Question 30.9: Fuel consumption in a reactor About 200 MeV of energy is rel...
Fuel consumption in a reactor
About 200 MeV of energy is released in the fission of a uranium atom. Let’s use this fact to estimate how much uranium has to undergo fission per unit time to provide 3000 MW of thermal power.
SET UP Each second, we need \mathrm{3000 MJ, or 3000 \times 10^6 J}. Each fission provides 200 MeV. From these numbers, we can find the number of fissions per second. Then, using the mass of a uranium atom, we can find the total mass of U needed per unit time.
SOLVE Each fission provides an amount of energy
200 MeV = (200 MeV)(1.6 \times 10^{-13} J/MeV)
= 3.2 \times 10^{-11} J.
The number of fissions needed per second is
\mathrm{\frac{3000 \times 10^6 J}{3.2 \times 10^{-11} J} = 9.4 \times 10^{19}. }
Each uranium atom has a mass of about
(235) (1.67 \times 10^{-27} kg) = 3.9 \times 10^{-25} kg,
so the mass of uranium needed per second is
(9.4 \times 10^{19})(3.9 \times 10^{-25} kg) = 3.7 \times 10^{-5} kg = 37 mg.
REFLECT The total consumption of uranium per day (86,400 seconds) is
(3.7 \times 10^{-5} kg/s)(86,400 s/d) = 3.2 kg/d.
For comparison, note that the 1000 MW coal-fired power plant we described in Section 16.9 burns 14,000 tons (about 10^7 kg) of coal per day. Combustion of one carbon atom yields about 2 eV of energy, while fission of one uranium nucleus yields 200 MeV, 10^8 times as much.
Practice Problem: In this example, the calculated mass includes only the mass of the fissionable nuclide ^{235}_{92}U. If the reactor fuel has been enriched to 3% by isotope separation of the natural mix of uranium nuclides, what total mass of uranium is required per day? Answer: about 100 kg/d.