When $\widehat W_1=\widehat W_2=\widehat W$, both concentrations acquire the same decay factor. The total reduced gravity consequently obeys
$$
\dot g'=\frac{\widehat W L}{V_0}g'.
$$
Eliminating time with the front equation gives
$$
\frac{d\sqrt{g'}}{dL}
=\frac{\widehat W L^{3/2}}{2\operatorname{Fr}V_0^{3/2}}.
$$
Integration from the initial state yields
$$
\sqrt{g'(L)}
=\sqrt{\widetilde g'}
+\frac{\widehat W}{5\operatorname{Fr}V_0^{3/2}}
\left(L^{5/2}-L_0^{5/2}\right),
$$
where
$$
\widetilde g'
=\frac g{\rho_a}
\left[(\rho_1-\rho_a)\widetilde\phi_1
+(\rho_2-\rho_a)\widetilde\phi_2\right].
$$
The <particle-laden gravity current> reaches its <runout length of a gravity current> when $g'$ vanishes. Since $\widehat W<0$,
$$
L_\infty
=\left[
L_0^{5/2}
+5\operatorname{Fr}
\frac{(L_0h_0)^{3/2}\sqrt{\widetilde g'}}{|\widehat W|}
\right]^{2/5}.
$$
Thus $C=5\operatorname{Fr}$ for the stated front condition; the common normalization $\operatorname{Fr}=1$ gives $C=5$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 345 / 1 / c / ii / Solution

A particle-laden gravity current is driven by the excess density of suspended particles. Deposition decreases its <reduced gravity>, so a finite-volume current can approach a finite runout length.

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