# Peeter Joot's (OLD) Blog.

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## PHY454H1S Continuum Mechanics. Lecture 14: Non-dimensionality and scaling. Taught by Prof. K. Das.

Posted by peeterjoot on March 7, 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. Surfaces

We are considering a surface as depicted in (\ref{fig:continuumL14:continuumL14Fig13})

\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{continuumL13Fig13}
\caption{Variable surface geometries}
\end{figure}

With the surface height given by

\begin{aligned}z = h(x, t),\end{aligned} \hspace{\stretch{1}}(2.1)

where this describes the interface. Taking the difference

\begin{aligned}\phi = z - h(x, t) = 0,\end{aligned} \hspace{\stretch{1}}(2.2)

we define a surface. We considered a small displacement as in (\ref{fig:continuumL14:continuumL14Fig14}).

\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{continuumL13Fig14}
\caption{A vector differential element}
\end{figure}

Recall that if $\phi$ is a constant, then $\boldsymbol{\nabla} \phi$ is a normal to the surface. We showed this by considering the differential

\begin{aligned}0 &= d\phi \\ &= \frac{\partial {\phi}}{\partial {x}} dx+\frac{\partial {\phi}}{\partial {y}} dy+\frac{\partial {\phi}}{\partial {z}} dz \\ &=(\boldsymbol{\nabla} \phi) \cdot d\mathbf{r}.\end{aligned}

We can construct the unit normal by scaling. For our 1D example we have

\begin{aligned}\hat{\mathbf{n}} &= \frac{\boldsymbol{\nabla} \phi}{{\left\lvert{\boldsymbol{\nabla} \phi}\right\rvert}} \\ &= \frac{1}{{{\left\lvert{\boldsymbol{\nabla} \phi}\right\rvert}}} \left(\frac{\partial {\phi}}{\partial {x}},\frac{\partial {\phi}}{\partial {y}}\right) \end{aligned}

so that our unit normal is

\begin{aligned}\hat{\mathbf{n}} = \frac{1}{{ \sqrt{1 + (h')^2}}}\left( -\frac{\partial {h}}{\partial {x}}, 1 \right)\end{aligned} \hspace{\stretch{1}}(2.3)

A unit tangent can also be constructed by inspection

\begin{aligned}\hat{\boldsymbol{\tau}} = \frac{1}{{ \sqrt{1 + (h')^2}}}\left( 1, \frac{\partial {h}}{\partial {x}} \right).\end{aligned} \hspace{\stretch{1}}(2.4)

# Traction vector at the interface.

Recall that our stress tensor has the form

\begin{aligned}T_{ij} = - p \delta_{ij} + \rho \nu \left( \frac{\partial {u_i}}{\partial {x_j}}+\frac{\partial {u_j}}{\partial {x_i}}\right)\end{aligned} \hspace{\stretch{1}}(3.5)

(here we are switching notations for the stress since we will be using $\sigma$ for surface tension in this section)

The traction vector components are

\begin{aligned}t_i = T_{ij} n_j =- p n_i + \rho \nu \left( \frac{\partial {u_i}}{\partial {x_j}}+\frac{\partial {u_j}}{\partial {x_i}}\right) n_j\end{aligned} \hspace{\stretch{1}}(3.6)

Considering a control volume as illustrated in we can arrive at what we call the jump stress balance equation

figure (\ref{fig:continuumL14:continuumL14fig3})
\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{continuumL14fig3}
\caption{Control volume for liquid air interface}
\end{figure}

\begin{aligned}[\mathbf{T} \hat{\mathbf{n}}]^2_1 = \frac{2 \sigma}{R} \hat{\mathbf{n}} - \boldsymbol{\nabla}_I \sigma\end{aligned} \hspace{\stretch{1}}(3.7)

where

\begin{aligned}\sigma &= \text{surface tension} \\ R &= \text{radius of curvature} \\ \boldsymbol{\nabla}_I &= \text{gradient along the interface}\end{aligned} \hspace{\stretch{1}}(3.8)

and the suffix $2$ and prefix $1$ indicates that we are considering the interface between fluids labelled $1$ and $2$ (liquid and air respectively in the diagram).

For a derivation see Prof after class?

Force balance along the normal direction gives

\begin{aligned}\hat{\mathbf{n}} [\mathbf{T} \hat{\mathbf{n}}]^2_1 = \hat{\mathbf{n}} \cdot \frac{2 \sigma}{R} \hat{\mathbf{n}} - \not{{\hat{\mathbf{n}} \cdot (\boldsymbol{\nabla}_I \sigma)}}\end{aligned} \hspace{\stretch{1}}(3.11)

If you do this calculation, you will get

\begin{aligned}[-p]^2_1 = \frac{ 2 \sigma}{R}\end{aligned} \hspace{\stretch{1}}(3.12)

I think this was called the Laplace equation?

Question: How was $\sigma$ defined? A: Energy per unit area.

Figure (\ref{fig:continuumL14:continuumL14fig4}) was given as part of an explaination of surface tension and curvature, but I missed part of that discussion. Perhaps this is elaborated on in the class notes?

