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The Batchelor entrainment hypothesis makes the inward entrainment velocity proportional to the local plume speed:where $\alpha >0$ is the dimensionless entrainment coefficient. For an upward plume $w>0$, this is $u_e=\alpha w$. The hypothesis represents turbulent ingestion of ambient fluid; it is a closure rather than a consequence of mass conservation.

Let $A=b^2$, suppressing the common factor $\pi$ from cross-sectional integrals. In the Boussinesq approximation, use a constant reference density ${\rho }_r$ in the leading-order mass conservation equation. Dividing that equation by ${\rho }_r$ gives volume conservation:The factor two is the ratio of the perimeter $2\pi b$ to the suppressed area factor $\pi$. Density differences are retained in buoyancy, even though they are neglected in the leading inertial and volume terms.

To derive the density-deficit balance, first leave the small density differences explicit in the supplied mass equation. Because the ambient density ${\rho }_0(z)$ is time independent, the product rule givesThus the ambient mass entrained through the perimeter cancels exactly against its contribution to the volume budget, yielding the required buoyancy balance. It is understood to the accuracy of the underlying Boussinesq approximation; the volume equation is not an additional exact constraint on a fully variable-density fluid.

For later use define reduced gravity $g^{\prime }=g({\rho }_0-\rho )/{\rho }_r$ and the squared buoyancy frequency $N^2=-g{\rho }_{0z}/{\rho }_r$. With $B=A{g}^{\prime }$ and $Q=Aw$, the balance isA stably stratified ambient has $N^2>0$, so upward motion consumes the plume's positive density deficit; an unstable ambient reverses this sign.

For a steady upward plume with prescribed $N^2=-G^2<0$, the kinematic plume fluxes satisfyThe positive sign in the last equation expresses the gain of buoyancy flux as the plume rises through the prescribed unstable ambient. Seek a positive power law solution $Q=q{z}^p$, $M=m{z}^r$, $F=f{z}^s$, with $z>0$. Matching powers givesThe last relation gives $s=p+1$ and the first $r=2p-2$. Substitution into the middle relation yields $4p-5=2p+1$, henceThe amplitude equations areFrom the first relation $m=9q^2/(4{\alpha }^2)$, while the last two give $m^2=G^2{q}^2/16$. Taking positive amplitudes gives $m=Gq/4$, and consequentlyReconstructing the radius, speed and reduced gravity gives the top-hat plume in constant unstable stratification:As a direct check, $M_z={\alpha }^2{G}^2{z}^3/9=FQ/M$ and $F_z={\alpha }^2{G}^3{z}^3/9=G^{2}Q$, while $Q_z={\alpha }^{2}G{z}^2/3=2\alpha \sqrt{M}$.

The corresponding ambient gradient is ${\rho }_{0z}={\rho }_r{G}^2/g$. Thus $\rho (z)={\rho }_0(z)-{\rho }_r{G}^{2}z/(4g)$; for a linear ambient this is ${\rho }_0(0)+3{\rho }_r{G}^{2}z/(4g)$. The plume is locally lighter than the ambient even though both densities increase upward in this prescribed unstable stratification.

The coordinate $z$ can be measured from a virtual origin. This exact power law branch has fluxes tending to zero there and describes buoyancy generated by the maintained unstable ambient; it does not impose arbitrary nonzero source fluxes at that origin. The given idealization holds the ambient quiescent despite its static instability. Finally, the Boussinesq approximation limits the modeled height range to small relative density changes, here $G^{2}z/g\ll 1$. Within these assumptions, the displayed positive branch proves the requested existence.

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 $S^3=2C_H^2{A}_H^2{B}_{s}H$.

To determine the inequality, consider a fully processed room. Its steady reduced gravity is $g^{\prime }=B_s/S$. If $p_0$ is the interior-to-exterior pressure difference at the floor, source-volume blocking of fresh floor inflow requires $p_0\ge 0$. Both openings then exhaust, withThe right-hand side is increasing in $p_0$ and has its minimum at zero. Consequently a fully processed state exists exactly whenThis is source-volume blocking of displacement ventilation. At equality floor inflow vanishes; above it the floor can also exhaust and no fresh lower-layer supply is needed. Below it, the floor pressure must be negative, admitting fresh fluid and supporting a positive lower layer. In fact the steady interface equation has a unique positive root: its loss side increases with $z_{\ast }$ because $Q(z_{\ast })$ increases, while its buoyancy-head side decreases.

