2011-23.md

course	course_year	question_number	tags	title	year
Fluid Dynamics	IB	23	IB 2011 Fluid Dynamics	Paper 4, Section II, D	2011

Show that an irrotational incompressible flow can be determined from a velocity potential $\phi$ that satisfies $\nabla^{2} \phi=0$.

Given that the general solution of $\nabla^{2} \phi=0$ in plane polar coordinates is

$$\phi=\sum_{n=-\infty}^{\infty}\left(a_{n} \cos n \theta+b_{n} \sin n \theta\right) r^{n}+c \log r+b \theta$$

obtain the corresponding fluid velocity.

A two-dimensional irrotational incompressible fluid flows past the circular disc with boundary $r=a$. For large $r$, the flow is uniform and parallel to the $x$-axis $(x=r \cos \theta)$. Write down the boundary conditions for large $r$ and on $r=a$, and hence derive the velocity potential in the form

$$\phi=U\left(r+\frac{a^{2}}{r}\right) \cos \theta+\frac{\kappa \theta}{2 \pi}$$

where $\kappa$ is the circulation.

Show that the acceleration of the fluid at $r=a$ and $\theta=0$ is

$$\frac{\kappa}{2 \pi a^{2}}\left(-\frac{\kappa}{2 \pi a} \mathbf{e}{r}-2 U \mathbf{e}{\theta}\right)$$