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\newpage
\section{Dynamics of Inviscid Flow}

\subsection{Surface and Volume Forces}

Two types of force act on a fluid (liquid / gases)
\begin{enumerate}[(i)]
    \item Those proportional to the volume, like gravity
    \item Those proportional to the area, like
        \emph{pressure} or \emph{viscous stress}
        (friction between moving fluid and
        neighbouring material, whether a boundary
        or other fluid).
\end{enumerate}

\subsubsection*{(i) Volume / body forces}

We denote the force on a small volume element
$\bf{f}(\bf{x}, t) \delta V$. $\bf{f}$ is often
\emph{conservative} with potential energy / unit
volume $\chi$ and $\bf{f} = -\nabla \chi$. The
most common case for us $\bf{f} \delta V = \rho
\bf{g} \delta V$ with $\chi = \rho gz$ if $\bf{g}
= (0, 0, -g)$.

\subsubsection*{(ii) Surface Forces}

Consider a small element of area $\bf{n} \delta A$
\begin{center}
    \includegraphics[width=0.2\linewidth]
    {images/ebb5142e9d6211ed.png}
\end{center}
Denote the surface force exerted \emph{by the $+$
side on the $-$ side} by $\bf{\tau}(\bf{x}, t;
\bf{n}) \delta A$ where $\bf{\tau}$ is the
\emph{stress} acting on the area element. (A
stress is a force per unit area). Note, the stress
$\bf{\tau}$ depends on the orientation $\bf{n}$.
For example, by Newton III, the surface force
exerted by the $-$ side on the $+$ side is
\[ -\bf{\tau}(\bf{x}, t; \bf{n}) =
\bf{\tau}(\bf{x}, t, -\bf{n}) \]
We defer any discussion of viscous stresses
(friction) to Chapter 3.

\myskip
In many phenomena, the viscous stresses are
negligible and the fluid behaves as if it is
\emph{inviscid} (frictionless). For example it
takes over an hour to slow down a $10\mathsf{cm}$
layer of water, $10^{-3}$ effect when pouring tea.

\myskip
For \emph{inviscid (frictionless) fluids}, the
stress $\bf{\tau}$ acting across $\bf{n}\delta A$
has no tangential component and has magnitude
independent of the orientation:
\[ \bf{\tau}(\bf{x}, t; \bf{n}) = -p(\bf{x},
t)\bf{n} \]
where $p$ is the pressure. Note the $-$ sign! The
$+$ side pushes on the $-$ side in the direction
$-\bf{n}$:
\begin{center}
    \includegraphics[width=0.3\linewidth]
    {images/b114cd269d6411ed.png}
\end{center}

\subsection{The Euler Momentum Equation}

(Inviscid for the rest of the section)
Consider an arbitrary volume $V$, fixed in space,
bounded by surface $\delta V$ with an outward
normal $\bf{n}$. The momentum $\int_V p\bf{u} \dd
V$ inside $V$ can change due to
\begin{enumerate}
    \item Flow of momentum across the boundary
        $\partial V$.
    \item Volume / body forces.
    \item Surface forces. 
        \begin{center}
            \includegraphics[width=0.3\linewidth]
            {images/faca80b49d6411ed.png}
        \end{center}
        Volume out across $\delta A$ in time
        $\delta t$ is $\bf{n} \delta A \cdot
        \bf{u} \delta t$, so momentum out across
        $\delta A$ in time $\delta t$ is $\rho
        \bf{u} (\bf{u} \cdot \bf{n} \delta A
        \delta t)$. Hence
        \[ \boxed{\dfrac{}{t} \int_V \rho \bf{u}
        \dd V = -\int_{\partial V} \rho
        \bf{u}(\bf{u} \cdot \bf{n}) \dd A +
        \int_{\partial V} (-p \bf{n}) \dd A +
        \int_V \bf{f} \dd V} \]
        (Momentum integral equation). $\rho u_i
        u_j$ is the momentum flux (tensor). Or in
        components:
        \[ \dfrac{}{t} \int_V \rho u_i \dd V =
        -\int_{\partial V} (\rho u_i u_j) n_j \dd
        A - \int_{\partial A} p n_i \dd A + \int
        f_i \dd V \]
        \[ \int_V \pfrac{}{t} (\rho u_i) \dd V =
        -\int_V \pfrac{}{x_j} (\rho u_i u_j) \dd V -
        \int_V \pfrac{p}{x_i} \dd V + \int_V f_i
        \dd V \]
        Since $V$ is arbitrary,
        \[ \pfrac{}{t} (\rho u_i) + \pfrac{}{x_j}
        (\rho u_i u_j) = -\pfrac{p}{x_i} + f_i \]
        \[ LHS = u_i \left( \pfrac{\rho}{t} +
        \pfrac{}{x_j} (\rho u_j) \right) + \rho
        \pfrac{u_i}{t} + \rho u_j \pfrac{u_i}{x_j} \]
        But $\pfrac{\rho}{t} + \pfrac{}{x_j} (\rho
        u_j) = 0$ by mass conservation. So
        \[ LHS = \rho \left( \pfrac{}{t} + \bf{u} \cdot
        \nabla \right) u_i \]
        Therefore
        \[ \boxed{\rho \Dfrac{\bf{u}}{t} = -\nabla p +
        \bf{f}} \]
        Euler momentum equation.
\end{enumerate}
A fluid accelerates due to the difference in
pressure on either side and to the volume force.

\subsubsection*{Dynamic boundary condition}

The stress $\bf{\tau}$ exerted on the fluid by the
boundary is $p\bf{n}$ (the stress exerted on the
boundary by the fluid is $p\bf{n}$).
\begin{center}
    \includegraphics[width=0.2\linewidth]
    {images/dd2e06b89d6711ed.png}
\end{center}

\subsubsection*{Application of momentum integral
equation}

Bent hose pipe:
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/fe0b56ba9d6711ed.png}
\end{center}
Assume steady uniform flow $U$ in and out with
same cross-sectional area $A$. Neglect gravity.
($\bf{f} = \bf{0}$).
\[ \int_{\text{walls}} + \int_{\text{ends}} [\rho
\bf{u} (\bf{u} \cdot \bf{n}) + p \bf{n}] \dd A = 0 \]
by steady momentum integral equation.
\begin{align*}
    \int_{\text{walls}} [\rho \bf{u} (\bf{u} \cdot
    \bf{n}) + p\bf{n}] \dd A
    &= \int_{\text{walls}} p\bf{n} \dd A
    &&(\text{$\bf{u} \cdot \bf{n} = 0$ by the
    kinematic boundary condition}) \\
    &= \text{Force by the fluid on the pipe} \\
    \int_{\text{ends}} [\rho \bf{u} (\bf{u} \cdot
    \bf{n}) + p\bf{n}] \dd A
    &= A(p_1 \bf{n}_1 + \rho(-U \bf{n}_1) (-U) +
    p_2 \bf{n}_2 + \rho(U \bf{n}_2)U)
\end{align*}
$p_1 = p_2$ (see next section).
\[ \text{Force on pipe} = -A (p + pU^2)(\bf{n}_1 +
\bf{n}_2) \]
\begin{center}
    \includegraphics[width=0.3\linewidth]
    {images/0f074c709d6911ed.png}
\end{center}
$pU^2$ contribution comes from charge in momentum
flux: momentum in goes to momentum out due to
force from wall.