Please answer all parts and show/explain all work. (Sorry, I know there\'s a lot

Question

Please answer all parts and show/explain all work. (Sorry, I know there's a lot of parts to this post. I am running low on questions left to ask in my subscription).

At noon on a nice summer day in Boulder, each 1 cm2 of surface is exposed to perhaps 150 mW (milliWatts) of visible radiation.

a) 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 2 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 2 m x 8 m seem like a reasonable size, by which we mean, would it fit on a typical household roof?

b) If the solar panels were 10% 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 150 mW/ cm2 ). For how many hours could this amount of energy power a 60 W bulb?

c) 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.

d) An average American home uses about 500-1000 kWh of electrical energy each month. Will these panels be sufficient for "average Americans"?

e) The US uses about 4000 TW*hours each year (TW is for terawatt where tera means 10!"). 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

part A. You have an area of 8 m^2. Each m^2 has (100)^2 cm = 10000 cm^2
Thus your total area is 640000 cm^2. At 150 mW per cm^2 you have 640000 cm^2 times .15 W/cm^2. That would yield 9600 W of power.

Part B: 1 Watt is 1 J/s you seek Joules so 9600 J/s times 3600 seconds gives 34560000 J of energy or 34.56 MJ.
==> if you have 10% efficient solar panels the only 10% of the 34.56 MJ is captured, thus 34.56 MJ times .10 is 3456 kJ.

==> We need to convert kJ into Kw-h by dividing by 3600. Thus 3456000 J times 1hr/3600s = 960 W-h. That is 960 W-h/60 W will give you 960/60 hours of time which is 16 hours

part c. 24*365*960 = 8294400watts approximatley

part D. 30*24*960 = 691.2kwats hence suitable aproximately

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