If the average daily solar energy flux is 450 W/m2 , and the efficiency of the s

Question

If the average daily solar energy flux is 450 W/m2 , and the efficiency of the solar collectors is 10%, how many miles of land would be required to displace all of coal that was produced in the U.S. in 2011, which was roughly 22.2 Quadrillion BTUs? Assume that the solar collectors can only produce electricity for 12 hours per day but can power batteries to provide energy during the other 12 hours with no additional heat loss from charging or discharging the battery. In other words, the solar panels must displace all daily coal production with the energy that they can generate in the 12 hours of full daylight. Also note that 1 km2 = 1,000,000 m 2 .

Explanation / Answer

1 quadrillion BTU=293 billion kWh=293*10^9*1000*3600 J=1.0548*10^18 J

solar energy consumed per day=24*450=10800 J/m^2

solar energy converted to electrical energy per day=0.1*10800=1080 J/m^2

let area required is A m^2.

then 1080 * A=1.0548*10^18

==> A=9.76*10^14 m^2

now, 1 mile=1609.34 m

==> 1 mile^2=2.5899*10^6 m^2

hence A in mile^2=376.83*10^6 mile^2

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