Henry Dover answers:

Michelle, John is quite correct in his math. The only problem with the problem as it’s stated is that the average solar panel is closer to 11% efficient. There are some leading edge ones that claim to have reached as high as 18%, I don’t know if they have been independently tested yet, but most run between 10 and 13 %. So about 11 square meters should do it based on your weather forecast.

We actually live in a solar powered home. Our panels are advertised at 12% efficiency, they are multicrystaline. Our entire array takes up 155 square feet, or about 15 square meters. Our home uses about 300 kwh per month, and in the summer, we usually have a little excess, meaning by 2 pm on sunny days, our batteries are full and can’t hold any more charge, so for an hour or two, the panels only power the small loads in the house, and the rest is wasted. This is necessary though, so that after a few cloudy days, when the sun comes out, there is extra to bring the batteries back up as well as run the home.

Are you working on a homework assignment, or just doing research? There are some better places to get info on this subject than asking hacks like me online. Solar Energy International would be my first pick, and some of the renewable energy associations too. I will put some names and links below for you. Also, there is a great periodical that covers this stuff right down to the nuts and bolts, Home Power Magazine. It is actually where we got started 11 years ago, now we have made it here. Check out some of the websites and do some of your own research, you’ll be well informed in the end. Good luck, and take care, Rudydoo

Henry Dover answers:

Let’s say you have 1 square meter of panels. Then you get 8 hours x 1 kW/m^2 x .14 = 1.12 kWh of energy per day.

If you have 2 square meters, you get 2.24 kWh’s per day.

If you have N square meters, it’s 1.12N kWh’s per day.

Scale that up to a month (use 30 days for a month unless advised otherwise), and you should be able to solve for N.

Sanity check: the answer should be between 5 and 20 square meters.

Henry Dover answers:

About a million

Henry Dover answers:

If the sun shines for 8 hours a day and assuming there are 30 days in each month, then in a month the sun shines for 8×30 = 240 hours

If a typical family home uses 380kW-h per month, then the power they actually use per month = 380kW-h /240hr = 1.5833 kW

Sunlight has power density of 1 kW/m^2 & photovoltaic cells can convert this to electricity with 16% efficiency. This means that one actually gets 0.16 x 1 kW/m^2 = 0.16 kW/m^2 from the cells

so the square meters of cells needed to meet energy requirement of home = (1.5833 kW / 0.16 kW/m^2) = 9.896 m^2

Henry Dover answers:

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