At noon on a nice summer day in Boulder, each 1 cm^2 of surface is exposed to pe

Question

At noon on a nice summer day in Boulder, each 1 cm^2 of surface is exposed to perhaps 160 mW (milliwatts) of visible radiation. You are considering adding solar panels to your house and want to see whether they will provide enough energy for your household needs. If your solar panels cover an area of 3 meter by 8 meters (that's about a dozen typical solar panels) how much solar power would be hitting the surface of the panels at noon on a clear day? And, does 3 m Times 8 m seem like a reasonable size, by which we mean, would it fit on a typical household roof? If the solar panels were 12% efficient at converting the solar energy into electrical energy, how much electrical energy would have been harnessed in 1 hour? (Assume that the exposure is pretty steady for that hour at 160 mW/cm^2). For how many hours could this amount of energy power a 60 W bulb? The above problems considered peak solar intensity at noon on a nice summer day. Of course, that doesn't last all day, nor all year. Make some reasonable "guesstimates" to predict how much electrical energy these panels would likely produce over the course of a (Boulder) year. An average American home uses about 500-1000 kWh of electrical energy each month. Will these panels be sufficient for "average Americans"? The US uses about 4000 TW*hours each year (TW is for terawatt where tera means 10^12). How much surface area would need to be covered with comparable solar panels to provide this? Given your numbers, (and considering how Boulder sunshine compares to other US cities) what can you conclude about the role of solar energy in the US' future energy balance?

Explanation / Answer

a)

I = 160 mW per cm2 = 0.160 watt per cm2

A = area = 3 x 8 = 24 m2 = 24 x 100 x 100 cm2 = 2.4 x 105 cm2

Solar Power hitting = P = IA = 0.160 x 2.4 x 105 = 38400 Watt

b)

t = time = 1 h = 3600 sec

efficiency = 0.12

E = energy = 0.12 Pt = 0.12 x 38400 x 3600 = 1.66 x 107 J

d)

t = 1 month = 30 x 24 h

E = 0.12 x 38400 x 30 x 24 = 33177.6 Wh = 33.178 kWh

hence this energy will not be sufficient

e)

P = 0.16 A

t = 1 year = 365 x 24

energy converted = P t = 0.16A x 365 x 24

4000 x 1012 = 0.16A x 365 x 24

A = 2.85 x 1012 cm2

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