Contents

Heat Transfer Home Work

% Programmer: Aaron Klapheck
% Chapter 1
% 17-Feb-2009
clear, clc, home
fprintf('The date and time: %s \n', datestr(now))

The date and time: 17-Feb-2009 09:56:34 

Problem 2.8 Heat loss through a plane wall

% Problem
fprintf('\n  *** Problem ***   \n')
fprintf('\nHeat loss through a plane wall\n')

% Find:
fprintf('\n  *** Find ***   \n')
fprintf('\nHeat flux and fill out the table for the unknows.\n')

% Assumptions:
fprintf('\n  *** Assumptions ***   \n')
fprintf('\nSteady state 1-D heat flow')
fprintf('\nNo heat generation')
fprintf('\nConstant properties\n')

% Given:
fprintf('\n  *** Schematics/Given ***   \n')
fprintf('\nfor schematics see page 83 of the text\n')
L = 0.25 % m. Thickness of wall.
k = 50 % W/(m*K). Thermal conductivity.

Matrix = [1, 50, -20, 0, 0; 2, -30, -10, 0, 0; 3, 70, 0, 160, 0; ...
    4, 0, 40, -80, 0; 5, 0, 30, 200, 0];


fprintf('Matrix Given: \n')
fprintf('Case \t T1 \t T2 \t dT/dx \t heat flux\n')
fprintf(' %3.0f \t %3.0f \t %3.0f \t %3.0f \t %3.0f \n', Matrix')
fprintf('\n*Zeros represent unknown values\n \n')


% Analysis:
fprintf('\n  *** Analysis ***   \n')

fprintf('\nHeat Equation:')
fprintf('\nd^2T/dx^2 = 0 ... (eqn 1)\n')
fprintf('\nIntegrate (eqn 1) twice:\n')
fprintf('T(x) = c1*x + c2 ... (eqn 2)\n')

fprintf('\nBoundry conditions:\n')
fprintf('\nBC1. T1 = T(0) = c2\n')
fprintf('c2 = T1\n')

fprintf('\nBC2. T2 = T(L) = c1*L + T1\n')
fprintf('c1 = (T2-T1)/L\n')

fprintf('\nPlug BC1 and BC2 into (eqn 2):\n')
fprintf('T(x) = ((T2-T1)/L)*x + T1\n')

x = 0:.001:0.25;

  *** Problem ***   

Heat loss through a plane wall

  *** Find ***   

Heat flux and fill out the table for the unknows.

  *** Assumptions ***   

Steady state 1-D heat flow
No heat generation
Constant properties

  *** Schematics/Given ***   

for schematics see page 83 of the text

L =

    0.2500


k =

    50

Matrix Given: 
Case 	 T1 	 T2 	 dT/dx 	 heat flux
   1 	  50 	 -20 	   0 	   0 
   2 	 -30 	 -10 	   0 	   0 
   3 	  70 	   0 	 160 	   0 
   4 	   0 	  40 	 -80 	   0 
   5 	   0 	  30 	 200 	   0 

*Zeros represent unknown values
 

  *** Analysis ***   

Heat Equation:
d^2T/dx^2 = 0 ... (eqn 1)

Integrate (eqn 1) twice:
T(x) = c1*x + c2 ... (eqn 2)

Boundry conditions:

BC1. T1 = T(0) = c2
c2 = T1

BC2. T2 = T(L) = c1*L + T1
c1 = (T2-T1)/L

Plug BC1 and BC2 into (eqn 2):
T(x) = ((T2-T1)/L)*x + T1

Case 1

T1 = Matrix(1,2); T2 = Matrix(1,3);
Matrix(1,4) = ((T2-T1)/L);
fprintf('\ndT/dx = (T2-T1)/L\n')
fprintf('dT/dx = %g\n', Matrix(1,4))
fprintf('\nq = -k*dT/dx\n')
q = -k*Matrix(1,4)
Matrix(1,5) = q;
T_case1 = ((T2-T1)/L)*x + T1;
plot(x, T_case1), xlabel('Position x, (m)'), ...
    ylabel('Temperature T, (degrees C)'), ...
    title('Temperature Distribution in a Plane Wall')

