Past exam of the mathematics course of the University of Cambridge / 2020 / ib / Paper 1 / 17C (Fluid Dynamics)

For a velocity potential, $\mathbf{u}=\nabla \varphi$. The incompressible flow condition $\nabla \cdot \mathbf{u}=0$ therefore becomes Laplace equation ${\nabla }^2\varphi =0$. The Laplace equation in polar coordinates givesso positivity of $\gamma$ and $\lambda$ requiresThe velocity components areThe no-penetration boundary condition on both walls says $u_{\theta }=0$ at $\theta =0$ and $\theta =\alpha$, hence $\lambda \alpha =n\pi$. If $n\ge 2$, $\sin (n\pi \theta /\alpha )$ changes sign inside the wedge. The stated sign condition therefore selects $n=1$, and

A streamline satisfiesIntegration givesThe radius is smallest on the wedge bisector $\theta =\alpha /2$, where $\sin (\gamma \theta )=1$. Labelling that radius by $r_{\min }$ therefore givesEvery streamline arrives from infinity along one wall, reaches its closest point on the bisector, and returns to infinity along the other wall:

Using $\gamma =\lambda$ and the Pythagorean trigonometric identity,so the speed is independent of $\theta$. Since $\gamma =\pi /\alpha >1$,and the speed tends to zero at the vertex. The steady irrotational Bernoulli equation givesThus the fluid pressure is also independent of $\theta$, withIn particular, the pressure difference between the vertex and $r_0$ is