ch03
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ch03


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= 3 mm, N: 12 - 21, h = 250 W/m^2.K
t = 4 mm, N: 12 -17, h = 250 W/m^2.K
In contrast, significant improvement is associated with changing the sleeve material, and it is only
necessary to have ks \u2248 25 W/m\u22c5K (e.g. a boron sleeve) to approach an upper limit to the influence of ks.
0 20 40 60 80 100
Sleeve conductivity, ks(W/m.K)
2E6
2.4E6
2.8E6
3.2E6
3.6E6
4E6
H
ea
t g
en
er
at
io
n,
 q
do
t(W
/m
^3
)
t = 4 mm, N = 12, h = 250 W/m^2.K
For h = 250 W/m2\u22c5K and ks = 25 W/m\u22c5K, only a slight improvement is obtained by increasing N. Hence,
the recommended conditions are:
2
sh 250 W m K, k 25 W m K, N 12, t 4mm= \u22c5 = \u22c5 = = <
COMMENTS: The upper limit to q\ufffd is reached as the total thermal resistance approaches zero, in which
case T1 \u2192 T\u221e. Hence ( ) 2 6 3max o oq 4k T T r 4.5 10 W m\u221e= \u2212 = ×\ufffd .
PROBLEM 3.140
KNOWN: Geometrical and convection conditions of internally finned, concentric tube air heater.
FIND: (a) Thermal circuit, (b) Heat rate per unit tube length, (c) Effect of changes in fin array.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional heat transfer in radial direction, (3)
Constant k, (4) Adiabatic outer surface.
ANALYSIS: (a) For the thermal circuit shown schematically,
( ) 1conv,i i 1R h 2 r\u3c0 \u2212\u2032 = , ( )cond 2 1R ln r r 2 k\u3c0\u2032 = , and ( ) 1t,o o o tR h A\u3b7 \u2212\u2032 \u2032= ,
where
( )fo f
t
NA
1 1
A
\u3b7 \u3b7
\u2032
= \u2212 \u2212
\u2032
, ( )f 3 2A 2L 2 r r\u2032 = = \u2212 , ( )t f 2A NA 2 r Nt\u3c0\u2032 \u2032= + \u2212 , and f tanh mL
mL
\u3b7 = .
(b) ( ),i ,o
conv,i cond t,o
T T
q
R R R
\u221e \u221e
\u2212
\u2032 =
\u2032 \u2032 \u2032+ +
Substituting the known conditions, it follows that
( ) 12 3conv,iR 5000 W m K 2 0.013m 2.45 10 m K W\u3c0 \u2212 \u2212\u2032 = \u22c5 × × = × \u22c5
( ) ( ) 3condR ln 0.016m 0.013m 2 20 W m K 1.65 10 m K W\u3c0 \u2212\u2032 = \u22c5 = × \u22c5
( ) 12 3t,oR 0.575 200 W m K 0.461m 18.86 10 m K W\u2212 \u2212\u2032 = × \u22c5 × = × \u22c5
where f\u3b7 = 0.490. Hence,
( )
( ) 3
90 25 C
q 2831W m
2.45 1.65 18.86 10 m K W\u2212
\u2212
\u2032 = =
+ + × \u22c5
$
<
(c) The small value of f\u3b7 suggests that some benefit may be gained by increasing t, as well as by
increasing N. With the requirement that Nt \u2264 50 mm, we use the IHT Performance Calculation,
Extended Surface Model for the Straight Fin Array to consider the following range of conditions: t = 2
mm, 12 \u2264 N \u2264 25; t = 3 mm, 8 \u2264 N \u2264 16; t = 4 mm, 6 \u2264 N \u2264 12; t = 5 mm, 5 \u2264 N \u2264 10. Calculations based
on the foregoing model are plotted as follows.
Continued...
PROBLEM 3.140 (Cont.)
5 10 15 20 25
Number of fins, N
2000
3000
4000
5000
H
ea
t r
at
e,
 q
'(w
/m
)
t = 2 mm
t = 3 mm
t = 4 mm
t = 5 mm
By increasing t from 2 to 5 mm, f\u3b7 increases from 0.410 to 0.598. Hence, for fixed N, q\u2032 increases
with increasing t. However, from the standpoint of maximizing tq\u2032 , it is clearly preferable to use the
larger number of thinner fins. Hence, subject to the prescribed constraint, we would choose t = 2 mm
and N = 25, for which q\u2032 = 4880 W/m.
