If my school uses 88240kWh of electricity per month and I have 000 50W solar panels running for 6 hours a day, does it mean it will take 88240kWh / {{[(50W x 3600s)*000]/000}kWh x 6} number of hours to generate that much electricity (88240kWh)?
your first question: Does a 50W solar panel generate 540kWh? 540 kW-hr / 50W = 3600 hours it would take that long for the panel to generate that amount of energy. your second question is confused, as you both stipulate the number of hours at 6 hours per day, and you try to solve for the number of hours. 000 x 50w = 50 kW 50 kW x 6 hr/day x 30 day/mo = 27000 kW-hr/mo = 27 MW-hr/mo that is the amount of energy generated. That is nowhere close to 88240 kWh or 88 MW-hr. I would take 7 times more solar panels to generate that much energy. .
running six hours a day doesn't mean much. You need to look at the solar insulation charts for your school's geographical location to come up with a better factor. The easiest number for you to use is sun hours. For example, Washington DC averages 4.23 hours. Do a Yahoo search for sun hours and you should find lots of charts. Solar panels rated at 50W give this output at full sun near noon at full brightness (no clouds). The sun hour factor makes it easy to find the equivalent number of full brightness hours. So, using Washington DC as an example you have: 000 panels * 50W * 4.23 sunhours/day = 634kWh a day on average. You state your school uses 88240kWh/month which is 6274kWh a day. This would mean you need ten times more solar panels since there is no way to get more daylight. Be careful to put in all the units in your formula and cancel them out to make sure you don't end up with a nonsense result. The title of the question would be answered as followed: 50W/000 * 4.23 sunhours/day = 0.63kWh/day or 9kWh per month or 228kWh a year. These are annual averages. If you wanted a specific month, you would need the sun hours for that month. Hope this helps.
