Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 80 (Geological Fluid Mechanics)

Use the flat-surface Boussinesq approximation, with small thermal expansion, constant properties, negligible surface deformation, and constant surface tension. The top is impermeable and shear-free, while the solid walls are no-slip. Temperature-dependent surface tension would introduce a separate Marangoni stress, which is not included in this buoyancy-driven model. The stated time scale requires $\kappa$ to mean thermal diffusivity; if the material property given is a dimensional thermal conductivity $k$, use $\kappa =k/({\rho }_0{c}_p)$.

Take $z$ positive downwards and define the nondimensional streamfunction by $u={\psi }_z$, $w=-{\psi }_x$. Velocity is scaled by $\kappa /L$, streamfunction by $\kappa$, and temperature is measured from the cold-end surface value. Define the Prandtl number and horizontal-length Rayleigh number bywhere ${\alpha }_T$ is the thermal expansion coefficient. Put $J(\psi ,f)={\psi }_z{f}_x-{\psi }_x{f}_z$, and let ${\nabla }^2={\partial }_x^2+{\partial }_z^2$. Taking the curl of the Boussinesq equations eliminates pressure and gives the full nonlinear vorticity-temperature systemThe buoyancy sign follows from the downward coordinate: warmer fluid has body acceleration $-g{\alpha }_T\Delta TT{\mathbf{e}}_z$.

Choose the same streamfunction constant, zero, on the connected boundary. The mechanical and thermal boundary conditions areThe upper stress-free boundary condition reduces to ${\psi }_{zz}=0$, since $\psi =0$ there already gives ${\psi }_{xx}=0$. An arbitrary additive constant in $T$ changes only the reference hydrostatic pressure.

In steady infinite-Prandtl-number convection, momentum inertia disappears but thermal advection remains:For a shallow lake, with fixed $\mathrm{Ra}$ as $ϵ\to 0$, the leading interior temperature is $T_0=x$. This is an outer solution: it does not satisfy the insulated sidewall derivative right at $x=0,1$. The leading global conductive solution, which resolves those end regions, isIt is harmonic, has $T_c(x,0)=x$, and satisfies the insulated sides and bottom. Away from sidewall layers of horizontal width $O(ϵ)$ it reduces to $x$ with exponentially small end corrections. These layers also turn the flow and enforce the sidewall no-slip boundary condition.

In the shallow interior the dominant momentum derivatives are vertical, so ${\psi }_{0,zzzz}=-\mathrm{Ra}$. Integrating with the surface and bottom conditions gives the shallow horizontal convection with an insulated bottom solution, with $y=z/ϵ$:Surface flow is toward the colder end, with a deeper return flow. The depth-integrated volume flux is zero because ${\psi }_0(ϵ)={\psi }_0(0)=0$. The bulk expressions apply away from the turning regions, not unchanged on the vertical walls. A useful small-advection condition in the end layers is $\mathrm{Ra}{ϵ}^4\ll 1$, automatic for fixed Rayleigh number in the stated thin limit.

The heat flux in these units is $\mathbf{q}=(uT-T_x,wT-T_z)$. For the leading bulk temperature, the horizontal transport through a vertical section per unit out-of-plane width isThus leading advective horizontal transport vanishes: both layers have the same temperature at a given $x$, and their opposing volume fluxes cancel. The leading conductive transport is from the hot end toward the cold end. In dimensional form it is $-{\rho }_0{c}_p\kappa \Delta Tϵ=-k\Delta Tϵ$ per unit width. Heat enters through the warmer portion of the maintained upper surface and leaves through its colder portion, with the conductive end adjustments particularly important near the two insulated sidewalls. It cannot enter through the insulated sides or bottom.

