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ch03


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W/m) + 30
W/m2\u22c5K (0.137 m)(400 K) = (2380 + 1644) W/m = 4024 W/m.
PROBLEM 3.151
KNOWN: Internal and external convection conditions for an internally finned tube. Fin/tube
dimensions and contact resistance.
FIND: Heat rate per unit tube length and corresponding effects of the contact resistance, number of fins,
and fin/tube material.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional heat transfer, (3) Constant
properties, (4) Negligible radiation, (5) Uniform convection coefficient on finned surfaces, (6) Tube wall
may be unfolded and approximated as a plane surface with N straight rectangular fins.
PROPERTIES: Copper: k = 400 W/m\u22c5K; St.St.: k = 20 W/m\u22c5K.
ANALYSIS: The heat rate per unit length may be expressed as
g w
t,o(c) cond conv,o
T T
q
R R R
\u2212
\u2032 =
\u2032 \u2032 \u2032+ +
where
( )t,o(c) o(c) g tR h A\u3b7 \u2032= , f fo(c)
t 1
NA
1 1
A C
\u3b7\u3b7
\u2032
= \u2212 \u2212
\u2032
\uf8eb \uf8f6\uf8ec \uf8f7\uf8ed \uf8f8 , ( )1 f g f t,c c,bC 1 h A R A\u3b7 \u2032 \u2032\u2032 \u2032= + ,
( )t f 1A NA 2 r Nt\u3c0\u2032 \u2032= + \u2212 , f 1A 2r\u2032 = , ( )1/ 2f 1 1 gtanh mr mr , m 2h kt\u3b7 = = c,bA t\u2032 = ,
( )2 1
cond
ln r r
R
2 k\u3c0
\u2032 = , and ( ) 1conv,o 2 wR 2 r h\u3c0 \u2212\u2032 = .
Using the IHT Performance Calculation, Extended Surface Model for the Straight Fin Array, the
following results were obtained. For the base case, q\u2032 = 3857 W/m, where t,o(c)R\u2032 = 0.101 m\u22c5K/W,
condR\u2032 = 7.25 × 10
-5
 m\u22c5K/W and conv,oR\u2032 = 0.00265 m\u22c5K/W. If the contact resistance is eliminated
( t,cR\u2032\u2032 = 0), „q = 3922 W/m, where t,oR\u2032 = 0.0993 m\u22c5K/W. If the number of fins is increased to N = 8,
q\u2032 = 5799 W/m, with t,o(c)R\u2032 = 0.063 m\u22c5K/W. If the material is changed to stainless steel, q\u2032 = 3591
W/m, with t,o(c)R\u2032 = 0.107 m\u22c5K/W and condR\u2032 = 0.00145 m\u22c5K/W.
COMMENTS:
 The small reduction in „q associated with use of stainless steel is perhaps surprising, in
view of the large reduction in k. However, because gh is small, the reduction in k does not significantly
reduce the fin efficiency ( f\u3b7 changes from 0.994 to 0.891). Hence, the heat rate remains large. The
influence of k would become more pronounced with increasing gh .
PROBLEM 3.152
KNOWN: Design and operating conditions of a tubular, air/water heater.
FIND: (a) Expressions for heat rate per unit length at inner and outer surfaces, (b) Expressions for inner
and outer surface temperatures, (c) Surface heat rates and temperatures as a function of volumetric
heating q\ufffd for prescribed conditions. Upper limit to q\ufffd .
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state, (2) Constant properties, (3) One-dimensional heat transfer.
PROPERTIES: Table A-1: Aluminum, T = 300 K, ak = 237 W/m\u22c5K.
ANALYSIS: (a) Applying Equation C.8 to the inner and outer surfaces, it follows that
( ) ( ) ( )
2 2
2 s o i
i i s,o s,i2
o i s o
2 k qr r
q r q r 1 T T
ln r r 4k r
\u3c0
\u3c0\u2032 = \u2212 \u2212 + \u2212
\uf8ee \uf8f9\uf8eb \uf8f6\uf8ef \uf8ec \uf8f7 \uf8fa\uf8ec \uf8f7\uf8ef \uf8fa\uf8ed \uf8f8\uf8f0 \uf8fb
\ufffd
\ufffd <
( ) ( ) ( )
2 2
2 s o i
o o s,o s,i2
o i s o
2 k qr r
q r q r 1 T T
ln r r 4k r
\u3c0
\u3c0\u2032 = \u2212 \u2212 + \u2212
\uf8ee \uf8f9\uf8eb \uf8f6\uf8ef \uf8ec \uf8f7 \uf8fa\uf8ec \uf8f7\uf8ef \uf8fa\uf8ed \uf8f8\uf8f0 \uf8fb
\ufffd
\ufffd <
(b) From Equations C.16 and C.17, energy balances at the inner and outer surfaces are of the form
( )
( )
( )
2 2
o i
s s,o s,i2
s oi
i ,i s,i
i o i
qr rk 1 T T
4k rqr
h T T
2 r ln r r\u221e
\u2212 + \u2212
\u2212 = \u2212
\uf8ee \uf8f9\uf8eb \uf8f6\uf8ef \uf8ec \uf8f7 \uf8fa\uf8ec \uf8f7\uf8ef \uf8fa\uf8ed \uf8f8\uf8f0 \uf8fb
\ufffd
\ufffd <
( )
( )
( )
2 2
o i
s s,o s,i2
s oo
o s,o ,o
o o i
qr rk 1 T T
4k rqr
U T T
2 r ln r r\u221e
\u2212 + \u2212
\u2212 = \u2212
\uf8ee \uf8f9\uf8eb \uf8f6\uf8ef \uf8ec \uf8f7 \uf8fa\uf8ec \uf8f7\uf8ef \uf8fa\uf8ed \uf8f8\uf8f0 \uf8fb
\ufffd
\ufffd <
Accounting for the fin array and the contact resistance, Equation 3.104 may be used to cast the overall
heat transfer coefficient oU in the form
( )
( )
o t
o o(c) o
w t,o(c) ww s,o ,o
q r A1
U h
A R AA T T
\u3b7
\u221e
\u2032 \u2032
= = =
\u2032 \u2032 \u2032\u2032
\u2212
where o(c)\u3b7 is determined from Equations 3.105a,b and w oA 2 r\u3c0\u2032 = .
