14-100
14-100 A flat-plate solar collector tilted 40qC from the horizontal is exposed to the calm ambient air. The
total rate of heat loss from the collector, the collector efficiency, and the temperature rise of water in the
collector are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with constant properties. 3 The local
atmospheric pressure is 1 atm. 4 There is no heat loss from the back surface of the absorber plate.
Properties The properties of air at 1 atm and the film temperature of
(Ts+Tf)/2 = (40+20)/2 = 30qC are (Table A-22)
k 0.02588 W/m.qC
Absorber
2
5
Plate
Q 1.608 u 10 m /s
Solar
radiation
Pr 0.7282
650
W/m2
1
1
Outdoors,
E
0.0033 K -1
Tf = 20qC
(30 273)K
Tf
Tsky = -40qC
Analysis (a) The characteristic length in
1.5 m
this case is determined from
Lc
As
p
(1.5 m)(2 m)
2(1.5 m + 2 m)
G = 3 cm
0.429 m 2
Glass
Cover,
40qC
Then,
Ra
gE (Tf Ts ) L3c
Q2
(1.608 u10
Nu
0.15 Ra
h
k
Nu
Ls
As
and
Q conv
5
2
m /s)
0.15(1.103 u 10 )
8 1/ 3
2
(0.7282) 1.103 u10 8
71.94
0.02588 W/m.qC
(71.94)
0.429 m
(1.5 m)(2 m)
Insulation
Pr
(9.81 m/s 2 )(cos 40q)(0.0033 K -1 )(40 20 K )(0.429 m) 3
1/ 3
T =40q
4.340 W/m 2 .qC
3 m2
hAs (Ts Tf )
(4.340 W/m 2 .qC)(3 m 2 )(40 20)qC
260.4 W
Heat transfer rate by radiation is
Q rad HAs V (Tsurr 4 Ts 4 )
(0.9)(3 m 2 )(5.67 u 10 8 W/m 2 .K 4 )[(40 273 K ) 4 (40 273 K ) 4 ] 1018 W
and
Q total
260.4 1018 1278 W
(b) The solar energy incident on the collector is
Q incident DqAs (0.88)(650 W/m 2 )(3 m 2 ) 1716 W
Then the collector efficiency becomes
Q incident Q lost 1716 1278
efficiency
1716
Q incident
0.255 25.5%
(c) The temperature rise of the water as it passes through the collector is
Q
(1716 1278) W
Q m c p 'T o 'T
6.3qC
m c p (1 / 60 kg/s)(4180 J/kg.qC)
14-101 ….. 14-103 Design and Essay Problems
KJ
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