A typical home may require a total of 2.10 x 10 3 kWh of energy per month. Suppo

Question

A typical home may require a total of 2.10 x 103 kWh of energy per month. Suppose you would like to obtain this energy from sunlight, which has an average daylight intensity of 1.00 103 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 37%

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photo below , please answer and show steps. thank you.

A typical home may require a total of 2.10.103 kWh of energy per month. Suppose you would like to obtain this energy from sunlight, which has an average daylight intensity of 00 x 103 w/m2 Assuming that sunlight is available 8.0 h per day, 25d per month (accounting for cloudy days, and tha you have a way to store energy from your collector when the Sunisnt shining, determine the smallest collector size that will provide the needed energk given a conversion efficiency of 37%.

Explanation / Answer

Given: Demand of energy per month = 2.10×103 kwh

Average daylight intensity of sunlight = 1×103 W/m2

Availability of sunlight = 8h per day, 25 day per month

Efficiency = 37% = 0.37

Have to find: smallest collector size in m2.

Step-1 : Recieved intensity from sun :

1000 W/m2× 0.37 = 370 W/m2

Step-2 :Per month receiving intensity :

370 W/m2×8h×25d = 74000 Wh/m2

Step-3: Size ofcollector in m2:

2.10×106Wh. ÷ 74000 Wh/m2

= 28.38 m2

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