CHAPTER3SOLUTIONSOFTHENEWTONIANVISCOUSFLOWEQUATIONSSURFACERESISTANCEWITHUNIFORMLAMINARFLOW(COUETTEFLOW)Inthissection,weconsiderthreetwo-dimensionallaminarflowswithparallelstreamlines.Theflowsaresteadyanduniform;thatis,thereisnochangeinvelocityalongastreamline.Usingthemomentumequation,wewillfirstderiveageneralequationfortheflowvelocityandthenapplyittothreespecificproblem.1. Flowproducedbyamovingplate2. Liquidflowdownaninclinedplane3. Flowbetweentwostationaryparallelplate

Considerthecontrolvolumeasshownabove,whichalignedwiththeflowindirections.Thestreamlinesare inclinedatanangle 𝜃withrespect to thehorizontalplane.Thecontrolvolumehasdimension∆𝑠×∆𝑦×𝑢𝑛𝑖𝑡𝑦;thatis,thecontrolvolumehasaunitlengthintothescreen.Byapplicationofthemomentumequation,thesumoftheforcesactinginthes-directionisequaltothenetoutflowofmomentumfromthecontrolvolume.

Theflowisuniform,sothatoutflowofmomentumisequaltothein-flowandthemomentumequationreducesto

𝐹_ = 0Therearethreeforcesactingonthematterinthecontrolvolume:theforcesduetopressure,shearstressandgravity.Thenetpressureforceis:

𝑝∆𝑦 − 𝑝 +𝑑𝑝𝑑𝑠∆𝑠 ∆𝑦 = −

𝑑𝑝𝑑𝑠∆𝑠∆𝑦

Thenetforceduetoshearstressis:

𝜏 +𝑑𝜏𝑑𝑦

∆𝑦 ∆𝑠 − 𝜏∆𝑠 =𝑑𝜏𝑑𝑦

∆𝑦∆𝑠

Thecomponentofgravitationalforceis𝜌𝑔∆𝑠∆𝑦 sin 𝜃.However,sin 𝜃canberelatedtotherateatwhichtheelevation,𝑧,decreaseswithincreasing𝑠andisgivenby− kl

k_.

Thus,thegravitationalforcebecomes:

𝜌𝑔∆𝑠∆𝑦 sin 𝜃 = −𝜌𝑔∆𝑠∆𝑦𝑑𝑧𝑑𝑠

Summingallforcestozeroanddividingthroughby∆𝑠∆𝑦resultsin:

𝑑𝜏𝑑𝑦

=𝑑𝑑𝑠

𝑝 + 𝜌𝑔𝑧 =𝑑𝑑𝑠

𝑝 + 𝛾𝑧

wherewenotethatthegradientoftheshearstressisequaltothechangeinpiezometricpressureintheflowdirection.Theshearstressisequalto𝜇 ko

kp,sotheequationbecomes:

𝑑q𝑢𝑑𝑦q

=1𝜇𝑑𝑑𝑠

𝑝 + 𝜌𝑔𝑧 =1𝜇𝑑𝑑𝑠

𝑝 + 𝛾𝑧 (1)

Wewillapplythisequationtothreeflowconfigurations.

FLOWPRODUCEDBYMOVINGPLATEConsidertheflowbetweenthetwoplatesasshownbelow.Thelowerplateisfixedandtheupperplate ismovingwithaspeed 𝑈.Theplatesareseparatedbyadistance 𝐿. In thissituation,thereisnopressuregradientintheflowdirection kx

k_= 0 andthestreamlines

areinthehorizontaldirection klk_= 0

Eq.(1)canbewrittenas:

𝑑q𝑢𝑑𝑦q

= 0

Thetwoboundaryconditionsare:𝑢 = 0at𝑦 = 0

𝑢 = 𝑈at𝑦 = 𝐿

Integratingthisequationtwicegives:

𝑢 = 𝐶z𝑦 + 𝐶qApplyingtheboundaryconditionsgives:

𝑢 = 𝑈𝑦𝐿

which shows that the velocity profile is linear between two plates. The shear stress isconstantandequalto:

𝜏 = 𝜇𝑑𝑢𝑑𝑦

= 𝜇𝑈𝐿

ThisflowisknownasaCouetteflowafteraFrenchscientist,M.Couette,whodidpioneeringworkontheflowbetweenparallelplatesandrotatingcylinders.

