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

Take $z$ increasing upwards and write $s=\varphi /{\varphi }_{\max }$. The particles have downward settling velocity; their upward volume flux is $\varphi V$. Conservation of particle volume in any interval givesThus the kinematic sedimentation equation and its characteristic speed areFor $0<s<1$ this is a scalar hyperbolic partial differential equation: its one characteristic speed is real. The method of characteristics gives $ds/dt=0$ and $dz/dt=a(s)$. A concentration disturbance need not descend at the particle settling velocity; $a(s)$ is the derivative of the transported flux, not $V(s)$. In particular $a(s)$ can be positive even though every particle settles downwards. At $\alpha =0$, $a=-V_0$ and the equation is ordinary translation. At the packing endpoint $s=1$ the differentiability of the chosen constitutive law depends on $\alpha$; none of the initial values used below is at that endpoint.

For $\alpha =1$, $f=V_0(s^2-s)$, so the characteristic speed is $a=V_0(2s-1)$ and the conservation law flux is strictly convex. Let $\tau =V_{0}t$ and let $\xi$ be the initial vertical position of a characteristic curve.

For the increasing ramp, the characteristic curves within its sloping part satisfy $z=\xi (1+0.6\tau )$. They spread apart and never cross. The outer plateau characteristic speeds are $-0.6V_0$ and $0.6V_0$, so the complete entropy solution isThe transition widens linearly and its gradient decreases. At large time it approaches a rarefaction wave, with $s\simeq (1+z/\tau )/2$ inside its expanding transition.

For the decreasing ramp, $z=\xi (1-0.6\tau )$ and the characteristic curves converge. Before they meet,All the ramp characteristic curves meet at $z=0$ atThereafter a stationary sedimentation shock separates $s_L=0.8$ below from $s_R=0.2$ above. Indeed the Rankine-Hugoniot condition givesand the incoming characteristic speeds satisfy $a(0.8)=0.6V_0>0>a(0.2)=-0.6V_0$, which makes it an admissible entropy shock. The discontinuity transports no net excess particle flux because both plateau fluxes equal $-0.16V_0$. These are local solutions before a finite container's boundaries affect the transition.

The dependence on $\alpha$ is controlled by the hindered-settling flux inflection:For any smooth initial profile, the characteristic flow map has derivativeThis formula gives the first characteristic crossing: minimize the positive values of $-1/[f^{\prime \prime }(s_0)s_0^{\prime }]$. At $\alpha =0$ neither ramp changes shape; both move down at $V_0$. For $0<\alpha \le 3/2$, the flux is convex throughout $0.2\le s\le 0.8$: the increasing ramp spreads and the decreasing ramp steepens, although its entire ramp generally no longer focuses simultaneously. For $\alpha \ge 9$, the flux is concave throughout that interval and these roles reverse. At the endpoint values the curvature vanishes at one plateau; it does not reverse sign inside the interval.

For $3/2<\alpha <9$, the inflection $s_i=2/(\alpha +1)$ lies between the plateau concentrations. Below $s_i$ the increasing ramp spreads, while above it that ramp compresses; the decreasing ramp has the opposite local behavior. After characteristic crossing, the entropy solution can contain both a shock wave and a rarefaction wave. Quantitatively, an increasing step is selected by the lower convex envelope of $f$ on the interval between its states: curved portions supply rarefaction waves, and straight portions supply shock waves. A decreasing step uses the upper concave envelope. A join between a fan and a shock wave at a state $s_{\ast }$ satisfies the tangent-chord relation $f^{\prime }(s_{\ast })=[f(s_b)-f(s_{\ast })]/(s_b-s_{\ast })$, where $s_b$ is the other shock state. Thus curvature, not simply the magnitude of $\alpha$, decides the wave pattern; the envelope may reduce to one shock if the tangent lies outside the state interval.

Let $\theta =T-T_0>0$ and ${\theta }_1=T_1-T_0$. The aperture spans the enclosure's depth $H$, so the incoming volume flux, equal to the outgoing volume flux, isOnly one of these equal exchange fluxes multiplies the temperature difference in the heat budget: incoming fluid brings $q{T}_0$, outgoing fluid removes $qT$. With constant mass density and specific heat capacity, those factors cancel and the well-mixed ventilation temperature balance isIntegrating from ${\theta }_1$ givesThe cooling is algebraic rather than exponential because the buoyancy-driven single-opening exchange flow weakens as its driving reduced gravity decreases. The formula assumes negligible wall heat capacity, no heat input, the specified aperture geometry, and a uniform interior temperature maintained by mixing.

With a low-level opening, the dense incoming cold fluid tends to spread along the floor instead of falling through the warm interior. A cold lower layer and warm upper layer develop, giving stable density stratification. The incoming fluid can short-circuit back out through the low opening after that layer occupies it, leaving warm fluid above poorly ventilated. Thus the uniform-temperature assumption is generally inappropriate: cooling becomes vertically nonuniform and substantially slower for the upper warm fluid. If mixing were artificially maintained, the same lumped budget would still apply; it is the loss of that assumption, rather than a reversal of buoyancy, that changes the physical outcome.