A typical home may require a total of 2.19E+3 kW*hr of energy per month. Suppose you would like to obtain this energy from sunlight, which has an average daylight intensity of 1.03E+3 W/m2. Assuming that sunlight is available 8.03 hr per day, 24 days 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 27.4 %.
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With a 1 m^2 size collector we will be able to capture 1.03 x 10^3 x (27.4/100) J = 282.22 ...
The solution determines the smallest solar collector size for needed energy. A complete, neat and step-by-step solution is provided.