《数值传热学》研究生课程教学资源（课件讲稿）Chapter 1 Introduction Numerical Heat Transfer

热流科学与工程亚步文源大堂G教育部重点实验室2.Momentum conservationApplying the 2nd law of Newton (F=ma) to theelementalcontrolvolume（控制容积）in the three-dimensional coordinates:r[IncreasingrateofmomentumoftheCV]=[Summationofexternal（外部）forcesapplying on the CVAdopting Stokes assumption: stress is linearly proportionalto strain(应力与应变成线性关系），Wehavefollowingresultfor component u in x-direction :CFD-NHT-EHT中6/57CENTER

6/57 Applying the 2nd law of Newton （F=ma) to the elemental control volume（控制容积） in the three￾dimensional coordinates: 2. Momentum conservation [Increasing rate of momentum of the CV] = [Summation of external（外部） forces applying on the CV] Adopting Stokes assumption：stress is linearly proportional to strain(应力与应变成线性关系)，We have following result for component u in x-direction：

热流科学与工程西步文源大堂教育部重点实验室SourcetermaOua(pu)a(puu).(puv).a(puw)op(AdivU +2nataxaxOxaxoyozConvectionDiffusionTransienttermtermtermaaouovouowpFnTOzOzayaxayOxDiffusiontermn dynamic viscosity , α fluid 2nd molecular viscosity.2For gas, π=-=n3CFD-NHT-EHTΦ7/57CENTER

7/57  dynamic viscosity ， ( ) ( ) ( ) ( ) ( 2 ) [ ( )] [ ( )] x u uu uv uw p u divU t x y z x x x v u u w F y x y z z x                                                 fluid 2nd molecular viscosity. For gas, 2 =- 3   Transient term Convection term Diffusion term Diffusion term Source term

热流科学与工程西步文源大堂E教育部重点实验室It can be shown (see the notes) that the above eguationcan be reformulated as（改写为）following general form ofNavier-Stokes equation for u component:a(pu)+ div(puU)= div(ngradu)+SatDiffusionTransientConvectionSourceterm扩散项term源项term非稳态项term对流项u, y, w ----velocity components in three directions, respectivelydependent variable（因变量）to be solved;U ----fluid velocity vector; U=ui +vj+ wkS,----source term.中CFD-NHT-EHT8/57CENTER

8/57 ( ) ( ) ( ) u div U div grad S u u u t         It can be shown (see the notes) that the above equation can be reformulated as （改写为）following general form of Navier-Stokes equation for u component: Transient term 非稳态项 Convection term对流项 Diffusion term扩散项 Source term源项 u, v, w -velocity components in three directions, respectively, dependent variable（因变量） to be solved； -fluid velocity vector; Su -source term. U U ui v j wk =  

热流科学与工程亚步文源大堂CE教育部重点实验室Sourcetermin x-direction:ForincompressiblefluidaaaaowauOvap(adivU)+'pFnaxOzaxaxayaxaxOxSimilarly:aaaOuOvOwaop(adivU)+ pF,-(nn(noyoyaxOzayayayayaaaavowaOuap(adivU)+ pF(nnOzOzayOzOzOzazaxFor incompressible fluid with constant properties the sourceterm does not contain velocity-related partΦCFD-NHT-EHT9/57CENTER

9/57 Source term in x-direction： ( ) ( ) ( ) ( ) u x u v w p S divU F x x x y z x x x                            ( ) ( ) ( ) ( ) v y u v w p S divU F x y y y z y y y                            ( ) ( ) ( ) ( ) w z u v w p S divU F x z z y z z z z                            Similarly： For incompressible fluid with constant properties the source term does not contain velocity-related part. For incompressible fluid

热流科学与工程西步文源大堂G教育部重点实验室3.Energy conservationIncreasingrateof internal energyintheCVl-[Netheat going into the CVJ+[Work conducted by bodyforcesand surfaceforcesIntroducing Fourier's law of heat conduction andneglecting the work conducted by forces; Introducingenthalpy （焰) h = C,T,assuming C,= constant,We have:a(pT+ div(pTU)= div(一grad(T) + STat2anaT-aTaT4nKgrad(TOzaxayPrcc,nC,nD中CFD-NHT-EHT10/57CENTER

10/57 3. Energy conservation [Increasing rate of internal energy in the CV]= [Net heat going into the CV]+[Work conducted by body forces and surface forces] Introducing Fourier’s law of heat conduction and neglecting the work conducted by forces；Introducing enthalpy（焓） ( ) ( ) ( ( )) T p div U di T T T c v grad S t         ( ) Pr p p p c c c           ( ) Pr p p p c c c           ( ) Pr p p p c c c           ( ) Pr p p p c c c           h c T  p , assuming p c  constant, ( ) T T T grad T i j k x y z          We have: