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# ch09

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```1 atm): n = 16.90 ´ 10
-6
m
2
/s, k = 27.0 ´
10
-3
W/m×K, a = 23.98 ´ 10
-6
m
2
/s, Pr = 0.706, b = 1/Tf.
ANALYSIS: (a) For the vertical lateral (lat) and top horizontal (top) surfaces, the heat loss by
( ) ( ) ( ) ( )( )2lat top lat r s top r sq q q h h DL T T h h D / 4 T Tp p¥ ¥= + = + - + + -
where, from Eq. 1.9, the linearized radiation coefficient is
( ) ( )2 2r s sh T T T Tes ¥ ¥= + +
( ) ( )8 2 4 2 2 2 2rh 0.8 5.67 10 W / m K 320 300 K 320 300 K 5.41W/m K.-= ´ ´ × + + = ×
The free convection coefficient for the lateral and top surfaces is:
Lateral-vertical plate: Using Eq. 9.26 with
( ) ( )( ) ( )33 2s 8
L 6 2 6 2
g T T H 9.8m/s 1/310K 47 27 K 0.5m
Ra 1.950 10
16.90 10 m / s 23.98 10 m / s
b
na
¥
- -
- -
= = = ´
´ ´ ´
( )
L
2
1/6
L
8/279/16
0.387Ra
Nu 0.825
1 0.492/Pr
ì ü
ï ïï ï= +í ý
ï ïé ù+ï ïê úë ûî þ
( )
( )
L
2
1/68
8/279/16
0.387 1.950 10
Nu 0.825 74.5
1 0.492/0.706
ì ü
´ï ïï ï= + =í ý
ï ïé ù+ï ïê úë ûî þ
L
2
lath Nu k / H 74.5 0.027W/m K/0.5m 4.02W/m K.= × = ´ × = ×
Continued \u2026..
PROBLEM 9.33 (Cont.)
Top-horizontal plate: Using Eq. 9.30 with
2
c s
D / 4
L A / P D / 4 0.075m
D
p
p
= = = =
( ) ( )( ) ( )323s c 5
L 6 2 6 2
g T T L 9.8m/s 1/310K 47 27 K 0.075m
Ra 6.598 10
16.90 10 m / s 23.98 10 m / s
b
na
¥
- -
- -
= = = ´
´ ´ ´
( )L 1/41/4 5LNu 0.54Ra 0.54 6.598 10 15.39= = ´ =
L
2
top ch Nu k / L 15.39 0.027W/m K/0.075m 5.54W/m K.= × = ´ × = ×
Hence, the heat loss by convection and radiation is
( ) ( ) ( )
( ) ( )( )
2
2 2 2
q 4.02 5.41 W / m K 0.30m 0.50m 47 20 K
5.54 5.41 W / m K 0.30 m / 4 47 20 K
p
p
= + × ´ ´ -
+ + × ´ -
( )q 88.9 15.5 W 104W.= + = <
(b) The effect of adding the vertical fins is to increase the area of the lateral surface to
( ) ( )wfA DH 30 t H 30 2 w Hpé ù= - × + ´ ×ë û
( ) ( )2 2wfA 0.30m 0.50m 30 0.005 0.500 m 30 2 0.075 0.500 mpé ù= ´ - ´ + ´ ´ê úë û
[ ] 2 2 2wfA 0.471 0.075 m 2.25m 2.646m .= - + =
where t and w are the thickness and width of the fins, respectively. Hence, the heat loss is now
( ) ( )lat top lat r wf s topq q q h h A T T q¥= + = + - +
( ) 2 2q 4.02 5.41 W / m 2.646m 20K 15.5W 515W.= + ´ ´ + = <
Adding the fins to the lateral surface increases the heat loss by a factor of five.
COMMENTS: Since the fins are not likely to have 100% efficiency, our estimate is optimistic.
Further, since the fins see one another, as well as the lateral surface, the radiative heat loss is over
predicted.
PROBLEM 9.34
KNOWN: Surface temperature of a long duct and ambient air temperature.
FIND: Heat gain to the duct per unit length of the duct.
SCHEMATIC:
ASSUMPTIONS: (1) Surface radiation effects are negligible, (2) Ambient air is quiescent.
PROPERTIES: Table A-4, Air (Tf = (T¥ + Ts)/2 » 300K, 1 atm): n = 15.89 ´ 10
-6
m
2
/s, k =
0.0263 W/m×K, a = 22.5 ´ 10
-6
m
2
/s, Pr = 0.707, b = 1/Tf.
ANALYSIS: The heat gain to the duct can be expressed as
( )( )s t b s t b sq 2q q q 2h H h W h W T T .¥¢ ¢ ¢ ¢= + + = × + × + × - (1)
Consider now correlations to estimate s t bh , h , and h . From Eq. 9.25, for the sides with L º H,
( ) ( ) ( ) ( )33 2s 7
L 6 2 6 2
g T T L 9.8m/s 1/300K 35 10 K 0.2m
Ra 1.827 10 .
15.89 10 m / s 22.5 10 m / s
b
na
¥
- -
- - ´
= = = ´
´ ´ ´
(2)
Eq. 9.27 is appropriate to estimate sh ,
( )
( )
( )
L
1/471/4
L
4/9 4/99/16 9/16
0.670 1.827 100.670Ra
Nu 0.68 0.68 34.29
1 0.492/Pr 1 0.492/0.707
´
= + = + =
é ù é ù+ +ê ú ê úë û ë û
L
2
sh Nu k / L 34.29 0.0263W/m K/0.2m 4.51W/m K.= × = ´ × = ×
(3)
For the top and bottom portions of the duct, L º As/P » W/2, (see Eq. 9.29), find the Rayleigh number
from Eq. (2) with L = 0.1 m, RaL = 2.284 ´ 10
6
. From the correlations, Eqs. 9.31 and 9.32 for the top
and bottom surfaces, respectively, find
( ) ( )
1/31/3 6 2
t L
k 0.0263W/m K
h 0.15Ra 0.15 2.284 10 5.17W/m K.
