- #1
mjaisit
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Im trying to write a MATLAB code for 2d transient heat transfer. The problem is a titanium block in the shape of a rectangle is staring at a specified temperature and is being exposed to a much hotter surrounding temperature. The dimensions, coefficients of heat transfer, and temperatures are all given. We are suppose to write a code for the temperature of the block as a function of time. I understand the basics of the problem, but I am having trouble understanding how to relate the boundary conditions to the initial temperature. We worked a problem similar, but the boundary conditions were that the walls were held at a specified temperature. With this problem though, the wall temperatures are a function of time. I started writing the code below and setup the temperatures of the nodes on the wall and at the corners using nodal finite-difference equations from our textbook. My main question is even though i setup the temperature of the walls of the block, I am not sure how to relate it to the initial temperature and making it a function of time. I know the next step will be to write the equations for the interior nodes, but I want to get the temperature of the nodes on the wall first. I also attached the problem statement in case my description isn't clear. Thanks for your time.
%Dimensions of the block
L=1.2;
H=1;
%Heat Transfer Coeffcients
h=150;
%Distance between nodes
dx=.005;
%Number of nodes in each direction
m=L/dx+1;
n=H/dx+1;
%Given Temperatures
To=278;
Tinf=533;
%Biot number equation
Bi=(h*dx)/k;
for j=2:n-1
%Left Wall
T(1,j)=((2*T(i+1,j)+T(i,j+1)+T(i,j-1))+2*Bi*Tinf)/(2*(Bi+2));
%Right Wall
T(m,j)=((2*T(i-1,j)+T(i,j+1)+T(i,j-1))+2*Bi*Tinf)/(2*(Bi+2));
end
for i=2:m-1
%Bottom Wall
T(i,1)=((2*T(i,j+1)+T(i-1,j)+T(i+1,j))+2*Bi*Tinf)/(2*(Bi+2));
%Top Wall
T(i,n)=((2*T(i,j-1)+T(i-1,j)+T(i+1,j))+2*Bi*Tinf)/(2*(Bi+2));
end
%Bottom Left Corner
T(1,1)=((T(i,j+1)+T(i+1,j))+2*Bi*Tinf)/(2*(Bi+1));
%Bottom Right Corner
T(m,1)=((T(i,j+1)+T(i-1,j))+2*Bi*Tinf)/(2*(Bi+1));
%Top Left Corner
T(1,n)=((T(i,j-1)+T(i+1,j))+2*Bi*Tinf)/(2*(Bi+1));
%Top Right Corner
T(m,n)=((T(i,j-1)+T(i-1,j))+2*Bi*Tinf)/(2*(Bi+1));
%Dimensions of the block
L=1.2;
H=1;
%Heat Transfer Coeffcients
h=150;
%Distance between nodes
dx=.005;
%Number of nodes in each direction
m=L/dx+1;
n=H/dx+1;
%Given Temperatures
To=278;
Tinf=533;
%Biot number equation
Bi=(h*dx)/k;
for j=2:n-1
%Left Wall
T(1,j)=((2*T(i+1,j)+T(i,j+1)+T(i,j-1))+2*Bi*Tinf)/(2*(Bi+2));
%Right Wall
T(m,j)=((2*T(i-1,j)+T(i,j+1)+T(i,j-1))+2*Bi*Tinf)/(2*(Bi+2));
end
for i=2:m-1
%Bottom Wall
T(i,1)=((2*T(i,j+1)+T(i-1,j)+T(i+1,j))+2*Bi*Tinf)/(2*(Bi+2));
%Top Wall
T(i,n)=((2*T(i,j-1)+T(i-1,j)+T(i+1,j))+2*Bi*Tinf)/(2*(Bi+2));
end
%Bottom Left Corner
T(1,1)=((T(i,j+1)+T(i+1,j))+2*Bi*Tinf)/(2*(Bi+1));
%Bottom Right Corner
T(m,1)=((T(i,j+1)+T(i-1,j))+2*Bi*Tinf)/(2*(Bi+1));
%Top Left Corner
T(1,n)=((T(i,j-1)+T(i+1,j))+2*Bi*Tinf)/(2*(Bi+1));
%Top Right Corner
T(m,n)=((T(i,j-1)+T(i-1,j))+2*Bi*Tinf)/(2*(Bi+1));