COMMENTS: (1) The air side resistance makes the dominant contribution to the total resistance, and
efforts to increase q\u2032 by reducing t,oR\u2032 are well directed. (2) A fin thickness any smaller than 2 mm
would be difficult to manufacture.
PROBLEM 3.141
KNOWN: Dimensions and number of rectangular aluminum fins. Convection coefficient with and without fins.
FIND: Percentage increase in heat transfer resulting from use of fins.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction, (3) Constant properties, (4)
Negligible radiation, (5) Negligible fin contact resistance, (6) Uniform convection coefficient.
PROPERTIES:
 Table A-1, Aluminum, pure: k \u2248 240 W/m\u22c5K.
ANALYSIS: Evaluate the fin parameters
cL L+t/2 0.05025m= =
-3 -6 2
p cA L t 0.05025m 0.5 10 m=25.13 10 m= = × × ×
( ) ( )
1/ 221/ 2 3/ 23/2
c w p
-6 2
30 W/m KL h / kA 0.05025m 
240 W/m K 25.13 10 m
\uf8ee \uf8f9
\u22c5
= \uf8ef \uf8fa\uf8ef \uf8fa\u22c5 × ×\uf8f0 \uf8fb
( )1/ 23/2c w pL h / kA 0.794=
It follows from Fig. 3.18 that \u3b7f \u2248 0.72. Hence,
f f max w bq q 0.72 h 2wL \u3b7 \u3b8= =
( ) ( )2f b bq 0.72 30 W/m K 2 0.05m w 2.16 W/m K w \u3b8 \u3b8= × \u22c5 × × × = \u22c5
With the fins, the heat transfer from the walls is
( )w f w bq N q 1 Nt w h \u3b8= + \u2212
( ) ( ) ( )4 2w b bWq 250 2.16 w 1m 250 5 10 m 30 W/m K w 
m K
\u3b8 \u3b8\u2212= × + \u2212 × × × \u22c5
\u22c5
( ) ( )w b bWq 540 26.3 w 566 w .
m K
\u3b8 \u3b8= + =
\u22c5
Without the fins, qwo = hwo 1m × w \u3b8b = 40 w \u3b8b. Hence the percentage increase in heat
transfer is
( ) bw wo
wo b
566 40 w q q 13.15 1315%
q 40 w 
\u3b8
\u3b8
\u2212
\u2212
= = = <
COMMENTS: If the infinite fin approximation is made, it follows that qf = (hPkAc)1/2 \u3b8b
=[hw2wkwt]1/2 \u3b8b = (30 × 2 × 240 × 5×10-4)1/2 w \u3b8b = 2.68 w \u3b8b. Hence, qf is
overestimated.
PROBLEM 3.142
KNOWN: Dimensions, base temperature and environmental conditions associated with rectangular and
triangular stainless steel fins.
FIND: Efficiency, heat loss per unit width and effectiveness associated with each fin.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction, (3) Constant properties,
(4) Negligible radiation, (5) Uniform convection coefficient.
PROPERTIES: Table A-1, Stainless Steel 304 (T = 333 K): k = 15.3 W/m\u22c5K.
ANALYSIS: For the rectangular fin, with Lc = L + t/2, evaluate the parameter
( ) ( ) ( )( )
1/ 221/ 2 3/ 23/ 2
c p
75 W m KL h kA 0.023m 0.66
15.3W m K 0.023m 0.006m
\uf8ee \uf8f9
\u22c5
= =\uf8ef \uf8fa
\u22c5\uf8ef \uf8fa\uf8f0 \uf8fb
.
