESSIBLE) ............................................................................................ 70 

1 GENERAL EXTERNAL FLOW CHARACTERISTICS .................................................................................................................. 70 2 LIFT AND DRAG CONCEPT ............................................................................................................................................ 71 

2.1 Definitions ..................................................................................................................................................... 71 2.2 Drag for different shapes .............................................................................................................................. 72 

2.3 Drag coefficient, for a sphere in stokes flow ( 1Re ) ............................................................................. 72 

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2.4 Transport of micro‐particles .......................................................................................................................... 73 3 CHARACTERISTICS OF FLOW PAST AN OBJECT .................................................................................................................. 74 4 BOUNDARY LAYER CHARACTERISTICS .............................................................................................................................. 76 

4.1 Boundary Layer Structure and Thickness on a Flat Plate .............................................................................. 76 4.2 Boundary layer thickness .............................................................................................................................. 76 4.3 Momentum‐Integral Boundary Layer Equation for a Flat Plate .................................................................... 77 4.4 Prandtl/Blasius Boundary Layer Solution ...................................................................................................... 80 

5 TURBULENT BOUNDARY LAYER...................................................................................................................................... 84 6. PRESSURE GRADIENT EFFECT ON FLOW SEPARATION ......................................................................................................... 85 7. EXERCISES ............................................................................................................................................................... 88 8. SOLUTIONS .............................................................................................................................................................. 94 9. REFERENCES ............................................................................................................................................................ 96 

CHAPTER 6 ........................................................................................................................................................ 97 

RHEOLOGY OF BLOOD ....................................................................................................................................... 97 

1. RHEOLOGY OF BLOOD AND NON‐NEWTONIAN EQUATIONS ................................................................................................ 97 2. EXERCISES ............................................................................................................................................................... 97 3. SOLUTIONS ............................................................................................................................................................ 101 

CHAPTER 7 ...................................................................................................................................................... 102 

INTRODUCTION TO FLUID MACHINERY ............................................................................................................ 102 

1. INTRODUCTION TO FLUID MACHINERY ......................................................................................................................... 102 2. EXERCISES ............................................................................................................................................................. 102 3. SOLUTIONS ............................................................................................................................................................ 108 

FORMULA ....................................................................................................................................................... 111 

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Chapter1

CardiovascularPhysiologyTextbook: K.B. Chandran et al. BiofluidMechanics:TheHumanCirculation.Taylor&Francis2dedition.:Chapter3p69‐108

SuggestedExercisesfromthetextbook:3.1;3.2;3.3;3.4;3.5;3.6

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Chapter2

Fundamentalsoffluidmechanics

1.IntrinsicpropertiesoffluidTextbook: K.B. Chandran et al. Biofluid mechanics: the human circulation. Taylor&Francis 2dedition.:Chapter1pp4‐8

2Conservationlaws2.1.Mathematicaltools2.1.1.DeloperatorDell is a vector operator (or Nabla operator). This operator makes the equation easier tounderstandandwrite.Calculationsarejustlikewithvectors,exceptthattheyactuallyoperateonwhatfollows(notjustmultipliesthem).

zyx e

ze

ye

x

i)Gradient:DelappliedtoascalarGradientofscalarisavector

x

p

ppgrad

y

p

z

p

Gradientisthevectorfieldtodescribeascalarfield.

7

Direction:directionofsteepestascentMagnitude:rateofascentScalarfield:__________Gradient:__________

ii)DivergenceScalarproductofDeloperator(orNabla)andavector:

z

u

y

u

x

uu zyx

Examplesofcalculus:Thephysicalmeaning isharder tounderstand than for thegradient.For themoment, rememberthatdivergenceofthevelocitydescribesareductionoranexpansionofVolume.Wewillprovethatlater.

8

Examples:

iii)CurlVectorproductofDeloperatorandavector:

z

u

y

u yz

uucurl

x

u

z

u zx

y

u

x

uxy

CurlisusedtodescribearotationExampleTherotationcouldbedonebythedifferenceof2adjacentvectors:Uy=x Ux=‐yorifthesuccessivevectorsdonothavethesamedirection:Uy=x2 Ux=y1/2

9

iv)Laplacienofascalar

2

2

2

2

2

22.

z

S

y

S

x

SSSS

ofavector:

2

2

2

2

2

2

z

u

y

u

x

u xxx

uu 2

2

2

2

2

2

2

z

u

y

u

x

u yyy

2

2

2

2

2

2

z

u

y

u

x

u zzz

iv)RulesE.g.scalarfunctionsofposition;A,Bvectorfunctionsofpositions

Gradient fg f g g f

A B A B B A A B B A

Divergence fA f A A f

A B B A A B

Curl fA f A A f

A B B A A B A B B A

Secondderivatives

2 2 2

22 2 2x y z

Laplacianoperator

0curl grad

0A div curl A

2.1.2MaterialderivativeThisderivativerepresentsthetimederivationofascalarorofavector.The equation applies to a fluid element which is a small “blob” of fluid that contains the samematerialatalltimesduringfluidmovement.Fluidelementsaredeformedastheymovebuttheyarenotbrokenup.Note that themassof a fluid element is constant.Material derivative considers aproperty(e.g. temperature,density,velocitycomponent)of the fluidelement. Ingeneral, thiswilldependonthetime,t,andontheposition(x,y,z)ofthefluidelementatthattime.So

10

γ=γ(x,y,z,t)=γ(r,t)Example:Tailpipe:Followthemassandmeasurehowthepropertieschange

t

z

zt

y

yt

x

xtdt

d

Bydefinition,thevelocityofthefluidelementis

t

z

t

y

t

xuuuU zyx ,,),,(

Hence

zu

yu

xu

tdt

dzyx

d/dtistherateofchangemovingwiththefluidelement.

t / istherateofchangeatafixedpointinspace.In fluiddynamics, thetimerateofchange fora fluidelement isusuallydenotedbyD/Dt.Thus inCartesianexpansion

zu

yu

xu

tDt

Dzyx

Ingeneralform

.

utDt

D

Example:forthetemperature(scalar):

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Materialderivativeofavector:

z

uu

y

uu

x

uu

t

uu

zu

yu

xu

t

u xz

xy

xx

xxzyx

x

.

uut

u

Dt

uD

.

yzyxy u

zu

yu

xu

t

u.

zzyx

z uz

uy

ux

ut

u.

Example:Velocitywhenyouarewalkingonamountain:

Uut

U

Dt

DU.

➀➁ ➀Howthethingsarechangingatafixedlocation➁Howthingschangeastheymovewiththefluid

2.1.3EinsteinnotationTheEinsteinnotationisanotational.ItwasintroducedbyAlbertEinsteinin1916.Accordingtothisconvention,whenanindexvariableappearstwiceinasingleterm,itimpliesthatwearesummingoverallofitspossiblevalues.Intypicalapplications,theindexvaluesare1,2,31,2,3representingthethreedimensionsofphysicalEuclideanspace(x,y,z)Examples

Amatrixoraijor

333231

232221

131211

aaa

aaa

aaa

xavectororxior(x1,x2,x3)forthreedimensionsofthespace

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Uavectororuior(u1,u2,u3)Einsteinnotation:aii=

i

i

x

uUdiv )(

xiui=aijxj=

2.1.4.Kronecker’sdeltaTheKroneckerisafunctionthatreturns1iftheindexisequaland0.Otherwise:

Examples:

ji

j

i

x

u

Fx

uA ji

j

iij

13

2.2MassConservation2.2.1MassConservation:Integralform

Considerafixedvolume .Fluidmovesintooroutof acrossthesurfaceA.

dA is the element of surface, with itsmagnitude denoting the area of theelement and direction of the normal

pointingoutof

U isthevelocityvectoratthepositionoftheelement.

Note:thecomponentofU parallelto dA denotingthetransferfluidoutof

dAU . istheaverageofmassfluxthoughthesurfaceelementleavingtheVolume .Where isthefluiddensity.

So,dAU . istherateoflossmassfrom .

Notethatisnegativeifthemassisincreasingin .

Alsod isthetotalmassin

thetotalconservationmassis:‐(rateofmassin) rateofchangeor + ofmass = 0

Rateofmassout in

0. ddt

ddAU

SincetheVolumeisfixedwecanwrite(Leibniz’sformula)

0.

dt

dAU (2.1)

14Thisistheconservationmassinintegralform.

Rememberthere isbalancebetweenaccumulatingmass inside and lossofmassacrosstheboundaryoftheVolume.Wewouldliketoexpressthefluidequationindifferentialformi.e.intermsofthederivate

ofetcU ,,

forexampleetc

t

v

x

u

t

,,

Wewanttofindthederivationequationatafixedpointinspace.

2.2.2Massconservation:differentialformSmallVolumeinCartesiancoordinates:

WewanttowritedAU . thefluxforeachface(6times!)

RecallTaylorseries:

...2

2

2

2

af

af

ffaa

a

Rightface:Position 2

xx

Taking 2

xx

and xa

...2

22

xx

x

fff

x

xx

x

Finally,since

x

f

x

xx

x

fx

x

f

2

2

...22

x

x

fff

xx

x

15LetuswriteUandρforthisfaceRightface:

Position 2

xx

...22

x

x

uuu

xx

x

...22

x

xxx

x

zydA

()22

ozyx

x

uzy

x

xuzyuudA

()

2

1ozyx

x

uzyu

Leftface:

Position 2

xx

...22

x

x

uuu

xx

x

...22

x

xxx

x

zydA

()

2

1ozyx

x

uzyuudA

Forthe6faces:‐x

x

‐y

y

‐z

z

Sum:dAU . =

zyx

z

w

y

v

x

u

WegetthefirstpartofthemassconservationequationThesecondpartisveryeasy:

zyxt

dt

Puttogether:

16

0

y

w

y

v

x

u

t

and 0

Ut

Wegetouranswer!Howdoweexpressthatwiththematerialderivative?1‐Expand…2‐Sort…3‐Wegetanotherformofthemassconservation:

0 UDt

D

(2.4)Divergence:

Weknowthat Dt

D

givesthedensitychangesinafluidelement.Butwhatabout U ?

Tohaveanidea,wecanconsideranuncompressiblefluid0

Dt

D

becausethedensityoffluiddoesnotchange.Inthiscondition

U =0foruncompressiblefluid(Divergenceofthevelocity)We saw that divergence refers to the Volume change of an element of fluid (with fixedmass).Wecanrewritetheequationoftheconservationofmassas:

0Dt

Dm

Itcouldalsobewritten:

0

Dt

D

Dt

D

Dt

D

Dt

Dm

17

Writing:Massconservationform1= (massconservationform2)weget:

0

Dt

D

Dt

DU

Dt

D

Dt

DU

1

(2.5)

It means that the normalized rate of change of Volume occupied by the lump of fluid isconnectedwiththevelocity.divUreflectstheVolumechange!

2.2.3Summaryofdifferentialstatements(2.1)AccumulationinafixedVolume+netfluxesoutacrossthesurfacecontrol=0(2.3)AccumulationinafixedVolume+netfluxesacrossthesurface=0(2.4)Rateofdensitychangefollowingafluidelement+changeofVolumeofthefluidelement=0! All three forms are equivalent. They give the same information fromdifferent points ofview!

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2.3.Conservationofmomentum2.3.1IntegralformTheprincipleofconservationofmomentumwasinitiallyformulatedfromNewton’ssecondlawofmotion,whichstatesthat“thesumoftheforces(ΣF)actingonanobjectisequaltoitsmass(m)timesitsacceleration(a)”

amF Rewriting(a)asDU/dtandbringingthemass(m)inthedifferential:

dt

UdmF

dt

Udm

isnowthetimerateofchangeofthemomentum(mU).ForacontrolVolume,thetimerateofchangeofthemomentumoftheVolumeisthesumof:TimerateofchangeintheVolume+Changethroughthesurface

dAUUdUt

.

(Demo,yourbookfluidIchap4)

F is the sum of forces acting on the Volume (body forces) and forces acting on thesurfaceofthesurfacecontrol(surfaceforces)

dAtdB

Exampleofbodyforces:

Forceofgravity g Accelerationduetothechoiceofthereferenceframe–Example:Forcesduetotherotation(coriolis)ElectromagneticforceSurfacesforces:Pressure(p)Viscousforce(τ)

19Finallywegettheconservationofmomentuminintegralform:

dAtdBdAUUdUt

.(2.6)

2.3.2DifferentialformWe examine the fluid mechanical equivalent of Newton’s second law, ΣF=ma, called themomentumequation.Sincewearefollowingafluidelementoffixedmassm,m=ρ.Volume,wecanwrite:Σf=ρaWherefaretheforcesperunitofVolume(f=F/Volume).

Aswefollowafluidelement Dt

uDa

(Materialderivative)RecallinCartesiancoordinate:

z

uu

y

uu

x

uu

t

uu

zu

yu

xu

t

u xz

xy

xx

xxzyx

x

.

uut

u

Dt

uD

.

yzyxy u

zu

yu

xu

t

u.

zzyx

z uz

uy

ux

ut

u.

Nowwehavetocheckforcesactingonthefluidelement.Thisincludes:BodyforceHydrostaticpressureViscousforceTwotypesofforcesareconsidered:surfaceforcesactingonthesurfaceoftheunitelementorVolumeandbodyforceswhicharedistributedthroughouttheelementorVolume.

a‐BodyforceBodyforcecouldbeForceduetothegravity:f=ρg

b‐HydrostaticpressureInthecoordinatedirectionswehavebodyforces

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UnitVolumeoffluidshowingthatpressureforcesactnormaltotheVolumeTaylorexpansiongivesus:

Pressurerightface:...

22

x

x

PPP

xx

x

Forcerightface:

2

x

x

PPzy

x

Pressureleftface:...

22

x

x

PPP

xxx

x

Forceleftface:

2

x

x

PPzy

x

Netforceinthexdirection:

zyxx

Px

x

PPzy

x

x

PPzy

xx

22

So,netforceperunitofVolumeinx‐directionis x

P

Similarly,wecandemonstratethatthenetforceperunitofVolumeinyandzdirectionsare

respectively: y

P

and z

P

VectorformoftheforceperunitVolumeis:

Pez

Pe

y

Pe

x

Pf zyx

δx

δy

δz

P

21c‐ViscousforcesForcesactingonaunitarea,A,arereferredtoasstresses.Therearestressesnormaltothesurfaceandstressestangentialtothesurface.Viscousstressesopposerelativemovementsbetweenneighboringfluidparticles.Thetangential,shearstresscouldbevisualizedbyconsideringtwoparallelplatesseparatedbyafluid.Eachplatehasadifferentvelocity.Becausethefluiddoesnot‘slip’onthesurfaceplate,thefluidelementwillbesubjectedtoashearStress.

