College of Engineering and Computer Science
Mechanical Engineering Department
Mechanical Engineering 375
Heat Transfer
Spring 2007 Number 17629 Instructor: Larry Caretto
Exercise Two Solutions
1. The inner and outer surfaces a spherical shell with radii of 1 cm and 2 cm are 300°C and
100°C, respectively. If the thermal conductivity of the shell is 2 W/m·K, determine the heat
transfer through the shell.
The answer is found by plugging the data into the equation for heat transfer through a spherical
   4k T2  T1   
shell: Q
1 r1  1 r2
4


2W
100 o C  300 o C
mK
 100.5 W
1
1

0.01 m 0.02 m
2. The inner and outer surfaces of a cylindrical shell with radii of 1 cm and 2 cm are 300oC
and 100oC, respectively. If the thermal conductivity of the shell is 2 W/m·K, determine the
heat transfer through the shell per unit length of the cylinder.
The answer is found by plugging the data into the equation for heat transfer through a cylindrical


2W
2
100 o C  300 o C

2k T2  T1 
Q
W
m

K
shell:


 3626
L
ln r2 r1 
m
 0.02 m 

ln 
 0.01 m 
3. What would your answers to problems 1 and 2 be if the thermal conductivity were given by
the equation k = a – b/T where a = 2 W/m·K and b = 100 W/m. Note that you must use T in
kelvins when computing the thermal conductivity.
T2
The average thermal conductivity is given by the equation, k avg
1
k 
 kdT . For the
T2  T1

T1
T2
1
b

equation k = a – b/T, the average thermal conductivity is k 
 a  dT
T2  T1 
T
T

1

 
1
aT  b ln T TT12  a  b ln  T2  . Plugging in the given values for a, b, T1, and
T2  T1
T2  T1  T1 
100 W
 T2  2 W
 573.15 K  1.785 W
b
m

T2, gives k  a 
ln   

ln 
T2  T1  T1  m  K 573.15 K  373.15 K  373.15 K 
m K
This average thermal conductivity is 0.8927 times the value of 2 W/m·K used in problems 1 and
2. Thus the value of the heat transfer will decrease by 0.8927 times giving an answer of 89.72 W
for problem 1 and an answer of 3237 W/m for problem 2.
Jacaranda (Engineering) 3333
E-mail: lcaretto@csun.edu
Mail Code
8348
Phone: 818.677.6448
Fax: 818.677.7062
Exercise Two Solutions
ME 375, L. S. Caretto, Spring 2007
Page 2
4. Consider a homogeneous spherical piece of radioactive material of radius,R = 0.04 m that
is generating heat at a constant rate of e gen = 4 x 107 W/m3. The heat generated is
dissipated to the environment steadily. The outer surface of the sphere is maintained at a
uniform temperature of 80°C and the thermal conductivity of the sphere is k = 15 W/m·°C.
Find: (a) the temperature at the center of the sphere, (b) the heat flux at the outer surface
of the sphere.
(a) The temperature is found by plugging the data into the equation for heat generation in a solid
sphere and setting r = 0 to get the temperature at the center of the sphere:
T  Touter surface 

e gen R  r
2
2
6k
  80 C 
o
4 x10 7 W
m
3
6
0.04 m  0
2
15 W
 791o C
mo C
Applying the given data to the equation for the heat flux in a solid sphere with heat conduction
and setting r = 0.04 m to get the heat flux at the outer surface gives.
q 
re gen
3

5.33x10 5 W
1 4 x10 7 W


0
.
04
m

3 m3
m2
Note that the formula for the heat flux at the outer surface satisfies a simple energy balance. The
heat flux leaving the surface equals the total heat generated divided by the outer surface area.
5. What would your answer to problem 4 be if you did not know the outer-surface
temperature of the sphere, but you were told that there was a convection between the
outer surface of the sphere and a coolant that was at 700F with a heat transfer coefficient
of 5000 W/m2·oC?
From problem 4 we know that the outer surface heat flux is 5.33x105 W/m2. This will be the same
if the heat generation and radius are the same. In order for this heat to be convected to a fluid at
70oC with a heat transfer coefficient of 80 W/m 2·oC the surface temperature of the sphere given
by the equation for convection heat transfer. (Convert 70oF to 21.1oC.)
q  hTs  T 
5.33 x10 5 W
q
m2
 Ts  T   21.1o C 
 127.8 o C
5000 W
h
m 2 o C
With this known surface temperature, we find the temperature at the center of the sphere as we
did in problem 4. (The heat flux at the outer radius remains at a value of 5.33x105 W/m2.)
T  Touter surface 

e gen R  r
2
6k
2



4 x10 7 W
0.04 m 2  0
3
o
m
 127.8 C 
 769 o C
15 W
6 o
m C