A typical home may require a total of 2.00×103kWh of energy per month. Suppose y

Question

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 1100 W/m2 .

Part A

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 % .

Express your answer using two significant figures.

Explanation / Answer

Here ,

total energy required , E = 2 *10^3 kWh

intensity of sunlight , I = 1100 W/m^2

let the area of collector is A

total energy = I * A * time

2 *10^3 * 1000 * 3600 J = 1100 * A * 8 * 3600 * 25

solving for A

A = 9.091 m^2

the smallest collector size that is needed is 9.091 m^2

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