CHAPTER4LAMINARBOUNDARYLAYERINVISCIDFLOWPASTWEDGESANDCORNERSFALKNER-SKANSOLUTIONInfluiddynamics,theFalkner–Skanboundarylayer(namedafterV.M.FalknerandSylviaW.Skan)describesthesteadytwo-dimensionallaminarboundarylayerthatformsonawedge,i.e.flowsinwhichtheplateisnotparalleltotheflow.ItisageneralizationoftheBlasiusboundarylayer.FalknerandSkan(1931)foundthatsimilaritywasachievedbythevariable𝜂 = 𝐶𝑦𝑥^,whichisconsistentwithapower-lawfreestreamvelocitydistribution.

𝑈 𝑥 = 𝐾𝑥a

𝑚 = 2𝑎 + 1Theexponent𝑚maybetermedtheFalkner-Skanpower-lawparameter.Theconstant𝐶mustmake𝜂dimensionlessbutisotherwisearbitrary.Thebestchoiceis𝐶g = h ija

gk,whichisconsistentwithitslimitingcasefor𝑚 = 0,theBlasius

variable.

Thus;

𝜂 = 𝑦𝑚 + 12

𝑈 𝑥𝜐𝑥

= 𝑦𝑚 + 12

𝑈𝜐𝑥

Substitutingthisparticular𝐶intoEq.4-67,givesthemostcommonformoftheFalkner-Skanequationforsimilarflows:

𝑓ttt + 𝑓𝑓tt + 𝛽 1 − 𝑓tg = 0where

𝛽 =2𝑚1 +𝑚

EXAMPLEEstimatethevariationofsurfacevelocityalongthewallifangleofwedgeis10°.

Knownthat:𝑈 𝑥 = 𝐾𝑥aand𝑚 = 2𝑎 + 1yzg= 10°,sothat𝛽 = i

{

𝛽 = ga

ija= i

{,sothat𝑚 = i

i|

Velocityprofileis𝑈 𝑥 = 𝐾𝑥

}}~

𝛽 𝑚 Descriptionofflow

−2 ≤ 𝛽 ≤ 0 −12≤ 𝑚 ≤ 0 Flowaroundanexpansioncornerofturningangleyz

g

𝛽 = 0 𝑚 = 0 Theflatplate

0 ≤ 𝛽 ≤ +2 0 ≤ 𝑚 ≤ ∞ Flowagainstawedgeofhalfangleyzg

𝛽 = 1 𝑚 = 1 Theplanestagnationpoint(180°wedge)

𝛽 = +4 𝑚 = −2 Doubletflownearaplanewall

𝛽 = +5 𝑚 = −53 Doubleflowneara90°corner

𝛽 = +∞ 𝑚 = −1 Flowtowardapointsink

THEPLANELAMINARJETFLOWConsideraplanejetemergingintoastillambientfluidfromaslotat𝑥 = 0,asshownbelow:

Inthissituation,theconservationofmomentumwasapplied.

Momentumflow,𝑚𝑉 = 𝜌 𝑉 ∙ 𝑛 𝐴𝑉 = 𝜌𝑉�𝐴𝑉

Themomentumfluxisdefinedasthemomentumflowperunitarea.Wecouldsimplifyitas:

Momentumflux,𝐽 = 𝜌 𝑢gj�

��𝑑𝑦 =

169𝜌 𝜐i/g 𝑎�

𝑎 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡

Themaximumvelocitycanbeconcludedas:

𝑢a^� = 0.4543𝐽g

𝜌𝜇𝑥

i/�

Wemaydefinethewidthofthejetastwicethedistance𝑦where𝑢 = 0.01𝑢a^�:

𝑊𝑖𝑑𝑡ℎ = 2𝑦 i% = 𝑏 = 21.8𝑥g𝜇g

𝐽𝜌

i/�

Themassflowrateacrossanyverticalplaneisgivenby:

𝑚 = 𝜌 𝑢j�

��𝑑𝑦 = 36𝐽𝜌𝜇𝑥 i/� = 3.302 𝐽𝜌𝜇𝑥 i/�

whichisseentoincreasewith𝑥i/�asthejetentrainsambientfluidbydraggingitalong.Thisresultiscorrectatlarge𝑥butimpliesfalselythat𝑚 = 0at𝑥 = 0,whichistheslotwherethejetissues.ThereasonisthattheboundarylayerapproximationsfailiftheReynoldsnumberissmall,andtheappropriateReynoldsnumberhereis:

𝑅𝑒 =𝑚𝜇

Thus,thesolutionisinvalidforsmallvaluesofReynoldsnumberof:

𝑅𝑒 =𝐽𝜌𝑥𝜇g

EFFECTOFPRESSUREGRADIENTSEPARATIONANDFLOWOVERCURVEDSURFACESAllsolidobjectstravelingthroughafluid(oralternativelyastationaryobjectexposedtoamovingfluid)acquireaboundarylayeroffluidaroundthemwhereviscousforcesoccurinthelayeroffluidclose to the solid surface. Boundary layers can be either laminar or turbulent. A reasonableassessmentofwhethertheboundarylayerwillbelaminarorturbulentcanbemadebycalculatingtheReynoldsnumberofthelocalflowconditions.Flowseparationoccurswhentheboundarylayertravelsfarenoughagainstanadversepressuregradientthatthespeedoftheboundarylayerrelativetotheobjectfallsalmosttozero.Thefluidflowbecomesdetachedfromthesurfaceoftheobject,andinsteadtakestheformsofeddiesandvortices.Boundarylayerseparationisthedetachmentofaboundarylayerfromthesurfaceintoabroaderwake.Boundarylayerseparationoccurswhentheportionoftheboundarylayerclosesttothewallorleadingedgereversesinflowdirection.Theseparationpointisdefinedasthepointbetweentheforwardandbackward flow,where theshear stress is zero.Theoverallboundary layer initiallythickenssuddenlyattheseparationpointandisthenforcedoffthesurfacebythereversedflowatitsbottom.

Wehavesofarconsideredflowinwhichthepressureoutsidetheboundarylayerisconstant. If,however, thepressure varies in thedirectionof flow, thebehaviourof the fluidmaybe greatlyaffected.Let us consider flow over a curved surface as illustrated below. The radius of curvature iseverywherelargecomparedwiththeboundarylayerthickness.

Asthefluidisdeflectedroundthesurface,itisacceleratedovertheleft-handsectionuntilatpositionC,thevelocityjustoutsidetheboundarylayerisamaximum.Here,thepressureisaminimum.Thus,fromAtoC,thepressuregradient� 

��isnegative.

Thenetpressureforceonanelementintheboundarylayerisintheforwarddirection.BeyondpointC,however,thepressureincreases,andsothenetpressureforceonanelementintheboundarylayeropposestheforwardflow.AtpointD,thevalueof�¡

�¢atthesurfaceiszero.

Furtherdownstream,atpointE,theflowclosetothesurfacehasactuallybeenreversed.Thefluidnolongerabletofollowthecontourofthesurface,breaksawayfromit.Thisbreakawaybeforetheendofthesurfaceisreachedisusuallytermedseparation.Itfirstoccursattheseparationpointwhere �¡

�¢ ¢£¤becomezero.

Itiscausedbythereductionofvelocityintheboundarylayer,combinedwithapositivepressuregradient.

Separationcanthereforeoccuronlywhenanadverse(positive)pressuregradientexists.Flowoveraflatplatewithzeroornegativepressuregradientwillneverseparatebeforereachingtheendoftheplate,nomatterhowlongtheplate.Inan ideal fluid,separationfromacontinuoussurfacewouldneveroccur,evenwithanadversepressuregradientbecausetherewouldbenofrictiontoproduceaboundarylayeralongthesurface.Thelineofzerovelocitydividingtheforwardandreverseflowleavesthesurfaceattheseparationpoint,and isknownas theseparationstreamline.Asaresultof thereverse flow, large irregulareddiesareformedinwhichmuchenergyisdissipatedasheat.Separationoccurswithboth laminarand turbulentboundary layer.However, laminarboundarylayersaremuchmorepronetoseparationthatturbulentones.Thisisbecauseinalaminarboundarylayer,theincreaseofvelocitywithdistancefromthesurfaceislessrapid,andtheadversepressuregradientcanmorereadilyhalttheslow-movingfluidclosetothesurface.

