Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 345 (Environmental Fluid Dynamics)

Write $h(t)$ for the height of the filling-box first front above the floor and $P(h)$ for the upward plume volume flux crossing it. Neglect the plume's horizontal area compared with $A$, use the Boussinesq approximation, and measure buoyancy relative to the initially unmodified ambient. The first front separates fluid already modified by plume discharge from the unmodified fluid below; the entire upper region need not be uniformly mixed during an unventilated filling box model transient.

A finite injected volume flux $Q$ cannot enter a hermetically sealed, fixed-volume room under the incompressible flow assumption. If $Q>0$, an equal overflow or exhaust is needed. Taking that exhaust at the ceiling, volume conservation of the region above the front givesThe source volume flux is subtracted because only entrained room fluid is removed from the initially unmodified region. The printed pure plume law has $P(0)=0$, so it cannot be an exact finite-$Q$ source law down to the physical floor. In the far field where $Q\ll P(h)$, or for a heat source whose injected volume flux is negligible, it givesHere $t=0$ is when the plume discharge first reaches the ceiling. Retaining $Q$ while using the far-field law gives $-\dot{h}=[\lambda B^{1/3}{h}^{5/3}-Q]/A$ only while that law remains a valid approximation; it must not be extrapolated to predict negative plume entrainment. A plume virtual origin below the floor is one way to match a nonzero source volume flux.

For displacement ventilation, let $F$ be the total ceiling exhaust, $S$ the total buoyant-source volume flux, and $L=F-S$ the separate ambient inflow into the lower layer. Volume conservation and the upper-layer buoyancy flux balance giveThe ambient inflow has zero reference buoyancy. In the ideal two-layer closure the upper layer is uniformly mixed with reduced gravity $b>0$, and the lower layer has zero reference buoyancy. The plume supplies the upper layer while lighter upper-layer air is removed at the ceiling. An interface is driven downward when $P(h)>F$ and upward when $P(h)<F$; since $P^{\prime }(h)>0$, its equilibrium is stable.

In the negligible-source-volume pure plume limit, identifying $F=Q_v$ gives the intended one-source answerThe prescribed inequality makes $h_1/H\ll 1$, so both layers fit in the room. It does not, by itself, justify neglecting a finite source $Q$: the pure plume source region must also be small compared with $h_1$, normally requiring $Q\ll Q_v$. If $Q_v$ denotes additional background ventilation rather than total exhaust, replace $Q_v$ in both formulas by $Q+Q_v$. If it is total exhaust, the separate ambient inlet carries $Q_v-Q$. These conventions must not be mixed.

For $n$ independent equal axisymmetric pure plumes, adding their volume fluxes givesThus the conventional multiple-plume displacement ventilation calculation with negligible source volume flux and separate ambient ventilation givesSplitting the buoyancy flux increases total fluid entrainment, so the same ventilation balances the plumes at a lower interface. The global buoyancy flux is unchanged, which leaves the upper-layer reduced gravity unchanged. Independence requires sufficiently separated sources and plumes that do not merge below the interface.

There is, however, a substantive problem with the literal source-volume specification in this paragraph of the PDF. It puts all $Q_v$ through the buoyant sources, so $S=F=Q_v$ and $L=0$. For every positive interface height, entraining plumes haveThere is no fresh ambient inflow to replace the lower-layer fluid being entrained. Consequently the stated positive steady two-layer configuration does not follow from those literal hypotheses. This is no steady displacement layer without ambient supply, not a small correction to the $n^{-2/5}$ scaling. For example, a matched plume virtual origin givesWhen $S=F$, this yields $h_{\ast }=0$. The unshifted answer $h_n$ equals the omitted $z_0$ in this case, so neglecting the source region cannot be justified. Also each source has reference buoyancy $(B/n)/(Q_v/n)=B/Q_v$, already equal to the proposed upper-layer value; a steady positive plume cannot entrain zero-buoyancy lower fluid and still deliver that same value. The conventional answer is valid for a corrected arrangement with negligible-volume buoyancy sources and a separate ambient inlet, or with source flux $S<Q_v$ and the remaining $Q_v-S$ supplied directly to the lower layer. A real room may have other circulation or continuous stratification, but that requires a different model.

