CHT - Homework 5
p. 4.9, pg. 192
Carl Roth
February 24, 2000
A solid sphere of radius r = b is initially at temperature F(r). For times t > 0 the boundary
at r = b is kept insulated. Develop an expression for the temperature distribution T(r,t) in
the sphere for times t > 0.
The solution to this problem is similar to example 4-1. The governing equation is:
1 2
1 T
(rT ) 
r r
 t
in 0  r < b
t>0
subject to the following boundary conditions:
T
 0 (H = 0),
r
@r=b
T = F(r)
for t = 0 in 0  r  b
t>0
Noting that the adiabatic boundary condition is a special case, as covered in Note 1
Using the transformation: U(r,t) = rT(r,t) and applying Note 1 (H = 0 @ r = b; insulated
sphere):
 2 U 1 U

r 2  t
in 0 < r < b, t > 0
U = 0 @ r = 0, t > 0
U 1
 U 0
r b
@ r = b, t > 0
U = rF(r)
for t = 0, in 0  r  b
Following note 1 eigenvalue problem, and solving the system yields:
3
T (r , t )  3
b
1  e  mt
r
F
(
r
)
dr

R(  m , r )  (  m ,r ' )r ' F (r ' )dr '

0
r m1 N (  m )
0
2
b
2
b
CHT - Homework 5
p. 12.20, pg. 501
Carl Roth
February 24, 2000
A 10 cm diameter solid steel bar of thermal conductivity k = 40 W/(mC) is heated
electrically by the passage of electric current that generates energy within the rod at a rate
of g = 4 x 106 W/m3. Heat is dissipated from the surface of the rod by convection with a
heat transfer coefficient h = 400 W/(m2C) into an ambient at temperature T = 20 C.
By dividing the radius into five equal parts, develop the finite-difference equations for
this heat conduction problem. Compare the finite-difference solution with the exact
analytical solution for the cases when the first order and second order accurate
differencing are used for the convection boundary condition.
This problem is identical to example 12-3 on page 463. The solution is as follows:
The problem involves six unknown node temperatures, Ti, i = 0, 1,..., 5, since the region o
 r  b is divided into five equal parts. The six finite-difference equations needed for
their determination are obtained as:
4(T1 - T0) + 10 = 0 , i = 0
1
1


1   Ti 1  2Ti  1   Ti 1  10  0 , i = 1,2,3,4
 2i 
 2i 
1st order accurate formula applied to boundary condition at i=M=5 yields: T5 
1
(T4  2)
11
.
2nd order accurate formula yields: T4  111
. T5  7.2  0 i = 5
The exact solution of this problem is given by
T (r )  T 
2
gb gb 2   r  

1    
2h 4 k   b  
or
  r2
T (r )  20  250  62.51    
  b  
The table below shows the comparison of finite-difference solutions with the exact results
for the cases when the first and second order accurate formulas are used for the
convective boundary condition. Gauss elimination method is used to solve the resulting
algebraic equations. For greater accuracy, subdivisions are increased from 5 to 10.
r/b
Exact
0.0
0.2
0.4
0.6
0.8
1.0
332.50
330.00
322.50
310.00
292.50
270.00
M=5
1st order
accurate
307.50
305.00
297.50
285.00
267.50
245.00
2nd order
accurate
332.50
330.00
322.50
310.00
292.50
270.00
M = 10
1st order
accurate
320.00
317.50
310.00
297.00
280.00
257.50