CH E 314 – Heat Transfer ASSIGNMENT 3 – Sample Solution
PROBLEM 2
A spherical research probe consists of two concentric hollow spherical layers, which surround an
instrument package. The inner most shell has an inside radius of 30 cm and is made from mild steel
with a k = 60.5 W/(mK). The outer shell is stainless steel with a k = 14.9 W/(mK). Each layer is 3
cm thick. There is a thermal contact resistance of 0.00022 (m2K)/W between the two shells. The
probe is surrounded by water at 4 °C. The convection heat transfer coefficient between the external
surface and the water is 1000 W/(m2K). The inside shell wall temperature is 50 °C. Calculate the
power output of the instrument package under these conditions.
Solution
Two concentric solid layers surround a heat source.
k1
Ti  50C
T  4C
k2
h  1000
W
m2 K
r1
r2
r3
Each layer has a uniform thickness. Therefore:
r2  33 cm
r1  30 cm
r3  36 cm
There is a thermal contact resistance at r2 . The interfacial area at
r2 is equal to:
A2  4r22  4  0.3 m   1.368 m2
2
ASSUMPTIONS:
- ONE
DIMENSIONAL
- STEADY STATE
- CONSTANT
PROPERTIES
The overall rate of heat transfer can be written in terms of
resistances in series:
q 
Ti  T
1 1
1  R
1 1
1
1
   tc 
 


4k1  r1 r2  A2
4k2  r2 r3  hA3
The outside surface area for heat transfer, A3 , is:
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2
A3  4r32  4  0.36 m   1.629 m2
2
Substitute all of the numbers.
50C  40C
q 
1
1  2.2 104
1
1 
1
 1
 1







4  60.5  0.30 0.33 
1.368
4 14.9   0.33 0.36  1000 1.629 
50C  40C
=q 
0.0252 K / W
Solving for q we obtain:
q  18239W 18.24 kW .
Problem 3
Consider a plane wall in which heat is generated at a uniform rate. The steady state temperature
profile is given by the following equation:
T  a  b x2
Where a = 200C and b = -2000C/m2. The thermal conductivity is 50 W/mK. The thickness of
the wall is 6 cm. Calculate the rate of heat generation in the wall and the heat flux at each face.
The wall coordinate system was defined as x = 0 on one face and x = L on the other face.
Solution
This problem deals with a plane wall in which there is a uniform heat generation. The situation is
illustrated below.
mm
650cm
k  50
x0
W
mK
ASSUMPTIONS:
1. ONE
DIMENSIONAL
2. STEADY
STATE
3. CONSTANT
PROPERTIES
xL
The temperature distribution is given by:
T  a  bx 2  200C - 2000 x 2
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3
with x in metres and temperature in Celcius. This solution must satisfy the energy balance
equation:
d 2T q
  0
k
dx 2
Taking the first and second derivatives of the temperature distribution gives:
d 2T
  4000
dx 2
dT
 2 b x   4000 x
dx
Substitute the second derivative into the energy balance:
4000
Solve for q  2  105
°C
q

 0
2
W
m
50
mK
W
.
m3
The heat flux at each face can be obtained from Fourier's law.
At x  0 :
q x0  k
dT
dx
 2b x  0
x 0
There is no heat transfer at x  0 .
At x  L :
q x L  k
dT
dx
xL


W  
C 
W

3
   50
 2  2000 2  60 10 m  12, 000 2
m K 
m 
m

NOTE: All of the thermal energy generated leaves through the face at x  L .
PROBLEM 4
A 6 cm diameter radioactive sphere at steady state generates a total of
40 W of thermal energy. The sphere is covered by a layer of
insulation 27.0 cm thick so that the total outside diameter is 60
cm The thermal conductivity of the radioactive material is 4.0
W/(mK) . The thermal conductivity of the insulation is 0.5
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ASSUMPTIONS:
- ONE
DIMENSIONAL
- STEADY STATE
- CONSTANT
PROPERTIES
- NEGLIGIBLE
RADIATION
4
W/(mK). The insulated sphere is surrounded by a gas at 25 °C with a convective heat transfer
coefficient of 20 W/(m2K). Compute the temperatures at the following points:
(a) Outside insulation surface.
(b) Interface between radioactive material and insulation.
(c) Centre of radioactive sphere
The volume of a sphere is V 
4 3
 r and the surface area of a sphere is A  4 r 2 .
3
Solution:
(a) All of the energy generated from the inside must be transported by convection from the surface.
q  h A TS  T 
The generated heat is 40 W and therefore:
q  h A TS  T   h 4  r 2 TS  T 
TS 
TS 
q
h 4  r2
 T
40W
(20W / m K ) 4  (30 103 m) 2
2
 25
TS  26.8C .
Solve for the unknown temperature of
(b) The temperature at the interface is computed using the conduction equation. For conduction
across a hollow sphere, the heat transfer rate is given by:
q 
TI  TS
TI  26.768
 40 
1 1
1
1
1
1


 





2

2
4  k  ri ro 
4  (0.5)  3.0 10
30 10 
TI  217.8C .
Solve for the unknown temperature of
(c) The temperature at the centre of the sphere is given by the solution of the conduction
equation with constant generation rate. The centreline temperature is:
T (0) 
q ro2
 TI
6k
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5
The heat generation rate is the total power generated divided by the volume:
q 
q
q
40W


 353677.65W / m3
4
4
V
 ro3
(0.03m)2
3
3
Substitute:
qro2
(353677.65W / m3 )(0.03m)2
T (0) 
 TI 
 217.8  231.1C
6k
6(4.0W / mK )
NOTE THAT
K
C so you may interchange the units in such calculations without
conversion
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