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Step2247 [10]
4 years ago
5

A typical home may require a total of 2.00*10^3kWh of energy per month. Suppose you would like to obtain this energy from sunlig

ht, which has an average daylight intensity of 1400 W/m^2 . Assuming that sunlight is available 8.0h 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 24% .
Physics
1 answer:
gayaneshka [121]4 years ago
5 0

Answer:

Smallest collector size≈30m^{2}

Explanation:

Daily sunlight intensity = 1400w/m^{2}.

For a collector of 24% efficiency, the absorption per day per m^{2}

= 0.24*1400 = 336w/m^{2}.

For 8 hours, 25days of available sunlight, energy absorbed by collector per metre square , will be = 8* 25 *336 = 67200wh.

This is the amount of sunlight energy which will be trapped by 1w/m^{2}. of collector in a month.

To find how many metre square of collector is required to get 2*10^{3}kwh, we simply divide the required energy by the amount absorbed per square metre.

Smallest size of collected needed = \frac{2.00*1000*10^{3}}{67200}

=29.76

≈30m^{2}

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