Example:IfthefluidbetweentheplatesisSAE30lubricatingoilat𝑇 = 38℃,theplatesarespaced0.3mmapart,theupperplateismovedatavelocityof1.0m/s,whatisthesurfaceresistancefor1.0m2oftheupperplate?

𝜏 = 𝜇𝑑𝑢𝑑𝑦

= 𝜇𝑈𝐿

= 0.11

0.0003= 333 𝑁/𝑚q

𝐹 = 𝜏𝐴 = 333𝑁

LIQUIDFLOWDOWNANINCLINEDPLANEConsiderthe flowdownan inclinedplaneasshownbelow.The liquid layer,whichhasadepth𝑑,hasafreesurfacewherethepressureisconstant,soalongthesurface kx

k_= 0 .

Thepressureishydrostaticacrossanysectioninthez-directionbutdoesnotchangeinthestreamlinedirection.Theshearstressatthefreesurfacebetweentheairandliquidissmallandthereforeneglected.

In this section, the gravitational force is balanced by the shear stress. The differentialequation,Eq.(1),reducesto:

𝑑q𝑢𝑑𝑦q

=1𝜇𝑑𝑑𝑠

𝑝 + 𝛾𝑧 = −𝛾𝜇sin 𝜃

Theboundaryconditionsare:

𝑢 = 0at𝑦 = 0

kokp= 0at𝑦 = 𝑑

Integratingthisequationoncegives:

𝑑𝑢𝑑𝑦

= −𝑦𝛾𝜇sin 𝜃 + 𝐶

Applyingboundaryconditionsat𝑦 = 𝑑shows:

𝐶 =𝛾𝑑𝜇

sin 𝜃

Theequationbecomes:𝑑𝑢𝑑𝑦

=𝛾𝜇sin 𝜃 𝑑 − 𝑦

Integratingthesecondtimeresultsin:

𝑢 =𝛾𝜇sin 𝜃 𝑦𝑑 −

𝑦q

2+ 𝐶

andtheconstantofintegrationissetequaltozerotosatisfytheboundaryconditionat𝑦 = 0Thisequationcanbewrittenas:

𝑢 =𝛾 sin 𝜃2𝜇

2𝑦𝑑 − 𝑦q =𝑔 sin 𝜃2𝜐

2𝑦𝑑 − 𝑦q

𝜐 = kinematicviscosity𝜇 = dynamicviscosity

Theresultingprofileisaparabolawithmaximumvelocityoccurringatthefreesurface.Themaximumvelocityis:

𝑢��� =𝛾𝑑q

2𝜇sin 𝜃

Thedischargeperunitwidthcanbeobtainedbyintegratingthevelocity𝑢overthedepthofflow:

𝑞 = 𝑢 ∙ 𝑑𝑦 =k

�

𝛾 sin 𝜃2𝜇

𝑑𝑦q −𝑦�

3 �

k

=13𝛾 sin 𝜃𝜇

𝑑�

Theaveragevelocity,𝑉,canbeobtainedby:

𝑉 =𝑞𝑑=13𝛾 sin 𝜃𝜇

𝑑q =𝑔𝑑q

3𝜐sin 𝜃

𝑑 = crosssectionalarea

𝑞 = discharge

The slope, 𝑆� = tan 𝜃 , is approximately equal to sin 𝜃 for a small slopes. Sometimes,equationforvelocitycanbewrittenas:

𝑉 =𝑔𝑑q𝑆�3𝜐

ExperimentshaveshownthatifReynoldsnumberbasedonthedepthoftheflow,

𝑅𝑒 =𝜌𝑉𝑑𝜇

=𝑉𝑑𝜐

is lessthan500,onecanexpect laminarflowinthissituation. If theReynoldsnumber isgreaterthan500,theflowbecometurbulentandtheresultsofthissectionarenolongervalid.

Example:Crudeoil,𝜐 = 9.3×10��m2/s,𝑆𝐺 = 0.92,flowsoveraflatplatethathasaslopeof𝑆� = 0.02. If the depth of flow is 6mm,what is themaximum velocity andwhat is thedischargepermeterofwidthofplate?DeterminetheReynoldsnumberforthisflow.

𝑢 =𝑔𝑆�2𝜐

2𝑦𝑑 − 𝑦q At𝑦 = 𝑑;𝑢��� = 0.038𝑚/𝑠Dischargepermeterofwidthis:

𝑞 =13𝛾 sin 𝜃𝜇

𝑑� =𝑔𝑆�𝑑�

3𝜐= 1.52×10��𝑚q/𝑠

Reynoldsnumber:

𝑅𝑒 =𝑉𝑑𝜐= 1.63

(Theflowislaminar)

FLOWBETWEENSTATIONARYPARALLELPLATESConsiderthetwoparallelplatesseparatedbyadistanceBasshownbelow.

