Past exam of the MPhil in nuclear energy course of the University of Cambridge / 2021 / Module NE1 / 2 / f

Differentiating the supplied relation givesThus the tangent approximation is $\rho \simeq \omega (\Lambda +\beta /\lambda )$ andIt is about $16.4\%$ longer than $60.58\mathrm{s}$ and therefore predicts slower growth. For this positive step it is not conservative as a growth-rate estimate. The local expansion needs $|\omega |\ll \lambda$; here the actual ratio is approximately $0.165$, so its error is appreciable.

The precursor exponential from part (e) has inverse period $\rho \lambda /(\beta -\rho )$. HenceThis is approximately $0.96\%$ shorter than the exact dominant period, so it slightly overestimates growth for these stipulated constants. Neglecting the small $\Lambda$ contribution is more accurate here than replacing the complete frequency dependence by its tangent at the origin.

With no delayed neutrons, set $\beta =0$ and omit the precursor equation. The source-free model reduces to $\dot{n}=(\rho /\Lambda )n$, givingThe delayed-neutron model's dominant e-folding time is about $121$ times larger. This comparison illustrates why a small delayed neutron fraction strongly alters the kinetics. It is a comparison of the supplied idealized models at the same imposed positive reactivity, not a control prescription or a claim that every transient has the same timescale.