7×7 nodes27×27 nodes2TEMP4.51.51.543.532.52refine1.510.50.50.5000.20.40.60.800.20.40.60.80-XXFig.2IsothermsfromTECPLOT11/91
11/91 Fig. 2 Isotherms from TECPLOT 77 nodes 2727 nodes refine
11-2 Steady heat conduction in a hollow cylinder---ASTMfor2ndand3rdboundaryconditions11-2-1PhysicalproblemanditsmathformulationKnown: Steady heat conduction in a hollow cylinder with variableproperty and source term shown in Fig. 1 has following boundaryconditions:几=0.2(1+T/100)Left boundary---given temperature:T=100(1+y)Right boundary---convective heat transfer:1=1S-100-0.5THeat transfer coefficient h = 5:hollowFluid temperature T= 100.R(1)= 1cylinderTop boundary---adiabatic;XBottom boundary---given heat flux: g = 5012/91
12/91 Known：Steady heat conduction in a hollow cylinder with variable property and source term shown in Fig. 1 has following boundary conditions: 11-2-1 Physical problem and its math formulation 11-2 Steady heat conduction in a hollow cylinder -ASTM for 2nd and 3rd boundary conditions Left boundary-given temperature: Right boundary-convective heat transfer: Top boundary-adiabatic; T =100(1+y) Heat transfer coefficient h = 5; Fluid temperature Tf = 100. Bottom boundary-given heat flux: q = 50 hollow cylinder
Variable thermo-properties:2=0.2(1+T/100)Thermal conductivity---for most region, a =1in a local region a =0.2(1 + T / 100)Sources term:1=1Entireregion---S=100-0.5TS-100-0.5TFind: temperature distribution in the domain.R(1)= 1Solution:Steady conduction problem with given Iand source term: I S..div(Togradp)+ S, = 013/91
13/91 Solution: Steady conduction problem with given and source term: Find: temperature distribution in the domain. S . Entire region- S =100 - 0.5T Thermal conductivity-for most region, in a local region = + 0.2(1 /100) T =1 div grad S ( ) 0 + = Variable thermo-properties: Sources term:
Steady heat conduction ina hollow cvlinderAdiabatic元= 0.2(1+ T /100)[2=1T=100(1+)h, T.h=5, T=100++↑+S-100-0.5TCylindricalq=50R(1)=coordinateXFig.1Diagram14/91
14/91 Fig.1 Diagram 2 x 1 1 Adiabatic y q = 50 T=100(1+y) r h, Tf h=5, Tf =100 = + 0.2(1 /100) T =1 x R(1)= S=100-0.5T Steady heat conduction in a hollow cylinder Cylindrical coordinate
ASTM(additional sourcetermmethod)ImplementingprocedureofASTMN(1) Determining Sc.ad, Spad for CV neighboring to boundary(2)Adding them into source term of related CVh,TSc Sc+Sc.adSp Sp+Sp,ad(8r）B(3) Setting the conductivity of boundary node as zeroABAy2nd BC.=0AyAV:01=0！awSxAy·T3rd BC.(x)B)(4) Solving the algebraic Eqs. for inner nodesAVThMB(5) Using Newton'law of cooling or Fourier eq. to getAySp.adtheboundarytemperatures when converged△V·[1/h+(Sx)/a,)15/91
ASTM (additional source term method) Implementing procedure of ASTM (1) Determining SC,ad, SP,ad for CV neighboring to boundary (2) Adding them into source term of related CV (3) Setting the conductivity of boundary node as zero (4) Solving the algebraic Eqs. for inner nodes (5) Using Newton’ law of cooling or Fourier eq. to get the boundary temperatures when converged SC SC +SC,ad SP SP +SP,ad W 0 y a x = 0 = = 2 nd BC. SP, ad = 0 3 rd BC. 15/91