Homework Set #6:
Due date: 10/27/06, by 11:00 am.
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
T
T
dr
k
dT
T
T
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
π
( )
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
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()
03
0
10
21
32
01
12
23
2
ln
ln
ln
TT
Q
rr
L
kkk
π
−
=
++
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
:
Solve problem 10.B.1 in BSL
Answer:
a.
Equation setup
Due to the motionless body of fluid, heat conduction is the only mechanism of energy transfer
accounting for the temperature distribution in the fluid. Obviously, this is a symmetric heat
transfer problem where 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.
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 Spring '08
 Peters
 Thermodynamics, Heat, Heat Transfer, Thermal conductivity

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