完整版传热学MATLAB温度分布大作业

1、东南大学能源与环境学院课程作业报告作业名称：传热学大作业一一利用 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：2021年11月9日一、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(入=1W/(m C)6. 计算结果的等温线图7. 计算小结题目：条件如以下图所示：10CC绝热T尸10.t) h=10W/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

2、)/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)/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=-

3、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

4、,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

5、,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.

6、6641 108.3119 108.152384.1429 67.9096 63.3793 62.421420.1557 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

7、,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,

8、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 200.0000100.0000 136.8905 146.9674 149.8587 150.744410

9、0.0000 102.3012 103.2880 103.8632 104.3496100.0000 70.6264 61.9465 59.8018 59.6008100.0000 19.0033 14.8903 14.5393 14.5117Jacobi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)>=epsx0=y;y=B*x0+f;n=n+1;end命令文件为：A=4,-1,0

10、,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,

11、-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,

12、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 14.7746四、不同初值时的收

13、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 ,即使前后所求误差到达 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,假设改为1:20 ,那么迭代20次.2、在同样的误差要求下,误差限制在 e的-6次方内,Gauss迭代用了 49次到达要求,

14、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次到达精确度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,可见存在一个最正确初始值,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 C, t =10 C,所以,热流量1=入* 200-100* x + 200-ta y 2 yx + 20 x + 3 x+0yyy 2=1* ( 100/2+(200- 139.6088)+(200-150.3312)+

15、(200-153.0517)+(200-153.5639)/2)=230.2264W下边界热流量 2=| 入* Ltm x+t x + Lp x + tL*x-y y y y 2h*(*tn-tx + lmjiL*y=|1*( 84.1429-20.1557)+(67.9096-15.4521)+(63.3793-14.8744)+(62.421414.7746 )/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根本思想是把在时间、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那么建立关于限制 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MATLA城者其他软件编程求解.