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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.

To connect the prescribed interfacial drag coefficient to the energy supply, state the lumped interfacial power model for entrainment explicitly:It estimates the power transmitted by lid-driven flow to the mixing interface. Constant geometric and transfer factors are included in $c_D$ or $C_W$. It is the local stress-power interpretation required for the depth laws in this question, not an identity equating arbitrary lid shaft power to interfacial power. With the potential energy of two-layer entrainment from (a),In model S, $u_I=\Omega R$ and $B$ are constant, soThus the depth increases at a constant rate proportional to the inverse bulk Richardson number, with $h=h_0+2C_W{c}_D\Omega Rt/{\mathrm{Ri}}_B$, until the layer reaches $H$. Here the prescribed $u_I$ must be read as a characteristic speed. Literal pure solid-body rotation has zero radial velocity, so the PDF's parenthetical description of $u_I$ as radial cannot simultaneously be a strict component statement and the assumption $u_I=\Omega R$.

If the phrase about lid work is taken literally as shaft power $P_{\mathrm{lid}}=\Omega {\mathcal{T}}_{\mathrm{lid}}$, the interfacial stress alone does not specify the lid torque or its energy transfer to the interface. The preceding closure is therefore a visible model assumption, not a hidden consequence of solid-body rotation.

Under the Boussinesq approximation, total upper-layer kinetic energy is $K={\rho }_{l}Sh{u}_U^2/2$, where $u_U$ is its volume root-mean-square speed. For solid-body rotation of angular speed $\omega (t)$, direct radial averaging gives $u_U^2={\omega }^2{R}^2/2$, consistent with that energy expression. A constant $K$ implies $h{u}_U^2=h_0{u}_U(0{)}^2$. Define $C_K=u_U(0)/(\Omega R)$ to obtain the fixed-energy turbulent mixed layer relationWith $u_I=C_I{u}_U$, the interfacial power model for entrainment used in (b) givesMultiply by $h^{3/2}$ and integrate from $h_0$. This yieldsThe fixed-energy two-layer entrainment law is thereforeHere ${\mathrm{Ri}}_{B,0}=B/({\Omega }^2{R}^2)$ uses the same lid-speed normalization as part (b). All numerical geometry factors depend on the chosen characteristic stress/speed convention. The formula ceases to apply once $h=H$.

The exponent $2/5$ requires the local power to decrease as $u_I^3\propto h^{-3/2}$. It does not follow from constant upper-layer kinetic energy and a fixed fraction of unspecified lid power alone. For example, an imposed constant shaft power together with a constant fraction going to potential energy would give constant $\dot{h}$, not this exponent. The appropriate distinction between local interface power and global forcing appears explicitly in the primary study Entrainment and mixed layer dynamics of a surface-stress-driven stratified fluid, equations (3.6)–(3.7); the calculations above use that local interpretation.