Past exam of the fluids course of the University of Oxford / 2021 (A16726W1)

Source paper: www.whatdotheyknow.com/request/oxford_physics_past_papers_2/response/2340696/attach/9/Physics%20Papers%202020.pdf?cookie_passthrough=1

The cover dates this paper to 11 June 2021 despite the attachment filename. Solutions identify a false printed biharmonic equation identity, retain the stipulated drop curvature approximation explicitly, and state the conditions needed for the final Joukowski map.

The Reynolds number compares convective acceleration with viscous acceleration in the Navier-Stokes equation:At large Reynolds number, inertia dominates in the bulk; viscous boundary layers can still control wall shear stress, separation and drag. Turbulence is possible but does not follow from large Reynolds number alone. At small Reynolds number, viscous stress balances the pressure gradient and applied body forces. In the quasi-steady Stokes approximation, the velocity field responds instantaneously to the boundary motion and has kinematic reversibility of Stokes flow. Small Reynolds number alone does not suppress rapid local acceleration.

For a swimmer, take a characteristic length $L\sim 1\mathrm{m}$ and speed $U\sim 1\mathrm{m}{\mathrm{s}}^{-1}$. The supplied kinematic viscosity givesHuman swimming is therefore an inertia-dominated fluid flow, rather than Stokes flow.

The divergence in polar coordinates and the given streamfunction convention yieldThus the incompressibility condition holds wherever the mixed partial derivatives commute. The vertical vorticity isNeglecting both inertial terms in the Navier-Stokes equation, for constant dynamic viscosity and mass density, gives the Stokes equation:Its curl eliminates the pressure gradient and the uniform gravitational body force, so $\mu {\nabla }^2\zeta =0$. Hence the biharmonic streamfunction satisfiesThis conclusion requires the quasi-steady Stokes approximation. If local acceleration were retained, the corresponding planar vorticity equation would instead be ${\partial }_t{\nabla }^2\psi =\nu {\nabla }^4\psi$.

Write ${\omega }_h=\dot{\alpha }$, positive on opening. The no-slip boundary condition equates the fluid velocity with each plate's velocity:The upper condition is a moving-wall no-penetration boundary condition, not zero normal velocity in the laboratory frame. At the hinge the velocity tends to zero. With the arbitrary constant of the streamfunction chosen so that $\psi (r,0)=0$, these conditions becomeA changing angle cannot give a literally steady flow in fixed coordinates. We interpret the stated steadiness as the quasi-steady Stokes approximation at each instantaneous $\alpha$.

The printed expression is not biharmonic. To test it without reproducing the source formula, write its bracket as $G_p(\theta )$ and its streamfunction as ${\psi }_p=-\Omega r^2{G}_p/2$. For any angular function $G$, the Laplacian givesAt the concrete angle $\alpha =\pi /4$, differentiation of the printed bracket givesFor example, at the interior point $\theta =\pi /8$,The printed bracket does satisfy $G_p(0)=G_p^{\prime }(0)=G_p^{\prime }(\alpha )=0$ and $G_p(\alpha )=1$, so its boundary velocities would require $\Omega ={\omega }_h$. Satisfying these boundary conditions does not cure its nonzero biharmonic operator. The stated series is likewise not a general biharmonic basis: its $r^2\theta \sin 2\theta$ term already fails the displayed test. The separated polar solutions of the biharmonic equation contain additional radial powers; for the required $r^2$ dependence, the angular basis is $1,\theta ,\sin 2\theta ,\cos 2\theta$.

The corrected slowly opening Stokes wedge similarity solution follows by setting $\psi =-{\omega }_h{r}^{2}G(\theta )/2$. The biharmonic equation gives $G^{(4)}+4G^{\prime \prime }=0$. The two lower-wall conditions reduce its general solution toThe upper-wall condition $G^{\prime }(\alpha )=0$ fixes the ratio of $A$ and $B$, and $G(\alpha )=1$ fixes their scale. An expression that also remains finite at $\alpha =\pi /2$ isFor $0<\alpha <\pi$, its denominator is positive: the derivative of $\sin \alpha -\alpha \cos \alpha$ is $\alpha \sin \alpha >0$. Direct differentiation verifies all four boundary conditions and the biharmonic equation. The resulting velocity components are $v_r=-\dot{\alpha }r{G}^{\prime }/2$ and $v_{\theta }=\dot{\alpha }rG$.

For a fixed angle of order one, $U(r)\sim |\dot{\alpha }|r$, so the local Reynolds number is ${\mathrm{Re}}_r\sim |\dot{\alpha }|r^2/\nu$. One breakdown length is thereforeA second is the geometric length $l_2\sim \min (L,W)$, where $L$ is the distance to the radial ends and $W$ the spanwise width. At $r\sim l_2$, matching to the surrounding bath or three-dimensional end flow matters. Thus $r\ll \min (l_1,l_2)$ is required for this local two-dimensional Stokes flow, with angle-dependent factors in the inertial estimate. Ideal infinite plates remove the geometric cutoff but not the inertial one.

There is also a temporal restriction: the relevant viscous diffusion time must be short compared with the time $T_s$ over which the shape and its velocity field change appreciably. Using $r$ as a conservative diffusion length gives $r\ll \sqrt{\nu T_s}$. For order-one angles and constant $\dot{\alpha }$, $T_s\sim |\dot{\alpha }{|}^{-1}$ gives the same scale as $l_1$; narrow angles require using their actual gap and angular factors rather than extrapolating this order-one-angle estimate.

Take outward radial volume flux per unit width as positive. The streamfunction in polar coordinates immediately givesFor spanwise width $W$, the volumetric flow rate is $Q=Wq$. The wedge inside radius $r$ has cross-sectional area $\alpha r^2/2$, whose rate of increase is $r^2\dot{\alpha }/2$. Opening therefore draws in exactly the volume needed to fill the expanding wedge; closing expels it. This is also a direct check of conservation of mass for the moving control volume. It follows from the wall boundary conditions and remains true even though the printed interior streamfunction is erroneous.

Given an admissible velocity field, recover the pressure gradient from the Stokes equation, $\nabla p=\mu {\nabla }^2\mathbf{v}$, and integrate it, fixing the additive constant by matching to the bath pressure near the open ends. The biharmonic equation ensures the compatibility of this integration. For a freely moving hinge, the laboratory velocity field also contains the unknown rigid translation and rotation. Solve the interior and exterior Stokes flows with these wall velocities, then impose zero total external force and torque on the complete apparatus if its motor supplies only internal actuation. The pinned solution alone cannot determine that free motion.

At a fixed angle, the actuation velocity scale is $U\sim L|\dot{\alpha }|$. The Stokes equation gives $\Delta p\sim \mu U/L\sim \mu |\dot{\alpha }|$. Integrating normal traction over a plate of length $L$ gives a force per unit width of scalewith a dimensionless signed coefficient depending on geometry and exterior matching. Normal viscous stress has the same dimensional scale and is included in $C$. The force reverses when the actuation reverses. A narrow gap can make $|C|$ large.

More specifically, the corrected local streamfunction gives, up to a reference pressure,Indeed, inserting $\mathbf{v}=(-\dot{\alpha }r{G}^{\prime }/2,\dot{\alpha }rG)$ into the vector Laplacian gives $p_r=-\mu \dot{\alpha }(G^{\prime \prime \prime }+4G^{\prime })/(2r)=\mu \dot{\alpha }K/r$ and $p_{\theta }=0$. For $\alpha <\pi /2$, opening produces reduced pressure toward the hinge when matched at $r_{\ast }\sim L$. The logarithmic singularity is integrable: ${\int }_0^L\log (L/r)dr=L$, consistent with the stated force scale. A real hinge regularizes the singularity; the local formula is not a full end-matched solution.

The scallop theorem applies to a single periodically actuated hinge in a stationary Newtonian fluid with negligible fluid and body inertia. At each shape, Linearity of Stokes flow makes its rigid velocity proportional to $\dot{\alpha }$. Closing retraces the sequence of shapes and reverses the corresponding motion, irrespective of different opening and closing rates. The hinge may move during each half-cycle, butA motor can generate instantaneous forces and consume energy through viscous dissipation without producing sustained swimming. Additional independently controlled shape changes, appreciable inertia, or a fluid with memory would require a different analysis.

Compressing a small region raises its mass density and, for positive compressibility, its pressure. The outward pressure gradient force accelerates neighbouring fluid elements. Their outward motion compresses fluid farther away, carrying the disturbance outward. The original compressed region relaxes; inertia allows the fluid motion to continue beyond zero excess pressure, producing a rarefaction and a restoring pressure gradient force. Elastic compression and inertia therefore support a propagating longitudinal wave.

In a local planar description with acoustic displacement $\xi (x,t)$,so ${\xi }_{tt}=c_0^2{\xi }_{xx}$. The speed of sound is a propagation speed, whereas the acoustic particle velocity is the small back-and-forth motion of the medium. In linear acoustics, energy travels with the disturbance without first-order net transport of the fluid over a wave period.

