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A dilute particle suspension can supply both excess mass density and heat. To first order its equation of state is $\rho /{\rho }_0=1+\gamma \varphi -\theta$, where $\varphi$ is the particle volume fraction and $\theta$ is fractional thermal expansion. Particle deposition flux reduces $\varphi$, while particle heating increases $\theta$. The competition can reverse the reduced gravity and create unstable density stratification.

The exact mixture equation of state is $\rho =(1-\varphi ){\rho }_f+\varphi {\rho }_p$. Keeping first-order terms in the small particle volume fraction and thermal expansion givesThe omitted term is ${\rho }_0\varphi \theta$. The downward solid-volume particle deposition flux is $J_v=W_s\varphi$; the corresponding particle mass flux is $J_m={\rho }_p{W}_s\varphi$. Uniform vertical mixing and constant layer depth implyFor $W_s>0$, set $r=W_s/H$. Thenand henceThe heated particle-laden layer loses its stable density stratification when this contrast vanishes. For $\beta >0$, solving for the neutral-buoyancy time givesThe layer is neutral at $T_s$ and statically unstable for $t>T_s$. The subsequent uniformly mixed lower-layer solution cannot represent the resulting overturning.

If $W_s=0$, then $\varphi ={\varphi }_0$, $\theta =\beta {\varphi }_{0}t$, and the continuous limit is $T_s=\gamma /\beta$. If $\beta =0$, the layer stays denser than its surroundings at every finite time and becomes neutral only asymptotically when $W_s>0$; thus $T_s=\infty$. These limiting cases must replace the printed formula when its denominator vanishes.

For a dilute particle suspension, take $W_s>0$ as the downward speed relative to the carrier fluid. The particle deposition flux through a horizontal absorbing bed is $W_s{\varphi }_b$, where ${\varphi }_b$ is the near-bed particle volume fraction. Without stirring, settling can create a particle-free upper layer and a descending concentration front, so uniform concentration throughout the remaining fluid is not automatic. Weak turbulence can maintain a nearly uniform suspended concentration while leaving the mean settling velocity approximately equal to the isolated-particle value; this justifies using $W_s\varphi$ in a mixed-layer balance. Strong stirring can produce resuspension and requires a separate erosion model.

For a bed inclined at angle $\vartheta$ to the horizontal, the normal settling speed is $W_s\cos \vartheta$, giving deposition flux $W_s{\varphi }_b\cos \vartheta$ per unit actual bed area. There is also downslope particle motion, and near-wall effects or resuspension can alter this simplest absorbing-boundary model. Integrating the normal projection over a sloping bed is equivalent to integrating $W_s{\varphi }_b$ over its horizontal projection.

The mixture mass density is $\rho ={\rho }_0+({\rho }_p-{\rho }_0)\varphi$, so its reduced gravity isThe Boussinesq approximation requires $({\rho }_p-{\rho }_0)\varphi /{\rho }_0\ll 1$, in addition to the dilution assumption.

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.

In a channel with width $2\lambda xz$ at height $z$, a well-mixed particle-laden gravity current of depth $h$ occupies area $\lambda x{h}^2$. Its volume conservation, momentum and particle deposition flux balances givewhere the reduced gravity is $g^{\prime }=d\varphi$. The $h/3$ coefficient is the cross-section-weighted mean of $h-z$ in the hydrostatic pressure force. The characteristic curves have speeds $u$ and $u\pm \sqrt{g^{\prime }h/2}$.