For the intended displacement ventilation model, reducing the exhaust to $F=Q_v/2$ makes the new equilibriumThe interface initially descends since its old plume volume flux exceeds the new exhaust. In ventilated filling-box relaxation, writing $V_{\ast }=A(H-h_{\ast })$ and taking the linearization of the two balances above givesAs $P_n^{\prime }(h_{\ast })=5F/(3h_{\ast })$, the two decay times areA dimensional analysis estimate of the full adjustment time is thereforeThe interface adjusts on the shorter lower-layer scale; replacing the buoyancy stored throughout the upper layer is slower. Exact equilibrium is approached asymptotically, and a specified small relative tolerance adds a logarithmic factor. For the literal all-inflow-through-sources arrangement there is no positive two-layer equilibrium to return to; $2AH/Q_v$ is only a flushing-scale estimate for a different, whole-room adjustment.

Use a two-sided top-hat line plume with full width $w(z)$, upward speed $U(z)$, and reduced gravity $b(z)$ uniform across the plume. Fluxes are measured per unit span along the line source: the volume flux is $V=wU$, the kinematic momentum flux is $M=w{U}^2$, and the buoyancy flux is $B=wUb$. If the inward edge speed is $eU$, where $e>0$ is the entrainment coefficient, the two exposed edges giveThese are the volume conservation, momentum conservation, and buoyancy flux balances for a Boussinesq approximation plume in an unstratified ambient. Entrained ambient fluid supplies neither vertical momentum nor reference buoyancy.

A pure plume has no persistent source length scale. Since a line-source $B$ has dimensions $L^3{T}^{-3}$, dimensional analysis gives constant $U\propto B^{1/3}$, $w\propto z$, and $V\propto z$. Substituting constant $U$ into $M=UV$ and the momentum balance determines the coefficients:Thus $\lambda =(2e{)}^{2/3}$ and $\mu =2e$ when width is full width and velocity and buoyancy have top-hat profiles. If width means half-width, $\mu =e$ instead. Other prescribed profile shapes change these numerical factors; the linear growth laws remain the same. The point-source $B$ in Question 1 has different dimensions, so its $z^{5/3}$ law must not be used here.

Let $C(x,z,t)$ be local contaminant concentration and let $c(z,t)=\int C(x,z,t)dx$ be horizontally integrated plume concentration. Its conserved amount per unit span is ${\int }_0^{\infty }cdz=V_c$. The dilute contaminant is a passive scalar; the maintained plume is unaffected by its impulsive release. A one-dimensional effective transport closure takes the integrated scalar flux to beThe constant advective speed and the eddy diffusivity scaling $K\sim Uw\propto B^{1/3}z$ follow from the line-plume scales. The entrainment coefficient alone does not determine the scalar dispersion coefficient: this Fickian closure for the integrated variable is an additional modelling assumption. Mass conservation $c_t+J_z=0$ then gives the displayed transport equation in the PDF. With this effective closure and advective speed chosen as $U$, $\alpha =(2e{)}^{-1/3}$; if $K=\kappa Uw$, then $\beta =\kappa (2e{)}^{2/3}$, with independent dimensionless mixing coefficient $\kappa$.

There is a second possible closure convention. If one instead applies local diffusive flux to horizontally averaged plume concentration $\overline{C}=c/w$, its integrated diffusive flux is $-Kw{\overline{C}}_z=-K(c_z-c/z)$. For $K=Dz$, the total scalar flux is then $(U+D)c-Dz{c}_z$. The same printed equation is obtained with $a=U+D$, rather than $a=U$. Thus the printed $\alpha$ and $\beta$ should be regarded as effective coefficients unless the averaging and closure conventions are specified; no equality between $\alpha$ and the velocity prefactor is universal.

