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Let $q_b$ be floor inflow and $q_t$ roof outflow. Use effective-area discharge $A\sqrt{\delta p/\rho }$, matching the factors in the supplied relations. With negligible source volume, both equal $q$. The lower region has exterior mass density while the warm upper layer has reduced gravity $g^{\prime }$ and depth $H-h$. hydrostatic pressure therefore supplies a total opening pressure head $g^{\prime }(H-h)$. Equal openings sharing equal discharge take half each, givingSteady lower-layer volume conservation sets this ventilation rate equal to the entrained turbulent plume flux at the interface, $q=\lambda B^{1/3}{h}^{5/3}$. Upper-layer buoyancy conservation gives $B=q{g}^{\prime }$. These are the required balances for displacement ventilation:Eliminating $g^{\prime }$ and $q$ gives the displacement-ventilation interface height equationThe left side is strictly increasing from zero to infinity, so there is one interface height. It depends on opening geometry and plume entrainment, not on $B$; the resulting $g^{\prime }=B^{2/3}/(\lambda h^{5/3})$ and throughflow do depend on $B$.

For finite source volume $Q$, conservation instead gives $q_t=q_b+Q$. The two pressure drops are no longer equal:Let $q_p(h;Q)$ be a physically entraining source plume: $q_p(0;Q)=Q$ and $q_p(h;Q)>Q$ for $h>0$. Its interface balance is $q_p=q_t=Q+q_b$. Therefore, as $q_b$ reaches zero, a positive-depth lower layer has no replenishment and cannot remain steady: this is no steady displacement layer without ambient supply. The limiting interface is $h=0$. At the threshold $q_t=Q$ and $g^{\prime }=B/Q$, and the entire hydrostatic head drives the roof discharge. HenceThis is source-volume blocking of displacement ventilation. The half-head factor of the negligible-source state must not be kept after the floor inflow vanishes. In the usual physical-area convention $q=A_{\mathrm{phys}}{C}_d\sqrt{2\delta p/\rho }$, the equivalent formula is $Q_c=(2C_d^2{A}_{\mathrm{phys}}^{2}BH{)}^{1/3}$.

The quoted pure-plume law cannot be used unmodified down to the origin for a source with nonzero volume: it would incorrectly give zero flux there. A compatible finite-source example uses a virtual origin,It recovers the supplied pure-plume limit when $Q\to 0$, and gives the same blocking threshold. The complete finite-$Q$ interface trajectory depends on that plume model, but the threshold needs only source-volume conservation and positive entrainment. For $Q>Q_c$, the assumed steady regime with a lower inflow no longer exists.

Use small Rossby number $U/(f_{0}l)\ll 1$, slow evolution on the advective time scale, small interface displacements relative to each layer depth, shallow hydrostatic layers, stable reduced gravity $g^{\prime }>0$, and inviscid unforced flow. On a beta plane, take $\beta l/f_0$ of the same small order as the Rossby number. The internal Burger number is retained at order unity so stratification and relative-vorticity effects can both enter the leading potential-vorticity anomaly.

The rigid-lid pressure in two-layer flow comes from neglecting the free-surface volume displacement in the rigid-lid approximation; it does not permit setting the common horizontal pressure gradient to zero. The small surface displacement multiplied by $g$ retains a finite lid-pressure multiplier. Write $h_1=H_1+\chi$, $h_2=H_2-\chi$, and let $\Pi$ denote this common pressure potential. Leading geostrophic balance giveswith ${\mathbf{u}}_i=(-{\psi }_{iy},{\psi }_{ix})$ at leading order. Literally setting $h_1+h_2$ to a spatial constant in the momentum gradients before taking the rigid-lid limit would suppress the upper-layer pressure field and fail to produce general two-layer QG dynamics.

Taking curl of each shallow-water momentum equation and using layer continuity gives material conservation of $(f+{\zeta }_i)/h_i$. Expanding it to first order and advecting the anomaly by the leading geostrophic velocity yieldsThe two-layer quasi-geostrophic potential vorticity equations areHere the common background $f_0/H_i$ has been removed and the anomaly multiplied by $H_i$. The advection term is retained at the same slow order as the time derivative even though the leading velocity/pressure balance was linear geostrophy. On an $f$ plane simply set $\beta =0$.

Use $F=F_s>0$ and constant entrainment parameter $\alpha >0$. The printed integrals are normalized by $\pi$, so a top-hat plume model has $Q=b^{2}w$, $M=b^2{w}^2$ and $F=b^{2}w{g}^{\prime }$; the physical area is still $\pi b^2$. Keeping this normalization avoids spurious factors of $\pi$ in the flux equations.

Eliminate height between the two flux balances. The plume flux-balance invariant follows fromsoFor positive fluxes, both $M$ and $Q$ increase. In particular $Q^{\prime }\ge 2\alpha \sqrt{M_s}>0$, so $Q\to \infty$ as $z\to \infty$. Therefore $C/(\lambda Q^2)\to 0$, whatever the finite source imbalance, and $M\sim {\lambda }^{2/5}{Q}^{4/5}$. Integrating $Q^{\prime }\sim 2\alpha {\lambda }^{1/5}{Q}^{2/5}$ gives the attracting similarity solutionThis proves attraction to pure plume similarity for physical positive source fluxes. Sources with a vanishing initial volume or momentum flux enter the positive regime immediately and can be obtained by the corresponding limiting solutions; a point buoyancy source has the same similarity form with zero virtual-origin offset. The statement assumes positive buoyancy flux: a jet with $F=0$ is a different asymptotic problem.

Define the plume balance parameterPure plume balance means $\Gamma =1$ at each height, or equivalently $C=0$: momentum, buoyancy and entrainment have their similarity relation, even when the source has finite radius. A forced plume has $\Gamma <1$, meaning excess momentum relative to this balance; a lazy plume has $\Gamma >1$, meaning insufficient momentum. The invariant ensures that the sign of the imbalance is preserved while $\Gamma \to 1$. In particular, a strictly positive source remains pure at every height precisely whenFor this source the exact solution is the same similarity form with $z$ replaced by $Z=z+z_0$. Writing $a=(9\alpha F_s/10{)}^{1/3}$,Thus the radius growth of an axisymmetric pure plume is $b^{\prime }=6\alpha /5$ and the plume virtual origin lies at $-z_0$, with $z_0=5b_s/(6\alpha )$. Equivalently, in terms of top-hat source properties, pure plume balance is $g_s^{\prime }{b}_s/w_s^2=8\alpha /5$.

For the stated source-to-source geometry, the relevant horizontal distance is $L_s$. Consequently the edge criterion is $b_m=L_s$, rather than the radius at which two expanding edges first touch. Since $b(z)=b_s+6\alpha z/5$,Here $b_s=Q_s/\sqrt{M_s}={\lambda }^{-1/5}{Q}_s^{3/5}$. A positive merging height requires $L_s>b_s$; equality puts it at the source, while an already wider source satisfies the geometric condition from the start. The no-interaction premise extrapolates independent plumes even though their edges would first touch earlier, when each radius is $L_s/2$.

Use $M^{5/2}=\lambda Q^2$ with $Q=b^{2}w$ and $M=b^2{w}^2$ to get $F_s=(8\alpha /5)b{w}^3$. The individual plume properties at the specified height are thereforeThe source volume flux affects the height through the virtual origin, but it does not affect these properties at a prescribed radius in an exactly pure plume.

For the stipulated merger of two equal pure plumes, the new radius is $b_c=\sqrt{2}{b}_m$. This factor is outside $b_m$, as required by the area. Since speed and reduced gravity are held fixed, the merged fluxes are $Q_c=2Q_m$, $M_c=2M_m$ and $F_c=2F_s$. Their balance parameter isThe immediate combined plume is lazy. The same result follows from $\Gamma =5g^{\prime }b/(8\alpha w^2)$: enlarging the radius at fixed $g^{\prime },w$ increases this parameter.

At fixed $w_m,g_m^{\prime }$, the radius required for pure plume balance isThat is half the stipulated combined area $2\pi L_s^2$. This fixed-speed pure-plume merger and flux conservation comparison exposes the limitation of demanding immediate pure balance with unchanged speed and reduced gravity: reducing the area to this value would also halve all three merged fluxes. A conservative merger must instead adjust its properties and begin out of pure balance. Under the given constant-entrainment equations, the resulting lazy plume subsequently approaches the attracting similarity solution with total buoyancy flux $2F_s$.

Let $a=({\rho }_b-{\rho }_t)/H={\rho }_0{N}^2/g$. The initial density is $\rho (z)={\rho }_t-az$. Conservation of mass in the material homogenized over $-h<z<0$ fixes its mean density to ${\rho }_u={\rho }_t+ah/2$, while the untouched lower layer has interfacial density ${\rho }_i={\rho }_t+ah$. Hence the reduced gravity difference across the interface isThe reference contribution involving ${\rho }_t$ cancels from the difference.

Only the upper layer changes its potential energy. The potential-energy cost of homogenizing a linear stratification issoThe positive sign reflects work against stable stratification.

The rotating boundary supplies energy on scales set by $\Omega$ and $R$. Turbulent drag and dissipation remove energy, and entrainment adds initially stationary fluid that must be accelerated. If upper-layer energy grows, the increasing turbulent loss provides a restoring tendency; after a short adjustment, input and losses can approximately balance while the interface moves on a slower time scale. This gives a physical rationale for a fixed-energy turbulent mixed layer, not a consequence of mass conservation alone. Its characteristic kinetic energy is $K\sim {\rho }_0\pi R^{2}h{u}_u^2/2$. With the stipulated $N$-independent energy closure,so the characteristic speed decreases as $h^{-1/2}$ rather than remaining constant.

The stress does work at a rate proportional to $\pi R^2\tau u_u=c_D{\rho }_0\pi R^2{u}_u^3$. Combine this with the energy increase and absorb fixed drag/geometrical factors into $C>0$:Here the exponents $\alpha ,\beta$ are unrelated to the plume entrainment coefficient in Question 2. The interfacial Richardson number isSet $q=\Omega /N$, $d=\delta /R$ and $r=h/R$. Then $u_u^2/(N^2{h}^2)=d/(2r{\mathrm{Ri}}_i)$ and $r=q(2A{\mathrm{Ri}}_i/d{)}^{1/2}$. Substitution givesThe assumption that $E$ depends on the local Richardson number alone excludes separate dependences on $q,d$. The entrainment exponent from a local interfacial Richardson closure is consequentlyThis comparison treats $q$ and $d$ as independently variable external controls; fitting only a single apparatus would not by itself determine the exponents.

Finally, use the fixed-energy closure again:Integrating from a positive initial mixed-layer depth $h_0$ at time $t_0$ givesWhen the growth term dominates the initial offset, the rotating-disc mixed-layer depth law isTaking the effective time origin to be zero gives the stated scaling. Its validity is limited to the regime of the closures and $h<H$; the singular velocity predicted at $h=0$ is not a model for the initial boundary-layer formation.

The Boussinesq approximation requires $\delta =(\rho -{\rho }_0)/{\rho }_0=\varphi ({\rho }_p/{\rho }_0-1)\ll 1$. The reference mass density ${\rho }_0$ can then be used in inertia, while the small density excess is retained in buoyancy. Dilute particle volume fraction $\varphi \ll 1$ alone does not suffice if ${\rho }_p/{\rho }_0$ is exceptionally large. Define $G=g({\rho }_p-{\rho }_0)/{\rho }_0$, so the reduced gravity is $g^{\prime }=G\varphi$.

