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How much uranium235 does a nuclear power generator consume to generate 1.5 GW? 
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#1
Dec309, 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 uranium235 does it consume in a year? ANSWER m(^{235}U)=kg 2. Relevant equations E=mc^{2} 3. The attempt at a solution E=mc^{2} 1.5*10^{9}J*60sec*60min*24hr*365days=m*(3*10^{8}m/s)^{2} Solving for m leaves me with .53 kg which come up incorrect in Mastering Physics. 


#2
Dec309, 05:57 AM

P: 477

Hi there,
0.53kg is the mass transformed into energy, not the amount of U235 needed to sustain a 1.5GW nuclear fission reaction. To find out how much U235 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 U235 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 U235. This calculation is a bit simplist but it would give a gross estimate of what is going on in a nuclear reactor. Cheers 


#3
Dec309, 05:57 AM

Sci Advisor
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.



#4
Dec309, 06:04 AM

Admin
P: 21,810

How much uranium235 does a nuclear power generator consume to generate 1.5 GW?
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 U235 produces ~200205 MeV/fission. (This is fine if one does not consider the contribution of Pu239/Pu240/Pu241 which builds up slowly during operation in commercial reactor.) 


#5
Dec309, 04:21 PM

P: 2

So total energy=1.5e9*60*60*24*365=4.73e16 J
Fission of one atom of u235=3.244e11 J Fissions needed: total energy/fission of one atom=1.46e27 weight of one u235 atom:3.9e25 kg weight of u235 used: 1.46e27*3.9e25 kg= 569.4 kg is this correct? 


#6
Dec309, 04:37 PM

Admin
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 U235 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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