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## PHY454H1S Continuum Mechanics. Lecture 17: Impulsive flow. Boundary layers. Oscillatory driven flow. Taught by Prof. K. Das.

Posted by peeterjoot on March 16, 2012

[Click here for a PDF of this post with nicer formatting and figures if the post had any (especially if my latex to wordpress script has left FORMULA DOES NOT PARSE errors.)]

# Disclaimer.

Peeter’s lecture notes from class. May not be entirely coherent.

# Review. Impulsively started flow.

Were looking at flow driven by an impulse, a sudden motion of the plate, as in figure (\ref{fig:continuumL17:continuumL17Fig1})
\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.3\textheight]{continuumL17Fig1}
\caption{Impulsively driven time dependent fluid flow.}
\end{figure}

where the fluid at the origin is pushed so that it is given the velocity

\begin{aligned}u(0, t) = \left\{\begin{array}{l l}0 & \quad \mbox{forlatex t < 0} \\ U(t) & \quad \mbox{for $t \ge 0$} \\ \end{array}\right.\end{aligned} \hspace{\stretch{1}}(2.1)

where $U \rightarrow 0$ as $y \rightarrow \infty$.

Navier-Stokes takes the form

\begin{aligned}\frac{\partial {u}}{\partial {t}} = \nu \frac{\partial^2 {{u}}}{\partial {{y}}^2}.\end{aligned} \hspace{\stretch{1}}(2.2)

With a similarity variable

\begin{aligned}\eta = \frac{y}{2 \sqrt{\nu t}},\end{aligned} \hspace{\stretch{1}}(2.3)

and

\begin{aligned}u = U f(\eta),\end{aligned} \hspace{\stretch{1}}(2.4)

we found that we needed to solve

\begin{aligned}f'' + 2 \eta f' = 0\end{aligned} \hspace{\stretch{1}}(2.5)

where

\begin{aligned}f' = \frac{df}{d\eta}\end{aligned} \hspace{\stretch{1}}(2.6)

with solution

\begin{aligned}u(y, t) = U(1 - \text{erf}(\eta)).\end{aligned} \hspace{\stretch{1}}(2.7)

Here, we’ve used the error function

\begin{aligned}\text{erf}(\eta) = \frac{2}{\sqrt{\pi}} \int_0^\eta e^{-s^2} ds,\end{aligned} \hspace{\stretch{1}}(2.8)

as plotted in figure (\ref{fig:continuumL17:continuumL17Fig2})
\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.3\textheight]{continuumL17Fig2}
\caption{Error function.}
\end{figure}

# Boundary layers.

Let’s look at spacetime points which are constant in $\eta$

\begin{aligned}\frac{y_1}{2 \sqrt{\nu t_1}} = \frac{y_2}{2 \sqrt{\nu t_2}},\end{aligned} \hspace{\stretch{1}}(3.9)

so that the speed at $(y_1, t_1)$ equals the speed at $(y_2, t_2)$. This is illustrated in figure (\ref{fig:continuumL17:continuumL17Fig3})
\begin{figure}[htp]
\centering
\def\svgwidth{0.6\columnwidth}
\caption{Velocity profiles at different times.}
\end{figure}

## Universal behavior

Looking at a plot with different viscosities for position vs time scaled as $\sqrt{\nu t}$ as in figure (\ref{fig:continuumL17:continuumL17Fig4}) we see a sort of universal behavior

\begin{figure}[htp]
\centering
\def\svgwidth{0.6\columnwidth}
\caption{Universal behaviour.}
\end{figure}

Characterizing this we introduce the concept of boundary layer thickness

\begin{definition}
\emph{(Boundary layer thickness)}

The length scale over which viscosity is dominant. This is the viscous length scale.
\end{definition}

This is similar to what we have in the heat equation

\begin{aligned}\frac{\partial {T}}{\partial {t}} = \kappa \frac{\partial^2 {{T}}}{\partial {{y}}^2},\end{aligned} \hspace{\stretch{1}}(3.10)

where the time scale for the diffusion can be expressed as

\begin{aligned}[\kappa_t] = \frac{d^2}{\kappa}.\end{aligned} \hspace{\stretch{1}}(3.11)

We could consider a scenario such as a heated plate in a cavity of height $\delta$ as in figure (\ref{fig:continuumL17:continuumL17Fig5})
\begin{figure}[htp]
\centering
\def\svgwidth{0.5\columnwidth}
\caption{Characteristic distances in heat flow problems.}
\end{figure}

with a temperature $T$ on the bottom plate. We can ask how fast the heat propagates through the medium.

