Reduced gravity ($g^{\prime }$)

For two Boussinesq fluids with density difference $\Delta \rho$ and reference density ${\rho }_0$, reduced gravity is $g^{\prime }=g\Delta \rho /{\rho }_0$.

An entraining shallow-water layer changes depth, density, and momentum as ambient fluid crosses its interface. Entrainment dilutes its reduced gravity and accelerates or decelerates it through the momentum needed to bring new fluid into the layer.

A gravity current is a predominantly horizontal flow driven by a density difference. Its front speed scales as the square root of reduced gravity times current depth.

For a one-sided fixed-volume gravity current in a unit-width channel with a closed rear wall, use front closure $\dot{L}=F\sqrt{g^{\prime }{h}_f}$ with $0<F<2$. An exact late-time similarity solution has $u=\dot{L}x/L$, $h=h_f[1-F^2/4+(F^2/4)(x/L{)}^2]$ and $h_{f}L=V/(1-F^2/6)$. Substituting into the shallow water equations proves the radius law. The Benjamin deep-ambient front condition has $F=\sqrt{2}$. The virtual time origin $t_v$ is fixed by matching to the finite release, not by volume alone.

A semi-infinite resting layer of depth $H$ and constant reduced gravity $b$ released into a deep ambient develops a rarefaction wave satisfying $u+2\sqrt{bh}=2\sqrt{bH}$. The head closure $u_f=F\sqrt{b{h}_f}$ gives $h_f=4H/(F+2{)}^2$ and $u_f=2F\sqrt{bH}/(F+2)$. A constant-depth shelf lies between the rarefaction tail speed $u_f-\sqrt{b{h}_f}$ and the front speed $u_f$; its width is positive for every finite positive $F$.

A rectangular lock of length ${\ell }_0$ and depth $h_0$ has rarefaction head speed $-c_0$, where $c_0=\sqrt{g^{\prime }{h}_0}$. It reaches a closed rear wall at $t_r={\ell }_0/c_0$. The first reflected positive characteristic curve follows $x={\ell }_0+2c_{0}t-3c_0{t}_r^{2/3}{t}^{1/3}$ inside the undisturbed rarefaction, then crosses the uniform shelf and reaches the front at the displayed time. This marks the end of the initial constant-speed front stage in the ideal fixed-front-Froude model.

In a prismatic triangular channel with constant reduced gravity, the wave speed is $c=\sqrt{g^{\prime }h/2}$ and the Riemann invariants are $u\pm 4c$. A dam release into a deep ambient produces a simple-wave rarefaction with $u+4c=4c_0$. The imposed gravity-current front condition $u_f=F\sqrt{g^{\prime }{h}_f}$ gives the displayed uniform nose-adjacent state. The fan has $c=(4c_0-x/t)/5$ and $u=4(c_0+x/t)/5$, extending from $x/t=-c_0$ to $u_f-c_f$; a uniform region then reaches the front $x/t=u_f$. The nose closure supplies physics absent from the one-layer interior equations.

For a prismatic channel of width $b(z)=z$, a Boussinesq current of depth $h$ occupies area $A=h^2/2$. Ambient fluid at rest enters across the interface at speed $w_e$, supplying volume but no horizontal momentum or excess density. The volume conservation, momentum and excess-density balances areHere $g^{\prime }$ is the reduced gravity. With $D={\partial }_t+u{\partial }_x$, these give $Dh+(h/2)u_x=w_e$, $Du+g^{\prime }{h}_x+(h/3)g_x^{\prime }=-2u{w}_e/h$, and $D{g}^{\prime }=-2g^{\prime }{w}_e/h$. The momentum source in primitive form describes acceleration of newly entrained fluid, even when bed drag vanishes.

For positive depth and reduced gravity, the characteristic speeds are $u,u\pm c$. Along the material characteristic, $D{g}^{\prime }=-2g^{\prime }{w}_e/h$. Along $D_{\pm }={\partial }_t+(u\pm c){\partial }_x$, the left-eigenvector compatibility equations areIf $w_e=0$ and $g^{\prime }$ is spatially constant, these reduce to the Riemann invariants $u\pm 4c$. Variable buoyancy or entrainment prevents that simplification.

