solar power electricity meter
Security question?
Modern solar panels are about 10% efficiency. That is 10% of energy in the panel's Sunlight actually become useful electricity. During the day, around 2.5×10 ^ 27 photons of visible light from the strike of a square kilometer of land every second (1 photon of visible light has an energy of about 4×10 ^ -19 joules). The Watt is a unit of power. 1 Watt is 1 joule per second. A 100 watt bulb light therefore requires 100 joules of energy per second to stay on. Based on this information how big a solar panel (in square meters) is required to provide 350 watts of electricity on a normal day? I have 35 square meters, but poor. What do you get?
Let's see … my way of solving this problem is as follows: the photon conversion to Joules: 2.5 * 10 ^ 27 photons / sq km * 4 * 10 (-19) joules / photon = 1 * 10 ^ 9 Joules/sq- Joules conversion of kilometers to square meters square kilometers, 1 * 10 ^ 9 / (1000 ^ 2) = 1 * 10 ^ 3 / sq-m incorporating the efficiency factor (10%) 1000 Joules / sq-m * .10 = 100 joules / m-m (usable energy) All these calculations are done on a per second basis. Thus, for a 350 watt (July / sec) determine the area 350 (J / s) / 100 (joules / sec (m) = 3.5 meters, so they're off by a factor of 10. I hope this helps.
Meter running backwards
