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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,n1) + T(m+1,n) +T(m1,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*((T1T2)/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,j1)+T(i+1,j)+T(i1,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,j1)+T(i+1,j)+T(i1,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,j1)+T(i+1,j)+T(i1,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,j1)+T(i+1,j)+T(i1,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,j1)+T(i+1,j)+T(i1,j))/4;
% end
% end
Q;
fprintf(1,'Press enter and type T for Temp. Distribution');pause;
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,n1) + T(m+1,n) +T(m1,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*((T1T2)/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,j1)+T(i+1,j)+T(i1,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,j1)+T(i+1,j)+T(i1,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,j1)+T(i+1,j)+T(i1,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,j1)+T(i+1,j)+T(i1,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,j1)+T(i+1,j)+T(i1,j))/4;
% end
% end
Q;
fprintf(1,'Press enter and type T for Temp. Distribution');pause;
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