Continued...
PROBLEM 3.152 (Cont.)
(c) For the prescribed conditions and a representative range of 107 \u2264 q\ufffd \u2264 108 W/m3, use of the relations
of part (b) with the capabilities of the IHT Performance Calculation Extended Surface Model for a
Circular Fin Array yields the following graphical results.
1E7 2E7 3E7 4E7 5E7 6E7 7E7 8E7 9E7 1E8
Heat generation, qdot(W/m^3)
300
340
380
420
460
500
Su
rfa
ce
 te
m
pe
ra
tu
re
s,
 T
s(K
)
Inner surface temperature, Ts,i
Outer surface temperature, Ts,o
It is in this range that the upper limit of s,iT = 373 K is exceeded for q\ufffd = 4.9 × 10
7
 W/m3, while the
corresponding value of s,oT = 379 K is well below the prescribed upper limit. The expressions of part
(a) yield the following results for the surface heat rates, where heat transfer in the negative r direction
corresponds to ( )iq r\u2032 < 0.
1E7 2E7 3E7 4E7 5E7 6E7 7E7 8E7 9E7 1E8
Heat generation, qdot(W/m^3)
-50000
-30000
-10000
10000
30000
50000
Su
rfa
ce
 h
ea
t r
at
es
, q
'(W
/m
)
q'(ri)
q'(ro)
For q\ufffd = 4.9 × 107 W/m3, ( )iq r\u2032 = -2.30 × 104 W/m and ( )oq r\u2032 = 1.93 × 104 W/m.
COMMENTS: The foregoing design provides for comparable heat transfer to the air and water streams.
This result is a consequence of the nearly equivalent thermal resistances associated with heat transfer
from the inner and outer surfaces. Specifically, ( ) 1conv,i i iR h 2 r\u3c0 \u2212\u2032 = = 0.00318 m\u22c5K/W is slightly
smaller than t,o(c)R\u2032 = 0.00411 m\u22c5K/W, in which case ( )iq r\u2032 is slightly larger than ( )oq r\u2032 , while s,iT
is slightly smaller than s,oT . Note that the solution must satisfy the energy conservation requirement,
( ) ( ) ( )2 2o i oir r q q r q r\u3c0 \u2032 \u2032\u2212 = +\ufffd .
	Problem 3.1
	3.2
	3.3
	3.4
	3.5
	3.6
	3.7
	3.8
	3.9
	3.10
	3.11
	3.12
	3.13
	3.14
	3.15
	3.16
	3.17
	3.18
	3.19
	3.20
	3.21
	3.22
	3.23
	3.24
	3.25
	3.26
	3.27
	3.28
	3.29
	3.30
	3.31
	3.32
	3.33
	3.34
	3.35
	3.36
	3.37
	3.38
	3.39
	3.40
	3.41
	3.42
	3.43
	3.44
	3.45
	3.46
	3.47
	3.48
	3.49
	3.50
	ch3_51-100.pdf
	Problem 3.51
	3.52
	3.53
	3.54
	3.55
	3.56
	3.57
	3.58
	3.59
	3.60
	3.61
	3.62
	3.63
	3.64
	3.65
	3.66
	3.67
	3.68
	3.69
	3.70
	3.71
	3.72
	3.73
	3.74
	3.75
	3.76
	3.77
	3.78
	3.79
	3.80
	3.81
	3.82
	3.83
	3.84
	3.85
	3.86
	3.87
	3.88
	3.89
	3.90
	3.91
	3.92
	3.93
	3.94
	3.95
	3.96
	3.97
	3.98
	3.99
	3.100
	ch3_101-152.pdf
	Problem 3.101
	3.102
	3.103
	3.104
	3.105
	3.106
	3.107
	3.108
	3.109
	3.110
	3.111
	3.112
	3.113
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	3.115
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	3.120
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	3.122
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	3.124
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	3.126
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	3.128
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	3.131
	3.132
	3.133
	3.134
	3.135
	3.136
	3.137
	3.138
	3.139
	3.140
	3.141
	3.142
	3.143
	3.144
	3.145
	3.146
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	3.148
	3.149
	3.150
	3.151
	3.152