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Combine a fixed-energy mixed-layer relation $u^{2}h=A{\Omega }^2{R}^3$, jump $g^{\prime }=N^{2}h/2$, and stress power proportional to $u^3$ with an energy-transfer factor $(\Omega /N{)}^{\alpha }(\delta /R{)}^{\beta }$. Eliminating depth in favor of the interfacial Richardson number gives $E\propto (\Omega /N{)}^{\alpha -1}(\delta /R{)}^{\beta +3/2}{\mathrm{Ri}}_i^{-3/2}$. A local-only entrainment closure requires $\alpha =1$, $\beta =-3/2$ and exponent $3/2$ on the Richardson number.

A local ratio of stabilizing buoyancy to shear across an interface of thickness $\delta$, using reduced gravity jump $g^{\prime }$ and characteristic velocity $u$. The dimensionless value is ${\mathrm{Ri}}_i=g^{\prime }\delta /u^2$. It is distinct from a whole-layer bulk Richardson number based on the total mixed depth. Local entrainment laws use this ratio when the transition thickness and velocity characterize the turbulence-producing region.

Let $h_0=h_1(0)$ and ${\rho }_{10}={\rho }_1(0)$, and write ${\rho }_L(t)$ for the mixed lower-layer density. PutUse the common Boussinesq reference/inertial density normalization from part (a). Here $AD$ is the integrated negative buoyancy of the lower layer relative to the upper one. Assume negligible source volume, fixed enclosure volume, small density contrasts, identical thermals with total buoyancy $B_0$, slow evolution during one rise, and deposition of their positive buoyancy in the lower layer until penetration begins. Entrainment of upper fluid carries zero density anomaly relative to ${\rho }_2$, while thermal injection reduces the negative anomaly. The lower-layer conservation of mass or equivalent buoyancy budget is thereforeIt would be inconsistent to conserve $h_1({\rho }_L-{\rho }_2)$ while also depositing the nonzero source buoyancy $f{B}_0$.

Quasi-steady point-source spherical thermal similarity at the interface gives $b^{\ast }=\alpha h_1$ and $u^{\ast }=C/h_1$. The Richardson number compares the potential-energy barrier associated with the density jump to the thermal's available kinetic energy:An entrainment coefficient $c$ must be positive for upward deepening. A positive exponent $n$ makes stronger stratification inhibit entrainment; $n=0$ removes this dependence and $n<0$ would perversely strengthen entrainment as the stabilizing barrier increases. The power law is intended for a suitable finite-$\mathrm{Ri}$ regime: as $\mathrm{Ri}\to 0$, its divergence must be limited by available turbulent motion. Its constants and validity are closure information, not consequences of dimensions alone. An additional kinetic-energy constraint is useful: the interfacial lifting cost scales as ${\rho }_0{g}^{\prime }{b}^{\ast }E{u}^{\ast }$, while the impinging turbulent power scales as ${\rho }_0{u}^{\ast 3}$. Hence $E\mathrm{Ri}=c{\mathrm{Ri}}^{1-n}$ must remain of order one or smaller. A law intended to hold at arbitrarily large $\mathrm{Ri}$ with no extra power source therefore requires $n\ge 1$; for $n=1$, $c$ measures a bounded conversion efficiency up to geometric factors. The requested $c=1$ represents an ideal order-one efficiency in this normalization.

For $c=n=1$, use the printed closure as a cycle-averaged interface law. ThenIntegrating from $h_0$ gives the thermal erosion of a two-layer interface resultThe thermal's density deficit relative to the lower layer at height $h_1$ is ${\rho }_{\mathrm{ref}}{B}_0/(gK{h}_1^3)$. Under the question's buoyancy-controlled penetration criterion, the threshold thermal is neutral relative to the upper layer. HenceThe time and height are determined implicitly by this threshold and the explicit $h_1(t)$ law. Equivalently,Use its first admissible root, provided it occurs before the upper layer is exhausted. Impingement initially requires $B_0<K{D}_0{h}_0^2$; otherwise penetration is immediate in this criterion. Depending on parameters, $D(t)h_1(t{)}^2$ can initially increase as entrainment deepens the layer, but it ultimately decreases to zero, so the first downward threshold crossing selects $t^{\ast }$.

The printed $E={\dot{h}}_1/u^{\ast }$ must be an already averaged law to use it for separated thermal arrivals; if $c$ describes only an isolated impingement, an encounter duration/area averaging factor is missing and the same numerical formula cannot be asserted. The present answer uses the supplied averaged closure. It also uses the natural inherited $B_0$ per thermal; neglecting deposition of source buoyancy would instead give the different no-source-budget approximation $h_1^4=h_0^4+4C^{3}t/(\alpha D_0)$. The two budgets are explicitly distinguished. Real inertial penetration can precede neutral buoyancy, but that requires an additional head/penetration model absent here.

