Jennifer Chen
Conduction Heat Transfer
03-02-2000
HW VI – Class Exercise 3
1. Consider a long cylindrical metallic rod of radius b = 0.1 m and thermal conductivity
k = 100 W/m K through which current is passed. Energy is generated internally at a rate
g = 400,000 W/m3. Its surface is maintained at T = 0.
a) Show that the steady state temperature inside the rod is given by
g
T ( r ) =  ( b2 – r2 )
4k
2T
1 T
1
1 T
 +   +  g(r, t) =  
r2
r r
k
 t
in
0  r  0.1 m,
t>0
T = 0 oC
at
r = 0.1 m,
t>0
T
 =0
r
at
r = 0,
t>0
To determine the appropriate Green’s function, we consider the homogeneous version of this problem as
2
1 
1 
 +   =  
r2
r r
 t
in
0  r  0.1 m,
t>0
 = 0 oC
at
r = 0.1 m,
t>0

 =0
r
at
r = 0,
t>0
For zero surface temperature, and heat generation of a constant rate of g o W/m3
2go 
Jo (m r)
2go

Jo (m r)
T (r, t) = 

 - 

 e^[- 2m t ]
kb m = 1 3m J1 (m b)
kb m = 1 3m J1 (m b)
For t  , the second term on the right-hand side vanishes and the first term must be equal to the steadystate temperature distribution in the cylinder:
2go

Jo (m r)
T (r, ) = 

 
kb m = 1 3m J1 (m b)
go ( b2 – r2 )

4k
b) Solve the problem numerically using a control volume method by subdividing the
radius of the rod into three control volumes (two "surface" volumes and one "interior"
volume; 3 nodes) and compare your results against the analytical solution.
For i = 0,
T1 – T0 1
- k  +  x go = 0
x
2
For i = 1, 2, . . . , m-1
Ti+1– Ti
Ti – Ti-1
k  - k  + x go = 0
x
x
For i = m,
Tm – Tm-1
1
k  +  x go = 0
x
2
For this problem, the temperature at the surface of the cylinder is 0 oC.
x = 0.05 m
k = 100 W/m K
g = 400,000 W/m3
Since the problem asks for the radius of the rod to be divided into 3 control volumes, i.e. the diameter of
the rod is divided into 5 control volumes, 5 nodes, because the middle node is a duplicate.
Let 0 be the node at the left surface of the cylinder, 4 be the node at the right surface of the cylinder.
Then, node 1 is 0.05m to the right of node 0, node 2 is the center of the cylinder, and node 3 is positioned
0.05m to the right of node 2.
Thus, solving the temperature equations above, I get
T0 = 0; T1 = 5; T2 = 10; T3 = 5; T4 = 0
Now using the steady state equation from part a), I get
For the center of the cylinder
T (0) = 10
For the interior volume
T (0.05) = 7.5
For the surface of the cylinder
T (0.1) = 0
Temperature at the surface and the center of the cylinder matches perfectly. But at the interior volume, the
temperature values are different by 33.3%.
c) Determine the mesh spacing required for the numerical solution to agree with the
analytical results within 1%
By trial and error, the mesh spacing, x = 0.067 m
2. Consider the quenching of a 0.2m thick copper slab initially at 1000K with its surface
maintained at T = 0 K.
a) Solve the problem numerically using a control volume method and an implicit scheme.
Subdivide the thickness of the slab into three control volumes (two "surface" volumes
and one "interior" volume; 3 nodes) and find the temperature at the center of the slab
after 90 seconds. Compare your results against the analytical solution.
The analytical solution is 100 K, as shown in previous homework problems.
The numerical method yields a solution of 110 K.
b) Determine the mesh spacing required for the numerical solution to agree with the
analytical results within 1%.
By trial and error, the mesh spacing, x = 0.01 m
c) Solve the problem numerically using a control volume method and an explicit scheme.
Subdivide the thickness of the slab into three control volumes (two "surface" volumes
and one "interior" volume; 3 nodes) and find the temperature at the center of the slab
after 90 seconds. Compare your results against the analytical solution.
The analytical solution is 100 K, as shown in previous homework problems.
The numerical method yields a solution of 110 K.
d) Determine the mesh spacing required for the numerical solution to agree with the
analytical results within 1%.
By trial and error, the mesh spacing, x = 0.01 m