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PHY454H1S Continuum Mechanics. Lecture 15: More on surface tension and Reynold’s number. Taught by Prof. K. Das.

Posted by peeterjoot on March 10, 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. Surface tension clarifications

For a surface like figure (\ref{fig:continuumL15:continuumL15Fig1})
\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{continuumL15Fig1}
\caption{Vapor liquid interface.}
\end{figure}

we have a discontinuous jump in density. We will have to consider three boundary value constraints

\begin{enumerate}
\item Mass balance. This is the continuity equation.
\item Momentum balance. This is the Navier-Stokes equation.
\item Energy balance. This is the heat equation.
\end{enumerate}

We have not yet discussed the heat equation, but this is required for non-isothermal problems.

The boundary condition at the interface is given by the stress balance. Denoting the difference in the traction vector at the interface by

\begin{aligned}[\mathbf{t}]_1^2 = \mathbf{t}_2 - \mathbf{t}_1 = -\frac{\sigma}{2 R} \hat{\mathbf{n}} - \boldsymbol{\nabla}_I \sigma\end{aligned} \hspace{\stretch{1}}(2.1)

Here the gradient is in the tangential direction of the surface as in figure (\ref{fig:continuumL15:continuumL15Fig2})
\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{continuumL15Fig2}
\caption{normal and tangent vectors on a curve.}
\end{figure}

In the normal direction

\begin{aligned}[\mathbf{t}]_1^2 \cdot \hat{\mathbf{n}}&= (\mathbf{t}_2 - \mathbf{t}_1) \cdot \hat{\mathbf{n}} \\ &= -\frac{\sigma}{2 R} \end{aligned}

With the traction vector having the value

\begin{aligned}\mathbf{t} &= \mathbf{e}_i T_{ij} n_j \\ &= \mathbf{e}_i \left( -p \delta_{ij} + \mu \left( \frac{\partial {u_i}}{\partial {x_j}}+\frac{\partial {u_j}}{\partial {x_i}}\right)\right)n_j\end{aligned}

We have in the normal direction

\begin{aligned}\mathbf{t} \cdot \mathbf{n} =n_i \left( -p \delta_{ij} + \mu \left( \frac{\partial {u_i}}{\partial {x_j}}+\frac{\partial {u_j}}{\partial {x_i}}\right)\right) n_j\end{aligned} \hspace{\stretch{1}}(2.2)

With \mathbf{u} = 0 on the surface, and n_i \delta_{ij} n_j = n_j n_j = 1 we have

\begin{aligned}\mathbf{t} \cdot \mathbf{n} = -p\end{aligned} \hspace{\stretch{1}}(2.3)

Returning to (\mathbf{t}_2 - \mathbf{t}_1) \cdot \hat{\mathbf{n}} we have

\begin{aligned}\boxed{-p_2 + p_1 = -\frac{\sigma}{2 R} }\end{aligned} \hspace{\stretch{1}}(2.4)

This is the Laplace pressure. Note that the sign of the difference is significant, since it effects the direction of the curvature. This is depicted pictorially in figure (\ref{fig:continuumL15:continuumL15Fig3})
\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{continuumL15Fig3}
\caption{pressure and curvature relationships}
\end{figure}

Question In [1] the curvature term is written

\begin{aligned}\frac{1}{{2 R}} \rightarrow \frac{1}{{R_1}} + \frac{1}{{R_2}}.\end{aligned} \hspace{\stretch{1}}(2.5)

Why the difference? Answer: This is to account for non-spherical surfaces, with curvature in two directions. Illustrating by example, imagine a surface like as in figure (\ref{fig:continuumL15:continuumL15Figq})
\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{continuumL15Figq}
\caption{Example of non-spherical curvature.}
\end{figure}

Followup required to truly understand things

While, this review clarifies things, we still don’t really know how the surface tension term \sigma is defined. Nor have we been given any sort of derivation of 2.1, from which the end result follows.