W / 2 0.1m
×
= ´ = ´ ´ = × (4)
( ) ( )
1/46 2
b
k 0.0263W/m K
h 0.27 2.284 10 2.76W/m K.
W / 2 0.1m
×
= ´ ´ ´ = × (5)
The heat rate, Eq. (1), can now be evaluated using the heat transfer coefficients estimated from Eqs.
(3), (4), and (5).
( )( )2 2 2q 2 4.51W/m K 0.2m 5.17W/m K 0.2m 2.76W/m K 0.2m 35 10 K¢ = ´ × ´ + × ´ + × ´ -
q 84.8W/m.¢ = <
COMMENTS: Radiation surface effects will be significant in this situation. With knowledge of the
duct emissivity and surroundings temperature, the radiation heat exchange could be estimated.
PROBLEM 9.35
KNOWN: Inner surface temperature and dimensions of rectangular duct. Thermal conductivity,
thickness and emissivity of insulation.
FIND: (a) Outer surface temperatures and heat losses from the walls, (b) Effect of insulation thickness
on outer surface temperatures and heat losses.
SCHEMATIC:
ASSUMPTIONS: (1) Ambient air is quiescent, (2) One-dimensional conduction, (3) Steady-state.
PROPERTIES: Table A.4, air (obtained from Properties Tool Pad of IHT).
ANALYSIS: (a) The analysis follows that of Example 9.3, except the surface energy balance must now
include the effect of radiation. Hence, cond conv radq q q\u2032\u2032 \u2032\u2032 \u2032\u2032= + , in which case
( )( ) ( ) ( )i s,1 s,2 s,2 r s,2 surk t T T h T T h T T\u221e\u2212 = \u2212 + \u2212
where ( )( )2 2r s,2 sur surs,2h T T T T\u3b5\u3c3= + + . Applying this expression to each of the top, bottom and
side walls, with the appropriate correlation obtained from the Correlations Tool Pad of IHT, the
following results are determined for t = 25 mm.
Sides: Ts,2 = 19.3°C, h = 2.82 W/m2\u22c5K, hrad = 5.54 W/m2\u22c5K
Top: Ts,2 = 19.3°C, h = 2.94 W/m2\u22c5K, hrad = 5.54 W/m2\u22c5K <
Bottom: Ts,2 = 20.1°C, h = 1.34 W/m2\u22c5K, hrad = 5.56 W/m2\u22c5K
With condq q\u2032\u2032 \u2032\u2032= , the surface heat losses may also be evaluated, and we obtain
Sides: q\u2032 = 2H q\u2032\u2032 = 21.6 W/m; Top: q\u2032 = w q\u2032\u2032 = 27.0 W/m; Bottom: q\u2032 = w q\u2032\u2032 = 26.2 W/m <
(b) For the top surface, the following results are obtained from the parametric calculations
Continued...
PROBLEM 9.35 (Cont.)
0 0.01 0.02 0.03 0.04 0.05
Insulation thickness, t(m)
15
25
35
45
Su
rfa
ce
te
m
pe
ra
tu
re
, T
s2
(C
)

0 0.01 0.02 0.03 0.04 0.05
Insulation thickness, t(m)
0
100
200
300
H
ea
t l
os
s,
q
'(W
/m
)
COMMENTS: Contrasting the heat rates of part (a) with those predicted in Comment 1 of Example 9.3,
it is evident that radiation is significant and increases the total heat loss from 57.6 W/m to 74.8 W/m. As
shown in part (b), reductions in Ts,o and \u2032q may be effected by increasing the insulation thickness above
0.025 W/m\u22c5K, although attendant benefits diminish with increasing t.
PROBLEM 9.36
KNOWN: Electric heater at bottom of tank of 400mm diameter maintains surface at 70°C with
engine oil at 5°C.
FIND: Power required to maintain 70°C surface temperature.
SCHEMATIC:
ASSUMPTIONS: (1) Oil is quiescent, (2) Quasi-steady state conditions exist.
PROPERTIES: Table A-5, Engine Oil (Tf = (T¥ + Ts)/2 = 310K): n = 288 ´ 10
-6
m
2
/s, k =
0.145 W/m×K, a = 0.847 ´ 10-7 m2/s, b = 0.70 ´ 10-3 K-1.
ANALYSIS: The heat rate from the bottom heater surface to the oil is
( )s sq hA T T¥= -
where h is estimated from the appropriate correlation depending upon the Rayleigh number RaL,
from Eq. 9.25, using the characteristic length, L, from Eq. 9.29,
2A D / 4 D 0.4msL 0.1m.
P D 4 4
p
p
= = = = =
The Rayleigh number is
( ) 3g T T LsRaL
b
na
- ¥=
( )2 3 1 3 39.8m/s 0.70 10 K 70 5 K 0.1 m 7Ra 1.828 10 .L 6 2 7 2288 10 m / s 0.847 10 m / s
- -´ ´ - ´
= = ´- -´ ´ ´
The appropriate correlation is Eq. 9.31 giving
( )1 / 3hL 1/3 7Nu 0.15Ra 0.15 1.828 10 39.5L Lk= = = ´ =
k 0.145W/m K 2h Nu 39.5 57.3W/m K.LL 0.1m
×= = ´ = ×```