利用matlab程序解决热传导问题

1、哈佛大学能源与环境学院课程作业报告作业名称：传热学大作业利用matlab程序解决热传导问题 院系：能源与环境学院 专业：建筑环境与设备工程 学号：5201314 姓名：盖茨比 2015年6月8日1、 题目及要求1. 原始题目及要求2. 各节点的离散化的代数方程3. 源程序4. 不同初值时的收敛快慢5. 上下边界的热流量（=1W/(m）6. 计算结果的等温线图7. 计算小结题目：已知条件如下图所示：二、各节点的离散化的代数方程各温度节点的代数方程ta=(300+b+e)/4 ; tb=(200+a+c+f)/4; tc=(200+b+d+g)/4; td=(2*c+200+h)/4te=(100+a+f+i)/4; tf=(b+e+g+j)/4; tg=(c+f+h+k)/4 ; th=(2*g+d+l)/4ti=(100+e+m+j)/4; tj=(f+i+k+n)/4; tk=(g+j+l+o)/4; tl=(2*k+h+q)/4tm=(2*i+300+n)/24; tn=(2*j+m+p+200)/24; to=(2*k+p+n+200)/24; tp=(l+o+100)/12三、

2、源程序【G-S迭代程序】【方法一】函数文件为：function y,n=gauseidel(A,b,x0,eps)D=diag(diag(A);L=-tril(A,-1);U=-triu(A,1);G=(D-L)U;f=(D-L)b;y=G*x0+f;n=1;while norm(y-x0)=eps x0=y; y=G*x0+f; n=n+1;end命令文件为：A=4,-1,0,0,-1,0,0,0,0,0,0,0,0,0,0,0; -1,4,-1,0,0,-1,0,0,0,0,0,0,0,0,0,0; 0,-1,4,-1,0,0,-1,0,0,0,0,0,0,0,0,0; 0,0,-2,4,0,0,0,-1,0,0,0,0,0,0,0,0; -1,0,0,0,4,-1,0,0,-1,0,0,0,0,0,0,0; 0,-1,0,0,-1,4,-1,0,0,-1,0,0,0,0,0,0; 0,0,-1,0,0,-1,4,-1,0,0,-1,0,0,0,0,0; 0,0,0,-1,0,0,-2,4,0,0,0,-1,0,0,0,0; 0,0,0,0,-1,0,-1,0,4,0,0,0,-1

3、,0,0,0; 0,0,0,0,0,-1,0,0,-1,4,-1,0,0,-1,0,0; 0,0,0,0,0,0,-1,0,0,-1,4,-1,0,0,-1,0; 0,0,0,0,0,0,0,-1,0,0,-2,4,0,0,0,-1; 0,0,0,0,0,0,0,0,-2,0,0,0,24,-1,0,0; 0,0,0,0,0,0,0,0,0,-2,0,0,-1,24,-1,0; 0,0,0,0,0,0,0,0,0,0,-2,0,0,-1,24,-1; 0,0,0,0,0,0,0,0,0,0,0,-1,0,0,-1,12;b=300,200,200,200,100,0,0,0,100,0,0,0,300,200,200,100;x,n=gauseidel(A,b,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1.0e-6)xx=1:1:4;yy=xx;X,Y=meshgrid(xx,yy);Z=reshape(x,4,4);Z=Zcontour(X,Y,Z,30)Z = 139.6088 150.3312 153.0517 153.5639 108.1040 108.6

4、641 108.3119 108.1523 84.1429 67.9096 63.3793 62.4214 20.1557 15.4521 14.8744 14.7746【方法2】 t=zeros(5,5);t(1,1)=100;t(1,2)=100;t(1,3)=100;t(1,4)=100;t(1,5)=100;t(2,1)=200;t(3,1)=200;t(4,1)=200;t(5,1)=200;for i=1:10t(2,2)=(300+t(3,2)+t(2,3)/4 ;t(3,2)=(200+t(2,2)+t(4,2)+t(3,3)/4;t(4,2)=(200+t(3,2)+t(5,2)+t(4,3)/4; t(5,2)=(2*t(4,2)+200+t(5,3)/4;t(2,3)=(100+t(2,2)+t(3,3)+t(2,4)/4; t(3,3)=(t(3,2)+t(2,3)+t(4,3)+t(3,4)/4; t(4,3)=(t(4,2)+t(3,3)+t(5,3)+t(4,4)/4; t(5,3)=(2*t(4,3)+t(5,2)+t(5,4)/4;t(2,4)=(10