Inthissituation,theflowvelocityiszeroatthesurfaceforbothplates.Theboundaryconditionsbecome:

𝑢 = 0at𝑦 = 0

𝑢 = 0at𝑦 = 𝐵

Thegradientinpiezometricpressureisconstantalongthestreamline.IntegratingEq.(1)twicegives:

𝑢 =𝑦q

2𝜇𝑑𝑑𝑠

𝑝 + 𝛾𝑧 + 𝐶z𝑦 + 𝐶q

Tosatisfytheboundaryconditionat𝑦 = 0,weneedtoset𝐶q = 0.Applyingtheboundaryconditionat𝑦 = 𝐵requiresthat𝐶zis:

𝐶z = −𝐵2𝜇

𝑑𝑑𝑠

𝑝 + 𝛾𝑧

Thefinalequationforthevelocityis:

𝑢 = −12𝜇

𝑑𝑑𝑠

𝑝 + 𝛾𝑧 𝐵𝑦 − 𝑦q = −𝛾2𝜇

𝐵𝑦 − 𝑦q𝑑ℎ𝑑𝑠

whichisparabolicprofilewiththemaximumvelocityoccurringonthecentrelinebetweentheplates 𝑦 = �

q.

Themaximumvelocityis:

𝑢��� = −𝐵q

8𝜇𝑑𝑑𝑠

𝑝 + 𝛾𝑧

orintermspiezometrichead:

𝑢��� = −𝐵q𝛾8𝜇

𝑑ℎ𝑑𝑠

Thefluidalwaysflowsinthedirectionofdecreasingpiezometricpressureorpiezometrichead,sok�

k_isnegative.

Thisgivingapositivevaluefor𝑢���.

Thedischargeperunitwidthcanbefindbyintegratingthevelocityoverthedistancebetweenplates:

𝑞 = 𝑢 ∙ 𝑑𝑦 =�

�−

𝐵�

12𝜇𝑑𝑑𝑠

𝑝 + 𝛾𝑧 = −𝐵�

12𝜇𝑑ℎ𝑑𝑠

Theaveragevelocityis:

𝑉 =𝑞𝐵= −

𝐵q

12𝜇𝑑𝑑𝑠

𝑝 + 𝛾𝑧 =23𝑢���

Aswasthecasewithunconfinedlaminarliquidflowoveraplanesurface,thevelocitydistributionisparabolic.However,inthissituation,themaximumvelocityoccursmidwaybetweentwoplates.Notethatflowisresultofachangeofthepiezometrichead,notjustchangeof𝑝or𝑧alone.ExperimentsrevealthatifReynoldsnumberislessthan1000,theflowislaminar.ForReynoldsnumbergreaterthan1000,theflowmaybeturbulentandtheequationsinthissectioninvalid.

Example:Oilhavingaspecificgravityof0.8andaviscosityof0.02N.s/m2flowsdownwardbetweentwoverticalsmoothplatesspaced10mmapart.Ifthedischargepermeterofwidthis0.01m2/s,whatisthepressuregradientkx

k_forthisflow.

Reynoldsnumber:

𝑅𝑒 =𝑉𝐵𝜐=𝜌𝑉𝐵𝜇

=𝜌𝑞𝜇= 400

(Flowislaminar)

𝜐 =𝜇𝜌= 0.000025𝑚q/𝑠

Thegradientinpiezometricheadcanbewrittenas:

𝑑ℎ𝑑𝑠

= −12𝜇𝑞𝐵�𝛾

= −12𝜐𝑞𝐵�𝑔

= −0.306

However,𝑑ℎ𝑑𝑠

=𝑑𝑑𝑠

𝑝𝛾+ 𝑧 = −0.306

Becausetheplatesareverticallyorientedand𝑠ispositivedownward,kl

k_= −1,

Thus,

𝑑 𝑝𝛾𝑑𝑠

= 1 − 0.306or

𝑑𝑝𝑑𝑠

= 5447𝑁/𝑚qIntheotherword,thepressureisincreasingdownwardatarateof5.45kPapermeteroflengthofplate.

