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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.

For potential energy of two-layer entrainment with constant $B=g^{\prime }h$, the fixed-energy turbulent mixed layer assumption gives $u_U^{2}h=C_K^2{\Omega }^2{R}^2{h}_0$. If $u_I=C_I{u}_U$ and the interfacial power model for entrainment applies with fixed coefficients, integration gives the displayed depth law with $A=5C_W{c}_D{C}_I^3{C}_K^3{\Omega }^2{R}^3/(B{h}_0)$. The relation applies only before the layer reaches the tank bottom. Its interfacial Richardson number $B{d}_I/(h{u}_I^2)$ stays constant for fixed interface thickness, whereas it decreases like $h^{-1}$ in the constant-speed entrainment model.

For stable shear-driven turbulence, let $P$ be the rate of production of turbulent kinetic energy by mean shear and $B_m>0$ the rate consumed by buoyancy work. The Flux Richardson number isWith an upward coordinate and vertical turbulent buoyancy flux $\overline{w^{\prime }{b}^{\prime }}$, $B_m=-{\rho }_l\int \overline{w^{\prime }{b}^{\prime }}d\mathcal{V}$, while shear production is $P=-{\rho }_l\int \overline{u^{\prime }{w}^{\prime }}{\partial }_{z}Ud\mathcal{V}$. In a locally steady balance without net turbulent-energy transport or storage, $P=B_m+\mathcal{D}$, so $0\le {\mathrm{Ri}}_f\le 1$ and the dissipation-normalized mixing coefficient is $B_m/\mathcal{D}={\mathrm{Ri}}_f/(1-{\mathrm{Ri}}_f)$. A global ratio $C_W={\dot{E}}_P/P_I$ need not equal this local production-normalized Flux Richardson number when energy transport, storage, or bulk acceleration is important.

The interfacial Richardson number in this tank isIn model S,Thus the relative importance of interface stratification changes continually. If mixing efficiency depends on interfacial Richardson number, constant $C_W$ is not generally self-consistent over substantial deepening; it can at best describe a regime where that dependence is weak. In model K,A local efficiency depending only on this ratio can therefore remain constant, provided the fixed interface thickness and the other empirical coefficients remain appropriate.

The energetic reason to expect intermediate behavior is that entrainment must both lift and homogenize dense fluid and mobilize it. The whole upper-layer kinetic energy budget has the form $P_{\mathrm{lid}}=\dot{K}+{\dot{E}}_P+\mathcal{D}$, apart from additional boundary losses. Keeping the characteristic speed constant makes $K\propto h$ and requires a continuing power supply to accelerate new fluid, together with dissipation in an ever larger layer. Keeping $K$ constant instead forces the speed to decline as $h^{-1/2}$. A finite lid forcing plausibly produces an initial regime near the former and a deeper regime near the latter, rather than indefinite constant-speed entrainment.

One can make this expected bracket quantitative after matching the models' initial interface speed $u_0$. Suppose the actual speed lies between $u_0\sqrt{h_0/h}$ and $u_0$, and the same local coefficients are used. Set $D_0=2C_W{c}_D{u}_0^3/B$. ThenIntegrating these inequalities givesEquivalently, if $K\propto h^q$ with $0\le q\le 1$, the local power closure yields a late-time depth exponent $2/(5-3q)$ between $2/5$ and $1$. This is the intended energetic interpolation between K and S, not a universal bound following from the printed assumptions alone. Different efficiencies or unmatched initial coefficients can invalidate pointwise ordering; quantitative prediction requires the lid-to-interface power transfer and the actual energy budget.

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.