 Solution

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.