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Suppose India had a target of producing by 2020 AD, 200,200 MW of electric power, ten percent of which was to be obtained from nuclear power plants. Suppose we are given that, on an average, the efficiency of utillization (i.e, conversion to electric energy) of thermal energy produced in a reactor was 25%. How much amount of fissionable uranium would our country need per year by 2020? Take the heat energy per fission of ""^(235)U to be about 200 MeV. |
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Answer» SOLUTION :Power requied from REACTOR = `2 xx 10^(10) W` For 1 year it is `2 xx 10^(10) xx 365.25 xx 86400` Let n kg of uranium is required. From 1 kg the energy released is `= (6.023 xx 10^23 xx 200 xx 10^6 xx 1.6 xx 10^(-19))/(0.235) J` `:.` Required energy = `25/100 xx m xx (6.023 xx 10^(23) xx 200 xx 10^6 xx 1.6 xx 10^(-19))/(0.235) J` `:. m = (2 xx 10^(10) xx 365.25 xx 86400 xx 100 xx 0.235)/(25 xx 6.023 xx 10^(23) xx 200 xx 10^(6) xx 1.6 xx 10^(-19)) = 3.08 xx 10^4 kg`. |
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