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 comentarios
Jan
el 15 de Abr. de 2013
We cannot try to solve the problem, when all we know is "something is wrong".
Respuesta aceptada
Yao Li
el 15 de Abr. de 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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