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Solutions for homework Set #6:
Problem #1
:
Solve problem 10.A.1 in BSL
Answer:
See Example 10.61 in BSL for the notation and the derivation of the energy equations in terms
of temperature for individual layers. The final forms of these equations are as follow:
10
0
1
00
0
1
01
21
1
2
1
2
12
32
23
0
0
2
3
0
0
23
Zone 01 (steel):
Zone 12
(magnesia):
Zone 23 (cor
k):
ln
ln
ln
rr
dT
k r
rq
T T
dr
k
dT
dr
k
dT
dr
k
Addition of these above equation gives
03
01
12
23
ln
ln
ln
TT
kkk
Thus, the heat loss over the inner surface of the pipe with an area of
0
2
Lr
0
01
12
23
2
2
ln
ln
ln
LT T
QL
r
q
Or the heat loss per hour per foot of the pipe
0
01
12
23
2
ln
ln
ln
Q
L
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View Full Document The r
i
for this problem are
1
2
3
2.067 / 2 1.0335 in.
1.0335 0.154 1.19 in.
1.19 2
3.19 in.
3.19 2
5.19 in.
o
r
r
r
r
Insertion of numerical values into the above formula gives:
24.58
Btu/hrfoot
o
Q
L
Problem #2
:
olve problem 10.B.1 in BSL
.
Equation setup
ue to the motionless body of fluid, heat conduction is the only mechanism of energy transfer
ccounting for the temperature distribution in the fluid. Obviously, this is a symmetric heat
here the heat is being steadily conducted in the r direction only. Therefore,
consider a control volume of thickness
∆
r over which we make the energy balance.
S
Answer:
a
D
a
transfer problem w
2
2
44
0
rr
r
rq
r
r q
or
22
0
r
know divide by
∆
r and then take the limit as
∆
r goes to zero
We
lim
r
0
r
r
r
0
We then use the definition of the first derivative to get
r
2
0
r
d
dr
Insertion of Fourier’s law of heat conduction with constant thermal conductivity
22
0
0
dd
T
d
d
T
rk
r
dr
dr
dr
dr
or
2
0
T
r
dr
dr
b.
Integration of this equation twice with respect to r gives
2
1
1
C
2
dT
rC
dr
r
TC
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This note was uploaded on 11/18/2010 for the course PGE 322K taught by Professor Dicarlo during the Spring '08 term at University of Texas at Austin.
 Spring '08
 dicarlo

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