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It is proposed to use nuclear fusion reaction;
2H1 + 2H1 -------> 4He2
in a nuclear reactor 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 2H1 and 4He2 are 2.0141 atomic mass units and 4.0026 atomic mass units respectively)
Mass defect in the given nuclear reaction:
Dm = 2(mass of deuterium) – (mass of helium)
= 2(2.0141) – (4.0026)
= 0.0256
Therefore, energy released
DE = (Dm) (931.48) MeV = 23.85 MeV
= 23.85 × 1.6 × 10–13 J
= 3.82 × 10–12 J
Efficiency is only 25%, therefore,
25% of ?E = (25/100) (3.82 × 10–12) J = 9.55 × 10–13 J
i.e., by the fusion of two deuterium nuclei, 9.55 × 10–13 J energy is available to the nuclear reactor.
Total energy required in one day to run the reactor with a given power of 200 MW:
ETotal = 200 × 106 × 24 × 3600 = 1.728 × 1013 J
\ Total number of deuterium nuclei required for this purpose:
n = Etotal/?E/2 = 2 x 1.728 x 1013 / 9.55 x 1013= 0.326 × 1026
\ Mass of deuterium required
= (Number of gm-moles of deuterium required) × 2 g
= (0.362 x 1016)/(6.02 x 1023)× 2 = 120.26 g.
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