David Wasserman
MEAE 6630 Conduction Heat Transfer
HW # 4
In class exercise 2:
a) This is a 1-D transient heat conduction problem with an infinite ability to extract heat
at the surface. The generic solution obtained by separation of variables is
2
2T 
J 0 (  m r)
T ( r, t )  0  e mt
b m1
 m J 1 (  m b)
b) From Table 3-1, case 3 we get that the eigenvalues are the positive roots of
J  (  m b)  0
and by taking  = 0, b = 0.1, and using Fig 3-1 we get the first few eigenvalues to be:
x
 m  , x  2.5,5.5,8.5,...
b
c) From the Heisler charts (Fig 3-9, p148) with 1/Bi = 0,  = 11.234e-5*90/.1^2, we get
the dimensionless temperature to be 0.004 and thus the center temperature is 4
Celsius.
d) From the second Heisler chart (Fig 3-9, p149) with r/b = 0.6, we get the
dimensionless temperature to be 0.55 and thus the temperature at 0.4 from the surface
is 2.2 Celsius.
e) If the slab were made out of brick we start with the center temp of 4 Celsius, calculate
the dimensionless temp to be 0.004, use Fig 3-9, p148 to obtain the dimensionless
time of 1, and calculate t using  = 5.2e-7 and b = 0.1. This gives t = 5.34 hours.
f) The temperature there would be 2.2 Celsius
g) This is a 2-D transient problem with infinite heat extraction ability The solution is:


T ( r, z, t )   cmp R0 (  m , r ) Z ( p , z )e
 (  m2 2p ) t
m1 p 1
This solution applies for both copper and brick.
h) The eigenvalues in the radial direction will be the same as before. The eigenvalues
for the z direction can be obtained from Table 2-2, case 6 assuming the midplane of
the cylinder in the z direction corresponds to z = 0. The first eigenvalue is:

1 
.1
This applies for both copper and brick.
i) Using the concept of product solutions and the Cylindrical coordinates and Cartesian
coordinates Heisler Charts, the temperature at the center of the cylinder for copper
after 1.5 minutes is  = 0.004*0.1 = 0.4 Celsius. The temperature for the brick
cylinder at the center after 1.5 minutes is approximately 1000 Celsius.
j) Using Heisler chart in Fig 3-9, p149 with  = 11.234e-5, 1/Bi = 0, b = 0.1, the
temperature at the center of the copper cylinder at 0.04 m from the cylindrical surface
along the z midplane after 90 seconds is 0.22 Celsius. For brick we would repeat the
same calculation but with  = 5.2 e-7 and we get a temperature of 550 Celsius.
k) Using Heisler chart in Fig 2-13, p 93 with  = 11.234e-5, 1/Bi = 0, L = 0.1, the
temperature along the axis of the copper cylinder at 0.04 m from the flat surface after
90 seconds is 0.24 Celsius. For brick we would repeat the same calculation but with 
= 5.2 e-7 and we get a temperature of 600 Celsius.
4.5)
We have a solid sphere of radius r = b initially at a specified temperature
distribution whose surface is kept insulated for t > 0. The problem formulation is:
 2T 2 T 1  
T  1 T

 2
(1   2 )

, 0  r  b, t  0
2

r
r r r  
   t
T
0
@ r  b, t  0
r
T  F ( r,  )
t0
inside
If we introduce a new dependent variable V(r,,t) as defined in equation 4-82 of the
textbook, we can transform the problem formulation into that of equation 4-83a,c.
Equation 4-83b will be different due to the insulated boundary condition and is given by
equation 4-101b. The elementary solutions of equation 4-83a are given by equations 411. However, the Legendre functions of the second kind of degree n are infinite at  =
+/- 1 and the Bessel functions of the second kind of order n + 1/2 are infinite at r = 0.
Therefore, they are inadmissible as solutions. The final solution can be constructed as:


V ( r,  , t )   cnp e
n 0 p 1
2npt
J n1/ 2 (np r ) Pn (  )
The coefficient cnp can be determined from equation 4-87a, while the eigenfunctions,
norm, and eigenvalues associated with the r variable can be obtained from Table 3-1, case
2 with  = n + 1/2.