CHT-Homework 2
P 2.8, p. 95
Carl Roth
February 3, 2000
1. A region x > 0, y > 0, z > 0 is initially at a uniform temperature T0. For times t > 0,
all boundary conditions are kept at zero temperature. Using the product solution,
obtain an expression for the temperature distribution T(x,y,z) in the medium.
Solution: The heat conduction equation with no generation is given:
 2T  2T  2T 1 T



x 2 y 2 y 2  t
for x > 0, y > 0, z > 0, t > 0
Homogeneous boundary conditions of the first kind are given as:
@ x = 0, T = 0 for t > 0
@ y = 0, T = 0 for t > 0
@ z = 0, T = 0 for t > 0
T = T0 for t = 0
Assume the initial condition is equal to the product: T0 = T0*1*1
Using the product solution method, this 3-Dimensional problem can be split into three
individual 1-Dimensional problems as follows:
1. For the x direction:
 2T1 1 T1

x 2  t
for x > 0, t > 0
@ x = 0, T1 = 0 for t > 0
T1 = T0 for t = 0
2. For the y direction
 2T2 1 T2

y 2  t
for y > 0, t > 0
@ y = 0, T2 = 0 for t > 0
T2 = 1 for t = 0
3. For the z direction
 2T3 1 T3

z 2  t
for z > 0, t > 0
@ z = 0, T3 = 0 for t > 0
T3 = 1 for t = 0
Following one dimensional homogeneous solution of the heat equation, given as:

T(x,t) =

e
 2t
 0
Where
1
N  
X (  , x)

 X (  , x' ) F ( x' )dx' d
x ' 0
X(,x) and
1
are given in Table 2-3
N ( )
the boundary conditions of this problem correspond to case 3 in Table 2-3:
X(,x) = sinx
1
2

N ( ) 
by substitution we have,
T(x,t) =
2



 t
 F ( x' )  e sin x sin x' ddx'
2
x ' 0
 0
the integration with respect to  is evaluated by making use of the following relation:
2sinxsinx’ = cos(x-x’)-cos(x+x’)
and from the Text, which references Dwight [17, #861.20], we have


e
 2t
 0


 ( x  x' ) 2 

cos  ( x  x' )d 
exp 

4t
4t 

e  t cos  ( x  x' )d 
2
 0
 ( x  x' ) 2 

exp 

4t
4t 

Then
2


e
 2t
 0
  ( x  x' ) 2 
 ( x  x' ) 2 
1
  exp  

sin xx' d 
exp  
4t 
4t 
(4t )1 / 2  

and the solution becomes
T(x,t) =
1
(4t )1 / 2
  ( x  x' ) 2 
 ( x  x' ) 2 


 
 dx'
F
(
x
'
)
exp


exp
 

4t 
4t 

x ' 0
 

For the constant initial temperature, F(x) = T0, the equation now becomes:
T(x,t) =


 ( x  x' ) 2 
 ( x  x' ) 2  
T ( x, t )
1



exp

dx
'

exp

x '0   4t dx' dx'

T0
4t 
(4t )1 / 2  x '0


Introducing the following new variables:
 

x  x'
4t
x  x'
4t
, dx '  4t d for the first integral
, dx '  4t d for the second integral
The equation becomes:
 



T ( x, t )
1 
 2
 2

e d   e d 


T0
  x '  x
x

x '
4t
 4t

Since e  is symmetrical about  = 0, the above equation can be written in the
following form:
2
x
T ( x, t )
2

T0

4t
e
 2
d
0
The right hand side of this equation is called the error function of argument
the solution in the x direction is expressed in the form:
 x 
T ( x, t )
 erf 

T0
 4t 
and similarly, the solutions in the y and z directions are:
 y 
T ( y, t )  erf 

 4t 
 z 
T ( z , t )  erf 

 4t 
Assuming a solution in the form:
T(x,y,z,t) = T1(x,t) T2(y,t) T3(z,t)
We combine the above results for the final solution:
 x   y   z 
erf 
erf 

T(x,y,z,t) = T0 erf 
 4t   4t   4t 
x
4t
, and