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PROBLEM 4.14
KNOWN:
Heat generation in a buried spherical container.
FIND:
(a) Outer surface temperature of the container, (b) Representative isotherms and heat flow
lines.
SCHEMATIC:
ASSUMPTIONS:
(1) Steadystate conditions, (2) Soil is a homogeneous medium with constant
properties.
PROPERTIES:
Table A3
, Soil (300K): k = 0.52 W/m
⋅
K.
ANALYSIS:
(a) From an energy balance on the container,
g
qE
=
&
and from the first entry in Table
4.1,
( )
12
2D
q
k
TT.
l
D/4z
π
=
−
−
Hence,
( )
q 1
D/4z
500W
1
2m/40m
T
T
20 C+
92.
7C
W
k
2 D
2
2m
0.52
mK
ππ
−−
=
+
==
⋅
oo
<
(b) The isotherms may be viewed as spherical surfaces whose center moves downward with
increasing radius.
The surface of the soil is an isotherm for which the center is at z =
∞
.
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View Full Document PROBLEM 4.15
KNOWN:
Temperature, diameter and burial depth of an insulated pipe.
FIND:
Heat loss per unit length of pipe.
SCHEMATIC:
ASSUMPTIONS:
(1) Steadystate conditions, (2) Onedimensional conduction through insulation,
twodimensional through soil, (3) Constant properties, (4) Negligible oil convection and pipe wall
conduction resistances.
PROPERTIES:
Table A3
, Soil (300K):
k = 0.52 W/m
⋅
K;
Table A3
, Cellular glass (365K):
k
= 0.069 W/m
⋅
K.
ANALYSIS:
The heat rate can be expressed as
12
tot
TT
q
R
−
=
where the thermal resistance is R
tot
= R
ins
+ R
soil
.
From Eq. 3.28,
( )
( )
21
nD / D
n 0.7m/0.5m
0.776m K/W
R.
2 Lk
2 L
0.069 W/m
KL
ππ
⋅
=
==
×⋅
ll
From Eq. 4.27 and Table 4.1,
( )
( )
( )
1
1
2
soil
soil
soil
cosh
2z/D
cosh
3/0.7
1
0.653
R
m K/W.
SK
2
0.52 W/mK
LL
=
=
=
=
⋅
Hence,
( )
( )
12
00C
W
q
8
4L
1
mK
m
0.776
0.653
LW
−
=
=
×
⋅
+
o
q
q/L
84 W/m.
ʹ
<
COMMENTS:
(1) Contributions of the soil and insulation to the total resistance are approximately
the same.
The heat loss may be reduced by burying the pipe deeper or adding more insulation.
(2) The convection resistance associated with the oil flow through the pipe may be significant, in
which case the foregoing result would overestimate the heat loss.
A calculation of this resistance may
be based on results presented in Chapter 8.
(3) Since z > 3D/2, the shape factor for the soil can also be evaluated from S = 2
π
L/ n
l
(4z/D) of
Table 4.1, and an equivalent result is obtained.
PROBLEM 4.16
KNOWN:
Operating conditions of a buried superconducting cable.
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This note was uploaded on 01/24/2012 for the course MECHE ALL taught by Professor Any during the Spring '11 term at Carnegie Mellon.
 Spring '11
 Any

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