For the advection correction in shallow horizontal convection, set $T=x+\Theta (z)$ in the bulk. Since $w_0=0$ and $T_x=1$, the steady heat equation gives ${\Theta }_{zz}=u_0={\psi }_{0,z}$. The conditions $\Theta (0)=0$ and ${\Theta }_z(ϵ)=0$ imply ${\Theta }_z={\psi }_0$ and henceFor example the bottom bulk temperature is $T(x,ϵ)=x-\mathrm{Ra}{ϵ}^5/320$. The correction makes the returning deeper fluid slightly colder. Its horizontal advective heat transport isThus the transport including this bulk correction is $\boxed{{\mathcal{H}}_x=-ϵ-19{\mathrm{Ra}}^2{ϵ}^9/1451520}$, both terms directed toward the colder end.

Vertical transport must distinguish local flux from net transport. In the leading interior $w_0=T_{0,z}=0$, whereas the temperature correction gives a local downward conductive flux $q_z=-{\Theta }_z=-{\psi }_0>0$. The end circulation gives an upward advective transport: with impermeable sidewalls and $T_0\simeq x$, integration by parts gives ${\int }_0^{1}w{T}_{0}dx=-{\int }_0^{1}x{\psi }_{x}dx={\int }_0^1\psi dx\simeq {\psi }_0(z)<0$. It compensates that downward conduction. More generally, steady heat conservation and insulated sides give $d[{\int }_0^1{q}_{z}dx]/dz=0$, and the insulated impermeable bottom sets the constant to zero. ThereforeThere is no net vertical heat transport through the lake, despite its local recirculating heat flux and its nonzero horizontal heat transport.

Use the steady axisymmetric top-hat plume model, with plume radius $b(z)$, uniform mean upward speed $w(z)>0$ and reduced gravity $g^{\prime }(z)=g[{\rho }_a(z)-{\rho }_p(z)]/{\rho }_0$. The Boussinesq approximation uses constant reference density ${\rho }_0$ in inertia and retains the small density difference only in buoyancy. Assume a quiescent hydrostatic ambient, a slender fully turbulent plume, negligible molecular transport and an entrainment velocity $\alpha w$, with constant positive entrainment coefficient $\alpha$. Define the ambient buoyancy frequency by $N^2=-(g/{\rho }_0)d{\rho }_a/dz$; positive $N^2$ is stable stratification.

The integrated volume, momentum and buoyancy balances of a Boussinesq top-hat plume in a stratified ambient areThe first follows from side entrainment, the second from the integrated buoyancy force, and the third accounts for the changing ambient density. Common factors of $\pi$ are cancelled. Equivalently, the requested radius, speed and reduced-gravity equations areThese equations stop describing an upward light plume once $w$ vanishes, and their light-plume interpretation requires $g^{\prime }>0$.

For nonzero $C$, seek plume similarity in power-law stratification in the form $b=\beta z$, $w=A{z}^n$ and $g^{\prime }=G{z}^m$. The first balance gives $\beta =2\alpha /(n+2)$. Momentum gives $m=2n-1$ and $G=2(n+1)A^2$. The buoyancy balance then requires $3n=2+n+p$, soChoosing $A>0$ gives the exact power-law family, including its normalization:These are the amplitudes and physical branches of a power-law stratified plume. No source flux was specified, so an arbitrary finite-source solution is not determined by $C,p$ alone; the displayed family is an exact similarity solution, rather than the general solution for arbitrary initial fluxes.

With the specific flux convention $Q=b^{2}w$, $M=b^2{w}^2$ and $B=b^{2}w{g}^{\prime }$, their values areThe physical volume flux is $\pi Q$, mass flux is ${\rho }_0\pi Q$, momentum flux is ${\rho }_0\pi M$, and kinematic buoyancy flux is $\pi B$. Thus the “specific mass flux” is the mass flux divided by reference density, with or without the explicitly stated factor $\pi$.