A turbulent boundary layer can survive an adverse pressure gradient for some distance beforeseparating.Foranyboundarylayer,however,thegreatertheadversepressuregradient,thesoonerseparationoccurs.Theboundarylayerthickensrapidlyinanadversepressuregradient,andtheassumptionthat𝛿issmallmaynolongerbevalid.

Inalmostallcasesinwhichflowtakesplaceroundasolidbody,theboundarylayerseparatesfromthesurfaceat somepoint.Oneexception isan infinitesimally thin flatplateparallel to themainstream.Downstreamoftheseparationpositiontheflowisgreatlydisturbedbylarge-scaleeddies,andthisregionofeddyingmotionisusuallyknownasthewake.Asaresultoftheenergydissipatedbythehighlyturbulentmotioninthewake,thepressuretherereducedandthepressuredragonthebodyisthusincreased.Themagnitudeofthepressuredragdependsverymuchonthesizeofthewakeandthis,inturn,dependsonthepositionofseparation.Iftheshapeofthebodyissuchthatseparationoccursonlywelltowardstherear,andthewakeissmall,thepressuredragisalsosmall.Suchabodyistermedastreamlinedbody.Forbluffbody,ontheotherhand,theflowisseparatedovermuchofthesurface,thewakeislargeandthepressuredragismuchgreaterthantheskinfriction.

DEVELOPMENTOFWAKEBEHINDCYLINDERTheflowpatterninthewakedependsontheReynoldsnumberoftheflow.

Figure(a):ForverylowReynoldsnumber 𝑅𝑒 < 0.5 ,theinertiaforcesarenegligible,andthestreamlinescometogetherbehindthecylinder.Figure(b):IfReynoldsnumberincreasedtotherange2-30,theboundarylayerseparatessymmetricallyfromthetwosidesatthepositionsS.Twoeddiesareformedwhichrotateinoppositedirections.AttheseReynoldsnumber,theyremainunchangedinposition,theirenergybeingmaintainbytheflowfromtheseparatedboundarylayer.Behindtheeddies,however,themainstreamlinescometogether.Thelengthofthewakeislimited.Figure(c):WithincreaseofReynoldsnumber,theeddieswaselongatedbutthearrangementisunstable.Figure(d):AtReynoldsnumber40-70,foracircularcylinder,aperiodicoscillationofthewakeisobserved.Then,atacertain limitingvalueofReynoldsnumber,usuallyabout90 foracircularcylinder inunconfined flow, theeddiesbreakoff fromeachsideof thecylinderalternatelyandarewasheddownstream.ThislimitingvalueofReynoldsnumberdependsontheturbulenceoftheoncomingflow,ontheshapeofthecylinderandonthenearnessofothersolidsurfaces.

In a certain range of Reynolds number above the limiting value, eddies are continuously shedalternatelyfromthetwosidesofthecylinderand,asaresult,theyformtworowsofvorticesinitswake,thecentreofavortexinonerowbeingoppositethepointmidwaybetweenthecentresofconsecutivevorticesintheotherrow.Thisarrangementofvorticesisknownasavortexstreetorvortextrail.Von Karman considered the vortex street as a series of separate vortices in an ideal fluid anddeducedthattheonlypatternstabletosmalldisturbance,andthenonlyif:

ℎ𝑙=1𝜋𝑎𝑟𝑐𝑠𝑖𝑛ℎ1 = 0.281

Avaluelaterisconfirmedexperimentally.

WAKEOFANAIRFOILAwakeisthedefectisstreamvelocitybehindanimmersedbodyinaflow,ashownbelow.

