A typical home may require a total of 200103kWh of energy pe

A typical home may require a total of 2.00×103kWh of energy per month. Suppose you would like to obtain this energy from sunlight, which has an average daylight intensity of 1300 W/m2. Assuming that sunlight is available 8.0 h per day, 25 d per month (accounting for cloudy days), and that you have a way to store energy from your collector when the Sun isn\'t shining, determine the smallest collector size that will provide the needed energy, given a conversion efficiency of 30 % .

Solution

P = E / t = 2 * 10³ kWh / 25 d * 8.0 hr

= 10 KW

A = P / 0.30 I

= (10 KW) / (0.30 * 1300)

A = 25.64 m²