Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 83 / 3

Write $\nu =\mu /\rho$ and average the horizontal Navier-Stokes equation, with zero-mean fluctuations and incompressible flow. The mean flow is stationary, has no vertical velocity and has no horizontal pressure gradient. The mean advection of $U(z)$ is therefore zero. Incompressibility rewrites the averaged nonlinear fluctuation term as a divergence:These correlations transport mean momentum and produce Reynolds stress. Horizontal statistical homogeneity makes the horizontal derivatives of the averaged correlations vanish. It does not require the instantaneous turbulent derivatives themselves to vanish. Only the vertical divergence remains, givingMultiplying by density and integrating once gives constant-stress Reynolds-averaged wall flow:Here ${\tau }_d$ is physical shear stress. Positive mean shear commonly has $\overline{u^{\prime }{w}^{\prime }}<0$, representing downward transport of positive horizontal momentum. The Reynolds term then contributes positively to the total stress.

The printed target formula has an extra vertical derivative in the Reynolds contribution. That term has dimensions of stress per length, while $\mu U_z$ has dimensions of stress, so the formula as printed is inconsistent. The derivative belongs in the stress-divergence equation above; the boxed integrated stress is the corrected result. Part b also uses a kinematic rather than physical stress, so the density factor must be tracked explicitly.

Define the friction velocity $u_{\ast }=\sqrt{{\tau }_d/\rho }$ for positive wall shear. In the near-wall region turbulence is suppressed and molecular viscosity carries the stress. With no slip, $U\simeq u_{\ast }^{2}z/\nu$. The viscous wall coordinate $z^{+}=z{u}_{\ast }/\nu$ becomes order one at the characteristic viscous sublayer lengthThe numerical extent of viscosity dominance depends on the crossover convention and wall condition; the scaling is the required viscous length.

An eddy at height $z$ cannot extend arbitrarily far through the ground, so its largest local vertical scale is proportional to $z$. Choose $\ell \sim z$ and absorb its constant of proportionality into the coefficient. With the prescribed constant turbulent velocity scale, writeThe mixing-length closure is $-\overline{u^{\prime }{w}^{\prime }}={\nu }_T{U}_z$. Thus the full kinematic stress isFor $z\gg \delta$, neglect the molecular term. With no internal momentum source, $T$ is constant, soIntegrating between two heights givesEquivalently define an extrapolated roughness length $z_0$ and write $U=(T/C)\log (z/z_0)$. If $C=\kappa u_{\ast }$, the familiar law of the wall is $U/u_{\ast }=(1/\kappa )\log (z/z_0)$. The logarithmic profile is valid outside the inner viscous region and below scales where the constant-stress or wall-distance assumptions fail. It cannot be continued to $z=0$ to impose the smooth-wall no-slip condition.

The part b notation $\tau ={\nu }_T{U}_z$ must therefore mean $\tau =T$, a stress divided by density. Using the physical stress from part a on that right-hand side would omit a factor of $\rho$.

The turbulent Schmidt number compares turbulent momentum and concentration diffusivities:The mean fluid has no vertical velocity. Taking $z$ positive upward and $W_s>0$, the settling flux is $-W_s\varphi$, while the given turbulent particle flux is $-{\kappa }_T{\varphi }_z$. Local particle-volume conservation therefore givesConsequently the evolution equation isFor constant settling speed this is ${\varphi }_t=W_s{\varphi }_z+{\partial }_z({\kappa }_T{\varphi }_z)$. In the logarithmic region, ${\kappa }_T=Cz/2$ from the preceding part. The sign is consistent with downward settling: without diffusion, a profile evolves as $\varphi (z,t)=\varphi (z+W_{s}t,0)$. The dilute suspension approximation neglects feedback on the mean flow and particle interactions; deposition or erosion must be imposed through the bed boundary flux.

Particles settle toward the bed under their submerged weight. A concentration increasing downward gives an upward turbulent diffusive flux, which can oppose that settling. If turbulence cannot sustain the particles, they enter the weakly mixed near-wall region and can deposit. Sediment resuspension requires bed forces strong enough to entrain grains, not merely a specified diffusivity far above the bed.

For noncohesive grains, an appropriate criterion uses the Shields parameterwhere $d$ is grain size, ${\rho }_s$ its density and ${\Theta }_c$ the sediment entrainment threshold expressed dimensionlessly. The threshold depends on grain geometry, contacts and viscous effects; cohesion changes the criterion. Sustained suspension additionally requires sufficiently vigorous upward turbulent motions relative to $W_s$. Bed motion and well-mixed suspension are therefore not identical conditions.

At sedimentation-resuspension equilibrium, the total vertical particle flux is constant. Set ${\kappa }_T=Kz$ with $K=C/2$. The stationary equation givesSince $W_s>0$ and $\varphi \to 0$ at infinity, necessarily $J=0$. Settling and upward diffusion then balance pointwise. Taking a reference height $z_1=1$ in the specified length units gives the equilibrium settling-diffusion profileIt is nonnegative and decays to zero. The profile describes the region $z\gg \delta$; it does not assert that this power law continues through the viscous layer down to the physical bed. The supplied reference concentration summarizes the bed exchange and near-wall physics that the outer diffusion closure does not resolve.