传热学数值计算大作业.doc

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1、一、任务分析阅读4-21题干，确定本题可视为无限大平板非稳态传热问题解决。根据教材4.4关于非稳态传热数值解法的内容，将墙壁剖面划分为（N-1）个部分，共有N个节点。将内壁侧视为左边界（left），外壁侧视作右边界（right），建立方程如下：对于内壁边界节点，即n=0处：对于外壁边界节点，即n=N处:对于中间的节点，即0

2、序号，k代表时间节点序号。#include#includeusingnamespacestd;constintn=3;constintk=24;doubledt=3600;//时间步长doubledr=0.24/(n-1);//空间步长doubleFo=0.81/1800/880dt/dr/dr;doubleBil=6dr/0.81;doubleBir=10dr/0.81;inttl=15;//内壁边界条件doubletr[24]={-5.9,-6.2,-6.6,-6.7,-6.8,-6.9,-7.2,-7

3、.7,-7.6,-7,-4.9,-2.3,-1,-2.4,1.8,1.8,1.6,0.5,-1.6,-2.8,-3.5,-4.3,-4.8,-5.3};//外壁边界条件voidheat(doublea[n][k]){for(intj=0;j

4、a[i][23];}else{a[n-1][0]=a[n-1][23](1-2FoBir-2Fo)+2Foa[n-2][23]+2FoBirtr[0];}cout<

5、2Foa[n-2][j-1]+2FoBirtr[j];}cout<>x;if(x='Y')select=true;elsesele

6、ct=false;}cout<

7、7.15078-1.24674714.34887.00998-1.54407814.2516.85463-1.65087914.14956.71283-1.502351014.04936.61331-0.1113.96066.630410.1213.90916.792471.524071313.9187.028751.384711413.9847.187942.98511514.06647.519463.610791614.20317.856743.84851714.37568.155653.604961814.56078.369042.7

8、96911914.73218.448252.101562014.86048.440221.602492114.93938.386841.114982214.97548.29487