Dragforcecanbewrittenas:

𝐹« = 𝐶« ∙12𝜌𝐴𝑈g

Thewakevelocitymaybewrittenas:

𝑢i𝑈¤

= 𝐶«𝑅𝑒¬16𝜋

i/g 𝐿𝑥

i/g𝑒𝑥𝑝 −

𝑈¤𝑦g

4𝑥𝜐

𝑢i = Defectvelocity

𝑈¤ = Freestreamvelocity𝜐 = Kinematicviscosityoffluid

THECORRELATIONMETHODOFTHWAITESFromVonKarmanintegralrelation,wecanrewritethemomentumrelationinthemorecompactformas:

𝜏°𝜌𝑈g

=𝐶±2=𝑑𝜃𝑑𝑥

+ 2 + 𝐻𝜃𝑈𝑑𝑈𝑑𝑥 (Eq.4-122)

𝐻 =𝛿∗

𝜃

𝜆 =𝜃g

𝜐∙𝑑𝑈𝑑𝑥

𝜐 = Kinematicviscosity

Multiplythemomentum-integralrelationby¶·k

𝑈𝜃𝜐

𝜏°𝜌𝑈g

=𝑈𝜃𝜐

𝑑𝜃𝑑𝑥

+𝑈𝜃𝜐

2 + 𝐻𝜃𝑈𝑑𝑈𝑑𝑥

𝜏°𝜃𝜇𝑈

=𝑈𝜃𝜐

𝑑𝜃𝑑𝑥

+𝜃g

𝜐𝑑𝑈𝑑𝑥

2 + 𝐻 (Eq.4-133)

Now,𝐻andtheleft-handsideofthisequationaredimensionlessboundarylayerfunction.Thus,byassumption,arecorrelatedreasonablybyasingleparameter(𝜆inthiscase).Thusweassume,afterHolsteinandBohlen(1940),that:

𝜏°𝜃𝜇𝑈

≈ 𝑆 𝜆 Shearcorrelation

𝐻 =𝛿∗

𝜃≈ 𝐻 𝜆 Shape-factorcorrelation

Andfurthernotethat:

𝜃𝑑𝜃 = 𝑑𝜃g

2

Eq.4-133maythusberewrittenas:

𝑈𝑑𝑑𝑥

𝜆𝑈′

≈ 2 𝑆 𝜆 − 𝜆 2 + 𝐻 = 𝐹 𝜆 (Eq.4-135)

Whereas earlier workers would have proposed a family of profiles to evaluate the parametricfunctionsinEq.4-135,Thwaites(1949)abandonedthefavorite-familyideaandlookedattheentirecollectionofknownanalyticandexperimentalresultstoseeiftheycouldbefitbyasetofaverageone-parameterfunctions.AsshowninFigure4-22,hefoundexcellentcorrelationforthefunction𝐹 𝜆 andproposedasimplelinearfit.

𝐹 𝜆 ≈ 0.45 − 6.0𝜆 (Eq.4-136)

If𝐹 = 𝑎 − 𝑏𝜆,Eq.4-135hasaclosed-formsolutionwhichthereadermayverifyasanexercise:

𝜃g

𝜐= 𝑎𝑈�» 𝑈»�i ∙ 𝑑𝑥 + 𝐶

�

�¼ (Eq.4-137)

If𝑥¤isastagnationpoint,theconstant𝐶mestbezerotoavoidaninfinitemomentumthicknesswhere𝑈 = 0.Thus,Thwaiteshasshownthat𝜃 𝑥 ispredictedveryaccurately(±3%),foralltypesoflaminarboundarylayers,bythesimplequadrature.

𝜃g ≈0.45𝜐𝑈¾

𝑈¿ ∙ 𝑑𝑥�

¤ (Eq.4-138)

Thwaites’suggestedcorrelationsfor𝑆 𝜆 and𝐻 𝜆 asfollows:

𝑆 𝜆 ≈ 𝜆 + 0.09 ¤.¾g (Eq.4-140)

𝐻 𝜆 ≈ 2.0 + 4.14𝑧 − 83.5𝑧g + 854𝑧� − 3337𝑧Â + 4576𝑧¿ (Eq.4-141)

𝑧 = 0.25 − 𝜆