Question

It is proposed to use the nuclear fusion reaction
${}_{1}^{2}H + {}_{1}^{2}H \to {}_{2}^{4}{He}$
in a nuclear reactor of 200 MW rating. If the energy from the above reaction is used with a 25% efficiency in the reactor, how many grams of deuterium fuel will be needed per day?
(The masses of ${}_{1}^{2}H and {}_{2}^{4}{He}$ are 2.0141 atomic mass unit and 4.0026 atomic mass unit (u) respectively.)

Solution

The given nuclear fusion reaction is,

{}_{1}^{2}H + {}_{1}^{2}H \to {}_{2}^{4}{He}

The 'mass defect' for this reaction is,

∆ m = (2 \times 2.0141 - 4.0026) u = (4.0282 - 4.0026) u = 0.0256 u

Therefore,
Energy released is,

∆ E' = 0.0256 \times 931 MeV = 23.8336 MeV = 23.8336 \times 1.6 \times 10^{- 13} J = 38.134 \times 10^{- 13} J

Since the reaction has a utilization efficiency of 25%, the energy utilized per reaction is,

∆ E = 38.134 \times 10^{- 13} \times \frac{25}{100} J = 9.533 \times 10^{- 13} J

The reactor rating is 200 MW.

Hence, the total energy required per day is,

E = 200 \times 10^{6} \times 24 \times 60 \times 60 s

= 172.8 \times 10^{11} J

Now, 2 deuterium atoms provide us with an 'available energy' of 9.533 x 10^-13J.

Hence, the number of deuterium atoms needed per day is,

n = \frac{E}{∆ E} = \frac{172.8 \times 10^{11} \times 2}{9.533 \times 10^{- 13}} = 36.25 \times 10^{24}

Now, 1 mole

= 6.02 \times 10^{23} atoms

of deuterium has a mass of 2.0141 g.

Thus, the mass of deuterium needed per day is,

m = \frac{2.0141 \times 36.25}{6.02 \times 10^{23}} \times 10^{24} g = 121.3 g .

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