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An eddy viscosity parametrizes turbulent momentum transport by $\overline{u^{\prime }{v}^{\prime }}=-{\nu }_T{U}_y$. The mixing length model gives ${\nu }_T=l^2|U_y|\ge 0$. This coefficient depends on the mean flow and position; it is distinct from molecular kinematic viscosity. It describes momentum transport rather than molecular energy dissipation.

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 mixing length is the characteristic transverse displacement over which a turbulent parcel retains its streamwise momentum before mixing with its surroundings. Taylor expansion of a mean velocity profile gives a fluctuation of magnitude $l|U_y|$. Combining comparable transverse fluctuations with down-gradient Reynolds stress transport gives the closure $\overline{u^{\prime }{v}^{\prime }}=-l^2|U_y|U_y$. This is a phenomenological approximation, not a consequence of the Navier-Stokes equation. In a free jet, a prescribed local mixing length must also be compatible with centreline regularity and vanishing outer momentum flux.

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.

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.

Write $q={\partial }_z\bar{\rho }>0$ for the unstable, horizontally averaged density gradient. A turbulent displacement over a mixing length of order $W$ encounters a density contrast $\Delta \rho \sim Wq$. Balancing its buoyancy acceleration with the inertial acceleration givesA turbulent diffusivity for density is therefore $D_T=K{u}_{\ast }W=K{W}^2\sqrt{gq/\bar{\rho }}$, absorbing dimensionless factors into $K>0$. The upward density flux is $J_{\rho }=⟨w^{\prime }{\rho }^{\prime }⟩=-D_{T}q$: turbulent exchange carries denser water downwards and lighter water upwards. Defining $F_{\rho }=-J_{\rho }$ as the positive downward flux yields the buoyancy-driven turbulent density diffusion lawThis is a flux per unit horizontal area, with units of mass per area per time. The total density transport across a horizontal section is $W^2{F}_{\rho }$.

The assumptions are established high-Reynolds number turbulence, width-limited eddies, an instantaneous local balance between buoyancy and inertia, approximately constant $K$, negligible molecular diffusion, and zero mean vertical volume flux. The last condition follows from incompressibility and impermeable sidewalls in a closed tube. Density variations and the eddy statistics must also vary slowly enough for a local mixing-length closure to be meaningful.

Horizontal averaging of mass conservation, with upward $z$, gives ${\bar{\rho }}_t=-{\partial }_z{J}_{\rho }={\partial }_z{F}_{\rho }$. Differentiating once more supplies the requested evolution of the density gradient:Under the Boussinesq approximation, replace $\bar{\rho }$ in the inertial prefactor by the constant reference density ${\rho }_0$; retain the changing density gradient inside the flux.

Let $L$ be the streamwise evolution scale, $\delta$ the jet width, and $U_c$ its velocity scale. A slender jet has $\epsilon =\delta /L\ll 1$; mean incompressibility gives $V=O(\epsilon U_c)$. Both mean advection terms are then $O(U_c^2/L)$. For turbulent parcels crossing a transverse distance $l$, retention of their original streamwise momentum gives a fluctuation $\hat{u}\sim -l{U}_y$; a comparable transverse fluctuating speed gives the down-gradient mixing length closureThe mixing length is an effective distance of momentum retention, with statistical coefficients absorbed into $l$; it is not the net transverse displacement of a mean streamline. This closure models the Reynolds stress rather than deriving it exactly.

In a slender equilibrium jet, the fluctuation magnitude inferred from momentum balance is $u^{\prime 2}\sim U_c^2\delta /L=\epsilon U_c^2$. The transverse Reynolds stress gradient then has size $u^{\prime 2}/\delta \sim U_c^2/L$, whereas the streamwise normal-stress gradient is $u^{\prime 2}/L\sim \epsilon U_c^2/L$. At sufficiently high Reynolds number, molecular transverse diffusion is smaller providedStreamwise molecular diffusion is smaller still by ${\epsilon }^2$. A free jet has ambient mean pressure constant to leading order; transverse momentum corrections imply only a higher-order mean streamwise pressure force. Thus high speed must be accompanied by the slender-jet ordering, not used alone to discard every other term.

For a symmetric jet decreasing away from the centre, $U_y<0$ on $y>0$ and $U_y>0$ on $y<0$. Accordingly $\overline{\hat{u}\hat{v}}=\mathrm{sgn}(y)l^2(U_y{)}^2$ off the centreline, and the leading equation isAt the centreline the signed form requires the stress to tend continuously to zero; otherwise differentiation would add a distributional term. The undifferentiated Reynolds stress form is the appropriate global equation.

