Topics in fluid mechanics course of the University of Oxford (C5.7)

Original solutions to past examinations of the topics in fluid mechanics course of the University of Oxford.

Solutions to all three questions of the topics in fluid mechanics course of the University of Oxford examination, sat on 9 June 2023. Source: original released PDF, inclusive bundle pages 329–334; paper pages 1–6. All 24 lettered or Roman-numbered leaves across the three questions are covered. The original PDF labels the final part of Question 1 as (c), which is the hierarchy used here.

Take $z$ positive away from the wall into the liquid, and write $p_a$ for atmospheric pressure. For a constant-mass density Newtonian fluid, the two-dimensional Navier-Stokes equation and continuity equation areThe stationary wall imposes the no-slip boundary condition and no-penetration boundary condition, $u=w=0$ at $z=0$. For a flat free surface with negligible air shear stress, the stress boundary condition is $u_z=0$, $p=p_a$ at $z=h_0$; the kinematic boundary condition is $w=0$ there.

For steady flow independent of $x$, these equations give $w=0$, $\mu u_{zz}=-\rho g\sin \alpha$ and $p_z=-\rho g\cos \alpha$. Two integrations and the boundary conditions give the uniform gravity-driven Newtonian film:Its surface velocity is $U_0=\rho g{h}_0^2\sin \alpha /(2\mu )$, while its mean velocity is $2U_0/3$.

Define the aspect ratio and the thickness-based Reynolds number byUse the nondimensionalizationThe small transverse velocity scale follows from incompressibility. Dropping stars, the base fields become $\bar{u}=2z-z^2$ and $\bar{p}=(1-z)\cot \alpha$. Before subtracting the base state, the dimensionless Navier-Stokes equation readsFor example, division of the downslope momentum equation by $\mu U_0/h_0^2$ makes the inertial coefficient $\rho U_0{h}_0^2/(\mu L)=\mathrm{Re}\epsilon$ and the pressure gradient coefficient $2\epsilon$. The transverse equation has been divided instead by $\mu U_0/h_0^2$ after using its smaller velocity scale; this gives the displayed $\mathrm{Re}{\epsilon }^2$ coefficient.

Now substitute $u=\bar{u}+\hat{u}$, $w=\hat{w}$, $p=\bar{p}+\hat{p}$. Since ${\bar{u}}_{zz}+2=0$, ${\bar{p}}_z=-\cot \alpha$, and the base fields have no $x$ or $t$ dependence, subtraction givesThese are nonlinear equations for the deviations; small amplitude has not been assumed. The later asymptotic expansion takes $\mathrm{Re}=O(1)$ as $\epsilon \to 0$, or more generally requires the retained inertial correction to remain small.

The two wall boundary conditions areDefine the capillary parameter for a falling filmRetain $S=O(1)$ in the long-wave approximation. Thus surface tension is retained in the leading normal stress balance even though geometric slope corrections are small. For $h_x=O(1)$ in the rescaled coordinates, the dimensional surface slope is $\epsilon h_x$.

The three free surface boundary conditions, accurate through $O(\epsilon )$, areThe first is the kinematic boundary condition; the second is the zero tangential traction condition. For the third, the dimensional liquid-side pressure is, to the required order,Division by $\rho g{h}_0\sin \alpha$ gives $p=-3S{h}_{xx}/2+\epsilon w_z$ in dimensionless variables. Subtracting $\bar{p}(h)=(1-h)\cot \alpha$ gives the stated perturbation condition. The additional slope-times-shear contribution to normal stress vanishes at this order because the tangential traction is zero. The omitted geometric and tangential-stress corrections are $O({\epsilon }^2)$ under this ordering.

