Power Density of Solar Cell Question

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In direct sunlight, solar cells provide a power density of about 15 mW/cm2. To power a sensor node that consumes 10 mW, a solar cell area of at least 0.67 cm2 is required, assuming continuous sunlight. However, considering that sunlight is available for only 12 hours, the total energy needed for 24 hours is 864,000 mJ, necessitating a larger array. The calculation suggests that an area of approximately 28,800 cm2 may be needed to account for energy storage for nighttime use. Therefore, both the size of the solar array and the efficiency of energy storage are critical for meeting the sensor node's power requirements.
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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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