A thin layer of very viscous liquid floating on an effectively inviscid denser liquid has negligible shear traction on both faces. Its leading horizontal velocity is a plug, and incompressibility gives vertical strain $w_z=-u_x$. Normal-stress balance shifts the pressure by $-2\mu u_x$, so the longitudinal excess stress is $4\mu u_x$. Combining depth-integrated momentum balance with hydrostatic reduced gravity gives the displayed extensional equations. This is distinct from a no-slip, shear-driven thin-film gravity current.

With fixed inlet thickness $h_0$ and velocity $u_0$, and nose tension $4\mu h_N{u}_x=\rho g^{\prime }{h}_N^2/2$, integrating momentum gives $u_x=\rho g^{\prime }h/(8\mu )$. For a slice injected at $t_0$, write $\tau =t-t_0$ and $T=8\mu /(\rho g^{\prime }{h}_0)$. Its thickness, velocity and position are $h=h_0/(1+\tau /T)$, $u=u_0(1+\tau /T)$, and $x=u_0[\tau +{\tau }^2/(2T)]$. Eliminating its age gives the steady profile behind the nose, while the nose is the slice with the earliest injection time.

A shallow radially spreading viscous drop on a horizontal plane has lubrication flux proportional to $-h^3{h}_r$. Absorbing the positive dimensional coefficient into time gives the displayed equation. For a finite-volume drop, $M={\int }_0^{R}rhdr$ is constant when the origin and front carry zero volume flux. A similarity form $h=t^{-\beta }f(r/t^{\alpha })$ has $2\alpha =\beta$ from volume and $\beta +1=4\beta +2\alpha$ from the equation, giving $\alpha =1/8$ and $\beta =1/4$.

For a fixed half-angle $0<\alpha <\pi /2$, a slow slender gravity current in a V-shaped channel has an almost horizontal free surface across each section. Its area is $h^2\tan \alpha$, and its axial volume flux is $-gQ(\alpha )h^4{h}_x/\nu$. Here $Q$ is the integral of the dimensionless cross-sectional Poisson equation solution: $\Delta \varphi =-1$, with zero velocity on the two rigid slopes and zero normal velocity derivative on the horizontal shear-free surface. Conservation of mass gives $C=gQ/(\nu \tan \alpha )$. This treats the full two-dimensional section and does not require a shallow V angle.

For a shallow V-channel with floor $z=\alpha |y|$, $0<\alpha \ll 1$, let the free surface be $z=h(x,t)$. Hydrostatic pressure gives $p_x=\rho g{h}_x$. A vertical strip has thickness $H_y=h-\alpha |y|$ and, with no slip below and zero shear above, flux $-\rho g{H}_y^3{h}_x/(3\mu )$. Integrating across $|y|<h/\alpha$ givesConservation of volume is $(h^2{)}_t=C(h^4{h}_x{)}_x$, $C=\rho g/(6\mu )$. Here $\alpha$ is the wall slope to the horizontal, not the opening half-angle measured from the vertical in a general V-section. Vertical shear dominates transverse shear by the shallow-slope approximation.

The variable $w=h^2$ converts the shallow V-channel lubrication flux equation into a porous medium equation:Indeed $h^4{h}_x=w^{3/2}{w}_x/2=(w^{5/2}{)}_x/5$. Its fixed-mass source-type solution gives $h\propto t^{-1/7}$ and support length proportional to $t^{2/7}$. Integrating the similarity equation once gives $C{h}^4{h}_x=-2x{h}^2/(7t)$, and hence $h=[3(X(t{)}^2-x^2)/(7Ct){]}_{+}^{1/3}$. The constant-volume similarity in a V-shaped channel fixes $X$ from the positive integral $J={\int }_{-1}^1(1-s^2{)}^{2/3}ds$: $X=(\alpha V/J{)}^{3/7}(7Ct/3{)}^{2/7}$.

The symmetric similarity solution for a localized fixed volume $V$ in a V-shaped channel gravity current has $h(x,t)=[3(X^2-x^2)/(7Ct){]}_{+}^{1/3}$. Integrating the similarity conservation of mass equation once gives $C{h}^4{h}_x=-2x{h}^2/(7t)$, and vanishing thickness at the two noses gives the profile. Its volume fixes $X=[V/(I\tan \alpha ){]}^{3/7}(7Ct/3{)}^{2/7}$, where $I={\int }_{-1}^1(1-s^2{)}^{2/3}ds$. At any fixed nonzero position, the thickness first rises after arrival and then falls; its maximum occurs when $X^2=7x^2/3$.