The normal shear stress due to viscosity is more difficult to visualize. It is acting whenviscous fluid (like honey) does not fall with gravitational acceleration, because of theviscousinteraction.Thesenormalviscousstressescanbeviewedasduetothe“stickyness”ofthefluid.

InrelationtoourunitVolumeintheCartesiancoordinatesystem,thenormalstresstothex‐directionisdefinedbyτxx=Fxx/A

Velocity = u

Velocity is proportional to z

u2

u1

Honey

22Thestressactingtangentialtothex‐normalbutinthey‐directionisτxy=Fxy/AThestressactingtangentialtothex‐normalbutinthez‐directionisτxz=Fxz/AThepressureandshearstressesforallthreecoordinatedirectionsareshownintheFigure:

UnitVolumeoffluidshowingpressureandshearstressesinthethreecoordinatedirections.

ij=viscousstressinthejdirectiononthefacenormaltoiaxisis:

zzyzxz

zyyyxy

zxyxxx

Note that under conditions of equilibrium (τxy= τyx, τxz=τzx, τyz=τzy), we can use thesurfaceforcesactingontheunitelementoffluid.Indevelopingtheequationsofmotionforafluid,weusethedifferentialVolumeagainandconsiderthatstressesvaryfrompointtopointinthefluid.WethusexpressthestressesonthevariousfacesofthedifferentialVolumeintermsofthestressactingononefaceoftheelementandthecorrespondingchangeinthestressforagivencoordinatedirection:

xx

xy

xz

zz

yy

yz

yx

zxzy

23

LookingattheunitVolumeoffluidshowingshearstresses(τ)inthexcoordinatedirectiononly,wehavethesumoftheforcesgivenby

zyxzyx

F zxyxxxx

Thesumoftheforcesfortheyandzdirectionsfollowssimilarly

zyxzyx

F yzyyyxy

zyxzyx

F zzzyzxz

ThenetviscousforceperunitVolumeis:

zyxviscousfx zxyxxx

,

zyxviscousfy yzyyxy

,

zyxviscousfz zzyzxz

,

Orwecanwrite:

j

ijij x

viscousf

x y

z

xxxx

x dx δyδz

1 2

x

x

x dx δyδz

1 2

-

zxzx

zdz δxδy

1

2

yx

y

y dy δxδz

1 2

y

y

y dy δxδz

1 2

-

zxzx

zdz δxδy

1

2-

24Where,arepeatedindexmeansthatweaddoverthatindex.(Einsteinnotation)Ingeneral, thestress issecondordertensor.Therateofdeformationof the fluid isalsoasecond order tensor (also called rate of strain tensor).We expect to have a relationshipbetweentheappliedstressandtheratedeformationofafluidelement.CaseofNewtonianfluid:TheNewtonianapproximationassumesthatthestressislinearlydependentontherateofstrain,whereproportionalcoefficientsarethecharacteristicofthesubstance.TherelationshipbetweenviscousstressandthedeformationrateofthefluidelementforaNewtonianfluidisgivenby:

Ux

uj

x

uij

ij

iij

Wherei,jstandforx,y,z.μistheviscosityandλiscalledsecondviscosity,or‘coefficientof

bulkviscosity’orasthe‘Lamé’constantasinlinearelastictheory. ijisanextrasymbol,it

isausefulnotation,itmeans:

ji

ji

ij

ij

0

1

Notethatμ,theviscosityiseasiertogetexperimentallythanλbecauseλappearscoupled

with U ,sofortheimpressiblefluid U =0!

Sothedevelopedformof

Ux

uj

x

uij

ij

iij

willbe:

z

u

y

u

x

u

x

u

x

u zyxxx

z

u

y

u

x

u

y

v zyxyy

2

z

u

y

u

x

u

z

w zyxzz

2

x

v

y

uyxxy

y

w

z

vzyyz

z

u

x

wxzzx

ThenwecanwritethenetviscousforceperunitVolumewiththeNewtonapproximation:

zyxviscousfx zxyxxx

,

MGC3143,M.Fenech.

25

z

u

x

w

zx

v

y

u

yz

u

y

u

x

u

x

u

xzyx

2

Andsimilarlyforthe2othercomponents

ForaNewtonianincompressiblefluid( U =0)andconstantviscosityweget:

2

2

2

2

2

2

,z

u

y

u

x

uviscousfx

Note: Blood does not have a constant viscosity at low stress. However, we can assumeNewtonian forhighshearratewhen theredbloodcellsaredisaggregate ina largearterywithoutdiseases.Butwehave tobe carefulwith that because shear stress is involved insomerelevantbiologicalphenomenaashaemolysis,arteriosclerosisformationorchangingshapeofendothelialcells.

2.4formoffluidmotionsequations2.4.1Generalform(Cauchyequations)

xbxzxyxxx fgzyxx

p

Dt

Du

ybyzyyyxy fgzyxy

p

Dt

Dv

zbzzzyzxz fgzyxz

p

Dt

Dw

Orinvectornotation:

bfgP

Dt

UD

(2.7)(i) (ii) (iii) (iv)(v)(i)m/Volume.a(ii)Pressureforce(iii)NetviscousforceperunitVolume(iv)NetweightforceperunitVolume(v)OtherbodyforceperunitVolumeNote,forthisgeneralcase,therearetoomanyunknownsforthenumberofequationsgiven!

2.4.2NavierStokesequationsNavier&Stokesequationsrefer to thecombinationofmassconservationandmomentumconservationforviscous,Newtonian(constantviscosity),anduncompressibleflow.Viscousflowwithconstantviscosityconservationofmomentumisfoundby:

MGC3143,M.Fenech.

26

2

2

2

2

2

2

z

u

y

u

x

ug

x

p

Dt

Dux

2

2

2

2

2

2

z

v

y

v

x

vg

y

p

Dt

Dvy

2

2

2

2

2

2

z

w

y

w

x

wg

z

p

Dt

Dwz

Orinvectornotation:

UgPDt

UD 2 (2.8)

Unknown:u,v,w,ρ,andwehave4equations(3formomentumand1forcontinuity)wecansolve given the initial and boundary condition! If the flow is compressible, then energyconservationneedstobesolvedatthesametimeascontinuityandconservationequations.Thisisbecausechangeinvelocitymayaffectthetemperature,andvise‐versa.These equations were named in honor of French mathematician, L.M.H. Navier (1758‐1836),andEnglishmathematician,SirG.G.Stokes(1819‐1903),whowereresponsible fortheirformulation.Becausetheseequationsaresecondorder,nonlinearpartialdifferentialequations,thereareonlyafewexactsolutionsavailable.The Navier‐Stokes equations are a set of second order, nonlinear partial differentialequations that are developed from the principal of conservation of mass (continuityequation)andfromtheconservationoflinearmomentum.Thefewexactsolutionstotheseequations thatexistare inverycloseagreementwithexperiments.Thesolutionsof theseequationsincludeturbulence,tornadoes,waves,boundarylayerandothercomplicatedfluidflowphenomena.

2.4.3Eulerequation:inviscidSoifthepressureforceonlyisinvolved,conservationofmomentumisdonebytheEULEREQUATION.Notethatassumeaninviscidfluid:

pDt

UD

Example:

MGC3143,M.Fenech.

27

2.4.4.StokesflowStokes flow is a type of fluid flowwhere inertial forces are small comparedwith viscousforces.TheReynoldsnumberislow,i.e.Re<1.Thisisatypicalsituationinflowswherethefluidvelocitiesareveryslow,theirviscositiestendtobeverylarge.Forthistypeofflow,theinertialforcesareassumedtobenegligibleandtheNavier‐StokesequationssimplifytogivetheStokesequations:Forthistypeofflow,theinertialforcesareassumedtobenegligibleandconservationofmomentumissimplifiedandisgivenby:

bfgP ij 0

In the common case of an incompressible Newtonian fluid, the Stokes equations are(momentumandcontinuity):

0

2

U

UgP

(2.9)Examples:‐Lubrification–Darcyequation

2.5.DimensionalanalysisDimensional analysis is a tool used to understand the properties of physical quantitiesindependentoftheunitsusedtomeasurethem.WhatdowegainbyusingDimensionalAnalysis? Anyconsistentsetofunitswillwork Wedon’thavetoconductanexperimentontherealsize Ourresultswillevenworkfordifferentfluids Ourresultsareuniversallyapplicable Wecanassesstherelativeimportanceoftermsinthemodelequations Itavoidsround‐offduetomanipulationswithlarge/smallnumbers

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Note:CFDcodeusesdimensionlessequations.

2.5.1Reynoldsnumber

The Reynolds number may be described as the ratio of inertial forces 2

2V to viscous

forces D

V

and, consequently, it quantifies the relative importance of these two types offorcesforgivenflowconditions.

VD

Re

Whereρisthefluiddensity,μisthefluiddynamicviscosity,Disthepipediameter,andVisthefluidvelocity.LaminarflowoccursinaflowenvironmentwhereRe<2000.Consequently,turbulentflowispresentifRe>4000;thetransitionrangeisbetweenthesecriticalvalues.TheRenumber isalsouseful forpredicting theentrance length inpipe flow.TheratioofentrancelengthXEandthepipediameterforlaminarsteadyflowisgivenby:

0.65XE D forRe<50

Re0.06XE D forlaminarflowRe>50

4/1E Re0.693X

D forturbulentflow

Mostof thebloodflowinthehumancirculation is laminar,havingaReof300or less.AnestimationofthetimeaverageReynoldsnumberReinthehumanaortaisabout1500.Thisvalueisbelowthecriticalvalue(2000).ButatthepeakflowrateduringthesystoletheRecanreach5000!However,theaortaisdistensible,sothecriticalRenumberdeterminedinarigid straight pipe is not applicable in this situation. In‐vivo experiments don’t showevidenceofsustainedturbulenceinthehumancirculation(inabsenceofdisease!).

2.5.2WomersleynumberTheWomersley number, or alpha parameter, is another dimensionless parameter. It is adimensionlessexpressionofthepulsatileflowfrequencyinrelationtoviscouseffects.i.e.itisthecomparisonbetweenunsteadyinertialforceandviscousforces.TheWomersleynumber,usuallydenotedα,canbewrittenas:

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r

risthevesselradiusωisthefundamentalfrequency,typicallytheheartrate,theunitmustberad/sρisthefluiddensityμthedynamicviscosityofthefluidWhenαissmall(1orless),itmeansthatthefrequencyofpulsationsissufficientlylowthataparabolicvelocityprofilehastimetodevelopduringeachcycle.Italsomeansthattheflowwill be very nearly in phase with the pressure gradient, and can be approximated byPoiseuille'slaw,usingtheinstantaneouspressuregradient.Itisaquasi‐steadyflow.Whenαis large (10 or more), it means the frequency of pulsations is sufficiently large that thevelocity profile is not parabolic. Inertial forces become more important and start to bedominate.Sometypicalvalueofα:Humanaortaα=20Canineaortaα=14Felineaortaα=8Rataortaα=3.

2.5.3SimilarityThe velocity field of a flow could be investigated using a larger or smaller model forconvenience. However, it must conserve hydrodynamic similarity (i.e. in order to obtainsimilaritybetweena flowinnatureandthesimulationmodel flow, theratiosofactuatingforceshavetobethesame:Reandαhavetobeconserved).

2.5.4DimensionalequationsBasedon theprinciples of dimensional analysis, the variablesU, p , x,y, z, and t couldbewrittenasafunctionofdimensionlessvariables:U*,p*,x*,y*,z*,andt*.Forexemple:u*=u/V∞ v*=v/V∞ w*=w/V∞x*=x/L y*=y/Lz*=z/Lt*=tV∞/LP*=P/(ρV∞2)orP*=PL/(μV∞)Note that depending on the flow characteristicwewant to emphasize, the dimensionlessvalueofPcouldbedifferent.Forexample:P*=P/(ρV∞2)willbeusedforflowwheretheinertialforceisdominatedP*=PL/(μV∞)willusedinflowwhereviscosityisdominatedExample:RewritingNS:2Dsteady

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IncompressibleTwo‐dimensionalNewtonianFluid

Dimensionvariableswillbesubstitutedintheequationtoobtainanequationwithphysicalquantitiesindependentoftheunits.u=u*.V∞ v=v*.V∞ w=w*.V∞x=x*.L y=y*.L z=z*.Lt=t*.V∞/LP=P*.(ρV∞2)Usingdimensionlessvariables:

LargeReWhen1/Reisneglected,wefindtheEulerequation.In Inviscid flow,wemust be careful because close to thewall in the boundary layer thesecondderivativeislargeandwecannotneglecttheeffectoftheviscosity.

LowReStokesfloworcreepingflowviscositydominate.Astheviscousforcedominates,canwejustkeeptheviscousterm?Toverifythatitislogicaltotakekeeptheviscousforcewhenwescalepressure,letustake:P*=P/(μV∞/L)Thenwecanshowthat,

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***

*Re 2UP

Dt

DU

Nowthefirst termbecomesnegligibleandweobtainthestokeequation indimensionlessform.Theflowisreversible!!!!(video)

NoteonWomersleynumberWhatabouttheWomersleynumberwhichisspecifictothepulsatileflowasflowinartery?TointroduceWomersleynumberinthepartialderivationequationwilltakealittlelonger.Youwilldothatinyourfirstassignment(withguidanceofcourse!)ToscaleexperimentationwithapulsatileflowyouneedtokeepthesameReynoldsnumberandthesameWomersleynumberasyoudidinthefirsttutorial.

2.6Energyconservations&BioheatEquationofMammalianTissue2.6.1FirstLawofthermodynamicsforaclosedsystemThefirstlawofthermodynamicsisusuallywrittenas:

WQem tot (I) (II) (III)Thevariationinenergy=Heattransfer+WorkdoneWhereQisheat;Wisworkandthetermetotincludesnumeroustypesofenergy:(I)TotalenergyInternalenergy CvT It is the formof storedenergywhich

can be directly influenced by a heattransfer.

