MCG3143: Biofluid mechanics
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Transcript of MCG3143: Biofluid mechanics
1
FacultyofEngineeringDepartmentofMechanicalEngineering
MCG3143:BiofluidmechanicsLecturesnotes
Summer2015
MarianneFenech
2
3
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
4
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
5
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
6
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):
11
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
12
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!
18
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
20
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
MGC3143,M.Fenech.
28
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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48
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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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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51
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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55
r
V
rr
V
z
p
t
V 112
2
becomes:
4‐SolvetheEq.W5forn=0.
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56
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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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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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