Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 65 / 3

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.

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.

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.

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.