# Document 15637596 ```HW#8:
March 16, 2000
Dianbo Li
5-1:
A slab, 0  x  L is initially at zero temperature. For times t &gt; 0, the boundary
surface at x = 0 is subjected to a time-varying temperature f(t) = b +ct, while the
boundary surface at x = L is kept at zero temperature. Using Duhamel’s theorem,
develop an expression for the temperature distribution T(x, t) in the slab for times t
&gt; 0.
Solution: The mathematical formulations are:
 2T ( x, t ) 1 T ( x, t )
in 0 &lt; x &lt; L, t &gt; 0;

 t
x 2
T ( x,t ) = f(t) = b+ ct
at x = 0, t &gt; 0;
T=0
at x = L, t &gt; 0;
T=0
at t = 0, in 0  x  L
The corresponding auxiliary problems becomes:
 2 ( x, t ) 1 ( x, t )
in 0 &lt; x &lt; L, t &gt; 0;

 t
x 2
( x,t ) = 1
 (x,t) = 0
(x,t) = 0
at x = 0, t &gt; 0;
at x = L, t &gt; 0;
at t = 0, in 0  x  L
m

x 2
 m2 ( t  )  1
the solution is ( x, t   )    e
sin( mx / L)
L L m1
m / L
here I use t -  replace , and m = m/L
t
df ( )
Since T ( x, t )   ( x, t   )
and df()/d = c
d
d
 0
so
t
 x 2   m2 ( t  )  1m

T ( x, t )      e
sin( mx / L) cd
L m 1
m / L
 0  L

final solution is :
2
 1 sin( mx / L)
x
2 
T ( x, t )  c t  c  (1  e  m (t  ) )
L
L m 1
(m / L) 3
m
7-4:A slab, , 0  x  L, is initially at uniform temperature To. For time t &gt; 0, the
boundary surface at x = 0 is kept insulated and the boundary surface at x = L is
kept at zero temperature. Obtain an expression for the temperature distribution
T(x,t) in the slab valid for very small times.
Solution: the mathematical formulation of this problem is given as:
 2T ( x, t ) 1 T ( x, t )
in 0 &lt; x
t &gt; 0;

 t
x 2
T
0
at x = 0, t &gt; 0;
x
T=0
at x = L, t &gt; 0;
T = To
at t = 0, in 0  x  L
The Laplace transform of these equations are
_
T
d 2 T ( x, s ) s _
 T ( x, s )   o
2


dx
in 0  x  L
_
dT
0
dx
at x = 0;
_
T 0
at x = L
_
T ( x, s ) 1 cosh( x s /  )
 
To
s s cosh( L s /  )
simplify the equation and use binomial method (P278)
so the solution is:
_

T ( x, s ) 1
 (1   (1) n e [ L (1 2 n ) x ]
To
s
n 0
s /

  (1) n e [ L (1 2 n ) x ]
s /
)
n 0
From Table 7-1, inverting term by term:

 L(1  2n)  x  
 L(1  2n)  x 
T ( x, t )
 1   (1) n erfc
   (1) n erfc

T0
4t
4t
n 0

 n 0


9-2. A semiinfinite medium x &gt;0 is intially at a uniform temperature Ti. For times t &gt;
0 , the boundary surface at x=0 is subjected to a prescribed heat flux, that is k(T/x) = f(t) at x = 0, where f(t) varies with time. Obtain an expression for the
temperature distribution T(x,t) in the medium using the integral method and a
cubic polynomial representation for T(x,t).
Solution: the mathematical formulation of this problem is given as:
 2T 1 T

x 2  t
T

 f (t )
x
T = Fi
Therefore,
in 0 &lt; x
t &gt; 0;
at x &gt; 0 , t &gt; 0;
at t = 0, in x  0
T
x
x 
T

x
x 0


1 d 

    Tdx   T
  dt  0

x 

d 

dt 
In view of the conditions:
T
T

T x  Ti
x   0
x  0  f (t )
x
x
then


d 
f (t )      Tdx  Ti  
------------------ ( 1 )
dt  x 0

Now apply a cubic polynomial representation for T(x,t),
 2T
x 2
x 
T(x,t) = a1 + a2x + a3x^2 + a4x^3
T ( x, t )  Ti
x
 (1  ) 3
T (t )  Ti

it turns out that
2
4
d  T  Ti  
f (t )  

3
dt  f (t ) 
Using equation ( 1 ) &amp; ( 2 ):
or
 f ' (t )T  Ti 
4

3
f (t ) 3
2
----------------------- ( 2 )
f or t &gt; 0,
0
```