Homework #4 Solutions
An array of photovoltaic (PV) panels, as shown, can be represented as a long, flat plate. During
operation, the surface of the plates is 50 °C. A wind with velocity 10 m/s and temperature 10°C is
blowing parallel to the PV array. What is the rate of heat loss from the array? If the wind shifts such
that it strikes the PV array normally to the surface, what is the heat loss?
PV Array
2m
Wind
Normal Flow
20 m
Parallel Flow
Solution Part 1:
This is forced, external flow over a flat plate. So heat loss by convection is
Q = h A (TS - T∞)
A = (2) m (20) m = 40 m2
TS = 50 °C; T∞ = 10 °C.
To solve for the convective heat transfer coefficient, solve ReL, then use a correlation for NUL, and then
solve h.
Find air properties at the average temperature: T 
TS  T 50  10

 30C
2
2
ν = 15.9 * 10-6 m2/s; k = 26.3 * 10-3 W/m K; Pr = 0.71
Re L 
V L


(10) (20)
 12.6 *10 6
6
15.9 *10
This Re give mixed laminar/turbulent boundary layer flow, so use Eq. 7.38
Nu L  (0.037 Re 0.8  871) Pr
1
3
 15,000
h = Nu (k/L) = (15,000)*(.0263/20) = 19.7 W/m2 K
Q = (19.7) W/m2 K (40) m2 (50 – 10)K = 31,600 W
Solution Part 2:
This is forced, external flow normal to a vertical plate. So heat loss by convection is
Q = h A (TS - T∞)
A = (2) m (20) m = 40 m2
TS = 50 °C; T∞ = 10 °C.
Using Table 7.3 for the correlation for a vertical plate
Nu D  C Re D Pr
m
1
3
From the table, C = 0.228 and m = 0.731
Re D 
V D


(10) (2.0)
 12.6 *10 5
6
15.9 *10
Note: This Reynolds number is beyond the recommended range for this correlation, but it will give a
reasonable value since no other correlation for flow normal to a flat plate is available.
Nu D  0.228 12.6 *10 5 
0.731
.71
1
3
 5850
h = Nu (k/D) = (5850)*(.0263/2.0) = 76.9 W/m2 K
Q = (76.9) W/m2 K (40) m2 (50 – 10)K = 123,000 W
So flow normal to the surface removes almost four times as much heat.