Past exam of the mathematics course of the University of Cambridge / 2016 / ib / Paper 2 / 7C (Fluid Dynamics)

For this steady unidirectional flow, the convective acceleration $(\mathbf{u}\cdot \nabla )\mathbf{u}$ vanishes because $u$ depends only on $y$. The axial Navier-Stokes equation reduces to $0=-p_x+\mu u^{\prime \prime }$, with $p_x=-G$. The no-slip boundary conditions therefore giveThis is a superposition of Couette flow and pressure-driven Poiseuille flow.

Integrate the axial ordinary differential equation twice: $u=-G{y}^2/(2\mu )+C_{1}y+C_2$. The no-slip boundary condition at zero gives $C_2=0$, and the one at $h$ gives $C_1=U/h+Gh/(2\mu )$. HenceThe first term is Couette flow; the second is Poiseuille flow driven in positive $x$ when $G>0$. Their addition is valid because the reduced Stokes equation is linear.

The bottom-wall shear stress vanishes exactly when $u^{\prime }(0)=0$. Differentiating the velocity profile givesWith this value the velocity profile becomes $u(y)=-G{y}^2/(2\mu )$, so the volume flux per unit transverse width isFor $G>0$, the plate must move against the pressure-driven direction, and the resulting signed volume flux is negative. Zero bottom-wall shear does not mean zero flow.