CHAPTER2:FUNDAMENTALEQUATIONSOFCOMPRESSIBLEVISCOUSFLOWCoordinatesystemTherearetwocommoncoordinatesystem:1. Cartesiancoordinate2. Polarcoordinate

Cartesiancoordinatesystem:A Cartesian coordinate system is a coordinate system that specifies each pointuniquely in a plane by a pair of numerical coordinates, which are the signeddistancestothepointfromtwofixedperpendiculardirectedlines,measuredinthesame unit of length. Each reference line is called a coordinate axis or just axis(pluralaxes)ofthesystem,andthepointwheretheymeetisitsorigin,atorderedpair (0, 0). The coordinates can also be defined as the positions of theperpendicular projections of the point onto the two axes, expressed as signeddistancesfromtheorigin.

ExampleofCartesiancoordinate

Motionofoneelementcanbeshownasbelow:

𝑢 = velocityatx-axis𝑣 = velocityaty-axis𝑞 = resultantvelocity

𝑞 = 𝑢[ + 𝑣[

Polarcoordinatesystem:In mathematics, the polar coordinate system is a two-dimensional coordinatesysteminwhicheachpointonaplaneisdeterminedbyadistancefromareferencepointandananglefromareferencedirection.Thereferencepoint(analogoustotheoriginofaCartesiancoordinatesystem)iscalledthepole,andtherayfromthepoleinthereferencedirectionisthepolaraxis.Thedistancefromthepoleiscalledtheradialcoordinateorradius,andtheangleiscalledtheangularcoordinate,polarangle,orazimuth.

𝑢^ = velocityatradialdirection𝑢_ = velocityattangentialdirection

𝑞 = resultantvelocity

𝑞 = 𝑢^ [ + 𝑢_ [

Thebasicequationsconsideredinthischapterarethethreelawsofconservationforphysicalsystems:1. Conservationofmass(continuity)2. Conservationofmomentum(Newton’ssecondlaw)3. Conservationofenergy(firstlawofthermodynamics)

CONSERVATIONOFMASSTHEEQUATIONOFCONTINUITYMassflowrateintotheelementinx-andy-directionisshowninthefigurebelow.

Thenetfluxofmassenteringtheelementequaltotherateofchangeofthemassoftheelement.

𝑚in − 𝑚out =𝜕𝜕𝑡𝑚element

𝜌𝑢𝑑𝑦𝑑𝑧 + 𝜌𝑣𝑑𝑥𝑑𝑧 − 𝜌𝑢 +𝜕 𝜌𝑢𝜕𝑥

𝑑𝑥 𝑑𝑦𝑑𝑧 − 𝜌𝑣 +𝜕 𝜌𝑣𝜕𝑦

𝑑𝑦 𝑑𝑥𝑑𝑧 =𝜕𝜕𝑡

𝜌𝑑𝑥𝑑𝑦𝑑𝑧

Simplifyingtheaboveexpression:

𝜕 𝜌𝑢𝜕𝑥

+𝜕 𝜌𝑣𝜕𝑦

+𝜕𝜌𝜕𝑡

= 0

Ifthez-directionisexist,itwillbecome:

𝜕 𝜌𝑢𝜕𝑥

+𝜕 𝜌𝑣𝜕𝑦

+𝜕 𝜌𝑤𝜕𝑧

+𝜕𝜌𝜕𝑡

= 0

Then,thedifferentialcontinuityequationcanbewrittenas:

𝜕𝜌𝜕𝑡+ 𝑢

𝜕𝜌𝜕𝑥

+ 𝑣𝜕𝜌𝜕𝑦

+ 𝑤𝜕𝜌𝜕𝑧

+ 𝜌𝜕𝑢𝜕𝑥

+𝜕𝑣𝜕𝑦

+𝜕𝑤𝜕𝑧

= 0

This is themost general form of the differential continuity equation expressedusingrectangularcoordinates.

Forthecaseofincompressibleflow,aflowinwhichdensityofafluidparticledoesnotchangeasittravelsalong,thecontinuityequationbecomes:

𝜕𝑢𝜕𝑥

+𝜕𝑣𝜕𝑦

+𝜕𝑤𝜕𝑧

= 0

Assume thatwe arediscussingonly2-D coordinate, and there is no changes indensity(incompressible),wemightexpressthecontinuityequationasfollows:

𝑑𝑢𝑑𝑥

+𝑑𝑣𝑑𝑦

= 0

CONSERVATIONOFMOMENTUMTHENAVIER-STOKESEQUATIONSTheNavier-Stokesequationiswidelyusedinboththeoryandinapplication.TheNavier-StokesequationrepresentsNewton’ssecondlawofmotionasappliedtoviscousflowofaNewtonianfluid. Inthisnotes,weassumeincompressibleflowandconstantviscosity.Similar to continuity equations, there are multiple ways to derive the Navier-Stokesequation.Thisnoteshowshowtoderive theequationbystartingwithafluid particle and applying Newton’s second law. Thus, the result will be thenonconservation form of the equation. Because the derive is complex,we omitsomeofthetechnicaldetails.