Use an asymptotic expansion at fixed $z$, evaluating the boundary conditions at the instantaneous $z=h(x,t)$. At leading order, the downslope equation gives ${\hat{u}}_{0,zz}=0$. The wall condition and the surface condition ${\bar{u}}_z(h)+{\hat{u}}_{0,z}(h)=0$ implyHere $u^{(0)}$ denotes the total leading velocity, to distinguish it from the dimensional scale $U_0$. The continuity equation and wall impermeability then giveThe transverse momentum equation gives ${\hat{p}}_{0,z}=0$. The normal stress condition therefore yieldsThe leading volume flux per unit width is $q^{(0)}={\int }_0^h{u}^{(0)}dz=2h^3/3$. Consequently mass conservation gives $h_t=-2h^2{h}_x+O(\epsilon )$. This relation eliminates $h_t$ inside the first inertial correction:Set $G=2h_x\cot \alpha -3S{h}_{xxx}$. The $O(\epsilon )$ downslope equation and boundary conditions areTwo integrations giveThe pressure correction ${\hat{p}}_1$ is not needed for this calculation because its downslope pressure gradient enters only at $O({\epsilon }^2)$.

Integrating the total velocity across the film givesIndeed the three inertial polynomial terms integrate to $\mathrm{Re}{h}^6{h}_x(1/30-1/6+2/3)=8\mathrm{Re}{h}^6{h}_x/15$. Integrating the continuity equation and using the kinematic boundary condition gives $h_t+q_x=0$. Hence the Benney equation isNormal gravity smooths height variations on the upper side of the slope, while the inertial correction can amplify them. Surface tension suppresses sufficiently short wavelengths.

For an inverted falling film, keep $z$ directed from the solid into the liquid. The downslope component of gravity is unchanged, but its normal component is now positive. Thus the hydrostatic pressure changes from $(h-z)\cot \alpha$ to $-(h-z)\cot \alpha$. The resulting Benney equation isOnly the normal-gravity contribution changes sign; the surface tension and inertial contributions retain their signs.

Define $A=2\cot \alpha /3+8\mathrm{Re}/15$. The linearization about $h=1$ givesThe Galilean transformation $\xi =x-2t$, $\tau =t$ has ${\partial }_t={\partial }_{\tau }-2{\partial }_{\xi }$ and ${\partial }_x={\partial }_{\xi }$. It removes the leading advection, leavingFor a normal mode proportional to $e^{\sigma \tau +ik\xi }$, the second and fourth derivatives contribute $-k^2$ and $k^4$. The dispersion relation is thereforeFor $A>0$ and $S>0$, the unstable wavenumbers satisfy $0<|k|<\sqrt{A/S}$. In the laboratory coordinates, the temporal exponent is $\sigma -2ik$, so this linear stability calculation predicts growth together with downslope transport at dimensionless speed $2$.

For $A=2\cot \alpha /3+8\mathrm{Re}/15>0$, maximize the growth rate $\sigma (k)=\epsilon (A{k}^2-S{k}^4)$ over real wavenumbers. Its derivative is $2\epsilon k(A-2S{k}^2)$, soThe second derivative at either nonzero maximizer is $-4\epsilon A<0$. If $A\le 0$, there is no positive growth rate in this model.

To compare growth with transport, restore dimensions. With $\epsilon =h_0/L$ and $S=2\gamma /(3\rho g{L}^2\sin \alpha )$, the maximal dimensional growth rate isThe corresponding dimensional wavenumber is $k^{\ast }/L=\sqrt{3\rho g\sin \alpha A/(4\gamma )}$, independent of the arbitrary reference length $L$. Let $d$ be the physical slope length. A disturbance convected at the leading kinematic wave speed $2U_0$ has residence time $d/(2U_0)$ and accumulates the growth exponentThus $d\ll \ell$ implies $N\ll 1$, and the amplitude multiplier $e^N$ remains close to one while the disturbance passes down the slope. This is the requested convective amplification length criterion. With the exact coefficients used here, one e-fold of convected growth occurs at $d=2\ell /3$; the stated much-less-than criterion is unaffected by this order-one factor. It is a transport-versus-growth estimate within the long-wave approximation, rather than a boundary-value stability theorem for every possible inlet and outlet condition.

