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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.

In a constant-volume gravity-current box model without settling, $\varphi ={\varphi }_0$ and $g^{\prime }=g{\varphi }_0(\gamma -\beta t)$. A constant Froude number then gives finite runout length of a gravity current at the neutral-buoyancy time $\gamma /\beta$. Further motion requires a different physical model once the stable density stratification is lost.

A heated particle-laden layer released along a bed forms a gravity current while $g^{\prime }=g(\gamma \varphi -\theta )>0$. In a gravity-current box model, heating can make $g^{\prime }$ vanish in finite time, whereas settling velocity alone can give finite runout length of a gravity current approached asymptotically.

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 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.

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.

For dilute sediment volume concentration $\varphi$, define $G=g({\rho }_p-{\rho }_0)/{\rho }_0$. Neglecting particle volume in the fluid-volume balance,This applies the Boussinesq approximation to the bulk mixture, requiring $G\varphi /g\ll 1$, and uses a deep ambient layer $h\ll H$. A gravity-current front condition closes the unresolved nose. Write it as $\dot{L}=F\sqrt{g^{\prime }h}$, with a prescribed order-one front coefficient $F$. If the chosen Froude number is instead $\mathrm{Fr}=\dot{L}/c$, then $F=\sqrt{2/3}\mathrm{Fr}$.

Use a parabolic-channel sediment-current box model for one side of the dam: initially its length is $L_0$, depth $h_0$, and uniform concentration ${\varphi }_0$. Its conserved fluid volume isFor vertical settling onto the channel sides, the projected deposition width is $b(h)=\sqrt{h}$. Thus the total particle deposition flux is $W_s\varphi \sqrt{h}L$ andThe geometric factor $3/2$ follows from $b(h)/A(h)$; using a rectangular deposition formula would give a different result.

Eliminate time and use $h^{3/2}=h_0^{3/2}{L}_0/L$:The runout length of a gravity current is the limit where suspended concentration vanishes:The model approaches this length asymptotically in time. It assumes negligible entrainment, erosion and water loss, and does not resolve the initial release or nose. If the initially disturbed region supplies both sides, apply the formula to each side using its share of the initial volume and length.

At the gravity current nose, the depth and velocity change over a distance comparable with the depth, with appreciable vertical acceleration and mixing. The hydrostatic approximation and the depth-uniform interior model do not resolve that structure. A suitable gravity-current front condition iswith a positive, order-one Froude number determined by the nose closure and geometry. It is not fixed by the interior equations alone. This definition uses $\sqrt{g^{\prime }h}$; if the front condition is expressed using $c=\sqrt{g^{\prime }h/2}$, its numerical coefficient changes by $\sqrt{2}$.

For a late-time gravity-current box model, take the depth and particle volume fraction to be uniform over $0<x<X(t)$, and neglect entrainment and resuspension. The initial volume is $\mathcal{V}=\beta h_0^{2}L/2$, soThe total particle volume is $\mathcal{V}\varphi$. Deposition over the projected boundary width gives $d(\mathcal{V}\varphi )/dt=-\beta hXw\varphi$. With the specified hindered settling factor, the prismatic triangular-channel gravity-current box model isInitializing this approximate model with $X(0)=L$ and $\varphi (0)={\varphi }_0$ matches the original volume and concentration; it does not assert that the immediate dam-break flow is a uniform box.

Eliminate time to obtainSince the left primitive is $2\mathrm{artanh}\sqrt{\varphi }$, the hindered-settling runout invariant in a prismatic triangular channel isThis relation, the volume constraint and $\dot{X}$ determine the late-time evolution. If desired, the time is given by the quadrature $t={\int }_L^{X(t)}[\mathrm{Fr}\sqrt{G\varphi (s)h_0\sqrt{L/s}}{]}^{-1}ds$. The physical branch has nonnegative right-hand side in the boxed relation; it ends when that expression reaches zero.

