A photovoltaic panel of dimension 2m×4m is installed on the

roof of a home. The panel is irradiated with a solar flux of GS =700W /m2, oriented normal to

the top panel surface. The absorptivity of the panel to the solar

Irradiation is s =0.83, and the efficiency of conversion of the absorbed flux to electrical power

is P /sGsA =0.553−0.001Tp, where Tp is the panel temperature expressed in Kelvins and A is

the solar panel area. Determine the electric power generated for

a. A still summer day, in which Tsur =T[infinity]=35C, h=100W /m2 â‹…K, and

b. A breezy winter day, for which Tsur =T[infinity]=−15oC, h=30W /m2 ⋅K. The panel Emissivity is 0.90.

Explanation:

Simply put, a solar panel works by allowing photons, or particles of light, to knock electrons free from atoms, generating a flow of electricity. Solar panels actually comprise many, smaller units called photovoltaic cells. (Photovoltaic simply means they convert sunlight into electricity. The attached diagram give an ilustsration of the photovotaic pannel mounted on a roof top.

Solution

To Determine the electric power generated for

a) A still summer day.

E = A * r * H * PR

E = Total Amount of Energy in kilowatt

A = Total Surface Area

r = efficiency Rating

H = global radiation value

PR = Performance Ratio

kwh = watt * Time/1000

kwh = 100 * 35/1000

3.5

b)

kwh = watt * Time/1000

kwh = 30 *15/1000

4.5

answer:

not sure

explanation: