Past exam of the fluids course of the University of Oxford / 2022 (A16726W1)

Source: www.whatdotheyknow.com/request/oxford_physics_past_papers_2/response/2340696/attach/10/Physics%20Papers%202021.pdf?cookie_passthrough=1

Source bundle pages 123–128. The paper cover establishes 2022; the attachment filename uses 2021. Solutions use the symbols of the source, with corrections and approximation limits identified where needed.

Take $x$ along a horizontal gap and $s$ perpendicular to its plates, with $0<s<b$. Assume a Newtonian fluid, stationary plates, a steady flow, a laminar flow and a fully developed flow away from the entrances and side edges. The no-slip boundary condition holds on both plates. Any transverse gravity is absorbed in the hydrostatic pressure; there is no imposed transverse through-flow. Translational symmetry and the no-penetration boundary condition then give $\mathbf{u}=(u(s),0,0)$. The Navier-Stokes equation reduces toTwice integrating gives the plane Poiseuille flow:The volumetric flow rate through one gap isThere are $n-1$ internal gaps. Their total cross-sectional area is $(n-1)wb$, because the plate thickness is negligible, so the Darcy velocity isThus the permeability of a porous medium has the dimensions of squared length. If finite plate thickness were included, the Darcy velocity averaged over the whole manifold would also contain its porosity.

Both the local acceleration and the convective acceleration vanish: ${\partial }_t\mathbf{u}=0$ and $(\mathbf{u}\cdot \nabla )\mathbf{u}=u_x{\partial }_x\mathbf{u}=0$. The dominant balance is therefore between the pressure gradient and viscous stress, with no inertia contribution in this fully developed solution. This exact cancellation does not guarantee that arbitrary large-Reynolds number realizations remain laminar flows; entrances, disturbances and turbulence can invalidate the assumptions.

Use a fixed representative volume $V$ containing many pores. Let $\phi$ be the porosity and let $\mathbf{q}$ be the Darcy velocity, defined per unit total area. The conservation of mass isThe solid matrix is rigid and fixed, and the liquid is fully saturating and has constant mass density, so the stored mass on the left is constant. Microscopic internal liquid-solid boundaries contribute no through-flux by the no-penetration boundary condition. Applying the divergence theorem to the external flux givesSince the representative volume is arbitrary, the macroscopic continuity equation isThere is no extra factor of $\phi$ in the flux: that factor is already incorporated in the definition of the Darcy velocity. A compressible liquid or deformable matrix would require retaining the storage term.

Under spherical symmetry, the continuity equation and Darcy's law giveHence $q_r=C/r^2$, and integrating the pressure gradient from infinity givesThe boundary condition at $r=a$ fixes $C={\Pi }_{1}a\Delta p/\eta$. Therefore spherical Darcy injection hasHere $q_r$ is the superficial Darcy velocity, not the pore velocity $q_r/\phi$.

For the comparison, assume geometrically similar pore networks, equal porosity, the same source radius and far boundary geometry, and equal pump efficiency. Dimensional analysis then gives ${\Pi }_1=C_{\mathrm{geom}}{d}^2$ with the same dimensionless factor. ConsequentlyAt a fixed desired volumetric flow rate, the required excess pressure and pumping power areAquifer A requires half the pumping power at equal throughput, or half the energy to inject the same volume over the same time. This compares the excess pressure-volume work dissipated by the Darcy flow; any different baseline compression or other plant costs need separate accounting.

The imposed comparison matters. At fixed $\Delta p$, A takes twice the volumetric flow rate and twice the instantaneous power, but the same excess energy per injected volume. The pore spacing alone does not determine permeability of a porous medium if network geometry or porosity changes; without the geometric assumptions the preference cannot be inferred from $d$ alone.

For the common constant mass density, the background hydrostatic pressure satisfies $d{p}_0/dz=-\rho g$. Define the excess pressure $P=p-p_0(z)$. In Darcy's law, the background pressure gradient cancels gravity:If the source imposes the same excess pressure relative to the local hydrostatic pressure everywhere on its spherical surface, $P(a)=\Delta p$, the spherical Darcy injection result survives exactly:It is important to specify what is maintained on the source. A literally uniform absolute pressure $p_s=p_0(0)+\Delta p$ on a finite sphere centered at $z=0$ instead gives $P(a,\theta )=\Delta p+\rho ga\cos \theta$, where $\theta$ is measured from the upward vertical. The decaying harmonic function outside the sphere is thenand thereforeThis supplies a concrete exception to an unqualified claim that gravity leaves the radial flow unchanged. The additional dipolar contribution has zero net volumetric flow rate, so $Q=4\pi {\Pi }_{1}a\Delta p/\eta$ still holds. Equal injected and resident mass densities are essential to the background cancellation.

At a fixed point beside a river, the Eulerian coordinate description measures $\mathbf{u}(\mathbf{x},t)$. Its local acceleration ${\partial }_t\mathbf{u}$ measures change at that fixed point. Following a floating parcel gives a Lagrangian trajectory $\mathbf{X}(t)$ satisfying $\dot{\mathbf{X}}=\mathbf{u}(\mathbf{X},t)$. The chain rule gives its physical material acceleration:The extra term is the convective acceleration. For a river that narrows steadily, an observer at one fixed position measures a constant velocity, yet a parcel speeds up as it reaches the narrowing. A parcel following a steady bend also experiences a centripetal acceleration even when its speed is constant. An Eulerian calculation of the physical acceleration must include the same convective acceleration; the two descriptions do not predict different forces.

Choose the stream function convention $u=-{\psi }_y$, $v={\psi }_x$, so that this is an incoming stagnation point flow rather than the reversed flow. The velocity field isAt the plate $v(x,0)=0$, satisfying the no-penetration boundary condition. The nonzero tangential velocity $u(x,0)=Ax$ is permitted for an inviscid flow; the no-slip boundary condition is not imposed on the Euler equations for an inviscid fluid.

The steady convective acceleration is $(A^{2}x,A^{2}y)$. Thus the Euler equations for an inviscid fluid requireIntegrating givesThe printed formula omits $\rho$ from its quadratic term. The source uses physical pressure, not pressure divided by mass density, as is clear from its $\rho gy$ term and its Navier-Stokes equation. The omitted factor is also required by dimensional analysis. For a concrete substitution check, the printed expression gives $-p_x/\rho =A^{2}x/\rho$ rather than the required $A^{2}x$ when $\rho \ne 1$ in the chosen numerical units.

A stationary viscous wall imposes the no-slip boundary condition, $u=v=0$, whereas the outer inviscid flow has $u=Ax$. At large Reynolds number, the transition is confined to a thin viscous boundary layer, where normal velocity gradients amplify the otherwise small viscous term.

Take a horizontal scale $L$, an outer speed $U=AL$, and a boundary layer thickness $\delta \ll L$. The continuity equation gives $v=O(U\delta /L)$. Write $ϵ=\delta /L$ and ${\mathrm{Re}}_L=UL/\nu$. The tangential Navier-Stokes equation has scalesBalancing convective acceleration with normal viscous diffusion gives $ϵ=O({\mathrm{Re}}_L^{-1/2})$. The leading tangential equation and incompressible flow constraint areHere and below $\nu$ is the kinematic viscosity, while $u$ is the horizontal velocity.

For the normal equation, first remove the hydrostatic pressure by writing $P=p+\rho gy$. Normal inertia and normal viscous terms are $O(\rho U^2\delta /L^2)$, so the change of $P$ across the layer is only $O(\rho U^2{ϵ}^2)$. Matching to the outer pressure therefore makes the leading $O(\rho U^2)$ excess pressure independent of $y$:The outer expression similarly has $(p_{\infty }{)}_y=-\rho g-\rho A^{2}y$. Thus the printed conditionis a leading-order boundary layer matching statement, with the smaller normal dynamic terms discarded on both sides. It is not an exact identity at the next order. In particular, retaining the outer $-\rho A^{2}y$ while dropping normal inertia and normal viscous terms inconsistently mixes orders. The tangential driving term is unambiguous: $-(p_{\infty }{)}_x/\rho =A^{2}x$.

Use the same stream function convention and denote differentiation with respect to the dimensionless variable $\eta =y/\delta$ by a prime. The proposed similarity solution givesIts continuity equation holds identically. Substitution in the tangential Prandtl boundary-layer equation givesChoose the constant scale $\delta =\sqrt{\nu /A}$. The Hiemenz flow equation and its boundary conditions areThe first two conditions impose the no-penetration boundary condition and no-slip boundary condition. The last matches the tangential outer velocity. One also has $f^{\prime \prime }\to 0$ and $f(\eta )/\eta \to 1$. It is too strong to impose $f-\eta \to 0$: the displacement thickness generally permits a constant offset, $f\sim \eta -B$.

The normal Navier-Stokes equation makes the approximation in part (c) explicit. With this similarity solution, it givesAt the wall this differs from $(p_{\infty }{)}_y=-\rho g$ by $-\rho A^2\delta f^{\prime \prime }(0)$. This is a concrete next-order counterexample to interpreting the printed normal pressure gradient equality exactly; the tangential Hiemenz flow equation remains valid at its stated order.

The Hiemenz flow profile changes on $\eta =O(1)$, so its boundary layer thickness isA specified edge convention, such as $u/(Ax)=0.99$, introduces an order-one numerical multiplier requiring the profile to be solved. Scaling alone does not determine that multiplier. The result has no separate dependence on $x$, $y$, $\rho$ or $p_0$ at fixed kinematic viscosity. At fixed dynamic viscosity, it is $\sqrt{\eta /(\rho A)}$.

The viscous diffusion time across a thickness $\delta$ is ${\delta }^2/\nu$; the imposed strain has time scale $A^{-1}$. Their balance sets the thickness. Increasing kinematic viscosity spreads the layer, while increasing the imposed velocity gradient presses the layer toward the wall. The outward tangential velocity increases as $Ax$, so the decreasing travel time per unit distance offsets the usual downstream broadening.

For a conventional estimate of transition to turbulence, form the local horizontal Reynolds number:If transition is assumed to occur at an empirical value ${\mathrm{Re}}_{\mathrm{crit}}$, thenA rough borrowed threshold of order $10^5$–$10^6$ would give $l$ of order $10^2$–$10^3\sqrt{\nu /A}$. This is a heuristic estimate, not a universal stability prediction: the accelerating outer flow and favorable pressure gradient can delay transition, and disturbance amplitudes and wall roughness matter. Simply setting ${\mathrm{Re}}_x\sim 1$ locates failure of the thin-layer approximation near the stagnation point, not onset of turbulence.

Each layer has constant mass density, initially negligible velocity and hence zero vorticity. Its Euler equations for an inviscid fluid have a conservative gravitational body force, so Kelvin's circulation theorem preserves zero bulk vorticity. In a simply connected layer, the irrotational flow therefore has ${\mathbf{v}}_i=\nabla {\varphi }_i$. The incompressible flow condition givesThis applies while the ideal layerwise solution is smooth. It allows an interfacial vortex sheet; it does not assert that the whole two-fluid flow has zero vorticity or that a smooth interface must exist for arbitrarily long times.

The rigid wall no-penetration boundary conditions arewith boundedness and ${\partial }_r{\varphi }_i=0$ at $r=0$ for regular axisymmetric fields. For a small interface displacement and small slope, linearization of the kinematic boundary condition about $z=0$ givesThe small-slope condition is an additional requirement for high-wavenumber modes: $|\alpha \eta |\ll 1$, not just $|\eta |\ll h,a$.