Take $z$ vertically upward and the porous skeleton stationary. Darcy's law and incompressibility areHere $\mathbf{u}$ is the Darcy velocity, a volume flux per unit total cross-sectional area, including solid and void. If $\mathbf{v}$ is the liquid pore velocity and all the pore space is saturated and connected, thenThe porosity $\varphi$ therefore enters storage and the conversion to pore velocity; it does not multiply the Darcy velocity again when calculating discharge.

Let $H\ll L$ be the characteristic saturated thickness and let $\delta =H/L$. Scale pressure by $\rho gH$. The horizontal part of Darcy's law gives the scaleBy incompressibility, $w\sim \delta u\sim K{\delta }^2$. The vertical part of Darcy's law isIts left side is smaller than the scale $K$ by ${\delta }^2$. Therefore $p_z+\rho g=O(\rho g{\delta }^2)$, and the leading pressure is hydrostatic pressure:Here the water table has pressure $p_a$. This is the Dupuit approximation, valid away from the small region near an outlet where the surface slope need not remain small.

The hydrostatic pressure gives $p_x=\rho g{h}_x$, independent of depth. Thus Darcy's law gives $u=-K{h}_x$ and the horizontal volume flux per unit width isThe coefficient $K$ is the hydraulic conductivity, with units of velocity. For a column of unit horizontal area, the stored water volume is $\varphi h$, and rainfall supplies volume flux $R$. The local mass conservation equation is $\varphi h_t+q_x=R$. Therefore the Boussinesq equation for an unconfined aquifer isThis assumes rainfall is entirely converted into recharge and neglects unsaturated storage and losses before the water reaches the water table.

Set $x=L\tilde{x}$, $h=H\tilde{h}$, and $t=T\tilde{t}$. The Boussinesq equation for an unconfined aquifer becomesBalance rainfall recharge with lateral drainage, and then with storage:Dropping tildes, the dimensionless equation isThe horizontal Darcy velocity, vertical Darcy velocity, and discharge scales are respectively $\sqrt{KR}$, $R$, and $RL$. Thus the vertical thickness scale is set by rainfall balanced against hydraulic conductivity over the finite drainage length $L$, rather than by porosity. The latter determines the storage time $T$. The Dupuit approximation requires $H/L=\sqrt{R/K}\ll 1$ for these scales. The initially dry initial condition is $h(x,0)=0$.

Use the dimensionless equation from part (a). For the similarity ansatz $h=t^{a}f(\eta )$, $\eta =x/t^b$, its three terms have time factorsA nontrivial similarity solution retaining storage, lateral drainage and nonzero rainfall requires the exponents to agree:Unlike a source-free diffusion equation, the nonzero prescribed rainfall fixes an amplitude-versus-time scale directly, so no additional conserved quantity or boundary condition is required to determine these exponents. Degenerate spatially uniform solutions are an exception to this all-terms balance. The rain-fed aquifer similarity solution satisfiesIf the dimensional equation is used directly without scaling out coefficients, its similarity profile instead obeys $\varphi (f-\eta f^{\prime })=K(f{f}^{\prime }{)}^{\prime }+R$. The normalization above is the one in which the subsequent condition $f(\infty )=1$ applies.

The condition $f(0)=0$ means the river maintains the water table at the impermeable base level: $h(0,t)=0$. The liquid is exposed to atmospheric pressure there. It does not imply zero drainage, because $h_x$ may be singular at the outlet.

Far from the river, lateral drainage has not significantly affected storage. Rainfall simply fills pore volume, so the dimensional height is $h\sim Rt/\varphi$. In the dimensionless variables this is $h\sim t$, giving $f(\infty )=1$.

Although the physical aquifer is finite, these boundary conditions describe its early similarity solution before the drainage-affected region reaches $x=L$. In dimensional variables its horizontal extent is of order $\sqrt{KR}t/\varphi$, so the semi-infinite approximation requires $t\ll T=\varphi L/\sqrt{KR}$.

