Past exam of the viscous flow course of the University of Oxford / 2023 / 3 / c / Solution

Let the constant in the integrated Reynolds equation be $C$:Both endpoint pressures are zero, so the integral of $d{p}^{\prime }/d{x}^{\prime }$ on the real line vanishes. DefineThe substitution $x^{\prime }=\sqrt{2}\tan \theta$ givesThe second value follows by integrating the supplied identity for ${\cos }^4\theta$; the nonconstant cosine terms have zero integral over this interval. Hence $12C{I}_3-6I_2=0$, givingThe pressure gradient is nowDirect differentiation showsThe candidate tends to zero at both ends, so the additive constant is zero andTogether with the local Couette-Poiseuille flow profile, this solves the sliding-plate flow through a parabolic gap.

To restore dimensions, integrate $u=U{u}^{\prime }$ across $dz=h_{0}d{z}^{\prime }$. The signed dimensional volume flux per unit width, positive in the $+x$ direction, isThus the magnitude drawn in the direction of the moving plate is $2U{h}_0/3$. For a plate width $B$ large enough to neglect side-wall effects, the total volumetric flow rate is $-2BU{h}_0/3$; without a specified width the answer is per unit width.

The Couette flow and plane Poiseuille flow contributions to the dimensional volume flux per unit width are respectivelyThey point in the same direction precisely when $Q_P<0$, givingAt the endpoints the plane Poiseuille flow contribution is zero. Outside that interval it opposes the Couette flow contribution, while their sum remains the constant negative volume flux per unit width $-2U{h}_0/3$. Thus pressure assists the moving plate near the narrowest gap and drives a compensating return contribution farther away.