Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 84 (Fluid Dynamics of Energy)

Assume the dense current lies on the bottom, $\Delta \rho >0$, and its height is small compared with the confined aquifer depth $w$. This permits the hydrostatic approximation and neglect of leading-order resistance from the return flow of the original fluid. The dense-layer pressure anomaly on the basal seal is $\Delta \rho gh$. The along-layer Darcy flux, integrated over the current thickness, and the downward leakage flux are respectivelyVolume conservation requires $\varphi h_t+{\partial }_x{q}_x=-q_{\ell }$. Thus the porous gravity current with hydrostatic leakage through a basal seal obeysHere permeability of a porous medium has units of area, and $\Omega$ has units of inverse time. The original PDF instead prints the sink coefficient $\lambda k\Delta \rho g/(\mu b)$. With $\lambda$ defined there as a permeability, that coefficient has units of area per time and cannot multiply $h$ to give $h_t$. It also omits the division of leakage by porosity. This is a genuine source error, not a Darcy law convention. The remaining parts solve the stated linear-drainage equation for a constant $\Omega >0$; the physically consistent coefficient is the one above. Formally treating the printed coefficient as an independently prescribed inverse-time constant leaves those mathematical solutions unchanged.

For a concentrated line release, take $V$ to be actual fluid volume per unit source length and allow symmetric spreading to both sides of $x=0$. PutSubstitution in the porous medium equation gives the exponential drainage transform for porous-medium diffusion:The Barenblatt solution has conserved area ${\int }_{-\infty }^{\infty }Hdx=V/\varphi$. A quadratic cap satisfies this equation:Indeed, inside the current $H_x=-x/(3Ds)$ and $R\propto s^{1/3}$; differentiating verifies $H_s=D(H{H}_x{)}_x$. Its area is $\int Hdx=2R^3/(9Ds)=V/\varphi$, fixing the prefactor. The flux vanishes at the two moving noses because $H=0$ there. Hence the remaining physical volume and occupied width areFor a one-sided current on a reflecting half-line, replace the $2$ in $R^3$ by $1$ when $V$ denotes the total release into that side. Volume alone does not select this exact solution for an arbitrary finite initial patch; it describes an impulsive line release. Its initial central height is singular, so a finite injection region and the constraint $h\ll w$ regularize the near-source early-time behavior.

The transformed time tends to a finite limit, $s\to {\Omega }^{-1}$. Therefore the symmetric current's maximum distance from the source isand its maximum total width is $2R_{\max }$. With the corrected physical leakage coefficient, this becomes $R_{\max }=[9kbV/(2\varphi \lambda ){]}^{1/3}$. The porous gravity current approaches this limiting distance asymptotically, while its height and mobile fluid volume tend to zero. With no leakage, $\Omega \to 0$, the limiting distance diverges and the ordinary $t^{1/3}$ spreading is recovered.

Write $R=C{s}^{1/3}$, where $C^3=9DV/(2\varphi )$. At a fixed $x$ inside the current,Since $ds/dt>0$, instantaneous constancy of depth is equivalent to ${\partial }_{s}h=0$. Differentiation givesThus the lateral locations at which the depth is neither increasing nor decreasing areAt a fixed nonzero position reached by the front, the height first increases from zero, attains this maximum, and later decreases. The invasion envelope of an exponentially draining porous current is therefore the parametric curveWithin the aquifer, the previously invaded region lies between the basal boundary and this maximum-height envelope, for $|x|<R_{\max }$. The envelope tends to zero height at the limiting lateral nose. Its central singularity belongs to the ideal impulsive release, so it cannot be extended physically above the finite aquifer thickness.

Let $k_1>k_2$ and let both original and injected fluids initially have dynamic viscosity $\mu$. Assume sharp displacement fronts, constant porosity and negligible exchange between layers. The imposed pressure difference gives Darcy velocities $q_i^0=k_i\Delta P/(\mu L)$ and interstitial velocities $u_i^0=q_i^0/\varphi$. The two arrival times are thereforeIf $A_i$ is the flow cross-section of layer $i$, its volume discharge is $Q_i^0=A_i{q}_i^0$. Equal viscosities make the hydraulic resistance independent of front position, so the total fluid discharge $Q_1^0+Q_2^0$ stays constant. The injected-fluid component at the outflow well isIt jumps from zero to the high-permeability discharge at the first breakthrough, then to the total discharge at the second. Before a layer breaks through, its outflow is original reservoir fluid. Cross-sections are needed to state an absolute volume discharge or its partition; the arrival times do not require them.

