Reactivity Example

keff = 1.00
N     =  500 generations

k^N = (1,003)⁵⁰⁰ = 4.47

or that is to say:

N_N = n₀ ·4.47

Differentiating eq. (11-1) with respect to dN and integrating:

If the reactivity is small, ln k@ (k-1) =Dk, and then:

ln n = N · Dk + const

n = n₀ when N = 0, thus, n = ln n₀. That is to say:

Then:

(11-2)

Returning to eq. (11-1), we had that in 500 generations, the gain is 4.47 times, and in 1000 generations it is about 20 times, and after 1500 generations, it will be more than 500 times. This type of gain is known as the exponential increase and is represented by eq. (11-2).

As long as reactivity remains small, as is generally the case, expressions (11-1) and (11-2) are equivalent. That is to say:

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