Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 345 (Fluid Dynamics of the Environment)

Let $u=ps{i}_z$ and $w=-ps{i}_x$, so mass conservation for the two-dimensional incompressible flow is automatic. Write $b$ for buoyancy and introduce the diffusive operatorsThe Linearized Boussinesq equations areTaking the curl of the momentum conservation equations eliminates the pressure and givesApply $D_{\kappa }$ and use $D_{\kappa }b=N^2{\psi }_x$. Since the constant-coefficient linear partial differential operators commute,This is the viscous-diffusive internal gravity wave equation.

Let the group velocity make an angle $\theta$ above the horizontal. A convenient right-handed choice of rotated unit vectors isThe first vector is parallel to the wavevector, the second is parallel to the group velocity, and their perpendicularity is the defining geometry of an internal-wave phase and group velocity. ThusWriting ${\Delta }_{\xi \zeta }={\partial }_{\xi }^2+{\partial }_{\zeta }^2$, the equation becomesFor an inviscid plane wave proportional to $e^{i(k\xi -\omega t)}$, its dispersion relation is $\omega =N\sin \theta$.

Put $Z=ϵ\zeta$ and seek the slowly attenuating wave envelope $A(Z)$ inAt order ${ϵ}^0$, the equation gives the internal gravity wave dispersion relationAt order ${ϵ}^1$, retaining one derivative of the slowly varying amplitude givesUsing $\omega =N\sin \theta$ therefore yieldsHence the leading viscous attenuation of an internal-wave beam isThe wave energy density, being quadratic in the amplitude, decays twice as rapidly in the exponent.

Setwhere $r$ is the magnitude of the internal-wave ray slope and $s$ is the bottom slope. A rightward ray remains in the triangular basin precisely when its bottom reflection is subcritical internal-wave reflection, namely $r>s$. Therefore

Conservation of frequency and of the component of the wavevector tangent to the slope gives the focusing power of internal-wave reflectionwhere $\tan \alpha =s$. In this range $\gamma >1$, so the reflected wavelength is shorter by the factor $\gamma$. The reflected normal group velocity is smaller by ${\gamma }^2$; conservation of normal energy flux therefore increases the wavelength-averaged energy density by

The incident internal-wave ray has equation $z=-rx$. Intersecting it with the bottom $z=-H_0+sx$ givesIts distance to the slope is consequentlyThe subcritically reflected ray rises through the same vertical distance at angle $\theta$, so its next free-surface reflection is at

On the first leg, viscous attenuation of an internal-wave beam multiplies the energy density by $e^{-2k\ell /\mathrm{Re}}$. The bottom reflection multiplies it by ${\gamma }^2$ and changes the wavenumber to $\gamma k$. Since the corresponding Reynolds number is $\mathrm{Re}/{\gamma }^2$, attenuation on the second leg contributes $e^{-2{\gamma }^{3}k\ell /\mathrm{Re}}$. Ignoring boundary-layer enhancement as requested,

The shallow-water approximation requires $h/L\ll 1$, a nearly hydrostatic pressure, negligible vertical acceleration, and approximately depth-independent horizontal velocity, concentration, and density. The large Reynolds number allows viscous stress to be neglected away from thin boundary layers, while the deep ambient is taken to remain stationary.

For unit channel width, volume conservation, chemical conservation, and mass conservation are respectivelyandEntrained ambient fluid contains no chemical and enters with density ${\rho }_0$; detrained fluid carries the local concentration and density. The ambient has no horizontal momentum, whereas detrained fluid carries horizontal momentum $\rho u$ per unit volume. The depth-integrated momentum conservation law is thereforeso $M=-\rho u{w}_d$.

Define the concentration-dependent reduced gravityUnder the Boussinesq approximation, the three shallow water equations can be written in advective form asThe coefficient matrix of this quasilinear system has characteristic speedsThus this is a hyperbolic system when $g^{\prime }h>0$. Along the intermediate characteristic curve $dx/dt=u$, the concentration obeysIt decreases because ambient entrainment dilutes the chemical; detrainment does not change the concentration of a well-mixed parcel.

