
Duct Temperature Distribution
Posted:
Nov 29, 2011 3:22 AM


Hello guys, i am trying to find the temperature distribution in a dukt that is insulated from both sides, constant temperature on one side, and const. heat flux from the top. in the inner duct steam flows at constant temperature. The following is the code that i used but i know something is not working. I am trying to validate my code by changing the values of the heat flux but the temperature distribution seems to stay the same. Also, if i run the code at 10 seconds and 100 seconds, the results are the same. If anybody understand such problems please help. Thank you.
clear clc
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 h=[1000000;0;0;0]; dt = .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 = 0.005; % 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)*(14*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)*(14*Fo)+2*Fo*(T(1,19) + T(2,20)+ qT*l/K); % 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)*(14*Fo) + Fo*(T(y1,x) + T(y+1,x) + 2*T(y,x1)); end; end; % Bottom left Node  insulated with left BC
T(20,1) = T(20,1)*(14*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)*(14*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)*(13*Fo) + Fo*((qT*l/K) + T(y,x1) + 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)*(14*Fo) + Fo*(T(y,x1) + T(y,x+1) + T(y1,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)*(14*Fo) + Fo*(T(y,x1) + T(y,x+1) + T(y1,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)*(14*Fo) + Fo*(T(y,x1) + T(y,x+1) + 2*T(y1,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=(n1)*.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)
[val,ind] = max(Tfinal(:)); [max_x,max_y] = ind2sub(size(Tfinal),ind);