Kineticenergy UU 2

1

Due to the velocity of the fluidparticle

Potentialenergy rg

Minus because g is pointing indirection of decreasing potentialenergy.risthepositionvector

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x

z

y (x,y,z

fx fx+dx

fy

fy+dy fz

fz+dzdxdz

dy

You can add other terms (electromagnetic, chemical, …) to the expression of energy.However,wewillusuallylimitourselvestothethreetermsabove.Note:Therelativemagnitudeofthethreecomponentsofenergyisoftenquitedifferent.Thestatementofaproblemshouldgiveaquickcluetowhichtypesoftermswillpredominate.Modestvelocitieswillresultinnegligiblechangesinkineticenergy.Similarly,smallchangesinelevationwillresultinnegligiblechangesinpotentialenergy.Toapply the firstprinciple,weassumethat there is thermodynamicequilibriumbetweentwostatesfollowinga fluidVolume.Wecanexpressthefirst lawundertheinstantaneoustimerateformandperunitVolumeas:

rgUUCvT

Dt

D

Dt

Detot

2

1 (2.6.1)

(II)RateofheatinputintotheelementConsider a cubic element; wewant the net heat transfer rate into it. The local heat fluxvector:fNetheatintotheelement:(fx‐fx+dx)dydz+(fy‐fy+dy)dxdz+(fz‐fz+dz)dxdyUsingTaylorseries:fx+dx=fx+∂fx/∂x+…

So,theNetheatintotheelement:dxdydzf-)dxdydz

z

f

y

f

x

f (- zyx

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Also,thelocalheatfluxisrelatedtothelocaltemperaturegradientbytheFourier’slawofheatconduction:

Tkf wherekisthelocalconductivity.So,Heat into fluid element by conduction from neighboring elements per unit Volume

z

Tk

zy

Tk

yx

Tk

xTk

(2.6.2)(III)Rateofworkdoneontheflowbysurroundingfluid:Pressureandviscosity

Netrateofworkdonebypressureforces:(pxux‐px+dxux+dx)dydz+(pyvy‐py+dyvy+dy)dxdz+(pzwz‐pz+dzwz+dz)dxdyUsingtaylorseries: Px+dx=Px+∂Px/∂x+…Ux+dx=Ux+∂Ux/∂x+…Netrateofworkdonebypressureforces:

)dxdydzz

pw

y

p

x

pu (-

v

Netrateofworkdonebypressureforcesperunitofvolume:

)(- UppW

(2.6.3)Proceedingaswedidbefore,weget:Netrateofworkdonebyviscousforcesperunitofvolume:

).( UviscW

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zzzyzxyzyyyxxzxyxx wvuz

wvuy

wvux

viscW

(2.6.4)(2.6.1)becomes:Conservationofenergyperunitofvolume:

).()(-2

1 UUpTkrgUUCvTDt

D

(2.6.5)(A) (B) (C)(D) (A)rateofchangeofthetotalenergyfollowingafluidelement(B)rateofheattransferintothefollowingelement(C)workdonebytheneighboringfluidviapressure(D)workdonebytheneighboringfluidviaviscousforce Wecanobtainanalternateformbychangingtheworktermofpressure:

pUUpUp )(

Frommassconservation: Dt

DU

1

pUDt

DpUp

)(

(2.6.6)

Also:

p

Dt

D

Dt

Dp

Dt

Dp

(2.6.7)

( Dt

Dp

Dt

Dp

Dt

D

D

Dp

Dt

Dp

Dt

Dp

Dt

Dpp

Dt

D

2

11)/1(1)/1(1)(

)

The(2.6.6)equationbecomespU

p

Dt

D

Dt

DpUp

)(

(2.6.8)

With PU

t

P

Dt

DP

.(2.6.7)becoming

p

Dt

D

t

pUp )(

(2.6.9)Substitutingtheresultin(2.6.5)weget:

).(2

1

Up

Dt

D

t

pTkrgUUCvT

Dt

D

(2.6.10)

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This result for steady, inviscid and adiabatic flow is the Bernouilli’s equation forcompressibleflow.

02

1

rgUU

pCvT

Dt

D

.2

1ConstrgUU

pCvT

.2

1ConstrgUUCpT

with.

p

CvTCpT theenthalpy

Alongastreamline(risthedirectionoftheflow) (2.6.11)

NOTE: p

isbecausetheflowworks:

2.6.2AlternativeformofenergyequationTo have a pure thermodynamic formulation, we will combine the momentum and theenergyequation:(D)fromequation(2.6.5)canbewrittenas:

j

iij x

UUU

).(.

ThelasttermiscalledviscousdissipationΦ,forNewtonianfluiditis:

2222222

222 Uz

u

x

w

y

w

z

v

x

v

y

u

z

w

y

v

x

u

x

U

j

iij

Sotheenergyconservationbecomes:

UpU

Dt

DpTkrgUUCvT

Dt

D

2

1

Nowregarding

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U momentumequationgives

gUUpUDt

UUD

Dt

UDU ij ...

.

2.

gUrgDt

D.

Theenergyconservationbecomes:

Dt

DpTkCvT

Dt

D

Anotherusefulstatementoftheenergyequation: (2.6.12)

2.6.3Secondlawofthethermodynamics.ConsideringthethermodynamicGibb’sequationabouttheentropys:

dp

CvdTpdCvTds2

dp

TdsCvdT2

Then Dt

Dp

Dt

DsTCvT

Dt

D

Intermsofentropy,s,(2.6.12)becomes:

TkDt

Dp

Dt

DTCv

Dt

DsT

(2.6.13)

Whatdoesitmean?Theenergydissipationfromheatandviscousareresponsiblefortheentropychangesofthefluidelement.(2.6.13)forincompressiblegivesus:Energyequationforincompressibleflow:

TkDt

DTCv

z

Tk

xy

Tk

xx

Tk

xz

Tu

y

Tu

x

Tu

t

TCv zyx

(2.6.14)Note:Inthecaseofincompressibleflow,ifweknowU,wecansolveenergyequationwithheattransfer.

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Isentropicflow

Bydefinition,isentropicflowmeans 0Tds

So,0

2

Dt

Dp

Dt

DcvT

Writing

Dt

pD

Dt

Dp

Dt

Dp

/12

Weget:

Dt

Dp

Dt

pD

Dt

DCvT

1/

(2.6.10)becomes:

).(2

1 Ut

prgUU

Dt

D

Dt

Dp

With Tk =0becausenoheatisexchangedwhenisentropic

).(2

1

2

1 Ut

prgUUUrgUU

tpU

t

p

Soforsteady,inviscidflow:

02

1

rgUUUpU

02

1

rgUUd

dp

TheBernouilliequationforisentropic,compressible,steady,inviscidflowis:

constrgUUdp

2

1

AlongastreamlineWerecognize:Bernouilliequationforisentropic,incompressible,steady,inviscidflowis:

constrgUUp 2

1

Alongastreamline

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2.6.4“bioheatequationofmammaliantissue”(pennes,1948)FromBiofluidDynamic,CKleinstreuer,TaylorandFrancisThebloodtemperatureintheheart’sventriclesandthemajorarteriesremainsessentiallyconstant, i.e., when body parts are suddenly being overheated or subcooled, tissuetemperature equilibration occurs as the blood passes through the smaller arteries. Bothlocalbloodandtissuetemperaturesarethesameuntilbloodmixesatvariousconfluencesaswellasinthevenacavaandtheheart’srightatrium.Amathematicaldescriptionofthethermalexchangeintissueiscomplicatedbytwosetsofblood vessels in the millimeter to micrometer range, sharply varying material propertyvalues,geometricirregularities,metabolicactivity,etc.Nevertheless,energyequation3.13canbewrittenas:

StTky

Tw

y

Tv

x

Tu

t

TCv

Where St here is the heat source (herewe gain heat, compared to viscous dissipationΦwhereweloseheat)Pennes,1948writethisequationinthefollowingforminaonedimensionaldirection:

StTkTTCvt

TCv Abbb

)(

(I)(II)(III)(IV)(I)accumulation(II)convection(III)conduction(IV)heatsourceWhere isthevolumetricflowrateofbloodperunitVolumeoftissue,TAthearterialbloodtemperature,andTthetissuetemperature.Thisequationisknownasthe“bioheatequationof mammalian tissue”. Its underlying assumptions include constant material properties,uniform distribution of blood capillaries in the tissue Volume, constant metabolic heatgeneration, andconstantarterialblood temperature.Assuminga idealized tissueVolume,“bioheatequationofmammaliantissue”hasbeenusedtopredictthetissuetemperatureinspace and time due to excessive body surface cooling (e.g. cryosurgery or frost bites),surfaceheating(e.g.skinburningorhyperthermia),andwholebodyfreezing.Hint:Topassfrom3.13tothebioheatequationofmammaliantissue:AssumingTchangesjustinthex‐direction(perpendiculartothevessel):

)( ATTdTdx

dT

dydz

Q

dx

dTu

x

Tu

y

Tw

y

Tv

x

Tu

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3ExercisesExercise2.1*Alveolarsurfacetension.UsetheLaplace’slawtocalculatethepressuredifferencepi‐po(airpressure‐liquidfilmpressure)inasinglealveolawiththenumericalvalues:R=150μmandσ=72dynescm‐1

Exercise2.2*Bernoulli’sequation.Sometimes,inhemodynamicfield,Bernoulli’sequationisreducedtop1–p2=4(v22–v12)whenpisexpressedinmmHgandvisexpressedinm/s.Usingthe“correct”formoftheBernoulliequation,p1–p2=ρ(v22/2–v12/2),whichapplieswhenpressure,densityρ,andvelocityareexpressedinanyconsistentsetofunits,computetheexactvalueofthecoefficientinthefirstequation.Thedensityofbloodis1.05g/cm3.(FromMortonH.Friedmancoursesnotes)

Exercise2.3*Stenosis.Consideracasewherethereisafocalstenosisof6mmdiameterfemoralarteryinwhichthecrosssectiondiameterisreducedtoone‐thirdofnormal.HowisthevelocityV2at thestenosiscompared to theupstreamvelocityV1? DetermineV2 ifV1 isequal to50cm/min.Determine thepressure at the stenosis if thepressure at theupstreamwas100mmHg.

Exercise2.4Reynoldsnumber.EstimatetheReynoldsnumberforthebloodflowforeachtypeofvesseldescribedinthefollowingtable:Vessel Diameter(cm) Velocity(cm)Aorta 2.5 48Largearteriole 0.05 1.4Arteriole(retinalmicrocirculation)

0.008 3

Capillary 0.0008 0.7Doyouthinkthatsomeoftheseflowsareturbulent?

Exercise2.5*Wormersleynumber.RecallthemeaningoftheWormersleynumber.Theheartrateofa400‐kghorse isapproximately36beatsperminute(bpm); theviscosityofhorseblood is0.0052Ns/m2.Weassumethesamebloodviscosityacrossthemammalspecies.Theheartrateofa3‐kgrabbitisapproximately210beatperminute(bpm).Therabbitbloodviscosityis0.0040Ns/m2.Allometricstudiesofmammalsshowthatthattheaorticdiametergrowsinacertainrelationshipwiththesizeoftheanimal.FromLi,1996:D=0.48W0.34WhereDistheaorticdiameter,givenincentimetreoftheaorta,andWtheweightgiveninkilograms.ComparetheWormersleynumberinthehorseaortatothatinarabbitaorta.Whatdoesthismean?

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Exercise2.6*Similarity.Anaspiringgraduatestudentwantstostudytheflowofbloodinthecoronaryarteries, but because a typical vessel is only 3 mm in diameter and too small to makemeasurements in,shemakesareplicathat is fourtimesthesizeofrealcoronaryarteries.Based on published data for coronary artery dimensions and flow, and the knownproperties of blood, she calculates that the time‐averageReynolds number of the in‐vivoflowis90andtheWomersleynumberis3.1.Theflowsystemsheisusinghasafixedperiodof3sec.(a).Whatshouldbethekinematicviscosityofherworkingfluid?(b).Whatshouldtheaveragevelocitybeattheinlettothemodeltoobtainsimilaritytotheinvivocase?Note: Theconvention inpulsatile flowcalculations is tousethetubediameter(D)as thecharacteristiclengthincalculatingtheReynoldsnumber(Re=ρDU/μ=DU/υ)andthetuberadiusasthecharacteristiclengthincalculatingtheWomersleynumber[α=(D/2)(ω/υ)0.5].(FromMortonH.Friedmancoursesnotes)

Exercise2.7A saline solution (density 1050 kg/m3) is ejected from a large syringe, through a smallneedle, at steady velocity of 0.5 m/s. Estimate the pressure developed in the syringe.Neglect viscous effect. Assume that the velocity of the fluid in the large syringe isapproximatelyzero,whencomparedwiththevelocityintheneedle.

Exercise2.8HagenPoisseuille.A100‐cm‐longcatheter,withaninsidediameterof0.4mmisconnectedto a syringe. In a typical infusionpump, the plunger is driven at a constant velocity. Thediameterof thesyringeis25mm.Foravelocityof50mm/min,whatvolumerateof flowwilldischarge through theCatheter?Assumethe fluidhasaviscosityof0.002Ns/m2anddensity of 1000 kg/m3. Estimate the pressure developed in the syringe. Neglect viscouseffect.Assumethatthevelocityofthefluidinthelargesyringeisapproximatelyzero,whencomparedwiththevelocityintheneedle.¸

Exercise2.9HagenPoisseuille.Whatpressurewillberequiredtoforce1cc/sofbloodserumthroughanintravenoustubeofradius0.5mmandlength3cmintoanarterywithameanpressureof100mmHg?(Assume:bloodserumviscosity,7cP)

Exercise2.10QuizW2010Salinesolution,withthesamedensityaswaterandfivetimestheviscosity of water, is to be administered continuously into a vein through aneedle,inaperfusion.Theinsidediameteroftheneedleis0.4mmanditis50

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mminlength.Togenerateflow,thebagoffluidishunghigherthanthepatient.Giventhat1cmH2O≈100Pa,determinehowhighshouldthebagbehungtogenerateaflowrateof1mlperminute. Note: you canneglect venous bloodpressure (4 cmH2O). State clearly yourassumptions.

Exercise2.11*Quiz W2010 Similarity. Human Spermatozoa. Some BiomedstudentsarepreparingademonstrationforUOttawaDay.Theywantto show the particular way that spermatozoa swim; with aunidirectional rotation of the tail. They plan to copy the rubbermechanismshownintheirbiofluidclass,butsinceaspermcellissosmall, they have to make a bigger model. They found in theliteraturethataspermcellis50µmlong,anditsvelocityisaround200µm/sinwater.(a)ComputetheReynoldsnumber.(b) Which concentration of glycerine should they use to get theproperviscosityiftheymakeamodel1000timesbigger,swimmingwithavelocityof1mmpersecond?(Assumeρglycerine=ρwater)(c)Isitrealistic?

Exercise2.12Hangen‐PoiseuilleQuizS2010Apatienthasatherosclerosis,whichproducesastenosisofhisaortaof16%diameterreduction.a‐Whatisthereductioninflowrate(assumetheheartdeliversthesamepressure)b‐Assuminglaminarsteadyflow,howmuchpressureincreaseisnecessarytocompensateforthisreduction?

