Past exam of the mathematics course of the University of Cambridge / 2015 / ii / Paper 2 / 35E (Fluid Dynamics II)

Near the closest point, in $X=x-Ut$, the gap is $h(X)\simeq h_0+X^2/(2a)$. The horizontal length scale is $\ell =\sqrt{2a{h}_0}$, up to an inessential constant. Thus the aspect ratio $h_0/\ell \ll 1$ follows from $h_0\ll a$. The lubrication approximation assumes an incompressible Newtonian fluid, no slip on the surfaces, slow variation of the gap, and negligible inertia relative to transverse viscous stress:This is the reduced Reynolds-number condition; it is weaker than requiring a Reynolds number based on $\ell$ to be small. In the cylinder frame the geometry and flow are steady. Pressure is uniform across the narrow gap to leading order, and equal ambient pressures far to either side close the model. The calculation assumes a full liquid film; cavitation would change that pressure condition.

Reflection about the closest point leaves the gap geometry unchanged and reverses the direction of translation. The linear Stokes equations make the excess pressure change sign when the imposed velocity changes sign. Combining reflection with this linearity shows that pressure is odd in $X$. Its leading normal force on the stationary plane therefore cancels between the two sides, giving ${\mathbf{e}}_y\cdot \mathbf{F}=0$ in the lubrication limit. Normal viscous stresses are of smaller order in the aspect ratio. This uses symmetry only, rather than a pressure calculation.

In the laboratory frame the lower wall has velocity zero and the cylinder surface has velocity $U$. Solving $\mu u_{yy}=p_X$ with these no-slip conditions givesThe relative velocity in the cylinder frame is $u-U$. Its flux is constant by incompressibility:Equal pressures at the two infinities require $\int p_{X}dX=0$. Put $I_j=\int h^{-j}dX$; then $q=-(U/2)I_2/I_3$. With $X=\ell s$ and $h=h_0(1+s^2)$, the given integrals yield $I_2/I_3=4h_0/3$. Hence $q=-2U{h}_0/3$ andIndeed $p=2\mu UX/h^2$ has this derivative and tends to ambient pressure at both ends. The distinction between the constant comoving flux and the varying laboratory flux is necessary because the gap translates.

The fluid's tangential traction on the lower stationary surface is $\tau =\mu u_y{|}_{y=0}$ in the positive $x$ direction. Using the pressure gradient found above givesThe drag on a plane beneath a translating cylinder follows by integrating along the wall:per unit cylinder-axis length. This is the force on the lower surface, so its direction follows the cylinder's translation.