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Suppose India a target of producing by 2020 AD, 2,00,000 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 utilization 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 :Total targeted power `= 2xx10^(5)MW` Total Nuclear power = 10% of `2xx10^(5)MW` `= 2xx10^(4)MW` Energy produced in fission = 200 MeV EFFECIENCY of power plant = 25% `therefore` Energy converted into electrical energy per fission `= (25)/(100)xx200=50 MeV` `= 50xx1.6xx10^(-13)` Joule. Total electrical energy to be produced `= 2xx10^(4)MW = 2xx10^(4)xx10^(6)` Watt `= 2xx10^(10)` Joule/Sec `= 2xx10^(10)xx60xx60xx24xx365` Joule/year No. of FISSIONS in one year `= (2xx10^(10)xx60xx60xx24xx365)/(50xx1.6xx10^(-13))` `=2xx(36xx24xx365)/(8)xx10^(24)` Mass of `6.023xx10^(23)` ATOMS of `U^(235)= 235 gm = 235xx10^(-3)kg` Mass of `(2xx36xx24xx365)/(8)xx10^(24)` atoms `= (235xx10^(-3))/(6.023xx10^(23))xx(2xx36xx24xx365xx20^(24))/(8)` `= 3.08xx10^(4)` Kg. Hence mass of Uranium needed per year `= 3.08xx10^(4) Kg`. |
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