14-57
14-67 Two glasses of a double pane window are maintained at specified temperatures. The fraction of heat
transferred through the enclosure by radiation is to be determined.
Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas
with constant properties. 3 The air pressure in the enclosure is 1 atm.
Air
Properties The properties of air at 1 atm and the average temperature of
Q
(T1+T2)/2 = (280+336)/2 = 308 K = 35qC are (Table A-22E)
k 0.02625 W/m.qC
336 K
280 K
Q 1.655 u 10 5 m 2 /s
L=0.4 m
Pr 0.7268
1
1
E
0.003247 K -1
H = 1.5 m
308 K
Tf
Analysis The characteristic length in this case is the distance
between the two glasses, Lc = L = 0.4 m. Then,
Ra L
gE (T1 T2 ) L3c
Q2
Pr
(9.81 m/s 2 )(0.003247 K -1 )(336 280 K )(0.4 m) 3
(1.655 u 10 5 m 2 /s) 2
(0.7268)
3.029 u 10 8
The aspect ratio of the geometry is H/L = 1.5/0.4 = 3.75. For this value of H/L the Nusselt number can be
determined from
Nu
§ Pr
·
0.22¨
Ra ¸
0
.
2
Pr
©
¹
As
H uW
0.28
§H·
¨ ¸
©L¹
1 / 4
§ 0.7268
·
0.22¨
(3.029 u 10 8 ) ¸
0
.
2
0
.
7268
©
¹
0.28
§ 1.5 ·
¨
¸
© 0.4 ¹
1 / 4
35.00
Then,
(1.5 m)(3 m)
4. 5 m 2
T1 T2
(336 280)K
(0.02625 W/m.qC)(35.00)(4.5 ft 2 )
L
0.4 m
The effective emissivity is
1
1
1
1
1
1
1 6.778 
o H eff 0.1475
0.15 0.90
H eff H 1 H 2
Q conv
kNuAs
578.9 W
The rate of heat transfer by radiation is
Q
H A V (T 4 T 4 )
rad
eff
s
1
2
2
(0.1475)(4.5 m )(5.67 u 10 8 W/m 2 .K 4 )[(336 K) 4 (280 K ) 4 ]
248.4 W
Then the fraction of heat transferred through the enclosure by radiation becomes
Q rad
248.4
f rad
0.30
Qconv Q rad 578.9 248.4
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