传热学大作业（第四章）姓名：张宝琪学号：03110608一、题目及要求1.各节点的离散化的代数方程2.源程序3.不同初值时的收敛快慢4.上下边界的热流量（λ=1W/(m℃））5.计算结果的等温线图6.计算小结题目：已知条件如下图所示：二、方程及程序(1)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)/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 (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)>=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;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 153.5639108.1040 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=1:10t(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)=(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'endcontour(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)>=epsx0=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 =97Z =139.6088 150.3312 153.0517 153.5639108.1040 108.6641 108.3119 108.152384.1429 67.9096 63.3793 62.421420.1557 15.4521 14.8744 14.7746三、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。

四、上下边界热流量：上边界t=200℃,∞t =10℃，所以，热流量Φ1=λ*[2*100-200x y∆∆+x ya∆∆t -200+x y∆∆bt -200+x y∆∆ct -200+2*t -200dx y∆∆]=1*（100/2+(200-139.6088)+(200-150.3312)+(200-153.0517)+(200-153.5639)/2) =230.2264W 下边界热流量Φ2=|λ*[x y∆∆m i t -t +x y∆∆o j t -t +x y∆∆p k t -t +2*t -t q l x y∆∆]-h*(2*10-100x y∆∆+x *t -t n ∆∆∞y+x *t -t o ∆∆∞y+x *t -t m ∆∆∞y+2*t -t p x y∆∆∞)|=|1*((84.1429-20.1557)+(67.9096-15.4521)+(63.3793-14.8744)+(62.4214-14.7746)/2)-10*(90/2+(20.1557-10)+(15.4521-10)+(14.8744 -10)+(14.7746-10)/2)| = |-489.925|W =489.25W五、温度等值线Gauss:Yacobi:六、计算小结导热问题进行有限差分数值计算的基本思想是把在时间、空间上连续的温度场用有限个离散点温度的集合来代替，即有限点代替无限点，通过求解根据傅里叶定律和能量守恒两大法则建立关于控制面内这些节点温度值的代数方程，获得各个离散点上的温度值。