A photovoltaic panel of dimension 2 m × 4 m is installed on the roof of a home. The panel is irradiated with a solar flux of GS = 700 W/m2, oriented normal to the top panel surface. The absorptivity of the panel to the solar irradiation is αs  0.83, and the efficiency of conversion of the absorbed flux to electrical power is Ƞ  P/αsGSA  0.553  0.001 K -1Tp, where Tp is the panel temperature expressed in kelvins and A is the solar panel area. Determine the electrical power generated for a) A still summer day, in which Tsur  T[infinity]  35C, h  10 W/m2⋅ K b) A breezy winter day, for which Tsur = T[infinity] = −15°C, h = 30 W/m2 ⋅ K The panel emissivity is   0.90

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tatendagota tatendagota · Answer 1 · 2020-02-17T15:09:06+01:00

Answer:

a. 992.348 W

b. 1291 W

Explanation:

An appropriate sketch will be useful for the problem.

Next, we make some simple assumptions:

1. Steady state heat transfer'.

2. There is uniform convention over the panel

3. There is no conduction involved

a.) Area of the surface = 2 × 4

= 8 m²

Heat absorbed by the panel = 700 × 0.3 × 8

= 4 648 W

Assuming that temperature of the panel, Tp must be between 300 and 400 ⁰C, then:

0.553 - 0.001Tp = 0.83 × 700 × 8 (0.553 = 0.001 × 339.5)

= 992.348 W

b. This part is more or less like part (a) above, so:

Tp must be somewhere between 275 - 275.5 K.

Assuming Tp = 275.25, then

0.8 × 700 × 8 (0.553 = 0.001 * 275.25)

= 1 291 W