传热学MATLAB温度分布大作业完整版

东南大学能源与环境学院

作业名称：

课程作业报告 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： 姓名： 2014年11月9日

一、题目及要求

1.

2.

3.

4.

5.

6.

7. 原始题目及要求 各节点的离散化的代数方程 源程序 不同初值时的收敛快慢 上下边界的热流量（λ=1W/(m℃）） 计算结果的等温线图 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)/4

te=(100+a+f+i)/4; tf=(b+e+g+j)/4; tg=(c+f+h+k)/4 ; th=(2*g+d+l)/4

ti=(100+e+m+j)/4; tj=(f+i+k+n)/4; tk=(g+j+l+o)/4; tl=(2*k+h+q)/4

tm=(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、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,-1,0,0,0;

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=Z'

contour(X,Y,Z,30)

Z =

139.6088 150.3312 153.0517 153.5639

108.1040 108.6641 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=1:10

t(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)=(100+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;

t'

end

contour(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 14.8903 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=D\b;

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,-1,0,4,0,0,0,-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=Z'

contour(X,Y,Z,30)

n =97

Z =

139.6088 150.3312 153.0517 153.5639

108.1040 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 …… 此处隐藏：3109字，全部文档内容请下载后查看。喜欢就下载吧 ……