Exercise2.13MidtermW2010.Similarity .Youwanttostudytheflowofbloodinthefemoralarteries.However,becauseatypicalvesselistoosmalltomakemeasurements,youhavetomakeareplica that is four times the size of real femoral arteries. Based on published data youknow that femoral arteries have a diameter Da, a time average velocity Va, and that theheartfrequencyisfhandthepropertiesofbloodυb.a)Express the time‐averageReynoldsnumberReaand theWomersleynumberαa in therealfemoralarteryasafunctionofDa,Va,fhandυb.b)Theavailableexperimentalflowsystemhasafixedfrequencyoffe.Whatshouldbethekinematicviscosityυeofyourworkingfluidtoobtainasimilarflowtotheinvivocase?c) What should the input average velocity Ve be in the experimental model to obtain asimilarflowtothein‐vivocase?d) Numerical application. Given: Da=2.2mm, Va=1mm/s, μa=4 cP, ρa=1.05 g/cm3,fh=1s‐1,andfe=0.1s‐1;computeυeandVe

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Exercise2.14Demonstratesomeofthefollowingrules:f,gscalarfunctionsofposition;A,Bvectorfunctionsofpositions

Gradient fg f g g f

A B A B B A A B B A

Divergence fA f A A f

A B B A A B

Curl fA f A A f

A B B A A B A B B A

Secondderivatives

2 2 2

22 2 2x y z

Laplacianoperator

0curl grad

0A div curl A

Exercice2.15Expressthematerialderivativeforone‐dimensionalflowandgiveaninterpretationofeachterm.

Exercice2.16

Computethematerialderivativeofzyx2

ee3e1

2U xt

t

xt

Exercice2.17Continuitylaw(i)Expressthecontinuitylawforaone‐dimensionalflowandgiveaninterpretationofeachterm.(ii) Explain why ⋅U = 0 refers to an incompressible fluid.What is this term for a onedimensionalflow?Explainwhatitrepresentsphysically.

Exercice2.18Equationofcontinuity.Writethespecialcasesoftheequationofcontinuityfora)steadycompressibleflowintheyzplane.b)unsteadyincompressibleflowinthexzplane.

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c)unsteadycompressibleflowinthexdirectiononly.d)steadycompressibleflowinr,θcoordinates.

Exercice2.19RadialflowIncompressible:Foraradialflowintherθplane,Vr=f(r),Vθ=0.Findf(r)forincompressibleflow.

Exercice2.20Navier‐Stokesequation:What is thephysicalmeaningofeachtermintheNavier‐Stokesequation?

upfuut

uv

2)..(

Whatweretheassumptionsinvolvedinthederivationofthelastterm?Simplifytheequationforasteadyflowinthexyplaneintheabsenceofanybodyforce.

Exercice2.21Stock’sequation.ShowthatthemomentumequationforNewtonianincompressibleflow,withoutbodyforce,isreducedtothefollowingformwhenviscositydominates:

**0 2UP Where:P*=P/(μUinf/L),U*=U/Uinf.Givesituationsinthebiomedicalfieldwherethisequationcouldbeused.

Exercice2.22TheReynoldsnumberWhatisthephysicalmeaningoftheReynoldsnumber?ComputetheReynoldsnumbers forvariousfluiddynamicexamples ineveryday life(flowaroundyour car, flowaroundyour car’s antenna, a bird flying, a fish swimming, a spoonstirringyourcoffee,insectflying,bloodflow,flowaroundyourbodywhileyouarewalkingtoschool,etc…)andconcludewhichexamples fall in theHighReynoldsnumbercategory,andwhichonesfallintheLowReynoldsnumbercategory.

Exercisesfromthetextbook:1.1;1.2;1.3;

4.SolutionsSolution2.17.2mmHg

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Solution2.2P1‐P2=3.93(V12‐V22)

Solution2.3V2=450cm/minP1‐P2=0.02mmHg

Solution2.5αhorse=32;αrabit=17

Solution2.6ν=7.8310‐6m2/secV=0.059m/s

Solution2.11a)Re=0.01b)5000cPc)no

5.ReferencesM.Radulescu,UniversityofOttawa,lecturenotes2008K.B. Chandran et al. Biofluid mechanics: the human circulation. Taylor and Francis. 2dedition.BiofluidDynamic,CKleinstreuer,TaylorandFrancis‘THENABLAOPERATOR’onlineresource

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Chapter3

Mathematicalsolutionsforbiofluidproblems

NavierandStokesequationsare called theOnemilliondollarsequations.Theyare calledOnemilliondollarsbecauseonemillioniswaitingtobewonbyanyonewhocansolveoneofthegrandmathematicalchallengesofthe21stcentury:Navier‐Stokesequations.NSareexceptionallyusefulbecausetheydescribethephysicsofmanythingsofacademicandeconomicinterest.Theymaybeusedtomodeltheweather,oceancurrents,waterflowin a pipe, the air's flow around awing, andmotion of stars inside a galaxy. TheNavier–Stokesequationsintheirfullandsimplifiedformshelpwiththedesignofaircraftsandcars,the studyofblood flow, thedesignofpower stations, theanalysisofpollution, andmanyotherthings.However,onlyfewproblemscanbesolvedmathematicallybecauseofthecomplexityoftheequations.Letussolvetheseequationsforproblemsrelatedtobiofluidstuff.

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1Howtosolveaproblem?

2BoundaryconditionsTypically, ifanysolidboundaryexists(i.e.wall,anobject intheflow…)thevelocityatthesurfaceisequaltozero.Thisaccountsfortheviscousstresses(friction)‐Velocitynormaltosurface=0(exceptforporoussurface)‐Velocitytangentialtosurface=velocityofthesurface=0ifthesurfacedoesnotmove.Itisthe“NOSLIP”condition.

3Mathematicalsolutionsforbiofluidproblems3.1ShearstressonarterialendothelialcellsExperimentalsituationsarisewhere it isnecessarytousetheNavier‐Stokesequationsforpredictingshearstressimpartedtotheboundary.Onesuchexperimentistodeterminetheeffectoffluidshearstressonhumanarterialendothelialcellsthatareculturedonflatplates[2]. Anexperimental flowrig is constructedsimilar to theexampleproblemaboveandafluidwithaknownviscosityispumpedataconstantvelocityacrossthecells.Shearstressiscalculatedbymultiplyingtheslopeofthevelocityprofileatthewalltimestheviscosityofthefluid.Theresultsofsuchastudyareshownbelow

video

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Humanarterialendothelialcellsareculturedinaparallelplateflowrigtodeterminetheeffectoffluidshearstressoncellstructure.Atphysiologicshearstress,thecellselongateinthedirectionofflow. These kinds of fluid mechanics studies reveal that human endothelial cells willchangetheirshapeiftheydonotexperienceacertainmagnitudeoffluidshearstress.Thishas been shown to affect their function in a negative way. Studies like this point to agrowing body of evidence that fluid mechanics plays a role in the formation ofatherosclerosis,thenumberonekillerofadultsinthewesternworld.Ithasbeenfoundthatthe disease forms lesions only in specific locations in the body. Experimental andcomputational flow studies in these regions point to lowmagnitude and oscillating wallshear stress as a common fluid mechanical characteristic in regions where the diseaseforms[3].AnexampleofanexactsolutiontotheNavier‐Stokesequationsfollows.Thisisalsoaverypractical one. We will consider the example of viscous flow between two fixed parallelplates:

x

y

z

h

h u

g

Viscousflowbetweentwofixedparallelplatesshowingaparabolicvelocityprofile.LetuswriteNSinCartesiancoordinatefor:2DsteadyIncompressibleTwo‐dimensional

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NewtonianFluid

In thisproblem, there is steady laminar flowbetween theplates, fluid ismoving in thex‐directiononly(v=0,w=0).Wewillalsoignoregravitysinceitwillhaveverylittleinfluenceonthe flowscenario. Giventhatweknowtheviscosityof the fluid,wewant toknowthevelocityprofileacross theplates. So,u=u(y)and theseconditionsareused in theNavier‐Stokesequationssothattheysimplifyto

2

2

0y

u

x

p

ygy

p

0

z

p

0

We’ve set gx = gy = 0 and gy = ‐g. These conditions make the equations simplify tosomethingmanageable.Thesecond‐orderequationfromabovecanberewrittenas

x

p

dy

ud

1

2

2

andintegratedtogive

1

1cy

x

p

dy

du

andintegratedyetagaintogiveasolutionforvelocityinthexdirection.

212

2

1cycy

x

pu

Theboundaryconditionsdeterminec1andc2.Ifthetwoplatesarefixed,thenu=0aty=+hduetothefactthataviscousfluidhaszerovelocityatthewall. Thisconditionissatisfiedwhenc1=0and

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49

2R

r

z

22 2

1h

x

pc

Therefore,thevelocitydistributionbecomes

)(2

1 22 hyx

pu

The observed experimental velocity profile between parallel plates very closelymatchesthatpredictedbytheNavier‐Stokesequations.Alsothemaximumvelocityis:

2max 2

1h

x

pu

Then

)1(2

2max h

yuu

Alsomeanvelocity:

max3

2

2u

h

dyu

u

h

hmean

Andtheshearstressisgivenby(Newtonianfluid):

yx

phy

yx

p

y

u

x

v

y

uxy

)(

2

1 22

Note:Theotherstermsarenull

Closethebottomwall:h

x

pxy

3.2NSinapipeNow we come to derive the most popularapplication of the internal flows, commonlyknown as Hagen Poiseuille Flow or, simplypipe flows. Since pipes have cylindricalgeometry,weusethecylindricalformofthemomentum equations. Let us assume anincompressible, steady flow through acircular pipe without any appreciable bodyforces. Assuming a parallel flow in the z‐

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50

direction, 0Vz ,but 0VVr .

Continuityequation 0

z

VV

r

1Vr

rz

r

0

z

Vz

As in the case of Plane Poiseuille flow,writing out themomentum equations in and r

directionwillsimplyresultin0

p

r

p

.Therefore,letusfocusonz‐direction.

z

p

=

2

2

2

2

2

11

z

uu

rr

ur

rrzzz

Wecanfurtherassume0

Vz

becauseofthecylindricalsymmetry.

z

p

=

r

ur

rrz1

or,integratingtwiceover“r”,weget

21

2

ln4

)( CrCdz

dprrVz

(C1,C2=Constants)

SincethepiperadiusisR,theboundaryconditionsmaybewrittenas 0)Rr(Vr

and0)0r(

dr

dVz .

Thesecondboundarycondition isdueto flowsymmetryatr=0,whereasthe firstone isdueto“no‐slip”condition.SolvingtheconstantsC1andC2weget

2

22

14

)(R

r

dz

dpRrVz

AsinthecaseofPlanePoiseuilleflows,0

dz

dp

forthisflowtoexist(i.e.,Q>0).

0 0

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x

yd

Someadditionalresultsare:

dz

dpRQ

8

4

Hagen‐Poisseuillerelationship L

pRQ

8

4

dz

dpRV

8

2

,

VVZ 2

max

,

dz

dpr

dr

dVzrz 2

[Note:Youmustuseanannularareaelement zedrr2Ad

toderiveVandQresults.]

3.21ValidityoftheHagen‐PoisseuillerelationshipinthecardiovascularsystemThesimplestmodelforbloodflowthroughavesselwouldbe:LaminarflowSteadyNewtonianfluidStraighttubewithaconstantcircularcrosssectionInthiscondition,Hagen‐Poisseuillegivestherelationship:

L

pRQ

8

4

WhereQ is the flow rate, Δp is the dropof pressure, L is the length of the tube, μ is theviscosity,andRisthetuberadius.Weshouldcriticallyexaminethevalidityoftheseassumptionsinmodelsdescribingbloodflowinarteries.NewtonianfluidAswepreviouslydiscussed,thebloodviscositydependsontheshearrateespeciallyatlowSR. But for high shear stress higher than 100s‐1 , the viscosity coefficient approaches aconstantvalue.Thus,forflowinlargebloodvessels,wherelowSRcanbeexpectedduringsystole,aNewtoniandescriptionappearstobereasonable.Laminarflow Aswesawpreviously,theassumptionoflaminarflowinthemodelalsoappearsreasonable.

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SteadyflowFlow in thehumanarteries is clearlypulsatile, consistingof systolicanddiastolicphases;therefore,theassumptionofsteadyflowsisNOTvalidinthemajorpartofthecirculatorysystem.RigidwallThearterialwallsarevisco‐elasticanddistendwiththepulsepressure.Theinteractionofthe wall and the fluid is an important factor in hemodynamics. This assumption is NOTvalid. However, in certain cases of steady flowmodels, the distensability don’t affect thesolution.Constantcircularcross‐sectionThis is a good approximation formost of the arteries in the systemic circulation.But theveinandthepulmonaryarteriesaremoreellipticalinshape.

3.3Pulsatileflow

3.3.1EffectofpulsatilityThis§isadaptedfromBiofluidmechanics,thehumancirculation,KChandran,p191The previous section focused on steady flow. However, we know that the blood flow inheart and arteries is pulsatile. When the heart contracts during systole, a pressure isgeneratedbytheleftventricle,andthewavetravelsduetotheelasticityofthesystem.Thepulsatilenatureoftheflowaffects:PressuredistributionVelocityprofilesAccordingtotheinstantaneousReduringthepickofsystole,weareexpectingaturbulentflow(5000).However,ithasbeenobservedthataorticbloodflowremainslaminarandwellstreamlinedundertheseconditions.Thereasonispartiallyduetothestabilizingeffectthatsystolicaccelerationhasontheflowandalsobecauseisthereisnotsufficienttimeavailableforflowtobecometurbulent.Inahealthyartery,despitea largeRe, the flowisstill laminar.Foranothealthycase, theflowwillbecometurbulentbecausetheReisnotappropriatewithapulsatileflow!TheWormerleynumberαisalsousetocharacterizetheperiodicnatureofbloodflow.

r

Asαincreases,theinertialforcesbecomemoreimportantandstarttodominate,initiatingatthecenterofthetube.Asresult,adelaycanbeobservedinthebulkflow,andthevelocityprofilebecomesmoreflatinthecentralregionofthetube.