I’m assuming that \sigma is a property of the two fluids at the interface, so if you have, for example, oil and vinegar in a bottle, we have surface tension and curvature that’s probably related to how the two of these interact. If there is still a mixing or settling process occurring, I’d imagine that this could even vary from point to point on the surface (imagine adding soap to a surface where stuff can float until the soap mixes in enough that things start sinking in the radius of influence of the soap).

Surface tension gradients.

Now consider the tangential component of the traction vector

\begin{aligned}\mathbf{t}_2 \cdot \hat{\boldsymbol{\tau}} - \mathbf{t}_1 \cdot \hat{\boldsymbol{\tau}} = - \not{{ \frac{\sigma}{2 R} \hat{\mathbf{n}} \cdot \hat{\boldsymbol{\tau}}}} - \hat{\boldsymbol{\tau}} \cdot \boldsymbol{\nabla}_I \sigma\end{aligned} \hspace{\stretch{1}}(2.6)

So we see that for a static fluid, we must have

\begin{aligned}\boldsymbol{\nabla}_I \sigma = 0\end{aligned} \hspace{\stretch{1}}(2.7)

For a static interface there cannot be any surface tension gradient. This becomes very important when considering stability issues. We can have surface tension induced flow called capillary, or mandarin (?) flow.

Reynold’s number.

In Navier-Stokes after making non-dimensionalization changes of the form

\begin{aligned}x \rightarrow L x'\end{aligned} \hspace{\stretch{1}}(3.8)

the control parameter is like Reynold’s number.

In NS

\begin{aligned}\rho \frac{\partial {\mathbf{u}}}{\partial {t}} + \rho ( \mathbf{u} \cdot \boldsymbol{\nabla} ) \mathbf{u} = - \boldsymbol{\nabla} p + \mu \boldsymbol{\nabla}^2 \mathbf{u}\end{aligned} \hspace{\stretch{1}}(3.9)

We call the term

\begin{aligned}\rho ( \mathbf{u} \cdot \boldsymbol{\nabla} ) \mathbf{u}\end{aligned} \hspace{\stretch{1}}(3.10)

the inertial term. It is non-zero only when something is being “carried along with the velocity”. Consider a volume fixed in space and one that is moving along with the fluid as in figure (\ref{fig:continuumL15:continuumL15Fig4})
\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{continuumL15Fig4}
\caption{Moving and fixed frame control volumes in a fluid.}
\end{figure}

All of our viscosity dependence shows up in the Laplacian term, so we can roughly characterize the Reynold’s number as the ratio

\begin{aligned}\text{Reynold's number}&\rightarrow\frac{{\left\lvert{\text{effect of inertia}}\right\rvert}}{{\left\lvert{\text{effect of viscosity}}\right\rvert}}  \\ &=\frac{ {\left\lvert{\rho ( \mathbf{u} \cdot \boldsymbol{\nabla} ) \mathbf{u} }\right\rvert} }{{\left\lvert{ \mu \boldsymbol{\nabla}^2 \mathbf{u}}\right\rvert}} \\ &\sim\frac{ \rho U^2/L }{\mu U/L^2} \\ &\sim\frac{ \rho U L }{\mu }\end{aligned}

In figures (\ref{fig:continuumL15:continuumL15Fig5}), (\ref{fig:continuumL15:continuumL15Fig6}) we have two illustrations of viscous and non-viscous regions the first with a moving probe pushing its way through a surface, and the second with a wing set at an angle of attack that generates some turbulence. Both are illustrations of the viscous and inviscous regions for the two flows. Both of these are characterized by the Reynold’s number in some way not really specified in class. One of the points of mentioning this is that when we are in an essentially inviscous region, we can neglect the viscosity (\mu \boldsymbol{\nabla}^2 \mathbf{u}) term of the flow.
\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{continuumL15Fig5}
\caption{continuumL15Fig5}
\end{figure}
\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{continuumL15Fig6}
\caption{continuumL15Fig6}
\end{figure}

References

[1] L.D. Landau and E.M. Lifshitz. A Course in Theoretical Physics-Fluid Mechanics. Pergamon Press Ltd., 1987.

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