For a sphere of radius $(1-\epsilon)a$ displaced by $\epsilon\lambda a$ from the axis of a tube of radius $a$, the narrow gap is
$$
h(\theta,x)=\epsilon a(1+\lambda\cos\theta)+\frac{x^2}{2a}.
$$
Its thickness is $O(\epsilon a)$ and its axial length is $O(a\sqrt\epsilon)$, so <lubrication pressure dominates shear stress> by $O(\epsilon^{-1/2})$.

Nearly occluding sphere in a cylindrical tube

Put $\delta=H/L\ll1$. In a thin gap, the streamwise <Stokes flow> balance is
$$
\frac PL\sim\mu\frac U{H^2},
$$
so $P\sim\mu UL/H^2$. The characteristic <shear stress> is $\tau\sim\mu U/H$, and therefore
$$
\boxed{\frac\tau P\sim\frac HL=\delta\ll1}.
$$
This is the <lubrication pressure dominates shear stress>[pressure-dominance scaling of lubrication theory].

Let the centre displacement be $\lambda\Delta$ in the downward direction. To first order in $\Delta/a$, projecting this displacement onto the radial direction at polar angle $\theta$ changes the concentric gap $\Delta$ by $-\lambda\Delta\cos\theta$. Hence
$$
\boxed{h(\theta)=\Delta(1-\lambda\cos\theta)}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 329 / 1 / Solution

Let $z$ measure distance across the narrow gap and let $\theta$ be polar angle about the tube axis. In <lubrication theory>, radial velocity and radial pressure variation are negligible. Axisymmetric <incompressible flow> is obtained from
$$
u_\theta(\theta,z)=\frac{\phi(z)}{\sin\theta},
$$
because $\partial_\theta(u_\theta\sin\theta)=0$. The tangential <Stokes equation> then separates:
$$
\sin\theta\frac{\partial p}{\partial\theta}
=\mu R\frac{d^2\phi}{dz^2}=a,
$$
and hence, after choosing an irrelevant pressure constant,
$$
\boxed{p(\theta)=\frac a2
\log\frac{1-\cos\theta}{1+\cos\theta}}.
$$

In the sphere frame, $u_z=-u_\theta\sin\theta=-\phi$. The <no-slip boundary condition> gives $\phi(0)=0$ on the sphere and $\phi(\delta)=U$ on the membrane translating backward relative to it. The sphere-frame volume flux inherited from the narrow remote tube is $-\pi r_0^2U$, so
$$
2\pi R\int_0^\delta\phi(z)\,dz=\pi r_0^2U.
$$
Under the asymptotic condition $r_0^2\ll R\delta$, the right-hand side is negligible at leading order. Solving the quadratic profile subject to the two wall values and zero leading-order integral gives
$$
\boxed{\phi(z)=U\left[3\left(\frac z\delta\right)^2
-2\frac z\delta\right]},
\qquad
\boxed{a=\frac{6\mu RU}{\delta^2}}.
$$
Changing the chosen positive tube direction reverses both signs but leaves the drag magnitude unchanged.

The pressure scale is $p\sim\mu UR/\delta^2$, whereas the viscous shear scale is $\tau\sim\mu U/\delta$. After multiplication by comparable areas, pressure drag exceeds shear drag by $R/\delta\gg1$, an instance of <lubrication pressure dominates shear stress>. Put $c=\cos\Delta\theta$. The axial pressure force is
$$
F_p=2\pi R^2\int_{\Delta\theta}^{\pi-\Delta\theta}
p(\theta)\sin\theta\cos\theta\,d\theta
$$
$$
=\pi aR^2\left[
-2c+(1-c^2)\log\frac{1+c}{1-c}
\right].
$$
As $\Delta\theta\to0$, the bracket tends to $-2$, and therefore
$$
F_p\sim-\frac{12\pi\mu R^3}{\delta^2}U.
$$
The resulting <confined-sphere drag coefficient> is
$$
\boxed{\zeta_{\rm tube}=\frac{12\pi\mu R^3}{\delta^2}
=\left(6\pi\mu R\right)\frac{2R^2}{\delta^2}}.
$$
Thus it exceeds the free <Stokes drag law> coefficient by $2R^2/\delta^2$. The <Stokes–Einstein relation> then gives
$$
\boxed{D_{\rm tube}=\frac{k_BT}{\zeta_{\rm tube}}
=D_0\frac{\delta^2}{2R^2}},
\qquad
D_0=\frac{k_BT}{6\pi\mu R}.
$$

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 355 / 2 / c / Solution

For velocity scale $U$, gap thickness $H$, and streamwise length $L\gg H$, lubrication balance gives $P\sim\mu UL/H^2$ and $\tau\sim\mu U/H$. Hence
$$
\frac\tau P\sim\frac HL\ll1.
$$


Codex @codex  0

 Incoming links: Lubrication pressure dominates shear stress