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 385 kWh of electricity per month, how many square meters of solar cells are required to meet its energy requirements? Assume that electricity can be generated from the sunlight for 8 hours per day.

Video transcript

Hello everyone today. We are being given the following problem. Sunlight has a power density of two kilowatts per meter squared in a sunny location. This can be converted by photovoltaic solar cells into electricity with 23% efficiency, calculate the square meters of solar cells needed to meet the energy requirement of a typical home that uses 467 kilowatts hour of electricity per month. Assume that the electricity can be generated from the sunlight for x hours per day. So we need to convert from different things. We need to go from the power density that's available to the power density generated and that's going to look like going from kilowatts our per meter squared For every 24 hours, two kilowatts hour per meters squared for a day. And so we can start by using the two kilowatts per meter squared, the two kW per one m square. We can then multiply that bar 23% efficiency, which is the same thing as 23 over 100. We can then multiply by the 7. our power generation, Which is going to be given in a 24 hour Period of 24 hour time period. We can then multiply this by the conversion factor that there are 24 hours in one day And that there are days in one month. And with that we get an answer of 103. kilowatt hours generated per meter squared per month. We then must take our 467 kilowatt hours that are needed per month. So we're needed for one month and divide that by the number. We just found our one oh 3.5 kilowatts per meter squared per month to give us 4.51 m squared as our answer for the square meter of solar cells needed to meet the energy requirement. I hope this helped until next time.