For one-dimensional inviscid flow, conservation of mass and the Euler equation areSubstitute $u=U_0+v_1$ and the stated background-plus-perturbation expressions. Dropping products of perturbations gives the linearizationThe Taylor expansion of the equation of state givesSince the background coefficients are constant, $D_0$ commutes with ${\partial }_x$. Applying $D_0$ to the linearized continuity equation and using the linearized Euler equation therefore yieldsThis is the convected acoustic wave equation: in the coordinate $x-U_{0}t$, it becomes the ordinary one-dimensional wave equation. Positive $c_0^2$ is required for propagating sound waves.

Use a plane wave proportional to $e^{i(kx-\omega t)}$. Substitution into the convected acoustic wave equation gives the dispersion relationThe two branches have signed phase velocities and group velocities $U_0\pm c_0$. On a branch with $\omega -U_{0}k=s{c}_{0}k$, $s=\pm 1$, the linearized Euler equation givesIn subsonic flow with $0\le U_0<c_0$, the absence of wave reflection selects the outgoing branch on each side:For $U_0=0$, take $A$ real and positive by choosing the source phase, and put $k_0=\omega /c_0$. At $\omega t=2n\pi$, the requested spatial profiles areFor complex $A$, retain $\mathrm{Re}[A{e}^{i{k}_0|x|}]$ instead. The normalized sound pressure and acoustic density perturbation lie on the same even cosine curve. The acoustic particle velocity follows that curve on the right and its negative on the left, as in the figure in part 2(a). Compressed fluid moves away from the source on both sides. These are two outgoing travelling waves, not a standing wave. The ideal point loudspeaker supplies the discontinuity of velocity at $x=0$; the source-free equations apply on either side of it.

Take the wind to blow toward positive $x$, so $U_0\ge 0$. In subsonic flow, both stationary observers record the source angular frequency $\omega$ and, for the imposed density boundary data, the same acoustic density perturbation amplitude ${\rho }_0|A|$. Their spatial records distinguish the wavelengths and directions:The downstream pattern moves at $c_0+U_0$ and the upstream pattern at signed velocity $U_0-c_0$. Thus the downstream observer sees longer wavelengths while the upstream observer sees shorter ones. The interval containing multiple wavelengths lets each measure the spatial spacing and its direction of travel. The arrival delays at $\pm L$ are $L/(c_0+U_0)$ and $L/(c_0-U_0)$ respectively; after arrival their temporal traces differ by these propagation phases, not by temporal frequency.

In supersonic flow, both characteristic curves have positive slopes $dx/dt=U_0\pm c_0$. With an initially undisturbed upstream region, the observer at $-L$ sees no source-generated disturbance. The observer at $+L$ can see two downstream travelling waves, withThe faster signal arrives first, at $L/(U_0+c_0)$; the slower one arrives at $L/(U_0-c_0)$. Once both are present, their different wavenumbers can produce spatial modulation, while the time dependence at a fixed position remains harmonic with angular frequency $\omega$.

There is a qualification to the source data: the single prescribed value of ${\rho }_1(0^{+},t)$ fixes only $A_{+}+A_{-}=A$. It does not determine the two complex amplitudes separately. A second source condition, such as its downstream acoustic particle velocity with amplitude $c_0(A_{+}-A_{-})$, or a specified mass/force source model, is needed. Consequently two downstream branches are allowed; the stated density data alone do not require both to have nonzero amplitude. Extending a formally upstream-going subsonic formula into $x<0$ when $U_0>c_0$ would impose a signal arriving from upstream infinity and violate the causal source interpretation.

As $U_0\to c_0$, one laboratory phase velocity tends to zero. Its wavelength tends to zero, its wavenumber diverges, and its arrival time diverges; a finite-time experiment never attains the assumed established upstream field at fixed $L$. Finite source size, viscosity, thermal losses and eventually molecular scales invalidate this singular inviscid flow limit. With effective acoustic diffusivity ${\nu }_a$, a loss estimate is ${\nu }_a{k}^2/\omega \sim {\nu }_a\omega /(U_0-c_0{)}^2$, which diverges. Moreover the finite-amplitude change in the characteristic speed is of order $c_0|A|$, so uniform linearization requires $|U_0-c_0|\gg c_0|A|$. At exactly sonic background speed, the zero-speed branch cannot support a nonzero-$\omega$ harmonic solution. Nonlinear acoustics and losses must resolve the near-sonic regime; an unlimited spatial compression of the ideal wave is not physical.