A model must track polymer age as well as its position. Polymer enters during $s\in [t_a,t_a+\Delta t]$ and activates at age $\tau$. The first activation in laboratory time is $t_a+\tau$, not $\tau$ measured from initial water injection. Assume activation permanently changes the local dynamic viscosity from $\mu$ to $\lambda \mu$, with $\lambda >1$, and neglect polymer diffusion, retention and shear-dependent rheology.

Let $X_i(t)={\int }_0^t{u}_i(r)dr$ be cumulative pore-fluid displacement. A fluid particle entering at time $s$ has position $X_i(t)-X_i(s)$ until it exits. At time $t\ge t_a+\tau$, the youngest activated polymer was injected at $s_y=\min (t_a+\Delta t,t-\tau )$. The length of activated polymer still inside layer $i$ is exactlySet ${\ell }_i=0$ when $t<t_a+\tau$. This accounts both for gradual activation of the slug and its later exit. Hydraulic resistances add in series, so delayed polymer diversion in parallel porous layers is described byThese history-dependent equations close the model and can be integrated forwards, retaining previously computed values of $X_i$. The overall volume discharge and low-permeability inflow fraction areThe original displacement front reaches the distal well when $X_i(t)=L$. Consequently the injected-fluid outflow is ${\sum }_i{Q}_i(t)1_{X_i(t)\ge L}$, and the complementary sum gives original-fluid production.

The mechanism is especially transparent if $\Delta t<\tau$, so the entire slug is injected before activation begins, and it activates before leaving the high-permeability layer. Before activation, its lengths are $u_i^0\Delta t$. While each complete activated slug remains in its layer, those lengths stay fixed andThe fractional resistance increase is larger in the fast layer. An effective selective regime is $\Delta t<t_a$ and $\Delta t<\tau <t_a$, with $(\lambda -1)\Delta t/t_a$ appreciable while $(\lambda -1)\Delta t/t_s$ is small; the full delay model handles the activation ramps and the later exits. The relative share entering the slow layer then increases, although its absolute discharge cannot increase at fixed pressure. The total discharge decreases during polymer occupancy and returns to its original value once the activated slugs leave. A fixed-total-injection-rate experiment would have a different pressure response and cannot be substituted for this fixed-pressure model.

Before the unmodified low-permeability breakthrough, the only original-fluid discharge is from layer 2. Polymer raises its resistance, so its rate is never greater than $Q_2^0$ at the same fixed pressure. Selective diversion therefore does not increase the instantaneous original-fluid discharge during this initial interval.

It can increase original-fluid production at a later specified time by delaying completion of the sweep. Let $T_2$ be the treated low-permeability breakthrough time. Integrating its interstitial velocity giveswhenever activated polymer occupies that layer for a positive time before breakthrough. For $t_s<t<T_2$, the unmodified reservoir produces no original fluid, whereas the treated reservoir still produces $Q_2(t)>0$. Thus the strategy can give greater original-fluid flux at a given time after unmodified breakthrough, through delayed sweep rather than increased absolute slow-layer flow. A sufficient activation condition is $t_a+\tau <t_s$ with a nonzero slug that has not already exited the slow layer. If polymer only affects layer 1 after its breakthrough and never affects layer 2 before its own breakthrough, this particular benefit does not occur. Integrated recovery over a chosen finite operating window must be distinguished from a comparison at one time.

Take $Q$ as the fluid volume flux per unit source length into one side of the aquifer. For a symmetric release with total rate $Q$, use $Q/2$ on each side. Introduce a constant porosity $\varphi$ to distinguish Darcy velocity from the advecting interstitial velocity; if the given flux is already normalized to pore volume, this factor is absorbed in its definition. With a pressure gradient $G=-p_x>0$, Darcy law givesThus, with $\xi =z/h$, the pore velocity and its depth average areThe tracer concentration initially obeys the two-dimensional advection-diffusion equationReflecting transverse boundaries and constant effective molecular diffusion are assumed. No independent radioactive decay rate is supplied; it can be added separately if it is significant over the experiment.

