CFD考试题答案参考

1、1. 采用时间前差，空间中差t =0.005tU0U1U2U3U4U5U6U7U8U9U100000000.1000000.00510000.0500.0500010.0110.500.02500.0500.02500.510.01510.50.262500.037500.037500.26250.510.0210.63120.250.1500.037500.150.250.631210.02510.6250.39060.1250.093800.09380.1250.39060.62510.0310.69530.3750.24220.06250.09380.06250.24220.3750.6

2、95310.003510.68750.46880.21880.1680.06250.1680.21880.46880.687510.0410.73440.45310.31840.14060.1680.14060.31840.45310.734410.04510.72660.52640.29690.24320.14060.24320.29690.52640.726610.0510.76320.51170.38480.21880.24320.21880.38480.51170.76321t=0.01t U0 U1 U2 U3 U4 U5 U6 U7 U8 U9 U10 0. 0.000.000.0

3、00.000.000.100.000.000.000.00000.1.1.01000.000.000.000.10-0.100.100.000.000.00000.1.1.02001.000.000.10-0.200.30-0.200.100.001.00000.1.1.03000.001.10-0.300.60-0.700.60-0.301.100.00000.1.1.04002.10-1.402.00-1.601.90-1.602.00-1.402.10000.1.1.0500-2.505.50-5.005.50-5.105.50-5.005.50-2.50000.1.-13.0-15.6

4、-15.6-13.01.06009.00016.00016.10016.0009.00000.1.-21.0-44.6-47.3-44.6-21.01.0700038.00047.70047.70038.000000.1.-103.130.3-139.142.7-139.130.3-103.1.080060.006006006006060.00000.1.-162.293.9 -373.412.6-421.412.6-373.293.9 -162.1.090060050090050060000.1.457.5-830.1080.-12081247.-12081080.-830.457.51.1

5、00000000.0010.000000000从时间步长为t=0.005情况看，速度逐步衰减，是收敛的；时间步长为t=0.01情况下，速度存 在突变，是发散的。由此可以看出选择时间步长对计算结果收敛情况起决定作用。分析：采用Von Neuman方法，分别计算T=0.005s和T=0.01s时的扩散因子d，可发现T=0.005s时0 d 1/2，计算结果稳定，而T=0.01s时d 1/2，计算结果不稳定。1.FDM不采用虚拟节点时：巴11/ 3节点 2： U3 2U22 U1 2u2 f2(1/3)222节点3：山爲吐2U3 f3(1/3)代入u41解得：u30.3516采用在节点1给出虫的二阶

6、精度公式3u1加2山1dx2 x2可解得：ua-9FVM节点1:U2 U1x1f x2 2节电2：U3U2U2U12u2 xf2gxx节点3：U4 U3U3U22U3 xf3gxx代入u41, x1/3, f238/ 9,f3解得：U-2/9精确解U2x2xxx32/9对比可得FVM得出的解等于精确解，即Neuman边界条件在FVM中精确实现。而FDM勺解有一 定误差，因为在FDM中难以实现Neuman边界条件，但可采用虚拟节点法，二阶精度法等来改 进FDM以得到精确解。2.前向差分xau bUi i eg 2 du 3（X3）1131列方程解得：a一, b3,c-, d-623后向差分令（_

7、u）au bUi - eUi 2 dq 3（ x3）x ix解得：a , b3, e ,d 623UiUi(U5解得:n4.U-中心差分UiUi(2(3U8 )s5,8aUi 3X)(上)iXUx)()iX(2 x)2(2!(3 x)22!(bq 2 cq（5a 0.0167, b 0.15,cd 0,e0.75, fn 1Ui2 t(Uin1 2Uinx22J2)ix2u-)ixU8) S7,80.15,gn Ui 1)假定任一个时间步n的数值解un(in12 in分解为傅立叶级数eU 1_n lijeN式代入（2）式得n iid (e12 e 1 )(2 x)3(3!(3 x)33!f8 A8fui 20.750.0167nUiin1)3U(Txgq 3(2 x)4 (药x4!4U川(T4! x(3 x)46(X )(1)(2 x)x5!(3 x)55!5U(Tx(2 x)66!(3 x)66!6U(6)i 令x令误差放大因子fn 代入（4）得2d (1 cos )(5)fn 12d(1 cos )这里采用反证法，假设其有条件稳定则由（5）（6）可推出d 0, fn 0