ConsolidatedEdison'sIndianPointNo.2 2,758 MW. Assuming that all fissions occur in 235U, calculate in grams per day the rate at which 235U is (a) fissioned, (b) consumed.

a) the rate at which 235U is fissioned is 2895.9 grams per day

b) the rate at which 235U is consumed is 3385.3071 gram/day

Explanation:

Given the data in the question;

a)

designed operation thermal power = 2758 MW.

we know that, the burn up rate fission rate of 235U is 1.05 grams per MW-day.

so, the rate at which 235U is fissioned will be;

⇒ 2758 × 1.05 = 2895.9 grams per day

Therefore, the rate at which 235U is fissioned is 2895.9 grams per day

b)

Consumption rate is given as;

Cr = 1.05 × ( 1 + ∝ )P gram/day

where ∝ is 0.169 for U-235

so,

Cr = 1.05 × ( 1 + 0.169 )2758 gram/day

Cr = 1.05 × 1.169 × 2758 gram/day

Cr = 3385.3071 gram/day

Therefore, the rate at which 235U is consumed is 3385.3071 gram/day