Thus the limiting runout length of a gravity current, measured from the closed end, isThe advance beyond the dam is $X_{\infty }-L$. In the dilute limit, $\mathrm{artanh}\sqrt{{\varphi }_0}\simeq \sqrt{{\varphi }_0}$. The ideal model approaches its finite runout length of a gravity current only as $t\to \infty$: near the limit $\varphi$ decays exponentially with rate $2w_0/h_{\infty }$ and the front speed tends to zero. The exponent $7/4$ uses the prismatic triangular geometry; a channel widening with $x$ has a different volume constraint and runout law.

Let $\mathcal{V}=H{L}_0$ be the layer volume per unit transverse width. The gravity-current box model imposes $hL=\mathcal{V}$; it approximates the front and scalar balances, rather than an exactly uniform solution of the local momentum equation. For a specified constant front Froude number $\mathrm{Fr}$, the heated particle-laden gravity current satisfiesThe initial data are $L=L_0$, $\varphi ={\varphi }_0$, and $\theta =0$. The gravity-current front condition applies only while $\gamma \varphi -\theta \ge 0$.

For heating without settling, $\varphi ={\varphi }_0$ and $\theta =\beta {\varphi }_{0}t$. Integration of the front equation givesHence the heated gravity-current runout isIt is reached at the neutral-buoyancy time $\gamma /\beta$ within this front model. A finite-momentum current could subsequently coast or lift off: the formula is the maximum predicted by the prescribed gravity-current front condition, which contains no independent front inertia.

For settling without heating, $\theta =0$. Set $K=\mathrm{Fr}\sqrt{g\gamma \mathcal{V}}$, so $\dot{L}=K\sqrt{\varphi }{L}^{-1/2}$. Eliminating time yieldsThus the runout length of a gravity current isThis length is approached as $t\to \infty$, because positive concentration cannot disappear at a finite time under the settling law. If distance means displacement from the removed barrier, the answer is $L_{\max }-L_0$ in either case.

Use the gravity-current front condition $\dot{L}=\mathrm{Fr}\sqrt{d\varphi h}$, with a specified order-one front Froude number. In a triangular-channel gravity-current box model, the depth and concentration are uniform over $0<x<L(t)$. The volume isThe horizontal projected deposition area is ${\int }_0^{L}Tdx=\lambda h{L}^2$. Consequently the integral model isThis front speed is a closure for the integral model, rather than the assertion that its uniform interior is an exact solution of the spatially varying shallow water equations.

Eliminating time and integrating givesThe limiting runout length of a gravity current is thereforeThe additional distance beyond the release gate is $L_{\infty }-L_0$. The limit is approached asymptotically as $\varphi \to 0$ and $\dot{L}\to 0$; the idealized box model does not predict a finite-time stop. Its numerical runout depends on the adopted front Froude number.

Use a well-mixed gravity-current box model of length $L(t)$ and depth $h(t)=V_0/L(t)$. Its fixed volume requires $w_e=w_d$. The integrated chemical balance and the standard gravity-current front condition areThese two ordinary differential equations are the required integral model.

First suppose $R_1>0$ and putEliminating time givesWithintegration from $(L,\varphi )=(L_0,1)$ yieldsAs $t\to \infty$, the concentration tends to zero and the runout length of a gravity current isThe formula has a regular $a\to 0$ limit. If $R_1=0<R_2$, direct integration instead giveswhose value at $\varphi =0$ gives $L_{\infty }$.

When ${\hat{W}}_1={\hat{W}}_2=\hat{W}$, both concentrations acquire the same decay factor. The total reduced gravity consequently obeysEliminating time with the front equation givesIntegration from the initial state yieldswhereThe particle-laden gravity current reaches its runout length of a gravity current when $g^{\prime }$ vanishes. Since $\hat{W}<0$,Thus $C=5\mathrm{Fr}$ for the stated front condition; the common normalization $\mathrm{Fr}=1$ gives $C=5$.

A finite-volume gravity-current box model in a widening triangular channel has volume proportional to $h^2{L}^2$, so $hL=K$ is constant. A gravity-current front condition and absorbing-bed particle deposition flux giveEliminating time makes $\sqrt{\varphi }$ affine in $L^{5/2}$, producing a finite limiting runout length of a gravity current despite an infinite idealized stopping time.