Let $H$ be the typical thickness, ${\ell }_r$ the radial variation length, $U$ the horizontal speed, and $W$ the vertical speed. For a thin smooth profile, $\epsilon =H/{\ell }_r\ll 1$. The axisymmetric continuity equation isAway from singular boundary regions, balancing its terms gives $U/{\ell }_r\sim W/H$, hence $\boxed{W/U\sim \epsilon \ll 1}$. Near $r=0$, regularity gives $v_r=O(r)$ and the same conclusion follows without a divergent $v_r/r$ term.

Vertical viscous derivatives dominate horizontal ones by ${\epsilon }^{-2}$. The ratio of convective acceleration to vertical viscous acceleration isFor thickness-evolution time $T$, the local inertial ratio is $H^2/(\nu T)$. When both ratios are small, lubrication theory reduces the Navier-Stokes equation toThinness must therefore accompany a slow enough motion; it is not a justification for dropping arbitrarily rapid imposed acceleration. The small vertical velocity also makes the leading vertical balance hydrostatic balance, $p_z=-\rho g$. The neglected normal viscous stress at the interface is smaller than the characteristic lubrication pressure by order ${\epsilon }^2$.

Use the source's curvature sign: $p_{\mathrm{air}}-p_{\mathrm{liquid}}=-\gamma \kappa$. The Young–Laplace equation and atmospheric pressure $p_0$ thus give $p(r,h,t)=p_0+\gamma \kappa$. Integrating the hydrostatic balance yieldsConsequently $p_r=\gamma {\kappa }_r+\rho g{h}_r$, independent of $z$ at this order. The radial Stokes equation and boundary conditions areThe first is the no-slip boundary condition; the second is the leading stress-free boundary condition, since the atmosphere has negligible dynamic viscosity and the surface tension is constant. Integrating twice givesThis is the horizontal velocity only. A smaller vertical velocity is generally required by the continuity equation.

Integrating the radial velocity over the thickness gives the radial volume flux per unit width:The full volumetric flow rate across a cylindrical surface, positive outward, isAn annulus of radial thickness $dr$ contains volume $2\pi rhdr$ to leading order. Its rate of gain equals incoming minus outgoing volume flux, soDividing and taking $dr\to 0$ yields the axisymmetric thin-film equation:The equation assumes no evaporation or injection through the boundaries. Integrating it over the whole drop also verifies conservation of mass when the edge carries zero volume flux.

The gravity-capillary drop with a meridional curvature approximation uses the stipulated $\kappa =-h^{\prime \prime }$ for this calculation. At equilibrium the thin-film equation givesRegularity at $r=0$ forces $C=0$. In the wetted interior $h>0$, the equilibrium pressure gradient therefore vanishes, givingThis is the capillary length. Integrating the constant-coefficient ordinary differential equation givesThe central boundary condition $h^{\prime }(0)=0$ sets $C_2=0$, and contact at $r=R$ gives $C_0=-C_1\cosh (R/l)$. Write $C_1=-\alpha l$. The prescribed contact angle fixesHere $\alpha$ is the dimensionless coefficient of this drop solution, unrelated to the hinge angle in question 1.

The axisymmetric volume integral is $V=2\pi {\int }_0^{R}rh(r)dr$. Substituting $\xi =r/l$ givesFor numerical evaluation, with $s=R/l$, the integral isTogether with $\alpha =\tan {\theta }_c/\sinh s$, this determines $R$ for fixed $V$ and ${\theta }_c$.

The Bond number $R^2/l^2=\rho g{R}^2/\gamma$ distinguishes the dominant balances. For $R\ll l$, surface tension dominates gravity and the surface is nearly a small-slope constant-curvature cap:For $R\gg l$, gravity flattens the interior, while surface tension bends the edge over the capillary length. The stipulated model givesup to relative corrections of order $l/R$. The contact angle fixes the edge slope and therefore the height scale; conservation of mass then fixes the lateral extent. Consistency of lubrication theory requires small slopes, in particular $\tan {\theta }_c\ll 1$.

The stated approximation differs from axisymmetric free-surface curvature. For an axisymmetric graph with the same sign convention, the sum of principal curvatures isThe omitted $h^{\prime }/r$ term is generally of the same order as $h^{\prime \prime }$. For example, $h=H(1-r^2/R^2)$ gives stipulated $\kappa =2H/R^2$ but axisymmetric small-slope $\kappa =4H/R^2$. Thus the required hyperbolic-cosine profile solves the explicitly prescribed reduced model, not the full axisymmetric capillary equilibrium. Keeping both small-slope terms instead giveswhere $I_0,I_1$ are modified Bessel functions. The qualitative capillary length and Bond number balances remain valid.

