Convert sunlight to electricity in a given area

AI Thread Summary
The discussion revolves around calculating the potential electricity generation from solar photovoltaic cells covering Lake Mead's area. Given the lake's average area of 100,000 acres and sunlight intensity of 250 W/m², the initial calculation assumes 100% efficiency, yielding approximately 1.012 x 10^11 W. However, to account for the 15% efficiency of the solar panels, the correct approach is to multiply by 0.15, resulting in a final output of about 1.51 x 10^10 W. Clarifications were made regarding unit conversions and the proper application of efficiency in calculations. The importance of accurate unit handling and efficiency application in energy calculations was emphasized.
Revicted
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Homework Statement


Question: Lake Mead is formed by the Hoover Dam, and while it actually changes in area and volume, consider it has an average area of 100,000 acres. Assume that 250 W/m^2 of sunlight falls on Lake Mead, how much electricity could be produced if this lake area was covered by solar voltaic cells that convert sunlight directly to electricity with 15% efficiency?

Homework Equations


1 Acre = 4046.85642 Square Meter

The Attempt at a Solution


I am trying to find the amount of electricity that could be produced in this area, so I know that the final answer needs to be in Watts. This means I need to find a way to cancel out the m^2 in a conversion.

Here is the conversion I attempted to cancel out the acres and square meters:
(100,000 acres)*(4046.85642 m^2)*(250W) = 1.012*10^11W

That number is assuming 100% efficiency. So to account for the 15% efficiency of the solar panels:
(1.012*10^11)/.15 = 6.74*10^11W

I don't feel very confident in this answer, but I'm not quite sure what I am missing or what I am doing wrong. Any nudges in the right direction would be greatly appreciated!
 
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You are essentially correct with one minor issue in a middle step: The units in the middle of your computation. The conversion factor should be 1 = 4046 m^2/acre and the power per area should have units of 250 W/m^2. In general you can treat units exactly like you do variables to find out the unit of a product or division. Since 4046 m^2 is 1 acre, 4046 m^2/1 acre = 1 acre / 1 acre = 1 and you can multiply anything with it and get the same thing.
 
Oh yes, sorry, I left that out on here. That is actually how I have it written out in the conversion on my paper in order to cancel out the units. Otherwise everything looks correct?
 
Revicted said:
I am trying to find the amount of electricity that could be prodThat number is assuming 100% efficiency. So to account for the 15% efficiency of the solar panels:
(1.012*10^11)/.15 = 6.74*10^11W

Sorry, I missed this at first read through. The 15% efficiency means you only get 15% of the power out of the system. You therefore need to multiply by 0.15 rather than divide by it. Otherwise it is what I would have done assuming the actual numbers are correct.
 
Makes sense. Not sure why I divided there. Thanks so much!
 
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