Past exam of the topics in fluid mechanics course of the University of Oxford / 2023 / 3 (Turbulent mixing layer)

For a constant-mass density incompressible flow with no imposed body force, the two-dimensional Navier-Stokes equation isA conservative body force, if retained, may be incorporated into a modified pressure. Using scales $L,U_0,L/U_0$ and $\rho U_0^2$ givesHere $\nu =\mu /\rho$ is the molecular kinematic viscosity. At high Reynolds number, inertia and the pressure gradient dominate on the outer length scale, while molecular viscous diffusion is small there. Thin boundary layers, narrow shear flows and fine turbulent scales can still have sufficiently large velocity gradients for viscous effects to matter. High Reynolds number permits instability and turbulence, but does not alone prove that every flow is turbulent. Molecular viscous dissipation remains essential at the small end of the turbulent cascade.

Write $u_i=U_i+u_i^{\prime }$ and $p=P+p^{\prime }$, with $\overline{u_i^{\prime }}=\overline{p^{\prime }}=0$. Assume a statistically stationary mean and sufficient regularity for Reynolds averaging to commute with spatial derivatives. Thenprovided the endpoint difference grows sublinearly. Averaging the continuity equation givesIn Einstein notation, the nonlinear term expands asThus the Reynolds-averaged Navier-Stokes equation isThe extra term survives because the average of a product of zero-mean fluctuations need not vanish; it represents turbulent transport of mean linear momentum.

The product rule and fluctuation incompressibility give the pointwise identityAveraging therefore givesThe fluctuation term is consequently the divergence of a Reynolds stress. The eddy viscosity model represents this momentum transport through a coefficient multiplying the mean strain-rate tensor:for its shear or deviatoric part. For example, $-\rho \overline{u^{\prime }{v}^{\prime }}\simeq {\mu }_T{U}_y$ in a thin parallel shear flow. The dynamic viscosity ${\mu }_T$ is an effective flow-dependent transport coefficient, rather than a molecular material property. Large eddies carry linear momentum between regions of different mean velocity.

There is a necessary qualification to the printed tensor equation (4). Taking its trace literally gives $0=-\rho \overline{u_i^{\prime }{u}_i^{\prime }}$, which would exclude nonzero turbulence. A consistent deviatoric eddy-viscosity closure in $d$ fluctuating dimensions isThe isotropic term is absorbed into the effective pressure $P_{\mathrm{eff}}=P+2\rho e_t/d$. Here $d=2$ for a strictly two-dimensional fluctuation model. The subsequent equations use $P$ for this effective pressure, or equivalently use (4) as a shear-stress closure. This preserves the intended thin-layer model without asserting an impossible full covariance identity.

For turbulent motions of scale $l$ and velocity scale $u_l$, the estimate ${\mu }_T\sim \rho u_{l}l$ gives ${\mu }_T/\mu \sim u_{l}l/\nu$. It is much greater than one on energetic large scales and in the inertial range well above the Kolmogorov length scale. At the molecular dissipation scales it is no longer large, and molecular transport also dominates very near a smooth solid wall.

Let $\delta \ll L$ be a representative mixing-layer thickness, ${ϵ}_m=\delta /L$, and let $U_s$ be a representative streamwise velocity. Use the scalesThe transverse velocity scale follows from the mean continuity equation. For turbulent transverse transport to balance streamwise inertia,With this ordering, the streamwise eddy-stress divergence isThus transverse diffusion of streamwise linear momentum is retained, while streamwise diffusion and the $V_x$ contribution are smaller by ${ϵ}_m^2$. Molecular kinematic viscosity is neglected where ${\nu }_T\gg \nu$.

In the transverse momentum equation, inertia and the retained eddy stresses have scale ${ϵ}_m{U}_s^2/L$, whereas an order-one transverse pressure variation would give $P_y/\rho \sim U_s^2/({ϵ}_{m}L)$. Hence the leading dimensionless transverse balance is $P_y=0$. The resulting boundary layer equations for the planar turbulent mixing layer areHere ${\nu }_T={\mu }_T/\rho$. The effective pressure convention explained in part (a)(iii) is understood. Since the outer streams have uniform horizontal velocities, their inviscid momentum equations give $P_x=0$; choosing the far-field gauge pressure as zero gives $P=0$ throughout this approximation.

Assume $U_c=(U_a+U_b)/2>0$ and set $\Delta U=U_a-U_b>0$. With ${\nu }_T=C\Delta U\delta (x)$ independent of $y$, balancing streamwise transport with transverse mixing gives a layer width proportional to $x$. Write its constant slope asHere $S>0$ is a spreading constant associated with the chosen definition of thickness; it is unrelated to the capillary $S$ in Question 1. The origin $x=0$ selects zero virtual-origin offset.

Define $a=\sqrt{S/C}$, $\xi =ay/\delta$, and introduce the stream functionThis automatically satisfies incompressibility and yieldsFor the momentum equation, use ${\xi }_x=-\xi {\delta }^{\prime }/\delta$ and ${\xi }_y=a/\delta$:Because $U_c^2{\delta }^{\prime }=S\Delta U{U}_c$, the common coefficient cancels, giving the turbulent mixing-layer similarity equationThe boundary layer approximation also requires $S\Delta U/U_c\ll 1$, consistently with the assumed slow spreading.

The two external stream velocities imposeLet $r=\Delta U/U_c\ll 1$ and seek $F=O(r)$ over $\xi =O(1)$. The nonlinear term $F{F}^{\prime \prime }$ is then $O(r^2)$, so the leading ordinary differential equation isFor $G=F^{\prime \prime }$, this is the first-order linear differential equation $G^{\prime }+\xi G=0$. Its integrating factor is $e^{{\xi }^2/2}$, giving $F^{\prime \prime }=M{e}^{-{\xi }^2/2}$. An integration using the error function givesThe two far-field boundary conditions require $E=0$ and $M=r/\sqrt{2\pi }$. A second integration usesConsequentlyThe relation between $A$ and $B$ cancels the extra $\xi e^{-{\xi }^2/2}$ term when differentiating. It gives the familiar error function mean velocity profileto first order in $r$. The additive constant $D=O(r)$ does not affect this velocity profile at the retained order; its interpretation comes from the transverse velocity.

For the approximate similarity profile, $F^{\prime }=A\mathrm{erf}(\xi /\sqrt{2})$, and thereforeFor the prescribed undisturbed horizontal streams, a suitable choice is $D=0$, which gives zero transverse velocity at both far-field limits. Then $V=-\Delta U\sqrt{CS}B{e}^{-{\xi }^2/2}$ inside the layer. The nonzero internal crossflow is consistent with the streamwise widening of the velocity transition.

Another possible convention is to make the line $y=0$ a streamline, imposing $V(0)=0$. This requires $D=-B$, equivalently $F(0)=0$, and gives $V(\pm \infty )=\Delta U\sqrt{CS}B$. It therefore requires a uniform outer crossflow relative to the coordinates used for the centred error function profile. A single constant cannot impose both zero far-field crossflow and $V(0)=0$ unless $\Delta U=0$. The physical lateral boundary conditions, or the chosen material centreline, determine which convention is appropriate. Since $B,D=O(r)$, these transverse velocities are of order $\Delta U^2/U_c$ for fixed $C,S$.