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ID: past-exam-of-the-mphil-in-nuclear-energy-course-of-the-university-of-cambridge/2020/paper-ne2/1/c/solution

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Past exam of the MPhil in nuclear energy course of the University of Cambridge / 2020 / Module NE2 / 1 / c / Solution by

Codex  0 2026-10-20

The two outward surface normals point in opposite directions. With positive heat flux defined as heat leaving each face, Fourier's law gives

q_{left}^{''} = k T^{'} (0), q_{right}^{''} = - k T^{'} (a) .

(1)

Using the derivative from part(a), the removed fluxes per unit face area are

q_{left}^{''} = \frac{q _{0}}{μ} [1 - \frac{1 - e ^{- μ a}}{μ a}] = \frac{q _{0}}{μ} (1 - C),

(2)

q_{right}^{''} = \frac{q _{0}}{μ} [\frac{1 - e ^{- μ a}}{μ a} - e^{- μ a}] = \frac{q _{0}}{μ} (C - e^{- β}) .

(3)

Both are positive because

e^{- β} < C < 1

. Their sum satisfies the independent conservation of energy check

q_{left}^{''} + q_{right}^{''} = \frac{q _{0}}{μ} (1 - e^{- μ a}) = \int_{0}^{a} q_{0} e^{- μx} d x .

(4)

The illuminated face removes more heat because heating is concentrated nearer it. Equivalently, weighting each source slice by the fraction of its heat conducted to either equal-temperature boundary gives

q_{left}^{''} = \int_{0}^{a} q^{'''} (x) \frac{a - x}{a} d x, q_{right}^{''} = \int_{0}^{a} q^{'''} (x) \frac{x}{a} d x .

(5)

A decreasing source weights the left face more strongly. In the uniform-heating limit, each face removes $q_{0} a /2$ . If a heat removal rate rather than a flux is wanted, multiply the relevant expression by the face area.

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