 Solution

In the torus frame the two planes translate with velocity $U{\mathbf{e}}_x$. Away from the $O(a)$ neighborhood of the torus, the depth-averaged Hele–Shaw flow between planes separated by $2a$ isNegligible leakage imposes ${\overline{u}}_r=0$ at $r=R$. The harmonic pressure that decays at infinity in the exterior and the regular harmonic pressure in the interior are thereforeup to a common constant. The interior velocity is zero, while the exterior flow is the uniform stream diverted around a circular obstacle. In plan view the inside has high pressure on the $+x$ side and low pressure on the $-x$ side; the immediately adjacent exterior has the opposite signs, producing the pressure jump across the torus.

The jump at $r=R$ isIntegrating it over the projected vertical area $2aRd\theta$ gives the global pressure resistanceThere are two narrow gaps, so their local resistance is twice the one-plane result from part b:Consequently the local gap resistance dominates when $a\ll R\ll a{\epsilon }^{-1/2}$, whereas the global Hele–Shaw pressure resistance dominates when

To interpret the upper bound, the pressure jump has scale $\Delta p\sim \mu UR/a^2$. Each narrow gap has thickness $a\epsilon$ and streamwise lubrication length $a\sqrt{\epsilon }$. Its pressure-driven leakage flux per unit centreline length therefore scales asThe blocked Hele–Shaw flux has scale $Ua$. Leakage is negligible precisely when $UR{\epsilon }^{5/2}\ll Ua$, or