Question

A photovoltaic panel of dimensions 6 m x 5 m is located on top of the roof of a house. Solar irradiation Gs = 900 W/m2 is incident on the panel. The panel has an absorptivity to solar irradiation, αs of 0.92. The freestream air temperature is Tinf and the surrounding temperature for radiation exchange with the sky is Tsurr. For this particular problem, it is given that Tinf= Tsurr. The convective heat transfer coefficient of air blowing over the panel is h W/m2-K.

Answer 1

2.7 W/m^2K

Step-by-step explanation:

Area of pane = 5 m x 6 m = 30 m^2

Solar irradiation Gs = 900 W/m2

Heat rate on panel = Gs x area = 900 x 30 = 27000 W

absorptivity to solar irradiation αs = 0.92

Therefore, absorbed heat is

0.92 x 27000 = 24840 W

For heat gain,

From E = §AT^4

Where § = stefan's constant = 5.7x10^-8 Wm^-2K^-1

T = temperature of panel

24840 = 5.7x10^-8 x 30 x T^4

24840 = 1.71x10^-6 x T^4

1.453x10^10 = T^4

T = 347.167 K

For net heat gain,

From E = §A(T^4 - T^4sur)

24840 = 5.7x10^-8 x 30 x (T^4 - T^4sur)

24840 = 1.71x10^-6 x (1.453x10^10 - T^4sur)

24840 = 24846.3 - 1.71x10^-6(T^4sur)

-6.3 = -1.71x10^-6(T^4sur)

3684210.526 = T^4sur

Tsur = 43.81 K

Also for convective heat,

E = Ah(T - Tsur)

24840 = 30h(347.167 - 43.81)

24840 = 30h x 303.357

81 = 30h

h = 2.7 W/m^2K