FULLYDEVELOPEDFLOWBETWEENPARALLELPLATESUSINGNAVIER-STOKESEQUATIONSTheflowfieldbetweenparallelplateswillbederivedhereusingthecontinuityandNavier-Stokesequations.Referenceismadetoabovementionfigure,where𝑦isthecoordinatenormaltotheplatesand𝑥istheflowdirection(sameas𝑠-direction).

Theflowfieldisfullydeveloped(uniform),sothederivatives o

 �and ¡

 �arezero.

Alsotheflowissteady,so o

 ¢and ¡

 ¢arezero.

Thecomponentofthegravityvectorinthe𝑥-directionis𝑔 sin 𝜃andThecomponentofthegravityvectorinthe𝑦-directionis−𝑔 cos 𝜃.Thecontinuityequationforplanarincompressibleflowis:

𝜕𝑢𝜕𝑥

+𝜕𝑣𝜕𝑦

= 0

Since o

 �= 0,thecontinuityequationreducesto:

𝜕𝑣𝜕𝑦

= 0

or𝑣 = constant

Atthesurfaceoftheplate(𝑦 = 0),thevelocityiszero.So,

𝑣 = 0everywhereintheflowfield.

TheNavier-Stokesequationinthe𝑦-directionis:

𝜌𝜕𝑣𝜕𝑡+ 𝜌𝑢

𝜕𝑣𝜕𝑥

+ 𝜌𝑣𝜕𝑣𝜕𝑦

= −𝜕𝑝𝜕𝑦

+ 𝜇𝜕q𝑣𝜕𝑥q

+𝜕q𝑣𝜕𝑦q

− 𝜌𝑔 cos 𝜃

Because𝑣iszeroeverywhere,thereisnoaccelerationofthefluidinthe𝑦-direction,sothiscomponentoftheNavier-Stokesequationreducesto:

𝜕𝑝𝜕𝑦

= −𝜌𝑔 cos 𝜃

Integratingover𝑦,onehas

𝑝 = −𝑦𝜌𝑔 cos 𝜃 + 𝑝p¥� 𝑥 where𝑝p¥� 𝑥 isthepressuredistributionalongthelowerwall.Thisequationshowsthatthepressuredecreaseswithelevationintheduct,asexpected.Infact,𝑦 cos 𝜃isequalto𝑧,sothisequationcanbewrittenastheequationforhydrostaticpressurevariation,namely:

𝜌 + 𝜌𝑔𝑧 = 𝑝p¥� 𝑥

Thepressuregradientinthe𝑥-directionis:

𝜕𝑝𝜕𝑥

=𝜕𝑝p¥�𝜕𝑥

=𝑑𝑝𝑑𝑥

andisthesameforallvaluesof𝑦acrosstheductforanyvalueof𝑥.TheNavier-Stokesequationinthe𝑥-directionis:

𝜌𝜕𝑢𝜕𝑡+ 𝜌𝑢

𝜕𝑢𝜕𝑥

+ 𝜌𝑣𝜕𝑢𝜕𝑦

= −𝜕𝑝𝜕𝑥

+ 𝜇𝜕q𝑢𝜕𝑥q

+𝜕q𝑢𝜕𝑦q

+ 𝜌𝑔 sin 𝜃

Forsteady,fullydevelopedflowtheleft-hand-sideofthisequationreducestozero(noaccelerationinthe𝑥-direction),andtheequationbecomes:

𝜕𝑝𝜕𝑥

− 𝜌𝑔 sin 𝜃 = 𝜇𝜕q𝑢𝜕𝑦q

Because𝑢isafunctionof𝑦only  o �= 0 and x

 �isafunctiononlyof𝑥,thisequation

becomes:=𝑑𝑝𝑑𝑥

− 𝜌𝑔 sin 𝜃 = 𝜇𝑑q𝑢𝑑𝑦q

Theslopeoftheductcanbeexpressedas:

sin 𝜃 = −𝑑𝑧𝑑𝑥

Previously,weknowthat:

𝜇𝑑q𝑢𝑑𝑦q

=𝑑𝑝𝑑𝑥

− 𝜌𝑔 −𝑑𝑧𝑑𝑥

= 𝑑𝑑𝑥

𝑝 + 𝜌𝑔𝑧

or

𝜌𝑔𝑑ℎ𝑑𝑠

= 𝛾𝑑ℎ𝑑𝑠

= 𝜇𝑑q𝑢𝑑𝑦q

whichisthesameequationasusedintheprevioussectiontoobtainthevelocitydistribution.