How to calculate how many kilowatt hours of photovoltaic energy production are produced per day?

1. A 1 KW module is effectively exposed to sunlight for 6 hours and generates 6 kWh of electricity per day, regardless of losses. The loss of independent systems is generally 30%.

2. Taking into account the intensity of solar radiation, 6 hours of effective sunshine, 6*0.7=4.2 kW/h. It produces 4.2 kilowatt hours of electricity per day. Effective daylight hours refer to the intensity of the radiation.

3. Daily solar power output = sunshine duration * total photovoltaic field power * power generation efficiency

4. The main principle of photovoltaic power generation is the photoelectric effect of semiconductors. When a photon irradiates a metal, all of its energy can be absorbed by an electron in the metal. The energy absorbed by the electron is large enough to overcome the metal's internal gravity to do work, escapingpper from the metal surface and become. photoelectrons.

Detailed information:

1. The kilowatt hour is what we usually call the “degree”. ", is the unit of electrical power, symbol: kWh, and the calculation formula is power multiplied by time. Suppose the power of an energy-consuming device is 2,500 watts, i.e. -say its electricity consumption per hour is 2.5 kilowatt hours, or 2.5 kilowatt hours of electricity per hour

The units of work include joules and kilowatt hours, and the relationship between them is. the following:

1 joule = 1 watt × second

1 kilowatt hour = 1 KW =1kW×h=1000W×h=1000W×3600s=3600000J;

2. Mitigation efficiency

After the crystalline silicon photovoltaic modules are installed and exposed to sunlight for 50. at 100 days, the effectivenesscited will ease by approximately 2-3%. , mitigation will slow significantly and stabilize at an annual mitigation of 0.5 to 0.8%, mitigated by about 20% in 20 years. Monocrystalline components have less attenuation than polycrystalline components. The attenuation of amorphous optical components is approximately lower than that of crystalline silicon.

Improving the conversion rate and reducing the cost per watt will therefore remain the two major themes for the future development of photovoltaics. Regardless of the method used, if large-scale application can increase the conversion rate to 30% and the cost is less than 5,000 yuan per kilowatt (the same as hydropower), then humanity will have the largest and cleanest energy production by nuclear fusion. before the search succeeds, the cheapest and almost unlimited new reliable energy.

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