# Physics revision help?

If 100 kWh are required to power a single home every day (24 hours), how

large must a solar panel be, if it is 20% efficient and illuminated for 8 hours per

day ? Assume an average insolation of 800 W m

-2

at ground level

Trying to revise past exam papers and I cannot figure this out.

Any help appreciated,thanks.

Relevance

Since your insolation is 800 Watts/m^2 and you get 20% efficiency, the amount of enegy converted would be 800 W/m^2 * 0.20 = 160 W/m^2 [this is the calculated rating of the panel in sq meters].

(100 KW*h) / (8 hours / 24 hours) = 3 * 100 KW*h = 300 KW*h [as a solar system]

{notice that you need a system 3x larger because you only get 1/3 of the day in sunlight}

(300,000 W*h) / (160W/m^2) = 1,875 m^2 * h ; which is 43.3013 meters x 43.3013 meters. [142.0642 ft x 142.0642 ft]

Your solar panel size is based upon the area of the panel and the conversion efficiency of the panel. Also keep in mind that the area of a panel is not 100% either ! there are many dead areas like the frame and between cells and the bonding areas as well. This is why panels are rated based upon emperical data testing, not calculated equations.

BTW, a typical American home consumes a daily average of 31KWh per day for 930Wh per month and 11MWh per year. So your system is at least 3X *larger* than necessary !!

http://www.eia.gov/tools/faqs/faq.cfm?id=97&t=3

Reworking the calculations with these numbers would be

(100,000 W*h) / (160W/m^2) = 625 m^2 * h = 25 meters x 25 meters [82.02075 ft x 82.02075 ft]

{which is much more realistic :-) ]

My home uses an average of 500KWh per month in Portland Oregon , but i have gas heating. :-b

Source(s): ALOT of engineering experience.
• If the area is A, then the power generated is 800A Watt. With 20% efficiency that results in 160A Watt. In 24 hours the amount of usable energy then is 24*0.160* A kWh .

So 3.84A = 100

A = 100 /3.84 = 26 m^2 ( approximately a 5m x 5m square)

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