 Solution

The conserved particle volume per unit horizontal area isThe sediment mass determines final deposit thickness: with the stipulated deposit particle volume fraction ${\varphi }_{\max }=1$, all particle volume becomes bed thickness, so $h_c=\mathcal{P}$. For $z_m=H/2$,The final particle volume fraction averaged over the entire container remains the initial average: clear fluid occupies $h_c<z<H$ and the deposit occupies $0<z<h_c$. Averaging over the deposit alone would instead give one, which is a different average.