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ID: past-exam-of-the-mphil-in-nuclear-energy-course-of-the-university-of-cambridge/2019/paper-ne1/2/d/solution

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Past exam of the MPhil in nuclear energy course of the University of Cambridge / 2019 / Module NE1 / 2 / d / Solution by

Codex  0 2026-10-20

The minimum-volume aspect ratio is $H / R ≃ 1.847$ . Keep the material parameter

B_{m}^{2} = (η - 1) Σ_{a} / D

fixed and positive. Write

a = 2.405

and

x = H / R

. The constraint from part (c) gives

\frac{a ^{2}}{R ^{2}} + \frac{π ^{2}}{H ^{2}} = B_{m}^{2}, R^{2} = \frac{a ^{2} + π ^{2} / x ^{2}}{B _{m}^{2}} .

The cylinder volume is consequently

V (x) = \frac{π x}{B _{m}^{3}} (a^{2} + \frac{π ^{2}}{x ^{2}})^{3/2}, x > 0.

Its logarithmic derivative is

\frac{d l n V}{d x} = \frac{1}{x} - \frac{3 π ^{2}}{x ( a ^{2} x ^{2} + π ^{2} )} = \frac{a ^{2} x ^{2} - 2 π ^{2}}{x ( a ^{2} x ^{2} + π ^{2} )} .

The denominator is positive. The derivative changes from negative to positive exactly once, and

V

diverges as

x \to 0

or

x \to \infty

. Hence this stationary point is the global minimum:

\frac{H}{R} = x_{*} = \frac{2 π}{2.405} = 1.84735.

Equivalently

H / (2 R) ≃ 0.924

. This is an aspect ratio in the supplied idealized boundary value problem; it does not determine an absolute size without a specified material parameter.

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