Peeter Joot's (OLD) Blog.

Math, physics, perl, and programming obscurity.

 Adam C Scott on avoiding gdb signal noise… Ken on Scotiabank iTrade RESP …… Alan Ball on Oops. Fixing a drill hole in P… Peeter Joot's B… on Stokes theorem in Geometric… Exploring Stokes The… on Stokes theorem in Geometric…

• 300,689

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.