Past exam of the fluids course of the University of Oxford / 2021 / 1

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.