Question

According to the Department of Energy (DOE), 99 U.S. reactors have generated 805 billion kilowatt hours of electricity in 2016. That is 20% of the U.S. energy generated that year and enough to power 73 million homes. How many tonnes (metric tons) of natural uranium have the U.S. needed that year? In your calculations, ignore the energy gained from the fissioning of Pu239 produced from the fertile U238 (i.e., assume all the energy generated from the fissile U235). Also ignore the efficiency of heat to electricity conversion (about 30%). Recall that the percentage of U235 in natural uranium is 0.7%. (Hint: first determine how many atoms of U235 must have undergone fission to produce the above mentioned energy given that each fission generates 200 MeV, then convert that into mass)

Answer 1

Given

Energy from the fission of one atom E = 200 MeV

Energy produced in 2006 U = 805 x 10⁹ kWh

U235 percentage in natural uranium = 0.7%

Solution

Energy from the fission of one atom E = 200 MeV

E = 200 x 10⁶ eV

E = 200 x 10⁶ x 1.6 x 10^-19

E = 3.2 x 10^-11 J

Energy produced in 2006 U = 805 x 10⁹ kWh

U = 805 x 10⁹ x 3.6 x 10⁶ J

U = 2.898 x 10¹⁸ J

Number of U235 atoms needed

N = U/E

N = 2.898 x 10¹⁸/3.2 x 10^-11

N = 9.05625 x 10²⁸

Atomic mass of U235 is 235.0439299 u

m = 235.0439299 u

m = 235.0439299 x 1.66053873 x 10^-27

m = 390.299549 x 10^-27 Kg

Mass of U235 needed

M = Nm

M = 9.05625 x 10²⁸ x 390.299549 x 10^-27

M = 35346.5 Kg

This is only accounts for the 0.7 percent of the U natural uranium

So the mass of natural uranium we need

M_natural = M x 100 / 0.7

M_natural = 35346.5 x 100 / 0.7

M_natural = 5049500 Kg

M_natural = 5049500 / 1000 tonne

M_natural = 5049.5 tonne

The US would have needed 5049.5 tonnes (metric tons) of natural uranium in 2016