For times long compared with $h^2/D$, transverse diffusion exchanges tracer between fast and slow streamlines. Write $c=C(x,t)+\chi (z)C_x+\cdots$, where $C=\overline{c}$ and $\overline{\chi }=0$. At the leading transverse-correction order,Integration yieldsIts mean and boundary derivatives are zero. Averaging the concentration equation gives $C_t+U{C}_x=[D-\overline{(u-U)\chi }]C_{xx}$. Integration by parts in the cell equation shows that the enhancement is positive:Consequently Taylor dispersion in a parabolic porous-layer velocity profile givesInjection fixes the inlet concentration or solute-flux boundary condition. This depth-averaged equation requires transverse equilibration and longitudinal concentration variations slow enough for the cell expansion. The supplied PDF hint has both a different polynomial and a false numerical value: its integral evaluates exactly to $-83/2520$, not $0.3525$. The cell problem above supplies the coefficient independently.

After transverse equilibration, a localized release on a sufficiently long aquifer has the advection-diffusion heat-kernel solutionwhere $\mathcal{M}=\int Cdx$ is conserved. The actual tracer amount per unit source length is $\varphi h\mathcal{M}$ in this normalization. Its center moves at the mean interstitial velocity $U$, its variance is $2D_{\mathrm{eff}}t$, and its peak decreases as $t^{-1/2}$. A finite initial packet is convolved with this kernel.

Before times of order $h^2/D$, this effective Gaussian description need not hold. The parabolic velocity profile initially shears a packet, and particles initially at different depths acquire different along-flow displacements. Transverse diffusion subsequently erases depth memory and produces the long-time Taylor dispersion law. A short pulse can therefore have an initially asymmetric or strongly distorted longitudinal shape before entering the averaged regime.

Interpret absorption as irreversible capture of radioactive tracer by the formation at the stated first-order rate. With stored amount $S$ measured per the same reference pore-storage volume as $C$, irreversible capture of an advecting tracer givesAdding the equations conserves dissolved plus captured tracer, apart from advective and diffusive transport. A different solid-volume normalization inserts the corresponding porosity factor in $S_t$. The averaged equation has the same transverse-equilibration limitation as in part (a).

For a maintained inlet concentration $C(0,t)=C_0$, a bounded downstream stationary water profile satisfies $D_{\mathrm{eff}}{C}^{\prime \prime }-U{C}^{\prime }-\lambda C=0$. Its characteristic roots are $(U\pm \Delta )/(2D_{\mathrm{eff}})$, where $\Delta =\sqrt{U^2+4\lambda D_{\mathrm{eff}}}$. Rejecting the growing root givesIf the inlet instead prescribes solute flux $J_0=UC-D_{\mathrm{eff}}{C}_x$, replace $C_0$ by $J_0/(U+D_{\mathrm{eff}}r)$. In the advection-dominated limit, $r\simeq \lambda /U$, giving an attenuation length $U/\lambda$.

The corresponding capture-rate profile is $\lambda C_{\infty }(x)$. There is an important qualification to the requested adsorbed profile: with a continuous source, irreversible adsorption and no solid decay or saturation, the water concentration can be stationary but the accumulated solid concentration cannot. At every point with positive $C_{\infty }$,Thus the deposited amount grows asymptotically linearly with a spatially exponential rate profile. A time-independent deposit would require additional desorption, finite capacity or radioactive decay in the solid; for example $S_t=\lambda C-{\gamma }_{s}S$ would give $S_{\infty }=\lambda C_{\infty }/{\gamma }_s$ if ${\gamma }_s>0$.

If the source is the localized release in part (b), there is instead a finite final deposit and no nonzero steady water concentration. On the full line, multiply the moving Gaussian by $e^{-\lambda t}$. Integrating its governing equation over all time gives $D_{\mathrm{eff}}{J}^{\prime \prime }-U{J}^{\prime }-\lambda J=-\mathcal{M}\delta (x-x_0)$ for $J={\int }_0^{\infty }Cdt$. Continuity and the derivative jump at $x_0$ then giveIts integral is $\mathcal{M}$, so all the released tracer is eventually captured in this infinite-domain model. This distinguishes the finite-pulse deposited profile from the continuously accumulating deposit under sustained injection.

The K-epsilon turbulence model evolves turbulent kinetic energy $k$ and its turbulent kinetic energy dissipation rate $ϵ$, rather than resolving turbulent fluctuations. It closes the Reynolds stress with an isotropic eddy viscosity, ${\nu }_T=C_{\mu }{k}^2/ϵ$, and models transport of $k$ and $ϵ$ by gradient diffusion with coefficients ${\nu }_T/{\sigma }_k$ and ${\nu }_T/{\sigma }_{ϵ}$. Production is tied to mean strain; dissipation-equation production and destruction are modeled terms with empirical constants. The characteristic velocity and length are $\sqrt{k}$ and $k^{3/2}/ϵ$. These assumptions are primarily for high Reynolds number, and strongly anisotropic flow, sharp interfaces and molecular-viscosity-dominated regions need additional treatment. The dissipation equation is a closure, not a second exact energy-conservation law.

Put $\xi =x/{\delta }_0$ and $n=2p-q$. The eddy viscosity is ${\nu }_T=C_{\mu }\sqrt{k_0}{\delta }_0{\xi }^n$, since ${\delta }_0=k_0^{3/2}/{ϵ}_0$. In the energy equation, advection scales as ${\xi }^{p-1}$, turbulent diffusion as ${\xi }^{n+p-2}$, and destruction as ${\xi }^q$. A leading advection–diffusion K-epsilon turbulence front requires $n=1$, henceComparing the leading coefficients then gives $U_0=C_{\mu }p\sqrt{k_0}/{\sigma }_k$. Similarly, the leading coefficients of the dissipation equation give $U_0=C_{\mu }q\sqrt{k_0}/{\sigma }_{ϵ}$. Their equality implies $q=\sigma p$, where $\sigma ={\sigma }_{ϵ}/{\sigma }_k$. Combining this with $q=2p-1$ yieldsThe assumed positive exponents require $0<\sigma <2$. Under these values, the omitted destruction terms are indeed higher order, as checked in part (d), so the leading balance is not circular.

The K-epsilon turbulence front hasIts characteristic integral length and the specified frequency scale areThe ratio $ϵ/k$ is an inverse eddy-turnover time. The PDF calls it turbulent vorticity, but it should not be identified with the exact root-mean-square vorticity at dissipative scales, which involves molecular viscosity as well.

The dominant balance is convection of small turbulent energy and dissipation by the steady mean flow against their modeled turbulent diffusion. Diffusion supplies turbulent quantities from the turbulent side to the front; there is no local turbulent production in this problem. For $q=2p-1$, the energy-destruction term divided by its advection term is proportional to ${\xi }^{q-(p-1)}={\xi }^p$. In the dissipation equation, the destruction term has exponent $2q-p$, so its ratio to advection is likewise ${\xi }^{(2q-p)-(q-1)}={\xi }^p$. Both tend to zero for $p>0$. Thus advection balances turbulent diffusion, while both modeled destruction terms are asymptotically smaller.

A continuous non-turbulent edge should have vanishing $k$, $ϵ$, characteristic turbulent length $L$ and turnover frequency $\omega$. With $p,q>0$ and $\sigma <2$, the length in part (c) vanishes exactly when $\sigma <3/2$, while the frequency vanishes exactly when $\sigma >1$. Hence the requested formal consistency interval isAt the endpoints one of these characteristic quantities remains finite rather than vanishing. This is consistency of the ideal turbulent closure and its edge limits, not a claim of validity arbitrarily close to $x=0$ in a real fluid: ${\nu }_T\to 0$ eventually makes neglected molecular viscosity important. Nor does it establish the stronger local scale separation $L\ll x$; the displayed solution has $L/x\to \infty$, another reason to treat the exact edge as a formal model limit.

Since $2p-q=1$, the leading turbulent viscosity is linear:Using $U_0=C_{\mu }p\sqrt{k_0}/{\sigma }_k$ and $1/p=2-\sigma$ gives

Let $P$ denote total gravitational potential energy, $P_b$ the background potential energy of the adiabatically sorted density field, and $A=P-P_b$ the available potential energy. Define $B$ positive for reversible transfer from kinetic energy to $P$, $ϵ$ as viscous kinetic-energy loss, $F$ as imposed mechanical work, $W$ as molecular transfer from internal energy to potential energy, and $M$ as the irreversible conversion of $A$ to $P_b$. With boundary contributions removed, the stratified mixing energy budget isFor an unforced flow set $F=0$. In an alternative notation, if ${\Phi }_d={\dot{P}}_b$ is the full diffusive increase of background energy and ${\Phi }_i=W$, then $M={\Phi }_d-{\Phi }_i$ and $\dot{A}=B+{\Phi }_i-{\Phi }_d$. This distinguishes irreversible density mixing from reversible buoyancy exchange. The internal reservoir obeys $\dot{I}=ϵ-W$, so summing the kinetic, potential and internal budgets gives $d(K+P+I)/dt=F$ as required by conservation of energy.

An instantaneous mixing efficiency defined as a fraction of irreversible mechanical-energy loss isThe cumulative mixing efficiency over $[0,t]$ isIt is weighted by the loss rate, rather than being the arithmetic time average of $\eta$. The flux coefficient $\Gamma =M/ϵ$ is also sometimes called mixing efficiency; it obeys $\eta =\Gamma /(1+\Gamma )$, and the corresponding cumulative coefficient is ${\Gamma }_c=\int M/\int ϵ$. These conventions should be stated rather than interchanged. Only when $\dot{A}$ is negligible does $M\simeq B$, permitting the frequently used buoyancy-flux form $B/(B+ϵ)$. Input-power efficiency $M/F$ is another quantity and generally differs during changes in stored energy.

Use the model's normalization in which $K$ has velocity-squared units: it is kinetic energy per fixed reference mass, or a correspondingly normalized total energy. LetFor $K>0$, setting $s=\sqrt{K}$ reduces the equation to $2\dot{s}=a-s^2/L_v$. Suppose first that $L_f<L_{\rho }$, so $a>0$. Put $s_{\infty }=\sqrt{a{L}_v}$, $y=s/s_{\infty }$, $y_0=\sqrt{K_0/(a{L}_v)}$ and $\gamma =s_{\infty }/(2L_v)$. Separating variables gives $\dot{y}=\gamma (1-y^2)$ and therefore the forced stratified mixing with square-root power input solutionThis expression satisfies $K(0)=K_0$ and the differential equation directly.

For $0<K_0<K_{\infty }$, one may equivalently write $y=\tanh [\gamma t+\mathrm{artanh}{y}_0]$; the energy increases monotonically to $K_{\infty }$. For $K_0>K_{\infty }$, $y=\coth [\gamma t+\mathrm{arccoth}{y}_0]$, and the energy decreases monotonically to the same value. $K_0=K_{\infty }$ is stationary. For a very large initial energy, the initial evolution is approximately $s=s_0/[1+s_{0}t/(2L_v)]$, until forcing becomes comparable with dissipation. For a very small positive seed, $s$ initially grows approximately linearly, before leveling off. Thus both paragraphs labelled (c) in the PDF are answered by the solution and its two branches.

The positive-seed condition is material. If $K_0=0$, the original equation also admits $K\equiv 0$, and its square-root right-hand side is not Lipschitz at zero. It admits delayed-start solutions $K=0$ for $t\le t_d$ followed by $K=K_{\infty }{\tanh }^2[\gamma (t-t_d)]$. Therefore convergence to the positive steady state is not guaranteed from exactly zero turbulence without a startup prescription. For completeness, if $a=0$ a positive seed decays as $K=K_0/[1+\sqrt{K_0}t/(2L_v){]}^2$. If $a<0$, it reaches zero in finite time, $t_{\ast }=2\sqrt{L_v/|a|}\arctan \sqrt{K_0/(|a|L_v)}$, and the nonnegative continuation is at rest.

With negligible internal-to-potential exchange and negligible change in available potential energy, the budgets give $M=B$. The model supplies $B=\sqrt{K}{U}_0^2/L_{\rho }$ and $ϵ=K^{3/2}/L_v$, so for $K(t)>0$ the instantaneous mixing efficiency isFor a positive initial seed and $L_f<L_{\rho }$, both loss rates approach positive constants. Their time integrals grow linearly at those rates, implyingIf mixing efficiency instead denotes the flux coefficient, then $\Gamma (t)=L_v{U}_0^2/[L_{\rho }K(t)]$ and ${\Gamma }_{c,\infty }=L_f/(L_{\rho }-L_f)$. During the transient, $\int (B+ϵ)dt=\int Fdt+K_0-K(t)$; equating irreversible loss with the instantaneous forcing would omit changing kinetic-energy storage. A permanently resting zero-seed branch has zero numerator and denominator, so it has no defined loss-based efficiency.

Keep the forcing ratio $r=L_f/L_{\rho }$ fixed in $(0,1)$. ThenFor $L_v\ll L_{\rho }$, dissipation is strong at a given $K$ and $K_{\infty }\ll U_0^2$ when $r$ is not extreme. If the initial energy is of order $U_0^2$, it decays toward a small steady value and starts with a small instantaneous mixing efficiency. This is a regime of short velocity-fluctuation scales and efficient viscous loss relative to the modeled buoyancy conversion at that initial energy.

For $L_v\gg L_{\rho }$, dissipation is weak at that same energy and $K_{\infty }\gg U_0^2$. A moderate initial energy grows and initially has a large instantaneous mixing efficiency; increasing stored energy eventually makes dissipation comparable with the remaining forcing. The energy-relaxation time is of order $L_v/\sqrt{K_{\infty }}$, which grows as $\sqrt{L_v}$ at fixed forcing and density lengths.

