Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 77 / 3

Choose the axial origin at the sphere's centre. Reflection $x\mapsto -x$ leaves the straight tube, the spherical geometry and the transverse displacement of the centre unchanged. It reverses the prescribed axial Hagen-Poiseuille flow, while it leaves a hypothetical transverse particle velocity unchanged. On the other hand, kinematic reversibility of Stokes flow says that reversing the imposed flow reverses every particle velocity, including that transverse component. The force-free and torque-free constraints are preserved by both operations. Uniqueness of the coupled Stokes flow and rigid-body motion therefore makes the transverse velocity equal to its own negative. It must vanish.

Consequently $\dot{c}=0$ as the particle is advected. This is no lateral migration of a single sphere in a uniform tube in Stokes flow. The argument uses the fore-aft symmetry of a rigid sphere and a uniform tube; it does not extend automatically to inertial migration or deformable particles.

Integrate axial force balance $\nabla \cdot \sigma =0$ over the fluid between the two remote tube sections, excluding the particle. The wall's axial traction contributesAt each remote section the Hagen-Poiseuille flow has $u_x$ independent of $x$, so ${\sigma }_{xx}=-p$. The downstream and upstream caps therefore contribute $-\pi a^{2}p(x_2)+\pi a^{2}p(x_1)$. The particle contributes zero net axial force because it is force-free. Adding the three contributions proves the wall-shear and pressure-drop balance in a tube:Here $p$ is the physical pressure. The sign follows from the outward normals on the two caps; no convention for a positive pressure-drop magnitude has yet been introduced.

In this continuation $\lambda$ measures eccentricity, not the viscosity ratio used in Question 2. Take $ϵ\to 0$ with $0\le \lambda <1$ fixed. For the nearly occluding sphere in a cylindrical tube, the narrow gap has thickness $H\sim ϵa$ and axial length $\ell \sim a\sqrt{ϵ}$, obtained by balancing $h_0$ and $x^2/(2a)$. In the sphere frame the tangential velocity scale is $U$. The lubrication approximation then gives the gap scalesThe large local pressure excursion is distinct from the eventual net pressure jump. Away from the sphere the velocity varies across a radius $a$, so the far Hagen-Poiseuille flow hasThese scales also show why the gap pressure can be much larger than its shear stress while the leading net jump can be smaller than that pressure.

Use $y=a-r$, so $y=0$ is the moving tube wall and $y=h$ is the locally stationary sphere. The no-slip boundary conditions are $u(0)=-U$, $u(h)=0$. Integrating $\mu u_{yy}=p_x$, with pressure independent of $y$ to leading order, gives the Couette-Poiseuille flow in a thin gapDifferentiate at both surfaces and eliminate $p_x$ using this moving-boundary lubrication flux:The coordinate reversal means ${\sigma }_{rx}=-{\sigma }_{xy}$ at the tube wall.

At leading order $q$ is independent of $x$ in each angular strip, though it may depend on $\theta$. Both boundaries are stationary as shapes in the sphere frame and impermeable, so the integrated continuity equation is $q_x+a^{-1}(q_{\theta }^{\mathrm{circ}}{)}_{\theta }=0$. The pressure changes along $x$ over $a\sqrt{ϵ}$ but along the circumference over $a$. Thus circumferential velocity is only $O(U\sqrt{ϵ})$, and its flux is $O(Ua{ϵ}^{3/2})$. Circumferential flux divergence is smaller than the possible leading axial divergence by a factor $ϵ$. Consequently $q_x=0$ to the order needed. This is not an assumption of equal flux at different angular positions.

For fixed $\theta$ define the lubrication resistance integrals for a nearly occluding sphereIn particular, $J_1=\pi \sqrt{2a/h_0}$, $J_2=J_1/(2h_0)$ and $J_3=3J_1/(8h_0^2)$. The local pressure relation givesA putative generic value $q=O(U{h}_0)$ makes this jump $O(\mu U/(a{ϵ}^{3/2}))$. However, the integrated wall shear in part (ii) is only $O(\mu Ua/\sqrt{ϵ})$, so the force-free global balance permits a common net pressure jump only of order $\mu U/(a\sqrt{ϵ})$. The outer flow has circumferentially uniform pressures on both sides. Therefore the $O({ϵ}^{-3/2})$ pressure jump must cancel in every angular strip, not merely after angular averaging. The pressure recovery condition in lubrication flow determinesThis explains how the leading-order part of the global force balance selects the local flux.

The total sphere-frame volume flux is nowTransforming back to the laboratory adds $\pi a^{2}U$, so $Q_{\mathrm{lab}}=\pi a^{2}U(1-4ϵ/3)$ to this order. For physical pressure, Hagen-Poiseuille flow gives $Q_{\mathrm{lab}}=-\pi a^4{p}_x/(8\mu )$. ThusThe printed positive expression is the pressure-drop gradient in the direction of the laboratory flow; the Poiseuille hint omits the minus sign when $p_x$ is read as the derivative of physical pressure. With the specified wall velocity $-U$, the laboratory flow is positive and its physical pressure gradient is negative.

Next insert the leading flux into the wall shear stress:Let $\Delta p>0$ be the localized extra pressure drop, after subtracting the regular far Poiseuille gradient. Contributions outside the neck are only $O(\mu U/a)$ and are smaller than the $O({ϵ}^{-1/2})$ jump. Since ${\sigma }_{rx}=-{\sigma }_{xy}$, part (ii) gives $\pi a^2\Delta p=a\int \int {\sigma }_{xy}(0)dxd\theta$. Hence the leading pressure drop isFor example, the concentric limit is $\Delta p\sim 4\pi \mu U\sqrt{2/ϵ}/a$. The integral grows as the minimum gap shrinks; the fixed-$\lambda <1$ qualification prevents treating a further near-contact limit as uniformly covered.

On the sphere the leading shear stress is ${\sigma }_{xy}(h)=-2\mu U/h+4\mu U{h}_0/h^2$, soPositive and negative tangential stresses cancel within each angular strip. Thus the leading tangential resultant, and its leading torque contribution, vanish: negligible rotation is consistent with the torque-free condition. The shear is not zero pointwise, and the pressure forces are not zero. Pressure on a sphere cannot produce a torque about its centre because its normal is radial; its force contributions and the matched far pressure jump remain essential to the force-free transport of a nearly occluding sphere.