Calculating Initial Mass for Nuclear Reactor Operation

[Rath123]

Homework Statement

What initial mass of 23592U is required to operate a 350 MW reactor for 3 yrs? Assume 46% efficiency.

Homework Equations

I used Mass* C* efficiency as a decimal= power (e^-6) * time (in seconds)
and got 800 as the mass, however, this was incorrect

The Attempt at a Solution

(M)*(3e8)^2* .46= 350e6* 9.14 i think it was (whatever 3 years in seconds is) and i was getting 800 as the mass, or 799Kg specifically however, this is incorrect . Where am i going wrong?

 

[Rath123]

the three years in seconds was : 9.461e+7

 

[SteamKing]

the three years in seconds was : 9.461e+7

Your calculations are a scramble.

You should try to lay them out in some logical fashion, so others can follow.

In a nuclear reactor, only a small amount of the mass of uranium is converted to energy. The rest winds up as nuclear waste.

The fission of U-235 in the reactor is governed by the reaction discussed in this article:

https://en.wikipedia.org/wiki/Uranium-235

 

[andrevdh]

These processes usually involve radioactive decay: http://www.nuclear-power.net/nuclear-power/reactor-physics/atomic-nuclear-physics/radioactive-decay
although I am not sure what the efficiency factor entails. I would think that it means that only 46% of the nuclear energy becomes available as electric energy due to the conversion process.

 

[SteamKing]

These processes usually involve radioactive decay: http://www.nuclear-power.net/nuclear-power/reactor-physics/atomic-nuclear-physics/radioactive-decay
although I am not sure what the efficiency factor entails.

I think the efficiency is for the conversion of the energy released by the fission into the electric power which comes out of the plant. There is a limit to the amount of energy which can be extracted from the steam turbines, generators, etc. which are all used with the nuclear reactor to turn the energy of fission into electricity.

 

[andrevdh]

Yes. I realized it and edited my post.