Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 90 / 3

Hindered settling is the reduction of the particle settling velocity by displaced-fluid backflow and interactions with other particles. In the specified model, the downward speed is $V_s$ at vanishing concentration and falls to zero at the maximum packing fraction ${\varphi }_{\max }$. The characteristic speed of a concentration disturbance will differ from this individual-particle speed.

Take $z$ positive upwards. For a locally uniform horizontal suspension without diffusion or bulk fluid motion, the upward particle flux isParticle volume conservation gives the scalar conservation lawThis is the quadratic hindered-settling flux. Along its characteristic curves,For a smooth initial profile ${\varphi }_0(\zeta )$, the map is $z=\zeta +c({\varphi }_0(\zeta ))t$. Its Jacobian is $1+(2V_s/{\varphi }_{\max }){\varphi }_0^{\prime }(\zeta )t$. A shock develops where this first reaches zero: the compressive condition is ${\varphi }_0^{\prime }<0$, andwhen the minimum is negative and no earlier boundary discontinuity is imposed. A discontinuous compressive initial state can instead contain a shock from $t=0$.

Let $z=s(t)$ separate states ${\varphi }_L$ below the shock and ${\varphi }_R$ above it. Integrating the conservation law across a moving thin interval gives the Rankine-Hugoniot conditionEquivalently, the flux relative to the shock agrees on both sides. For this quadratic flux,Since $j^{\prime \prime }=2V_s/{\varphi }_{\max }>0$, a physical compression shock has ${\varphi }_L>{\varphi }_R$ and satisfies $c({\varphi }_L)>\dot{s}>c({\varphi }_R)$: characteristic curves enter it from both sides. These conditions select the shock over a nonphysical expansive discontinuity.

Use the upward coordinate of the preceding kinematic sedimentation calculation, with the given smooth profile on $0\le \zeta \le h$ and clear fluid above. This asks for the first characteristic breaking of that profile; a separately imposed packed-bed discontinuity would be a different boundary problem.

Substitution of the initial concentration into the characteristic speed givesThe characteristic Jacobian isIt first vanishes at the upper edge $\zeta =h$. That edge has zero concentration and descends initially at $V_s$, soAlthough the derivative of the initial profile jumps to zero in the clear region, the concentration itself is continuous there. The shock starts with vanishing amplitude. Both limiting concentrations at onset are zero, and the Rankine-Hugoniot condition consequently givesThis is a downward speed $V_s$.

After formation the upper state remains clear, ${\varphi }_R=0$, and the lower state is positive. The parabolic-profile sedimentation shock hasIt therefore initially slows down in magnitude: its signed upward-coordinate speed increases from $-V_s$. One can also verify that its captured left state increases. Write $r={\zeta }_L/h$, where ${\zeta }_L$ labels the incident lower characteristic. Conservation of the particle volume initially above that characteristic givesThe derivative of the right-hand side with respect to $r$ is $-(r+3)/(r+1{)}^3<0$. Thus $r$ decreases as time increases, and ${\varphi }_L={\varphi }_{\max }(1-r^2)/3$ increases. The downward shock speed continues to decrease on this branch until a bed or another boundary interacts with it. No later boundary behavior is needed for the requested conclusion.

Assume the carrier fluid has ambient mass density ${\rho }_0$, the particles have mass density ${\rho }_p>{\rho }_0$, and volumes add. The well-mixed current then hasHere $g_0^{\prime }$ is the particle-fluid reference reduced gravity per unit volume fraction, which is the convention needed in the requested expansion. The initial current reduced gravity is $g_0^{\prime }{\varphi }_0$; it should not be confused with $g_0^{\prime }$ itself. The Boussinesq approximation requires $\varphi ({\rho }_p-{\rho }_0)/{\rho }_0\ll 1$.

For particles to remain well mixed, the turbulent vertical mixing time must be short compared with the settling time. In a turbulent diffusion description with diffusivity $K_z$, require $h^2/K_z\ll h/[V_s(1-\varphi )]$, or $K_z\gg h{V}_s(1-\varphi )$. Equivalently, characteristic turbulent vertical speeds should substantially exceed the particle settling velocity. Turbulence can maintain an approximately uniform concentration in the body while particles still settle relative to the fluid. At the impermeable bed, fluid has no normal transport into the solid boundary, but particles can reach and be captured by it. Assume no resuspension after capture.

Use a gravity-current box model of length $L(t)$, uniform depth $h(t)$ and uniform concentration $\varphi (t)$. Neglect ambient entrainment and the small volume change caused by depositing dilute solids, so $Lh=M$ remains constant. Take a constant front Froude number $F>0$ and the gravity-current front condition $\dot{L}=F\sqrt{g^{\prime }h}$. With ${\varphi }_{\max }=1$, the deposition flux is $V_s\varphi (1-\varphi )$ per unit bed area. The particle-volume budget isHence the integral model isThe model assumes a fixed rear boundary, one advancing front, a deep stationary ambient, negligible drag during the inertial spreading regime, and a perfectly absorbing bed. The supplied hindered-settling law is retained in the deposition closure.

For $0<\varphi <1$, eliminating time givesThe concentration integral is $2\mathrm{artanh}\sqrt{\varphi }$. ThereforeAs the particle concentration tends to zero, the current reduced gravity and front speed tend to zero. The hindered-settling runout in a rectangular channel isIt is a finite limiting distance; in this idealized model, the exponential tail of particle loss makes complete stopping asymptotic in time. For $L_{\infty }\gg L_0$, omit $L_0^{5/2}$. Using $\mathrm{artanh}x=x+x^3/3+O(x^5)$ givesThe original PDF has $V_s^2$ in this expression; the TeX aid's $V_s^3$ is a transcription error. Dimensions independently confirm the square: $M^3{g}_0^{\prime }/V_s^2$ has dimensions of length to the fifth power. If one instead names the initial current reduced gravity $G_{\mathrm{init}}=g_0^{\prime }{\varphi }_0$, the same expansion is $25F^2{M}^3{G}_{\mathrm{init}}[1+2{\varphi }_0/3+O({\varphi }_0^2)]/V_s^2$. Late loss of turbulent mixing, viscous resistance or bed resuspension can change the physical runout beyond this specified box model.