The parameter restrictions follow directly from positivity. Positive radius needs $p>-6$. Positive $g^{\prime }$ and real positive $w$ further require $p>-4$ and $-C/(3p+8)>0$, givingFor stable $C>0$, buoyancy flux decreases with height, while volume and momentum fluxes increase. However, $B\propto z^{4+3p/2}$ diverges at $z=0$ throughout this interval. Also the ambient density gradient and plume speed are singular there. Thus this stable similarity branch can describe a region away from a virtual source, but cannot represent a finite-buoyancy-flux point source extending all the way to $z=0$. For ordinary regular stable stratifications, including positive $p$, a finite-source rising plume instead loses buoyancy and eventually reaches neutral buoyancy and a finite rise limit; the pure-power ansatz does not supply such a solution.

For unstable $C<0$ and $p>-8/3$, the buoyancy flux grows with height and tends to zero at the origin. The plume draws its increasing buoyancy from an unstable ambient rather than a finite positive source buoyancy flux. Such an ambient must be maintained or treated locally, since it is itself convectively unstable. If $p\le -1$, its density profile is singular at the source, while for $p>-1$ its increasing density contrast eventually violates the Boussinesq approximation at sufficiently large height. These are formal local similarity solutions with corresponding physical limits. For $C<0$ and $-6<p<-4$, the algebra also gives a real positive speed but negative $g^{\prime }$: this is a heavy upward flow, not the specified light plume.

At nonzero $C$, the values $p=-6,-4,-8/3$ do not give finite coefficients. Respectively they would require zero entrained volume-flux growth, zero momentum-flux growth despite nonzero buoyancy, or constant buoyancy flux despite nonzero ambient stratification. They are not omitted regular endpoints.

When $C=0$, the ambient is unstratified regardless of the nominal value of $p$. The nontrivial positive-buoyancy point-source solution is the Boussinesq point-source plume:Here $B_0>0$ is the constant specific buoyancy flux with $\pi$ removed. Its $Q\propto z^{5/3}$ and $M\propto z^{4/3}$ vanish at the point source, while $B=B_0$. Its near-source singularity is the usual ideal point-source limit; a finite vent replaces that limit by finite source data or a virtual origin.

Use an inertial gravity-current box model in a horizontal prismatic V-shaped valley. Let its section have area $A(h)=\lambda h^2$ and width $2\lambda h$, where the dimensionless constant $\lambda >0$ is half-width divided by depth. Let $X(t)$ be front position and $h(t)$ the representative uniform current depth. Assume negligible entrainment and volume loss, hydrostatic cross-sections, large Reynolds number, and negligible bed friction and axial bed slope. These assumptions describe the bulk spreading after the short dam-break adjustment; they do not give a universal law for a steep or friction-dominated valley.

For water of fixed volume $W$, conservation gives $W=\lambda h^{2}X$. Close the front speed by the gravity-current front condition $\dot{X}=\mathrm{Fr}\sqrt{gh}$, where $\mathrm{Fr}$ is a constant order-one Froude number for the chosen inertial front model. ThereforeIntegration gives $X^{5/4}=X_0^{5/4}+5K_{w}t/4$. Once the initial reservoir length $X_0$ is negligible, the inertial dam-break current in a triangular valley satisfiesThe time exponent differs from a flat-bottomed channel because a V-shaped section stores volume proportional to $h^2$. A viscous V-shaped channel gravity current obeys a different spreading law, so retaining the inertial assumption is essential.

For the short-lived eruption, interpret $W$ as the bulk mixture volume released during the event. Let $\varphi (t)$ be the well-mixed particle volume fraction, initially ${\varphi }_0$, let ${\rho }_p$ and ${\rho }_a$ be particle and ambient carrier densities, and assume a dilute Boussinesq approximation for the mixture. The particles provide reduced gravityAssume thermal buoyancy, compressibility, erosion and continued injection are negligible, and that the carrier volume remains $W$. Turbulent mixing keeps the particles suspended uniformly while they settle vertically at a constant terminal settling velocity $w_s$. For small spherical particles of radius $a_p$ in carrier viscosity ${\mu }_a$, the Stokes settling velocity is $w_s=2({\rho }_p-{\rho }_a)g{a}_p^2/(9{\mu }_a)$ when the particle Reynolds number is small. If an eruption volume is quoted as solid particle volume rather than mixture volume, replace the $W$ used here by that volume divided by ${\varphi }_0$.