Assume $a,D>0$, $z>0$, an initial impulse $V_c{\delta }_0$ at the origin, no further contaminant input, and zero endpoint scalar flux for $t>0$. For a similarity solution, putSubstituting into the advection-diffusion equation givesDecay at infinity makes the integrated constant zero. Hence $f^{\prime }/f=r/\eta -1$, and normalization with the gamma function gives $f(\eta )={\eta }^r{e}^{-\eta }/\Gamma (1+r)$. The resulting gamma impulse solution for linearly increasing diffusivity isIts integral over $z>0$ is $V_c$. Its flux is $J=(z/t)c$, which vanishes at both endpoints for $t>0$. The scale $Dt$ shrinks to zero as $t\downarrow 0$, so the normalized solution has weak convergence of probability measures to the required unit source impulse. The mass at the boundary is a full unit impulse on the half-line, not half of a whole-line impulse.

For the integrated quantity, ${\partial }_z\log c=r/z-1/(Dt)$, and consequentlyThis verifies the printed location for horizontally integrated plume concentration. It is a maximum because the derivative changes from positive to negative there. The normalized profile is a gamma distribution with shape $1+r$: its expected value is $(a+D)t$ and its variance is $(a+D)Dt$, so neither the mean position nor the spread should be mistaken for the modal position.

The PDF changes from “integral” to “averaged” concentration in its final request. A genuine horizontally averaged plume concentration divides by the growing width $w=\mu z$, and is thereforeIt is normalized by ${\int }_0^{\infty }w\overline{C}dz=V_c$, not by $\int \overline{C}dz$. When $a>D$, its interior maximum isWhen $a=D$ it decreases from a finite boundary supremum; when $0<a<D$ it is singular at the ideal point source and has no positive interior maximum. A finite source regularizes that singularity. Thus the final printed maximum is correct for the integrated variable $c$, or for an “average” using a fixed reference width, but is not generally the maximum of the local mean $c/w(z)$. Both quantities have been given rather than silently identifying them. If $D=0$, the smooth similarity formula is replaced by the advected impulse $V_c\delta (z-at)$; positive eddy diffusivity is essential to the gamma profile.

Take the current to be dilute, vertically well mixed and Boussinesq approximation, with constant channel width $w$, a horizontal absorbing bed, negligible ambient fluid entrainment, negligible bed friction, and no resuspension. The shallow-water approximation neglects vertical inertia and gives hydrostatic pressure. Particle loss alters reduced gravity but does not alter the leading-order carrier-fluid volume or inertia; otherwise additional mass and momentum terms would be necessary. The current occupies depth $h(x,t)$, has depth-uniform speed $u(x,t)$, and has particle volume fraction $c(x,t)$, with $g^{\prime }=g_{o}c$ and constant $g_o>0$.

Volume conservation of a short channel segment gives $h_t+(uh{)}_x=0$. Its suspended particle amount per unit bed area is $hc$, and the downward particle deposition flux is $v_{s}c$. Thus $(hc{)}_t+(uhc{)}_x=-v_{s}c$, which becomes $(h{g}^{\prime }{)}_t+(uh{g}^{\prime }{)}_x=-v_s{g}^{\prime }$ on multiplying by $g_o$.

At vertical coordinate $y$ above the bed, excess hydrostatic pressure divided by reference mass density is $g^{\prime }(h-y)$. Integrating across the depth gives the depth-integrated force flux $g^{\prime }{h}^2/2$. The horizontal momentum per unit bed area is $hu$ and the advected kinematic momentum flux is $h{u}^2$. The resulting variable-buoyancy shallow water equations areThe derivative of the pressure term acts on both $h$ and $g^{\prime }$; treating reduced gravity as horizontally constant while particles settle would lose part of the force.

For a steady interior, let $q=Q/w=uh>0$ denote volume flux per unit width. Then the particle balance is a first-order linear differential equation,Here $g_0^{\prime }=g^{\prime }(0)>0$, $v_s\ge 0$ is constant, and deposited particles are not returned to the current. In particular the decay length is $q/v_s=Q/(w{v}_s)$ for $v_s>0$, and the cancellation of depth follows from $uh=q$: dilution and transit time compensate in the deposition balance.