The proposed mixing length can be motivated as a local phenomenological estimate: on either half of a self-similar jet, the centreline supplies a distinguished reference plane, and an eddy at distance $|y|$ might retain momentum over a fixed fraction $C_1|y|$ of that transverse distance. This givesIt is compatible with the dilation $x\mapsto ax$, $y\mapsto ay$ and introduces no independent outer length. In the turbulent plane jet similarity regime its typical value at $|y|\sim \delta$ is $C_1\delta$, so the scaling from part (b) has $\delta /L\sim C_1^2$.

This argument is a modeling assumption, not a physical necessity: a jet centreline is not a wall that prevents turbulent eddies from crossing it. Thus a mixing length vanishing there cannot automatically be justified throughout the core. Parts (e) and (f) exhibit the precise difficulty: the resulting finite-velocity similarity profile has singular centreline shear and zero shear at a finite free edge, contradicting the printed edge definition. The local dimensional motivation does not make all the stipulated global jet properties consistent.

First derive the formal turbulent plane jet similarity equations, then check the printed edge conditions. Define a streamfunction by $U={\psi }_y$, $V=-{\psi }_x$, so mean incompressibility is automatic. Momentum conservation gives $U_c^2\delta \sim M_0$. With $l=C_1|y|$, the closure balance $U_c^2/x\sim C_1^2{U}_c^2/\delta$ gives $\delta \propto x$. Hence $U_c\propto M_0^{1/2}{x}^{-1/2}$ and the consistent self-similar ansatz isAn odd $f$ gives symmetric $U$ and antisymmetric $V$. Direct differentiation givesThe advection and Reynolds stress terms becomeThus the two required consistency equations, the first for $\eta >0$ and the second for a symmetric finite-width profile, areThe second is momentum normalization; it is the equation referenced as (2) in the PDF, although its display in part (d) has no printed number. Use an infinite upper limit for a decaying profile of infinite support. Symmetry sets $f(0)=0$. A finite centreline velocity and continuous zero centreline stress give the integrated first equationThese equations establish the similarity reduction. They also give a direct finite-edge stress condition for a mixing-length jet. At a finite edge ${\eta }_w>0$, $U=0$ requires $f^{\prime }({\eta }_w)=0$. For finite $f({\eta }_w)$, the integrated equation forces $f^{\prime \prime }({\eta }_w)=0$. A nontrivial decreasing velocity profile has nonzero shear somewhere inside, so zero edge shear cannot be its maximum. Allowing a one-sided nonzero edge shear instead produces nonzero edge stress and an unbalanced jump in the conservative momentum equation.

For a concrete diagnostic, take $f^{\prime }=1-{\eta }^2$ on $0\le \eta \le 1$, so $f=\eta -{\eta }^3/3$. It has zero edge velocity and its largest one-sided shear magnitude at the edge, matching that descriptive condition. Yet at $\eta =1$ the integrated equation has left-hand side zero and right-hand side $8C_1^2>0$: such a profile is not a solution of the stated momentum closure.

The centreline is also problematic for a smooth interpretation. If $f^{\prime }(0)=a>0$, then $f\sim a\eta$, and the decreasing branch of the integrated equation givesA continuous formal profile can have this cusp, but it is not a differentiable symmetric velocity field, and molecular diffusion cannot be neglected uniformly in that core. Therefore there is no nontrivial regular similarity solution satisfying all the printed edge requirements. The boxed consistency equations are valid; the additional coincidence of a zero-velocity edge with maximum shear is not. A consistent jet model must relax that edge definition and regularize the core or use a different mixing length.

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.

For a slender plane jet with conserved specific momentum flux $M_0=\int U^{2}dy$ and a mixing length proportional to transverse position, dimensional balance gives width proportional to $x$ and centreline speed proportional to $\sqrt{M_0/x}$. The streamfunction $\psi =\sqrt{M_{0}x}f(\eta )$, $\eta =y/x$, yields $U=\sqrt{M_0/x}{f}^{\prime }$ and $V=\sqrt{M_0/x}(\eta f^{\prime }-f/2)$. With $l=C_1|y|$, the upper-half Reynolds-averaged momentum equation integrates to $f{f}^{\prime }=2C_1^2{\eta }^2(f^{\prime \prime }{)}^2$ for a finite centreline velocity and zero centreline stress. The momentum normalization is $2{\int }_0^{{\eta }_w}(f^{\prime }{)}^{2}d\eta =1$ for a compact jet, or the same integral to infinity for a decaying jet. These formal equations do not justify an independently imposed edge condition.