Choose the unperturbed hydrostatic pressures as ${\bar{p}}_i(z)=p_{\ast }-{\rho }_{i}gz$. The linearized Bernoulli equation gives the pressure perturbation $p_i^{\prime }=-{\rho }_i{\partial }_t{\varphi }_i$, after absorbing spatially constant gauges into ${\varphi }_i$. Since surface tension and viscous normal stress vanish, pressure continuity on $z=\eta$ givesorThe initial velocity is zero, so ${\eta }_t(r,0)=0$. Separate conservation of mass in each closed layer also requires ${\int }_0^a\eta (r,t)rdr=0$ relative to the initially horizontal interface.

Write $\eta =H{e}^{\sigma t}{J}_0(\alpha r)$ using the Bessel function of the first kind. The wall boundary conditions at $z=\pm h$ selectBecause $J_0^{\prime }(s)=-J_1(s)$, the sidewall no-penetration boundary condition requireswhere $j_{1,n}$ are the positive zeros of $J_1$. These radial eigenfunctions automatically preserve layer volume becauseThe constant radial mode $\alpha =0$ cannot change the interface height in a closed tank: the Laplace equation and the impermeable lids then give zero vertical velocity, and a uniform displacement would change each layer's volume.

The kinematic boundary condition fixesThus the interface values are ${\varphi }_1(0)=\sigma \eta \coth (\alpha h)/\alpha$ and ${\varphi }_2(0)=-\sigma \eta \coth (\alpha h)/\alpha$. Substitution in the normal stress condition gives the finite-depth Rayleigh-Taylor dispersion relation:soHere $\mathcal{A}$ is the Atwood number. Every allowed mode has a positive growth rate and is subject to Rayleigh-Taylor instability. Since $d[\alpha \tanh (\alpha h)]/d\alpha =\tanh (\alpha h)+\alpha h{\mathrm{sech}}^2(\alpha h)>0$, shorter radial scales grow faster. The limits areThere is no fastest finite wavenumber in the ideal sharp-interface model without surface tension, viscosity or diffusion. The zero-initial-velocity condition uses both exponential branches: an initial mode $H{J}_0({\alpha }_{n}r)$ evolves as $H\cosh ({\sigma }_{n,+}t)J_0({\alpha }_{n}r)$. If the mass density ordering were reversed the same calculation would give oscillatory stable modes; equal mass densities give neutral modes.

For an inviscid flow, the force per unit volume is $\mathbf{F}=-\nabla p-\rho g\hat{\mathbf{z}}$. The gravitational acceleration has zero curl. Using the curl product rule,The supplied two-dimensional vorticity equation therefore becomesThe out-of-plane vorticity has no stretching term in this two-dimensional setting. The source is baroclinic vorticity generation: differently directed mass density and pressure gradients generate rotation even from zero initial vorticity.

For an explicit sign check, set ${\omega }_y={\partial }_{z}u-{\partial }_{x}w$, with $z$ upward. A thin smoothed interface has $\rho \simeq \bar{\rho }(z-\eta (x,t))$, with ${\bar{\rho }}^{\prime }>0$. If $P(x,t)$ denotes the total pressure evaluated on the interface, then within the thin layer $p_x+{\eta }_x{p}_z\simeq P_x$ andIntegrating through that layer and taking its thickness to zero gives the vortex sheet strength $\gamma =u_2-u_1$ and its linear evolution:For a growing Fourier mode, the potentials from the preceding calculation with $\alpha$ replaced by $k$ giveThus the right-hand descending slope of a light-fluid crest produces positive ${\omega }_y$, while the left-hand ascending slope produces negative ${\omega }_y$. Their induced velocity raises the light-fluid crest and lowers the heavy-fluid trough, increasing the slope and generating more vorticity. This is the positive feedback underlying Rayleigh-Taylor instability.