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3.3.2WormersleysolutionWormesley solution is one way to estimate unsteady velocity in a straight pipe. Othersolutionsthataremorecomplexincludetheelasticityofthetube.Thesearealsoproposedintheliterature.HerewewillfocusontheWormersleysolutiononly,aftersolvingtogetherthe equation, you will have to use matlab to plot the velocity profile in a tube for anunsteady flow. Changing α, you will have the opportunity to see the importance of theinertialforces.Let us write continuity and Navier Stokes equations without body force in a cylindricalcoordinate for an unsteady flow of an incompressible Newtonian fluid, in z‐direction.Vesselsareassumednon‐elastic(asstraightpipe)Forthispulsatileflowcase,weassumearigidwall.Thentherewillalsobenoradialmotionofthewall.Therefore,wecanassumethattheradialvelocitycomponentwillbezero.Alsotheflowisaxisymmetric.Justifyingeachsimplification,showthat:

r

V

rr

V

z

p

t

V 112

2

Eq.W1

Sincepisafunctionofzandt,dp/dzwillbeafunctionoftonly.ItispossibletowritepwithaFourierseries:

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nti

n neaez

p

0 Eq.W2

Thenwecanwriteeachcomponentof thepressuregradientasa complexexponentialasfollowing:

ntin

n

eaz

p

Eq.W3Where an is a constant representing the amplitude of the harmonic n of the pressuregradient,ωisthefundamentalfrequency,andiistheunitcomplexnumber.InthiscontextofFourierseries,theformvelocityVis:

0n nVeV

ntinn erfV )( Eq.W4

ShowthatEq.1canbewrittenforaharmonicn,independentlyofthetimeasfollows:

Brfdr

rfd

rdr

rfdnn

nn )()(1)( 2

2

2

Eq.W5

where 2n andBareconstantsthatyouhavetodetermine.

n

z

t

u

n

z

r

u

n

z

r

u2

2

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r

V

rr

V

z

p

t

V 112

2

becomes:

4‐SolvetheEq.W5forn=0.

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5‐DeterminedaparticularsolutionoftheEq.W5.forn≠0

Hint:trytheeasyone: .)( Constparticularrfn 6‐.ThehomogenousdifferentialequationofEq.5forn≠0withoutthesecondmemberisthewellknownBessel’sequationofthefirstkindandzero‐order:

0)()(1)( 2

2

2

rfdr

rfd

rdr

rfdnn

nn Eq.W6

InthiscasethesolutionofEq.6forn>0is:

)(1)( rJoCrf nHomogenousn

Where Jo is the Bessel function of the first kind (you do not need to know the exactexpressionofJoforthefollowing).Withtheappropriateboundaryconditionprovethat

1

)(

)()(

RJo

rJo

ni

arf

n

nnn

Eq.W7

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57

7‐DeducetheexpressionofVasafunctionofrandtusingthe3firstharmonicsi.e.n=0,1,2(becarefulthesolutionforn=0comesfrompoint4).

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8‐IntegratenumericallythevelocityV(r,t)overtheentirecross‐sectionofthevessel.Theoutputistheflowrate,Q(t).9‐ShowthatthenormalizationofEq.5usingr*=r/Randα2=R2ωρ/μis:

****

1

*2

2

2

Bfdr

fd

rdr

fdnn

nn Eq.8

where 2*n andB*areconstantsthatyouhavetodetermine.

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10‐Whatisthephysicalmeaningoftheparameterα?Howwillthevelocityprofilelooklikeifα<<1?

4.ExercisesBecausethematerialisdifficult,problemsolvingistheonlywayyouwillbeabletoretainit.Thesolutionswillnotbegivenforallproblemsandyouareresponsibleforknowinghowtosolve theproblems. Similarproblemswill be alsogivenon themidtermand final, soyoushouldknowhowtodothemefficiently.Problem Solving Procedure: The following procedure should be used in formulating allwrittenproblemsolutionsinexamsandhomework.1.Clearlyformulatetheassumptionsyoumake.2. Clearly formulate the analytic solution as far as possible using symbolic forms beforesubstitutingnumericalvalues.“plugandchug”willbepenalized.3.Onlyoncetheentireproblemissolved,youcansubstituteappropriatenumericalvalues.4.Concludebydiscussingthevalidityofyoursolutioninviewoftheassumptionsyouhavemade.

Exercise3.1MidtermWinter 2010 ‐ ‘Planar’ Couette flow. Consider blood as an incompressibleNewtonianisothermalfluidinalaminar,steady,fullydevelopedflowbetweentwoparallel

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60

plates.Oneplate is fixed,theothermoveswithaconstantvelocityU.Gravity isneglected.Endeffectsmaybeneglected.

a)Givethesimplifiedcontinuityandmomentumequationsthatmodelthisflow.Stateyourassumptionsconciselyandfullyjustifyyoursimplifications.b)Writetheboundaryconditions.c)Showthatthevelocitydistributioninthefluidis

)(2

1 2 byyx

p

b

yUu

Where x

p

andUareconstantsd)Drawtheshapesofthevelocityprofilesforthefollowingcases:

(i)0

x

p

;U>0

(ii)0

x

p

;U=0

(iii)Deducetheprofileshapeforthefollowingcase0

x

p

;U>0e)Computetheshearstressτxyinthegap.

f)Assumeforthisquestionthat0

x

p

andU≠0.Theredbloodcellsinasalinesolutionaredamaged(haemolysis)whentheyexperienceashearstressaboveacriticalvalue c.Given the blood viscosity μa=3.5 cP, b= 1mm, and c = 1500 dynes/cm2, propose aconditiononthevelocityUtoavoidhaemolysis.

Exercise3.2Navier‐Stokesequationsincylindricalcoordinate.Thebloodflowinanextracorporealline is assumed laminar, fully developed, and steady; the blood viscosity is assumedconstantandthevesselcrosssectioncircular.Considerthebloodinahorizontaltube.EndeffectsmaybeneglectedbecausethetubelengthLisrelativelylargecomparedtothetuberadiusR.ThefluidflowsundertheinfluenceofbothapressuredifferenceΔpandgravity.

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a)GivethecontinuityandNavier‐Stokesequationssimplifiedtomodeltheflowofthefluid.b)Determinethesteady‐statevelocitydistributioninthefluid.c)Determinetherelationshipbetweentheflowrateandthepressuredrop.d)DeterminethemaximumvelocityUmax.e)DeterminethevelocitydistributionUz(directionofthetube)intermofUmax,randR.

Exercise3.3Inclinedplatesurface.(MiddtermS2010).Aliquidflowssteadilydownaninclinedplaneforminga laminar filmof thicknessh.The inclinationangleof theplate issmallsuchthatyoucansafelyassumefullydevelopedflow(∂u/∂x=0)andnegligibleaccelerationinthex‐direction.The fluid canbe assumed incompressiblewith a constant viscosityμ. Since thepressureon thesurfaceof the film isconstant,wecanassume∂ρ/∂x=0for this thin free‐surfaceflow.

a) Simplify the continuity and Navier‐Stokes equations for this flow field. State yourassumptionsconciselyandfullyjustifyyoursimplifications.b)Giveaphysicalexplanationforwhyyoucanassumethattheshearstressonthesurfaceofthefilmcanbeassumednegligible?c)Showthatthevelocityprofileisgivenby

d)Showthattheshearstressdistributioninthefluidisgivenby

e)Showthattheaverageshearstressisgivenby

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62

2

sin hgxy

f)Astudenthastostudythesizeofredbloodcellaggregatefunctionoftheaverageshearrate. She designed an experimental set‐up with which it is possible to watch the bloodflowingoveraslopingslideunderamicroscope.Changingtheinclinationangle,willchangetheaverageshearstressandthenshewillbeabletowatchthesizeofaggregatesthroughthemicroscope.Givethefunctionthatwillallowhertocomputethemeanshearstress inthe blood film as a function of the angle of inclination of the slide. Blood is assumedNewtonian.Thesystemisfed,usingasyringepump,withaconstantflowrateQ.ThebloodisflowinginachannelwithawidthL.AccordingtoNusselthethicknessofafilmneglecting

thesurfacetensionforceis:

3/1

sin

3

gh N

where LQN / isthefeedrateperunitofwidth.

Given:Q=1ml/hL=1mmBlooddensity1050kg/m3g=10m/s2Bloodviscosity=4cPoise

Exercise3.4*Stock flow: lubrication. Flowbetween two concentric rotating spheres: This could be thefluidlubricationinhumanprosthesishipjoints.

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Consider an incompressible Newtonian, isothermal fluid in laminar flow between twoconcentric spheres, whose inner and outer wetted surfaces have radii of kR and R,respectively.Theinnerandouterspheresarerotatingatconstantangularvelocities iando, respectively. The spheres rotate slowly enough that the creeping flow assumption is

valid.a)GivethecontinuityandNavier‐Stokesequationssimplifiedtomodelthisflowfiled.b)Determine thesteady‐statevelocitydistribution in the fluid(forsmallvaluesof iando,wecanassumeDU/Dt=0).

c)DeterminetheShearstresswith o=0.d)RegardingtheShearstress,discusswhythesynovialfluidisashear‐thinningfluid

Exercise3.5*RadialFlowbetween2disks.Steady,laminarflowoccursinthespacebetweentwofixedparallel,circulardisksseparatedbyasmallgap2b.Thefluidflowsradiallyoutwardduetoapressure difference (P1 − P2) between the inner and outer radii r1 and r2, respectively.Neglect end effects and consider the region r1 ≤ r ≤ r2 only. Such a flow occurswhen alubricantflowsincertainlubricationsystems.

Figure.Radialflowbetweentwoparalleldisks.a)Simplifytheequationofcontinuitytoshowthatrvr=f,wherefisafunctionofonlyz.b)SimplifytheequationofmotionforincompressibleflowofaNewtonianfluidofviscosityμanddensityρ.c)Obtainthevelocityprofileassumingcreepingflow.d)Sketchthevelocityprofilevr(r,z)andthepressureprofileP(r).e)Determineanexpressionforthemassflowratebyintegratingthevelocityprofile.f) Derive the mass flow rate expression in e) using an alternative short‐cut method byadaptingtheplanenarrowslitsolution.

Exercise3.6*Diskviscosmeter. A parallel ‐ disk viscometer consists of two circular disks of radius RseparatedbyasmallgapB(withR>>B).Afluidofconstantdensityρ,whoseviscosityμistobemeasured,isplacedinthegapbetweenthedisks.Thelowerdiskatz=0isfixed.ThetorqueTz,necessarytorotatetheupperdisk(atz=B)withaconstantangularvelocityΩ,ismeasured.ThetaskhereistodeduceaworkingequationfortheviscositywhentheangularvelocityΩissmall(creepingflow).

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Figure.Parallel‐diskviscometer.a)Simplifytheequationsofcontinuityandmotiontodescribetheflowintheparallel‐diskviscometer.b) Obtain the tangential velocity profile after writing down appropriate boundaryconditions.c) Derive the formula for determining the viscosity μ of a Newtonian fluid frommeasurements of the torque Tz and angular velocity Ω in a parallel ‐ disk viscometer.Neglectthepressureterm.

Exercise3.7Applicationofthebioheadequation (fromBiofluiddynamics,C.Kleinstreuer,Taylor&Francis)Considerbloodperfusionofatissuelayerofthicknesshwhereatthefat‐tissueinterfaceT=T(x=0)=T1andat the tissue‐core interfaceT=T(x=h)=T2.Theblood(ρ,Cp)enters thetissuewithaconstantflowrate andtemperatureTa<T1<T2Sketch: Assumptions:

Steady1‐DflowuniformflowNegligiblemetabolicrateConstantproperties

Approach:Reduced bioheadequationDirectintegration

a)Basedonthestatedassumptionsreducethebioheadequationofmammaliantissue:

StTkTTCpt

TCv a

)(

Where istheflowrateandStistheheatsourceb)Showthatthefollowingpropositionisthesolutionoftheenergyequationinthepresentcaseandthatitrespectstheboundarycondition:

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65

)sinh()coth()cosh()sinh(

)sinh(

1

2

1mxmhmx

mh

mx

TaT

TaT

TaT

TaT

Where k

Cpm

2

c)Plot forTa=32°C,T1=34.5°C,T2=37°C,Cp=3.6103J/(kg.K),k=0.37W/(m.K)and =400ml/min.Plotagainfor =2400ml/min

Exercisesfromthetextbook4.1

5.SolutionsSolution3.4http://www.syvum.com/cgi/online/serve.cgi/eng/fluid/fluid302.html

Solution3.5http://www.syvum.com/cgi/online/serve.cgi/eng/fluid/fluid306.html

Solution3.6http://www.syvum.com/cgi/online/serve.cgi/eng/fluid/fluid305.html

6.AssignementTheWormersleyequations:bloodpulsatileflowinfemoralartery.Eachstudentshoulddotheirownassignment,althoughyoumayworktogether.YoumayNOTshare electronic copiesof the solution.Working togethermeans that youmay lookat eachother’swork,askquestions,discusssolutions,butpleasenotcopypaste!Objectives•ToapplytheNavier‐Stokesequationstounsteadyflow•TogainpracticeusingaFouriercoefficientsrepresentationtogenerateapressurewaveforminMaple.•Togenerateaflowwaveformfromapulsatilepressurewaveform

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PART1:ResolutionofNavier‐StokesequationsinclassPART2:Visualisation:MatlabThediameteroffemoralarteryis2.5mm.Weassumeaconstantbloodviscosityμ=0.0035Ns/m2andρ=1060kg/m3.Question1‐TheFouriercoefficientsshownbelowdescribethepressure‐gradient‐versus‐time curve in femoral artery. The frequency is 1 Hz (Recall ω=2π/Period). Plot thepressure‐gradient‐versus‐timecurve.n 0 1 2ancoefficient(Pa/m)

‐621 1250i

531

Matlabhint:Youcanuseiorjfortheunitcomplexnumber,butbecarefultonotusethesameletterforanothervariable.Youcantaketherealpartofanumberusingreal().Question2.UsingtheWomersley’sNavier‐Stokessolution,on4differentfigures,plotthe3termsof the velocity (n=0, n=1, n=2) and the total velocity as a function of the radius att=0.5sMatlabhint: theBessel functioncommand for a zeroorderBessel function isBesselJ(0,argument).Question3‐On4differentfigures,plotthe3termsofthevelocity(n=0,n=1,n=2)andthetotalvelocityprofileasafunctionofradiusfort=0tot=1(20plotspergraph).Matlabhintusethecommands‘holdon’’holdoff’.Question4‐ Integratenumerically thevelocityV(r,t) over theentire cross‐sectionof thevessel.Theoutputisflowrate,Q(t).Pleaseshowtheflowrateinliter/minute.Question5‐ Insomediseases, thebloodviscositycouldreach8cP.Plot thevelocityasafunctionoftheradiusasinquestion3usingthishyperviscosity.

eθ

Vez

er

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67

Question6‐ Discuss the results obtained Question 3 and 5 highlighting α value in eachcase.Question7‐Toconclude,discusstheassumptionsmadeinthismodelandproposehowtoassessthemodel.Youhavetosubmitahardcopyofyourreportincludinganintroductionandaconclusion.Foreachquestionmathematicalformofequation,extractofmatlabcode,figuresanddescriptionofresultsareexpected.

