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Green Sal
Green Sal 2011-12-2
关闭： MATLAB Answer Bot 2021-8-20
Hey guys. So now i know the time at which the duct reach steady state condition but i would like to plot the isotherms at only t=0.25*t(ss), t=0.50*t(ss) and t(ss) is the time at which it reaches steady state.
Please Help.
Thank you!
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Green Sal
Green Sal 2011-12-2
Jan, my code is the following and the it is for a square duct with heat steam passing inside of it.
T = zeros(20,20);
% Initial condition
% Set all material to 20C ambient temp
for y = 1:20;
for x = 1:20;
T(y,x) = 20;
end
end
dt = 0.1; %(Time increment, Seconds)
p = 9000; %(Density of Copper, kg/m^3)
Cp = 380; %(Specific heat of Copper, J/kg*K)
qT = 500; %(Solar heat flux, W/m^2)
TL = 50; %(Temperature of the left side of the block, Celsius)
Tsteam = 400; % (Temperature of steam inside the duct, Celsius)
l = 0.01; % 0.005 Meters
K = 400; % (Thermal conductivity of Coppy,W/m*C)
Alpha = K/(p*Cp); % (Thermal diffusivity)
Fo = (Alpha*dt)/(l^2); % Seconds
%Bi=(h*l)/K;
b(1)=det(T);
fprintf('Enter Time (s):\n')
t = input(' ');
A=t/dt; %time/step increment size
%instant steam at center
for y = 6:15;
for x = 6:15;
T(y,x) = Tsteam;
end;
end;
%for whole matrix
for n = 1:A
% BOUNDRY CONDITIONS
% Node 1 - Top left node
T(1,1) = T(1,1)*(1-4*Fo)+2*Fo*(T(1,2) + T(2,1) + (2/K)*TL+ qT*l/K);
% Node 21 - Top right node
T(1,20) = T(1,20)*(1-4*Fo)+2*Fo*(T(1,19) + T(2,20));
% Left nodes - Left side BC
for x = 1;
for y = 2:19;
T(y,x) = TL;
end
end
% Right nodes - Right surface BC (Insulated section (no qT))
for x = 20;
for y = 3:19;
T(y,x) = T(y,x)*(1-4*Fo) + Fo*(T(y-1,x) + T(y+1,x) + 2*T(y,x-1));
end;
end;
% Bottom left Node - insulated with left BC
T(20,1) = T(20,1)*(1-4*Fo) + 2*Fo*(T(19,1) + T(20,2) + (2*TL)/K);
% Bottom right Node - insulated with right BC
T(20,20) = T(20,20)*(1-4*Fo) + 2*Fo*(T(19,20) + T(20,19));
% Top nodes calcs - Top surface (Heat Flux addition)
for y = 1;
for x = 2:19;
T(y,x) = T(y,x)*(1-3*Fo) + Fo*((qT*l/K) + T(y,x-1) + T(y,x+1) + T(y+1,x));
end;
end;
% Calculating the in between nodal temperatures at each delta(time)
for y = 2:19; %inside left slab and right slab
for x = [2:5,16:19]
T(y,x) = T(y,x)*(1-4*Fo) + Fo*(T(y,x-1) + T(y,x+1) + T(y-1,x) + T(y+1,x));
end;
end;
%central nodes
for x = 6:19;%center slab portions above and below duct
for y = [2:5,16:19];
T(y,x) = T(y,x)*(1-4*Fo) + Fo*(T(y,x-1) + T(y,x+1) + T(y-1,x) + T(y+1,x));
end;
end;
% Bottom Nodes - Insulated section (no qT)
for y = 20;
for x = 2:19;
T(y,x) = T(y,x)*(1-4*Fo) + Fo*(T(y,x-1) + T(y,x+1) + 2*T(y-1,x));
end;
end;
%this part below kills the initial for loop if the duct has reached steady
%state, and displays the time in seconds
b(n+1)=det(T);
if b(n)== b(n+1);
s=(n-1)*.1;
fprintf('The Duct reached Steady State at: %4.1f seconds \n',s);
break
end
end;
Tfinal = flipud(T);
% Graphing the temperature isotherm
Spacing = (20:5:400);
[C,h]=contourf(Tfinal,Spacing);
%clabel(C,h)
colorbar;
Green Sal
Green Sal 2011-12-3
does anybody know how to deal with this?

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