Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 65 / 4

Let $h_0=h(0)$ and $S=\pi R^2$. The closed tank has constant mass, soTherefore $({\rho }_l-{\rho }_u)h=({\rho }_l-{\rho }_u(0))h_0$. Under the Boussinesq approximation, define $g^{\prime }=g({\rho }_l-{\rho }_u)/{\rho }_l$ and obtainThe gravitational potential energy is especially easy to evaluate with the stipulated downward depth coordinate: the gravitational potential per unit mass is $-gz$. HenceThis proves the potential energy of two-layer entrainment relationThe positive sign expresses the lifting and homogenization of dense entrained fluid within the upper layer. The fixed thin interface is neglected in this leading two-layer energy integral.

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.

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.