If a solar cell has power density of15 mW/cm^2 and a device consumes 10mW of power on average ... continued?
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Answer:
(10mW*24h)/(15mW/cm^2*12h)=1.3... cm^2. The above post about solar cell efficiency doesn't matter. 15 mW/cm^2 is the measured power output of the cell and not the rated value. I work with solar cells that can produce that amount of power, so that is a realistic value.
you need 220 mW of power for a 24 hrs day and you are getting a 170 with 12 hrs of sunlight(you arent really because that 15 is at peak power output level and you only get that for about 3 hrs a day.on a good cell you might get 60-70 % rated power for the other 8-9 hrs a day...so allow for non peak output production Id say you would need somewhere around 400-450 mW output but you didnt state what type of storge capacity you will be using for non output hrs..deep cell batteries for storage allows you to fudge some on the output...its impossible to give you an accurate answer w/o knowing your storage capacity
We know the power density (intensity) j = 15 mW/cm2 and the power P = 10 mW, we also know that j = P / S where S is the surface area that is perpendicular to the direction of the heat power. We solve for S:
S = P / j = 10 mW / 15 mW/cm2 = 0.67 cm2 = 67 mm2
Since there's sunlight only 12 hours, which is the half of the day, the cell has to be twice this large, so S = 1.3 cm2
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