Given dw =100,Φ, = 200P =2for P =0,1,2,4200the calculated results-areP=0Dshown in the figure精确角150Physically and accordingto the analytical solution211exp100Φ-dΦ, -doexp50the value of @bshould beOTLlarger than zero.P=4puoxPr16/49
16/49 Given 100, 200 W E = = Physically and according to the analytical solution for P 0,1,2,4 = 0 0 exp( ) 1 L exp( ) 1 uL x L uL − − = − − the value of should be larger than zero. p u x P = P 4 = P 2 = the calculated results are shown in the figure. p
Thus when P is larger than 2, numerical solutions areunrealistic （不现实的）：Φp is less than its two neighboringgrid values, which is not possible for the case without source.Thereason is P >2 ag=(1-P)<0, ie the east infuencingcoefficient is negative,which is physically meaningless.4.2.3Firstorderupwind(FUD)schemeofconvectionterm1. Definition in FD-i+1WEadadd. -dd+1 -Φu<oxAxaxOx2.Definition in FV-interpolation of interface always takesupstream grid value17/49
17/49 Thus when is larger than 2，numerical solutions are unrealistic（不现实的）： is less than its two neighboring grid values, which is not possible for the case without source. P P The reason is <0，i.e. the east influencing 1 1 (1 ) 2 2 E a P = − coefficient is negative，which is physically meaningless. 4.2.3 First order upwind (FUD) scheme of convection term 2. Definition in FV－interpolation of interface always takes upstream grid value 1. Definition in FD－ P 1 1 ) , 0; ) , 0 i i i i i i u u x x x x − + − − = = P 2