Use a well-mixed gravity-current box model of length $L(t)$ and depth $h(t)=V_0/L(t)$. Its fixed volume requires $w_e=w_d$. The integrated chemical balance and the standard gravity-current front condition areThese two ordinary differential equations are the required integral model.

First suppose $R_1>0$ and putEliminating time givesWithintegration from $(L,\varphi )=(L_0,1)$ yieldsAs $t\to \infty$, the concentration tends to zero and the runout length of a gravity current isThe formula has a regular $a\to 0$ limit. If $R_1=0<R_2$, direct integration instead giveswhose value at $\varphi =0$ gives $L_{\infty }$.

Write $\hat{\rho }(z)$ for the ambient density far from the wall and $\rho (z)$ for the density at the wall. Across the plume, let $y=x/b$ and use the prescribed triangular profilesDirect integration gives the volume flux, mass flux, momentum flux, and density-weighted buoyancy flux, all per unit radiator length:These coefficients distinguish the triangular-profile wall line plume from a top-hat plume model.

The Batchelor entrainment hypothesis takes the inflow speed through the plume's exposed outer edge to be $\alpha W$, where $\alpha$ is the entrainment coefficient. A wall plume has only one such edge, so $dV/dz=\alpha W$.

The Boussinesq approximation replaces density by a constant reference value ${\rho }_0$ in inertia and mass flux while retaining the small density deficit in buoyancy. It requires $|\hat{\rho }-\rho |/{\rho }_0\ll 1$. A sufficiently hot radiator can violate this near the source, where thermal expansion is large and the developed-plume description may also fail.

Put $B_0=F_0/{\rho }_0$ for the kinematic buoyancy flux per unit length. Dimensional analysis for a line plume givesThe plume rise time is therefore $t_{\mathrm{rise}}\sim H/B_0^{1/3}$. Changing the room stratification requires a plume volume comparable with $XH$, so $t_{\mathrm{room}}\sim XH/V(H)\sim X/B_0^{1/3}$. HenceThe plume consequently follows the slowly changing ambient through a quasi-steady approximation when $X\gg H$.

Choose ${\rho }_0$ as a representative room density, for example the fresh-air density, and neglect relative density variations everywhere except in buoyancy. Withthe triangular profiles giveSolving these algebraic relations,

The Batchelor entrainment hypothesis, vertical momentum conservation, and mass conservation giveAn ascending parcel entrains ambient fluid from progressively lower ambient density. With the buoyancy frequencythe change of ambient reference density subtracts $Q{N}^2$ from its density-weighted buoyancy flux, so

The source supplies the kinematic buoyancy flux $B_0=F_0/{\rho }_0$. The unstratified line plume velocity scale is $B_0^{1/3}$. Comparing this with the stratification time scale $N_0^{-1}$ gives the stratified line-plume height scaleIf $z_s\gg H$, the initial plume is only weakly affected by the stable density stratification and reaches the ceiling much like an unstratified wall plume. If $z_s\sim H$, its buoyancy flux falls substantially during the rise; it approaches neutral buoyancy near the upper room, overshoots because of its momentum flux, and spreads as a horizontal buoyant intrusion. If $z_s\ll H$, the plume reaches neutral buoyancy low in the room and forms a low intrusion after a modest overshoot. In each sketch the plume widens by entrainment, while increasing $N_0$ lowers the neutral-buoyancy and overshoot heights.

The final state is displacement ventilation: fresh air of density ${\rho }_0$ forms a cool lower layer, radiator-heated air forms a well-mixed upper layer, and the wall plume crosses their interface at height $h$. Steady volume conservation requires the plume volume flux there to equal the imposed ventilation flux,

Below the interface the ambient is uniform, so $B_0=F_0/{\rho }_0$ is constant. Put $J=M/{\rho }_0$. The triangular-profile wall line plume equations reduce toThe pure plume conditions $V,J\to 0$ at the radiator select the similarity solutionConsequently the displacement-ventilation interface height isThe upper-layer reduced gravity follows from its steady buoyancy balance as $g_u^{\prime }=B_0/A=F_0/({\rho }_{0}A)$. The two-layer solution applies when the calculated interface satisfies $0<h<H$.