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Nuclear fission and amount of U-235 used

  1. Nov 6, 2007 #1
    [SOLVED] Nuclear fission and amount of U-235 used

    1. The problem statement, all variables and given/known data

    At the fission of a U-235 atom about 200 MeV of energy is released. How many fissions occur in 1 second in an power plant which has the heat energy capacity of 2050 MW? How much U-235 is consumed in one year (330 days)?


    2. Relevant equations

    U-235 -> 8,2 * 10^13 J/kg
    q = 200MeV
    Ø = 2050 MW
    Elementary charge = 1,602*10^-19 C
    kWH -> J 1 kWH = 3,6 * 10^6 J


    3. The attempt at a solution

    Ø = n' * q => n' = Ø / q = 2050 * 10^6 / ((200*10^6) * (1,602*10^-19) =
    6,4*10^19 fissions / second

    2050000 * 3,6 = 7,38 * 10^12 J/h

    7,38 *10^12 J/h * 24 * 330 = 5,85 * 10^16 J/year

    (5,85 * 10^16) / (8,2 * 10^13) = 713,4 kg/year

    Are these correct? I'm not sure if the way I calculated the consumption of U-235 per year is correct.
     
  2. jcsd
  3. Nov 6, 2007 #2

    dynamicsolo

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    Homework Helper

    I agree with this result.

    I think you dropped a few zeroes in typing your 2,050,000,000 W (2050 MW), but the figure on that line is right. But on the next line, why are there 330 days in a year? (EDIT: Sorry, I missed the specification in the problem.)

    Another way to get to the answer is this:

    You found that there are 6.4 · 10^19 fissions per second. Therefore, that many U-235 nuclei are fissioning in one second; so there are (6.4 · 10^19)(3600)(24)(330) = 1.82 · 10^27 nuclei fissioning per year.

    This is (2.02 · 10^27)/(6.02 · 10^23) = 3030 moles of U-235 consumed in a year, which is 3030 · 235 = 712,300 gm or 712.3 kg of U-235 per year.
     
    Last edited: Nov 6, 2007
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