The Nuclear Reactor — Question 9

Section 43-3 · Physics pages 7208–7209

Question 9 A neutron-absorbing product approaches a source-removal equilibrium A generic absorbing fission product has inventory X obeying dX/dt = S −(λ + a)X, with constant production S=1200/s, radioactive removal λ = .010 s−1 and neutron-absorption removal a = .020 s−1. Initially X=0. product X S λX aX Schematic; labels, not drawing lengths, specify the data. (a) Derive the equilibrium inventory and the time-dependent approach to it. (b) Find X after one relaxation time and the fractions of equilibrium removals due to decay and absorption.
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Question 9 - Solution A neutron-absorbing product approaches a source-removal equilibrium product X S λX aX (a) Method and calculation Xeq = S/(λ + a) = 1200/.030 = 40000 . X(t) = Xeq(1 −e−(λ+a)t), τ = 33.333 s . (b) Interpretation and further calculation X(τ) = 40000(1 −e−1) = 25284.8 . fdecay = λ/(λ + a) = 1/3, fabsorption = 2/3 . Physical check The absorption rate coefficient is held fixed here. In a coupled reactor model it depends on the neutron field, so this solution is conditional.

Question 9

A neutron-absorbing product approaches a source-removal equilibrium

A generic absorbing fission product has inventory X obeying dX/dt=S−(λ+a)XdX/dt=S-(\lambda+a)X, with constant production S=1200/s, radioactive removal λ=.010s−1\lambda=.010\,\mathrm{s^{-1}} and neutron-absorption removal a=.020s−1a=.020\,\mathrm{s^{-1}}. Initially X=0.

Diagram shown in the original SVG worksheet.

  1. Derive the equilibrium inventory and the time-dependent approach to it.

  2. Find X after one relaxation time and the fractions of equilibrium removals due to decay and absorption.

Question 9 - Solution

A neutron-absorbing product approaches a source-removal equilibrium

Diagram shown in the original SVG worksheet.

(a) Method and calculation

Xeq=S/(λ+a)=1200/.030=40000.\boxed{X_{\mathrm{eq}}=S/(\lambda+a)=1200/.030=40000}.X(t)=Xeq(1−e−(λ+a)t),τ=33.333s.\boxed{X(t)=X_{\mathrm{eq}}(1-e^{-(\lambda+a)t}),\quad\tau=33.333\,\mathrm s}.

(b) Interpretation and further calculation

X(τ)=40000(1−e−1)=25284.8.\boxed{X(\tau)=40000(1-e^{-1})=25284.8}.fdecay=λ/(λ+a)=1/3,fabsorption=2/3.\boxed{f_{\mathrm{decay}}=\lambda/(\lambda+a)=1/3,\quad f_{\mathrm{absorption}}=2/3}.

Physical check

The absorption rate coefficient is held fixed here. In a coupled reactor model it depends on the neutron field, so this solution is conditional.