Matlab in applying Finite Difference for Temp. distribution/ rate of heat flow

In summary, the conversation involves a person seeking help and feedback on a Matlab program that they are working on for calculating and outputting the steady state space distribution of temperature and resulting rate of heat flow in an I Beam made of Duralumin using the method of finite difference. They have divided the I Beam into a nodal grid and have written a partial code, but are unsure if they are on the right path. They have inputted all the variables and created an outline of the beam with boundaries and initial conditions. They have also applied equations and FOR loops for the interior nodes to calculate temperature and rate of heat flow. However, they are unsure if they are applying the correct equations and methods as they are getting incorrect results. They have attached
  • #1
mejia198021
3
0
Hi:

I need some assistance or feedback on a Matlab program that I working on by applying the method of finite difference to calculate and output the steady state space distribution of temperature and resulting rate of hear flow in an I Beam made of Duralumin. I have written a partial code but I am not sure if I am on the right path in solving this problem.

First: I divided the I Beam into a nodal grid, which would include 289 interior nodes (not including boundaries). The beam measures 2.4cm(.024m) x 3cm (.03m) and dx=dy=.001m.

Basically in my code I inputted all the variables given, which includes the upper/lower surface temperatures, thermoconductivity (K). Next, I created a outline of the beam by inputing the boundaries/ initial conditions. Then I applied an interior node equation that I looked up in my Heat Transfer Text:

T(m,n+1) + T(m,n-1) + T(m+1,n) +T(m-1,n) -4*T(m,n)=0

I applied this equation in 3 FOR loops for the interior nodes: Upper, center and lower sections of the I Beam. Next, I used a heat flux equation to calculate the rate of heat flow by applying: qx=k*((T1-T2)/L)

I used the same method as above to calculate this equation by stating the boundaries/ initial conditions and then apply 3 FOR loops for the three sections of the I Beam.

At this point I am not sure if I am applying the correct equations or methods. I get Temp. outputs for each interior nodes but at some pts. it seems that it is not correct. For the Heat Flux I obtain section that are 0 and neg., so that it wrong.

Any help or tips would be appreciated. Below is my code and I have attached the problem (NMProb.doc) if you would like to take a look at it. Thank you for your time.


clc
clear all;

T1 = input('Fixed Temperature at the upper surface Tt[60 deg C] : '); % deg C fixed Temperature at the upper surface.
T2 = input('Fixed Temperature at the lower surface. Tl[15 deg C] : '); % deg C fixed Temperature at the lower surface.
k = input('Enter the thermal conductivity k[Duralumin 164 W/m deg C] : '); % W/m deg C (thermal conductivity).



%Intilializing Constant Variables
k=164; %Thermoconductivity
Dn=2787; %Dn is defined as Density
Qi=0;
Qw=0;

%Outline dimensions of I beam
Wt=0.024; %Width
Ht=0.03; %Height

%Temperature Conditions
Tt=60; %Upper face Temperature (333.15K)
Tb=15; %Lower face Temperature (288.15K)
Ti=20; %Initial Temperature of interior nodes (293.15K)

%Conditions of nodal points
dx=.001; %Delta x - distance between each node
dy=.001; %Delta y - distance between each node

M=31; %Number of nodes in the Y direction -i
N=25; %Number of nodes in the X direction -j

%Applying Boundaries/ Initial Conditions to I Beam(Temperature)
T=zeros(31,25);
T(1,:)=T1; %Top Surface
T(31,:)=T2; %Bottom Surface
%Left
T(1:6,1)=0; %Upper Top Left
T(6,1:10)=0; %Bottom Top Left
T(6:26,10)=0; %Inner Center Left
T(26,1:10)=0; %Lower Bottom Left
T(26:31,1)=0; %Lower Left
%Right
T(1:6,25)=0; %Upper Top Right
T(6,16:25)=0; %Bottom Top Right
T(6:26,16)=0; %Inner Center Right
T(26,16:25)=0; %Lower Bottom Right
T(26:31,25)=0; %Lower Right

T(2:5,2:24)=Ti; %Upper Section
T(6:26,11:15)=Ti; %Center Section
T(27:30,2:24)=Ti; %Lower Section

T;

%Temperature Distribution of Interior Nodes (289 Nodal Points)
for(i=2:5)
for(j=2:24)
%Temperature of Nodal Points for Upper Portion
T(i,j)=(T(i,j+1)+T(i,j-1)+T(i+1,j)+T(i-1,j))/4;
end
end

for(i=6:26)
for(j=11:15)
%Temperature of Nodal Points for Center Portion
T(i,j)=(T(i,j+1)+T(i,j-1)+T(i+1,j)+T(i-1,j))/4;
end
end

for(i=27:30)
for(j=2:24)
%Temperature of Nodal Points for Bottom Portion
T(i,j)=(T(i,j+1)+T(i,j-1)+T(i+1,j)+T(i-1,j))/4;
end
end





Q=zeros(31,25);
Q(1,:)=Qw; %Top Surface
Q(31,:)=Qw; %Bottom Surface
%Left
Q(1:6,1)=Qw; %Upper Top Left
Q(6,1:10)=Qw; %Bottom Top Left
Q(6:26,10)=Qw; %Inner Center Left
Q(26,1:10)=Qw; %Lower Bottom Left
Q(26:31,1)=Qw; %Lower Left
%Right
Q(1:6,25)=Qw; %Upper Top Right
Q(6,16:25)=Qw; %Bottom Top Right
Q(6:26,16)=Qw; %Inner Center Right
Q(26,16:25)=Qw; %Lower Bottom Right
Q(26:31,25)=Qw; %Lower Right

