Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 345 (Fluid Dynamics of the Environment)

With buoyancy perturbation $b=-g{\rho }^{\prime }/{\rho }_0$, the inviscid Boussinesq equations areForincompressibility is automatic. Define the Jacobian determinantThe material derivative is $D/Dt={\partial }_t+J(\psi ,\cdot )$. Taking the $y$ component of the curl of momentum gives the exact finite-amplitude system

If the disturbance amplitude is small enough that each Jacobian is asymptotically smaller than its corresponding time derivative, the system can be linearized. Eliminating $b$ then gives

For a single phase $\varphi =kx+mz-\omega t$, every disturbance field is a function of $\varphi$ alone. Its velocity is tangent to the phase surfaces:Consequently both nonlinear Jacobians vanish exactly, irrespective of amplitude. Substitution givesA nonzero amplitude therefore requires the internal gravity wave dispersion relationThe wave is an exact nonlinear solution of the inviscid Boussinesq equations. The parameter $ϵ$ merely sets its amplitude; it does not change its frequency or waveform. Very large $ϵ$ can nevertheless invalidate the Boussinesq model or make the total stratification locally overturn.

At $O(ϵ\gamma )$, the slowly varying wave envelope obeyswhere the group velocity obtained from $\omega =Nk/(k^2+m^2{)}^{1/2}$ isIt satisfies $\mathbf{k}\cdot {\mathbf{c}}_g=0$. If initially $\mathbf{k}\cdot \hat{\nabla }\tilde{\psi }=0$, the envelope structure varies across the wavevector and translates unchanged in the direction ${\mathbf{c}}_g$ at speedThus internal-wave crests propagate with phase velocity parallel to $\mathbf{k}$, while the packet and its energy propagate perpendicular to $\mathbf{k}$ with the group velocity.

The shallow-water approximation requires current depth $h$ to be much smaller than its horizontal radius, hydrostatic vertical momentum balance, and horizontal velocity and temperature that are nearly depth-uniform. The radial scale and cooling time must also be long compared with the vertical adjustment scales.

The density contrast isso the inertia may use the common reference density while this small difference is retained in buoyancy: the flow is Boussinesq. Because the current is lighter, it occupies the top of the lake. Its free-surface displacement is smaller than its internal-interface displacement by the ratio of reduced gravity to gravity. A rigid-lid reduced-gravity model is therefore valid only when $g^{\prime }\ll g$; otherwise the external free-surface mode must be retained.

SetThe axisymmetric reduced-gravity equations areThe extra $h{g}_r^{\prime }/2$ in momentum is the depth average of the hydrostatic pressure gradient caused by horizontal density variation.

In variables $(u,c,\theta )$ these becomeThe three characteristic speeds areAlong $dr/dt=u$, temperature obeysAlong $dr/dt=u+sc$, $s=\pm 1$, a left-characteristic projection gives

The shallow-water equations cease to apply inside the narrow nose, so a gravity-current front condition is needed to relate its speed to the depth immediately behind it. Usewhere $Fr$ is the front Froude number; the ideal deep-ambient von Kármán condition gives $Fr=\sqrt{2}$.

In a uniform axisymmetric box model, conservation of contaminated volume givesThe total nondimensional heat content is $V\theta$, while cooling acts over area $\pi R^2$, soEliminating time and taking the initial release radius as negligible givesSpreading stops as $\theta \to 0$, atHence the maximum covered area is

Batchelor entrainment models a turbulent plume as drawing ambient fluid inward at a speed proportional to its characteristic vertical velocity. In a top-hat axisymmetric plume, letbe volume, momentum, and buoyancy flux. The Boussinesq approximation uses a common density in inertia and volume conservation while retaining the small density difference in $g^{\prime }=g({\rho }_0-\rho )/{\rho }_0$. It requires $g^{\prime }\ll g$ and becomes inaccurate for very hot source fluid or near openings with large density changes.

For a point source in an unstratified lower layer, the integral plume equations arewhere $\alpha$ is the entrainment coefficient. Their pure-plume solution iswith

Outside, pressure is hydrostatic with slope $-{\rho }_{0}g$. Inside it has the same slope below $h_1$ and the shallower slope $-{\rho }_{1}g$ above $h_1$. Thus $\Delta p=p_{\mathrm{in}}-p_{\mathrm{out}}$ is a negative constant below the interface, rises linearly above it with slope ${\rho }_0{g}_{10}^{\prime }$, and is positive at the ceiling. The floor vent admits air and the ceiling vent exhausts it.

Let $C_d$ be a common discharge coefficient and define the effective opening areaThe ventilation flow isSteady volume and buoyancy balances requireTherefore $h_1=H/2$ precisely when the effective area is chosen asThe individual areas must additionally realize this $A^{\ast }$; their ratio fixes how the total pressure drop is divided between the floor and ceiling vents.

The weaker plume detrains into the middle layer at $h_2$, while the stronger plume passes through it and feeds the upper layer at $h_1$. Middle-layer volume balance isUsing $Q_i(z)=C_Q{F}_i^{1/3}{z}^{5/3}$ gives

The middle layer receives buoyancy flux $F_2$ and loses volume $Q_2(h_2)$ by entrainment into the strong plume. The upper layer ultimately receives both source buoyancy fluxes and loses volume $Q_1(h_1)$ through the ceiling. Hence

The interior-minus-exterior pressure is constant below $h_2$, rises with slope ${\rho }_0{\hat{g}}_{10}^{\prime }$ in the middle layer, and with slope ${\rho }_0{\hat{g}}_{20}^{\prime }$ in the upper layer. PutEqual ideal openings split the total pressure difference equally, and the common ventilation rate isThus the required equal vent areas are