Calculate power output of 92235 U reactor, if it takes 30 days to use up 2 kg of fuel, and if each fission gives 185 MeV of useable energy. Avogadro's number =6 × 1023 / mol ?

Question

Calculate power output of 92235 U reactor, if it takes 30 days to use up 2 kg of fuel, and if each fission gives 185 MeV of useable energy. Avogadro's number =6 × 1023 / mol ? (A) 56.3 MW (B) 60.3 MW (C) 58.3 MW (D) 54.3 MW

Tardigrade Admin · Accepted Answer

Mass of 23592U per second in the reactor is m=(2×103/30×24×60×60)=7.72×10-4 g/sec ∴ Number of fissions reaction per second =(6×1023/235)× m=(6×1023×7.72×10-4/235) =1.97×1018/sec Power of nuclear reacto =1.97×1018×185 MeV/s =1.97×1018× 106× 1.6× 10-19 J/s. =58.3 MW.

Q. Calculate power output of 92235​U reactor, if it takes 30 days to use up 2kg of fuel, and if each fission gives 185MeV of useable energy. Avogadro's number =6×1023/mol ?

56.3 MW

60.3 MW

58.3 MW

54.3 MW

Mass of 92235​U per second in the reactor is m=30×24×60×602×103​=7.72×10−4 g/sec ∴ Number of fissions reaction per second =2356×1023​×m=2356×1023×7.72×10−4​ =1.97×1018/sec Power of nuclear reacto =1.97×1018×185 MeV/s =1.97×1018×106×1.6×10−19J/s. =58.3 MW.

Q. Calculate power output of $_{92}^{235} U$ reactor, if it takes 30 days to use up $2 k g$ of fuel, and if each fission gives $185 M e V$ of useable energy. Avogadro's number $= 6 \times 1 0^{23} / m o l$ ?