Past exam of the fluids course of the University of Oxford / 2022 / 4

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.