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badaspie
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BEST ANSWER  chosen by asker   |  badaspie  |  June 26, 2009 09:25 AM
The sun's energy output can be calculated from the solar constant (incoming solar radiation per unit area at a distance of 1 AU), which averages 1.366 kW/m^2, or 1.74 x 10^14 kW for the entire earth. Given the known diameters of the sun and the earth, and the angular diameter of the sun as seen from the earth, the solid angle of the earth as seen from the sun can be calculated. This gives the fraction of the sun's sky that the earth covers, which equals the fraction of the sun's radiation that the earth intercepts (about one two-billionth), so the sun's total energy output is about 3.9 x 10^23 kW. One watt equals one joule (one newton-meter) per second, so the sun emits energy at a rate of 1.4 x 10^30 joules per hour.
tags: sun, energy
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davepamn
davepamn  |  June 26, 2009 02:03 PM
How much area would be required to capture 90,000 megawatts of electricity

Assume you have a solar panel could capture 60 percent of the radiant energy hitting the earth.
davepamn
davepamn  |  June 26, 2009 02:13 PM
Goal
90,000,000,000.00 watts of solar energy
Given = 1.366 kW/m^2
90X10^9/1.336=65,885,797,950.22 square meters
Assume 4,046 square meters per acre
65,885,797,950.22/4046 = 16,284,181.40 acres
ratio for the inefficiency of the solar cell
1.4 factor
22,797,853.96 acres could provide 90,000 megawatts

You need an area the size of Nevada.
papyrus
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papyrus  |  June 26, 2009 05:14 AM
The rate at which the Sun produces energy is equal to the rate at which it emits energy from its surface (its luminosity), which is around 3.8 x 1026 Watts -- this number can be determined from measurements of how bright the Sun appears from Earth as well as its distance from us.
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