A thin dense gravity current on a rigid no-slip substrate can be carried downstream by imposed upper shear stress $\tau$. For density contrast $\Delta \rho$, lubrication theory gives $\mathbf{q}=A{h}^2{\mathbf{e}}_x-D{h}^3\nabla h$, where $A=\tau /(2\mu )$ and $D=\Delta \rho g/(3\mu )$. The continuity equation gives the displayed equation. Besides small slopes and reduced inertia, the hydrostatic reduction requires $\tau /(\Delta \rho gL)\ll 1$ to keep shear-related normal stress small. Finite zero-thickness fronts are formal outer profiles and need a separate local description when their slopes are large.

For a steady line-source shear-driven viscous gravity current with downstream constant thickness, $h_d^2=2\mu Q_{2d}/\tau$. Zero upstream volume flux per unit width gives $h{h}_x=3\tau /(2\Delta \rho g)$, so $h^2=h_d^2+3\tau x/(\Delta \rho g)$ until it vanishes at $x=-x_N$. Unlike the point-source scaling estimate, this reduced line-source model fixes the exact coefficient $2/3$. The square-root nose has an unresolved steep edge.

For a steady point-source shear-driven viscous gravity current far downstream, $h\ll y_N\ll x$ permits dropping the downstream gravity flux. With $w=h^2$, the equation is $w_x=D(w^2{)}_{yy}/(4A)$, a quadratic porous medium equation. Conserving $A\int wdy=Q$ yields $y_N=(3\mu \Delta \rho gQx/{\tau }^2{)}^{1/3}$ and $h^2=\tau (y_N^2-y^2)/(2\Delta \rho gx)$ inside the current. The central height decays as $x^{-1/6}$. Near the source, balancing shear and gravity gives an upstream reach of order $\sqrt{\mu \Delta \rho gQ}/\tau$, whose numerical coefficient is not determined by scaling.

In a water layer confined between horizontal rock and a movable constant-pressure ice roof, hydrostatic pressure and two wall stresses $\mu u/{\delta }_v$ give $u=-Dh{h}_x$, $q=-D{h}^2{h}_x$ and $h_t=D(h^2{h}_x{)}_x$, with $D={\rho }_{w}g{\delta }_v/(2\mu )>0$. This closure uses a constant viscous boundary layer thickness, not a generic quadratic turbulent drag law.

A shallow current receiving meltwater over its wetted area obeys $h_t+q_x=s_m$, with $q=-D{h}^2{h}_x$. The source is zero ahead of the advancing front. The Stefan condition determines normal ice-retreat speed $v_m$; the water-volume source is $s_m=({\rho }_i/{\rho }_w)v_m$, or simply $v_m$ in the equal-density approximation. Water volume and geometric excavation must be distinguished if densities differ.

For constant water-production rate per wetted area $s_m$, dry zero-flux noses and footprint length $\mathcal{L}(t)$, the moving-boundary conservation of mass gives $\dot{\mathcal{V}}=s_m\mathcal{L}$ per out-of-plane span. Consequently $\ddot{\mathcal{V}}=s_m\dot{\mathcal{L}}$. Equivalently the melt volume is $s_m{\int }_0^t\mathcal{L}(t\prime )dt\prime$; a point begins producing melt only after the front arrives.

For steady flow with entrainment coefficient $E$, bottom-drag coefficient $c$ and slope angle $\theta$, a constant buoyancy flux $B$ gives $h=h_0+Ex$, $g^{\prime }=B/(uh)$ and $u^3=B(\sin \theta -E\cos \theta /2)/(c+E)$. The half factor comes from hydrostatic pressure, while the denominator includes both bed drag and acceleration of entrained fluid.

A viscous current with density between those of two ambient fluids spreads along their interface. Vertical buoyancy divides its thickness above and below the undisturbed interface. When its internal shear and in-plane extension are both negligible, resistance from the two ambient Stokes flows controls its spreading.

When vertical diffusion equation controls depth, $H\sim \sqrt{Dt}$, while conserved excess solute mass gives $\Delta \rho R^{2}H\sim \Delta {\rho }_0{V}_0$. Ambient viscous resistance then leads to a $t^{1/2}$ spreading law. The enriched-fluid volume grows, so the fixed-volume $t^{1/5}$ law does not apply.

