It is found that, in a certain thermal reactor, fueled with partially enriched u

Question

It is found that, in a certain thermal reactor, fueled with partially enriched uranium, 13% of the fission neutrons are absorbed in resonances of 238U and 3% leak out of the reactor, both while these neutrons are slowing down; 5% of the neutrons that slow down in the reactor subsequently leak out; of those slow neutrons that do not leak out, 82% are absorbed in fuel, 74% of these in 235U. (a) What is the multiplication factor of this reactor?

Note that the 74% of neutrons absorbed in 235U refers to 74% out of those absorbed in the fuel.

Explanation / Answer

Multilpication factor

K = production of neutrons/(absorption+Leakage neutrons)

Let production of electrons = 100

now 13 neutrons are absorbed in resonance and 3 are leak out.

Remaining 84 neutrons

5% are leak out = 84*5/100= 4.2 neutrons(leak out)

remaining = 84-4.2 = 79.8 neutrons

now 82% of 79.8 = 65.436 neutrons are absorbed

so multiplication factor

K = 100/(13+65.436+3+4.2)

K = 1.16

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.