Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 3 / 38C (Fluid Dynamics II)

Let $H$ be a representative positive gap thickness and setthe larger-order horizontal speed after including the squeeze-induced flux. The lubrication-limit scaling for a moving thin gap requiresEquivalently, the inertial condition containsThe gap must also remain open, $h(x)>0$. These conditions make streamwise derivatives small relative to transverse derivatives, keep the pressure uniform across the gap to leading order, and reduce the Navier-Stokes equation to Stokes flow locally.

The leading streamwise momentum equation in lubrication theory iswith $p=p(x)$. The no-slip boundary condition isTwice integrating in $y$ gives the local Couette-Poiseuille flow in a thin gapIts volume flux per unit span is

Integrating the incompressible continuity equation $u_x+v_y=0$ across the moving gap, with $v(x,0)=0$ and $v(x,h)=-V$, gives the Reynolds lubrication equationHencefor a constant $C$, and the flux formula from part (b) givesDefine the spatial meanBecause both ends meet fluid at pressure $p_0$, the pressure recovery condition in lubrication flow isIt determinesTherefore the requested pressure gradient in a translating and squeezing finite gap is

When $U=0$, putThe pressure gradient from part (c) reduces toBecause $h^{-3}>0$, $x_{\ast }$ is a weighted mean of points in $[0,L]$ and therefore belongs to $[0,L]$; it lies strictly inside unless the weight is singularly concentrated at an endpoint. Thus $p_x(x_{\ast })=0$, proving that the pressure gradient vanishes somewhere between the ends.

The leading shear stress exerted by the fluid on the lower surface isThe horizontal force per unit span is ${\int }_0^L\tau dx$. Substituting the pressure gradient from part (c) givesUsing the value of $C$ and setting this force to zero yieldsConsequently the zero-shear lower wall in a translating and squeezing finite gap condition is