The gravity force per interfacial area is $\Delta \rho g{H}^2/R$, while ambient viscous traction is $\mu u/R$. Their balance gives the radial velocity scale independently of current viscosity within the appropriate plug-flow window.

Linearity of Stokes flow and Uniqueness of Stokes flow identify the ambient resistance to an expanding thin plug with the reversed perturbation around a stationary disc in an axisymmetric straining flow. The imposed straining flow has zero tangential stress on the disc plane. Each face then resists radial velocity $Er$ with traction $-8\mu Er/[\pi \sqrt{R^2-r^2}]$.

For fixed volume and equal ambient viscosities, radial volume conservation gives $u=r/(5t)$. Summing both ambient-face tractions gives $H^2=32\mu R/(5\pi \Delta \rho gt)$ and $R^5=125\Delta \rho g{V}^{2}t/(512\pi \mu )$. The edge slope is singular, so the profile is a leading bulk thin-current approximation.

The lower bound makes internal vertical velocity variation negligible; the upper bound makes internal in-plane extensional stress negligible compared with ambient traction. Both are needed for an ambient-controlled interfacial plug current.

Hydrostatic pressure continuity across both interfaces fixes the partition of current thickness. The modified current pressure is $\Delta \rho gh$, with $\Delta \rho =({\rho }_0-{\rho }_1)({\rho }_2-{\rho }_0)/({\rho }_2-{\rho }_1)$. The equivalent radial body force is $-\Delta \rho g{h}_r{\mathbf{e}}_r$.

The Froude number compares inertial speed with a gravity-wave speed. For a shallow gravity current of depth $h$ and reduced gravity $g^{\prime }$, $\mathrm{Fr}=U/\sqrt{g^{\prime }h}$.

For a steady slowly varying rectangular channel, mass conservation and the Bernoulli equation give $(W/w{)}^2=((F+2)/F)s^2-(2/F)s^3$, where $s=h/H$. The positive cubic has its maximum at $s=(F+2)/3$, with value $(F+2{)}^3/(27F)$. This gives the displayed minimum possible width for the prescribed upstream flux and energy. Smooth subcritical and supercritical branches connected to upstream data remain distinct unless the critical width is attained.

A hydraulic control is a critical section where a shallow water flow passes smoothly between subcritical flow and supercritical flow. One long-wave characteristic speed relative to the bed vanishes there. Regularity supplies an extra relation between discharge, depth and channel geometry, allowing the control to set the discharge or upstream state. A critical depth alone is not sufficient for a regular transcritical solution: the channel forcing must also be compatible there.

A weir is an obstruction across a channel that can regulate volume flux by creating a hydraulic control. In a shallow water model a sufficiently smooth crest has a critical section, with upstream flow subcritical and downstream flow supercritical under free discharge. Losses, submergence and geometry determine whether this ideal control applies.

For reservoir head $\delta$ above a broad crest, negligible approach velocity, constant reduced gravity, and no head loss, hydraulic control gives $\delta =3h_c/2$ and the displayed discharge. Reflecting the vertical coordinate gives the same result for a light layer escaping beneath a lowered roof. An orifice discharge coefficient is a separate closure and should not be silently inserted into this inviscid weir law.

A light shallow layer flows below a roof of height $H(x)$ and width $W(x)$ through a deeper approximately motionless ambient. Hydrostatic equilibrium and inviscid steady Bernoulli equation give the displayed specific energy for constant volume flux $Q$ and reduced gravity $g^{\prime }$. At fixed position, $E_h=g^{\prime }(1-{\mathrm{Fr}}^2)$ and $h_c=[Q^2/(g^{\prime }{W}^2){]}^{1/3}$. A roof depression raises the minimum admissible energy, just as a bed elevation does for a heavy lower layer.

For two same-height throats carrying the same light-layer volume flux and reduced gravity, the narrower throat has the larger critical minimum energy. A narrower downstream throat can back up and submerge an upstream throat. If the first throat is narrower, it can control discharge and feed a supercritical current through the second; a hydraulic jump followed by a second control is also possible when the available energy drop and intermediate geometry permit it. Equal throat widths are a marginal lossless case. Relative width alone does not fix jump position or mixing in the intervening reservoir.

