Question 7: Certain areas in Arizona, Nevada, California and Texas can have up to 4000 sunny hours a year. The fact that these vast lands are less populated and agriculturally not suitable makes them ideal for solar power plants. Though most commercial solar panels have efficiencies from 15% to 20%, researchers have now developed solar cells with efficiencies approaching 50%. If you are asked to plant a solar energy system in these areas using the newest technology, how much land do you need to produce enough energy for the US. Make the conservative assumptions that the average light energy landing to the Earth' surface is 1.0 cal/(cm2 min). An estimate for the US energy consumption by major sources is given below. U.S. primary energy consumption by major sources, 1950-2019 quadrillion British thermal units 110 100 90 80 70 60 50 40 30 20 10 1950 1955 1960 1965 1970 1975 1980 1985 1990 1995 2000 2005 2010 2015 renewables nuclear petroleum natural gas coal

Question 7: Certain areas in Arizona, Nevada, California and Texas can have up to 4000 sunny hours a year. The fact that these vast lands are less populated and agriculturally not suitable makes them ideal for solar power plants. Though most commercial solar panels have efficiencies from 15% to 20%, researchers have now developed solar cells with efficiencies approaching 50%. If you are asked to plant a solar energy system in these areas using the newest technology, how much land do you need to produce enough energy for the US. Make the conservative assumptions that the average light energy landing to the Earth' surface is 1.0 cal/(cm2 min). An estimate for the US energy consumption by major sources is given below. U.S. primary energy consumption by major sources, 1950-2019 quadrillion British thermal units 110 100 90 80 70 60 50 40 30 20 10 1950 1955 1960 1965 1970 1975 1980 1985 1990 1995 2000 2005 2010 2015 renewables nuclear petroleum natural gas coal

College Physics

1st Edition

ISBN:9781938168000

Author:Paul Peter Urone, Roger Hinrichs

Publisher:Paul Peter Urone, Roger Hinrichs

Chapter7: Work, Energy, And Energy Resources

Section: Chapter Questions

Problem 43PE: (a) Calculate the power per square meter reaching Earth's upper atmosphere from the Sun. (Take the...

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(a) Calculate the power per square meter reaching Earth's upper atmosphere from the Sun. (Take the power output of the Sun to be 4.001026 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 km 2 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 (1.051020J) ? Australia's energy needs (5.41018J) ? China's energy needs (6.31019J) ? (These energy consumption values are from 2006.)
At a certain location, the solar power per unit area reaching Earth's surface is 200 W/ m^2, averaged over a 24-hour day. If the average power requirement in your home is 3 kW and you can convert solar power to electric power with 10 % efficiency, how large a collector area will you need to meet all your household energy requirements from solar energy? (Will a collector fit in your yard or on your roof? ).
At a certain location, the solar power per unit area reaching Earth’s surface is 200 W/m2, averaged over a 24-hour day. If the average power requirement in your home is 3 kW and you can convert solar power to electric power with 10% efficiency, how large a collector area will you need to meet all your household energy requirements from solar energy? (Will a collector fit in your yard or on your roof?)
At a certain location , the solar power per unit area reaching Earth’s surface is 200 W/m square , average over a 24 hours day. If the average power requirement in your home 4.0 KW you can covert solar power to electric power with 13% efficiency. How large a collector area will you need to meet all your household energy requirements from solar energy?
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The average electricity consumption of a house in Gainesville is known to be 1,007 kWh in a month (One month = 30 days). They would like to install solar panels of 29 % efficiency to generate this electricity. Given that the average solar power density in Gainesville is 5.47 kWh/m2/day, how much surface area must the panels occupy? Calculate the result in m2 but do not write the unit. Round off %3D you answer to a whole number (zero decimal place.)
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