Let $\mathcal V=HL_0$ be the layer volume per unit transverse width. The <gravity-current box model> imposes $hL=\mathcal V$; it approximates the front and scalar balances, rather than an exactly uniform solution of the local <momentum> equation. For a specified constant front <Froude number> $\operatorname{Fr}$, the <heated particle-laden gravity current> satisfies
$$
\boxed{h=\frac{\mathcal V}{L},\qquad
\dot L=\operatorname{Fr}\sqrt{\frac{g\mathcal V(\gamma\phi-\theta)}{L}},\qquad
\dot\phi=-\frac{W_sL}{\mathcal V}\phi,\qquad
\dot\theta=\beta\phi}.
$$
The initial data are $L=L_0$, $\phi=\phi_0$, and $\theta=0$. The <gravity-current front condition> applies only while $\gamma\phi-\theta\geq0$.

For heating without settling, $\phi=\phi_0$ and $\theta=\beta\phi_0t$. Integration of the front equation gives
$$
\boxed{L(t)=\left[L_0^{3/2}+\frac{\operatorname{Fr}\sqrt{g\mathcal V\phi_0}}{\beta}
\left\{\gamma^{3/2}-(\gamma-\beta t)^{3/2}\right\}\right]^{2/3}},\qquad
0\leq t\leq\frac\gamma\beta.
$$
Hence the <heated gravity-current runout> is
$$
\boxed{L_{\max}=\left[L_0^{3/2}+\frac{\operatorname{Fr}\sqrt{g\mathcal V\phi_0}\,\gamma^{3/2}}{\beta}\right]^{2/3}}.
$$
It is reached at the neutral-buoyancy time $\gamma/\beta$ within this front model. A finite-<momentum> current could subsequently coast or lift off: the formula is the maximum predicted by the prescribed <gravity-current front condition>, which contains no independent front inertia.

For settling without heating, $\theta=0$. Set $K=\operatorname{Fr}\sqrt{g\gamma\mathcal V}$, so $\dot L=K\sqrt\phi\,L^{-1/2}$. Eliminating time yields
$$
\frac{d\sqrt\phi}{dL}=-\frac{W_sL^{3/2}}{2\mathcal V K},\qquad
\sqrt\phi=\sqrt{\phi_0}-\frac{W_s}{5\mathcal V K}(L^{5/2}-L_0^{5/2}).
$$
Thus the <runout length of a gravity current> is
$$
\boxed{L_{\max}=\left[L_0^{5/2}+
\frac{5\operatorname{Fr}\mathcal V^{3/2}\sqrt{g\gamma\phi_0}}{W_s}\right]^{2/5}}.
$$
This length is approached as $t\to\infty$, because positive concentration cannot disappear at a finite time under the settling law. If distance means displacement from the removed barrier, the answer is $L_{\max}-L_0$ in either case.

Solution

Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 345 / 2 / c / Solution

A <heated particle-laden layer> released along a bed forms a <gravity current> while $g'=g(\gamma\phi-\theta)>0$. In a <gravity-current box model>, heating can make $g'$ vanish in finite time, whereas <settling velocity> alone can give finite <runout length of a gravity current> approached asymptotically.

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