7.ReferencesBiofluidmechanics,thehumancirculation,KChandranandal.,Taylor&Francis2dedition.p41‐43Fuidmechanics,Munson‐Young–Okiishi.Ebook.

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Chapter4

ComputationalFluidDynamic(CFD)andmeasurementtechniquesinbiomedical

1.ComputationalfluiddynamicsTextbook:Biofluidmechanics,thehumancirculation,KChandranandal.,Taylor&Francis2dedition.Chapitres11.

2.FlowmeasurementinthecardiovascularsystemTextbook:Biofluidmechanics,thehumancirculation,KChandranandal.,Taylor&Francis2dedition.Chapitres10

3.ExercicesExercise3.1Apatient’scardiacoutputis5500ml/minwhilehisarterialoxygenconcentrationis0.2ml/mlandhervenousoxygenconcentration0.15ml/ml.Findthisperson’sspirometeroxygenconsumption

Exercise3.2UsingacontinuouswaveDopplerwithacarrierfrequencyof7Mhz,α=45°,thespeedofsound=1500m/s,andDopplershiftedfrequency5000Hzfindthebloodvelocity.

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Exercise3.3Usingthethermodilutionmethodformeasuringcardiacoutput,10mlofinjectateisinjectedover2.5secondsoveraperiodof5seconds.Thecardiacoutputis4.0l/min.Usethe

followingdatatoestimatethevalueof 1

0

t

b dtT

Volumeofinjectate=10ml,dTofinjectate=‐30KDensityofinjectate=1005kg/m3

Heatcapacityofinjectate=4170J/(kgK)Densityofblood=1060kg/m3

Heatcapacityofblood=3640J/(kgK)

SuggestedExercisesfromthetextbook:10.4;10.5;10.6;10.7

4.CFDsimulationassignmentTextbookpagep413

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Chapter5

Flowoverimmersedbody(incompressible)AdaptedfromFundamentalsofFuidMechanics,Munson‐Young–Okiishi(FFM)AndfromMultimediaFluidmechanics,CambridgeUniversitypress(MFM)Goal:StudyhowtheforcesduetothefluidflowactonabodyExamplesin:BiofluidmechanicsShearstressesaffectendothelialcellShearstressesaffectredbloodcell(hemolysis)andothercellsFishswim/BirdflyAirintherespiratorysystem/ParticletransportExamplesin:ClassicalfluidmechanicsAirfoilCarsPlanesEct…

1GeneralExternalFlowCharacteristicsA body immersed in a moving fluid experiences a resultant force due to the interactionbetweenthebodyandthefluidsurroundingit.2Cases:ThefluidisstationaryandthebodymovesthroughthefluidwithvelocityU.ThebodyisstationaryandthefluidflowspastthebodywithvelocityU.►Inanycase,wecanfixthecoordinatesysteminthebodyandtreatthesituationasfluidflowingpastastationarybodyStreamlinedorbluntStreamlinedbodies:airfoils,racingcars,…►littleeffectonthesurroundingfluid,

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Bluntbodies:parachutes,buildings…►StrongeffectonthesurroundingfluidItissometimesdesirabletomakeanobjectasstreamlinedaspossible.Inothersituationsabluntobjectisdesired.Typically,streamlinedobjectshavelessdragthanbluntobjects.Akayakisastreamlinedobjectthatmovesthroughthewaterwithminimalresistanceanddisturbancetothefluid. Itrequiresarelativelysmallpropulsiveforce.Thepurposeofthepaddleistoimpartthepropulsiveforcetothekayak.Todosoitmustgeneratearelativelylarge resistance to motion through the water. A paddle is a blunt object. The Reynoldsnumbersforthepaddleandthekayakareintheorderof100,000to1,000,000.

2LiftandDragConcept2.1DefinitionsForces at the fluid–body interface due to the interaction between the body and the fluidoccurs:Wallshearstresses,duetoviscouseffects wNormalstressesduetothepressure,p

Totalsurfaceforce:

dAdAPFwFpSurface

w

Surface

DFDdrag:resultantforceinthedirectionoftheupstreamvelocityLlift:resultantforcenormaltotheupstreamvelocityDragCoefficient=Dragforce/characteristicinertiaforce

oftenCDisanexperimentallyexpressedfunctionofReynoldsnumberCD=f(Re)(seetable1)LiftCoefficient=Liftforce/characteristicinertiaforce

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2.3Dragfordifferentshapes

Table5.1:LowReynoldsNumberDragcoeficient[FFM]

Figure5.1:Exempleofdrag[FFM]

2.2Dragcoefficient,forasphereinstokesflow( 1Re )

FromDragcoefficientdefinition:42

22 D

UCD Dsphere

Fromthetable: UDCD

24

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73

So,42

242

2 DU

UDDsphere

Usefulresult: UD

Dsphere 26

Note:Utherelativevelocityofthesphereinthefluid(Ufluid‐Usphere)

2.4Transportofmicro‐particlesBiomedical applications: red blood cell sedimentation, particles in the upper air way,aerosol…Assumingsphericaldilutemicro‐particles insuspensionwithanegligiblerotation,wecanwritetheNewtonlaw:

oBBrgDp

p FFFFFdt

dvm

Mass.acceleration=drag+gravity+Brownian+Buoyant+other(electrostatic,magnetic,…)

Dragforce: Apuu

CF pff

dD

2)(2

Forasphereinstokesflow)(6 pfD uuRF

uf:fluidvelocityup:particlevelocityR:particleradiusρf:fluiddensityρp:particledensity

Gravityforce: gRF pG 3

34

Buoyantforce: gRF fG 3

34

Brownianforcehastobeconsideredforasub‐micrometricparticle.ParticlesedimentationIf we assume spherical a micrometric particle, neglecting Brownian force, quasi static(sedimentation)weget:

gRgRuR fpp 33

34

34)0(60

gR

u fpp )(9

2 2

Dragforcedominates(everyshape)

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ppffdp

p AuuC

dt

dum 2)(

2

Wegetadifferentialequation.Thisequation isused insimulationofparticledeposition/aerosolintheupperairway.However,turbulencehastobeconsidered.Therefore,thetermof the fluid velocity is actually more complex, composed of an average velocity and afluctuatingvelocitytomodeltheturbulence.

3CharacteristicsofFlowPastanobjectExternal flow past objects encompasses an extremely wide variety of fluid mechanicsphenomena.Clearly,thecharacteristicoftheflowfieldisafunctionoftheshapeofthebody.There canbeawidevariety in the sizeof aboundary layer and the structureof the flowwithinit.Partofthisvariationisduetotheshapeoftheobjectonwhichtheboundarylayerforms.Flowspastrelativelysimplegeometricshapes(i.e.asphereorcircularcylinder)areexpectedtohavelesscomplexflowfieldsthanflowsthatpassacomplexshapesuchasanairplaneoratree.However,eventhesimplest‐shapedobjectsproducerathercomplexflows.In this section we consider the simplest situation, one in which the boundary layer isformedonaninfinitelylongflatplatealongwhichflowsaviscous,incompressiblefluidandanotherinwhichtheboundarylayerisformedaroundacircularcylinderoranairfoil.

Figure 5.2 :Thenature of the flowpast a body depends strongly onwhether Re<<1orRe>>1[FFM]

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Figure5.3Flowseparationmayoccurbehindbluntobjects.[FFM]

Figure5.4:Effectonthedragcoefficientonacircularcylynder[FFM]

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4Boundarylayercharacteristics4.1BoundaryLayerStructureandThicknessonaFlatPlate

Figure5.5:1BoundaryLayerStructureandThicknessonaFlatPlate[FFM]

4.2Boundarylayerthickness

Figure5.6:1BoundaryLayerThickness[FFM]Firstdefinition:Thickness

: Boundarylayerdisplacementthickness

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MomentumThickness,θ

Detailofcalculation:

4.3Momentum‐IntegralBoundaryLayerEquationforaFlatPlate

Figure5.10:Controlvolumeusedinthederivationofthemomentumintegralequationforboundarylayerflow.[FFM]Oneoftheimportantaspectsoftheboundarylayertheoryisthedeterminationofthedragcausedbyshearforcesonabody.Weconsidertheuniformflowpastaflatplateandthefixedcontrolvolume.Inagreementwithadvancedtheoryandexperiments,weassumethatthepressureisconstantthroughout

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theflowfield.Theflowenteringthecontrolvolumeattheleadingedgeoftheplate[section1] is uniform, while the velocity of the flow exiting the control volume varies from theupstreamvelocityattheedgeoftheboundarylayertozerovelocityontheplate.Thefluidadjacenttotheplatemakesupthelowerportionofthecontrolsurface.Theuppersurfacecoincideswiththestreamlinejustoutsidetheedgeoftheboundarylayeratsection2. It need not (in fact, does not) coincidewith the edge of the boundary layer except atsection2.Bydefinitionthedragforceis

Disthedragthattheplateexertsonthefluid.Notethatthenetforcecausedbytheuniformpressuredistributiondoesnotcontributetothisflow.If we apply the momentum equation in x direction for steady flow of fluid within thiscontrol volume we obtain Since the plate is solid and the upper surface of the controlvolumeisastreamline,thereisnoflowthroughtheseareas.Thus,

WhereforaplateofwidthbFlowratethroughsection1mustequalthatthroughsection2:

MultiplybyρUb: Sowegetforthedrag:

Recallthemomentumthickness: ,wecanwrite:

On an experimental point of view, if you canmeasure the u, you can compute θ and thendeducethedrag.Notethatthisequationisvalidforlaminarorturbulentflows.

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Asxincreases,increasesδ,andthedragincreases.Thethickeningoftheboundarylayerisnecessarytoovercomethedragoftheviscousshearstressontheplate.Thisiscontrarytohorizontalfullydevelopedpipeflowinwhichthemomentumofthefluidremainsconstantandtheshearforceisovercomebythepressuregradientalongthepipe.The shear stress distribution can be obtained from this equation by differentiating bothsideswithrespecttoxtoobtain

The increase in drag per length of the plate occurs at the expense of an increase of themomentumboundarylayerthickness,whichrepresentsadecreaseinthemomentumofthefluid.

Since (fromdragdefinition)itfollowsthat

Andfinally On an experimental point of view, you need tomeasure the field of velocity to be able tocomputeθatanypoint;thiscouldbedoneusingPIVforexample.ApproximationoftheBLthickness:Theusefulnessofthisrelationshipliesintheabilitytoobtainapproximateboundarylayerresults easily by using rather crude assumptions. For example, if we knew the detailed

velocityprofileintheboundarylayer,wecouldevaluatethedragfrom and

theshearstressfrom. Example: assuming linear velocity profile in theBL u=Uy/δ, estimate the boundary layerthicknessfunctionoftheposition

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4.4Prandtl/BlasiusBoundaryLayerSolutionFrom mass conservation in an incompressible fluid, thestreamwisevelocityvariationmustbethesamesizeasthecross‐streamvelocity variation.This equality givesus anestimate for the cross stream velocity, V. Since we haveassumed that the boundary layer thickness, δ, is smallcompared to thebody length,L,wecansee that ismuchsmallerthanUFigure5.7:1BoundaryLayerThickness[MFM]

Massconservation

Exactequation: 0

y

v

x

u

Approximateequation: V

L

U

Meansthat U

LV

ForthestreamwiseNSequation,withintheBL,thechangeinUinthestreamwisedirectionx,canbeneglectedcomparetothechangeinthecross‐streamdirection.UsingBernouilli’sequation,wecanreplace thepressuregradientby thevelocitygradient in theouter flow.Thismeanswecansimplifythestreamwisemomentumequation.Note,thatforsteadyflowsince the velocity is constant, and fromBernouilliwe know the pressure is constant, thevariationofpressurecouldcancelled.Momentumequationforsteadyflowx‐direction

Exactequation:

2

2

2

21

y

u

x

u

x

p

y

uv

x

uu

Dimensionalanalysis:22

2

0

U

L

UUU

LL

U

SimplifyBLequation:2

2

y

u

y

uv

x

uu

Momentumequationforsteadyflowy‐direction

Exactequation:

2

2

2

21

y

v

x

v

y

p

y

vv

x

vu

Dimensionalanalysis: L

U

L

U

L

U

L

U

32

2

2

2

?

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2

2

U

L

U

Thus 0

y

p

Finally,Boundarylayerequations(Newtonian,incompressible,steady):

Continuity:0

u v

x y

Momentum:2

2

y

u

y

uv

x

uu

BC: u(x,0)=0u(x,inf)=Uv(x,0)=0This x‐ dimensional analysis give us themagnitude of the BL thickness at the length x :

U

x

Ingeneral,thesolutionsofnonlinearpartialdifferentialequationssuchastheboundarylayerequations,areextremelydifficulttoobtain.However,byapplyingaclevercoordinatetransformationandchangeofvariables,Blasiusreducedthepartialdifferentialequationstoanordinarydifferentialequation.Blasiuscoordinatetransformationisthefollowing:

Description of this process can be found in standard books dealingwith boundary layerflow.SolutionofBLequationisgiveinthefollowingtable:

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Figure5.9:Boundarylayerequation,Blasiussolution[FFM]TheBlasiussolutioncouldbesummarizedas:

With the velocity profile known, it is an easymatter to determine thewall shear stress,

wherethevelocitygradientisevaluatedattheplate.Thevalueof aty=0canbeobtainedfromtheBlasiussolutiontogive:

Indimensionlessform,wecanhavethelocalfrictioncoefficientcf:

x

wf

Uc

Re

644.0...

2

1 2

Orfortheallsurface,thedragfrictioncoefficient,CDf

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x

l

wfDf

lbU

dxb

AU

DC

Re

328.1

2

1.

2

1 2

0

2

5Turbulentboundarylayer

Figure5.12:Turbulentflow[FFM]Theanalyticalresultsarerestrictedtolaminarboundarylayerflowsalongaflatplatewithazero pressure gradient. They agree quite well with experimental results up to the pointwhere the boundary layer flow becomes turbulent, whichwill occur for any free streamvelocityandanyfluidprovidedtheplateislongenough.ThevalueoftheReynoldsnumberatthetransitionlocationisarathercomplexfunctionofvariousparametersinvolved,including:theroughnessthecurvaturedisturbancesintheflowoutsidetheboundarylayer.Example:Onaflatplatewithasharpleadingedgeinatypicalairstream,thetransitiontakesplaceata distance x from the leading edge of a Re around 5 .105 . Actually, the transition fromlaminar to turbulent boundary layer flowmay occur over a region of the plate, not at aspecific single location. This occurs, in part, because of the spottiness of the transition.Typically,thetransitionbeginsatrandomlocationsontheplateThecomplexprocessoftransitionfromlaminartoturbulentflowinvolvestheinstabilityoftheflowfield.Typical profiles obtained in theneighbourhoodof the transition location are indicated inthenextfigure.Theturbulentprofilesareflatter,havelargervelocitygradientsatthewall,andproducealargerboundarylayerthicknessthanthelaminarprofiles.

