(a) Calculate the power per square meter reaching Earth’s upper atmosphere from the Sun. (Take the power output of the Sun to be 4.00×1026 W.) (b) Part of this is absorbed and reflected by the atmosphere, so that a maximum of 1.30 kW/m2 reaches Earth’s surface. Calculate the area in km2 of solar energy collectors needed to replace an electric power plant that generates 750 MW if the collectors convert an average of 2.00% of the maximum power into electricity. (This small conversion efficiency is due to the devices themselves, and the fact that the sun is directly overhead only briefly.) With the same assumptions, what area would be needed to meet the United States’ energy needs (1.05×1020 J)? Australia’s energy needs (5.4×1018 J)? China’s energy needs (6.3×1019 J)? (These energy consumption values are from 2006.)
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In part (a) we need to calculate the power per unit area that reaches earth’s upper atmosphere from the sun, using the power out of the sun as .
In this case, the power per unit area that reaches earth’s upper atmosphere from the sun is nothing but the intensity of solar energy that reaches upper atmosphere of earth. It is given by,
Taking power of source as power output of sun () ,The surface area is the area of the sphere that is formed by the upper atmosphere and is given by , where is the radius of sphere formed by the upper atmosphere and it is given by .