The floor-opening area affects the positive-interface state and its approach, but drops out of the blocking threshold because floor flux is zero at the threshold. Equal splitting of pressure head, or using a source-free effective opening area at this transition, would miss this distinction. If the floor opening is sealed, the fresh-inflow argument no longer applies: a finite source with only roof exhaust instead cycles the room asymptotically as in the ceiling-opening case of part (c).

The shallow-water approximation assumes a horizontal length scale $\ell$ much greater than the current depth, $h/\ell \ll 1$. Vertical acceleration is then small compared with gravity and pressure is given by the hydrostatic approximation. A depth-averaged model also assumes that the horizontal velocity and excess mass density can be represented by nearly uniform cross-sectional values. Here the ambient fluid is deep or sufficiently undisturbed that its motion and interface-pressure perturbations can be neglected.

The Boussinesq approximation assumes $|\rho -{\rho }_0|/{\rho }_0\ll 1$: mass density differences are retained in gravitational forces but a common reference mass density is used for inertia. A dense bottom current has the positive reduced gravityThus the pressure excess over the ambient hydrostatic value at height $z$ within the current is ${\rho }_0{g}^{\prime }(h-z)$. We neglect bed drag, molecular diffusion and horizontal turbulent stress in the leading balances, but retain the prescribed turbulent entrainment across the interface. Negligible bed drag does not imply that entrainment has no effect on the current's velocity or mass density.

The occupied cross-sectional area and interface width areThe incoming volume flux per unit channel length is $h{w}_e$. Volume conservation therefore gives $A_t+(Au{)}_x=h{w}_e$. Ambient fluid carries no excess mass density, so integrating the excess-mass density balance gives $(A{g}^{\prime }{)}_t+(Au{g}^{\prime }{)}_x=0$. These two equations imply dilution of the reduced gravity, even though total integrated buoyancy is conserved.

The integrated horizontal hydrostatic pressure force, divided by the reference mass density, isThe prismatic walls contribute no axial component of pressure force. Since ambient fluid enters with zero horizontal velocity, it supplies no horizontal momentum flux. Thus the entraining shallow-water current in a triangular channel obeys the conservative equationsTo put them in primitive form, let $D={\partial }_t+u{\partial }_x$ be the material derivative. The volume equation gives $Dh+(h/2)u_x=w_e$. Subtracting $u$ times the volume equation from the momentum equation, and $g^{\prime }$ times it from the buoyancy equation, yieldsThe momentum source is the acceleration of newly entrained ambient fluid, not bed drag. The source in the reduced gravity equation describes mixing with ambient fluid of zero excess mass density. In particular, the rectangular-channel coefficients cannot be used: the triangular area produces both $h/2$ in the depth equation and $h/3$ in the buoyancy-gradient force.

Assume the carrier fluid has ambient mass density ${\rho }_0$, the particles have mass density ${\rho }_p>{\rho }_0$, and volumes add. The well-mixed current then hasHere $g_0^{\prime }$ is the particle-fluid reference reduced gravity per unit volume fraction, which is the convention needed in the requested expansion. The initial current reduced gravity is $g_0^{\prime }{\varphi }_0$; it should not be confused with $g_0^{\prime }$ itself. The Boussinesq approximation requires $\varphi ({\rho }_p-{\rho }_0)/{\rho }_0\ll 1$.

For particles to remain well mixed, the turbulent vertical mixing time must be short compared with the settling time. In a turbulent diffusion description with diffusivity $K_z$, require $h^2/K_z\ll h/[V_s(1-\varphi )]$, or $K_z\gg h{V}_s(1-\varphi )$. Equivalently, characteristic turbulent vertical speeds should substantially exceed the particle settling velocity. Turbulence can maintain an approximately uniform concentration in the body while particles still settle relative to the fluid. At the impermeable bed, fluid has no normal transport into the solid boundary, but particles can reach and be captured by it. Assume no resuspension after capture.

Use a gravity-current box model of length $L(t)$, uniform depth $h(t)$ and uniform concentration $\varphi (t)$. Neglect ambient entrainment and the small volume change caused by depositing dilute solids, so $Lh=M$ remains constant. Take a constant front Froude number $F>0$ and the gravity-current front condition $\dot{L}=F\sqrt{g^{\prime }h}$. With ${\varphi }_{\max }=1$, the deposition flux is $V_s\varphi (1-\varphi )$ per unit bed area. The particle-volume budget isHence the integral model isThe model assumes a fixed rear boundary, one advancing front, a deep stationary ambient, negligible drag during the inertial spreading regime, and a perfectly absorbing bed. The supplied hindered-settling law is retained in the deposition closure.

