CHT - In Class Exercise 2
Harold Haugeto
2/14/00
0c
T0 = 1000 c
0c
0.1 m
A 0.2 m diameter cast copper cylindrical rod, which is imuch longer than its
diameter, isi initially at 1000 C and its surface is maintained at 0 c
a) Write down the solution to this problem in terms of linear coordinates of
characteristic functions
On Pg. 118, equation (3-65b)
2 HTo
T (r , t ) 
b

 e 

2
mt
m 1
J o ( m r )
( m2
 H 2 ) J o ( m b)
b) Give the value of the first few eigenvalues
From Table 3-1, pg. 108, No. 3
J v (  mb)  0
Therefore the eigenvalues are 2.3, 5.2, 8.5
c) What would be the temperature after 1.5 minutes?
From Fig 3-9, pg. 148
where  = 11.234 x 10-5

t
b
2

11234
.
* 10 5 * 90
01
.2
 1011
.
1
0
Bi
  0.004  0.004 * 1000  4c
where
d) What would be the temperature 0.04 m from the surface at the same time?
From Fig 3-9, pg 149
x 0.06

 0.6;   0.58
L
01
.
temp  0.58 * 4c  2.32c
e) If the slab were a brick, How long would it take for it’s center to reach the
temperature of the center of the copper slab reaches 90 seconds?
Brick  = 5.2 x 10-7
1011
.

5.2 * 10 7 * t
01
.2
t  19230.8 sec onds
t  5hours,20 min,31 sec onds
f) What would be its temperature 0.04 m from the surface after that time?
x 0.06

 0.6;q  0.58
L
01
.
temp  0.58*40c  2.32c
Consider now instead a short copper cylinder 0.2 m in diameter and 0.2 m in height
initially at 1000 C and whose all surfaces are then maintained at 0 C
g) Write down the solution to this problem in terms of a linear combination of
coordinate functions
From Pg. 130, equation 3-21

T (r , z, t ) 


m1 n  0
e  ( m  n ) t
R (  , r ) Z (n, z)dn
N ( m ) N (n) o m
2
2
b

 
r ' Ro ( m , r ' ) Z (n, z' ) F (r ' , z' )dz' dr '
r ' 0 z ' 0
h) Give the value of the first few eigenvalues
where H1 = h1/k1 = 
From Table 2-3, pg 48
 m cos  m 01
.   H1
J v (  m , b)  0
Therefore the eigenvalues are , 2, 3, 2.3, 5.2, 8.5
i) Perform calculation similar to (c) to (f) above
1. What would be the temperature after 1.5 minutes?
From Fig 2-13, pg 92
where  = 11.234 x 10-5

t
2

11234
.
* 10 5 * 90
L
01
.2
where
1
 0,   01
.
Bi
 1011
.
From Part c) above,  = 0.004
1000*0.004*0.1 = .4 C
From Part d) above,
.4*0.58 = .232 C
From Part e) above,
t = 5 hours, 20 minutes
j) What would its temperature be 0.04 m from the outer cylinder surface after that time?
= 2.32 C
k) What would its temperature be 0.04 m from the flat surface along the axis after that
time?
= 2.4 C
l) For Brick:
The same temperature would be realized after the equivalent
amount of time (i.e. 5 hours, 20 minutes instead of the 90 seconds)
CHT - Homework # 3-1
Harold Haugeto
2/14/00
Obtain an expression for the temperature distribution T(r,,t) in a solid sphere of radius r =b that initially at
temperature F(r,) and for times t>0 the boundary surface ar r = b is kept insulated.
Using the transformed heat conduction equation (4-5) on Pg. 155. This equation is simplified due to the
lack of dependence on the azimuth angle .
 2V 1 V 1 V
1  
V  1 V

In 0<r<b, -1<<1, t>0


 2
1  2   

2
2
r r 4 r
   r
r
r  
Subject to
V 1
 V  0 At r = b, t>0
r b
1
V  r 2 F (r ,  ) For T=0, in the sphere
The separation of the variables can be found on Pg. 156 in equations (4-10a), (4-10b), (4-10c):
The elementary solutions of these equations are:
(t ) : e  t
2
R(r ) : J n 1
2 (r ) andYn 1 2
M (  ) : Pn (  )andQn (  )
The solutions Qn() become infinite at  = + or – 1 and Yn+1/2(r) become infinite at r = 0
From The previous Chapter, identified on Table 3-1 on Pg. 108, the eigenvalues for the r direction are
J ' n1 2 (r )
Due to the boundary condition being insulated at r = b, an additional term is required to account for the
mean initial temperature over the volume of the sphere. Therefore, Equation (4-88) on Pg 176 is modified
to include this additional term.
The complete solution is:
T (r ,  , t ) 
2


3 b 2
1
  np t  1
e
r

 r F ( r ) dr 
3
n  0 p  1 N ( n) N ( np )
b 0
2
J'
n 1 2
(
np
b
1
3 2
r )P ( ) 
r'
( r ' ) P (  ' ) F ( r ' ,  ' ) d ' dr'

n
np
n
r '  0  '  1