Nuclear fission and amount of U-235 used

[FinSanity]

[SOLVED] Nuclear fission and amount of U-235 used

Homework Statement

At the fission of a U-235 atom about 200 MeV of energy is released. How many fissions occur in 1 second in an power plant which has the heat energy capacity of 2050 MW? How much U-235 is consumed in one year (330 days)?

Homework Equations

U-235 -> 8,2 * 10^13 J/kg
q = 200MeV
Ø = 2050 MW
Elementary charge = 1,602*10^-19 C
kWH -> J 1 kWH = 3,6 * 10^6 J

The Attempt at a Solution

Ø = n' * q => n' = Ø / q = 2050 * 10^6 / ((200*10^6) * (1,602*10^-19) =
6,4*10^19 fissions / second

2050000 * 3,6 = 7,38 * 10^12 J/h

7,38 *10^12 J/h * 24 * 330 = 5,85 * 10^16 J/year

(5,85 * 10^16) / (8,2 * 10^13) = 713,4 kg/year

Are these correct? I'm not sure if the way I calculated the consumption of U-235 per year is correct.

 

[dynamicsolo]

The Attempt at a Solution

Ø = n' * q => n' = Ø / q = 2050 * 10^6 / ((200*10^6) * (1,602*10^-19) =
6,4*10^19 fissions / second

I agree with this result.

2050000 * 3,6 = 7,38 * 10^12 J/h

7,38 *10^12 J/h * 24 * 330 = 5,85 * 10^16 J/year

(5,85 * 10^16) / (8,2 * 10^13) = 713,4 kg/year

I think you dropped a few zeroes in typing your 2,050,000,000 W (2050 MW), but the figure on that line is right. But on the next line, why are there 330 days in a year? (EDIT: Sorry, I missed the specification in the problem.)

Another way to get to the answer is this:

You found that there are 6.4 · 10^19 fissions per second. Therefore, that many U-235 nuclei are fissioning in one second; so there are (6.4 · 10^19)(3600)(24)(330) = 1.82 · 10^27 nuclei fissioning per year.

This is (2.02 · 10^27)/(6.02 · 10^23) = 3030 moles of U-235 consumed in a year, which is 3030 · 235 = 712,300 gm or 712.3 kg of U-235 per year.