Q(2:5,2:24)=Qi; %Upper Section
Q(6:26,11:15)=Qi; %Center Section
Q(27:30,2:24)=Qi; %Lower Section


%Rate of Heat Flow of Interior Nodes (289 Nodal Points
% for(i=2:5)
% for(j=2:24)
% %Temperature of Nodal Points for Upper Portion
% Q(i,j)=k*((T(i,j+1)-T(i,j))/dx);
% end
% end

for(i=6:26)
for(j=11:15)
%Temperature of Nodal Points for Center Portion
T(i,j)=(T(i,j+1)+T(i,j-1)+T(i+1,j)+T(i-1,j))/4;
end
end
%
% for(i=27:30)
% for(j=2:24)
% %Temperature of Nodal Points for Bottom Portion
% T(i,j)=(T(i,j+1)+T(i,j-1)+T(i+1,j)+T(i-1,j))/4;
% end
% end


Q;

fprintf(1,'Press enter and type T for Temp. Distribution');pause;
 

Attachments

  • NMProb.doc
    178.5 KB · Views: 423
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  • #2

fprintf(1,'Press enter and type Q for Rate of Heat Flow');pause;

%T=Temperature Distribution Matrix
%Q=Rate of Heat Flow Matrix
 
  • #3
disp(' ')
fprintf(1,'Press enter and type Q for Rate of Heat Flow');pause;disp(' ')
fprintf(1,'Press enter and type ALL for both T and Q');pause;disp(' ')

input('Enter your choice:','s');

if ans== 'T'
disp(' ');
disp('Temperature Distribution: ');
disp(T);
elseif ans== 'Q'
disp(' ');
disp('Rate of Heat Flow: ');
disp(Q);
elseif ans== 'ALL'
disp(' ');
disp('Temperature Distribution: ');
disp(T);
disp(' ');
disp('Rate of Heat Flow: ');
disp(Q);
else
disp('Invalid Choice!');
end



Hi there,

Thank you for sharing your code and the problem statement. From what I can see, you are on the right track in terms of applying the method of finite difference to solve for the steady state temperature distribution and heat flow rate in an I Beam. However, there are a few areas that could be improved upon to ensure the accuracy of your results.

Firstly, I recommend double checking your input values for the dimensions of the I Beam and the thermal conductivity. In your code, you have defined the thermal conductivity as 164 W/m deg C, but in the problem statement, it is listed as 164 W/m K. This may be causing errors in your calculations. Also, make sure that the dimensions of your I Beam are consistent with the problem statement. In your code, the width is listed as 0.024m, but in the problem statement it is listed as 2.4cm, which would be 0.024m. These small discrepancies can have a big impact on your results.

Next, I would suggest using a more accurate equation for the interior nodes. The equation you have used, T(m,n+1) + T(m,n-1) + T(m+1,n) +T(m-1,n) -4*T(m,n)=0, is accurate for a 2D square grid, but since the I Beam is not a perfect square, a more accurate equation would be T(m,n+1) + T(m,n-1) + T(m+1,n) +T(m-1,n) -4*T(m,n) = (dx^2 * dy^2 * Q(m,n)) / (2 * k * (dx^2 + dy^2)), where Q
 

1. What is Matlab and how is it used in applying Finite Difference for Temperature distribution/ rate of heat flow?

Matlab is a high-level programming language and interactive environment that is commonly used in scientific and engineering applications. It allows for easy manipulation and visualization of data, making it a useful tool for solving complex problems such as temperature distribution and heat flow. Using Matlab, you can write code that implements the finite difference method to solve these types of problems.

2. What is the finite difference method and how does it relate to temperature distribution/ rate of heat flow?

The finite difference method is a numerical technique used to approximate solutions to differential equations. It involves dividing a continuous domain into a discrete grid and approximating derivatives using difference equations. This method is commonly used in problems involving temperature distribution and heat flow, as it allows for the calculation of temperature at discrete points in a given system.

3. What are the advantages of using Matlab for solving problems related to temperature distribution/ rate of heat flow?

One of the main advantages of using Matlab for these types of problems is its ability to handle large amounts of data and complex mathematical operations. It also has built-in functions and toolboxes specifically designed for solving scientific and engineering problems, making it a powerful tool for solving problems related to temperature distribution and heat flow. Additionally, Matlab has a user-friendly interface, making it easier for scientists and researchers to implement the finite difference method and analyze their results.

4. What are some common challenges when using Matlab for temperature distribution/ rate of heat flow problems?

One challenge when using Matlab for these types of problems is ensuring the accuracy of the results. Since the finite difference method involves approximating derivatives, small errors in the calculation can lead to significant discrepancies in the final results. It is important to carefully choose the grid size and step size to minimize these errors. Another challenge is the computational time required for solving large and complex problems, which can be addressed by optimizing the code and utilizing parallel computing.

5. Can Matlab be used to model real-world temperature distribution/ rate of heat flow problems?

Yes, Matlab can be used to model real-world temperature distribution and heat flow problems. However, it is important to note that the accuracy of the results will highly depend on the input parameters and assumptions made in the model. It is crucial to carefully validate and verify the model against experimental data to ensure the reliability of the results. Additionally, incorporating real-world factors such as material properties, boundary conditions, and external sources of heat is necessary for accurate modeling.

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