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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12一、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（λ=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+
2、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)/412tm=(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]=gause
3、idel(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;whilenorm(y-x0)>=epsx0=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;120,0,-2,4,0,0
4、,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,
5、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,
6、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=139.6088150.3312153.0517153.563912108.1040108.6641108.3119108.152384.142967.909663.379362.421420.155715.452114.874414.7746【方法2】>>t=zeros(5,5)
7、;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;fori=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(
8、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;12t
