Question

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

Answer 1

Total energy used in nuclear reactor per day $E=Pt$
$= \left(200 \times10^{6}\right) \times\left(24 \times3600\right)$
$= 1.72 \times10^{12}J$
Energy used in reactor per reaction
$= \frac{25}{100}\left(2 \times2.0141 -4.0026\right) \times931MeV$
$= 5.9584 MeV$
$= 9.5334 \times10^{-13} J$
The mass of deuterium used per reaction
$=2 \times2.0141 =4.0282$ amu
$= \frac{4.0282}{6.02 \times10^{23}} g =0.6691 \times10^{-23}g$
Thus mass of deuterium required to produce $1.72 \times 10^{12}J$ of energy
$= \frac{\left(0.6691 \times10^{-23}\right) \times\left(1.72 \times10^{12}\right)}{9.5334 \times10^{-13}}$
$ =120g$