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Figure5.12:Boundarylayerprofiles[FFM]

6.PressuregradienteffectonflowseparationTheboundarylayerdiscussionsinthepreviouspartshavedealtwithflowalongaflatplateinwhichthepressureisconstantthroughoutthefluid.Ingeneral,whenafluidflowspastan

objectotherthanaflatplate,thepressurefieldisnotuniform.Physically, in the absence of viscous effects, afluidparticletravelingfromthefronttothebackof the cylinder coasts down the “pressure hill”fromtoAtoCandthenbackupthehilltofrompointCtoFwithoutanylossofenergy.Thereisan exchange between kinetic and pressureenergy,buttherearenoenergylosses.The same pressure distribution is imposed onthe viscous fluid within the boundary layer.However,becauseoftheviscouseffectsinvolved,theparticle intheboundarylayerexperiencesalossofenergyas it flowsalong.This lossmeansthattheparticledoesnothaveenoughenergytocoastallofthewayupthepressurehillfromCtoFandtoreachpointFattherearofthecylinder.

Figure5.13:Pressuregradiant[FFM]

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ThiskineticenergydeficitisseeninthevelocityprofiledetailatpointC.Becauseoffriction,theboundarylayerfluidcannottravelfromthefronttotherearofthecylinder.

Thesituation is similar toabicyclist coastingdownahill andup theother side of the valley. If therewere no friction, startingwith zerospeed, the rider could reach the same height from which he or shestarted. Clearly, friction causes a loss of energymaking it impossiblefortheridertoreachtheheightfromwhichheorshestartedwithoutsupplyingadditionalenergy

Figure5.14:Pressuregradiant:Bicycleanalogy[MFM]The fluidwithin theboundary layerdoesnothave suchanenergy supply.Thus, the fluidflows against the increasing pressure as far as it can, atwhich point the boundary layerseparatesfromthesurface.Attheseparationlocation,thevelocitygradientatthewallandthewallshearstressarezero.Beyondthatlocation,fromDtoE,thereisreverseflowintheboundarylayer.

Figure5.15:Boundarylayerprofile,Separationpoint[FFM]

Compared with a laminar boundary layer, aturbulent boundary layer flow has more kineticenergyandmomentumassociatedwith it.Thus, theturbulent boundary layer can flow farther aroundthe cylinder before it separates than the laminarboundarylayercan.

Figure5.16:Separationpointpictures[MFM]AvoidingflowseparationA loss of pressure in the separated flow regionbehindblunt bodies causes an imbalancebetween the upstream and downstream pressure forces, contributing greatly to anincreasednetdragforce.IninternalflowBLseparationhasdrasticconsequencestopressurelosses.

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Shapedesignlimitstheadversepressuregradient

Figure5.17:Separationpoint,effectoftheshape[MFM]Energizetheflowbybypass

Figure5.18:Separationpoint,bypass[MFM]Energizetheflowbyturbulence

Figure5.19:Separationpoint,turbulence[MFM]EnergizetheBlowing

Figure5.20:Separationpoint,blowing[MFM]

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SuctionEffect

Figure5.21:Separationpoint,suction[MFM]

7.ExercisesExercise5.1*(from FuidMechanics,Munson ‐ Young – Okiishi) A 0.23m diameter soccer ballmovesthroughtheairwithaspeedof10m/s.Wouldtheflowaroundtheballbeclassifiedaslow,moderate,orlargeReynoldsnumberflow?Explain.Exercise5.2*(fromFuidMechanics,Munson ‐ Young –Okiishi)A small 15mm long fish swimswith aspeedof20mm/s.Wouldaboundary layertype flowbedevelopedalongthesidesof thefish?Explain.Exercise5.3*(fromFuidMechanics,Munson‐Young–Okiishi)Airflowoveraflatplateoflengthl=2ftsuchthattheReynoldsnumberbasedontheplatelengthisRe=2105.Plottheboundarylayerthicknessδ,for0<x<lExercise5.4*(fromFuidMechanics,Munson‐Young–Okiishi)AlaminarboundarylayerformedononesideofaplateoflengthLproducesadragD.HowmuchmusttheplatebeshortenedifthedragonthenewplateistobeD/4?Assumetheupstreamvelocityremainsthesame.Explainyouranswerphysically.Exercise5.5*(fromFuidMechanics,Munson‐Young–Okiishi)Howmuchlesspowerisrequiredtopedalaracing‐stylebicycleat20mphwitha10‐mphtailwindthanatthesamespeedwitha10mphheadwind?

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Exercise5.6A vertical wind tunnel (Quiz winter 2010) A vertical wind tunnel can be used forskydivingpractice.Estimatetheverticalwindspeedneededifa150‐lbpersonistobeableto“float”motionlesswhentheperson(a)curlsupasinacrouchingpositionor(b)liesflat.

g=32.2ft/s2Airdensity=0.745lb/ft3Exercise5.7(Quizwinter2010)Inthephotographsshown,indicateapproximatelytheregioninwhichyouexpectviscouseffectstobeimportantandtheregionwheretheflowcanbeassumedtobeinviscid.InwhichregioncanyouapplyBernoulli’sequation?

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Exercise5‐8GossamerCondorversusAlbatros(FinalW2010)In1977theGossamerCondorwontheKremerprizebybeing the first human‐powered aircraft to complete aprescribed figure‐eightcoursecoveringatotalofamile(1.6kilometres).

Figure4:GossamerCondor Figure5:AlbatrosliftanddragcoefficientsThefollowingdatapertainstothisaircraft:

DeterminetheliftcoefficientfortheGossamerCondorDeterminethelift‐to‐dragratiooftheGossamerCondorDeterminethepowertoovercomethedragDeducethepower,requiredbythepilotoftheGossamerCondorDeterminethemeanlift‐to‐dragratiooftheAlbatrosfromtheFigure5(reddottedline)Compareresultsobtainedinb)ande)withthelift‐to‐dragratiogivenintable2andexplainwhyahigherlift‐to‐dragratioistypicallyoneofthemajorgoalsinaircraftdesign.

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Flightarticle

Boeing747

Cessna

Concorde

Housesparrow

L/Dratio

17 7 4‐7 4

Table2:RepresentativeL/Dratios

Exercise5.9FlowoveraHummer(FinalS2008).TheflowaroundaHummerH2automobileisbeingstudied using a 1/18 scale model in a water tunnel at the University of Ottawa. TheexperimentisintendedtosimulatetheairflowaroundarealHummerdrivinginastraightlineat30km/houronastillday(nowind).Figure4:Hummermodel’sshapeanddimensionsa) In the photograph shown Figure A‐1, indicate approximately the region inwhich youexpectviscouseffectstobeimportantandtheregionwheretheflowcanbeassumedtobeinviscid.InwhichregioncanyouapplyBernoulli’sequation?b)DeterminetheReynoldsnumbercorrespondingtotheflowaroundtheHummerH2.GIVEN:Densityofair1.2kg/m3. Viscosityofair1.78×10−5kg/(m·s). Densityofwater998kg/m3. Viscosityofwater8.90×10−4kg/(m·s).c) The experiment is performed by mounting themodel Hummer H2 upside‐down on astationary plate in the water tunnel and by pumping water by the model at a constantvelocity.WhatwatervelocityshouldbeusedfortheexperimenttobedynamicallysimilartothetrueflowaroundarealHummerH2ontheroad?

Figure5:HummerH2experimentsetupinthewatertunnel.d)AdragforceofFD=282Nismeasuredonthemodel.Computethedragcoefficientofthecar.Whatforcedoesthiscorrespondtoontherealcar?e)Doyou foresee anyproblemsassociatedwith this setup?Discuss any sourcesof errorthatmaybepresent.f)Bonus:ReportthedragcoefficientontheFigureA‐2.Whywedonotbuythiskindofcar?Thinkgreen!

0.26m 0.11m

0.11

m

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FigureA‐1:Flowoveracar.

FigureA‐2:Thehistoricaltrendofstreamliningautomobilestoreducetheiraerodynamicdragandincreasetheirmilespergallon.FromfundamentaloffluidmechanicsMunsonedt.

Exercise5.10Redbloodcellsedimentation.1)Howfastdoesaredbloodcellof90μm3(assumeitisasphere)fallthroughtheplasma?Thedragforasphereinstokeflowisgivenby6πμRU.Theplasmaviscosityis1.4cPandtheredbloodcelldensity1.09g/cm3,plasmadensity1.03g/cm32)Theredblood inclines tocluster (aggregation).Compute thehowfastdoesasphericalclusterwithadiameterof20redbloodcellfallthroughtheplasma?Note: If you do a blood test sometime the sedimentation velocity is measured to haveinformationabouthowyourredbloodcellsinclinetoaggregate.Exercise5.11Flow of a white blood cell in amicro channel. (Midterm for undergraduate Frenchstudents, UTC). For the first approximation of the microcirculation, the capillary vesselscouldbemodelledascylinderswithradiusbandthecellsassolidbodiescenteredontheaxisofthecylinder.Weconsiderastretchedwhitebloodcell(WBC)asacylinderoflengthLand radiusb.TheWBC ismoving in the capillarywitha constant velocityUbecauseof apressuredropp1‐p2,wherep1isthepressureattheentranceofthefilm,p2pressureattheexit.

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Weassume:h=b‐a,L>>hWeneglect:EndeffectsGravitationaleffect1)Showthattheequationofconservationincylindricalcoordinateforthisproblemisgivenby:

0r

pet 0)(

1

dr

dur

dr

d

rx

p Withu=ux(r).

2)Writetheboundaryconditionsforu(r=a)andu(r=b).3)Determinerthevelocitydistributioninthefilmasafunctionofthepressuredropp/xandU.Youcanusethefollowingnotation:

G=constant=p/xand

22

4)ln(

1ba

GU

ba

A

4)Showthattheshearstressdistributiononthewhitebloodcellatr=aisgivenby:

a

Aa

Grx

2 5)Deducethedragforceduetotheshearstressappliedonthewhitebloodcell.6)What are the forcesapplyingon theWBC?Deduce the relationshipbetween thewhitebloodcellvelocityandthepressuredropG.7)Whathappenstothevelocityprofileinthefilmwhenh<<b?

Hint:saw,inthefilm1

b

r

8) Discuss this model for the flow of a white blood cell in a capillary (as shown on thefigure)Exercise5‐12Boundarylayerovertheearth’ssurface.Anatmosphericboundarylayerisformedwhenthewindblowsovertheearth’ssurface.Typically,suchvelocityprofilescanbewrittenasapower law: where the constants a and n depend on the roughness of the terrain. As isindicated in theFig.Below.Typicalvaluesare forurbanareas, forwoodlandorsuburbanareas,andforflatopencountry.Ifthevelocityis20ft_satthebottomofthesailonyourboat,whatisthevelocityatthetopofthemastIftheaveragevelocityis10mphonthetenthfloorofanurbanbuilding,whatistheaveragevelocityonthesixtiethfloor?

CapillaryWall Whitebloodcell

U a b x

r

p1 p2

L

film

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8.SolutionsSolution5.1

Solution5.2

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Solution5.4

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Solution5.5ANS:0.375hp

9.ReferencesFundamentalsofFuidMechanics,Munson‐Young–Okiishi(FFM)MultimediaFluidmechanics,Cambridgeuniversitypress(MFM)

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Chapter6

Rheologyofblood

1.Rheologyofbloodandnon‐NewtonianequationsTextbook:Biofluidmechanics,thehumancirculation,KChandranandal.,Taylor&Francis2dedition.Chapter4

2.ExercisesExercise6.1Compute the apparent viscosity of blood flowing through a tubewith a 100umdiameterusingthefreemarginalcelllayertheory.Freemarginalcelllayerforhumanblood:assumevaluesofplasmaviscosityof1.2CPandwholebloodviscosityof3.3CPat37oC.Assumeacellfreelayerthicknessof3um.

Exercise6.2(QuizW2010)

ThebloodasaNon‐Newtonianfluidcanbecharacterisedbythepowerlaw:nK

Usingaln‐lngraph,determinedgraphicallyKandnwiththefollowingdata:

)ln( 0.0 1.4 2.3 3.0 3.7 4.6 5.3 6.2)ln( ‐2.3 ‐1.2 ‐0.7 ‐0.2 0.3 1.0 1.5 2.2

)ln(

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)ln(

Exercise6.3(FinalW2010)–CapillaryViscometer

Figure6.2:PrincipaldesignofcapillaryviscometerGiven cylindrical coordinates and pressure driven laminar flow, analyze the capillaryviscometer.Theflowispressuredrivenbyapressuredropof PthroughacapillarywithalengthL(Figure3).Showthat incylindricalcoordinatestheonlynon‐zerocomponentsoftheconservationequationis:(justifyeachsimplificationyoumake)

)(1

rrrrz

Prz

Integratethisequationbypartsusingtheappropriateboundaryconditionandshowthat:

2)(

r

L

Prrz

( 0P )AssumingaNewtonia,nfluidshowthat:

)(4

1 22 rRL

Pu

ForaNewtonianfluid,computetheflowrateanddeducetherelationshipbetweenviscosityandQ, P,RandL:

L

P

Q

R

8

4

Assumingapowerlawfluidmodel,

n

rz r

uK

showthat:

)(21

11/1

n

n

n

nn

rRKL

P

n

nu

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Explainhowtoobtainthefollowingexpressionfortheflowrate(Givesomekeypointsbutdonotdothefullcalculation):

n

KL

PR

n

RnQ

/13

213

Explain qualitatively theway to getK andn, in order to characterise thenon‐Newtonianfluid(nocomputationisneeded).Exercice6.4QUIZsummer2010

x

y

z

h

h u

g

Weconsiderbloodflowinachannelwithw>>h. The relationship between and

y

u

ofthebloodfollowthepowerlawthatisgivenby:

n

y

uK

Foranyfluidthroughtwoplates,thebalancebetweenthepressuregradientandtheshearstressisgivenby:

x

py

Wheredoestheequation x

py

comefrom?

Computethevelocityprofile.Stateclearlytheboundaryconditionused.