The shallow water equations require depth small compared with the horizontal evolution scale, weak vertical acceleration and approximately hydrostatic pressure. Here the ambient is deep and quiescent, so its leading contribution is a hydrostatic reference pressure; the excess pressure in the layer is ${\rho }_{0}G\varphi (h-z)$. The model also needs nearly uniform streamwise velocity across the section and the stated absence of secondary circulation. The advancing gravity current nose is a separate region where these assumptions can fail.

A nearly uniform particle volume fraction can be maintained by turbulent mixing with eddy diffusivity $K_z$, provided $h^2/K_z\ll h/w$ and the mixing time is short compared with horizontal advection and bulk evolution. Equivalently, $wh/K_z\ll 1$ keeps the bulk settling-induced gradient small. The particles should respond rapidly enough to follow the mixing motions. A thin boundary layer adjacent to the absorbing bed or walls need not be well mixed.

Uniform bulk concentration does not imply zero sedimentation. The relative downward particle velocity still supplies a particle deposition flux of approximately $w\varphi$ to an absorbing boundary. Mixing redistributes the remaining particles and maintains the bulk concentration profile while its overall level decreases; it need not suspend every particle indefinitely. Deposition requires negligible resuspension in the model.

Seek a steady downslope solution with $u>0$ constant. The volume conservation equation reduces to $h_x=E$, so the constant-speed entraining gravity current on a slope hasThe conserved buoyancy flux per unit width is $B=u{g}^{\prime }h$, givingSubstitute these expressions into the momentum equation. Its advective term is $E{u}^2$ and its hydrostatic term is $\frac{1}{2}E(B/u)\cos \theta$. The balance is thereforewhich givesThe assumed slope inequality makes ${\beta }_{\ast }>0$. The increasing thickness dilutes reduced gravity while leaving $g^{\prime }h$ and the buoyancy flux constant. The denominator includes both bottom drag and the momentum needed to accelerate newly entrained ambient fluid.

For identical suspended particles, reduced gravity is proportional to particle volume fraction. If particles settle through the bed at speed $v_s$ and are not resuspended, the downward particle deposition flux removes a buoyancy-equivalent amount $v_s{g}^{\prime }$ per unit bed area. The depth-integrated scalar mass conservation equation isFor a steady particle-laden gravity current, $B_x=-v_{s}B/(uh)$. If $v_s/u\ll 1$, with fixed positive $E$ and drag/slope coefficients, the locally adjusted leading solution of the momentum and volume equations is $u\sim {\beta }_{\ast }{B}^{1/3}$ and $h\sim h_0+E(x-x_0)$. The weak-settling attenuation of a sloping gravity current then givesand henceFor an inlet at $x_0=0$ this has the requested logarithmic form. If the inlet is genuinely at $x_0\ne 0$, the argument must contain $x-x_0$ in order to satisfy the specified inlet condition.

The speed factor must be the ${\beta }_{\ast }$ derived in the previous part. The printed $\beta$ omits the factor one-half multiplying $E\cos \theta$ and uses an undefined uppercase $C$ instead of the governing equation's $c$. Taking $C=c$ still leaves a nonzero leading momentum residual $-\frac{1}{2}E(B/u)\cos \theta$ for the printed speed; this discrepancy cannot be repaired by a correction of order $v_s/u$. With the given hydrostatic momentum equation, the logarithmic law is valid with ${\beta }_{\ast }$, not the printed speed factor.

A controlled local approximation also needs $v_s/(Eu)\ll 1$ if $E$ is allowed to become small. Indeed the exact volume balance is $h_x=E-h{u}_x/u$; under $u\sim {\beta }_{\ast }{B}^{1/3}$ its second term is $v_s/(3u)$. The neglected depth and acceleration corrections become significant as $B$ approaches zero.

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.