Contents

Heat Transfer Home Work

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

The date and time: 03-Feb-2009 10:47:56 

Problem 1.2 Heat loss through a wall

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

% Find:
fprintf('\n  *** Find ***   \n')
fprintf('\nGraph of heat loss vs. outer wall temperature for three thermal conductivities\n')

% Assumptions:
fprintf('\n  *** Assumptions ***   \n')
fprintf('\nSteady state 1-D heat flow')
fprintf('\nLinear tempterature distribution')
fprintf('\nk is constant\n')

% Given:
fprintf('\n  *** Given ***   \n')
A = 20 % m^2. Surface area of concrete wall.
L = 0.3 % m. Thickness of wall.
k1 = 1 % W/(m*K). Thermal conductivity of material 1 (concrete).
T1 = 25 % degrees C. Inner surface temperature of wall.

% (a)
T2 = [-15:.01:38]; % degrees C. Temperate range of outer wall surface.

% (b)
k2 = 0.75 % W/(m*K). Thermal conductivity of material 2.
k3 = 1.25 % W/(m*K). Thermal conductivity of material 3.


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

fprintf('\nHeat Equation:')
fprintf('\nq = (k*(T1-T2)/L)*A\n')
q1 = k1.*(T1 - T2)./L.*A; % W. Heat loss for material 1
q2 = k2.*(T1 - T2)./L.*A; % W. Heat loss for material 1
q3 = k3.*(T1 - T2)./L.*A; % W. Heat loss for material 1


plot(T2, q1, '--', T2, q2, ':', T2, q3), grid, ...
    xlabel('Outside Surface Temperature, (degrees C)'), ...
    ylabel('Heat Loss, (W)'), legend('k = 1', 'k = 0.75', 'k = 1.25'), ...
    title('Heat Loss vs. Outer Wall Temperature for Three Thermal Conductivities')


% Comments:
fprintf('\n  *** Comments ***   \n')
fprintf('\nThe heat loss is greater when k is larger.')
fprintf('\nHeat transfer becomes smaller as the temperature difference becomes smaller.\n')

  *** Problem ***   

Heat loss through a wall

  *** Find ***   

Graph of heat loss vs. outer wall temperature for three thermal conductivities

  *** Assumptions ***   

Steady state 1-D heat flow
Linear tempterature distribution
k is constant

  *** Given ***   

A =

    20


L =

    0.3000


k1 =

     1


T1 =

    25


k2 =

    0.7500


k3 =

    1.2500


  *** Analysis ***   

Heat Equation:
q = (k*(T1-T2)/L)*A

  *** Comments ***   

The heat loss is greater when k is larger.
Heat transfer becomes smaller as the temperature difference becomes smaller.

Published with MATLAB® 7.4