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A two-equation closure transporting turbulent kinetic energy $k$ and its turbulent kinetic energy dissipation rate $ϵ$. Its eddy viscosity is $C_{\mu }{k}^2/ϵ$. In a high-Reynolds-number wall-normal reduction, turbulent diffusion has coefficients ${\nu }_T/{\sigma }_k$ and ${\nu }_T/{\sigma }_{ϵ}$; the energy equation has shear production $P$ and destruction $ϵ$, while the dissipation equation has modeled production $C_{ϵ1}Pϵ/k$ and destruction $C_{ϵ2}{ϵ}^2/k$. The dissipation production/destruction terms are closure terms, not a second conservation of energy.

The work of the Reynolds stress against the mean velocity gradient transfers kinetic energy from the mean flow into fluctuations. In a plane shear flow $U(y)$ this is $P=-\overline{u^{\prime }{v}^{\prime }}{U}_y$. The eddy viscosity closure gives $P={\nu }_T{U}_y^2\ge 0$. Buoyancy production and transport are separate terms in the turbulent kinetic energy budget.

The K-epsilon turbulence model evolves turbulent kinetic energy $k$ and its turbulent kinetic energy dissipation rate $ϵ$, rather than resolving turbulent fluctuations. It closes the Reynolds stress with an isotropic eddy viscosity, ${\nu }_T=C_{\mu }{k}^2/ϵ$, and models transport of $k$ and $ϵ$ by gradient diffusion with coefficients ${\nu }_T/{\sigma }_k$ and ${\nu }_T/{\sigma }_{ϵ}$. Production is tied to mean strain; dissipation-equation production and destruction are modeled terms with empirical constants. The characteristic velocity and length are $\sqrt{k}$ and $k^{3/2}/ϵ$. These assumptions are primarily for high Reynolds number, and strongly anisotropic flow, sharp interfaces and molecular-viscosity-dominated regions need additional treatment. The dissipation equation is a closure, not a second exact energy-conservation law.

Here $k$ is turbulent kinetic energy per unit mass, $k=\frac{1}{2}\overline{u^{\prime 2}+v^{\prime 2}+w^{\prime 2}}$, rather than the permeability of a porous medium used in the porous-flow questions. The K-epsilon turbulence model represents the high-Reynolds-number transport of this energy and its turbulent kinetic energy dissipation rate $ϵ$.

In the $k$ equation, $d[({\nu }_T/{\sigma }_k)k_y]/dy$ is the divergence of modeled turbulent diffusion, $\mathcal{P}$ is mean-shear production, and $-ϵ$ is molecular conversion of fluctuation energy into heat. Statistical stationarity and fully developed flow have removed storage and streamwise advection from this reduction.

In the $ϵ$ equation, $d[({\nu }_T/{\sigma }_{ϵ}){ϵ}_y]/dy$ transports dissipation intensity; $C_{ϵ1}\mathcal{P}ϵ/k$ models its generation associated with energy production; and $-C_{ϵ2}{ϵ}^2/k$ models its destruction. The time scale $k/ϵ$ supplies the source-rate dimension. The $ϵ$ equation is a closure, not another energy-conservation equation. Finally, ${\nu }_T=C_{\mu }{k}^2/ϵ$ has dynamic viscosity dimensions and parametrizes Reynolds stress transport; ${\sigma }_k,{\sigma }_{ϵ}$ set the relative diffusion strengths, and the remaining constants calibrate the closure.

In the Boussinesq approximation, put $\gamma =K{W}^2\sqrt{g/{\rho }_0}$. The buoyancy-driven turbulent density diffusion equation gives, wherever $q={\bar{\rho }}_z>0$,The coefficient outside the bracket is positive, so the local density gradient increases precisely whenAt a point where $q=0$, use the original degenerate diffusion equation rather than dividing by $\sqrt{q}$.

For the proposed self-similar solution, $\bar{\rho }={\rho }_0[1+F(t)G(z)]$ and $F(0)=1$. Substitution into ${\bar{\rho }}_t={\partial }_z{F}_{\rho }$ givesAn odd power profile $G(z)=C\mathrm{sgn}(z)|z{|}^p$ has right-hand spatial power $(3p-5)/2$. Matching it to $p$ requires $p=5$. Thus choose $G=C{z}^5$ with $C>0$, an odd, increasing profile. Since $G^{\prime }=5C{z}^4$, its flux is proportional to $z^6$ on both sides of the midplane. The resulting quintic density-profile similarity in an unstable tube satisfiesIntegration gives $F^{-1/2}=1-At/2$. ConsequentlyAlthough mixing transports density downwards, the local density gradient can grow because spatial variations of the flux redistribute density. In this example the downward flux increases towards each end, driving the formal amplification of the odd profile.

This self-similar solution deliberately ignores the end conditions: its nonzero end flux cannot satisfy impermeable ends. Moreover, the Boussinesq approximation requires $FC|z{|}^5\ll 1$, which fails before the formal divergence. A closed tube has only finite available potential energy, so it cannot sustain indefinitely growing turbulent kinetic energy or the assumed instantaneous flux law. End effects, changing eddy statistics and decay of the unstable stratification must eventually replace this local model. The conflicting endpoint labels in the source do not enter the calculation because end conditions have been expressly excluded.

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.

Define the positive total shear stress by $\tau =\rho [\nu U_z-\overline{u^{\prime }{w}^{\prime }}]$, so the physical wall traction on the fluid opposes the positive flow. The steady mean momentum equation givesFor a driving pressure gradient $P_x<0$, the stress decreases upward. It is approximately constant whileAn outer boundary-layer height can impose a stricter limit. In the turbulent region outside the viscous or roughness sublayer, the only local neutral scales are $u_{\ast }$ and $z$, so $U_z=C{u}_{\ast }/z$. Conventionally write $C=1/\kappa$, where $\kappa$ is the von Kármán constant. Equivalently, a mixing length $\ell =\kappa z$ and stress closure $u_{\ast }^2={\ell }^2{U}_z^2$ give the same result. Integration yields the law of the wallThe roughness length $z_0$ is an extrapolation scale, not a statement that the logarithm is valid at the actual solid surface. For a smooth boundary, the viscous inner scales are $z^{+}=z{u}_{\ast }/\nu$, $U^{+}=U/u_{\ast }$. The viscous sublayer has $U^{+}\simeq z^{+}$; farther out $U^{+}={\kappa }^{-1}\log z^{+}+C_s$, corresponding to $z_0=(\nu /u_{\ast })e^{-\kappa C_s}$. For a fully rough boundary, $z_0$ is proportional to a geometric roughness height and viscosity no longer determines the leading profile. The logarithmic region still needs roughness height $\ll z\ll {\delta }_{\tau }$.

In local equilibrium, production of turbulent kinetic energy by the shear is $P=-\overline{u^{\prime }{w}^{\prime }}{U}_z\simeq u_{\ast }^2{U}_z$. Neglecting its transport relative to production and viscous dissipation,Dimensional similarity alone gives $ϵ\propto u_{\ast }^3/z$; the displayed coefficient additionally uses the local energy balance and the logarithmic shear.

The positive rate per unit mass at which molecular viscosity removes turbulent kinetic energy, with $s_{ij}^{\prime }$ the fluctuating symmetric velocity-gradient tensor. Its dimensions are velocity squared per time. The integral-scale estimate is $ϵ\sim q^3/L$, expressing transfer over turnover time $L/q$; numerical constants depend on the convention for the turbulent velocity and length scales.