5、0+t(2,3)+t(2,5)+t(3,4)/4; t(3,4)=(t(3,3)+t(2,4)+t(4,4)+t(3,5)/4; t(4,4)=(t(4,3)+t(4,5)+t(3,4)+t(5,4)/4;t(5,4)=(2*t(4,4)+t(5,3)+t(5,5)/4;t(2,5)=(2*t(2,4)+300+t(3,5)/24; t(3,5)=(2*t(3,4)+t(2,5)+t(4,5)+200)/24; t(4,5)=(2*t(4,4)+t(3,5)+t(5,5)+200)/24; t(5,5)=(t(5,4)+t(4,5)+100)/12;tendcontour(t,50);ans = 100.0000 200.0000 200.0000 200.0000 200.0000 100.0000 136.8905 146.9674 149.8587 150.7444 100.0000 102.3012 103.2880 103.8632 104.3496 100.0000 70.6264 61.9465 59.8018 59.6008 100.0000 19.0033 14.89

6、03 14.5393 14.5117【Jacobi迭代程序】函数文件为：function y,n=jacobi(A,b,x0,eps)D=diag(diag(A);L=-tril(A,-1);U=-triu(A,1);B=D(L+U);f=Db;y=B*x0+f;n=1;while norm(y-x0)=eps x0=y; y=B*x0+f; n=n+1;end命令文件为：A=4,-1,0,0,-1,0,0,0,0,0,0,0,0,0,0,0; -1,4,-1,0,0,-1,0,0,0,0,0,0,0,0,0,0; 0,-1,4,-1,0,0,-1,0,0,0,0,0,0,0,0,0; 0,0,-2,4,0,0,0,-1,0,0,0,0,0,0,0,0; -1,0,0,0,4,-1,0,0,-1,0,0,0,0,0,0,0; 0,-1,0,0,-1,4,-1,0,0,-1,0,0,0,0,0,0; 0,0,-1,0,0,-1,4,-1,0,0,-1,0,0,0,0,0; 0,0,0,-1,0,0,-2,4,0,0,0,-1,0,0,0,0; 0,0,0,0,-1,0,-1,0,4,0,

7、0,0,-1,0,0,0; 0,0,0,0,0,-1,0,0,-1,4,-1,0,0,-1,0,0; 0,0,0,0,0,0,-1,0,0,-1,4,-1,0,0,-1,0; 0,0,0,0,0,0,0,-1,0,0,-2,4,0,0,0,-1; 0,0,0,0,0,0,0,0,-2,0,0,0,24,-1,0,0; 0,0,0,0,0,0,0,0,0,-2,0,0,-1,24,-1,0; 0,0,0,0,0,0,0,0,0,0,-2,0,0,-1,24,-1; 0,0,0,0,0,0,0,0,0,0,0,-1,0,0,-1,12;b=300,200,200,200,100,0,0,0,100,0,0,0,300,200,200,100;x,n=jacobi(A,b,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1.0e-6);xx=1:1:4;yy=xx;X,Y=meshgrid(xx,yy);Z=reshape(x,4,4);Z=Zcontour(X,Y,Z,30)n =97Z = 139.6088 150.3312 153.0517 153.5639 108.1

8、040 108.6641 108.3119 108.1523 84.1429 67.9096 63.3793 62.4214 20.1557 15.4521 14.8744 14.7746四、不同初值时的收敛快慢1、方法1在Gauss迭代和Jacobi迭代中，本程序应用的收敛条件均为norm(y-x0)=eps，即使前后所求误差达到e的-6次方时，跳出循环得出结果。将误差改为0.01时，只需迭代25次，如下x,n=gauseidel(A,b,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0.01)运行结果为将误差改为0.1时，需迭代20次，可见随着迭代次数增加，误差减小，变化速度也在减小。方法2通过 i=1:10判断收敛，为迭代10次，若改为1:20，则迭代20次。2、在同样的误差要求下，误差控制在e的-6次方内，Gauss迭代用了49次达到要求，而Jacobi迭代用了97次，可见，在迭代中尽量采用最新值，可以大幅度的减少迭代次数，迭代过程收敛快一些。在Gauss中，初值为100，迭代46次达到精确度1.0e-6,初值为50时，迭代47次，初值为0时，迭代49次，初值为200时迭代50次，可见存在一个最佳初始值，是迭代最快。这一点在jacobi迭代中表现的尤为明显。五、上下边界的热流量：上边界t=200,=10，所以，热流量1=*+=1*（100/2+(200-139.6088)+(200-150.3312)+(200-153.0517)+(200-153.5639)/2)=230.2264W下边界热流量2=|*+-h*(+)|=|1*(84.1429-20.1557)+(67.9096-15.4521)+(63.3793-14.8744)+(62.4214- 14.7746

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