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How much uranium-235 does a nuclear power generator consume to generate 1.5 GW?

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swilson31
#1
Dec3-09, 03:30 AM
P: 2
SOLVED
thank you for the help


1. The problem statement, all variables and given/known data
The total thermal power generated in a nuclear power reactor is 1.5 GW.
How much uranium-235 does it consume in a year?

ANSWER
m(235U)=-----kg

2. Relevant equations
E=mc2


3. The attempt at a solution
E=mc2
1.5*109J*60sec*60min*24hr*365days=m*(3*108m/s)2

Solving for m leaves me with .53 kg which come up incorrect in Mastering Physics.
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fatra2
#2
Dec3-09, 05:57 AM
P: 477
Hi there,

0.53kg is the mass transformed into energy, not the amount of U-235 needed to sustain a 1.5GW nuclear fission reaction.

To find out how much U-235 you need, you must get the total amount of energy (which you seem to have correct in your equation). Then, how energy is liberated in each fission of U-235 atom can help you find out how many reactions are needed to sustain this power output. Having that number, you can evaluate the number of matter (moles) needed, and then the mass of U-235.

This calculation is a bit simplist but it would give a gross estimate of what is going on in a nuclear reactor.

Cheers
Matterwave
#3
Dec3-09, 05:57 AM
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P: 2,663
A fission reaction does not annihilate all the matter, only a small fraction of the mass of the uranium 235 will be converted into energy. You need to know the starting mass and the mass of the products.

Astronuc
#4
Dec3-09, 06:04 AM
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P: 21,810
How much uranium-235 does a nuclear power generator consume to generate 1.5 GW?

Quote Quote by swilson31 View Post
1. The problem statement, all variables and given/known data
The total thermal power generated in a nuclear power reactor is 1.5 GW.
How much uranium-235 does it consume in a year?

ANSWER
m(235U)=-----kg

2. Relevant equations
E=mc2


3. The attempt at a solution
E=mc2
1.5*109J*60sec*60min*24hr*365days=m*(3*108m/s)2

Solving for m leaves me with .53 kg which come up incorrect in Mastering Physics.
The approach is correct, but some steps are missing.

One must determine the energy E used in a year. E = Power (average) * time, so J = W * s.

Then one must realize the energy per fission, fission consumes 1 atom and the mass of 1 atom. Fission of U-235 produces ~200-205 MeV/fission. (This is fine if one does not consider the contribution of Pu-239/Pu-240/Pu-241 which builds up slowly during operation in commercial reactor.)
swilson31
#5
Dec3-09, 04:21 PM
P: 2
So total energy=1.5e9*60*60*24*365=4.73e16 J
Fission of one atom of u-235=3.244e-11 J
Fissions needed: total energy/fission of one atom=1.46e27
weight of one u-235 atom:3.9e-25 kg
weight of u-235 used: 1.46e27*3.9e-25 kg= 569.4 kg

is this correct?
Astronuc
#6
Dec3-09, 04:37 PM
Admin
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P: 21,810
Method is correct, and result seems to be correct.

This of course assumes that the 1.5 GW is thermal energy, which would be a small reactor.

If 1.5 GW is electrical energy, and the process is about 33% efficient, then the thermal energy would be about 4.5 GW, and the amount of U-235 would be 3 * 569 kg.

A large 3.5 GWt reactor has a core size of about 100 MT or 100,000 kg of fuel.


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