Fin heat transfer Matrix

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Yogesh Bhambhwani
Yogesh Bhambhwani 2020 年 12 月 16 日
編集済み: Alan Stevens 2020 年 12 月 17 日
I need help solving this matrix with the equations given to me:
For the first node: T_1 = T_b
for the internal nodes: (T_2,T_3,T_4) = -T_i-1 +(2+(mdeltax)^2)T_i - Ti+1 = (mdeltax)^2T_inf
for the 5th node: -T_4 + (1+((mdeltax)^2/2))T_5 = ((mdeltax)^2/2)T_inf
where m = sqrt((hp)/(KA_c))
T_inf = 900 degrees celsius
T_b = 400 degress celsius
also have to compare the matrix solution to:
T_analytic = (cosh(m(L-x))/cosh(mL))*(T_b-T_inf)+T_inf
I need some help with the matrix solution.
  1 件のコメント
Ive J
Ive J 2020 年 12 月 17 日
Share with us what you've tried so far and clearly explain how do you want to solve this heat transfer equation in particular?

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採用された回答

Alan Stevens
Alan Stevens 2020 年 12 月 17 日
編集済み: Alan Stevens 2020 年 12 月 17 日
% Construct the matrix
% M = [ 1 0 0 0 0;
% -1 (2+(m*dx)^2) -1 0 0;
% 0 -1 (2+(m*dx)^2) -1 0;
% 0 0 -1 (2+(m*dx)^2) -1;
% 0 0 0 -1 (2+(m*dx)^2)/2];
%
% and the column vector
% K = [T_b;
% (m*dx)^2*T_inf;
% (m*dx)^2*T_inf;
% (m*dx)^2*T_inf;
% (m*dx)^2/2*T_inf];
%
% then you have the matrix equation M*T = K
% where T is a column vector of values of T_1; T_2 ...T_5
% and you can solve for T using T = M\K (notice the backslash
% not forward slash)

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