# Another worked problem.

Consider an oscillating plate, driving the motion of the fluid, as in figure (\ref{fig:continuumL17:continuumL17Fig6})
\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.3\textheight]{continuumL17Fig6}
\caption{Time dependent fluid motion due to oscillating plate.}
\end{figure}

\begin{aligned}U(t) = U_0 \cos \Omega t = \text{Real}\left( U_0 e^{i \Omega t} \right).\end{aligned} \hspace{\stretch{1}}(4.12)

(we are thinking here about the always oscillating case, and not an impulsive plate motion).

We write

\begin{aligned}u(y, t) = \text{Real}\left( f(y) e^{i \Omega t} \right)\end{aligned} \hspace{\stretch{1}}(4.13)

with substitution into

\begin{aligned}\frac{\partial {u}}{\partial {t}} = \nu \frac{\partial^2 {{u}}}{\partial {{y}}^2},\end{aligned} \hspace{\stretch{1}}(4.14)

we have

\begin{aligned}i \Omega f(y) e^{i \Omega t} = \nu f'' e^{i \Omega t}\end{aligned} \hspace{\stretch{1}}(4.15)

or

\begin{aligned}i \Omega f(y) = \nu f'' \end{aligned} \hspace{\stretch{1}}(4.16)

This is an equation of the form

\begin{aligned}f'' = m^2 f\end{aligned} \hspace{\stretch{1}}(4.17)

where

\begin{aligned}m^2 = \frac{i \Omega}{\nu}.\end{aligned} \hspace{\stretch{1}}(4.18)

or

\begin{aligned}m = \sqrt{\frac{i \Omega}{\nu}} = \lambda (1 + i),\end{aligned} \hspace{\stretch{1}}(4.19)

where

\begin{aligned}\lambda = \sqrt{\frac{\Omega}{2 \nu}}.\end{aligned} \hspace{\stretch{1}}(4.20)

check:

\begin{aligned}m^2 &= \frac{\Omega}{2 \nu} (i + 1)^2 \\ &= \frac{\Omega}{2 \nu} (i^2 + 1 + 2 i) \\ &= \frac{\Omega}{\nu} i\end{aligned}

Considering the boundary value constraints we have

\begin{aligned}f(y) = A e^{\lambda (1 + i) y}+ B e^{-\lambda (1 + i) y}\end{aligned} \hspace{\stretch{1}}(4.21)

Since $u(\infty, t) \rightarrow 0$ we must have

\begin{aligned}f(\infty) = 0,\end{aligned} \hspace{\stretch{1}}(4.22)

so we must kill off the exponentially increasing (albeit also oscillating) term by setting $A = 0$. Also, since

\begin{aligned}u(0, t) = U(t)\end{aligned} \hspace{\stretch{1}}(4.23)

we must have

\begin{aligned}f(0) = U_0\end{aligned} \hspace{\stretch{1}}(4.24)

or

\begin{aligned}B = U_0\end{aligned} \hspace{\stretch{1}}(4.25)

so

\begin{aligned}f(y) = U_0 e^{-\lambda (1 + i) y}\end{aligned} \hspace{\stretch{1}}(4.26)

and

\begin{aligned}u(y, t) = \text{Real}\left(U_0 e^{-\lambda y} e^{ -i (\lambda y - \Omega t) }\right)\end{aligned} \hspace{\stretch{1}}(4.27)

or

\begin{aligned}u(y, t) = U_0 e^{-\lambda y} \cos\left( -i (\lambda y - \Omega t) \right).\end{aligned} \hspace{\stretch{1}}(4.28)

This is a damped transverse wave function

\begin{aligned}u(y, t) = f(y - c t),\end{aligned} \hspace{\stretch{1}}(4.29)

where

\begin{aligned}c = \frac{\Omega}{\lambda},\end{aligned} \hspace{\stretch{1}}(4.30)

is the wave speed.

Since we have an exponential damping here, the flow of fluid will essentially be confined to a boundary layer, where after distance $y = n/\lambda$, the oscillation falls off as

\begin{aligned}\frac{1}{{e^n}}.\end{aligned} \hspace{\stretch{1}}(4.31)

We can find a nice illustration of such a flow in [1].

# Flow over static object.

Next time we’ll start considering fluid flow around a fixed object as in figure (\ref{fig:continuumL17:continuumL17Fig7})

\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.3\textheight]{continuumL17Fig7}
\caption{Fluid flow around fixed object.}
\end{figure}

# References

[1] Wikipedia. Stokes boundary layer — wikipedia, the free encyclopedia [online]. 2011. [Online; accessed 16-March-2012]. http://en.wikipedia.org/w/index.php?title=Stokes_boundary_layer&oldid=466077125.