Finite difference temperature distribution with TDMA

68 次查看(过去 30 天)
A stainless steel (thermal conductivity, k = 20 W/mK) fin has a circular cross-sectional area (diameter, D=4 cm) and length L=16 cm. The wing is attached to a wall with a temperature of 200C. The ambient temperature of the fluid surrounding the fin is 20C and the heat transfer coefficient is h=10 W/m2K. Fin tip is insulated. Determine the following by applying finite differences with TDMA.
1. Temperatures inside the fin and the amount of heat radiated from the fin. Plot the temperature distribution.
2. Discuss the effect of the number of element dimensions for N = 5, 10, 50, and 100. Plot the temperature distribution.
The temperature distribution from the analitic solution is like this: T1=150.54, T2=111.107, T3=78.671, T4=50.741, T5=25.172, my code does not give these values.
clc
clear all
L=0.16; %lenght
N=6; %number of nodes
dx=L/(N-1); %lenght between nodes
T=zeros(N+1,1);
Tb=200; %the wall temperature (boundary condition)
k=6; %number of iterations
for j=1:1:k
T(1,1)=Tb;
T(5,1)=T(7,1); %the second boundary condition due to the isulation on the tip of the fin
for i=2:1:N
T(i,1)=(T(i+1,1)+T(i-1,1)+1.536)/2.0768; %FDE narrowed down
end
T(N,1)=T(N-1,1);
plot(T);
hold on
end
hold off
If I can get some help with my code.
  16 个评论
Cubii4
Cubii4 2022-11-30
clc
clear all
L=0.16;
N=6;
dx=L/(N-1);
T=zeros(N+1,1);
Tb=200;
k=1000;
for i=1:1:k
T(1,1)=Tb;
for j=2:1:N
T(j,1)=(T(j+1,1)+T(j-1,1)+1.024)/2.0512;
end
T(N+1)=T(N-1);
plot(T);
hold on
end
hold off
for j=1:N
disp(T(j,1));
end
I just corrected the code like this and it gave me the temperature results as the analytic solutions. I can't thank you enough, is there anything else I have to correct in my code.
Cubii4
Cubii4 2022-11-30
Thank you very much I really appreciated Your help. Can you write it as an answer so I can accept it.

请先登录,再进行评论。

采纳的回答

Torsten
Torsten 2022-11-30
移动:Torsten 2022-11-30
You solve the system iteratively. But you were told to solve it with the Thomas-Algorithm. So apply this algorithm to the matrix A below.
k = 20;
T_inf = 20;
h = 10;
D = 0.04;
L = 0.16;
T_at_wall = 200;
N = 6;
dL = L/(N-1);
A = zeros(N+1);
b = zeros(N+1,1);
A(1,1) = 1.0;
b(1) = T_at_wall;
for i = 2:N
A(i,i-1) = k/dL^2;
A(i,i) = -(2*k/dL^2 + 4/D*h);
A(i,i+1) = k/dL^2;
b(i) = -4/D*h*T_inf;
end
A(N+1,N-1) = -1.0;
A(N+1,N+1) = 1.0;
b(N+1) = 0;
T = A\b;
T = T(1:N)
T = 6×1
200.0000 171.3794 150.5095 136.3216 128.0894 125.3914
  2 个评论
EGE
EGE 2023-3-7
can you please explain the line after "end"? I could not figure out what it is. Thanks in advance.
Torsten
Torsten 2023-3-7
编辑:Torsten 2023-3-7
It's the discretized insulated Neumann condition at x=L: 0 = dT/dx ~ T_(n+1)-T_(n-1).
And the point (n+1) is a ghost node in the discretization.

请先登录,再进行评论。

更多回答(0 个)

类别

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

产品


版本

R2015a

Community Treasure Hunt

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

Start Hunting!

Translated by