Introduce the positive outlet constantFor a draining similarity profile, $J$ is finite and positive. A regular linear profile would have $f{f}^{\prime }\to 0$ and $(f{f}^{\prime }{)}^{\prime }\ge 0$ at the origin, incompatible with the rain-fed aquifer similarity solution $(f{f}^{\prime }{)}^{\prime }(0)=-1$. Instead there is a square-root drainage boundary. At leading order, $(f^2{)}^{\prime }=2f{f}^{\prime }\to 2J$, givingMore explicitly, the rain-fed aquifer similarity solution implies $(f{f}^{\prime }{)}^{\prime }=-1+O(\sqrt{\eta })$, hence $f{f}^{\prime }=J-\eta +O({\eta }^{3/2})$ and $f^2=2J\eta -{\eta }^2+O({\eta }^{5/2})$. The height vanishes but its derivative diverges; their product remains finite.

Use $t_d,x_d,h_d$ temporarily for dimensional quantities. The nondimensionalization in part (a) givesTaking the drainage flux to be positive out of the left-hand end, $Q_0=-q(0,t_d)$, givesThe printed expression in terms of $f(0)$ and $f^{\prime }(0)$ must be understood as this limit; a literal product of their separate endpoint values is undefined. This is discharge per unit transverse width. Very close to the river, the divergent slope marks the limit of the Dupuit approximation, while the outer solution still supplies the leading discharge.

Return to dimensional variables. The river at base level imposes the drained boundary condition, while the right-hand end is impermeable:As long as the right-hand end remains wet, the latter is $h_x(L,t)=0$. The initial condition for the drought is the height profile inherited at $t=t^{\ast }$; stopping rainfall changes the forcing but does not instantaneously remove stored water.

With $R=0$, substitute $h=X(x)T(t)$ into the dimensional Boussinesq equation for an unconfined aquifer:For a positive draining profile, take $\lambda >0$. Normalize $T(t^{\ast })=1$, so $X$ has the dimensions of height. The temporal ordinary differential equation integrates toThe spatial ordinary differential equation and boundary conditions areHere $\lambda$ has units of inverse length with this normalization. The scale freedom between $X$ and $T$ is fixed by the chosen value $T(t^{\ast })=1$; the amplitude of $X$, and hence $\lambda$, must be matched to the available water.

This separable aquifer drawdown is exact if the inherited height profile has the separated spatial shape. For a general rainfall-generated initial condition, it describes the late drought regime; it cannot reproduce an arbitrary profile at the instant rainfall stops merely by choosing one temporal constant.

Let $J_X={\lim }_{x\downarrow 0}X(x)X^{\prime }(x)>0$. The spatial ordinary differential equation gives $(X{X}^{\prime }{)}^{\prime }=-\lambda X$, so its leading outlet behavior isAgain this is a square-root drainage boundary. For the separable aquifer drawdown, the dimensional drainage flux isEquivalently, with ${\tau }_d=\varphi /(K\lambda )$, it is $Q_0(t^{\ast })[1+(t-t^{\ast })/{\tau }_d{]}^{-2}$ for an exact separated profile. Integrating the spatial ordinary differential equation gives $J_X=\lambda {\int }_0^{L}Xdx$. This verifies the global mass conservation checkThus the late drainage flux follows an inverse-square power law, while stored water decays inversely with time.

The aquifer drainage hydrograph starts at zero, rises linearly in the early rainfall similarity solution, and approaches the total recharge $RL$ if rainfall lasts long enough for a finite-length steady state. Indeed, steady mass conservation gives $q_x=R$ and $q(L)=0$, hence $Q_0=RL$. At $t^{\ast }$, the outlet flux stays continuous for a regular inherited flux profile, then recedes as water drains; the eventual decay is proportional to $(t-t^{\ast }+\text{constant}{)}^{-2}$. An arbitrary transition requires the full time-dependent partial differential equation, so matching an early rainfall formula directly to a separated drought formula is only a schematic approximation.

The curve illustrates these regimes and a possible $t^{\ast }$. Its numerical shape and the relative duration of the regimes are schematic; the early rainfall and late drought formulas apply in their respective asymptotic ranges.

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.