Assuming 800 w / m 2 solar irradiance and a 35 % efficient solar panel, how much roof area should be covered to supply 10 a at 120 v ? a = 4.286 m 2 (within three decimal places) given an average of 7 h o u r s of sunshine per day and a utility cost of $ 0.22 k w h , how much of the utility cost can such a solar panel save? ignore the initial cost or any maintenance cost of the solar panels

a) 4.286 m² b) $ 55.44/mo

Explanation:

If we assume that the sun is behaving as an isotropic radiator, the power density that is arriving to the house, is constant and equal to the quoted solar irradiance.

If the energy conversion capability of the solar panels were 100%, the roof area needed to supply the power required, would be simply the quotient between the power required and the solar irradiance, as follows:

A = P / SI = 10 A* 120 V / 800 W/m² = 1200 W / 800 W/m²= 1,5 m²

As the solar panels are only 35% efficient in converting the solar energy to useful electrical energy, we will need more roof area, according to this expression:

Ae = At / 0.35 = 1,5 / 0.35 = 4.286 m²

b) If we can get 1200 W during 7 hs/day, the energy supplied by the solar panels will be the product of the power times the time, as follows:

E= 1200 W* 7 hs = 8.4 Kwh

If the cost per Kwh, is $0.22, assuming 7 hs. of use in average during a month (assumed to be of 30 days), we can have savings as follows:

Cost = 0.22($/Kwh)* 8.4 (Kwh/day)*30 (days/mo) = $ 55.44

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explanation:

answer:

an example of solubility is salt or sugar dissolving in water.

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