For $0<\varphi <1$, eliminating time givesThe concentration integral is $2\mathrm{artanh}\sqrt{\varphi }$. ThereforeAs the particle concentration tends to zero, the current reduced gravity and front speed tend to zero. The hindered-settling runout in a rectangular channel isIt is a finite limiting distance; in this idealized model, the exponential tail of particle loss makes complete stopping asymptotic in time. For $L_{\infty }\gg L_0$, omit $L_0^{5/2}$. Using $\mathrm{artanh}x=x+x^3/3+O(x^5)$ givesThe original PDF has $V_s^2$ in this expression; the TeX aid's $V_s^3$ is a transcription error. Dimensions independently confirm the square: $M^3{g}_0^{\prime }/V_s^2$ has dimensions of length to the fifth power. If one instead names the initial current reduced gravity $G_{\mathrm{init}}=g_0^{\prime }{\varphi }_0$, the same expansion is $25F^2{M}^3{G}_{\mathrm{init}}[1+2{\varphi }_0/3+O({\varphi }_0^2)]/V_s^2$. Late loss of turbulent mixing, viscous resistance or bed resuspension can change the physical runout beyond this specified box model.

In a homogeneous ambient fluid, $N=0$ and $R={\rho }_0$ is constant. Use kinematic fluxes or, equivalently, the primitive form of the unsteady top-hat line-plume balances:Here $g^{\prime }=g({\rho }_0-\rho )/R>0$ is the positive buoyant reduced gravity. Seek a nonsteady separable power-law ansatz with both time differentiation and vertical transport retained. Matching powers in the volume equation forces $b\propto z$ with no time dependence. Matching the two momentum terms then gives $w\propto z/\tau$, and the buoyancy force has $g^{\prime }\propto z/{\tau }^2$, where $\tau =t+t_{\ast }>0$ is a shifted age. Thus writeSubstitution determines the constants rather than just the exponents. The volume equation gives $2BA=\alpha A$, so for a nonzero buoyant plume $B=\alpha /2$. The momentum equation gives $C=-A+3A^2$, while the buoyancy equation gives $C(-2+3A)=0$. The buoyant nontrivial branch has $A=2/3$ and $C=2/3$. Consequently the separable decaying top-hat line plume isThe Boussinesq approximation requires $2z/(3g{\tau }^2)\ll 1$. The corresponding mass density-weighted fluxes, to this order, areFor a direct check, $Q^2/M=R\alpha z/2$ is time independent and $FQ/M=R\alpha z^2/(3{\tau }^2)$. Hence $(Q^2/M{)}_t+Q_z=2R\alpha z/(3\tau )=R\alpha M/Q$, $Q_t+M_z=R\alpha z^2/(3{\tau }^2)=FQ/M$, and $(FQ/M{)}_t+F_z=0$. This verifies all three balances and the required width coefficient.

This nonsteady branch is not the steady, continuously supplied, constant-flux line plume. The latter has constant $w$ and half-width $b=\alpha z$ in the same top-hat convention. Also, the unshifted power law has $Q,F\to 0$ as $z\to 0$ and is singular at $\tau =0$. It is therefore an interior separable solution, not by itself a solution with a nonzero continuously operating point-source flux at the origin. A finite-width source boundary, a virtual origin, or a matching region is needed to connect it to physical source and initial data. For example, replacing $z$ by $z+z_{\ast }$ supplies a finite base width and time-varying positive base fluxes when $z_{\ast }>0$. No source or initial flux history is specified in the question to select that matching; the requested power-law width follows from the local balances above.

Let the undisturbed water surface be $z=0$, the oil's upper surface $z=\eta (x,t)$ and its lower surface $z=b(x,t)$, so $h=\eta -b$. Leading hydrostatic pressure in the nearly inviscid water is $p_w=-{\rho }_{w}gz$ relative to atmospheric pressure; in the oil it is $p_H=\rho g(\eta -z)$. Equality at the interface gives $\rho gh=-{\rho }_{w}gb$, henceThus the horizontal hydrostatic driving gradient isThis reduced gravity accounts for most of the oil thickness being below the undisturbed water level.

Because the oil is much more viscous than the water and air, tangential tractions on both faces are negligible. In a thin layer, the leading horizontal velocity is consequently independent of depth: the floating extensional viscous gravity current is a plug, rather than the parabolic profile of a no-slip gravity current. Integrating incompressibility $u_x+w_z=0$ between the moving material boundaries, and using their kinematic boundary condition, gives conservation of oil volume,To derive the factor $4$ in the extensional balance, incompressibility gives $w_z=-u_x$. The Newtonian fluid stress tensor has ${\sigma }_{zz}=-p-2\mu u_x$. Normal-stress balance relative to the hydrostatic part therefore requires $p=p_H-2\mu u_x$. The horizontal stress then satisfiesThus the longitudinal tensile force in an oil slice, relative to its hydrostatic pressure, is $4\mu h{u}_x$ per unit width. Balancing its difference across the slice with the horizontal driving force $-h{p}_{H,x}dx$ yieldsThe coefficient is the planar extensional Trouton ratio, arising from both the direct horizontal strain and the pressure correction required by the transverse normal-stress condition.

