CFD考试题答案参考

1.采用前的时间差异，空间差异T=0.005tU0U1U2U3U4U5U6U7U8U9U100000000.1000000.00510000.0500.0500010.0110.500.02500.0500.02500.510.01510.50.262500.037500.037500.26250.510.0210.63120.250.1500.037500.150.250.631210.02510.6250.39060.1250.093800.09380.1250.39060.62510.0310.69530.3750.24220.06250.09380.06250.24220.3750.695310.003510.68750.46880.21880.1680.06250.1680.21880.46880.687510.0410.73440.45310.31840.14060.1680.14060.31840.45310.734410.04510.72660.52640.29690.24320.14060.24320.29690.52640.726610.0510.76320.51170.38480.21880.24320.21880.38480.51170.76321T=0.01tU0U1U2U3U4U5U6U7U8U9U1000.000.000.000.000.000.100.000.000.000.000.000.011.000.000.000.000.10-0.100.100.000.000.001.000.021.001.000.000.10-0.200.30-0.200.100.001.001.000.031.000.001.10-0.300.60-0.700.60-0.301.100.001.000.041.002.10-1.402.00-1.601.90-1.602.00-1.402.101.000.051.00-2.505.50-5.005.50-5.105.50-5.005.50-2.501.000.061.009.00-13.0016.00-15.6016.10-15.6016.00-13.009.001.000.071.00-21.0038.00-44.6047.70-47.3047.70-44.6038.00-21.001.000.081.0060.00-103.60130.30-139.60142.70-139.60130.30-103.6060.001.000.091.00-162.60293.90-373.50412.60-421.90412.60-373.50293.90-162.601.000.101.00457.50-830.001080.00-1208.001247.10-1208.001080.00-830.00457.501.00如果时间点为t=0.005，则速度将逐渐减小并收敛。如果时间点为t=0.01，则速度有突变且发散。这表明选择时间点决定了计算结果的收敛情况。分析：使用Von Neumann方法，在T=0.005s和T=0.01s中分别计算扩散系数d，可以看到在T=0.005s中计算结果稳定，而在T=0.01s中计算结果不稳定。1.FDM如果没有虚拟节点：节点2:节点3:高考解决方案：使用给节点1的二次精度公式可解决：FVM节点1:省电2:节点3:高考解决方案：精确的解决方案比较FVM是正确的解决方案。这意味着FVM中的Neumann边界条件已正确实现。FDM中很难实现Neumann边界条件，但是FDM解决方案存在一些错误，因为为确保正确的解决方案，可以使用虚拟节点方法、辅助精度方法等改进FDM。2.正向差逮捕令解行方程式：反向差异逮捕令解决方案：中心差逮捕令解决方案：4.(1)假设任意时间步长n的数值解法高考.(2)傅里叶级数分解.(3)(3)表达式替换(2)逮捕令.(4)用(4)替换错误放大元件.(5).(6)假定反证法是有条件的稳定(5)(6)发