Past exam of the mathematics course of the University of Cambridge / 2002 / iii / Paper 52 / 4

Let $W\ge 0$ be the downward fluid suction speed and $V\ge 0$ the downward particle settling velocity relative to that fluid. Set $Q=uh$ and $g^{\prime }=g_0^{\prime }\varphi$. The carrier volume conservation and particle-volume balances per unit channel width areThe particle sink includes suction as well as relative settling: the particles follow the well-mixed withdrawn fluid, in addition to settling through it. Subtracting $\varphi Q^{\prime }$ yields $Q{\varphi }^{\prime }=-V\varphi$. Thus for $W>0$, as long as $Q>0$,For $W=0$ the continuous limit is $Q=Q_0$ and $\varphi =e^{-Vx/Q_0}$. In particular suction alone does not change the well-mixed particle fraction: $V=0$ implies $\varphi =1$.

Withdrawn fluid carries its local horizontal momentum, so the momentum flux and hydrostatic pressure force obeyExpand this and use $Q^{\prime }=-W$. The suction momentum terms cancel in the primitive acceleration equation, leavingThese are the governing equations for a particle-laden current with suction. Substituting $u=Q/h$ and the concentration equation gives the convenient depth equationThe steady carrier flux is exhausted at the formal length $\boxed{x_{\ast }=Q_0/W}$ when $W>0$. Positive $Q$ cannot continue beyond it. Near that limit a chosen depth branch, mixing or shallow-layer assumptions may fail, so this is the mass-budget extent, not a guarantee that every arbitrary inlet state has a regular shallow continuation to it. When $W=0$, settling makes the driving reduced gravity exponentially small but does not remove carrier fluid; this ideal steady model has no finite mass-budget runout.

At $V=0$, the particle budget gives $\varphi =1$, so $g^{\prime }=g_0^{\prime }$ is constant. The primitive momentum conservation equation integrates toIntroduce $q=Q/Q_0=1-Wx/Q_0$, $y=h/h_0$ and $\epsilon ={\mathrm{Fr}}_0^2=Q_0^2/(g_0^{\prime }{h}_0^3)\ll 1$. The exact relation on the subcritical flow branch isExpanding uniformly for $0\le q\le 1$ givesThe depth increases slightly because horizontal kinetic energy decreases as fluid is removed. In fact the depth equation has $h^{\prime }\ge 0$ on this branch, andThus the small-Froude number approximation becomes better downstream, including as the flux vanishes. At leading order $u=(Q_0-Wx)/h_0$.

With $W=0$ and $V>0$, $Q=Q_0$ and $\varphi =e^{-kx}$, where $k=V/Q_0$. No carrier momentum is withdrawn; hence the integrated momentum flux is exactly constant:On an initially small-Froude number branch, the pressure term dominates. At leading orderA slightly more accurate large-distance amplitude is $\sqrt{2K/g_0^{\prime }}$, so $h\sim \sqrt{2K/g_0^{\prime }}{e}^{kx/2}$. The Froude number satisfies ${\mathrm{Fr}}^2\sim {\mathrm{Fr}}_0^2{e}^{-kx/2}$ and remains small. The growing depth can eventually violate confinement or the shallow water assumption; this does not make the asymptotic momentum calculation inconsistent within its intended regime.

On an initially large-Froude number branch, horizontal momentum flux dominates instead. The leading depth is constant, $\boxed{h\sim h_0}$, while the decaying reduced gravity makes the Froude number still larger. To resolve the small change, put $\delta ={\mathrm{Fr}}_0^{-2}\ll 1$ and $y=h/h_0$:soThe depth decreases slightly and tends exactly to $h_{\infty }=Q_0^2/K=h_0/(1+\delta /2)$. Its Froude number grows like ${\mathrm{Fr}}_0{e}^{kx/2}$. Both approximations agree with the exact depth equation $h^{\prime }=kh/[2(1-{\mathrm{Fr}}^2)]$: the subcritical flow deepens, while the supercritical flow thins.