The <nearly occluding sphere in a cylindrical tube> has gap thickness $H=O(\epsilon a)$ and axial length $L=O(\sqrt{aH})=O(a\sqrt\epsilon)$. In the gap, <lubrication theory> therefore gives
$$
p_x=O\left(\frac{\mu U}{\epsilon^2a^2}\right),
\qquad
p=O\left(\frac{\mu U}{a\epsilon^{3/2}}\right),
\qquad
\sigma_{xy}=O\left(\frac{\mu U}{\epsilon a}\right).
$$
Far ahead of and behind the sphere, the length and width scales are both $a$, so
$$
p_x=O\left(\frac{\mu U}{a^2}\right),
\qquad
\sigma_{rx}=O\left(\frac{\mu U}{a}\right).
$$

In the sphere frame the tube wall at $y=0$ moves with velocity $-U$ and the sphere surface at $y=h$ is stationary. The local <Couette-Poiseuille flow in a thin gap> is
$$
u=-U+\frac Uh y+\frac{p_x}{2\mu}y(y-h).
$$
Its axial <volume flux> per unit circumferential width is
$$
\boxed{q=-\frac{h^3}{12\mu}p_x-\frac{Uh}{2}.}
$$
Differentiating the profile and eliminating $p_x$ gives the two wall stresses
$$
\boxed{
\left.\frac{\sigma_{xy}}\mu\right|_{y=0}
=\frac{4U}{h}+\frac{6q}{h^2},
\qquad
\left.\frac{\sigma_{xy}}\mu\right|_{y=h}
=-\frac{2U}{h}-\frac{6q}{h^2}.}
$$
The <continuity equation> integrated across the gap says that changes of $q$ along $x$ are balanced by circumferential flux divergence. Circumferential variations occur on scale $a$, much longer than the axial scale $a\sqrt\epsilon$, so $q=q(\theta)$ is independent of $x$ at leading order.

Put $h=h_0(1+\xi^2)$ with $x=\sqrt{2ah_0}\,\xi$. The leading $O(\epsilon^{-3/2})$ pressure jump must vanish in the global balance from part ii. Equivalently, the <pressure recovery condition in lubrication flow> gives
$$
0=\int_{-\infty}^{\infty}p_x\,dx
=-12\mu q\int h^{-3}\,dx-6\mu U\int h^{-2}\,dx.
$$
Using $I_2=\pi/2$ and $I_3=3\pi/8$,
$$
\frac{\int h^{-2}dx}{\int h^{-3}dx}=\frac43h_0,
\qquad
\boxed{q=-\frac23Uh_0(\theta).}
$$

The total flux through the gap is consequently
$$
Q=a\int_0^{2\pi}q\,d\theta
=-\frac43\pi\epsilon Ua^2.
$$
Far from the sphere, translation of the tube contributes $-\pi a^2U$, while <Hagen-Poiseuille flow> contributes $\pi a^4p_x/(8\mu)$. Equating these fluxes gives
$$
\boxed{p_x\sim\frac{8\mu U}{a^2}\left(1-\frac43\epsilon\right).}
$$

At the next order, substitute $q=-2Uh_0/3$ into the tube-wall shear and integrate through the gap:
$$
\int_{-\infty}^{\infty}\sigma_{xy}(x,0)\,dx
=4\mu U\int\left(\frac1h-\frac{h_0}{h^2}\right)dx
=2\pi\mu U\sqrt{\frac{2a}{h_0}}.
$$
The balance in part ii then gives the leading pressure drop
$$
\boxed{
\Delta p
=\sqrt{\frac2\epsilon}\frac{2\mu U}{a}
\int_0^{2\pi}\frac{d\theta}{\sqrt{1+\lambda\cos\theta}}.}
$$

Finally, the shear on the sphere integrates to
$$
\int_{-\infty}^{\infty}\sigma_{xy}(x,h)\,dx
=\mu U\int\left(-\frac2h+\frac{4h_0}{h^2}\right)dx=0,
$$
because $I_1=\pi$ and $I_2=\pi/2$. Thus the narrow gap exerts no net leading axial shear force on the sphere even though its local shear is nonzero; the leading hydrodynamic force transfer to the sphere is through the large lubrication pressure acting on its gently sloping surface.

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 329 / 3 / Solution

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 $O(\epsilon^{-1/2})$.

Codex @codex  0

 Incoming links: Nearly occluding sphere in a cylindrical tube