Hence, from Fig. 3.18, the fin efficiency is
f 0.79\u3b7 \u2248 <
From Eq. 3.86, the fin heat rate is f f f b f c b f c bq hA hPL h2wL\u3b7 \u3b8 \u3b7 \u3b8 \u3b7 \u3b8= = = or, per unit width,
( ) ( )2ff qq 0.79 75W m K 2 0.023m 80 C 218 W m
w
\u2032 = = \u22c5 =
$
. <
From Eq. 3.81, the fin effectiveness is
( ) ( )
f f
f 2
c,b b b
q q w 218W m 6.06
hA h t w 75 W m K 0.006m 80 C
\u3b5
\u3b8 \u3b8
\u2032 ×
= = = =
×
\u22c5
$
. <
For the triangular fin with
( ) ( ) ( )( )( )
1/ 221/ 2 3/ 23/ 2
c p
75 W m KL h kA 0.02m 0.81
15.3W m K 0.020m 0.003m
\uf8ee \uf8f9
\u22c5
= =\uf8ef \uf8fa
\u22c5\uf8ef \uf8fa\uf8f0 \uf8fb
,
find from Figure 3.18,
f 0.78\u3b7 \u2248 , <
From Eq. 3.86 and Table 3.5 find
( ) 1/ 222f f f b f bq hA h2 L t 2\u3b7 \u3b8 \u3b7 \u3b8\uf8ee \uf8f9\u2032 \u2032= = +\uf8ef \uf8fa\uf8f0 \uf8fb
( ) ( ) ( )1/ 22 22fq 0.78 75W m K 2 0.02 0.006 2 m 80 C 187 W m\uf8ee \uf8f9\u2032 = × \u22c5 × + =\uf8ef \uf8fa\uf8f0 \uf8fb $ . <
and from Eq. 3.81, the fin effectiveness is
( ) ( )
f
f 2b
q w 187 W m 5.19
h t w 75W m K 0.006m 80 C
\u3b5
\u3b8
\u2032 ×
= = =
×
\u22c5
$
<
COMMENTS: Although it is 14% less effective, the triangular fin offers a 50% weight savings.
PROBLEM 3.143
KNOWN: Dimensions, base temperature and environmental conditions associated with a triangular,
aluminum fin.
FIND: (a) Fin efficiency and effectiveness, (b) Heat dissipation per unit width.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction, (3) Constant properties,
(4) Negligible radiation and base contact resistance, (5) Uniform convection coefficient.
PROPERTIES: Table A-1, Aluminum, pure (T \u2248 400 K): k = 240 W/m\u22c5K.
ANALYSIS: (a) With Lc = L = 0.006 m, find
( )( ) 6 2pA Lt 2 0.006 m 0.002 m 2 6 10 m\u2212= = = × ,
( ) ( )
1/ 221/ 2 3 / 23 / 2
c p 6 2
40 W m K
L h kA 0.006 m 0.077
240 W m K 6 10 m\u2212
\u22c5
= =
\u22c5 × ×
\uf8eb \uf8f6\uf8ec \uf8f7\uf8ec \uf8f7\uf8ed \uf8f8
and from Fig. 3.18, the fin efficiency is
f 0.99\u3b7 \u2248 . <
From Eq. 3.86 and Table 3.5, the fin heat rate is
( ) 1/ 222f f max f f (tri) b f bq q hA 2 hw L t 2\u3b7 \u3b7 \u3b8 \u3b7 \u3b8= = = +\uf8ee \uf8f9\uf8ef \uf8fa\uf8f0 \uf8fb .
From Eq. 3.81, the fin effectiveness is
( )
( )
( )1/ 2 1/ 22 22 2f b ff
f
c,b b b
2 hw L t 2 2 L t 2q
hA g w t t
\u3b7 \u3b8 \u3b7
\u3b5
\u3b8 \u3b8
+ +
= = =
\u22c5
\uf8ee \uf8f9 \uf8ee \uf8f9\uf8ef \uf8fa \uf8ef \uf8fa\uf8f0 \uf8fb \uf8f0 \uf8fb
( ) ( ) 1/ 22 2
f
2 0.99 0.006 0.002 2 m
6.02
0.002 m
\u3b5
× +
= =
\uf8ee \uf8f9\uf8ef \uf8fa\uf8f0 \uf8fb <
(b) The heat dissipation per unit width is
( ) ( ) 1/ 222f f f bq q w 2 h L t 2\u3b7 \u3b8\u2032 = = +\uf8ee \uf8f9\uf8ef \uf8fa\uf8f0 \uf8fb
( ) ( ) ( )1/ 22 22fq 2 0.99 40 W m K 0.006 0.002 2 m 250 20 C 110.8 W m\u2032 = × × \u22c5 + × \u2212