2D Heat Conduction - MATLAB help

In summary: Summary: In summary, the conversation discusses a 2D conduction problem with convection heat transfer. The code written for this problem is not working and the person is seeking help and ideas for improvement. The code is very long and involves creating matrices and solving for unknown temperatures. One issue that may be causing problems is a mistake in one of the equations.
  • #1
abe_cooldude
15
0
Hi all, I am working on the problem below, and I wrote the code, but it's not working. Can anyone help me out? And even any ideas on how to improve the code to make it more succinct?

http://i25.tinypic.com/2v0zi8m.jpg
2v0zi8m.jpg


Basically, it is a 2D conduction problem with convection heat transfer on the top, insulated at the bottom edge, and temperature held constant at the left and right edge. The dimensions of the plate are 0.8x0.7 with dx=dy=dx=0.1. I used the symmetry, and the just worked on the left side of the symmetry line (nodes 1-40), wrote nodal equations (finite difference eqs) for each node, and then created a 40x40 matrix in MATLAB to solve system of unknown temperatures. However, my answer is not even remotely close. It doesn't matter what values I choose for k, h, Tb, q_dot, or any other constants. Here's my code. It's very long. I hope there's an easier way to implement this. IF there is, can someone help me or give me any ideas?

clc
clear

%Material Properties & Constants
k = 205;
h = 10;
qdot = 1000;
Tinf = 293.15;
Tb = 423.15;
dx = 0.1;
c = (-qdot*dx^2)/k;
d = 2*((h*dx)/k)*Tinf;
%Matrices
n = 5*8;
a = zeros(n,n);
b = zeros(n);

%node(1)
a(1,1)=-2*((h*dx)/k+1);
a(1,2)=1;
a(1,6)=1;
b(1)=c-d;

%node(2)
a(2,2)=-2*((h*dx)/k+2);
a(2,1)=1;
a(2,3)=1;
a(2,7)=2;
b(2)=c-d;

%node(3)
a(3,3)=-2*((h*dx)/k+2);
a(3,2)=1;
a(3,4)=1;
a(3,8)=2;
b(3)=c-d;

%node(4)
a(4,4)=-2*((h*dx)/k+2);
a(4,3)=1;
a(4,5)=1;
a(4,9)=2;
b(4)=c-d;

%node(5)
a(5,5)=-2*((h*dx)/k+2);
a(5,4)=2;
a(5,10)=2;
b(5)=c-d;

%node(6)
a(6,6)=Tb;
b(6)=0;

%node(7)
a(7,7)=-4;
a(7,2)=1;
a(7,6)=1;
a(7,8)=1;
a(7,12)=1;
b(7)=c;

%node(8)
a(8,8)=-4;
a(8,3)=1;
a(8,7)=1;
a(8,9)=1;
a(8,13)=1;
b(8)=c;

%node(9)
a(9,9)=-4;
a(9,4)=1;
a(9,8)=1;
a(9,10)=1;
a(9,14)=1;
b(9)=c;

%node(10)
a(10,10)=-4;
a(10,5)=1;
a(10,9)=2;
a(10,15)=1;
b(10)=c;

%node(11)
a(11,11)=Tb;
b(11)=0;

%node(12)
a(12,12)=-4;
a(12,7)=1;
a(12,11)=1;
a(12,13)=1;
a(12,17)=1;
b(12)=c;

%node(13)
a(13,13)=-4;
a(13,8)=1;
a(13,12)=1;
a(13,14)=1;
a(13,18)=1;
b(13)=c;

%node(14)
a(14,14)=-4;
a(14,8)=1;
a(14,13)=1;
a(14,15)=1;
a(14,19)=1;
b(14)=c;

%node(15)
a(15,15)=-4;
a(15,10)=1;
a(15,14)=2;
a(15,20)=1;
b(15)=c;

%node(16)
a(16,16)=Tb;
b(16)=0;

%node(17)
a(17,17)=-4;
a(17,12)=1;
a(17,16)=1;
a(17,18)=1;
a(17,22)=1;
b(17)=c;

%node(18)
a(18,18)=-4;
a(18,13)=1;
a(18,17)=1;
a(18,19)=1;
a(18,23)=1;
b(18)=c;

%node(19)
a(19,19)=-4;
a(19,14)=1;
a(19,18)=1;
a(19,20)=1;
a(19,24)=1;
b(19)=c;

%node(20)
a(20,20)=-4;
a(20,15)=1;
a(20,19)=2;
a(20,25)=1;
b(20)=c;

%node(21)
a(21,21)=Tb;
b(21)=0;

