Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 345 / 2

The shallow-water approximation requires $h/L\ll 1$, a nearly hydrostatic pressure, negligible vertical acceleration, and approximately depth-independent horizontal velocity, concentration, and density. The large Reynolds number allows viscous stress to be neglected away from thin boundary layers, while the deep ambient is taken to remain stationary.

For unit channel width, volume conservation, chemical conservation, and mass conservation are respectivelyandEntrained ambient fluid contains no chemical and enters with density ${\rho }_0$; detrained fluid carries the local concentration and density. The ambient has no horizontal momentum, whereas detrained fluid carries horizontal momentum $\rho u$ per unit volume. The depth-integrated momentum conservation law is thereforeso $M=-\rho u{w}_d$.

Define the concentration-dependent reduced gravityUnder the Boussinesq approximation, the three shallow water equations can be written in advective form asThe coefficient matrix of this quasilinear system has characteristic speedsThus this is a hyperbolic system when $g^{\prime }h>0$. Along the intermediate characteristic curve $dx/dt=u$, the concentration obeysIt decreases because ambient entrainment dilutes the chemical; detrainment does not change the concentration of a well-mixed parcel.

Use a well-mixed gravity-current box model of length $L(t)$ and depth $h(t)=V_0/L(t)$. Its fixed volume requires $w_e=w_d$. The integrated chemical balance and the standard gravity-current front condition areThese two ordinary differential equations are the required integral model.

First suppose $R_1>0$ and putEliminating time givesWithintegration from $(L,\varphi )=(L_0,1)$ yieldsAs $t\to \infty$, the concentration tends to zero and the runout length of a gravity current isThe formula has a regular $a\to 0$ limit. If $R_1=0<R_2$, direct integration instead giveswhose value at $\varphi =0$ gives $L_{\infty }$.