(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Current solar-panel technology has the efficiency of converting 12% of the solar energy to electricity. Given that on a sunny day in Arizona the average solar radiation energy flux is 1.0 kW/m^2 , and that the area of Arizona is 3.0 × 10^5 km^2 ,

1. What percentage of that area needs to be covered with solar panels in order to supply all of the electrical requirements of the United States, estimated at 5 × 10^20 J/yr? (Assume that each day has 12 hours of cloudless daylight, and round your answer to the nearest percent):

2. At the Niagara Falls water is falling an average of 52 m at a rate of r =dm/dt = 1.8 × 10^6 kg/s. If 50% of that potential energy could be converted into electricity, what is the solar-panel area equivalent to the Niagara Falls in km^2 ? (round to one decimal place):

2. Relevant equations

3. The attempt at a solution

What I did:

We get 1000 W/m^2 per second of input. As we run it only half the time, and the efficiency is 0.12, we get a net power of 0.12*500 W/m^2 = 60W/m^2.

A year has 365*86400 seconds, which are roughly 300*10^5 = 3*10^7 seconds. So the power needed by the US is

5*10^20 J/yr = 5*10^20 / (3*10^7) J/s = 1.66*10^13 W.

So

166*10^11 W / 60 W/m^2 =approx 3*10^11 m^2 = 3*10^5 km^2

Is it right? How do I convert to % after this?

2. Per second, m*g*h = 1.8*10^6 * 9.81 * 52 * 0.5 Joule =approx 450 MJ (megajoule),

Area=

450 MW / 60W/m² = 450/60 km² = 7.5km²

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# Solar energy cell question

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