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ConsolidatedEdison'sIndianPointNo.2reactorisdesignedtooperateatapowerof 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.

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Answer:

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

Step-by-step 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

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