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For a stationary, horizontally homogeneous mean flow without a horizontal pressure gradient, the averaged horizontal Navier-Stokes equation is $\nu U_{zz}-(\overline{u^{\prime }{w}^{\prime }}{)}_z=0$. Integrating gives the displayed physical shear stress. The derivative of the Reynolds covariance belongs in the stress-divergence equation, not inside the stress itself. The kinematic stress is ${\tau }_d/\rho$, which must be used when an eddy viscosity with units of kinematic viscosity multiplies the mean gradient.

For a symmetric localized turbulent plane jet similarity profile, integrated momentum balance gives $f{f}^{\prime }=2C_1^2{\eta }^2(f^{\prime \prime }{)}^2$ on $\eta >0$. At a finite free edge ${\eta }_w>0$, $f^{\prime }({\eta }_w)=0$ and finite $f({\eta }_w)$ force $f^{\prime \prime }({\eta }_w)=0$. Thus the mixing length model $l=C_1|y|$ cannot simultaneously impose zero velocity and a nonzero maximum velocity gradient at the edge. Its eddy viscosity vanishes there. Moreover, a nonzero finite centreline speed $f^{\prime }(0)=a$ forces $|f^{\prime \prime }|\sim a/(\sqrt{2}{C}_1\sqrt{\eta })$ near the centre, displaying a singular-shear defect of that local closure. The molecular-viscosity neglect cannot be uniform in this core.

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.

A mixing-length closure uses a prescribed mixing length $\ell$ to approximate turbulent transport. For mean shear ${\partial }_{z}U$, a standard model gives eddy viscosity ${\nu }_t={\ell }^2|{\partial }_{z}U|$ and down-gradient Reynolds stress. Scalar-diffusion versions replace momentum transport by an appropriate scalar flux. The choice of length and correlations is a modelling assumption rather than an exact closure of the Navier-Stokes equation.

For stationary horizontally homogeneous means, incompressibility gives ${\bar{w}}_z=0$ and the impermeable boundary makes $\bar{w}=0$ throughout. With no imposed horizontal mean pressure gradient, the balances reduce toThe last equation includes the vertical Reynolds stress; omitting it requires an additional small-normal-stress approximation. The total vertical buoyancy flux is $B_0$ and the horizontal momentum flux is a signed constant $J_M$ with $|J_M|=U_{\ast }^2$. These are fluxes per unit reference density.

The local down-gradient closure defines eddy viscosity and eddy diffusivity byA parcel moving upwards typically retains the smaller momentum or scalar value from its previous height in an increasing mean profile; its fluctuation then correlates negatively with upward motion. This motivates positive coefficients, but is a closure assumption, not a universal theorem about turbulent or convective fluxes. The mean gradients obeyFor positive applied traction on a fluid above a bottom boundary, $J_M=+U_{\ast }^2$ and the mean velocity decreases away from the forcing boundary. The usual atmospheric convention specifies the stress absorbed by the boundary and has the opposite sign of $J_M$; scalar and Richardson-number results are unchanged. At large Reynolds number and Péclet number, outside the molecular wall sublayers, $K_M\gg \nu$ and $K_B\gg \kappa$, so molecular transport can be neglected in these flux-gradient relations. Large bulk numbers do not justify doing this exactly at a smooth solid wall.

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.

Put $\xi =x/{\delta }_0$ and $n=2p-q$. The eddy viscosity is ${\nu }_T=C_{\mu }\sqrt{k_0}{\delta }_0{\xi }^n$, since ${\delta }_0=k_0^{3/2}/{ϵ}_0$. In the energy equation, advection scales as ${\xi }^{p-1}$, turbulent diffusion as ${\xi }^{n+p-2}$, and destruction as ${\xi }^q$. A leading advection–diffusion K-epsilon turbulence front requires $n=1$, henceComparing the leading coefficients then gives $U_0=C_{\mu }p\sqrt{k_0}/{\sigma }_k$. Similarly, the leading coefficients of the dissipation equation give $U_0=C_{\mu }q\sqrt{k_0}/{\sigma }_{ϵ}$. Their equality implies $q=\sigma p$, where $\sigma ={\sigma }_{ϵ}/{\sigma }_k$. Combining this with $q=2p-1$ yieldsThe assumed positive exponents require $0<\sigma <2$. Under these values, the omitted destruction terms are indeed higher order, as checked in part (d), so the leading balance is not circular.

Let the mean streamwise velocity be $U(y)$ and set $S=dU/dy>0$. The relevant covariance is $\overline{u^{\prime }{v}^{\prime }}<0$ on this half of the channel. The work of the Reynolds stress against the mean shear rate givesThe eddy viscosity relation is $-\overline{u^{\prime }{v}^{\prime }}={\nu }_{T}S$, soThese are the kinematic-stress conventions, with the covariance measured per unit mass; inserting a stress measured in force per area would require division by mass density.