The particle deposition flux per horizontal projected bed area is $w_s\varphi$. The current's projected settling area is $2\lambda hX$, soThe factor two reflects the triangular section's mean depth $h/2$. Put $s=\sqrt{\varphi /{\varphi }_0}$ and $K=\mathrm{Fr}\sqrt{g_0^{\prime }}(W/\lambda {)}^{1/4}$. The front and concentration balances becomeTaking the initial release length negligible compared with the final runout, integration yields the constant-settling runout in a triangular valley:In particular $R\propto W^{3/7}(g_0^{\prime }{)}^{2/7}{w}_s^{-4/7}$ at fixed valley shape and front coefficient. With a finite initial length, replace $R^{7/4}$ by $X_0^{7/4}+7K\sqrt{W/\lambda }/(4w_s)$ in the corresponding integrated relation.

The whole time dependence is determined analytically by the quadratureEarly spreading has $X\simeq (5Kt/4{)}^{4/5}$, the water law with $g$ replaced by $g_0^{\prime }$. At late times the concentration loss slows the flow, and $R-X$ decays exponentially with rate $7K/(4R^{5/4})=w_s/h_R$, where $h_R=\sqrt{W/(\lambda R)}$. The final length $R$ is a limiting runout reached as $t\to \infty$, not a finite-time stopping event in this ideal model.

To calculate the deposit profile of a finite triangular-valley particle current, let $m_d(x)$ be final deposited particle mass per unit downstream length, and $M_0={\rho }_p{\varphi }_{0}W$ the initial particle mass. Deposition at a fixed section starts at arrival time $t_a(x)$, when the front first reaches $x$. With the instantaneous front coordinate denoted $\xi$, use $dt={\xi }^{1/4}d\xi /[K(1-(\xi /R{)}^{7/4})]$ to obtainWriting $\eta =x/R$, this evaluates toand $m_d=0$ beyond $R$. The profile decreases downstream, vanishes quadratically at the runout, and obeys ${\int }_0^R{m}_d(x)dx=M_0$. This integral check ensures that all initially suspended particle mass is accounted for.

For an areal rather than a linear deposit density, the changing valley width must also be specified. At lateral coordinate $y_{\perp }$, the current covers the bed only while $|y_{\perp }|\le \lambda h$. Set $Z=\min (R,\lambda W/y_{\perp }^2)$, with $Z=R$ on the valley axis. The mass per horizontal projected bed area is zero if $Z<x$, and otherwise isOn the axis this reduces to ${\rho }_p{\varphi }_0{h}_R[\frac{7}{5}(1-{\eta }^{5/4})-\frac{7}{12}(1-{\eta }^3)]$. Divide projected areal density by $\sqrt{1+{\lambda }^{-2}}$ to obtain mass per actual sloping bed area. A specified packed deposit density then converts this areal mass to local deposit thickness; the spreading model itself predicts deposited mass, not a packing fraction.

Let $a(t)$ be the particle radius and $\Delta T=T_s-T_{\infty }>0$ the supercooling. Use spherical symmetry, constant thermal properties, equal phase densities $\rho$, and negligible fluid advection. Let $\mathcal{L}$ be latent heat per unit mass, $c_p$ the liquid specific heat, $k_{\ell }$ its thermal conductivity and $\kappa =k_{\ell }/(\rho c_p)$ its thermal diffusivity. The exterior temperature satisfies the radial heat equationwith $T(r,0)=T_{\infty }$ for $r>a_0$, $a(0)=a_0$ and $T(r,t)\to T_{\infty }$ as $r\to \infty$.

