Power Density of Solar Cell Question

[T-Man78]

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

 

[denverdoc]

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.

 

[m2003]

Hello! Thank you for reaching out for help with your homework question. Your thinking is on the right track, but there are a few things that need to be clarified and corrected.

Firstly, the unit for power density is mW/cm2, not just mW. This means that the power density of a solar cell is the amount of power (in milliwatts) that is produced per unit area (in square centimeters). So in this case, the power density of a solar cell in direct sunlight is 15 mW/cm2.

Next, the unit for time is missing in your equation. The question states that the sensor node consumes 10mW of power on average, so the total power consumption over a 12-hour period (43200 seconds) would be 10mW * 43200s = 432000mW*seconds, or 432000mJ.

To determine the size of the solar cell array needed, we need to divide the total power consumption by the power density of the solar cell. This will give us the total area of solar cells needed to produce enough power to meet the sensor node's requirements.

So the correct equation would be: (10mW * 43200s) / 15mW/cm2 = 28800 cm2, or 2.88 square meters.

This means that a solar cell array of at least 2.88 square meters would be needed to meet the sensor node's power requirements, assuming direct sunlight is available for 12 hours each day and all excess power can be stored and used at night.

I hope this helps clarify the question and provide a correct answer. Keep up the good thinking and good luck with your homework!