For loss of fluid carrying its local axial velocity, the inviscid Bernoulli equation head stays constant while discharge decreases. On a flat bed a regular control requires a positive local maximum of $q=Q/b$. In the parabolic-width channel $b=b_0(1+x^2/L^2)$, this maximum occurs at $x_c=-S{L}^2/(Q_0+\sqrt{Q_0^2+S^2{L}^2})$, with $q_c=(Q_0+\sqrt{Q_0^2+S^2{L}^2})/(2b_0)$. Its critical depth is $(q_c^2/g{)}^{1/3}$. A control exists only when the prescribed total head equals the maximum critical-head envelope and the reach includes that point while retaining positive discharge.

For a prescribed positive head-loss rate $\gamma (x)$, the steady energy balance is $(E+H{)}_x=-\gamma$. The accumulated loss is added to the bed elevation when applying hydraulic control compatibility: $H_x+\gamma =h_c{b}_x/b$. A local loss law alone does not determine the accumulated loss between a remote inlet and the control; that integral must also be specified.

For a smooth rectangular channel, conservation of discharge and Bernoulli equation head gives $(1-F^2)h_x=F^{2}h{b}_x/b-H_x$. A regular hydraulic control has $F=1$ and the displayed geometric compatibility condition. At fixed discharge the critical-head envelope is $T(x)=H(x)+3h_c(x)/2$, with $h_c=(Q^2/(g{b}^2){)}^{1/3}$. A simple smooth control requires a local maximum of $T$: $T_x=0$ and $T_{xx}<0$. A local minimum cannot be crossed at the critical head because the real depth branches disappear nearby.

A shallow gravity current is supercritical when its Froude number exceeds one: both gravity-wave characteristic curves move downstream relative to the bed.

A shallow gravity current is subcritical when its Froude number is less than one: a gravity disturbance can propagate upstream relative to the bed. This definition uses the characteristic speed $\sqrt{g^{\prime }h}$ appropriate to the stated model.

A common high-Reynolds number closure for a deep-ambient gravity current is $\dot{L}=\mathrm{Fr}\sqrt{g^{\prime }h}$, where $L$ is front position and the order-one Froude number depends on the chosen front model.

In an ideal Boussinesq gravity current advancing beneath a much deeper ambient, the local head Froude number is $u_f/\sqrt{g^{\prime }{h}_f}=\sqrt{2}$. The ambient must be displaced around the finite-depth head. This differs from the Saint-Venant dry-front condition, which neglects ambient inertia. The coefficient belongs to the ideal deep-ambient head model; confinement, mixing and drag can change it.

The shallow water equations at a dry bed, with negligible ambient mass density, end in a vanishing-depth edge, rather than a finite-depth gravity current head. In a prismatic triangular channel the Riemann invariant $u+4\sqrt{g^{\prime }h/2}$ gives the speed $2\sqrt{2g^{\prime }{h}_0}$ after release of a resting semi-infinite layer of depth $h_0$. In a rectangular channel the corresponding speed is $2\sqrt{g^{\prime }{h}_0}$; the channel geometry matters.

A gravity-current box model replaces the current by a well-mixed region with uniform depth and density. Integral volume, scalar, and front-speed balances then reduce the spreading problem to ordinary differential equations.

A gravity-current box model in a prismatic channel with width equal to height has volume $V_0=L{h}^2/2$. Closing the front speed by $\dot{L}=F\sqrt{g^{\prime }h}$ gives the displayed law and long-time spreading $L\propto t^{4/5}$. This is an integral approximation with constant volume and reduced gravity, not an exact uniform-depth solution of the local shallow water equations.

For a constant-volume unit-width particle-laden gravity current with $M=Lh$, define the particle-fluid reference reduced gravity $g_0^{\prime }=g({\rho }_p-{\rho }_0)/{\rho }_0$, so the current has $g^{\prime }=g_0^{\prime }\varphi$. A constant front Froude number and absorbing-bed hindered settling give $\dot{L}=F\sqrt{g_0^{\prime }\varphi M/L}$ and $\dot{\varphi }=-(V_{s}L/M)\varphi (1-\varphi )$. Eliminating time and integrating gives the displayed runout length of a gravity current. For large runout and dilute initial concentration, $L_{\infty }^5\sim 25F^2{M}^3{g}_0^{\prime }({\varphi }_0+2{\varphi }_0^2/3+\cdots )/V_s^2$. This $g_0^{\prime }$ is buoyancy per unit particle fraction, not the initial current reduced gravity $g_0^{\prime }{\varphi }_0$.