Within this closure, both limits have the same long-time loss fraction ${\eta }_{c,\infty }=L_f/L_{\rho }$. The steady energy changes with $L_v$ so as to compensate for its explicit effect on dissipation. Different initial conditions and finite observation windows give different transient efficiencies; one cannot infer a universal long-time efficiency from $L_v/L_{\rho }$ alone. In a real experiment the length scales need not remain independent of the changing flow, so this conclusion belongs to the stated model.

Use pi-normalized kinematic plume fluxes: physical volume, momentum and buoyancy fluxes are $\pi Q$, $\pi M$ and $\pi F$. For a top-hat plume model in a homogeneous ambient, constant entrainment coefficient $\alpha$ givesEliminating height between the first two equations yields $d(M^{5/2})/dQ=5FQ/(4\alpha )$. Hence the plume flux-balance invariant is $M^{5/2}-5F{Q}^2/(8\alpha )$. Pure plume balance means that this invariant vanishes, including at a finite-volume source:It does not mean that $Q_s$ and $M_s$ must individually be zero. The balanced solution has a plume virtual origin below the physical source:where $z_v=[Q_s/(C{F}_s^{1/3}){]}^{3/5}>0$. For given nonzero source fluxes, the displayed balance is a compatibility condition; they cannot be chosen independently and still describe an exactly balanced source.

Let $z_f(t)$ be the filling-box first front, with $z_f(0)=H$. Below it the original fluid has not yet been through the plume. In the filling box model, the turbulent plume adjusts rapidly and occupies a negligible fraction of the room's area. A point source with negligible source volume flux supplies $Q(z)=C{F}_s^{1/3}{z}^{5/3}$. All its volume crossing the front grows the processed region, so volume conservation givesPut $a_p=\pi C{F}_s^{1/3}/A_c$. Integrating $d(z_f^{-2/3})/dt=2a_p/3$ yieldsThe first front approaches the floor only as $t\to \infty$. A nonzero source volume flux would require an equal exhaust to keep the room volume fixed; it is treated in the following parts.

The balanced finite source has $Q(z)=C{F}_s^{1/3}(z+z_v{)}^{5/3}$ and $Q(0)=Q_s$. An ideal small opening exhausts the physical source volume $\pi Q_s$. When $z_f>h$, the opening is in the unprocessed lower region. Its outflow and the plume's entrainment both remove lower fluid, while all plume flux across the front enters the processed region. When $z_f<h$, the opening exhausts processed fluid, reducing the net growth of that region by $\pi Q_s$. Thus the finite-source filling-box front with a side opening obeysThe ideal exhaust switches fluid type as the front crosses its height. Above the opening the solution is $z_f=[(H+z_v{)}^{-2/3}+2a_{p}t/3{]}^{-3/2}-z_v$, and it reaches $h$ atBelow the opening, the implicit solution isThe model assumes an effective point opening with one-way source-volume exhaust; a tall opening supporting simultaneous density-driven exchange would need additional boundary-flow equations.

If $h>0$, the front enters the lower-opening branch before reaching the floor. Near $z_f=0$,The final approach is exponential and the time integral in part (c) diverges logarithmically. Thus no finite floor-arrival time exists. If the opening is at $h=0$, the first branch applies until arrival and has a finite nonzero downward speed at the floor. The arrival time isHence finite-time floor arrival occurs only for a floor opening, within the stated finite-source model. In the point-source limit $Q_s\to 0$, $z_v\to 0$ and this arrival time diverges, recovering part (b).

Assume both openings have positive area, no imposed wind pressure, a light buoyant plume, and a Boussinesq approximation. Write $S=\pi Q_s$ and $B_s=\pi F_s$ for physical source volume and buoyancy flux. Include discharge coefficients $C_H,C_0$; if ideal openings are intended, set these to one. In a steady two-layer displacement ventilation state with interface height $z_{\ast }$, let the floor inflow be $q_0$ and roof outflow $q_H$. Volume and buoyancy balances giveThe hydrostatic pressure head across the warm upper layer must equal the two opening losses:A positive unprocessed lower layer needs $q_0>0$. At the boundary where this layer disappears, $z_{\ast }=0$, the plume can entrain no lower ambient fluid, so $q_0=0$, $q_H=S$ and $g_u^{\prime }=B_s/S$. The critical condition is therefore $S^3=2C_H^2{A}_H^2{B}_{s}H$.