For the simple shear flow, the vertical vorticity isFor $\beta \ne 0$, this is not an irrotational flow. A complex potential of a two-dimensional potential flow $w=\varphi +i\psi$, analytic in the flow region, has complex velocity $w^{\prime }=v_x-i{v}_y$. The Cauchy-Riemann equations imply both incompressibility and zero vorticity, so this shear has no such complex potential. It does have a Cartesian streamfunction, $\psi =\beta y^2/2$. In the degenerate case $\beta =0$, the resting fluid has a constant complex potential.

Using $w=\varphi +i\psi$ and the complex velocity convention $w^{\prime }=v_x-i{v}_y$,The streamlines are $xy=\psi /(2\beta )$, a family of rectangular hyperbolas whose asymptotes are the coordinate axes. The origin is a stagnation point. Fluid moves inward along the $y$ axis and outward along the $x$ axis, with the same directions approached by the neighbouring streamlines. This is a planar extensional incompressible flow, illustrated in the left panel below.

Taking the imaginary part of the complex potential gives the streamfunctionThe corresponding velocity field isPut $\zeta =z{e}^{i\alpha /2}=\xi +i\eta$. Then $w=\beta {\zeta }^2$ and $\psi =2\beta \xi \eta$. Thus the streamlines are the previous rectangular hyperbolas rotated clockwise by $\alpha /2$. Their separatrices have angles $\theta =-\alpha /2+n\pi /2$; the extension axis is $\theta =-\alpha /2$ and the compression axis is perpendicular to it. The middle panel of the figure shows this rotation.

For pure advection, a marked fluid element obeysConsequently the dye circle becomes an ellipse with semiaxes $R{e}^{2\beta t}$ and $R{e}^{-2\beta t}$. Its area isStretching in one principal direction exactly compensates compression in the other. Equivalently the flow-map Jacobian determinant is one, as required by incompressibility. Negligible diffusion is needed to regard the sharp dye boundary as a material curve.

The coefficient of the growing term is fixed by the far-field complex potential: $A=\beta e^{i\alpha }$. On $z=a{e}^{i\theta }$, the no-penetration boundary condition makes the cylinder a streamline, so the imaginary part of $A{a}^2{e}^{2i\theta }+B{a}^{-2}{e}^{-2i\theta }$ must be constant. Choosing the two terms to be complex conjugates makes that constant zero, givingThis is the zero-circulation solution, also obtained by the Milne-Thomson circle theorem. Impermeability and the leading far field alone permit an additional circulation term $-i\Gamma \log z/(2\pi )$; the requested algebraic form implicitly selects $\Gamma =0$. No no-slip boundary condition is imposed on an inviscid flow.

The streamfunction in polar coordinates isIt gives $v_r(a,\theta )=0$ andThe liquid still exerts normal pressure on each surface element even though tangential viscous traction is absent. For this steady flow with zero vorticity, the Bernoulli equation has one constant $C$ throughout the connected fluid region:There is no finite uniform far-field pressure to substitute for $C$, because this straining velocity field grows with radius. Taking $\mathbf{n}=(\cos \theta ,\sin \theta )$ outward from the cylinder, the fluid force per unit axial width isIndeed $p(\theta +\pi )=p(\theta )$ while $\mathbf{n}(\theta +\pi )=-\mathbf{n}(\theta )$: diametrically opposite elements cancel exactly. Zero resultant force does not mean zero local pressure or zero deformation tendency.

Use the Joukowski map with $|c|<a$, in particular its intended range $0\le c<a/2$. On the original circle,Define the exterior inversewhere the holomorphic square root is chosen so that $\sqrt{Z^2-4c^2}\sim Z$ at infinity. Then $f(Z)\sim Z$ is the branch mapping the exterior of the ellipse to $|z|>a$. Composition of the zero-circulation complex potential from part 4(c) with this inverse givesThe ellipse is a streamline because its inverse image is the circle, and $W(Z)\sim \beta e^{i\alpha }{Z}^2$ at infinity. An additive constant in the complex potential has no effect on the velocity.

The transformed complex velocity obeys the chain rule:Since $|c|<a$, the denominator is nonzero on and outside $|z|=a$. The four stagnation points therefore come fromTheir physical locations areThey all lie on the ellipse, and their positions are independent of $\beta >0$, which only sets the overall speed scale.

The printed restriction that $c$ is real and $c<a/2$ is insufficient literally: $c=-2a$ satisfies it but makes $dZ/dz=0$ at $z=2a$ in the exterior, so there is no exterior conformal map of the claimed type; the stated minor semiaxis would also be negative. The intended positive-parameter range, or more generally $|c|<a$, is needed for the above solution. The reference to a sphere in the source is a wording slip: the geometry is two-dimensional flow around the mapped cylinder's elliptical cross-section.