Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 345 / 2 / c

Solution

The shallow-water equations cease to apply inside the narrow nose, so a gravity-current front condition is needed to relate its speed to the depth immediately behind it. Use

\dot{R} = F r g^{'} h = F r c,

where

F r

is the front Froude number; the ideal deep-ambient von Kármán condition gives

F r = 2

In a uniform axisymmetric box model, conservation of contaminated volume gives

h = \frac{V}{π R ^{2}} .

The total nondimensional heat content is

V θ

, while cooling acts over area

π R^{2}

, so

\dot{θ} = - \frac{K π R ^{2}}{V} θ, \dot{R} = \frac{F r}{R} \frac{g _{0}^{'} V}{π} θ .

Eliminating time and taking the initial release radius as negligible gives

θ (R) = 1 - \frac{K π R ^{4}}{4 F r V g _{0}^{'} V / π} .

Spreading stops as

θ \to 0

, at

R_{m a x}^{4} = \frac{4 F r V}{K π} \frac{g _{0}^{'} V}{π} .

Hence the maximum covered area is

A_{m a x} = π R_{m a x}^{2} = 2 π [\frac{F r V}{K π} \frac{g _{0}^{'} V}{π}]^{1/2} .

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