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Thomas Hammel
Conduction Heat Transfer
Homework #7
file = cht_hw07.doc
March 9, 2000
6-6 Solve the following heat conduction problem by using Green’s function approach:
1
1 T
2T
 g  x, t  
k
 t
 x2
T 0
at
T
 HT  0
x
T  F ( x)
in 0  x  L , t  0
x  0, t  0
at
x  L, t  0
for t  0 , 0  r  b
Solution: Consider only the homogeneous version of the problem given by
1 
 2

 t
 x2
in
0  x  L, t  0
 0
at
x  0, t  0

 H  0
x
  F ( x)
at
for t  0 ,
x  L, t  0
in 0  x  L
The solution of this problem is obtained from equation (2-36a) and Case 7 of Table 2-2
from the text, and rearranged in the form


 2m  H 2

2


sin  m x sin  m x  F  x  dx 
 x, t   2    e    m t
2

2
Lm H  H
x  0 m  1


L



 

The solution of the homogeneous problem in terms of Green’s function is given,
according to equation (6-13) of the text, as
 x, t  
L
 Gx, t | x,  |
x  0
0
F x  dx 
By comparing the two solutions, we find the Green’s function for  = 0.
 2m  H 2

2
sin  m x sin  m x 
G x, t | x ,0  2  e    m t
2  2 H
L

m 1
m H



 

and Green’s function Gx, t | x,   is obtained by replacing t by (t-) in the expression
 2m  H 2

2 t   


sin  m x sin  m x 

G x, t | x ,    2  e
m
2
2
L


H
m H
m 1



 
