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\subsection{Plane Couette Flow and Viscosity}

\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/513aa1e2a86211ed.png}
\end{center}
Steady flow between parallel plates driven only by
the motion of the top plate. Find experimatelly
for simple (Newtonian) fluids -- air, water, oild,
syrup, glycerol -- that
\begin{enumerate}[(i)]
    \item Fluid velocity is $U$ on the top plate
        and 0 on the bottom plate.
    \item The fluid velocity varies linearly
        between the top and bottom plates.
    \item Tangential force per unit area $\tau_3$
        is\ldots
\end{enumerate}
TODO
By considering a slab of fluid $a < y < b$ as a
Couette Flow, for example
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/284242d2a86611ed.png}
\end{center}
deduce: The tangential shear stress exerted by the
$+$ side on the $-$ side of a surface $y =
\text{const}$ is
\[\boxed{\tau_3 = \mu \pfrac{u}{n}} \]
For example, the fluid exerts a shear stress
$\mu \pfrac{u}{y}$ on the bottom plate and
$-\mu \pfrac{u}{y}$ on the top plate ($\bf{n}$
into fluid)

\subsection{2D Parallel Viscous Flow}

Steady case with $\bf{f} = \bf{0}$
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/fd57eef0a86511ed.png}
\end{center}
Steady $\implies$ no acceleration $\implies$
forces exerted by the surrounding fluid on the
dashed rectangle must balance.

\myskip
In the $x$-direction
\[ p(x) \delta y - p(x + \delta x) \delta y +
\mu \left. \pfrac{u}{y} \right|_{y + \delta y}
\delta x - \mu \left. \pfrac{u}{y} \right|_y \delta x
= 0 \]
Divide by $\delta x \delta y$ and take the limit:
\[ -\pfrac{p}{x} + \mu \pfrac[2]{u}{y} = 0 \]
and similarly in the $y$-direction
\[ -\pfrac{p}{y} = 0 \]

\subsubsection*{General Case}

For 2D unsteady parallel viscous flow $\bf{u} =
(u(y, t), 0, 0)$ with body force $\bf{f} = (f_x,
f_y, 0)$ (Sheet 2 Question 1)
\begin{equation*}
    \boxed{
    \begin{aligned}
        p \pfrac{u}{t}
        &= -\pfrac{p}{x} + \mu \pfrac[2]{u}{y}
        + f_x \\
        0
        &= -\pfrac{p}{y} + f_y
    \end{aligned}
    }
\end{equation*}
Note: $\bf{u}(u(y, t), 0, 0) \implies \nabla \cdot
\bf{u} = 0$ and $(\bf{u} \cdot \nabla) \bf{u} =
0$.

\subsubsection*{No-slip boundary condition}

It has been verified experimentally (down to
molecular scales) that at a rigid boundary viscous
fluids satisfy a \emph{no slip boundary
condition}:
\begin{center}
    the tangential component of the fluid must
    equal that of the boundary.
\end{center}
Combined with the mass conserving kinematic
boundary condition $\bf{u} = \bf{U}$ (contrast
with $\bf{u} \cdot \bf{n} = \bf{U} \cdot \bf{n}$
only for inviscid fluids).

\begin{example*}[Poiseuille Flow in a Channel]
    Steady flow in a channel driven by a pressure
    gradient.
    \begin{center}
        \includegraphics[width=0.6\linewidth]
        {images/f40bf32ca86611ed.png}
    \end{center}
    \[ \pfrac{p}{y} = 0 \implies p = p(x) \]
    Steady, so
    \[ \mu \pfrac[2]{u}{y} = \dfrac{p}{x} = -G
    \qquad \text{const} \]
    \[ u(0) = u(h) = 0 \]
    \[ \implies u = \frac{G}{2\mu}y(h - y) \]
    Flux (per unit width into page), $q = \int_0^h
    u \dd y = \frac{Gh^3}{12 \mu}$.
    \begin{center}
        \includegraphics[width=0.2\linewidth]
        {images/7c69c67ca86711ed.png}
    \end{center}
    Overall force
    balance $Gh - \mu \left. \pfrac{u}{y}
    \right|_0 + \mu \left. \pfrac{u}{y}
    \right|_h = 0$. ($Gh$ is pressure gradient,
    and the next two terms are the shear stress $R
    \to L$).
\end{example*}

\subsubsection*{Example Viscous Flow Down a Slope}

\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/0fe97cdea86911ed.png}
\end{center}
Assume $p_{\text{air}}$ is uniform (because
$p_{\text{air}} \ll p_{\text{syrup}}$). Shear
stress exerted by the air is negligible (because
$\mu_{\text{air}} \ll \mu_{\text{syrup}}$). Take
coordinates parallel and perpendicular to the
plane:
\[ \bf{f} = (\rho g \sin \alpha, -\rho g
\cos\alpha) \]
perpendicular:
\[ \pfrac{p}{y} = \rho g \cos\alpha, \quad p = p_a
\text{   at y = h} \implies p = p_a + \rho g
\cos\alpha (h - y) \implies \pfrac{p}{x} = 0 \]
Parallel:
\begin{align*}
    \mu \pfrac[2]{u}{y}
    &= -\rho g \sin\alpha \\
    u(0)
    &= 0
    &&\text{(no slip)} \\
    \mu \left. \pfrac{u}{y} \right|_h
    &= 0
    &&(\text{no shear stress from / on air}) \\
    \implies u
    &= \frac{\rho g \sin\alpha}{2\mu} y(2h - y)
\end{align*}
\begin{center}
    \includegraphics[width=0.2\linewidth]
    {images/1d827f62a86911ed.png}
\end{center}