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\subsubsection{Steady Translation of a Cylinder with Circulation}

Work in frame moving steadily with the cylinder.
From section 4.2, uniform flow past a cylinder
with circulation $\kappa$ given by
\[ \phi = U\cos\theta \left( r + \frac{a^2}{r}
\right) + \frac{\kappa}{2\pi}\theta, \qquad
\text{and on $r = a$}, u_r = 0, u_\theta = -2U
\sin\theta + \frac{\kappa}{2\pi a} \]
(For $\kappa > 4\pi Ua$, there is no streamline
connecting the cylinder to $\infty$.) Use
expression for pressure in potential flow (section
4.3) to deduce
\[ -\half \rho \left( -2U \sin\theta +
\frac{\kappa}{2\pi a} \right)^2 + p(a, \theta) =
\half \rho U^2 + p_\infty \]
Fluid force on the cylinder (per unit length) is
$-\int_{r = a} p(a, \theta) \bf{n} \dd A$.
\[ = -\int \left[ p_\infty + \half \rho U^2 -
\half \rho \left( 4U^2 \sin^2 \theta -
\frac{2U\kappa}{\pi a} \sin\theta + \frac{\kappa^2}{4\pi^2
a^2} \right) \right] (\cos\theta, \sin\theta) a
\dd \theta \]
The only term that doesn't integrate to $0$ is the
$-\frac{2U\kappa}{\pi a} \sin\theta \times (0,
\sin\theta)$ term. So it equals
\[ = \left( 0, -\half \rho \cdot \frac{2U
\kappa}{\pi a} \half 2\pi a \right) = (0, -\rho
U\kappa) \]
i.e. we get a \emph{lift} force (perpendicular to
$\bf{U}$) -- down for $\kappa > 0$, up for $\kappa
< 0$. Recall $\kappa = 2\pi Ua$:
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/2c3e7522b75211ed.png}
\end{center}
Can show (Sheet 3 Question 1) that uniform
potential flow past \emph{any} 2D body with
circulation $\kappa$ produces lift $-\rho
U\kappa$. For example Aerofoils -- wings,
propellors, wind turbines.
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/5c0ab8f6b75211ed.png}
\end{center}
Circulation is generated by the condition that the
flow separates smoothly at the back (see handout).

\subsection{Some Unsteady Potential Flows}

\subsubsection{Free Oscillations in a $U$-tube Manometer}

\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/22e858a2b75311ed.png}
\end{center}
Start from rest, displacement $\zeta(0) \neq 0$.
$\implies$ irrotational $\implies$ potential.
Assume long tubes of equal areas, and short, wide
junction. At the bottom,
\[ \Delta \phi = \int_L^R \bf{u} \cdot \dd \bf{x}
\simeq 0 \]
because $\dd \bf{x}$ is small (short) and $\bf{u}$
is small (wide). WLOG can take $\phi = 0$ at the
base of both tubes. By mass conservation, $u$ is
uniform and equals $\dot{\zeta}$. RHS
\[ \phi = uz, \quad \pfrac{\phi}{t} = \dot{u} z
\quad \left. \pfrac{\phi}{t} \right|_{h + \zeta} =
\ddot{\zeta}(h + \zeta) \]
(Note, this is not $\dfrac{}{t} (\phi(h +
\zeta))$.) LHS
\[ \phi = -uz \quad \pfrac{\phi}{t} = -\dot{u} z
\quad \left. \pfrac{\phi}{t} \right|_{h - \zeta} =
-\ddot{\zeta}(h - \zeta) \]
Apply $\rho \pfrac{\phi}{t} + \half \rho u^2 + p +
\chi$ independent of position.
\[ p\ddot{z}(h + \cancel{\zeta}) + \cancel{\half
\phi \dot{\zeta}^2} + \cancel{p_a} + \rho
g(\cancel{h} + \zeta) = -\rho \ddot{\zeta}(h -
\cancel{\zeta}) + \cancel{\half \rho
\dot{\zeta}^2} + \cancel{p_a} + \rho g(\cancel{h}
- \zeta) \]
\[ \rho \ddot{\zeta}(2h) = -\rho g(2\zeta) \]
implies SHM with frequency $\sqrt{\frac{g}{h}} =
\omega$. But nonlinear if tubes have different
areas.

\subsubsection{Oscillation / Expansion / Collapse of a Bubble}

Air is $15,000$ times more compressible than water
(compare $\frac{1}{v} \pfrac{V}{\rho}$). Consider
incompressible flow of fluid outside a bubble of
radius $a(t)$.
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/105985fab75611ed.png}
\end{center}
Neglect gravity. Spherical symmetry implies radial
flow.
$u \propto \frac{1}{r^2}$ by mass conservation.
(implies $\bf{\omega} = \bf{0}$).
$u = \dot{a}$ on $r = a$ implies $\bf{u} =
\frac{a^2 \dot{a}}{r^2} \bf{e}_r = \nabla \phi$
with $\phi = -\frac{a^2 \dot{a}}{r}$ (source
flow).
\[ \pfrac{\phi}{t} = -\frac{a^2 \ddot{a} + 2a
\dot{a}^2}{r} \qquad \left. \pfrac{\phi}{t}
\right|_{r = a} = -a\ddot{a} - 2\dot{a}^2 \]
(Not $\dfrac{}{t} \phi(a(t), t)$). So using $\rho
\pfrac{\phi}{t} + \half \rho u^2 + p = f(t)$
($\chi = 0$) at $r = a$, $r = \infty$ we obtain
\[ \boxed{-\rho a\ddot{a} - \frac{3}{2} \rho
\dot{a}^2 = p_\infty - p(a(t), t)} \]
Multiply by $a^2 \dot{a}$ implies $\dfrac{}{t}
\left( \half \rho a^3 \dot{a}^2 \right) = a^2
\dot{a}(p(a) - p_\infty)$. Interpret as /
motivated by $\dot{KE}$ equals rate of working by
the pressures at $a$ and $\infty$. (Note $4\pi a^2
\dot{a} = 4\pi r^2 ur$). Can integrate if $p(a,
t)$ given as $F(a)$ ad $F(a)$ known from interior
of bubble. For example under water explosion $p_a
\gg p_\infty$, collapse of a void $p_a \ll
p_\infty$.