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Collector Size of solar panel

  1. Mar 2, 2007 #1
    1. The problem statement, all variables and given/known data

    Your house requires 2000 kWh of energy per month and you want to power it using sunlight which has an average daylight intensity of 1 kW/m2. Assuming that sunlight is available 8 hours per day, 25 days per month (only 5 – 6 cloudy days per months!) and that you have a way to store energy when the sun isn’t shining, calculate the smallest collector size you would need if its conversion efficiency were R=28 % ___m^2

    2. Relevant equations


    3. The attempt at a solution

    First I found how many seconds that light was present in the month by dimensional analysis of the 25 days/month and the 8 hours light/day. I got 720,000 sec. Then, I converted 2000kWh to W/s and got 555.556 W/s. Then I multiplied 720,000s by 555.556 W/s and got 4*10^8 W as the power. Seeing that it was 28% conversion efficiency, I multiplied the power by .28 and got 1.12*10^8 W. Then, to find area, I used the above equation:

    1kW/m^2=1.12*10^8/Area, and solved for area, which was 112000m^2, but it was the wrong answer.
  2. jcsd
  3. Mar 2, 2007 #2


    User Avatar

    Staff: Mentor

    There is a problem with your units, and thus the calculation. You cannot convert kWhr to W/s, that makes no sense. kWhr is a measure of energy, and W is a measure of Energy/time.

    I like to carry units along in my calculations to be sure that I don't do the wrong multiplication or division. Try doing the calcs again, and carry the units along just like they were variables (I like to carry them along in square brackets [] to show that they are units).

    What do you get if you're careful to keep the units consistent for each term and on both sides of each equation?
  4. Mar 3, 2007 #3
    Keeping units in check, here's what i did:

    1. Covert 2000kWh to kWs = 7.2x10^9 Joules
    2. Because of efficiency, 7.2x10^9 J * .28 = 2.016x10^9 J
    3. Converted days to seconds: (25d)(8h/d)(3600s/h)=72000s
    4. Power= 2.016x10^9 J/72000s= 2800J/s= 2800 W
    5. Pave/Iave= Area so 2800 W/(10^3 W/m^2)=2.8m^2

    However, this isn't the correct answer. Did I approach it correctly?
  5. Mar 4, 2007 #4
    *bump for viewing/replies*
  6. Mar 5, 2007 #5


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    Staff: Mentor

    I think maybe step #2 was wrong. If your conversion efficiency is 0.28, and you want 2000kWhr output energy, how much input energy do you need?
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