I need help trying to code an mn by mn A matrix to be use to calculate temperature

2 次查看(过去 30 天)
I have a hard coded version of what the code sure do but it is returning the warning that the matrix is singular. I have started coding for an m by n however I getting confuse on how to code using if statements. The problem have the conditions that the left and bottom side of the red figure is insulated and the right and top side have the different h values and a cooling temperature of 25 deg C. I have provided an figure to help visual the problem:
%% Input
T_inf = 25; % C
kx = 5; % W/mk
ky = 10; % W/mk
q_dot = 200000; % W/m^3
hy = 10; % W/m^2K
hx = 100; %W/m^2k
% nodes
m = 5;
n = 5;
%Grid Setup
X = .025; %m
Y = .1; %m
% Spacing
dx = X/m;
dy = Y/n;
% Create A matrix
s = m*n;
A = zeros(s,s);
% Create b matrix
b = zeros(s,1);
%% Mohen
A11 = -3.*((hx.*dx./2) + (kx.*dy./(2.*dx)) + (ky.*dx./(2.*dy)));
A12 = kx.*dy./2;
A13 = ky.*dx./2;
A21 = -4 .* (kx.*dx + kx.*dy./(2.*dx) + ky.*dx./dy + kx.*dy./(2.*dx));
A22 = kx.*dy./(2.*dx);
A23 = ky.*dx./dy;
A24 = kx.*dy./(.2*dx);
A31 = -4.*(hx.*dx./2 + hy.*dy ./ 2 + ky.*dx./(2.*dx)+kx.*dy./(2.*dx));
A32 = ky.*dx./(2.*dy);
A33 = kx.*dy./(2.*dx);
A41 = -3.*(ky.*dx./(2.*dy)+ kx .* dy./dx + ky.*dx./(2.*dy));
A42 = ky.*dx./(2.*dy);
A43 = kx.*dy./dx;
A44 = ky.*dx./(2.*dy);
A51 = -4.*(ky.*dx./dy + kx.*dy./dx + ky.*dx./dy + ky.*dy./dx);
A52 = ky.*dx./dy;
A53 = kx.*dy./dx;
A54 = ky.*dx./dy;
A55 = kx.*dy./dx;
%% Alice
A61 = -4.*(ky.*dx./(2.*dy)+hy.*dy+ky.*dx./(2.*dy)+kx.*dy./dx);
A62 = ky*dx/2/dy;
A63 = ky*dx/2/dy;
A64 = kx*dy/dx;
A71 = -2*(ky*dx/2/dy+kx*dy/2/dx);
A72 = ky*dx/2/dy;
A73 = kx*dy/2/dx;
A81 = -3*(ky*dx/dy+kx/2/dx);
A82 = ky*dx/dy;
A83 = kx*dy/2/dx;
A84 = kx*dy/2/dx;
A91 = -3*(ky*dx/2/dy+hy*dy/2+kx*dy/2/dx);
A92 = ky*dx/2/dy;
A93 = kx*dy/2/dx;
%% Travion
b1 = -hx*(dx/2)*T_inf - q_dot*(dx/2)*(dy/2);
b2 = -hx*dx*T_inf - q_dot*dx*(dy/2);
b3 = -hx*(dx/2)*T_inf - q_dot*(dx/2)*(dy/2) - hy*(dy/2)*T_inf;
b4 = -q_dot*(dx/2)*dy;
b5 = -q_dot*dx*dy;
b6 = -q_dot*(dx/2)*dy-hy*dy*T_inf;
b7 = -q_dot*(dx/2)*(dy/2);
b8 = -q_dot*dx*(dy/2);
b9 = -q_dot*(dx/2)*(dy/2)-hy*(dy/2)*T_inf;
%% Hard Code
A(1,1) = A11;
A(1,2) = A12;
A(1,6) = A13;
A(2,2) = A21;
A(2,3) = A22;
A(2,7) = A23;
A(2,1) = A24;
A(3,3) = A21;
A(3,4) = A22;
A(3,8) = A23;
A(3,2) = A24;
A(4,4) = A21;
A(4,5) = A22;
A(4,9) = A23;
A(4,3) = A24;
A(5,5) = A31;
A(5,10) = A32;
A(5,4) = A33;
A(6,6) = A41;
A(6,1) = A42;
A(6,7) = A43;
A(6,11) = A44;
A(11,11) = A41;
A(11,6) = A42;
A(11,12) = A43;
A(11,16) = A44;
A(16,16) = A41;
A(16,11) = A42;
A(16,17) = A43;
A(16,21) = A44;
A(7,7) = A51;
A(7,2) = A52;
A(7,8) = A53;
A(7,12) = A54;
A(7,6) = A55;
A(8,8) = A51;
A(8,3) = A52;
A(8,9) = A53;
A(8,13) = A54;
A(8,7) = A55;
A(9,10) = A51;
A(9,5) = A52;
A(9,10) = A53;
A(9,14) = A54;
A(9,8) = A55;
A(12,12) = A51;
A(12,7) = A52;
A(12,13) = A53;
A(12,17) = A54;
A(12,11) = A55;
A(13,13) = A51;
A(13,8) = A52;
A(13,14) = A53;
A(13,18) = A54;
A(13,12) = A55;
A(14,14) = A51;
A(14,9) = A52;
A(14,15) = A53;
A(14,19) = A54;
A(14,13) = A55;
A(17,17) = A51;
A(17,12) = A52;
A(17,18) = A53;
A(17,22) = A54;
A(17,16) = A55;
A(18,18) = A51;
A(18,13) = A52;
A(18,19) = A53;
A(18,23) = A54;
A(18,17) = A55;
A(19,19) = A51;
A(19,14) = A52;
A(19,20) = A53;
A(19,24) = A54;
A(19,18) = A55;
A(10,10) = A61;
A(10,5) = A62;
A(10,15) = A63;
A(10,9) = A64;
A(15,15) = A61;
A(15,10) = A62;
A(15,20) = A63;
A(15,14) = A64;
A(20,20) = A61;
A(20,15) = A62;
A(20,25) = A63;
A(20,19) = A64;
A(21,21) = A71;
A(21,16) = A72;
A(21,22) = A73;
A(22,22) = A81;
A(22,17) = A82;
A(22,23) = A83;
A(22,21) = A84;
A(23,23) = A81;
A(23,18) = A82;
A(23,24) = A83;
A(23,22) = A84;
A(22,24) = A81;
A(22,19) = A82;
A(22,25) = A83;
A(22,23) = A84;
A(25,25) = A91;
A(25,20) = A92;
A(25,24) = A93;
b = [b1; b2; b2; b2; b3; b4; b5; b5; b5;
b6; b4; b5; b5; b5; b6; b4;
b5; b5; b5; b6; b7; b8; b8; b8; b9;];
T = A\b;
%% Actual Code for an N x M
[C,D] = sizes(A);
for i = 1:c
for j = 1:d
if i = 2:c-1
Incorrect use of '=' operator. Assign a value to a variable using '=' and compare values for equality using '=='.
if j = 2:d-1
A(i,j) = A51;
A(i,j-d) = A52;
A(i,j+1) = A53;
A(i,j+d) = A54;
A(i,j-1) = A55;
end
end
end

回答(1 个)

Walter Roberson
Walter Roberson 2024-3-11
if i >= 2 && i <= c-1
if j >= 2 && j <= d-1

类别

Help CenterFile Exchange 中查找有关 Programming 的更多信息

产品


版本

R2023b

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by