(adsbygoogle = window.adsbygoogle || []).push({}); 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

Iave=Pave/Area

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.

**Physics Forums - The Fusion of Science and Community**

# Collector Size of solar panel

Have something to add?

- Similar discussions for: Collector Size of solar panel

Loading...

**Physics Forums - The Fusion of Science and Community**