Calculating U-235 Requirements for a 707 MW Nuclear Fission Reactor

AI Thread Summary
To determine the kilograms of U-235 required to operate a 707 MW nuclear reactor for one year, the total heat energy needed is calculated to be approximately 22.3 trillion joules. Each fission event of U-235 releases about 2.95E-11 joules, and there are roughly 2.56E24 atoms in one kilogram of U-235. The total energy released per kilogram of U-235 is around 75.5 trillion joules. By dividing the total energy requirement by the energy produced per kilogram, the final amount of U-235 needed can be calculated. This approach simplifies the problem by assuming the reactor only uses U-235 and operates at a set efficiency.
haxxorboi
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Homework Statement


How many kilograms of U-235 would be needed to run a 707 MW reactor for 1 year?


Homework Equations


183.9 MeV released per reaction
7.57E26 reactions per year
(These numbers are from the prior problems)

The Attempt at a Solution


I thought maybe half life, but you'd never get 0kg for the final answer with that.

I can't seem to wrap my mind around this one, I've gotten the other parts but this one is throwing me...

Any help is appreciated
 
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haxxorboi said:

Homework Statement


How many kilograms of U-235 would be needed to run a 707 MW reactor for 1 year?
This is a complicated question to answer if you are working from first principles. I think you should make it simple and assume that the reactor fissions only U235 (no PU239), that it operates at a set efficiency (say 30%) and that the reactor produces 707 MW of electricity for one year. Also assume that the question is really asking how much U235 is consumed.

Set out your information and assumptions:

Total heat energy needed to produce 707 MW of electricity for one year (in Joules).
Heat Energy released in one fission event of U235 (in eV converted to Joules)
No. of U235 atoms in 1 Kg
Total energy released per kg. (in Joules)

You get the idea.

AM
 
haxxorboi said:
How many kilograms of U-235 would be needed to run a 707 MW reactor for 1 year?
…
183.9 MeV released per reaction
7.57E26 reactions per year
(These numbers are from the prior problems)
…
I can't seem to wrap my mind around this one, I've gotten the other parts but this one is throwing me...

Hi haxxorboi! :smile:

Hint: how much energy is 707 MW for 1 year? :wink:
 
Holy crap, THANK YOU!

22.29595E15 joules heat energy per year at 707 MW

2.9464E-11 Joules per event

2.5617E24 atoms per kg

75.478E12 Joules per kg

and then obviously divide to get final answer of kg.

That one was killing me, thanks so much both of you.
 

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