Step1:SelectafluidparticleSelectafluidparticleinaflowingfluid.Imaginethattheparticlehastheshapeofacube. Assume the dimensions are infinitesimal and that the particle is at theposition 𝑥, 𝑦, 𝑧 attheinstantshown.

Step2:ApplyNewton’ssecondlawRegardingtheforces,thetwocategoriesarebodyforcesandsurfaceforces.Theonlypossiblesurfaceforcesarethepressureforce 𝐹p andtheshearforce 𝐹q .Assumethattheonlybodyforceistheweight 𝑊 .

𝐹 = 𝑚𝑎Sumofforcesonaparticle =(mass)×(acceleration)

Bodyforce+Surfaceforce = 𝑚𝑎 = 𝜌∀𝑎

∀= volume

𝑊 + 𝐹v + 𝐹w = 𝜌∀𝑑𝑣𝑑𝑡

𝑊 = 𝑚𝑔 = 𝜌∀𝑔

𝜌∀𝑔 + 𝐹v + 𝐹w = 𝜌∀𝑑𝑣𝑑𝑡

Step3:AnalyzethepressureforceTobegin,considertheforcesonthex-facesoftheparticle.

Thenetforceduetopressureonthex-facesis:

𝐹v = 𝑃𝐴

𝐹v{| = 𝑃|{∆|[

𝐴 − 𝑃|~∆|[

𝐴 ∙ 𝑖

= 𝑃|{∆|[

− 𝑃|~∆|[

∙ ∆𝑥 ∙ ∆𝑦 ∙ 𝑖

Simplify above equation by applying a Taylor series expansion (twice) andneglectinghigherordertermtogive:

𝐹v{| =𝜕𝑝𝜕𝑥

∆𝑥∆𝑦∆𝑧 𝑖Repeatthisprocessforthey-facesandz-faces,andcombineresultstogive:

𝐹v{��� = −𝜕𝑝𝜕𝑥

∆𝑥∆𝑦∆𝑧 𝑖 +𝜕𝑝𝜕𝑦

∆𝑥∆𝑦∆𝑧 𝑗 +𝜕𝑝𝜕𝑧

∆𝑥∆𝑦∆𝑧 𝑘

Simplifyitandthenintroducevectornotationtogive:

𝐹v = −𝜕𝑝𝜕𝑥𝑖 +

𝜕𝑝𝜕𝑦𝑗 +

𝜕𝑝𝜕𝑧𝑘 ∆𝑥∆𝑦∆𝑧 = −∇𝑝 ∆𝑥∆𝑦∆𝑧

Itrevealsaphysicalinterpretationofthegradient:

Gradientofthepressurefieldatapoint =

NetpressureforceonafluidparticleVolumeoftheparticle

Step4:AnalyzetheshearforceTheshearforceisthenetforceonthefluidparticleduetoshearstresses.ShearstressiscausedbyviscouseffectsandisrepresentedmathematicallyasinFigurebelow.This figureshowsthateach faceof the fluidparticlehas three(3)stresscomponents.Forexample,thepositivex-faceshasthreestresscomponents,whichare𝜏||, 𝜏|�and𝜏|�.Thedoublesubscriptnotationdescribesthedirectionofthestresscomponentandthefaceonwhichthecomponentacts.Forexample:o 𝜏||istheshearstressonthex-facesinthe𝑥-directiono 𝜏|�istheshearstressonthex-facesinthe𝑦-directiono 𝜏|�istheshearstressonthex-facesinthe𝑧-direction

Shearstressisatypeofmathematicalentitycalledasecondordertensor.Atensorisanalogoustobutmoregeneralthanavector.Example:Azerothordertensorisa scalar, a first order tensor is a vector. A second order tensor hasmagnitude,directionandorientation(whereorientationdescribeswhichfacethestressactson)

Tofindthenetshearforceontheparticle,eachstresscomponentisbemultipliedbyarea,andtheforcesareadded.Then,aTaylorseriesexpansionisappliedandtheresultisthat:

𝐹w���^ = 𝐹w =𝐹|,q���^𝐹�,q���^𝐹�,q���^

=

𝜕𝜏||𝜕𝑥

+𝜕𝜏|�𝜕𝑥

+𝜕𝜏|�𝜕𝑥

𝜕𝜏�|𝜕𝑦

+𝜕𝜏��𝜕𝑦

+𝜕𝜏��𝜕𝑦

𝜕𝜏�|𝜕𝑧

+𝜕𝜏��𝜕𝑧

+𝜕𝜏��𝜕𝑧

∆𝑥∆𝑦∆𝑧

Thiscanbewrittenininvariantnotationas:

𝐹w���^ = 𝐹w = ∇ ∙ 𝜏 ∀= div 𝜏 ∀wherethetermsontherightsiderepresentthedivergenceofthestresstensortimesthevolumeofthefluidparticle.

Itrevealsthephysicsofthedivergencewhenitoperatesonthestresstensor.Notethat this is the third physical interpretation of the divergence operator. This isbecausethephysicsofamathematicaloperatordependonthecontextinwhichtheoperatorisused.

Divergenceofthestresstensor =

NetshearforceonafluidparticleVolumeoftheparticle

Step5:CombinetermsSubstitutetheshearforceandpressureforceintoNewton’ssecondlawofmotion.Then,dividebythevolumeofthefluidparticletogive:

𝑚𝑎 = 𝐹

𝜌 ∀𝑑𝑣𝑑𝑡 =𝜌∀𝑔 + 𝐹v + 𝐹w

= 𝜌∀𝑔 − ∇𝑝 ∆𝑥∆𝑦∆𝑧 + ∇ ∙ 𝜏 ∀

= 𝜌𝑔 ∀ − ∇𝑝 ∀ + ∇ ∙ 𝜏 ∀

Dividewithvolume, ∀

𝜌𝑑𝑣𝑑𝑡 = 𝜌𝑔 − ∇𝑝 + ∇𝜏

This is the differential form of the linear momentum equation without anyassumptionabout thenatureof the fluid.Thenext step involvesmodifying thisequationtothatitappliestoaNewtonianfluid.

Step6:AssumeaNewtonianfluidStokesin1845figuredoutawaytowritethestresstensorintermsoftherate-of-straintensoroftheflowingfluid.Thedetailsareomittedhere.AfterStokes’resultsareintroduced,assumeconstantdensityandviscosity,aboveequationbecomes:

𝜌𝑑𝑣𝑑𝑡

= 𝜌𝑔 − ∇𝑝 + ∇𝜏Aboveequationcanbespecificallywrittenas:

𝑥 − 𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚 𝜌𝜕𝑢𝜕𝑡+ 𝑢

𝜕𝑢𝜕𝑥

+ 𝑣𝜕𝑢𝜕𝑦

+ 𝑤𝜕𝑢𝜕𝑧

= 𝜌𝑔| −∂𝑝𝜕𝑥

+∂𝜏||𝜕𝑥

+∂𝜏�|𝜕𝑦

+∂𝜏�|𝜕𝑧

𝑦 −𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚 𝜌𝜕𝑣𝜕𝑡+ 𝑢

𝜕𝑣𝜕𝑥

+ 𝑣𝜕𝑣𝜕𝑦

+ 𝑤𝜕𝑣𝜕𝑧

= 𝜌𝑔� −∂𝑝𝜕𝑦

+∂𝜏|�𝜕𝑥

+∂𝜏��𝜕𝑦

+∂𝜏��𝜕𝑧

𝑧 − 𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚 𝜌𝜕𝑤𝜕𝑡

+ 𝑢𝜕𝑤𝜕𝑥

+ 𝑣𝜕𝑤𝜕𝑦

+ 𝑤𝜕𝑤𝜕𝑧

= 𝜌𝑔� −∂𝑝𝜕𝑧+∂𝜏|�𝜕𝑥

+∂𝜏��𝜕𝑦

+∂𝜏��𝜕𝑧

ItmaybenotedthatthelastthreeconvectivetermsontheRHSofabovementionequations make it highly non-linear and complicates the general analysis. Asimplification is possible for considering an incompressible flow of Newtonianfluidwheretheviscousstressesareproportional to theelementstrainrateandcoefficientofviscosity 𝜇 .Foranincompressibleflow,thesheartermsmaybewritten:

𝜏|| = 2𝜇𝜕𝑢𝜕𝑥 𝜏|� = 𝜏�| = 𝜇

𝜕𝑢𝜕𝑦

+𝜕𝑥𝜕𝑥

𝜏�� = 2𝜇𝜕𝑣𝜕𝑦 𝜏|� = 𝜏�| = 𝜇

𝜕𝑤𝜕𝑥

+𝜕𝑢𝜕𝑧

𝜏�� = 2𝜇𝜕𝑤𝜕𝑧 𝜏�� = 𝜏�� = 𝜇

𝜕𝑣𝜕𝑧+𝜕𝑤𝜕𝑦

Thus,thedifferentialmomentumequationforNewtonianfluidwithconstantdensityandviscosityisgivenby:

𝑥 − 𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚 𝜌𝐷𝑢𝐷𝑡 = 𝜌𝑔| −

∂𝑝𝜕𝑥

+ 𝜇∂[𝑢𝜕𝑥[

+∂[𝑢𝜕𝑦[

+∂[𝑢𝜕𝑧[

𝑦 − 𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚 𝜌𝐷𝑣𝐷𝑡

= 𝜌𝑔� −∂𝑝𝜕𝑦

+ 𝜇∂[𝑣𝜕𝑥[

+∂[𝑣𝜕𝑦[

+∂[𝑣𝜕𝑧[

𝑧 − 𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚 𝜌𝐷𝑤𝐷𝑡

= 𝜌𝑔� −∂𝑝𝜕𝑧+ 𝜇

∂[𝑤𝜕𝑥[

+∂[𝑤𝜕𝑦[

+∂[𝑤𝜕𝑧[

Itisasecondorder,non-linearpartialdifferentialequationandisknownasNavier-Stokesequation.Invectorform,itmayberepresentedas:

𝜌𝑑𝑉𝑑𝑡

= 𝜌𝑔 − ∇𝑝 + µ∇[𝑉where∇[𝑉isamathematicaloperatorthatiscalledtheLaplacianofthevelocityfield.Asaconclusion,thisistheNavier-Stokesequation.

𝜌𝑑𝑉𝑑𝑡

= 𝜌𝑔 − ∇𝑝 + µ∇[𝑉

Step7:InterpretthephysicsThephysicsoftheNavier-Stokesequationare:

𝜌𝑑𝑣𝑑𝑡 = 𝜌𝑔 − ∇𝑝 + µ∇[𝑉

Massoftheparticletimesaccelerationoftheparticledividedbythevolumeofthe

particle.

Weightoftheparticledividedby

itsvolume.

Netpressureforceontheparticledividedbyitsvolume.

Netshearforceontheparticledividedbyitsvolume.

Notethedimensionsandunits:

Dimension =ForceVolume

=𝑁𝑚� =

𝑘𝑔𝑚[ ∙ 𝑠[

Navier-Stokesequation(constantproperties)forCartesiancoordinatesis:

𝜌𝜕𝑢𝜕𝑡+ 𝑢

𝜕𝑢𝜕𝑥

+ 𝑣𝜕𝑢𝜕𝑦

+ 𝑤𝜕𝑢𝜕𝑧

= 𝜌𝑔| −𝜕𝑝𝜕𝑥

+ 𝜇𝜕[𝑢𝜕𝑥[

+𝜕[𝑢𝜕𝑦[

+𝜕[𝑢𝜕𝑧[

𝜌𝜕𝑣𝜕𝑡+ 𝑢

𝜕𝑣𝜕𝑥

+ 𝑣𝜕𝑣𝜕𝑦

+ 𝑤𝜕𝑣𝜕𝑧

= 𝜌𝑔� −𝜕𝑝𝜕𝑦

+ 𝜇𝜕[𝑣𝜕𝑥[

+𝜕[𝑣𝜕𝑦[

+𝜕[𝑣𝜕𝑧[

𝜌𝜕𝑤𝜕𝑡

+ 𝑢𝜕𝑤𝜕𝑥

+ 𝑣𝜕𝑤𝜕𝑦

+ 𝑤𝜕𝑤𝜕𝑧

= 𝜌𝑔� −𝜕𝑝𝜕𝑧

+ 𝜇𝜕[𝑤𝜕𝑥[

+𝜕[𝑤𝜕𝑦[

+𝜕[𝑤𝜕𝑧[

Navier-Stokesequation(constantproperties)forcylindrical(polar)coordinatesis:𝑟 − 𝑎𝑥𝑖𝑠:

𝜌𝜕𝑣^𝜕𝑡

+ 𝑣^𝜕𝑣^𝜕𝑟

+𝑣_𝑟𝜕𝑣^𝜕𝜃

+ 𝑣�𝜕𝑣^𝜕𝑧

−𝑣_[

𝑟= 𝜌𝑔^ −

𝜕𝑝𝜕𝑟

+ 𝜇1𝑟𝜕𝜕𝑟

𝑟𝜕𝑣^𝜕𝑟

+1𝑟[𝜕[𝑣^𝜕𝜃[

+𝜕[𝑣^𝜕𝑧[

−𝑣^𝑟[+2𝑟[𝜕𝑣_𝜕𝜃

𝜃 − 𝑎𝑥𝑖𝑠:

𝜌𝜕𝑣_𝜕𝑡

+ 𝑣^𝜕𝑣_𝜕𝑟

+𝑣_𝑟𝜕𝑣_𝜕𝜃

+ 𝑣�𝜕𝑣_𝜕𝑧

−𝑣^𝑣_𝑟

= 𝜌𝑔_ −1𝑟𝜕𝑝𝜕𝜃

+ 𝜇1𝑟𝜕𝜕𝑟

𝑟𝜕𝑣_𝜕𝑟

+1𝑟[𝜕[𝑣_𝜕𝜃[

+𝜕[𝑣_𝜕𝑧[

−𝑣_𝑟[+2𝑟[𝜕𝑣_𝜕𝜃

𝑧 − 𝑎𝑥𝑖𝑠:

𝜌𝜕𝑣�𝜕𝑡

+ 𝑣^𝜕𝑣�𝜕𝑟

+𝑣_𝑟𝜕𝑣�𝜕𝜃

+ 𝑣�𝜕𝑣�𝜕𝑧

= 𝜌𝑔� −𝜕𝑝𝜕𝑧

+ 𝜇1𝑟𝜕𝜕𝑟

𝑟𝜕𝑣�𝜕𝑟

+1𝑟[𝜕[𝑣�𝜕𝜃[

+𝜕[𝑣�𝜕𝑧[

EQUATIONOFCONTINUITYCartesiancoordinates:

0 =𝜕𝜌𝜕𝑡+𝜕 𝜌𝑢𝜕𝑥

+𝜕 𝜌𝑣𝜕𝑦

+𝜕 𝜌𝑤𝜕𝑧

Polarcoordinates:

0 =𝜕𝜌𝜕𝑡+1𝑟𝜕𝜕𝑟

𝜌𝑢^𝑟 +𝜕𝜕𝑧

𝜌𝑢� +1𝑟𝜕𝜕𝜃

𝜌𝑢_

Navier-Stokesequation(constantproperties)forCartesiancoordinatesis:

𝐷𝐷𝑡

=𝜕𝜕𝑡+ 𝑢

𝜕𝜕𝑥

+ 𝑣𝜕𝜕𝑦

+ 𝑤𝜕𝜕𝑧

∇[=𝜕[

𝜕𝑥[+𝜕[

𝜕𝑦[+𝜕[

𝜕𝑧[

𝜌𝜕𝑢𝜕𝑡+ 𝑢

𝜕𝑢𝜕𝑥

+ 𝑣𝜕𝑢𝜕𝑦

+ 𝑤𝜕𝑢𝜕𝑧

= 𝜌𝑔| −𝜕𝑝𝜕𝑥

+ 𝜇𝜕[𝑢𝜕𝑥[

+𝜕[𝑢𝜕𝑦[

+𝜕[𝑢𝜕𝑧[

𝜕𝑢𝜕𝑡+ 𝑢

𝜕𝑢𝜕𝑥

+ 𝑣𝜕𝑢𝜕𝑦

+ 𝑤𝜕𝑢𝜕𝑧 = 𝑔| −

1𝜌𝜕𝑝𝜕𝑥

+𝜇𝜌

𝜕[𝑢𝜕𝑥[

+𝜕[𝑢𝜕𝑦[

+𝜕[𝑢𝜕𝑧[

𝐷𝑢𝐷𝑡 = 𝑔| −

1𝜌𝜕𝑝𝜕𝑥

+ 𝜐 ∇[ 𝑢

𝜌𝜕𝑣𝜕𝑡+ 𝑢

𝜕𝑣𝜕𝑥

+ 𝑣𝜕𝑣𝜕𝑦

+ 𝑤𝜕𝑣𝜕𝑧

= 𝜌𝑔� −𝜕𝑝𝜕𝑦

+ 𝜇𝜕[𝑣𝜕𝑥[

+𝜕[𝑣𝜕𝑦[

+𝜕[𝑣𝜕𝑧[

𝜕𝑣𝜕𝑡+ 𝑢

𝜕𝑣𝜕𝑥

+ 𝑣𝜕𝑣𝜕𝑦

+ 𝑤𝜕𝑣𝜕𝑧 = 𝑔� −

1𝜌𝜕𝑝𝜕𝑦

+𝜇𝜌

𝜕[𝑣𝜕𝑥[

+𝜕[𝑣𝜕𝑦[

+𝜕[𝑣𝜕𝑧[

𝐷𝑢𝐷𝑡 = 𝑔� −

1𝜌𝜕𝑝𝜕𝑦

+ 𝜐 ∇[ 𝑣

𝜌𝜕𝑤𝜕𝑡

+ 𝑢𝜕𝑤𝜕𝑥

+ 𝑣𝜕𝑤𝜕𝑦

+ 𝑤𝜕𝑤𝜕𝑧

= 𝜌𝑔� −𝜕𝑝𝜕𝑧

+ 𝜇𝜕[𝑤𝜕𝑥[

+𝜕[𝑤𝜕𝑦[

+𝜕[𝑤𝜕𝑧[

𝜕𝑤𝜕𝑡

+ 𝑢𝜕𝑤𝜕𝑥

+ 𝑣𝜕𝑤𝜕𝑦

+ 𝑤𝜕𝑤𝜕𝑧 = 𝑔� −

1𝜌𝜕𝑝𝜕𝑧

+𝜇𝜌

𝜕[𝑤𝜕𝑥[

+𝜕[𝑤𝜕𝑦[

+𝜕[𝑤𝜕𝑧[

𝐷𝑤𝐷𝑡 = 𝑔� −

1𝜌𝜕𝑝𝜕𝑧

+ 𝜐 ∇[ 𝑤

Navier-Stokesequation(constantproperties)forcylindrical(polar)coordinatesis:

𝐷𝐷𝑡

=𝜕𝜕𝑡+ 𝑢^

𝜕𝜕𝑟+𝑢_𝑟𝜕𝜕𝜃

+ 𝑢�𝜕𝜕𝑧

∇[=𝜕[

𝜕𝑟[+1𝑟𝜕𝜕𝑟+1𝑟[

𝜕[

𝜕𝜃[+𝜕[

𝜕𝑧[

𝑟 − 𝑎𝑥𝑖𝑠:

𝐷𝑢^𝐷𝑡

−𝑢_[

𝑟 = 𝑔^ −

1𝜌𝜕𝑝𝜕𝑟

+ 𝜈 ∇[𝑢^ −𝑢^𝑟[+2𝑟[𝜕𝑢_𝜕𝜃

𝜕𝑢^𝜕𝑡

+ 𝑢^𝜕𝑢^𝜕𝑟

+𝑢_𝑟𝜕𝑢^𝜕𝜃

+ 𝑢�𝜕𝑢^𝜕𝑧

−𝑢_[

𝑟 = 𝑔^ −

1𝜌𝜕𝑝𝜕𝑟

+ 𝜈𝜕[𝑢^𝜕𝑟[

+1𝑟𝜕𝑢^𝜕𝑟

+1𝑟[𝜕[𝑢^𝜕𝜃[

+𝜕[𝑢^𝜕𝑧[

−𝑢^𝑟[+2𝑟[𝜕𝑢_𝜕𝜃

𝜃 − 𝑎𝑥𝑖𝑠:

𝐷𝑢_𝐷𝑡

−𝑢^𝑢_𝑟

= 𝑔_ −1𝜌𝑟𝜕𝑝𝜕𝜃

+ 𝜈 ∇[𝑢_ +2𝑟[𝜕𝑢^𝜕𝜃

−𝑢_𝑟[

𝜕𝑢_𝜕𝑡

+ 𝑢^𝜕𝑢_𝜕𝑟

+𝑢_𝑟𝜕𝑢_𝜕𝜃

+ 𝑢�𝜕𝑢_𝜕𝑧

−𝑢^𝑢_𝑟 = 𝑔_ −

1𝜌𝑟𝜕𝑝𝜕𝜃

+ 𝜈𝜕[𝑢_𝜕𝑟[

+1𝑟𝜕𝑢_𝜕𝑟

+1𝑟[𝜕[𝑢_𝜕𝜃[

+𝜕[𝑢_𝜕𝑧[

+2𝑟[𝜕𝑢^𝜕𝜃

−𝑢_𝑟[

𝑧 − 𝑎𝑥𝑖𝑠:

𝐷𝑢�𝐷𝑡

= 𝑔� −1𝜌𝜕𝑝𝜕𝑧

+ 𝜈 ∇[𝑢�

𝜕𝑢�𝜕𝑡

+ 𝑢^𝜕𝑢�𝜕𝑟

+𝑢_𝑟𝜕𝑢�𝜕𝜃

+ 𝑢�𝜕𝑢�𝜕𝑧

= 𝑔� −1𝜌𝜕𝑝𝜕𝑧

+ 𝜈𝜕[𝑢�𝜕𝑟[

+1𝑟𝜕𝑢�𝜕𝑟

+1𝑟[𝜕[𝑢�𝜕𝜃[

+𝜕[𝑢�𝜕𝑧[

Note:

1𝑟𝜕𝜕𝑟

𝑟𝜕𝑢�𝜕𝑟

=1𝑟𝑟𝜕𝜕𝑟

𝜕𝑢�𝜕𝑟

+𝜕𝑢�𝜕𝑟

𝜕𝑟𝜕𝑟

=1𝑟𝑟𝜕[𝑢�𝜕𝑟[

+𝜕𝑢�𝜕𝑟

1𝑟𝜕𝜕𝑟

𝑟𝜕𝑢�𝜕𝑟

=𝜕[𝑢�𝜕𝑟[

+1𝑟𝜕𝑢�𝜕𝑟

CONSERVATIONOFENERGYRecalltheintegralrelationofenergyequationforafixedcontrolvolume:

𝑄 −𝑊w −𝑊¢ =𝜕𝜕𝑡

𝑒𝜌𝑑∀£¢

+ 𝑒 +𝑝𝜌𝜌 𝑉 ∙ 𝑛

£w𝑑𝐴 (1)

Figure1

Ifthecontrolvolumehappenstobeanelementalsystemasshownabove,thentherewillbenoshaftworkterm 𝑊w = 0 .Denotingtheenergyperunitvolumeas:

𝑒 = 𝑢 +12𝑉[ + 𝑔𝑧

Thenetenergyflowacrossthesixcontrolsurfacecanbecalculatedasbelow:Face Inletenergyflow Outletenergyflow

𝑥 𝜌𝑢 𝑒 +𝑝𝜌𝑑𝑦𝑑𝑧 𝜌𝑢 𝑒 +

𝑝𝜌+𝜕𝜕𝑥

𝜌𝑢 𝑒 +𝑝𝜌𝑑𝑥 𝑑𝑦𝑑𝑧

𝑦 𝜌𝑣 𝑒 +𝑝𝜌𝑑𝑥𝑑𝑧 𝜌𝑣 𝑒 +

𝑝𝜌+𝜕𝜕𝑦

𝜌𝑣 𝑒 +𝑝𝜌𝑑𝑦 𝑑𝑥𝑑𝑧

𝑧 𝜌𝑤 𝑒 +𝑝𝜌𝑑𝑥𝑑𝑦 𝜌𝑤 𝑒 +

𝑝𝜌+

𝜕𝜕𝑤

𝜌𝑤 𝑒 +𝑝𝜌𝑑𝑧 𝑑𝑥𝑑𝑦

Hence,Eq.(1)canbewrittenas:

𝑄 −𝑊¢ =𝜕𝜕𝑡𝜌 𝑒 +

𝑝𝜌+𝜕𝜕𝑥

𝜌𝑢 𝑒 +𝑝𝜌+𝜕𝜕𝑦

𝜌𝑣 𝑒 +𝑝𝜌+𝜕𝜕𝑧

𝜌𝑤 𝑒 +𝑝𝜌

𝑑𝑥𝑑𝑦𝑑𝑧 (2)

Withthehelpofcontinuityequationandsimilaranalogyconsideredduringthederivationofmomentumequation,Eq.(2)canbesimplifiedas:

𝑄 −𝑊¢ = 𝜌𝑑𝑒𝑑𝑡+ 𝑉 ∙ ∇𝑝 𝑑𝑥𝑑𝑦𝑑𝑧 (3)

Ifoneconsiderstheenergytransferasheat 𝑄 throughpureconduction,theFourier’slawofheatconductioncanbeappliedtotheelementalsystem.

𝑞 = −𝑘∇𝑇 (4)Where𝑘isthethermalconductivityofthefluid.

