Moving-boundary lubrication flux

The flux of the local Couette-Poiseuille profile is

For a sphere falling with speed $U$ through a tube of radius $a$, let $Q$ be the downward laboratory volume flux. In the sphere frame the wall moves upwards and the narrow annulus carries upward flux per unit circumference $q=(\pi a^{2}U-Q)/(2\pi a)$. Couette-Poiseuille flow in a thin gap gives $q=Uh/2-h^3{p}_z/(12\mu )$. Thus the pressure below minus above the sphere is $\Delta P=6\mu (2q{I}_3-U{I}_2)$, using lubrication resistance integrals for a nearly occluding sphere. A nearly piston-like laboratory flux can coexist with a nonzero relative gap flux.

For a full Newtonian liquid film in the lubrication approximation, a cylinder translating parallel to a stationary plane with minimum gap $h_0\ll a$ exerts tangential force $2\pi \mu U\sqrt{2a/h_0}$ per unit axial length on the plane. The leading normal force vanishes by the odd symmetry of excess pressure.

Incompressible mass conservation in a moving gap givesCombined with the local Couette-Poiseuille flux, this is the Reynolds lubrication equation.

For a thin gap $h$ with tangential wall velocities ${\mathbf{v}}_0,{\mathbf{v}}_h$, lubrication theory and the no-slip boundary condition give the integrated volume fluxConservation of mass gives $h_t+{\nabla }_{\parallel }\cdot \mathbf{q}=0$. This is the Reynolds equation; it determines the lubrication pressure from wall motion and gap geometry.

A sphere of radius $a$ at gap $aϵ$ translates at $U{\mathbf{e}}_x$ and rotates at $\omega {\mathbf{e}}_y$ above a stationary wall in a right-handed coordinate system. At leading order its gap is $h=aϵ+r^2/(2a)$. In its translating frame the lower and upper tangential velocities are $-U{\mathbf{e}}_x$ and $-a\omega {\mathbf{e}}_x$. The Reynolds lubrication equation givesIntegrating the upper-boundary traction, including pressure on the sloping surface, and its moment about the sphere centre gives the sphere-on-fluid force and torqueThe off-diagonal symmetry is required by the Lorentz reciprocal theorem for Stokes flow. A torque-free sphere therefore has $a\omega =U/4$ and leading drag $3\pi \mu aU\log (1/ϵ)$. Compare the rigid-wall terms of Bertin et al., equations (4.4)–(4.5); those forces act on the sphere and have the opposite sign.

For a fixed outer radius $b\ll a$ and $h=aϵ+r^2/(2a)$,Indeed, for $a\sqrt{ϵ}\ll r\ll b$ the integrand is $(2a{)}^n/r$; integrating from a scale proportional to $a\sqrt{ϵ}$ to $b$ produces half of $\log (1/ϵ)$. The inner region contributes a bounded term after rescaling $r=a\sqrt{ϵ}s$. In particular, $\int h^{-1}dxdy$ and $\int x^2/(a{h}^2)dxdy$ have the same leading coefficient $2\pi a$.

For lower speed $U$, stationary upper tangential speed, upper normal speed $-V$, and equal endpoint pressures, define $⟨f⟩=L^{-1}{\int }_0^{L}f(x)dx$. Thenwhere

The horizontal force on the lower wall vanishes when

For a cylinder of radius $a$ in a coaxial container with gap $h\ll a$, global displacement forces reverse flow through the gap and createsThe pressure drag exceeds side shear by order $a/h$. Balancing it with excess weight gives $U\simeq \Delta \rho ,g{h}^3/(6\mu a)$.