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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.
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
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ch3_51-100.pdf
Problem 3.51
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ch3_101-152.pdf
Problem 3.101
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