Calculating the mass of consumed Uranium 235

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SUMMARY

The discussion focuses on calculating the mass of Uranium-235 (235U) consumed daily in a nuclear fission power plant producing 3.36 GW of electrical power with an efficiency of 44.6%. The initial calculations incorrectly converted energy units, using 1.6E-13 instead of the correct conversion factor of 6.24E12 MeV per Joule. The correct approach involves determining the total energy output, calculating the number of fission reactions, and then converting that to mass consumed per day, leading to a final result of 1.62538E-25 kg/day.

PREREQUISITES
  • Nuclear physics principles, specifically fission reactions
  • Energy conversion between Joules and MeV
  • Basic calculus for rate of reactions
  • Understanding of mass-energy equivalence (E=mc²)
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  • Explore the concept of mass-energy equivalence in detail
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Homework Statement



A typical nuclear fission power plant produces
about 3.36 GW of electrical power. Assume
that the plant has an overall efficiency of
44.6% and that each fission event produces
250 MeV of thermal energy.
Calculate the mass of 235U consumed each
day.
Answer in units of kg/d.


Homework Equations





The Attempt at a Solution



Okay, so first I converted the 3.36 GW in 3.36E^9 W and divded that by .446 To get the Etotal.
Multiply that times 1.6E^-13 to get MeV of 1.205E^-3.
Divide the MeV/# of reactions by 250 MeV to get 4.821E^-6 reactions per second.
Multiplied 4.821E^-6 * the mass of U235 which is 235.043925 and 1.66E^-27 to get 1.881228E^-30 kg consumed every second.

1.881228E^-30 (60)(60)(24) should give kg/ d, but for some reason It's not coming out as a correct answer. I get 1.62538E^-25 kg/day

Can anyone find a mistake in my work?
 
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Well I made it as far as your conversion from Watts=J/s to MeV/s. Why did you multiply it by 1.6E-13?

1 Joule = 6.24E12 MeV

So you should multiply it by 6.24E12.
 

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