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\subsection{Bernoulli's Equation for Steady Flow with Potential Forces}

Euler equation
\[ \rho \left( \pfrac{\bf{u}}{t} + (\bf{u} \cdot
\nabla)\bf{u} \right) = -\nabla p + \bf{f} \]
Steady $\equiv \pfrac{}{t} = 0$. Potential forces
$\implies \bf{f} = -\nabla \chi$. Use vector
identity
\[ \boxed{\bf{u} \times (\nabla \times \bf{u}) =
\nabla \left( \half u^2 \right) - (\bf{u} \cdot
\nabla )\bf{u}} \]
(check) (where $u = |\bf{u}|$).
Introduce the \emph{vorticity} $\bf{w} = \nabla
\times \bf{u}$.

\myskip
Euler equation reduces to
\begin{align*}
    \rho \left( 0 + \nabla \left( \half u^2
    \right)  - \bf{u} \times \bf{w} \right)
    &= -\nabla \rho - \nabla \chi \\
    \implies \nabla \left( \half \rho u^2 + p +
    \chi \right)
    &= \rho \bf{u} \times \bf{w}
    &&(\rho = \text{const}) \\
    \implies \bf{u} \cdot \nabla \left( \half \rho
    u^2 + p + \chi \right)
    &= 0
\end{align*}
$H = \half \rho u^2 + p + \chi$ is constant along
streamlines. Bernoulli (1738). So $H =
\text{const}$ implies $p$ low where $u$ is high
and vice versa.
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/7eb5052ca15011ed.png}
\end{center}
See Question 9 on Sheet 1 for interpretation on
conservation of energy.

\subsubsection*{Applications of Steady Bernoulli}

Emptying Tank
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/c8460dbaa15211ed.png}
\end{center}

\subsubsection*{Pitot Tube}

To measure air speed.
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/0b0273d2a15311ed.png}
\end{center}
Mount facing into the flow
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/156d0724a15311ed.png}
\end{center}
Bernoulli along the streamline
\begin{align*}
    \half \rho_{\text{air}} U^2 + p_0
    &= 0 + p_1
    \implies U \\
    &= \left[ \frac{2(p_1 -
    p_0)}{\rho_{\text{air}}} \right]^{1/2}
\end{align*}

\subsubsection*{Venturi Meter}

To measure flow rate in pipe without any moving
parts.
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/017b36a4a15411ed.png}
\end{center}
Assume steady flow, uniform across a cross-section
-- OK for gentle variation of $A$. Mass
conservation
\[ A_1 u_1 = A_2 u_2 = Q \]
Bernoulli:
\begin{align*}
    \half \rho u_1^2 + p_1
    &= \half \rho u_2^2 + p_2 \\
    \implies p_1 - p_2
    &= \half \rho u_1^2 \left(
    \frac{A_1^2}{A_2^2} - 1 \right)
    > 0
\end{align*}
Measure $h$ $\to$
\begin{align*}
    \rho gh
    &= p_1 - p_2 \\
    \implies
    &~~u_1 \\
    \implies Q
    &= \sqrt{2gh} \frac{A_1 A_2}{\sqrt{A_1^2 -
    A_2^2}}
\end{align*}
(check)

\begin{remark*}[Non-Examinable]
    Cannot use Bernoulli for sudden enlargement of
    pipe. See
    \begin{center}
        \includegraphics[width=0.6\linewidth]
        {images/8186e8aca15411ed.png}
    \end{center}
    Exercise: Assume that $p_{\text{jet}}
    \simeq p_{\text{still}}$ (because no sideways
    acceleration) and $p_{\text{jet}} \simeq p_0$.
    Apply the momentum integral equation to
    \begin{center}
        \includegraphics[width=0.6\linewidth]
        {images/e98622faa15511ed.png}
    \end{center}
    $\bf{f} = 0$, $\dfrac{}{t} \left( \int \rho
    \bf{u} \dd V \right)  \simeq 0$ on average.
    Show
    \[ p_1 = p_0 + \rho u_1^2 \left(
    \frac{A_1}{A_0} - 1 \right) \left(
    \frac{A_1}{A_0} \right) \]
\end{remark*}

\subsubsection*{Water jet hitting an oblique wall}

Two-dimensional version (neglect gravity). Assume
steady.
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/88627590a15611ed.png}
\end{center}
Suppose that the small angle between the two
sections is $\beta$.
Bernoulli on surface streamline where $p = p_a$,
constant implies speed is constant $=U$ along this
streamline.
\myskip
So far from impact where flow is uniform, and $p =
p_a$, (vorticity zero by section 2.5) have $u = U$
again. Mass conservation
\[ \implies aU = a_1 +
a_2 U \tag{1} \]