In a horizontal prismatic V-shaped valley with section area $A=\lambda h^2$, a fixed-volume high-Reynolds-number gravity-current box model has $h=(W/(\lambda X){)}^{1/2}$ and front speed $\dot{X}=\mathrm{Fr}\sqrt{gh}$. Integrating gives $X=[5\mathrm{Fr}\sqrt{g}(W/\lambda {)}^{1/4}t/4{]}^{4/5}$ after the initial release scale is neglected. This is an inertial closure, distinct from the viscous V-shaped channel gravity current, whose exponent is different.

A well-mixed particle-laden gravity current in a prismatic channel with $b(z)=C\sqrt{z}$ has volume $V=2C{h}^{3/2}L/3$. Negligible entrainment gives fixed $V$. Vertical particle deposition flux is $W_s\varphi b(h)L$, so $\dot{\varphi }=-3W_s\varphi /(2h)$. Close the front by $\dot{L}=F\sqrt{G\varphi h}$, where $G=g({\rho }_p-{\rho }_0)/{\rho }_0$. Eliminating time givesThe concentration tends to zero at a finite limiting runout length of a gravity current, although reaching that limit takes infinite time in this model.

A uniform-depth particle-laden gravity current in fixed triangular cross section has conserved volume proportional to $h^{2}X$. A gravity-current front condition and absorbing-boundary particle deposition flux give $\dot{X}=\mathrm{Fr}\sqrt{G\varphi h}$ and $\dot{\varphi }=-2w(\varphi )\varphi /h$. The box describes late-time bulk evolution; it does not resolve the nose or immediate dam-break flow.

With constant buoyancy and no settling or entrainment, the box volume is proportional to $h^2\ell$. Thus $h=h_0\sqrt{x_0/\ell }$ for an initial lock length $x_0$. A gravity-current front condition gives $\dot{\ell }=F\sqrt{g^{\prime }{h}_0}(x_0/\ell {)}^{1/4}$, so ${\ell }^{5/4}$ increases linearly with time. This is a late-time integral approximation; it does not reproduce the initial rarefaction or the detailed nose.

For a dilute well-mixed particle current of conserved bulk volume $W$, triangular section $\lambda h^2$, constant particle settling speed $w_s$ and initial reduced gravity $g_0^{\prime }$, put $s=\sqrt{\varphi /{\varphi }_0}$. The balances give $\dot{X}=Bs{X}^{-1/4}$ and $\dot{\varphi }=-2w_s\varphi /h$, where $B=\mathrm{Fr}\sqrt{g_0^{\prime }}(W/\lambda {)}^{1/4}$. Eliminating time gives $s=1-(X/R{)}^{7/4}$ and the displayed limiting runout. It is approached at infinite time. Early spreading has the inertial $t^{4/5}$ exponent, while settling limits the eventual reach.

Under the same box-model assumptions, $M_0={\rho }_p{\varphi }_{0}W$ is initial particle mass and $m_d$ is final deposited mass per downstream length. Integrate the vertical particle deposition flux across the instantaneous triangular-valley width $2\lambda h$ from arrival of the front at $x$ until settling is complete. This gives the displayed profile on $0\le x\le R$ and zero beyond the runout. Its integral is $M_0$, verifying particle mass conservation. Areal deposit density also depends on the lateral coordinate because the wetted valley width changes as the current thins.

// Target: geophysics.bigb

In the prismatic triangular-channel gravity-current box model, take $h=h_0\sqrt{L/X}$ and $w(\varphi )=w_0(1-\varphi )$. Eliminating time gives $d\varphi /[\sqrt{\varphi }(1-\varphi )]=-2w_0{X}^{3/4}dX/(\mathrm{Fr}\sqrt{G}{h}_0^{3/2}{L}^{3/4})$. Integration yields the displayed invariant with $K=4w_0/(7\mathrm{Fr}\sqrt{G}{h}_0^{3/2}{L}^{3/4})$. The positive-concentration branch approaches a finite runout length of a gravity current as time tends to infinity.

