Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 345 (Fluid Dynamics of the Environment)

Let the velocity perturbation be $(u,w)$, pressure perturbation be $p$, and density perturbation be ${\rho }^{\prime }$ about the hydrostatic background $\hat{\rho }(z)$. The linearized inviscid Boussinesq approximation givesStable stratification means ${\hat{\rho }}_z<0$, and the buoyancy frequency isDifferentiate the momentum equations to eliminate $p$, use incompressibility, and then use the density equation to eliminate ${\rho }^{\prime }$. This yieldsFor $w\propto e^{i(kx+mz-\omega t)}$, the internal gravity wave dispersion relation is ${\omega }^2=N^2{k}^2/(k^2+m^2)$.

A localized monochromatic cylinder emits four narrow beams, one in each quadrant, forming a St Andrew's cross. If $\alpha$ is the beam angle to the horizontal,equivalently, its angle $\theta =\pi /2-\alpha$ to the vertical obeys $\cos \theta =\omega /N$. As $\omega$ rises from zero to $N$, the beams rotate from horizontal toward vertical. For $\omega >N$ no freely propagating internal wave exists and the response is evanescent.

The wavevector and phase velocity are parallel. The dispersion relation is homogeneous of degree zero in $(k,m)$, so Euler's theorem gives $\mathbf{k}\cdot {\mathbf{c}}_g=0$: the group velocity is perpendicular to both the wavevector and phase velocity. Energy travels along the beams in the group-velocity direction.

Write $\mu =\tan \theta =m/k$. The incident down-right ray from $(0,s)$ has slope $-\cot \theta$ and meets the parabolic bottom $H(x)=2x^2$ at $x=x_0\in (0,1/2)$ providedAt the impact point the bottom slope is $S=H^{\prime }(x_0)=4x_0$. A stationary reflection preserves frequency and tangential wavenumber $k+Sm$. For the incident wavevector $(k,m)=a(1,\mu )$ and a reflected up-left ray $(k_r,m_r)=a_r(-1,\mu )$, tangential matching givesA positive reflected magnitude $a_r$ therefore exists exactly when $S\mu >1$. This is the supercritical-slope condition for reflection of an internal-wave ray and produces a group velocity in the second quadrant.

Since $x_0<1/2$, such a point exists whenFor any such $\theta$, chooseThe initial ray then hits the parabola at a supercritical point and its reflected ray travels up and left, as required.

Assume a hydrostatic, vertically well-mixed, inviscid shallow layer, negligible ambient horizontal velocity, and uniform channel width. Volume, mass, and horizontal momentum balances areCombining the first two givesDetrainment removes fluid with the layer's own density and velocity, so it cancels from the corresponding material density and velocity equations; entrainment dilutes and exerts a momentum load.

Define the reduced gravityand use the Boussinesq limit ${\rho }_1\simeq {\rho }_2$ except in buoyancy. The balances becomeThus, for $\mathbf{q}=(g^{\prime },h,u{)}^T$,which is the required entraining shallow-water layer system.

Put $c=\sqrt{g^{\prime }h}$. The three real eigenvalues areso the system is strictly hyperbolic for $g^{\prime }h>0$. Along $dx/dt=u$,Left eigenvectors for ${\lambda }_{\pm }$ are $(\pm h/(2c),\pm c/h,1)$, hencewhere $D_{\pm }={\partial }_t+(u\pm c){\partial }_x$.

When $w_e,w_d\to 0$, reduced gravity is materially conserved. If it is initially uniform, it remains constant and the other two relations integrate to the standard Riemann invariants

Taking the curl of the Boussinesq momentum equation and using incompressibility gives the vorticity equationThe second term on the right is baroclinic vorticity generation. It is nonzero where a horizontal density gradient crosses the vertical gravitational acceleration.

In steady inviscid two-dimensional flow the spanwise vorticity obeysIntegrating this equation over the front-frame control volume converts the left side to vorticity flux through its upstream and downstream faces. For a sharp interface, the baroclinic source integrates to the circulation generated by the hydrostatic pressure jump, $g^{\prime }h$. With plug flow downstream, the resulting balance isVolume conservation in the front frame gives $u_{1}H=u_2(H-h)$. The undisturbed indoor air is stationary in the laboratory, so $u_1$ is the front speed:

With a long uniform corridor, constant depth, and negligible entrainment, the pressure head $g^{\prime }h$ driving the gravity current does not change as the nose advances. There is no growing geometric length in the local front balance, so dimensional analysis gives the constant velocity $U_f\propto \sqrt{g^{\prime }h}$ after the short release transient.

The energy-conserving full-depth lock-exchange flow is symmetric between the cold lower current and warm upper return current, soPart (b) then gives $u_2=\sqrt{g^{\prime }H}$ and the laboratory front speedDuring one opening, the cold current displaces the volumeThe same volume of warm air exits in the upper layer.

For an ideal gas in the Boussinesq limit,Each opening removes ${\rho }_0{c}_p{V}_{\mathrm{ex}}(T-T_0)$ of heat. With $n$ openings per hour, the mean removal rate is $(n/3600)$ times this quantity. Equating it to $\dot{q}$ and writing $\Delta T=T-T_0$ givesTherefore

Mechanical mixing destroys the sharp density interface that supports efficient displacement ventilation. It entrains warm air into the incoming cold current and cold air into the outgoing current, reducing both reduced gravity and the heat removed per exchanged volume. Stopping fans or other mixing while the door is open therefore preserves the two-layer lock-exchange flow and improves cooling efficiency.

The Boussinesq approximation uses a constant reference density in inertia and continuity while retaining the small temperature-dependent density deficit in buoyancy. Here $({\rho }_0-\rho )/{\rho }_0=(T-T_0)/T_0$, so the plume remains incompressible but rises under $g(T-T_0)/T_0$.

The Batchelor entrainment hypothesis sets the ambient inflow speed at the exposed plume edge to $EW$, where $E$ is a dimensionless entrainment coefficient. For this one-sided heated wall plume, it gives $dQ/dz=EW$.

With thermal diffusivity ${\kappa }_T$, temperature obeysScale $z$ with $h_0$, $x$ with $b_0=ϵh_0$, and $w$ with $w_0$. Continuity gives the cross-plume velocity scale $u_0\sim (b_0/h_0)w_0=ϵw_0$, so both advective terms scale as $w_0\Delta T/h_0$. Horizontal and vertical diffusion scale as ${\kappa }_T\Delta T/b_0^2$ and ${\kappa }_T\Delta T/h_0^2$, respectively. Their ratio is

For top-hat profiles, $Q=bW$, $M=b{W}^2$, and $\mathcal{B}=Qg(T-T_0)/T_0$. One-sided entrainment, vertical momentum, and the wall heat input giveIntegrating the temperature equation across the plume givesCombining this with the definition of buoyancy flux yields

Far above the source, $\mathcal{B}\sim \beta z$. Suppose $Q\sim z^p$ and $M\sim z^r$. The first integral equation gives $p-1=r-p$, while the second gives $r-1=1+p-r$. Solving,Hence

Introduce the virtual-origin coordinateso that $\mathcal{B}=\beta \zeta$. Seek $Q=A{\zeta }^{4/3}$ and $M=C{\zeta }^{5/3}$. Substitution givesTherefore the exact similarity solution isand

At the source, $\zeta =z_0={\mathcal{B}}_0/\beta$. The initial width required to lie exactly on the similarity solution is thereforeThe corresponding $Q_0=A{z}_0^{4/3}$ and $M_0=C{z}_0^{5/3}$ are the compatible initial fluxes stipulated in the question.