The Nuclear Reactor — Question 5

Section 43-3 · Physics pages 7200–7201

Question 5 Delayed-neutron precursors continue emitting after production stops One hypothetical precursor group obeys dC/dt = SC −λC. Before shutdown it is in equilibrium at constant SC = 108 s−1; its half-life is 10.0 s. At t=0 production stops instantly. Each precursor decay emits one neutron. t Sn Schematic; labels, not drawing lengths, specify the data. (a) Find the equilibrium precursor inventory and the subsequent neutron-emission rate. (b) Calculate the number of delayed neutrons emitted in the first 20.0 s.
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Question 5 - Solution Delayed-neutron precursors continue emitting after production stops t Sn (a) Method and calculation ˙C = 0 ⇒C0 = SC/λ = 108(10) ln 2 = 1.44270 × 109 . C(t) = C0e−λt ⇒Sn(t) = λC(t) = 108 2−t/(10 s) s−1 . (b) Interpretation and further calculation Nn(0, T) = ∫T 0 Sn(t)dt = C0(1 −e−λT ). Nn(0, 20) = 3 4C0 = 1.08202 × 109 . Physical check The delayed source has a decay timescale even after new fissions cease. A real reactor has several precursor groups and coupled neutron dynamics.

Question 5

Delayed-neutron precursors continue emitting after production stops

One hypothetical precursor group obeys dC/dt=SC−λCdC/dt=S_C-\lambda C. Before shutdown it is in equilibrium at constant SC=108s−1S_C=10^8\,\mathrm{s^{-1}}; its half-life is 10.0 s. At t=0 production stops instantly. Each precursor decay emits one neutron.

Diagram shown in the original SVG worksheet.

  1. Find the equilibrium precursor inventory and the subsequent neutron-emission rate.

  2. Calculate the number of delayed neutrons emitted in the first 20.0 s.

Question 5 - Solution

Delayed-neutron precursors continue emitting after production stops

Diagram shown in the original SVG worksheet.

(a) Method and calculation

Ċ=0⇒C0=SC/λ=108(10)ln⁡2=1.44270×109.\dot C=0\Rightarrow\boxed{C_0=S_C/\lambda=\frac{10^8(10)}{\ln2}=1.44270\times10^9}.C(t)=C0e−λt⇒Sn(t)=λC(t)=1082−t/(10s)s−1.C(t)=C_0e^{-\lambda t}\Rightarrow\boxed{S_n(t)=\lambda C(t)=10^8\,2^{-t/(10\,\mathrm s)}\,\mathrm{s^{-1}}}.

(b) Interpretation and further calculation

Nn(0,T)=∫0TSn(t)dt=C0(1−e−λT).N_n(0,T)=\int_0^TS_n(t)dt=C_0(1-e^{-\lambda T}).Nn(0,20)=34C0=1.08202×109.\boxed{N_n(0,20)=\frac34C_0=1.08202\times10^9}.

Physical check

The delayed source has a decay timescale even after new fissions cease. A real reactor has several precursor groups and coupled neutron dynamics.