CFD考试题答案参考

精品文档精品文档..计算流体动力学及其应用——习题1、扩散(质量、动量、能量)现象的方程和边界条件为：2u

0tT 0xLx2初始条件：u(x,0)=0(除x=0.5外)；u(0.5,0)=0.1边界条件：u(0,t)=1，u(L,t)=1(除t=0外)；u(x,0)=0设μ=1,Δx=0.1,L=1,T=2,I=10；请在①Δt=0.005，②Δt=0.01两种情况下各进行10个时间步的数值计算，并找出某些规律。计算流体动力学及其应用——习题1、扩散(质量、动量、能量)现象的方程和边界条件为：2u

0tT 0xLx2初始条件：u(x,0)=0(除x=0.5外)；u(0.5,0)=0.1边界条件：u(0,t)=1，u(L,t)=1(除t=0外)；u(x,0)=0设μ=1,Δx=0.1,L=1,T=2,I=10；请在①Δt=0.005，②Δt=0.01两种情况下各进行10个时间步的数值计算，并找出某些规律。提示：采用时间前差空间中差格式。看两种情况是否稳定？提示：采用时间前差空间中差格式。看两种情况是否稳定？tntu0u1u2u3u4u5u6u7u8u u9 100000000.1000000.005110.01110.015110.02110.025110.03112、已知：2、已知：d2udx24x22x40<x<1边界条件：x=0, dx1；x=1,u=-1du利用FDM和FVM，求各节点的u值，并与精确解进行比较。提示：参考P6-8Neumann边界条件3、用一般方法3、用一般方法(前向差分、后向差分、中心差分)导出一阶微分具有三(六)阶精度的差分格式。p204、采用Von方法判断下式的稳定性。un1un1i2ti(u n2unuin)x2iiO(t2,x2)P37-39可采用反证法证明其无条件不稳定。552uf(xy)u2x2y2应用第一类边界条件，有：uuuuu02, 3, 6, 9uu848ff205 8采用有限体积法求22单位网格的数值解。p1126、利用CFD软件(如FLUENT或CFX或STAR-CD等)计算一个实例。并打印出模型、网格及计算结果。1. 采用时间前差，空间中差un1und(un 2unun)i i ii it＝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=0.01tU0U1U2U3U4U5U6U7U8U9U1000.000.000.000.000.000.100.000.000.000.000.000.011.000.000.000.000.10-0.100.100.000.000.001.000.021.001.000.000.10-0.200.30-0.200.100.001.001.000.031.000.001.10-0.300.60-0.700.60-0.301.100.001.000.041.002.10-1.402.00-1.601.90-1.602.00-1.402.101.000.051.00-2.505.50-5.005.50-5.105.50-5.005.50-2.501.000.061.009.00-13.0016.00-15.6016.10-15.6016.00-13.009.001.000.071.00-21.0038.00-44.6047.70-47.3047.70-44.6038.00-21.001.000.081.0060.00-103.60130.30-139.60142.70-139.60130.30-103.6060.001.000.091.00-162.60293.90-373.50412.60-421.90412.60-373.50293.90-162.601.000.101.00457.50-830.001080.00-1208.001247.10-1208.001080.00-830.00457.501.00t=0.00t0.0可以看出选择时间步长对计算结果收敛情况起决定作用。和时0d1/2时d1/2，计算结果不稳定。FDM

u u2 3 1u节点：

2u u2 (1/3)2

1/32u 2

2u 2uu节点4 3 (1/3)2代入u 14

2u f3 3解得0.35163d

3u

124u u12采用在节点1给出udx2

2x

31可解得3 9FVMu u节点2 1

1f xxuu

2 2u u节电：

2 2 12uxf xx x 2 2u u uu节点：

3

22u

xf xx x 3 3代入u 1,x1/3,f 38/9,f 32/94 2 3解得2/93精确解u2x2x对比可得FVMFVM中精确实现而FDMFDM中难Neuman边界条件，但可采用虚拟节点法，二阶精度法等来FM以得到精确解。

u (x)2 2u (x)3 3u (x)4 4uu ui1

x(x

( )2! x2

( )3! x3

( )4! x4iu (2x)2 2u (2x)3 3u (2x)4 4uiui2

u2x(x

( )2! x2i

( )3! x3i

( )4! x4iiu (3x)2 2u (3x)3 3u (3x)4 4uiui3

u3x(x

( )2! x2i

( )3! x3i

( )4! x4iu au

du令( ) xi

i1x

i2

i3

(x3)后向差分

11,b3,c3,d16 2 3u (x)2 2u (x)3 3u (x)4 4uu ui1

(x)(x

( )2! x2i

( )3! x3i

( )4! x4iiu (2x)2 2u (2x)3 3u (2x)4 4uiui2

u(2x)(x

( )2! x2i

( )3! x3i

( )4! x4iiu (3x)2 2u (3x)3 3u (3x)4 4uiui3

u(3x)(x

( )2! x2i

( )3! x3i

( )4! x4iu au

du令( ) xi

i1x

i2

i3

(x3)11 3 解得 ,b3,c ,d11 3 6 2 3

iu (3x)2 2u (3x)3 3u (3x)4 4u (3x)5 5u (3x)6 6uiui3

u(3x)(x

( )2! x2i

( )3! x3i

( )4! x4i

( )5! x5i

( )6! x6iiu (2x)2 2u (2x)3 3u (2x)4 4u (2x)5 5u (2x)6 6uiui2

u(2x)(x

( )2! x2i

( )3! x3i

( )4! x4i

( )5! x5i

( )6! x6iu (x)2 2u (x)3 3u (x)4 4u (x)5 5u (x)6 6uu ui1

(x)(x

( )2! x2i

( )3! x3i

( )4! x4i

( )5! x5i

( )6! x6iu (x)2 2u (x)3 3u (x)4 4u (x)5 5u (x)6 6uu ui1

(x)(x

( )2! x2

( )3! x3

( )4! x4

( )5! x5

( )6! x6iu (2x)2 2u (2x)3 3u (2x)4 4u (2x)5 5u (2x)6 6uiui2

u(2x)(x

( )2! x2i

( )3! x3i

( )4! x4i

( )5! x5i

( )6! x6iiu (3x)2 2u (3x)3 3u (3x)4 4u (3x)5 5u (3x)6 6uiui3

u(3x)(x

( )2! x2i

( )3! x3i

( )4! x4i

( )5! x5i

( ) 令6! x6i(uu5

)s5,8

(u9

u)s8

(u7

u)s8

7,8

fA8 8u au

fu

gu( ) i3xi

i2

i1x

i1

i2

i3

(x6)ia0.0167,b0.15,c0.75id0,e0.75,f0.15,g0.0167un1un1 (un

2unun)i

i i1

i i1

(t2,x2) 2t x2nununnii i iin1

(

2nn)i代入得ii

2t

i1

x2

i1

……（2）将n分解为傅立叶级数ini

N neIi ）jjNj将（3）式代入（2）式得n12t

eIi

(neI(i1)neIineI(i1))x2令d

2tx2n1

dn(eI2eI) n令误差放大因子fn1

(n1)

代入（4）得f n1 fn1

2d(1cos) f n fn1

2d(1cos) 这里采用反证法，假设其有条件稳定则由）可推d0,f 0再由）—）得n1 1f f n1 n

而该方程与假设f fn n1

n1

f,fn

fn1

）矛盾固假设不成立，即）式无条件不稳定。节点5的差分方程为(u u2

)s2,5

(u6

u)s5

(u8

u)s5

(u4

u)s5

4,5

fA5 5f