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1-8Rayleigh.tex
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\section{Rayleigh's Problem - solid wall as a source of vorticity }
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Owing to terms representing convective inertia, the Navier-Stokes equations are highly nonlinear. Explicit solutions are usually limited to a class of problems where inertia is identically zero. This happens when the flow is unidirectional and uniform. Flow quantities depend only on a transverse coordinate. We discuss one such example with a view to examining the role of viscosity.
Consider a two-dimensional flow in the upper half plane of $(x,y)$ bounded below by a rigid plate coinciding with the $x$ axis. At $t=0$ the plate suddenly moves in the tangential direction at constant velocity $U$. Find the development of the fluid motion in the region $y>0$.
Because the plate is infinite in extent, the flow must be uniform in $x$, i.e.
$\f{\p}{\p x} = 0 $. It follows from continuity that
\[ \f{\p v}{\p y} = 0, ~~~y>0 \]
implying that $v =$ constant in $y$. Since
$v(0, t) = 0$ , $v \equiv 0$ for all $y$.
Therefore, the only unknown is $u(y,t)$ which must satisfy the momentum equation,
\be \f{\p u}{\p t} = \nu \, \f{\p^2 u}{\p y^2} \label{Eq:(7.1)}\ee
where
\be \nu=\f{\mu}{\rho}\ee
denotes the kinematic viscosity. The boundary conditions are :
\be u = U, \,\, y = 0, \,\, t > 0 ; ~~~\mbox{no slip}\ee
\be u = 0, \, \, y \sim \infty, \, \, t > 0 \ee
The initial condition is
\be u = 0, \, \, t = 0, \, \, \forall y \ee
Mathematically this is the heat conduction problem for a semi-infinite rod. The solution is well-known (Carlaw \& Jeager, {\em Conduction of Heat in Solids} or Mei, {\em Mathematical Analysis in Engineering}) ,
\be u = U \lp 1 - \erf \f{y}{2\sqrt{\nu t}} \rp \label{Eq:(7.2)} \ee
where \be \erf \, \zeta = \f{2}{\sqrt{\pi}} \int^\zeta_0 \, e^{-\lambda^2} \, d\lambda . \label{Eq:(7.3)} \ee
is the error function. As shown in Figure \ref{fig:F1-8-1},
\begin{figure}
\begin{center} \includegraphics[scale=0.90]{F1-8-1.EPS}
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\caption{Velocity profile due to impulsive motion of $x$-plane}
\label{fig:F1-8-1}
\end{figure}
fluid
momentum is diffused away from the plane $y=0$. The region affected by viscosity (the boundary layer) grows in time as $\delta \sim \sqrt{\nu t}$. This observation can be confirmed, indeed anticipated, merely by a scaling argument based on the momentum equation (\ref{Eq:(7.1)}) without solving it. Let $U, t, \delta$ denote the scales of velocity, time and region of viscosity respectively. For viscosity to be important, the two terms in (\ref{Eq:(7.1)}) must be comparable in order of magnitude, i.e.,
\[ \f{U}{t}\sim \nu \f{U}{\delta^2}\]
It follows that
\[ \delta \sim \sqrt{\nu t}\]
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Let us use this simple example to study the role of vorticity \be \vec{\zeta} = \lp\f{\p v}{\p x}- \f{\p u}{\p y}\rp \vec{k} \label{Eq:(7.4)} \ee
In this problem there is only one vorticity component,
\be \zeta_3 = - \f{\p u}{\p y} = U \, \f{\p}{\p y} \, \erf \, \f{y}{2\sqrt{\nu t}} = \f{2U}{\sqrt{4 \pi \nu t}} \, e^{-y^2/4\nu t} . \label{Eq:(7.5)} \ee
which is just the velocity shear.
Mathematically (\ref{Eq:(7.5)}) is the solution to the diffusion equation
\be \f{\p \zeta}{\p t} = \nu \, \f{\p^2 \zeta}{\p y^2} . \label{Eq:(7.6)} \ee
which follows from (\ref{Eq:(7.1)}), and the initial condition that there is a plane source of at $y=0$:
\be \zeta_3(y,0) = 2U\delta(y).\label{initial-vorticity}\ee
Thus vorticity is diffused away from the solid wall which acts as a voriticity source. Note that the shear stress at the wall is
\be \tau_{xy}(0, t) = \mu \left. \f{\p u}{\p y}\right|_{y=0} = - \rho U \, \sqrt{\f{\nu}{\pi t}} . \label{Eq:(7.7)} \ee
which is initially infinite but decays with time.
Why is the wall a source of vorticity? Just after the plane started to move there is a velocity discontinuity at $y=0+$. The associated velocity gradient is $\p u/\p y= -U\delta(y) $ hence the vorticity is a highly concentrated function of $y$: $-\p u/\p y= U\delta(y) $. Furthermore the half space ($00$ and half to $y<0$. Thus, the solid wall is the source of vorticity.
The reader can verify the solution (\ref{Eq:(7.5)}) by assuming a similarity form,
\be\zeta_3(y,t) = \f{C}{\sqrt{t}}f\lp \f{y}{\sqrt{t}}\rp \ee
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