Help in plot!

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!

[Jan]

What is your question? What are your data? What is your current code? Please post the necessary details to give us a chance to answer.

[Green Sal]

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]

does anybody know how to deal with this?