someting is wrong but ı didnt find it , I uploaded this question's image
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k = 0.16 W/(cm oC)
H = 0.073 W/ (cm2 oC)
Te = 25 oC
Tp = Temperature values of the points on the upper edge.
Q = 27
Δx=Δy=0.5 cm
perform a direct solution of the problem by using finite differences formulas for the derivatives.
clear all ,close all, clc
Q=27;
delh=0.5;
k=0.16;
H=0.073;
te=298;
w=2;
for i=1:15
A(i)=-(Q*delh^2/(k*w))+30*delh;
A(136-i)=-(Q*delh^2/(k*w))-2*delh*H*te/k;
end
for i=16:120
A(i)=-(Q*delh*delh/(k*w));
end
for i=1:9
A(15*i)=A(15*i)-20;
A(15*i-14)=A(15*i-14)-20;
end
for i=1:120
t(i,i)=-4;
end
for i=121:135
t(i,i)=-4-2*delh*H/k;
end
for i=1:134
t((i+1),i)=1;
t(i,(i+1))=1;
end
for i=1:8
t((15*i+1),(15*i))=0;
t((15*i),(15*i+1))=0;
end
for i=1:15
t(i,(15+i))=2;
t((136-i),(121-i))=2;
end
for i=16:120
t(i,(i+15))=1;
t((136-i),(121-i))=1;
end
U=inv(t)*A
2 Comments
Accepted Answer
Yao Li
on 15 Apr 2013
The dimention of Matrix A is 1*135 and the dimention of inv(t) is 135*135. I don't think it's Ok for inv(t)*A. I'm not sure what you really want, but try
U=inv(t)*A'
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