传热学试验--墙角matlab导热问题

二维导热物体温度场的数值模拟姓名小明学号111111班级能动学院能动

一、问题描述有一墙角模型，尺寸如图1所示，导热系数0.53W/(m•K),墙角内外壁为第一类边界条件。求解该模型的温度分布及导热量。图1q=01.5m二、计算原理根据热平衡法列出节点方程，各方向导入单元体的热量之和为零。内节点和绝热边界点(图1点划线上的点)的方程形式不同。图2Ax内节点图2Ax内节点图2所示的内节点和绝热边界节点方程如下:内节点：

「N%Y-eY‘W二'小(%1-t,j)x(ti,ji-t|1j)x(t,ij-t,,j)y(t「N%Y-eY‘W二'小ILyyxx绝热边界点:力■力■：.：,■中■：.D=«1«NSEW*1-t—)x2y2y三、计算过程用Matlab7.1语百编写计算程序，初取网格步长Ax=Ay=0.1mfunctionI□fori=1;15J(i?1)=30;end-forj=l:12I(bj}=30;endforik=6:15forn=5:12I(叫n.=0;endend1(16,2:5)=124,18,12,6];T(2:5,12)t[24,13,12,6];while1fori=2:15for尸2:5eps=l:teihp=I(ijj)：J(if3)=(1(1-1,j)+T(i+bj)+T(i,j-C+Ki,j+l))/4;*ps=T11,j)-+emp;end-endifieps<le-8break;endend□whil«I\forj=2:5?ps=l;temp=I(16,j);|Itlfi,j>=(2*T(15,j)+1(lfi,r1)+1(lfi,j+l))/4;即后111jj-temp:endifeps<le-8break;endendnwhiieifori=2：5□forj=2:11融二I；te]Mp=T(i,j);Z(ifj)=(T(i-lJj)+T(i+Uj)+T(itj-l)+T(ifj+D)/4;eps=I,ijj)-temp;end-endifeps<le^Ebreak;end.endEwhile1□fori=2:5temp=I(i,12);Ki,12)=(2*r(i,ID+Tti-1,12)+r(i+l>12))/舒eps=I(i,12)-temp;*ndifeps<le-8break;end-end500101:(1*);shadinginterp-figtircontour(T?10);运行结果：

30,0030,0030.0030.0030.0030.0030,0030.00沁0030.0030,0030,0030.0028,9327,8926.90比26.9026.0225.3024.79244624.1324.0624.05范0C27.8&25-7223.702L8T20.3919.3918.7918-43IE-221&09IE.0830.0026.7223.4720.2917.3615.0013.6212.8612.4612.2312.1012.0830.0025.5921.1316.6412,308.637.216.596,306,156,066.05池0024.4918.821.2.856.550.000.000.000.000.000.000.0030.0C24.2710.4512.466.290.000.000,001000.000.CO0.9030,0024.1513.2512.256.160.000,000,000,000,000,000.0030.0024.0318.1312.146.08QJ0Q.OOQ.000.000.000,00范0C24,C&18-0712.07乳。50.000.000.001000-000,CO0.0030.0024.C213.0412.046.020.000.000.000.000.000.000.0030.0024,0118.0212.026,01CLOO0,000.000,000,000,000,00池0024.0113.011.2.016.010.000.000.000.000.000.000.0030.0C24.CO1S.0112.016.000.000.000.000.000.00O.COCL9030.0024.0018.0012.006.000.000,000.000.000.000,000.0030.0024,⑩13.0012.0Q6.0DQJ0Q.000.0。Ci000.000.000,00图1:各个点的温度数值图2:分层设色等温线分布T"-20--T"-20--15”-10图3:等温线分布（每两条线间隔为三度）20406080100120140长度四、小结本次数值模拟是运用matlab程序用于数