In a sunny location, sunlight has a power density of about 1 kW/m2. Photovoltaic solar cells can convert this power into electricity with 15% efficiency.
If a typical home uses 390 kWh of electricity per month, how many square meters of solar cells would be required to meet its energy requirements? Assume that electricity can be generated from the sunlight for 8 hours per day.
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{ 2 comments… read them below or add one }
I am bad at math so I hope someone else double checks this I just really wanted to see if I could solve it
so you have 1kw/m2 * 8 hours a day
so 8kwh/m2 * 15% efficiency
1.2kwh/m2 per day with 30 days in a month
36 kwh/m2 per month
so 390 kWh / 36 kwh/m2 = 10.8333
I’m not sure if the month is a part of it or not, if not I got 325 m2 which seems awfully high but maybe? I really hope someone with a talent for math does this.
So let me get this straight BillRussell I say that I’m not good at math, I ask someone else to double check this, and even though I did this correctly, you still give me a thumbs down? Wow man you must have a serious hard on for YA! points. I even gave you a thumbs up, because this is clearly your subject.
First of all, 390 kW-hour / month x 1 month/30 days x 1 day/24 hours = 540 watts average
That is quite a bit below average which is about twice that.
1 kW/m² x 0.15 = 150 watts/m²
540 watts / 150 watts/m² = 3.6 m²
but that is for 24 hours of sun a day. Multiply by 3 (24/8) to correct for that and you get
10.8 m²
.