The steady momentum equation also givesThe inlet depth selects $K=q^2/h_0+g_0^{\prime }{h}_0^2/2$. Depth and speed at every downstream position are determined by the continuous positive root of a polynomial inThis implicit expression also makes clear why the inlet condition is needed: there are generally two possible positive depths.

The depth branches of a steady depositing gravity current are distinguished by the Froude number ${\mathrm{Fr}}^2=q^2/(g^{\prime }{h}^3)$. Differentiating the momentum invariant, and using the exponential decay of reduced gravity, givesFor $v_s>0$, supercritical flow has $\mathrm{Fr}>1$ and decreasing depth, with $h\to q^2/K$ and $u\to K/q$ as $x\to \infty$. Subcritical flow has $\mathrm{Fr}<1$ and increasing depth, with $h\sim \sqrt{2K/g^{\prime }(x)}\propto e^{v_{s}x/(2q)}$. This indefinitely deepening branch eventually violates the assumed shallow or confined-channel geometry. Hence the depth does not universally decrease: that conclusion needs a supercritical inlet.

For fixed $g^{\prime }>0$, $q^2/h+g^{\prime }{h}^2/2$ has its unique minimum at $h_c=(q^2/g^{\prime }{)}^{1/3}$, with $K_c=3q^{4/3}(g^{\prime }{)}^{1/3}/2$. The two positive roots when $K>K_c$ lie below and above $h_c$. Since $K_c$ decreases downstream while $K$ is constant, an initially strict branch remains separate from the other branch and cannot cross a critical point. A critical inlet with $v_s>0$ has no finite depth derivative: at $\mathrm{Fr}=1$ the first differentiated equation would equate zero to a positive value. It requires a singular entrance adjustment or physics beyond this steady model. If $v_s=0$, both reduced gravity and the continuous depth branch are constant. These steady solutions describe the supplied current interior, not a moving finite-release front.

For two particle populations, write $g^{\prime }=g_1^{\prime }+g_2^{\prime }$ and $g_i^{\prime }=g_{0i}{c}_i$, allowing different constant particle-density conversion factors. The bidisperse particle-laden gravity current equations areFor the steady case,Thus the depth is obtained from the same cubic polynomial invariant with the sum of the two exponentially decreasing reduced gravity contributions.

The local deposit consists of the two particle deposition fluxes $v_1{c}_1$ and $v_2{c}_2$. For a positive total deposition rate, the bidisperse gravity-current deposition fraction by particle volume isIf the particles have a common material density, this is also their local mass fraction. If they have common $g_o$, it can be written in the source's buoyancy notation asFor unequal $g_{0i}$ replace each $g_{i0}^{\prime }$ by $g_{i0}^{\prime }/g_{0i}$. The exponent sign matters: with both populations present and $v_1>v_2>0$, the faster-settling population has decreasing local deposit fraction downstream, since $f_1^{\prime }=(v_2-v_1)f_1(1-f_1)/q<0$. Equal positive settling speeds give a constant fraction. A nonsettling population contributes no deposit; if the entire local deposition rate is zero the deposit fraction is undefined.

If “fraction of particles” means number fraction rather than particle-volume or mass fraction, let $a_i$ be a single particle's volume in population $i$. The correct weights are then $v_i{c}_i/a_i$, soSettling speeds alone do not determine $a_i$ without particle properties and a drag law. Finally, if the intended deposit is cumulative over the bed from $0$ to $x$, rather than local at $x$, integrating particle deposition flux gives the species volumes per unit channel width $D_i(x)=q{c}_{i0}(1-e^{-v_{i}x/q})$. The cumulative particle-volume fraction isFor both $v_i>0$, its limit at infinity is the incoming particle-volume fraction, whereas its limit as $x\downarrow 0$ is the local deposition fraction at the inlet. Local sorting and the cumulative inventory are different observables.