Mass of Fuel Rods for 760 MW Nuclear Reactor

[kraigandrews]

Homework Statement

The electric power generated by a single average size nuclear reactor block is between 400 and 800 MW. The fission of one 235U nucleus generates 185 MeV energy on average. What is the total mass of the fuel rods loaded into a 760 MW reactor for a one year period, if the fuel rods contain 3.50 percent of 235U? Assume that the overall thermodynamic efficiency of converting heat to electric energy is 33.0 percent.

The Attempt at a Solution

So to start out I multiply the 185MeV by 3.5% then from there I am lost.
Help in any way is appreciated, thanks.

 

[SteamKing]

You are starting at the wrong end of the process. You know the rated power output of the plant in MW. You know the efficiency of the conversion of heat energy into electric power output. You know the reactor is loaded with enough 235U to supply one year of electric power at the rated output of 760 MW. Find out how much energy the fuel rods in the reactor are required to produce for a year. Then apply the energy equivalent of a single fission to determine the amount of 235U, factoring in the ratio of 235U to the rest of the fuel rod material.

 

[kraigandrews]

So i multiplied 760MW by 31536000s to get the Energy output for one year, then do I take the energy of one fission and divide by that and multiply by 0.035 and then 0.33?

 

[SteamKing]

This is basically a unit conversion problem. You are not thinking it through clearly.

 

[paranormal]

I would approach this problem by first converting the given power output of the nuclear reactor (760 MW) into its equivalent energy output. This can be done by using the formula P = E/t, where P is power, E is energy, and t is time. Rearranging the formula to solve for energy, we get E = Pt. Plugging in the values given (760 MW for P and 1 year for t), we get an energy output of 6.29 x 10^15 Joules.

Next, I would use the given efficiency of 33.0% to calculate the total energy input needed to produce 6.29 x 10^15 Joules of electricity. This can be done by dividing the energy output by the efficiency percentage (6.29 x 10^15 J / 0.33 = 1.91 x 10^16 J).

Now, we can use the energy released from the fission of one 235U nucleus (185 MeV) to calculate the total number of 235U nuclei needed to produce 1.91 x 10^16 J of energy. This can be done by converting the energy into joules (1.91 x 10^16 J = 1.91 x 10^19 MeV) and then dividing by the energy released per nucleus (1.91 x 10^19 MeV / 185 MeV = 1.03 x 10^14 nuclei).

Finally, we can use the given percentage of 235U in the fuel rods (3.50%) to calculate the total mass of fuel rods needed. This can be done by multiplying the total number of nuclei by the percentage of 235U (1.03 x 10^14 nuclei x 0.035 = 3.61 x 10^12 nuclei). However, this answer is in terms of nuclei, so we need to convert it to mass by multiplying by the mass of one 235U nucleus (3.61 x 10^12 nuclei x 235 atomic mass units = 8.48 x 10^14 atomic mass units).

In conclusion, the total mass of fuel rods needed for a 760 MW nuclear reactor for one year is approximately 8.48 x 10^14 atomic mass units. This calculation assumes that all of the energy released from the fission of 235U nuclei is converted to electricity,