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MPhil in nuclear energy course of the University of Cambridge

Past exam of the MPhil in nuclear energy course of the University of Cambridge

2021

Module NE1

Differentiating the supplied relation gives
$$
\frac{d\rho}{d\omega}=\Lambda+\frac{\beta\lambda}{(\omega+\lambda)^2},
\qquad\left.\frac{d\rho}{d\omega}\right|_0=\Lambda+\frac\beta\lambda.
$$
Thus the tangent approximation is $\rho\simeq\omega(\Lambda+\beta/\lambda)$ and
$$
\boxed{T_{\rm tangent}=\frac{\Lambda+\beta/\lambda}{\rho}=70.5\ \mathrm s.}
$$
It 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.


Solution

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

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

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