Let $\pi$ be pressure divided by the reference density and $b$ the buoyancy perturbation. The linear, inviscid Boussinesq equations about $U(z)$ areFor a stationary Fourier component $e^{ikx}$, incompressibility gives $u=i{w}^{\prime }/k$. The horizontal momentum equation then gives $\pi =i(U^{\prime }w-U{w}^{\prime })/k$, while buoyancy conservation gives $b=i{N}^{2}w/(kU)$. Substitute these into the vertical momentum equation to obtain the stationary Taylor–Goldstein equation:Hence its local WKB approximation isFor $U^{\prime \prime }$ negligible, an outgoing upward wave has $mU/k>0$. In particular $m>0$ for $k,U>0$. In an upper evanescent region choose the branch with $\mathrm{Im}m>0$ in $e^{i\int mdz}$ so the solution decays upward. These choices express the radiation condition; a reflected wave has the opposite real branch and downward group velocity.

There are two different limiting levels. A critical layer in a shear flow occurs at $U=0$, where the intrinsic frequency vanishes, $|m|\to \infty$, and upward propagation slows while the vertical wavelength contracts. Dissipation or nonlinear effects can absorb the wave and transfer its momentum to the mean flow. For a stable, slowly varying critical layer with Richardson number $\mathrm{Ri}=N_0^2/(U^{\prime }{)}^2>1/4$, the usual causal inviscid continuation suppresses transmission; it is not the same as reflection at $m=0$. When the Richardson number is small, a general absorption-only conclusion needs extra stability assumptions.

Reflection instead occurs at the turning level of a stationary internal wave $k|U|=N_0$, beyond which $m^2<0$. It is sometimes called a critical reflection height, but the wind there is nonzero. Assuming $0<k{U}_0<N_0$, the specified profiles giveThe decreasing-wind profile never encounters $m=0$ before its zero-wind layer. Both profiles have $U^{\prime \prime }=0$; a well-separated WKB description requires $N_{0}H/U_0\gg 1$, and a turning-layer transition is still required for the increasing profile.

Let $a=({\rho }_b-{\rho }_t)/H={\rho }_0{N}^2/g$. The initial density is $\rho (z)={\rho }_t-az$. Conservation of mass in the material homogenized over $-h<z<0$ fixes its mean density to ${\rho }_u={\rho }_t+ah/2$, while the untouched lower layer has interfacial density ${\rho }_i={\rho }_t+ah$. Hence the reduced gravity difference across the interface isThe reference contribution involving ${\rho }_t$ cancels from the difference.

Only the upper layer changes its potential energy. The potential-energy cost of homogenizing a linear stratification issoThe positive sign reflects work against stable stratification.

The rotating boundary supplies energy on scales set by $\Omega$ and $R$. Turbulent drag and dissipation remove energy, and entrainment adds initially stationary fluid that must be accelerated. If upper-layer energy grows, the increasing turbulent loss provides a restoring tendency; after a short adjustment, input and losses can approximately balance while the interface moves on a slower time scale. This gives a physical rationale for a fixed-energy turbulent mixed layer, not a consequence of mass conservation alone. Its characteristic kinetic energy is $K\sim {\rho }_0\pi R^{2}h{u}_u^2/2$. With the stipulated $N$-independent energy closure,so the characteristic speed decreases as $h^{-1/2}$ rather than remaining constant.

The stress does work at a rate proportional to $\pi R^2\tau u_u=c_D{\rho }_0\pi R^2{u}_u^3$. Combine this with the energy increase and absorb fixed drag/geometrical factors into $C>0$:Here the exponents $\alpha ,\beta$ are unrelated to the plume entrainment coefficient in Question 2. The interfacial Richardson number isSet $q=\Omega /N$, $d=\delta /R$ and $r=h/R$. Then $u_u^2/(N^2{h}^2)=d/(2r{\mathrm{Ri}}_i)$ and $r=q(2A{\mathrm{Ri}}_i/d{)}^{1/2}$. Substitution givesThe assumption that $E$ depends on the local Richardson number alone excludes separate dependences on $q,d$. The entrainment exponent from a local interfacial Richardson closure is consequentlyThis comparison treats $q$ and $d$ as independently variable external controls; fitting only a single apparatus would not by itself determine the exponents.

Finally, use the fixed-energy closure again:Integrating from a positive initial mixed-layer depth $h_0$ at time $t_0$ givesWhen the growth term dominates the initial offset, the rotating-disc mixed-layer depth law isTaking the effective time origin to be zero gives the stated scaling. Its validity is limited to the regime of the closures and $h<H$; the singular velocity predicted at $h=0$ is not a model for the initial boundary-layer formation.