The Sun is the most important source of energy for life on Earth.
The other significant source of Earth’s energy is the store of fissionable materials trapped in the Earth`s crust. These fissionable materials give rise to geothermal energy, which drives the volcanism on Earth and also makes it possible to fuel nuclear reactors.
The diameter of the Sun is about 109 times that of Earth, and the mass is about 330,000 times that of Earth. About 75% of the Sun’s mass consists of hydrogen; the rest is mostly helium, with small quantities of elements such as oxygen, carbon, neon and iron.
The mean distance of the Sun’s center to Earth’s center is approximately 1 astronomical unit, about 150,000,000 km. This distance varies as Earth moves from perihelion in January to aphelion in July.
At this average distance, light travels from the Sun’s horizon to Earth’s horizon in about 8 minutes and 19 seconds. The energy of this sunlight supports almost all life on Earth by photosynthesis and drives Earth’s climate and weather
The amount of power that the Sun deposits per unit area that is directly exposed to sunlight is approximately 1368 W/m2 at a distance of one astronomical unit (AU) which is the distance between the sun an the Earth.
Since the sunlight have to pass through the Earth`s atmosphere it is attenuated such that the power reaching the surface of the Earth is around 1000 W/m2 provided clear conditions and the Sun being near to zenith (“directly above”).
The sunlight, before it reaches the Earth`s atmosphere is composed of around 50% infrared light, 40% visible light, and 10% ultraviolet light. Passing through the Eart`s atmosphere more than 70% of the ultraviolet light is filtered out.
The surface temperature of the sun is around 5500 oC.
Nuclear fusion processes, transforming hydrogen into helium, primarily generate the heat. Through these processes the sun´s mass is reduced by 4,3 million tons/second which is converted into radiation.
Example – Solar panels
We want to find out what area of solar panels (photovoltaic cells) that would be needed to provide 50 kWh of electric power per day.
Assuming standard conditions (STC) i.e.:
- 25 °C
- Irradiance (G) of 1000 W/m2
- Air mass 1.5 (AM1.5) spectrum.
This corresponds to a clear day with sunlight incident upon a sun-facing 37°-tilted surface with the sun at an angle of 41.81° above the horizon.
Furhter we assume that our photovoltaic cell can convert 20% of the sunlight into electrical power and that the actual conditions equals the above conditions for an average of 6 hours/day.
To deliver 50 kWh (kilo-watt-hours) per day we would need:
50000 Wh/1000 W/m2 x 0,2 x 6h = 41,7 m2 of solar panels.
More information on photovoltaic energy generation you will find here: Photovoltaic energy generation