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Collector Size of solar panel

  • Thread starter deenuh20
  • Start date
  • #1
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Homework Statement



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

Homework Equations



Iave=Pave/Area

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.
 

Answers and Replies

  • #2
berkeman
Mentor
56,825
6,793
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?
 
  • #3
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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?
 
  • #4
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*bump for viewing/replies*
 
  • #5
berkeman
Mentor
56,825
6,793
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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