%node(22)
a(22,22)=-4;
a(22,17)=1;
a(22,21)=1;
a(22,23)=1;
a(22,27)=1;
b(22)=c;

%node(23)
a(23,23)=-4;
a(23,18)=1;
a(23,22)=1;
a(23,24)=1;
a(23,28)=1;
b(23)=c;

%node(24)
a(24,24)=-4;
a(24,19)=1;
a(24,23)=1;
a(24,25)=1;
a(24,29)=1;
b(24)=c;

%node(25)
a(25,25)=-4;
a(25,20)=1;
a(25,24)=2;
a(25,30)=1;
b(25)=c;

%node(26)
a(26,26)=Tb;
b(26)=0;

%node(27)
a(27,27)=-4;
a(27,22)=1;
a(27,26)=1;
a(27,28)=1;
a(27,32)=1;
b(27)=c;

%node(28)
a(28,28)=-4;
a(28,23)=1;
a(28,27)=1;
a(28,29)=1;
a(28,33)=1;
b(28)=c;

%node(29)
a(29,29)=-4;
a(29,24)=1;
a(29,28)=1;
a(29,30)=1;
a(29,34)=1;
b(29)=c;

%node(30)
a(30,30)=-4;
a(30,25)=1;
a(30,29)=2;
a(30,35)=1;
b(30)=c;

%node(31)
a(31,31)=Tb;
b(31)=0;

%node(32)
a(32,32)=-4;
a(32,27)=1;
a(32,31)=1;
a(32,33)=1;
a(32,37)=1;
b(32)=c;

%node(33)
a(33,33)=-4;
a(33,28)=1;
a(33,32)=1;
a(33,34)=1;
a(33,38)=1;
b(33)=c;

%node(34)
a(34,34)=-4;
a(34,29)=1;
a(34,33)=1;
a(34,35)=1;
a(34,39)=1;
b(34)=c;

%node(35)
a(35,35)=-4;
a(35,30)=1;
a(35,34)=2;
a(35,40)=1;
b(35)=c;

%node(36)
a(36,36)=Tb;
b(36)=0;

%node(37)
a(37,37)=-4;
a(37,32)=2;
a(37,36)=1;
a(37,38)=1;
b(37)=c;

%node(38)
a(38,38)=-4;
a(38,33)=2;
a(38,37)=1;
a(38,39)=1;
b(38)=c;

%node(39)
a(39,39)=-4;
a(39,34)=2;
a(39,38)=1;
a(39,40)=1;
b(39)=c;

%node(40)
a(40,40)=-4;
a(40,35)=2;
a(40,39)=2;
b(40)=c;

T=a\b;

fprintf(' Forcing Function Vector \n ')
b(:,1)
fprintf(' NODE TEMPERATURE (K) \n ')
TSolution=T(:,1);

contourf(T)

for i=1:length(TSolution)
fprintf(' %4.0f \t \t %6.1f \n',i,TSolution(i))
end
 
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  • #2
One of the problems that you may be having is:

%node(16)
a(16,16)=Tb;
b(16)=0;

I think that this should be:

%node(16)
a(16,16)=1;
b(16)=Tb;

so the equation that you would get in this case would be:

1 x T16 = Tb
so T16 = Tb/1 = Tb

In your case:

Tb x T16 = 0
so T16 = 0/Tb = 0
 
  • #3
Question:Why u put C-D at node 1 until node 5?
 

1. What is 2D heat conduction?

2D heat conduction is a physical phenomenon where heat transfer occurs in two dimensions, typically in a plane or surface. It is governed by the laws of thermodynamics and can be described by mathematical equations.

2. How is MATLAB used in 2D heat conduction?

MATLAB is often used to simulate and solve equations related to 2D heat conduction. It provides a powerful platform for numerical computations and can handle complex equations and boundary conditions.

3. What are some common techniques used in 2D heat conduction analysis?

Some commonly used techniques include finite difference method, finite element method, and boundary element method. These methods involve discretizing the domain into smaller elements and solving the equations for each element to obtain a solution for the entire domain.

4. Can MATLAB help with visualizing 2D heat conduction?

Yes, MATLAB has built-in functions and tools that can help visualize the temperature distribution and heat flow in 2D heat conduction problems. This can aid in understanding the behavior of the system and validating the results.

5. Are there any limitations to using MATLAB for 2D heat conduction analysis?

While MATLAB is a powerful tool, it does have some limitations when it comes to 2D heat conduction analysis. It may not be suitable for problems with extremely complex geometries or those that involve transient heat transfer. In such cases, other software or analytical methods may be more appropriate.

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