Exercice6.5*Bingham fluid flow in a plane narrow slit. Consider a fluid (of density ρ) inincompressible,laminarflowinaplanenarrowslitoflengthLandwidthWformedbytwoflatparallelwallsthatareadistance2Bapart.EndeffectsmaybeneglectedbecauseB<<W<<L.ThefluidflowsundertheinfluenceofapressuredifferenceΔp,gravityorboth.

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Figure:Fluidflowinplanenarrowslit.

a) Determine the steady‐state velocity distribution for a non‐Newtonian fluid that isdescribedbytheBinghammodel.b)ObtainthemassflowrateforaBinghamfluidinslitflow.

Exercice6.6*Bingham fluid flow inacirculartube.Considera fluid(ofdensityρ) in incompressible,laminarflowinalongcirculartubeofradiusRandlengthL.EndeffectsmaybeneglectedbecausethetubelengthLisrelativelylargecomparedtothetuberadiusR.ThefluidflowsundertheinfluenceofapressuredifferenceΔp,gravityorboth.

Figure.Fluidflowincirculartube.

a) Determine the steady‐state velocity distribution for a non‐Newtonian fluid that isdescribedbytheBinghammodel.b)ObtainthemassrateofflowforaBinghamfluidinacirculartube.

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Exercise6.7Cell‐free Marginal layer. Assume that whole blood with a hematocrit of 45% flowsthroughasmalldiametertube.Thetotalflowrateis8µl/h,althoughinthecoreregionitis6 µl/h and in the peripherical region it is 4 µl/h. The blood cells accumulate in the coreregionwithavolumeof5µlandtherearenobloodcellspresentinthecell‐freeperiphericalregionwhichhasavolumeof3µl.a.)Drawafigureshowingthedifferentareas:cell‐freeperiphericalandcoreregionb.)Determinethehematocritinthecoreregion.c.)Thendeterminetheaveragehematocritinthewholetube.d.)Whatistheeffectofthecellfreelayerontheapparentviscosityofthebloodinthetube?

SuggestedExercisesfromthetextbook:4.3;4.4;4.5.4.6;4.7;

3.SolutionsSolution6.5BinghamfluidflowinaplanenarrowslitSolution:http://www.syvum.com/cgi/online/serve.cgi/eng/fluid/fluid806.html

Solution6.6BinghamfluidflowinacirculartubeSolution:http://www.syvum.com/cgi/online/serve.cgi/eng/fluid/fluid807.html

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Chapter7

Introductiontofluidmachinery

1.IntroductiontofluidmachineryChapterfromthebook‘Fuidmechanics(Munson‐Young‐Okiishi)’PPTarepostedonline

2.ExercisesExercise7.1*Centrifugal blood pumps (Final W2010). A centrifugal blood pump is designed toproduceavolumeflowrateof11L/minatanimpellerrotationalspeedof3610rpm.ThegeometricalspecificationsofthecentrifugalbloodpumpimpellergivenbytheconstructoraresummarizedinTable1.Theviscosityofbloodis1050kg/m3.

Figure1:Invivoanimaltestrig:(a)anatomicalanimalstudy;(b)viewoftheimpellerafterinvivotest;and(c)viewofthevolutecasingafterinvivotest.

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Figure2:Centrifugalpumpflowinandout

Inletdiameter(mm) 17.61Exitdiameter(mm) 30.00Inletwidth(mm) 2.70Exitwidth(mm) 2.70

Numberofblades 5Tipclearance(mm) 0.30Lengthinaxialdirection(mm) 5.70Bladeangleatinlet(o) 75.80Bladeangleatdischarge(o) 67.92

Table1:Specificationsofthecentrifugalbloodpumpimpeller.

a)Assuminguniform flowat the inlet and theoutlet, and that the flowenters and leavestangenttoblade:drawtheinletandoutletvelocitydiagram.

b)Showthat

cot

2 rb

QUVt

.c)Calculatethemechanicalpowerandthetheoreticalhead.d)AccordingtotheFigureA‐1intheappendix,whatisthepumpefficiencyfortheworkingconditiondescribedabove?e)Deducethepowerneededandthepumphead.PlotyourresultsontheFiguresA‐2andA‐3.Explainanydifferencesyouobtained.f)Calculatethespecificspeedandthespecificdiameterusinggivenequations.OnFigureA‐4, which presents data of efficient industrial turbomachines, plot the location of ourcentrifugalpumpusingastarsymbol.

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FigureA‐1:BloodcentrifugalpumpcharacteristicobtainedfromOhetal.withblood:Pumpefficiency

Figure A‐2: Blood centrifugal pumpcharacteristicobtainedfromOhetal.withblood:Pumpinputpower.

Figure A‐3: Blood centrifugal pumpcharacteristicobtainedfromOhetal.withblood:Staticpressurerise.

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Figure A‐4 : Data of efficient industrial turbomachines [4] Balje, O. E. Turbomachines: aguidetodesign,selection,andtheory,1981(JohnWiley,NewYork).Reference:HWOh,ESYoon,MRPark,KSun,andCMHwang.Hydrodynamicdesignandperformance analysis of a centrifugal blood pump for cardiopulmonary circulation. Proc.IMechEVol.219PartA:J.PowerandEnergy.p525‐532.Exercise72.

Heartpump‐head.Areyourfeetstillperfusedwhenyoudoahandstand?Neglecting the friction loss, estimate the elevation of the blood that ahealthy heart can achieve with a diastolic/ systolic pressure 80/120mmHg.

Exercise7.3Calculation of pump characteristics from test data‐ (From : Introduction to fluidsmechanicsFoxandall)

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Theflowsystemusedtotestacentrifugalpumpatanominalspeedof1750rpmisshown.The liquid is water at 80oF, and the suction and discharge pipe diameters are 6in. Datameasuredduringthetestaregiveninthetable.Theelectricmotorsupplied460V,3‐phase,andhasapowerfactorof0.875andaconstantefficiencyof90%.Calculatethenetheaddeliveredandthepumpefficiencyatavolumeflowrateof1000gpm.Plotthepumphead,powerinput,andefficiencyasfunctionofvolumeflowrate.

Exercise7.4*Centrifugalpump‐From:FluidmechanicsMunson‐Young‐OkiishiShown in theFig. are front and sideviewsof a centrifugalpumprotor or impeller. If thepumpdelivers200liters/sofwaterandthebladeexitangleisfromthetangentialdirection,

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determinethepowerrequirementassociatedwithflowleavingatthebladeangle.Theflowenteringtherotorbladerowisessentiallyradialasviewedfromastationaryframe.

Exercise75*SimilaritylawsFrom:FluidmechanicsMunson‐Young‐OkiishiWhen the shaft horsepower supplied to a certain centrifugal pump is 25 hp, the pumpdischarges700gpmofwaterwhileoperatingat1800rpmwithaheadriseof90ft.(a) If the pump speed is reduced to 1200 rpm, determine the new head rise and shafthorsepower.Assumetheefficiencyremainsthesame.(b)Whatisthespecificspeed,forthispump?

Exercise7.6*Velocitytriangles‐From:FluidmechanicsMunson‐Young‐OkiishiAnaxial‐flowturbomachinerotorinvolvestheupstream112anddownstream122velocitytrianglesshownintheFig. Is this turbomachineaturbineora fan?Sketchanappropriatebladesectionanddeterminetheenergytransferredperunitmassoffluid.

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3.SolutionsSolution7.1r1=0.0088r2=0.0150B1=75.8000B2=67.9200w=377.8467U1=3.3269U2=5.6677Q=1.8333e‐04b1=0.0027

b2=0.0027p=1050Vt1=3.0153Vt2=5.3748Ts=0.0104Wm=3.9329H=2.0848HmmHg=153.2698efficiency=0.5800

Wp=2.2811Hp=1.1850HpmmHg=87.1186headcoef=0.6360flowcoef=0.1438specificspead=0.8133specificdiameter=4.0901

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Solution7.4

Solution75SimilaritylawsANS:40ft,7.41hp;1630

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Solution7.6Velocitytriangles‐ANS:turbine;_36.9ft2_s2

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FormulaUNITS

SI Units

Metre/kg/second

CGS Units

Cm/gram/second

US customary Units

Inch/pound/second Length [L]

metre: m centimetre: 1 cm = 0.01m

inch: 1 in = 0.0254 m feet: 1 ft = 0.305 m

Mass [M]

kilogram: kg gram:1g = 0.001 kg pound:1lbm = 0. 454 kg

Time [T]

second: s

Surface [L2]

m2 1 cm2 = 0.0001m2 1 in2 = 0.000645 m2

Volume [L3]

m3 1 cm3 = 1ml =10-6 m3 1 in3 =0.004329 gallon = 1.64e-005 m3

Velocity [LT-1]

m.s-1 cm.s-1 1 mi.hr-1 = 0.447 m.s-1 = 1.467 ft.s-1

Acceleration [LT-2]

m.s-2 cm.s-2 ft.s-2

Force [LMT-2]

newton: N = m. kg.s-2 dyne : 1 dyn = 1 cm.g.s-2

= 1e-005 N

Pounds force : 1 lbf = 4.448 N

= 444822 dyn Frequency [T-1]

hertz: Hz = 1.s -1

Pressure, stress [L-1MT-2]

pascal: Pa = N.m-2 = kg.m-1 .s-2 1 cmH2O = 98 Pa 1 mmHg = 133.3 Pa

1 dyn.cm-2= 1 g. cm-1. s-2

= 0.1 Pa = 0.00102 cmH2O = 0.00075 mmHg

(called also barye)

Psi =1 lb.in-2

= 6895 Pa = 27.68 inH2O = 2.036 inHg

Energy, work, quantity of heat [L2MT-2]

joule: J = N m = m2 .kg.s-2

erg :1 erg = 1 g·cm2/s2 = 1e-007 J

BTU :1 btu = 1.055e10 erg =778 ft.lbf

kinematic viscosity [L2T-1]

m2·s-1 stokes: 1 St = 1 cm2·s-1 = 0.0001m2·s-1

1 ft2.s-1 = 900 St = 9e4 cSt

Density [L-3M]

Kg.m-3 1 g.cm-3= 1000 Kg.m-3 1 lbm.in-3= 16.02 Kg.m-3

dynamic viscosity [L-1MT-1]

Pa·s = N.m-2.s= m-1. kg.s-1

poise :1 P = 1 dyn.s.cm-2 = 100 cP = 0.1 Pa s

1 lbf.s-1. ft-1 = 10 P

= 1000 cP

Surface tension [MT-2]

N.m-1 = kg.s-2

lbm.s-2

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FLUID’S DENSITY AND VISCOSITY

air water plasma blood µ 18.3 µPa.s 0.89 cP 1.3 cP 3.5 cPρ (kg/m3) 1.2 1000 1025 1050

CONSERVATION LAWS: General form

Continuity equation (1 equation)

. 0U dA dt

Momentum conservation (3 equations) (+ other forces)

Ine .Ud U U dA Bd tdAt

rtial forces Pressure forces Viscous

forces Body forces

Energy Equation (1 equation)

Thermal energy Thermal expansion Viscous dissipation Heat conduction

Newtonian Fluid

Viscous stress components for a Newtonian fluid: Cartesian cylindrical

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Non-Newtonian Fluid (cylindrical)

Power law Bingham

Casson

Cauchy equation, Cartesian coordinate 

0)(2

z

w

y

v

x

u

zw

yv

xu

t

 

xxzxyxxx fgzyxx

p

z

uw

y

uv

x

uu

t

u

 

yyzyyyxy fgzyxy

p

z

vw

y

vv

x

vu

t

v

 

zzzzyzxz fgzyxz

p

z

ww

y

wv

x

wu

t

w

 

 Cauchy equation, Cylindrical coordinate 

zu

r

u

ru

t zr

 + 0

1)(

1

z

uu

rur

rrz

r  

r

u

z

uu

u

r

u

r

uu

t

u rz

rrr

r2

 = ‐

r

p

 +zrrr

r

rrzrrr

11

+ frgr  

r

uu

z

uu

u

r

u

r

uu

t

u rzr

=‐

p

r

1+

zrr

r

rzr

11 2

2+ fg  

z

uu

u

r

u

r

uu

t

u zz

zzr

z

=‐

z

p

 +zrr

r

rzzzzr

11

+ zz fg  

  

n

zrz dr

duK

yz

rz dr

duK yrz

yrz 0

dr

duz

0dr

duzyrz

yz

rz dr

duK yrz

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Navier Stokes equations, Cartesian coordinate  

0

z

w

y

v

x

u 

2

2

2

2

2

2

z

u

y

u

x

ug

x

p

z

uw

y

uv

x

uu

t

ux

 

2

2

2

2

2

2

z

v

y

v

x

vg

y

p

z

vw

y

vv

x

vu

t

vy

 

2

2

2

2

2

2

z

w

y

w

x

wg

z

p

z

ww

y

wv

x

wu

t

wz

 

  Navier Stokes equations, Cylindrical coordinate  

01

)(1

z

uu

rur

rrz

r  

r

u

z

uu

u

r

u

r

uu

t

u rz

rrr

r2

 = ‐

r

p

+

2

2

22

2

2

211

z

uu

r

u

rru

rrrrr

r + rg  

r

uu

z

uu

u

r

u

r

uu

t

u rzr

=‐

p

r

1+

ru

rz

uu

rru

rrr 22

2

2

2

2

211 + g  

z

uu

u

r

u

r

uu

t

u zz

zzr

z

=‐

z

p

 +

2

2

2

2

2

11

z

uu

rr

ur

rrzzz

+ zg  

 Stokes equations, Cartesian coordinate 

02

z

w

y

v

x

u 

2

2

2

2

2

2

z

u

y

u

x

ug

x

px

 

2

2

2

2

2

2

z

v

y

v

x

vg

y

py

 

2

2

2

2

2

2

z

w

y

w

x

wg

z

pz

 

    

Stokes equations, Cylindrical coordinate 

01

)(1

z

uu

rur

rrz

r  

r

p

=

2

2

22

2

2

211

z

uu

r

u

rru

rrrrr

r + rg  

p

r

1 =

ru

rz

uu

rru

rrr 22

2

2

2

2

211 + g

z

p

 =

2

2

2

2

2

11

z

uu

rr

ur

rrzzz

+ zg  

 

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LIFT AND DRAG

Drag for a sphere in stokes flow )(6 pfD uuRF

BL Thickness

BL displacement thickness

Momentum Thickness, θ

TURBOMACHINERY:PUMP Euler turbo-machine equation

Tangential out velocity

Mechanical Power

Head

Efficiency

Hydraulic power

Pump head

Net positive suction head: System equation

g

Pv

g

Vs

g

PsNPSH

2

2

flp HHzg

V

g

pz

g

V

g

p

1

21

11

2

22

22

22

2KQH fl