Equating the two expressions for eddy viscosity gives directlyFrom the preceding part, $u_{\ast }^2=\sqrt{C_{\mu }}k$. Replace one factor of $k$ using that identity, and then replace $u_{\ast }$ itself by $C_{\mu }^{1/4}{k}^{1/2}$. This proves all three equivalent dissipation expressions:Each is the same turbulent kinetic energy dissipation rate, with dimensions of velocity cubed divided by length. Its equivalent friction-velocity form is $u_{\ast }^3/(\kappa y)$.

The constant Reynolds stress and the logarithmic shear imply the eddy viscosity ${\nu }_T=u_{\ast }\kappa y$. On the other hand the K-epsilon model gives ${\nu }_T=C_{\mu }{k}^2/ϵ$. Insert $ϵ=u_{\ast }^3/(\kappa y)$:Canceling the positive factors yields $C_{\mu }{k}^2=u_{\ast }^4$. Since energy is nonnegative, the log-layer solution of the k-epsilon model hasindependent of $y$ to this leading overlap-layer accuracy. In particular $k_y=0$, and the $k$ diffusion term vanishes, confirming the local production–dissipation balance in the energy equation.

Let $E(k)$ be the one-dimensional turbulent energy spectrum, normalized so that $\frac{1}{2}⟨|{\mathbf{u}}^{\prime }{|}^2⟩={\int }_0^{\infty }E(k)dk$. Large energy-containing eddies receive energy from the forcing. In the inertial range, nonlinear transfer passes this energy towards smaller scales at a mean rate $ϵ$ per unit mass. Dimensional analysis of Kolmogorov 1941 theory givesThe dissipation range begins near the inverse of the Kolmogorov length scale, where kinematic viscosity removes the transferred kinetic energy. The large-scale spectrum depends on the forcing; the rising low-$k$ branch in this sketch is illustrative, rather than a universal power law.

The tube has neither homogeneous nor isotropic large-scale turbulence: gravity selects a vertical direction, its walls constrain the motion, and its density gradient varies with height. The width $W$ bounds the transverse size of the energy-containing eddies. At sufficiently high Reynolds number, much smaller eddies may become approximately locally isotropic and exhibit an inertial range, even though the full tube flow does not satisfy the homogeneous-isotropic assumptions.

To obtain the Reynolds stress, write $u_i=U_i+u_i^{\prime }$ and $p=P+p^{\prime }$, with $⟨u_i^{\prime }⟩=0$. Averaging the incompressible Navier-Stokes equations uses $⟨u_i{u}_j⟩=U_i{U}_j+⟨u_i^{\prime }{u}_j^{\prime }⟩$ and gives the Reynolds-averaged momentum equationHere the final term is the mean buoyancy force, with the hydrostatic reference removed. Thus averaging the quadratic momentum flux introduces the covariance $⟨u_i^{\prime }{u}_j^{\prime }⟩$; multiplied by ${\rho }_0$, it has the dimensions of a stress. Its divergence transfers momentum between the mean flow and the fluctuations.

An eddy viscosity model represents this Reynolds stress byThe isotropic term can be absorbed into the mean pressure, while the deviatoric term gives a diffusion of mean momentum with turbulent viscosity ${\nu }_T\sim u_{\ast }\ell$, where $u_{\ast }$ and $\ell$ are representative turbulent speed and mixing length. Unlike kinematic viscosity, this coefficient depends on the flow. The closure assumes a local relation between stress and mean strain; anisotropic, buoyancy-driven large eddies may require a more detailed Reynolds stress model.

The eddy viscosity in the prescribed mixing length model isThus at a fixed similarity coordinate $\eta$, every nonzero value grows as $x^{1/2}$. Its maximum occurs at a fixed coordinate maximizing ${\eta }^2|f^{\prime \prime }|$, so the maximum itself has the same downstream scaling.

The formal argument intended by the edge assumption is also clear. If one independently stipulates $|f^{\prime \prime }(\eta )|\le |f^{\prime \prime }({\eta }_w)|$ for $|\eta |\le {\eta }_w$, then ${\eta }^2|f^{\prime \prime }(\eta )|\le {\eta }_w^2|f^{\prime \prime }({\eta }_w)|$. This would put the maximum eddy viscosity at the edge and giveindependent of $x$, provided $f^{\prime \prime }({\eta }_w)\ne 0$. That is a conditional scaling calculation, not an admissible solution of the full momentum problem.

The printed edge claim fails for the reason established in part (e): $f^{\prime }({\eta }_w)=0$ forces $f^{\prime \prime }({\eta }_w)=0$, and thereforeA nontrivial profile has positive eddy viscosity in the interior; hence its maximum cannot occur at that finite free edge. Moreover $U(x,0)y_w/{\nu }_T(x,\pm y_w)$ has a zero denominator, so it is not a finite effective Reynolds number under the literal assumptions.

The intended downstream cancellation is valid if one uses a consistent interior reference point ${\eta }_{\ast }\in (0,{\eta }_w)$ with $f^{\prime \prime }({\eta }_{\ast })\ne 0$, for example the maximum of ${\nu }_T$. Since $U(x,0)=M_0^{1/2}{x}^{-1/2}{f}^{\prime }(0)$ and $y_w={\eta }_{w}x$,which is independent of $x$. This is a qualified correction of the reference point, not a proof of the false edge maximum.