Theheatflowpassingthroughx-faceisshowninFigure1(b),andforallthesixfaces,itissummarizedinthefollowingtable:Face Inletenergyflow Outletenergyflow

𝑥 𝑞|𝑑𝑦𝑑𝑧 𝑞| +𝜕𝜕𝑥

𝑞| 𝑑𝑥 𝑑𝑦𝑑𝑧

𝑦 𝑞�𝑑𝑥𝑑𝑧 𝑞� +𝜕𝜕𝑦

𝑞� 𝑑𝑥 𝑑𝑥𝑑𝑧

𝑧 𝑞�𝑑𝑥𝑑𝑦 𝑞� +𝜕𝜕𝑧

𝑞� 𝑑𝑥 𝑑𝑥𝑑𝑦

Thenetheatfluxcanbeobtainedbythedifferenceininletandoutletheatfluxes.

𝑄 = −𝜕𝜕𝑥

𝑞| +𝜕𝜕𝑦

𝑞� +𝜕𝜕𝑧

𝑞� 𝑑𝑥𝑑𝑦𝑑𝑧

= − ∇ ∙ 𝑞 𝑑𝑥𝑑𝑦𝑑𝑧

= ∇ ∙ 𝑘∇𝑇 𝑑𝑥𝑑𝑦𝑑𝑧 (5)Therateofworkdonebytheviscousstressesontheleftx-facesasshownintheFigure1(b)isgivenby:

𝑊¢,¥¦ = −𝑤|𝑑𝑦𝑑𝑧

= − 𝑢𝜏|| + 𝑣𝜏|� + 𝑤𝜏|� 𝑑𝑦𝑑𝑧 (6)

Inthesimilarmanner,thenetviscousratesareobtainedandisgivenby:

𝑊¢ = −𝜕𝜕𝑥

𝑢𝜏|| + 𝑣𝜏�� + 𝑤𝜏�� +𝜕𝜕𝑦

𝑢𝜏�| + 𝑣𝜏�� + 𝑤𝜏�� +𝜕𝜕𝑧

𝑢𝜏�| + 𝑣𝜏�� + 𝑤𝜏�� 𝑑𝑥𝑑𝑦𝑑𝑧

= −∇ 𝑉 ∙ 𝜏§¨ 𝑑𝑥𝑑𝑦𝑑𝑧 (7)

Then,substituteEq.(5)andEq.(7)intoEq.(3),

𝜌𝑑𝑒𝑑𝑡+ 𝑉 ∙ ∇𝑝 = ∇ ∙ 𝑘∇𝑇 + ∇ ∙ 𝑉 ∙ 𝜏§¨ (8)

ThesecondtermintheRHSofEq.(8)canbewritteninthefollowingterm,

∇ ∙ 𝑉 ∙ 𝜏§¨ = 𝑉 ∙ ∇ ∙ 𝜏§¨ + Φ (9)

Here, Φ is known as the viscous-dissipation function. For Newtonianincompressibleviscousfluid,thisfunctionasthefollowingform.

Φ = µ 2𝜕𝑢𝜕𝑥

[

+ 2𝜕𝑣𝜕𝑦

[

+ 2𝜕𝑤𝜕𝑧

[

+𝜕𝑣𝜕𝑥

+𝜕𝑢𝜕𝑦

[

+𝜕𝑤𝜕𝑦

+𝜕𝑣𝜕𝑧

[

+𝜕𝑢𝜕𝑧

+𝜕𝑤𝜕𝑥

[

(10)

SinceallthetermsinEq.(10)arequadratic,sotheviscousdissipationtermsarealways positive, the the flow always tends to lose its available energy due todissipation.WhenEq.(9)isusedinEq.(8),simplifiedusinglinear-momentumequationandthetermsare rearranged, then thegeneral formof energyequation isobtained forNewtonianviscousfluid.

ρ𝑑𝑢𝑑𝑡+ 𝑝 ∇ ∙ 𝑉 = ∇ ∙ 𝑘∇𝑇 + Φ (11)

Foranalysispointofview,thefollowingvalidapproximationscanbemadeforEq.(11)

𝑑𝑢 ≈ 𝑐¢𝑑𝑇

𝑐¢, 𝜇, 𝑘and𝜌areconstant

ρ𝑐¢𝑑𝑇𝑑𝑡 = 𝑘∇[𝑇 + Φ (13)

Where,

𝑑𝑇𝑑𝑡

=𝜕𝑇𝜕𝑡+ 𝑢

𝜕𝑇𝜕𝑥

+ 𝑣𝜕𝑇𝜕𝑦

+ 𝑤𝜕𝑇𝜕𝑧