dT/dx = (T2-T1)/L
dT/dx = -280

q = -k*dT/dx

q =

       14000

Case 2

T1 = Matrix(2,2); T2 = Matrix(2,3);
Matrix(2,4) = ((T2-T1)/L);
fprintf('\ndT/dx = (T2-T1)/L\n')
fprintf('\ndT/dx = %g\n', Matrix(2,4))
fprintf('\nq = -k*dT/dx\n')
q = -k*Matrix(2,4)
Matrix(2,5) = q;
T_case2 = ((T2-T1)/L)*x + T1;
plot(x, T_case2), xlabel('Position x, (m)'), ...
    ylabel('Temperature T, (degrees C)'), ...
    title('Temperature Distribution in a Plane Wall')

dT/dx = (T2-T1)/L

dT/dx = 80

q = -k*dT/dx

q =

       -4000

Case 3

T1 = Matrix(3,2); dTdx = Matrix(3,4);
Matrix(3,3) = dTdx*L + T1;
T2 = Matrix(3,3);
fprintf('\nT2 = dT/dx*L + T1\n')
fprintf('\nT2 = %g\n', Matrix(3,3))
fprintf('\nq = -k*dT/dx\n')
q = -k*Matrix(3,4)
Matrix(3,5) = q;
T_case3 = ((T2-T1)/L)*x + T1;
plot(x, T_case3), xlabel('Position x, (m)'), ...
    ylabel('Temperature T, (degrees C)'), ...
    title('Temperature Distribution in a Plane Wall')

T2 = dT/dx*L + T1

T2 = 110

q = -k*dT/dx

q =

       -8000

Case 4

T2 = Matrix(4,3); dTdx = Matrix(4,4);
Matrix(4,2) = T2 - dTdx*L;
T1 = Matrix(4,2);
fprintf('\nT1 = T2 - dT/dx*L\n')
fprintf('\nT1 = %g\n', Matrix(4,2))
fprintf('\nq = -k*dT/dx\n')
q = -k*Matrix(4,4)
Matrix(4,5) = q;
T_case4 = ((T2-T1)/L)*x + T1;
plot(x, T_case4), xlabel('Position x, (m)'), ...
    ylabel('Temperature T, (degrees C)'), ...
    title('Temperature Distribution in a Plane Wall')

T1 = T2 - dT/dx*L

T1 = 60

q = -k*dT/dx

q =

        4000

Case 5

T2 = Matrix(5,3); dTdx = Matrix(5,4);
Matrix(5,2) = T2 - dTdx*L;
T1 = Matrix(5,2);
fprintf('\nT1 = T2 - dT/dx*L\n')
fprintf('\nT1 = %g\n', Matrix(5,2))
fprintf('\nq = -k*dT/dx\n')
q = -k*Matrix(5,4)
Matrix(5,5) = q;
T_case5 = ((T2-T1)/L)*x + T1;
plot(x, T_case5), xlabel('Position x, (m)'), ...
    ylabel('Temperature T, (degrees C)'), ...
    title('Temperature Distribution in a Plane Wall')

T1 = T2 - dT/dx*L

T1 = -20

q = -k*dT/dx

q =

      -10000

Table filled out

fprintf('Matrix Given: \n')
fprintf('Case \t T1 \t T2 \t dT/dx \t heat flux\n')
fprintf(' %3.0f \t %3.0f \t %3.0f \t %3.0f \t %g \n', Matrix')

Matrix Given: 
Case 	 T1 	 T2 	 dT/dx 	 heat flux
   1 	  50 	 -20 	 -280 	 14000 
   2 	 -30 	 -10 	  80 	 -4000 
   3 	  70 	 110 	 160 	 -8000 
   4 	  60 	  40 	 -80 	 4000 
   5 	 -20 	  30 	 200 	 -10000 

Comments:

fprintf('\nThe dirrection of heat flux is from hot to cold.')
fprintf('\nLooking at the graphs, the heat flux is dirrected "down hill".')
fprintf('\nThe isotherms for this wall would be parallel to the wall surface.\n')

The dirrection of heat flux is from hot to cold.
Looking at the graphs, the heat flux is dirrected "down hill".
The isotherms for this wall would be parallel to the wall surface.

Published with MATLAB® 7.5