传热学MATLAB温度分布大作业完整版

1、东南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： 姓名： 2014年11月9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

2、; 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、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

3、=(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

4、,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,

5、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 108.1

6、040 108.6641 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)

7、=(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

8、,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

9、 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=Db;y=B*x0+f;n=1;while norm(y-x0)>=eps x0=y; y=B*x0

10、+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,

11、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;

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 153.5639 108.1040 108.6641 108.3119 108.1523 84.1429 67.9096 63.3793

13、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。

14、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)+(200-15

15、3.5639)/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)/2)-10*(90/2+(20.1557-10)+(15.4521-10)+(14.8744 -10)+(14.7746-10)/2)| = |-489.925|W =489.25W六、温度等值线Gauss:Yacobi:七、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。要先划分查分网格，在建立差分代数方程组，用MATLAB或者其他软件编程求解。高斯