An uranium 92U235 fission reaction produces 216Mev .the amount of uranium required to run a 100Mw nuclear power plant for a day with 100%efficiency is.
Answers
Answered by
1
If a uranium 92U235 fission reaction produces 216MeV, the amount of uranium required to run a 100MW nuclear power plant for a day with 100% efficiency is given as-
1. Let 92U235 be represented by U.
2. We know that, 1eV= 8.89*10^-9 Wsec
=> 216MeV = 216*10^6eV
= 216*10^6 * 8.89*10^-9 Wsec= 1.94Wsec
3. For 1 day, the energy produced by one reaction would be
1.94/86400 = 2.24418*10^-5
4. Thus the amount of U required to run the 100MW = 10*8W plant
= 2.24418*10^-5/10*8 = 4.45597*10^12.
Similar questions