Question

Consolidated Edison's Indian Point No. 2 reactor is designed to operate at a power of 2,758 MW (thermal). Assuming that all fissions occur in 235U, calculate in grams per day the rate at which 235U is

(a) fissioned,

(b) consumed.

what is the total accumulated activity of the fission products in the Indian Point No. 2 reactor I day after shutdown following 1 year of operation?

Answer 1

a) The rate at which ²³⁵U is fissioned is 2896 g/day

b) The rate at which ²³⁵U is consumed is 3385.31 g/day

Step-by-step explanation:

a) When ²³⁵U is fissioned:

The fission of burn up rate is equal to:

F = 1.05 gP = 1.05 * 2758 = 2896 g/day

b) When ²³⁵U is consumed:

The comsuption rate is:

C = 1.05*(1+α)*gp, where α = 16.9 for ²³⁵U

C = 1.05*(1 + (16.9/100)) * 2758 = 3385.31 g/day