One can close this mechanism quantitatively. For $\eta =H(t)\cos (kx)$, the kinematic boundary conditions give $\gamma =2\dot{H}\coth (kh)\sin (kx)$. Combining with the sheet-generation equation givesFor $kh\ll 1$, the lids confine motion throughout both layers and horizontal redistribution is slow, yielding $\sigma \sim \sqrt{\mathcal{A}gh}k$. For $kh\gg 1$, motion is localized within depth $k^{-1}$ on each side. Less fluid must accelerate at shorter wavelength, giving $\sigma \sim \sqrt{\mathcal{A}gk}$. Since $k_n=n\pi /b$, the ideal growth rate increases with $n$ with no short-wave cutoff.

The sharp-interface result requires $k\delta \ll 1$ for the smoothed layer. When $k\delta$ becomes order one, it no longer applies. A smooth adverse mass density gradient has a local overturning time of order $[g{\bar{\rho }}^{\prime }/\bar{\rho }{]}^{-1/2}$ rather than the unbounded sharp-interface rate. Finite molecular diffusion also relaxes small-scale mass density perturbations on a time of order $(D{k}^2{)}^{-1}$, and viscosity, if included, damps small-scale velocity. Weak diffusion motivates the thin-layer picture; it does not justify extrapolating the discontinuous dispersion relation to arbitrary $n$.

A small Reynolds number, $\mathrm{Re}=\rho UL/\eta \ll 1$, means that inertia is small compared with viscous forces. It can arise from small speed, small length or large dynamic viscosity. Examples are the locomotion of a microswimmer and the slow sedimentation of a small particle in a viscous liquid.

To neglect local acceleration as well, an externally imposed time scale $T$ must satisfy $\rho L^2/(\eta T)=L^2/(\nu T)\ll 1$. This holds for a steady flow or a sufficiently slowly varying quasi-steady Stokes approximation. Small Reynolds number alone is insufficient for that extra step. For an incompressible flow with constant material coefficients and uniform gravity, the Navier-Stokes equation then reduces to the Stokes equation:Taking the divergence and using commuting spatial derivatives givesTaking the curl, which annihilates the pressure gradient and uniform gravitational force, givesThese are the Harmonic pressure and vorticity in Stokes flow results.

A concrete counterexample to inferring both solely from small Reynolds number is Stokes' second problem: an oscillating plate with arbitrarily small velocity amplitude has negligible inertia, but ${\partial }_t\mathbf{\omega }=\nu {\nabla }^2\mathbf{\omega }$ is nonzero. Its vorticity is not a harmonic function. The quasi-steady qualification is therefore necessary.

For an axisymmetric field without azimuthal motion, the divergence in spherical coordinates isSubstituting the given axisymmetric stream function representation givesThus incompressible flow is automatic.

The velocity components are laboratory-frame components resolved about the instantaneous bubble center. The center itself translates with $U\hat{\mathbf{z}}=U\cos \theta \hat{\mathbf{r}}-U\sin \theta \hat{\mathbf{\theta }}$. The kinematic boundary condition for a spherical interface givesFor negligible internal dynamic viscosity, continuity of tangential traction imposes zero tangential shear stress, rather than a prescribed tangential velocity. In terms of the strain-rate tensor,Equivalently, ${\partial }_r{v}_{\theta }(a)=v_{\theta }(a)/a+U\sin \theta /a$. Axisymmetry gives $v_{\varphi }=0$, and the far-field boundary condition is $\mathbf{v}\to 0$. There is no no-slip boundary condition $v_{\theta }=-U\sin \theta$ on a clean bubble; the liquid can slip tangentially relative to the translating interface.

These are the conditions of the prescribed-sphere clean-bubble Stokes drag model. A genuinely free bubble also requires a normal stress balance. If one additionally insists on spatially uniform gas pressure, while removing both surface tension and the variation of hydrostatic pressure, that extra condition is incompatible with the translating spherical solution in part (c). The approximation must therefore be understood as computing drag on an imposed spherical interface, or as retaining the normal balancing effects when checking a free bubble.

Differentiating the axisymmetric stream function givesThe normal kinematic boundary condition fixes $C=Ua/2$, soBoth velocity components vanish at infinity, and the axisymmetric stream function is regular at the polar axes. The tangential component of the strain-rate tensor vanishes everywhere:Thus the stress-free boundary condition is satisfied. The vorticity is ${\omega }_{\varphi }=2C\sin \theta /r^2=Ua\sin \theta /r^2$, and the supplied dipolar pressure is consistent with the Stokes equation.

Use the Newtonian fluid stress tensor $\mathbf{\sigma }=-p\mathbf{I}+2\eta \mathbf{e}$. At the bubble surface,The vertical force exerted by the liquid is the integral of the vertical traction:Since the integral is $2/3$, the drag is downward and has magnitudeThe constant pressure integrates to zero force. The pressure perturbation contributes $4\pi \eta aU/3$ and the normal viscous stress contributes $8\pi \eta aU/3$; there is no tangential traction contribution.

For the free-interface qualification in part (b), the calculated normal traction at $\theta =0$ and $\theta =\pi$ differs by $6\eta U/a$. It cannot equal a single uniform gas pressure if every restoring effect is discarded. Restoring the external-minus-internal variation of hydrostatic pressure supplies a term $({\rho }_w-{\rho }_a)ga\cos \theta$. It cancels the dynamic normal traction at the terminal velocity $U=({\rho }_w-{\rho }_a)g{a}^2/(3\eta )$, consistent with a spherical clean bubble without surface tension. This distinguishes the drag calculation from the extra conditions needed for a freely rising bubble.

Retain the clean-bubble approximation ${\eta }_a\ll {\eta }_w$ used in parts (b) and (c), with constant mass densities and diffusion coefficient. For a spherical bubble,ConsequentlyThe buoyancy minus bubble weight is $4\pi ({\rho }_w-{\rho }_a)g{a}^3/3$. Balancing it against the clean-bubble Stokes drag gives the instantaneous terminal velocitySet $z=0$ at the initial position. Combining this with the dissolution equation gives $da/dz=-D_a/(K{a}^3)$ and hence the dissolution-limited bubble rise lawThe expression applies until the bracket vanishes. The rise speed decreases continuously as the bubble shrinks. Equivalently, $z(t)=K(a_0^{2}t-D_a{t}^2)$, providing a direct check on the elimination of time.

The phrase “constant terminal velocity” must here mean rapid adjustment to a force balance at each radius. A literally time-independent speed is incompatible with that balance because $U\propto a^2$. If instead an external mechanism maintained a fixed $U_{\ast }$, the distinct result would be $a^2(z)=a_0^2-2D_{a}z/U_{\ast }$ and $z_{\mathrm{diss}}=U_{\ast }{a}_0^2/(2D_a)$. In particular, holding the initial terminal speed fixed doubles the predicted rise distance; it is not the freely rising shrinking-bubble model.

Using the supplied values and $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$ givesEquality means dissolution exactly on arrival at the surface; dissolution strictly before arrival needs a greater initial depth. This is a formal low-Reynolds number estimate. The same values predict $U(a_0)\simeq 0.033\mathrm{m}{\mathrm{s}}^{-1}$ and a radius-based Reynolds number ${\rho }_{w}U(a_0)a_0/{\eta }_w\simeq 3.3$; the diameter-based value is about $6.5$. Neither is $\ll 1$. Thus the numerical input does not consistently satisfy the Stokes flow assumption initially, and the $0.82\mathrm{m}$ result should not be presented as an accurate oceanic prediction. Within the stated ideal model, bubble compression, contaminants and changes in dissolved-gas concentration are also excluded.