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// Display mode
mode(0);
// Display warning for floating point exception
ieee(1);
clc;
disp("Principles of Heat Transfer, 7th Ed. Frank Kreith et. al Chapter - 3 Example # 3.5 ")
//Thermal conductivity of alloy bus bar in W/m-K is given as
k = 20;
//Heat generation rate in W/m3 is given as
qg = 10^6;
//dimensions of the bar in m is given as
L = 0.1;//Length in m
b = 0.05;//Width in m
d = 0.01;//Thickness in m
//For top edge, heat transfer coefficient in W/m2K and ambient temperature
//in C are
h = 75;
Tinfinity = 0;
//We are taking a total of 11 nodes in the direction of length and 6 nodes
//in the direction of width
dx = 0.01; //dx in m
dy = 0.01; //dy in m
//Assigning a guess temperature of 25C to all nodes
for i = 1:6
for j = 1:11
//Old temp. in degree C
Told(i,j) = 25;
end;
end;
//Assigning temperature on the left and right hand side
for i = 1:6
//Old temp. in degree C
Told(i,1) = 40;
Told(i,11) = 10;
//New temp. in degree C
Tnew(i,1) = 40;
Tnew(i,11) = 10;
end;
//Intitalisation of looping parameter
p = 0;
//Iteration to find temperature distribution
while p<500
//Equation for all interior nodes
for i = 2:5
for j = 2:10
//New temp. in degree C
Tnew(i,j) = 0.25*(Told(i-1,j)+Told(i+1,j)+Told(i,j-1)+Told(i,j+1)+((qg*dx)*dx)/k);
end;
end;
//Equation for top wall
for j = 2:10
//New temp. in degree C
Tnew(1,j) = (h*Tinfinity+(qg*dx)/2+(k*(0.5*(Told(1,j-1)+Told(1,j+1))+Told(2,j)))/dx)/(h+(2*k)/dx);
end;
//Equation for bottom wall
for j = 2:10
//New temp. in degree C
Tnew(6,j) = 0.25*(Told(6,j-1)+Told(6,j+1))+0.5*Told(5,j)+((qg*dx)*dx)/(4*k);
end;
for i = 1:6
for j = 1:11
//Assigning Old Temp=New Temp
Told(i,j) = Tnew(i,j);
end;
end;
//New looping parameter incremented
p = p+1;
end;
disp("The temperature distribution in the bar in C is the following")
//Old temp. in degree C
Told
//Finding maximum temperature
Tmax = Told(1,1);
for i = 1:6
for j = 1:11
if Told(i,j)>Tmax then
Tmax = Told(i,j);
else
Tmax = Tmax;
end;
end;
end;
disp("The maximum temperature in C in the alloy bus bar is")
//maximum temperature in C
Tmax
//Finding heat transfer rate
dz = 0.01; //dz in m
//Defining areas
for i = 2:10
A(1,i) = dx*dz; //Area in m2
end;
A = mtlb_i(A,1,(dx*dz)/2);
A = mtlb_i(A,11,A(1));
for i = 1:11
//heat transfer rate in W
q(1,i) = (h*A(i))*(Tnew(1,i)-Tinfinity);
end;
disp("The heat transfer rate from the top edge in W is given by")
//heat transfer rate in W
q
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