 Thermal erosion of a two-layer interface

Suppose identical buoyant thermals deliver buoyancy $B_0$ at rate $f$ to a mixed lower layer of area $A$ before penetrating its interface. Neglect source volume and heat loss. Relative to the upper-layer density, let $D=gh({\rho }_L-{\rho }_2)/{\rho }_{\mathrm{ref}}$, $P=f{B}_0/A$, and $C=\sqrt{B_0/(2K)}$, where $K=4\pi {\alpha }^3/3$. The lower-layer buoyancy budget gives $D=D_0-Pt$. A cycle-averaged fluid entrainment closure $\dot{h}/u^{\ast }={\mathrm{Ri}}^{-1}$, with $u^{\ast }=C/h$, $b^{\ast }=\alpha h$, yields the displayed law. Neutrality of the thermal at the interface requires $D(t^{\ast })h(t^{\ast }{)}^2=B_0/K$. The closure must incorporate the encounter duty cycle; an isolated-event entrainment law needs an additional averaging prescription.