The Gibbs-Thomson effect gives curvature-induced melting-temperature depression. For a spherical interface writewhere $\gamma$ is solid-liquid interfacial energy per area and the factor two is the sphere's curvature $2/a$. Here $T_s$ in the thermodynamic formula is an absolute temperature. The notation $\Gamma$ also permits a given measured capillary coefficient without requiring its microscopic derivation.

In the one-phase Stefan problem, the solid has no internal heat flux and the Stefan condition isThe sign follows from heat balance: positive outward liquid heat flux releases latent heat as the solid grows. If the solid's thermal storage is retained, its temperature $T^{(s)}$ instead satisfies its own radial heat equation, ${\partial }_r{T}^{(s)}(0,t)=0$, and $T^{(s)}(a^{-},t)=T(a^{+},t)$. The two-phase condition is $\rho \mathcal{L}\dot{a}=k_s{T}_r^{(s)}(a^{-},t)-k_{\ell }{T}_r(a^{+},t)$. It reduces to the one-phase condition in the quasistatic limit because a regular harmonic temperature inside a sphere is spatially constant. An initial solid temperature profile is additional data for a fully time-dependent two-phase problem.

Two reciprocal Stefan number conventions are common. The large parameter appropriate to the requested slow-interface limit is the latent-to-sensible heat ratioIt measures the energy required for phase change relative to cooling or heating by the supercooling. The often-used sensible-to-latent definition is $\mathrm{Ste}=c_p\Delta T/\mathcal{L}=1/S$. Thus “large Stefan number” here means large $S$, not large $\mathrm{Ste}$ in that other convention.

The thermal diffusion time for scale $a_0$ is $a_0^2/\kappa$, whereas the interface time is $S{a}_0^2/\kappa$. More explicitly, with $R=r/a_0$, $\theta =(T-T_{\infty })/\Delta T$ and slow time $\tau =\kappa t/(S{a}_0^2)$, the liquid equation isFor $S\gg 1$, temperature therefore equilibrates before the radius changes appreciably. The quasistatic spherical Stefan problem with capillarity has a harmonic exterior temperature, fixed by its interface and far-field values:The small initial diffusion adjustment is omitted at this order. Inserting $T_r(a)=-\Delta T/a+\Gamma /a^2$ into the Stefan condition givesThis is the critical radius for a supercooled spherical seed. For $a_0>a_c$, the interface is warmer than the liquid and loses heat, driving growth. For $0<a_0<a_c$, its curvature-depressed interface temperature is colder than the liquid, so heat enters and the solid melts. At $a=a_c$, the uniform temperature $T_{\infty }$ supplies an equilibrium radius. It is unstable, since the radius-velocity derivative there is $\kappa /(S{a}_c^2)>0$. For a growing particle with $a\gg a_c$, $a^2$ increases asymptotically as $2\kappa t/S$.

For a subcritical particle the radius decreases monotonically and remains below $a_0$. HenceIt cannot remain positive beyond $S{a}_0^3/[3\kappa (a_c-a_0)]$, proving finite-time extinction of a subcritical spherical seed. The exact extinction time in the reduced model is obtained by separating variables:With ${\eta }_0=a_0/a_c\in (0,1)$, this isNear disappearance, the capillary term dominates and $\dot{a}\sim -\kappa a_c/(S{a}^2)$, so $\boxed{a^3\sim 3\kappa a_c(t_e-t)/S}$. For $a_0\ll a_c$, the lifetime is $t_e\sim S{a}_0^3/(3\kappa a_c)$; it diverges logarithmically as $a_0\to a_c^{-}$, consistently with the unstable equilibrium. The threshold and lifetime above are conclusions of the specified large-$S$ quasistatic continuum model, rather than an assertion that its approximation is uniform through every finite-$S$ shrinking-interface transient.