Past exam of the mathematics course of the University of Cambridge / 2023 / ib / Paper 1 / 16C (Fluid Dynamics)

Across the two vertical faces of a control rectangle, the mass fluxes arewith opposite signs, since $u$ is independent of $x$. There is no flux across the horizontal faces because the $y$-velocity is zero. The net mass flux is therefore zero, equivalently

The pressure force in the $x$-direction on a rectangle of dimensions $\delta x\times \delta y$ isThe shear stress is ${\tau }_{xy}=\mu u_y$, so the net viscous force isThere is no convective acceleration. Dividing the momentum balance by the area and taking the limit givesThe $y$-momentum balance gives $p_y=0$, so the pressure depends only on $x$.

For steady flow the momentum equation isApplying no slip, $u(0)=0$ and $u(h)=U$, gives the Couette-Poiseuille flow in a thin gap

The fluid shear stress isThe condition ${\tau }_{xy}(0)=0$ therefore gives

At $G=G_0$, the linear terms cancel andThe volume flux per unit width is

At the upper wall,This is the shear exerted by the plate on the fluid in the positive $x$-direction. By action and reaction, the flow exerts on the top plate the stress

The same result follows from a force balance on a rectangle of length $\delta x$ and full height $h$. The pressure force is $-G_{0}h\delta x{e}_x$, the bottom shear vanishes, and equilibrium requires the top plate to exert $G_{0}h\delta x{e}_x$ on the fluid. Thus the fluid exerts $-G_{0}h{e}_x=-2\mu U{e}_x/h$ on the plate. These results form the Couette-Poiseuille flow with a stress-free stationary wall.