David Wasserman
MEAE 6630 Conduction Heat Transfer
HW # 6
In class exercise 3:
1)a) The heat conduction equation governing this steady state problem with heat
generation is given by equations 3-187a-c in the text, with the addition of the Insulated
boundary condition at r = 0. The steady state portion of this problem is given by
equations 3-188a-b in the text. To solve this problem we can use a U substitution along
with an integrating factor. If we let U = dT/dr then dU/dr = d2T/dr2. Then the steady
state problem becomes a linear first order ODE like such:
1
1
U  U   g
r
k
This is of the form:
U   P( r )U  Q ( r )
We can solve for U using
1
U
 ( r )Q ( r )dr
 (r) 
P ( r ) dr
 (r)  e 
1
r
g
Q(r )  
k
We can solve for U and then integrate once more with respect to r in order to get T(r) :
1 gr 2
T (r)  
C
4 k
Using the boundary conditions of T = 0 at r = b we find C to be gb2/4k and thus T(r) is
g 2
b  r 2 
T (r) 
4k
b) We can also solve the problem numerically using three control volumes: two
boundary volumes and an internal volume. The finite difference equations for the r =
0 boundary node and the interior node are, respectively:
 2g
T0 
 T1
4k
1 3
1
 2g 

T1     Ti 1  Ti 1 
2 2
2
k 
We do not need an equation for the r = b node because that temperature is held at zero for
all times. Using the values for the k, g, and using a mesh spacing of 0.05 for delta, we
can iterate for the final solution. Using an initial guess of zero temperature for all three
nodes, after 10 iterations the temperatures at the nodes are
T = 9.9902 K for r = 0m
T = 7.4926 K for r = 0.05m
P( r ) 
T = 0 K for r = 0.1m
The analytic solution gives T = 10, 7.5, 0 for r = 0, 0.5, 0.1 respectively. As we can see,
the results agree very well.
c) To agree within 1% of the analytical results we would require temperatures of 9.9 K
and 7.425 K. Since our numerical solution has given even better results than that, our
mesh size is sufficient.
2)a) If we divide the slab into three volumes we get a node on the x = -L surface, a node
at x = 0, and a node at x = L. We know that initially the slab is at 1000 K. For times
thereafter, the nodes at the surfaces are maintained at 0 K. The implicit equation is:
Ti n1  Ti n
T n1  2Ti n1  Ti n11
  i 1
t
( x ) 2
If we let n = 0 we know that the slab is at 1000 K, however the n+1 terms at the i-1 and
i+1 nodes are equal to zero when the exponent is greater than 0. Therefore, they drop out
of the equation, which can be rearranged as:
( x ) 2
Ti
n 1

t
Ti 
 ( x ) 2


 2 
 t

If we use a time step of 1 second, a mesh spacing of 0.1 meter, and the alpha for pure
copper, after 90 seconds we get a center temperature of 132.3961 K. The analytical
solution can be obtained from the Heisler charts or by summing the first few terms of the
series solution. The Heisler charts yield a center temperature of 100 K after 90 seconds.
The more accurate method of summing the first four terms (m=1,2,3,4) of the series
solution yields a center temperature of 105 K. Therefore, the numerical solution for the
implicit method differs from the analytical solution by about 30%.
b)
c) The equation for the explicit method is:
Ti n 1  rTi n1  (1  2r )Ti n  rTi n1
r
t
( x ) 2
Again the slab is initially at 1000 K and the boundary nodes are held at 0 K for times
thereafter. For n > 0 the i-1 and i+1 nodes have temperatures of 0 K and thus drop out of
the equation. If we use a time step of 1 second and a mesh spacing of 0.1 meters, and
initial temperatures of 1000 K all nodes at n = 0, we get a center temperature of 129.3566
K after 90 seconds. Again, this differs from the analytic solution by about 30%.
d) If I use a time step of 0.1 seconds and seven nodes, the temperature at the center of the
plate is 105.9509 K, which agrees with the analytical solution within 1%.