Using the pressure gradient above, integration in $x$ givesThe stated nose stress condition sets $C(t)=0$, not an arbitrary imposed inlet tension. ConsequentlyFor a material slice injected at time $t_0$, let $\tau =t-t_0$ be its age. The thickness equation along its Lagrangian trajectory is $dh/dt=-h{u}_x=-k{h}^2$, with $h=h_0$ at injection. DefineThen $h=h_0/(1+\tau /T)$. The slice injected during $d{t}_0$ has volume $h_0{u}_{0}d{t}_0$ per unit width, so at fixed observation time $t$,Since a newly injected slice has $x(t,t)=0$, integration and subsequent differentiation along a material slice give the constant-flux floating extensional gravity current:The oldest slice forms the nose. Measuring time from the start of injection, $t_0=0$ there, soIts speed ${\dot{x}}_N=u_0(1+t/T)$ is exactly the material velocity at the nose, as required.

For a fixed position $x$ behind the nose, the age needed to reach it is determined by $x/u_0=\tau +{\tau }^2/(2T)$, independently of the observation time. Hence both thickness and velocity become stationary there once the nose has passed. Eliminating $\tau$ givesTheir product is $h_0{u}_0$, so the steady profile carries precisely the inlet volume flux; integrating it to $x_N(t)$ gives the total injected volume $h_0{u}_{0}t$.

The lengthis the horizontal distance over which reduced-gravity driving causes an order-one change in inlet speed and thickness by viscous extension. The associated age is $T=8\ell /u_0$; the exact profile contains $x/(4\ell )$. It is an extensional adjustment length, not a shear-diffusion scale. The result applies while the stated thin-layer and negligible-inertia approximations remain valid.

Take $z$ positive upward from the release and define the pressure scale height $H_p=p_0/({\rho }_{0}g)$. Neglecting the small ambient density variation in the hydrostatic pressure calculation,Conservation of gas mass and Boyle's law giveHere $V_b(0)\simeq {\varphi }_0{V}_0$ to leading order in the dilute bubble concentration. If the thermal's total volume is $V(z)$, its bubble fraction and bubble contribution to reduced gravity areThus the volume-integrated kinematic buoyancy contribution is $g\varphi V=g{\varphi }_0{V}_0/(1-z/H_p)$, and the corresponding upward force is approximately ${\rho }_0$ times this quantity. Bubble expansion increases this total contribution even while entrainment dilutes the local concentration.

Neglect bubble rise relative to the water when the bubbles' terminal slip speed is much smaller than the thermal rise speed, their response time is short compared with $a/W$, and they remain well dispersed and retained. Small bubbles are more likely to satisfy these conditions; large bubbles can escape through the thermal. The approximations also require ${\rho }_b\ll {\rho }_0$, $\varphi \ll 1$, negligible gas dissolution and no important surface-tension correction. Pressure must remain positive and the hydrostatic expression is used only below the ocean surface.

For constant buoyancy frequency in the Boussinesq approximation, the ambient density isConservation of liquid mass during fluid entrainment gives $d(V\rho )/dz={\rho }_a(z)V^{\prime }(z)$, with $\rho (0)={\rho }_0$. Integrating by parts,Since $a=a_0+\alpha z$ and $V\propto a^3$,Hence the water density within the thermal isThe entrained water is denser than its new surroundings because it retains a mixture of fluid drawn from lower levels.

The bulk density is $\rho (1-\varphi )$. Dropping products of the small density contrast with $\varphi$, its net reduced gravity isThe first term is the expanding-bubble contribution and the second is the restoring effect of stratification. In the quasi-steady Froude model, ascent stops when $g^{\prime }=0$. The neutral height of a bubbly thermal in a stratified fluid therefore satisfiesThis is a fifth-degree polynomial, with leading coefficient $-{\alpha }^4/H_p$ at finite release pressure. The relevant height is the first positive zero reached while the thermal is still below the surface, dilute and within the Boussinesq range. Depending on parameters there may be no such zero before the surface, or more than one mathematical zero; a later root beyond a first stopping point is not another continuous ascent solution. The model's stopping height is a neutral-buoyancy estimate, not a claim that inertial overshoot is absent.

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.