If a finite-volume gravity current loses or dilutes the scalar that supplies its reduced gravity, its front can approach a finite runout length of a gravity current. A box model couples the scalar balance to a gravity-current front condition; eliminating time then gives the front position directly as a function of the remaining scalar concentration.

A particle-laden gravity current is driven by the excess density of suspended particles. Deposition decreases its reduced gravity, so a finite-volume current can approach a finite runout length.

A steady well-mixed dilute particle-laden gravity current above an absorbing porous bed loses carrier fluid at speed $W$ and particles at total downward speed $W+V$, where $V$ is their relative settling velocity magnitude. Volume conservation and the particle budget give $Q=Q_0-Wx$ and $\varphi =(Q/Q_0{)}^{V/W}$ for $W>0$. The momentum conservation equation is $(Q^2/h+g_0^{\prime }\varphi h^2/2{)}^{\prime }=-Wu$ if the withdrawn fluid carries its local horizontal velocity. Consequently $u{u}^{\prime }+g_0^{\prime }\varphi h^{\prime }+g_0^{\prime }h{\varphi }^{\prime }/2=0$. At $W=0$, $\varphi =e^{-Vx/Q_0}$ and the momentum flux is constant.

With reduced gravity $b$ and settling velocity $V_s$, a mixed rectangular-channel current obeys $Dh=-h{u}_x$, $Du=-b{h}_x-h{b}_x/2$, $Db=-V_{s}b/h$, where $D={\partial }_t+u{\partial }_x$. The characteristic speeds are $u,u\pm c$, with $c^2=bh$. Along the latter curves,The density differential matters: $u\pm 2c$ are Riemann invariants only when the reduced gravity is constant without deposition.

An axisymmetric gravity-current box model with no entrainment and negligible suspended-particle volume loss has constant bulk volume $\mathcal{V}$, uniform height $h$, and uniform particle volume fraction $c$. For an absorbing horizontal bed and downward settling velocity $w_s>0$, particle mass conservation gives $\dot{c}=-w_{s}c/h$. A constant front Froude number closes spreading by $\dot{R}=\mathrm{Fr}\sqrt{g_{0}ch}$, where $g_0=g({\rho }_p-{\rho }_a)/{\rho }_a$ is buoyancy per unit volume fraction. This model describes the mixed bulk and its head closure, not the initial inertial blast or a resolved head shape.

In a mixed particle-laden gravity current, a size class with settling speed $w$ has concentration $c(w,t)=c_0(w)e^{-wJ(t)}$. The shared reduced gravity is $g^{\prime }(J)=\int g_0(w)c_0(w)e^{-wJ}dw$, and $d{R}^4/dJ=4\mathrm{Fr}(\mathcal{V}/\pi {)}^{3/2}\sqrt{g^{\prime }(J)}$. Finite runout therefore requires ${\int }_0^{\infty }\sqrt{g^{\prime }(J)}dJ<\infty$. A positive minimum settling speed guarantees this; a distribution with arbitrarily small settling speeds need not. For $g_0{c}_0(w)\sim C{w}^p$ at zero, with $p>-1$, the integral converges precisely when $p>1$.

With $A=\mathcal{V}/\pi$, $g_0^{\prime }=g_0{c}_0$ and initial radius $R_0$, eliminating time from the axisymmetric settling box model gives $\sqrt{c/c_0}=(R_{\infty }^4-R^4)/(R_{\infty }^4-R_0^4)$. Put $s_0=R_0^2/R_{\infty }^2$ and $\tau =4A/(w_s{R}_{\infty }^2)$. Then $R^2/R_{\infty }^2=\tanh [t/\tau +\mathrm{artanh}{s}_0]$. Thus the finite limiting runout is approached at infinite time; treating it as an exactly attained finite-time stopping point contradicts these balances.

For the monodisperse circular ash runout, integrate the local particle deposition flux ${\rho }_p{w}_{s}c(t)$ from first arrival to infinite time. Here $s=\max (s_0,r^2/R_{\infty }^2)$ for $0\le r<R_{\infty }$ and $s_0=R_0^2/R_{\infty }^2$. The displayed $\Sigma$ is deposited particle mass per horizontal area, and vanishes outside the limiting current. It is constant under the initial disk, then decreases to zero; integrating $2\pi r\Sigma$ gives the initial ash mass ${\rho }_p{c}_0\mathcal{V}$. A deposit thickness additionally requires its packing fraction or bulk deposit density.