Power Density of Solar Cell Question

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SUMMARY

The power density of solar cells in direct sunlight is approximately 15 mW/cm². To power a sensor node that consumes an average of 10 mW, a solar cell array of at least 28,800 cm² is required, assuming direct sunlight is available for 12 hours daily. This calculation accounts for the total energy consumption over a 24-hour period, factoring in the need for excess power storage for nighttime use. Additionally, to ensure continuous operation, the area may need to be doubled to accommodate energy losses during periods of no sunlight.

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T-Man78
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I have this homework question I can't quite figure out. Any help is greatly appreciated.

In direct sunlight, solar cells have a power density of approximately 15 mW/cm2. Suppose a sensor node consumes on average 10mW of power. How large of a solar cell array is needed to meet this requirement assuming direct sunlight is available for 12 hours each day? (Assume all excess power generated during the day can be stored and used during the night).

My thinking is this: maybe I'm right maybe I'm wrong
15mW/cm2 x Y cm2 = 10mW * 43200s
Y = 28,800 cm2
 
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Shooting from the hip, as I'm not sure what a sensor node is, but I figure its overhead, as it helps to keep the cells aligned. So in designing an array, you need to keep track of losses. In sun then you need enuf area to meets its demands=this is easy, you need 10/15 cm^2. Each cm^2 delivers 15 mW, and you need only 10. Now if it runs all the time, and it only gets lit 1/2 the time, then you have payback during the night. It is still consuming but not generating, in which case you need to figure in twice the area.
 

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