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=0.01tU0U1U2U3U4U5U6U7U8U9U1000.000.000.000.000.000.100

3、.000.000.000.000.000.011.000.000.000.000.10-0.100.100.000.000.001.000.021.001.000.000.10-0.200.30-0.200.100.001.001.000.031.000.001.10-0.300.60-0.700.60-0.301.100.001.000.041.002.10-1.402.00-1.601.90-1.602.00-1.402.101.000.051.00-2.505.50-5.005.50-5.105.50-5.005.50-2.501.000.061.009.00-13.0016.00-15

4、.6016.10-15.6016.00-13.009.001.000.071.00-21.0038.00-44.6047.70-47.3047.70-44.6038.00-21.001.000.081.0060.00-103.60130.30-139.60142.70-139.60130.30-103.6060.001.000.091.00-162.60293.90-373.50412.60-421.90412.60-373.50293.90-162.601.000.101.00457.50-830.001080.00-1208.001247.10-1208.001080.00-830.004

5、57.501.00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:，速度存从时间在突变，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VonNeuman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:U2 -U3 =1/3-节点 2： u3 2u2 +u1 八（1/3）2节点3：古-2u f3代入U4 = -1解得：出=-0.3516采用在节点1给出b的二阶精度公式3u1

6、+4u2-u3 = 1dx2Z可解得:FVM节点1：u_1x节电2:uu2口2 -qAx一2比& = fx节点3:u4u3u3u2-2T = f3x代入 u4 =1，心x = 1/3, f2 = 38/9, f-32/9解得:U3 =2/9精确解u = -2x2 +x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.前向差分令(昱).=aui +bur +cui七 +dui 书 ex+ O(Ax3)列方程解得:Z1131a = ,b

7、=3,c = ,d =623后向差分令（鱼）i旦ex+ bui4 +cu2+du2+o(Ax3)解得：a=11,b=-3,c='3,d6 2中心差分5七 7 +(2也x)(% +(2；x)(豊)i2!exCX(x)3 g3u) +(2Ax)4 严4u、(x)5 g5u) gx)66 3!cx3)i4!(cx4)i5! Qx5)i6!Ui +g）（¥）i + 呼（導）i + 畔（空）i + 呼（卑）i + 畔（空）i 十!（卑）i dx2! <2x3! ex4! ex5!ex34!ex46!ex6,、,、,、cUaui 3 +bUi2+cUi1+dUi +eUiH +fU

8、i42+gUij36（U5 -U8）S5,8 +（U9 U8）S9,8 中（U7 -U8）S7,8 = fgA （） = +0 （AX ）exx辰力作 a = 0.0167, b = 0.15,c = 0.75 解得：d =0,e =0.75, f = 0.15, g =0.0167n -1n 丄Ui -Ui4.2也t/ n c n n /（%2Ui +心+0（事2,加2）假定任一个时间步n的数值解Uin+gin代入得2也tn + n,n cn.n,寄 _a（寺十一2£（ +名i）Ax2（2）将Sin分解为傅立叶级数N工n p -n Iid> St =2Eje wj =N将（3）式代入（2）式得2n+YnJd?n（ed2+e 甥（4）令误差放大因子fn代入（4）得1fn+ = 2d（1 -cos ）Tn1n =-2d（1 -cos ）n -1这里采用反证法，假设其有条件稳定 则由（5）（6）可推出d aO, fn aO再由（5） （6）1 1fn+ - fn =