help The power output of the Sun is about 4.00 times 10^26 W. Calculate the powe

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The power output of the Sun is about 4.00 times 10^26 W. Calculate the power per unit area (intensity) in kilowatts per square meter, reaching Earth's upper atmosphere from the Sun. The radius of the Earth's orbit as 1.5 times 10^11 m. Part of this power is absorbed and reflected by the atmosphere, so that a maximum of 1.30 kW/m^2 reaches Earth's surface. Calculate the area, in square kilometers, of solar energy collectors needed to replace as 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 directly overhead only briefly.) With the same assumptions, what area (again in square kilometers) would be needed to meet the United States' energy needs (which is 1.05 times 10^20 J/ry)? What would this area be (in square kilometers) for Australia's energy needs (5.4 times 10^13 J/yr)? What about China's energy needs. (6.3 times 10^19 J/yr), in square kilometers?

Explanation / Answer

c)

total energy need = 1.05x10^20 J/yr = 3.3295x10^12 J/s

we know solar constant = 1400 watt /m^2

so, US area = 3.3295x10^12/1400 = 2.378x10^9 m^2 = 2378 km^2

d)

total energy need = 5.4x10^18 J/yr = 1.7123x10^11 J/s

we know solar constant = 1400 watt /m^2

so, AUSTRALIA area = 1.7123x10^11/1400 = 1.22309x10^8 m^2 = 122.31 km^2

E)

total energy need = 6.3x10^19 J/yr = 1.9977x10^12 J/s

we know solar constant = 1400 watt /m^2

so, china area = 1.9977x10^12/1400 = 1.42694x10^9 m^2 = 1426.9km^2

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