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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.

Let the Eulerian time average commute with space and time derivatives and let the statistically steady mean have no time derivative. Define $U=\overline{u}$, $V=\overline{v}$, $\hat{u}=u-U$ and $\hat{v}=v-V$, so $\overline{\hat{u}}=\overline{\hat{v}}=0$. Averaging instantaneous incompressibility givesand subtracting gives ${\hat{u}}_x+{\hat{v}}_y=0$ for the fluctuations. Start from the instantaneous streamwise Navier-Stokes equationUse incompressibility to write $u{u}_x+v{u}_y={\partial }_x(u^2)+{\partial }_y(uv)$. The averaged products are $\overline{u^2}=U^2+\overline{{\hat{u}}^2}$ and $\overline{uv}=UV+\overline{\hat{u}\hat{v}}$. The mean conservative terms reduce back to $U{U}_x+V{U}_y$ because $U_x+V_y=0$. The Reynolds-averaged momentum equation is consequentlyThe signs use the positive covariance convention for the Reynolds stress. In particular, both quadratic averages involve the fluctuations, not products of the mean $U$ with a fluctuation; the converted TeX loses the overbar/hat structure here.

Use mean incompressibility to convert advection into momentum-flux divergence:Therefore the reduced Reynolds-averaged momentum equation givesFor a localized free jet, $UV$ and the Reynolds stress tend to zero in the quiescent ambient as $y\to \pm \infty$. Assume sufficient decay to differentiate its momentum integral. Integrating across the jet givesThus $M_0=\int U^{2}dy$ is constant. It is a specific momentum flux per unit span: multiplication by density gives the momentum flux. A clipped profile with nonzero stress just inside its artificial edge would fail this boundary-flux assumption unless an additional external momentum source were supplied.

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.