1- Temperature distribution in a part of a heat exchanger in the steady state co

Question

1- Temperature distribution in a part of a heat exchanger in the steady state condition should be determined. The part is shown in Figure-1 has a thermal conductivity of k 25W/m.K and dimensions and boundary conditions are specified in Figure-1 and Figure-2 respectively. Develop a computer code to find the temperature distribution in the part (use Ax-Ay lmm Under operating conditions for which ho-1000 Wm K, T 1700 K, hi 200 Wm2 K, and T 400 K. Also, at what location is the temperature a maximum? 5mm 2mm 3mm 2mm Figure-1. Dimensions Tono.h Toni,h Figure-2. Boundary Conditions

Explanation / Answer

clear all
close all
%Specify grid size
Nx = 10;
Ny = 10;
%Specify boundary conditions
Tbottom = 100
Ttop = 150
Tleft = 250
Tright = 300
% initialize coefficient matrix and constant vector with zeros
A = zeros(Nx*Ny);
C = zeros(Nx*Ny,1);
% initial 'guess' for temperature distribution
T(1:Nx*Ny,1) = 100;
% Build coefficient matrix and constant vector
% inner nodes
for n = 2:(Ny-1)
for m = 2:(Nx-1)
i = (n-1)*Nx + m;

A(i,i+Nx) = 1;
A(i,i-Nx) = 1;
A(i,i+1) = 1;
A(i,i-1) = 1;
A(i,i) = -4;
end
end
% Edge nodes
% bottom
for m = 2:(Nx-1)
%n = 1
i = m;
A(i,i+Nx) = 1;
A(i,i+1) = 1;
A(i,i-1) = 1;
A(i,i) = -4;
C(i) = -Tbottom;
end

%top:
for m = 2:(Nx-1)
% n = Ny
i = (Ny-1)*Nx + m;
A(i,i-Nx) = 1;
A(i,i+1) = 1;
A(i,i-1) = 1;
A(i,i) = -4;
C(i) = -Ttop;
end
%left:
for n=2:(Ny-1)
%m = 1
i = (n-1)*Nx + 1;

A(i,i+Nx) = 1;
A(i,i+1) = 1;
A(i,i-Nx) = 1;
A(i,i) = -4;
end
%right:
for n=2:(Ny-1)
%m = Nx
i = (n-1)*Nx + Nx;
A(i,i+Nx) = 1;
A(i,i-1) = 1;
A(i,i-Nx) = 1;
A(i,i) = -4;
C(i) = -Tright;
end
% Corners
%bottom left (i=1):
i=1;

A(i,Nx+i) = 1;
A(i,2) = 1;
A(i,1) = -4;
C(i) = -(Tbottom + Tleft);
%bottom right:
i = Nx;
A(i,i+Nx) = 1;
A(i,i-1) = 1;
A(i,i) = -4;
C(i) = -(Tbottom + Tright);
%top left:
i = (Ny-1)*Nx + 1;
A(i,i+1) = 1;
A(i,i) = -4;
A(i,i-Nx) = 1;
C(i) = -(Ttop + Tleft);
%top right:
i = Nx*Ny;

A(i,i-1) = 1;
A(i,i) = -4;
A(i,i-Nx) = 1;
C(i) = -(Tright + Ttop);
%Solve using Gauss-Seidel
residual = 100;
iterations = 0;
while (residual > 0.0001) % The residual criterion is 0.0001 in this example
% You can test different values
iterations = iterations+1;
%Transfer the previously computed temperatures to an array Told
Told = T;
%Update estimate of the temperature distribution
for n=1:Ny
for m=1:Nx
i = (n-1)*Nx + m;

Told(i) = T(i);
end
end
% iterate through all of the equations
for n=1:Ny
for m=1:Nx
i = (n-1)*Nx + m;
%sum the terms based on updated temperatures
sum1 = 0;
for j=1:i-1
sum1 = sum1 + A(i,j)*T(j);
end
%sum the terms based on temperatures not yet updated
sum2 = 0;
for j=i+1:Nx*Ny
sum2 = sum2 + A(i,j)*Told(j);
end
% update the temperature for the current node
T(i) = (1/A(i,i)) * (C(i) - sum1 - sum2);
end
end

residual = max(T(i) - Told(i));
end
%compute residual
deltaT = abs(T - Told);
residual = max(deltaT);
iterations; % report the number of iterations that were executed
%Now transform T into 2-D network so it can be plotted.
delta_x = 0.03/(Nx+1);
delta_y = 0.03/(Ny+1);
for n=1:Ny
for m=1:Nx
i = (n-1)*Nx + m;
T2d(m,n) = T(i);
x(m) = m*delta_x;
y(n) = n*delta_y;
end
end

T2d;
surf(x,y,T2d)
figure
contour(x,y,T2d)

Hope will help you thankyou.

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