\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{continuumL14fig4}
\caption{Molecular gas and liquid interactions at a surface.}
\end{figure}

# Non dimensionalization and scaling

## Motivation.

By scaling we mean how much detail do you want to look at in the analysis. Consider the figure (\ref{fig:continuumL14:continuumL14fig5a}) where we imagine that we zoom in on something that appears smooth from a distance, but perhaps grandular close up as in figure (\ref{fig:continuumL14:continuumL14fig5b}). Picking the length scale to be used in this case can be very important.

\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{continuumL14fig5a}
\caption{Coarse scaling example.}
\end{figure}

\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{continuumL14fig5b}
\caption{Fine grain scaling example (a zoom).}
\end{figure}

FIXME: prof wrote:

\begin{aligned}y = A x^\alpha\end{aligned} \hspace{\stretch{1}}(4.13)

## Rescaling by characteristic length and velocity.

Suppose that a fluid is flowing with

\begin{itemize}
\item a characteristic velocity $U$, with dimensions $[U] \sim L T^{-1}$
\item a characteristic length scale $L$
\end{itemize}

Considering the dimensions of the terms in the NS equation

\begin{aligned}[\rho] = M L^{-3}\end{aligned} \hspace{\stretch{1}}(4.14)

\begin{aligned}[p] = M L T^{-2} L^{-2} = M L^{-1} T^{-2}\end{aligned} \hspace{\stretch{1}}(4.15)

\begin{aligned}[t] = T = \frac{L}{U}\end{aligned} \hspace{\stretch{1}}(4.16)

so

\begin{aligned}[p] = [\rho U^2] = M L^{-3} L^2 T^{-2} = M L^{-1} T^{-2}\end{aligned} \hspace{\stretch{1}}(4.17)

Now let’s alter the NS equation using some scaling to put it into a dimensionless form

\begin{aligned}\frac{\partial {\mathbf{u}}}{\partial {t}} + (\mathbf{u} \cdot \boldsymbol{\nabla}) \mathbf{u} = - \frac{1}{{\rho}} \boldsymbol{\nabla} + \nu \boldsymbol{\nabla}^2 \mathbf{u}\end{aligned} \hspace{\stretch{1}}(4.18)

\begin{aligned}\frac{\partial {\mathbf{u}}}{\partial {t}} \rightarrow \frac{\partial {(U \mathbf{u}')}}{\partial {\left(\frac{L}{U} t'\right)}} = \frac{U^2}{L} \frac{\partial {\mathbf{u}'}}{\partial {t'}}\end{aligned} \hspace{\stretch{1}}(4.19)

\begin{aligned}\boldsymbol{\nabla} = \hat{\mathbf{x}} \frac{\partial {}}{\partial {x}}+\hat{\mathbf{y}} \frac{\partial {}}{\partial {y}}+\hat{\mathbf{z}} \frac{\partial {}}{\partial {z}}\rightarrow \hat{\mathbf{x}} \frac{\partial {}}{\partial {L x'}}\hat{\mathbf{y}} \frac{\partial {}}{\partial {L y'}}\hat{\mathbf{z}} \frac{\partial {}}{\partial {L z'}}\end{aligned} \hspace{\stretch{1}}(4.20)

so that

\begin{aligned}\boldsymbol{\nabla} \rightarrow \frac{1}{{L}} \boldsymbol{\nabla}'\end{aligned} \hspace{\stretch{1}}(4.21)

\begin{aligned}(\mathbf{u} \cdot \boldsymbol{\nabla} ) \mathbf{u} \rightarrow \left( U \mathbf{u}' \cdot \frac{1}{{L}} \boldsymbol{\nabla}' \right) U \mathbf{u}' = \frac{U^2}{L} (\mathbf{u}' \cdot \boldsymbol{\nabla}') \mathbf{u}'\end{aligned} \hspace{\stretch{1}}(4.22)

\begin{aligned}\frac{1}{{\rho}} \boldsymbol{\nabla} p \rightarrow \frac{1}{{L}} \frac{\boldsymbol{\nabla}' (\not{{\rho}} U^2) }{\not{{\rho}}} p' = \frac{U^2}{L} \boldsymbol{\nabla}' p'\end{aligned} \hspace{\stretch{1}}(4.23)

\begin{aligned}\nu \boldsymbol{\nabla}^2 \mathbf{u} \rightarrow \frac{\nu}{L^2} \boldsymbol{\nabla}' U \mathbf{u}' = \frac{\nu U}{L^2} \boldsymbol{\nabla}' \mathbf{u}'\end{aligned} \hspace{\stretch{1}}(4.24)

Putting everything together, NS takes the form

\begin{aligned}\frac{U^2}{L} \frac{\partial {\mathbf{u}'}}{\partial {t'}} + \frac{U^2}{L} (\mathbf{u}' \cdot \boldsymbol{\nabla}') \mathbf{u}' = \frac{U^2}{L} \boldsymbol{\nabla}' p' + \frac{\nu U}{L^2} \boldsymbol{\nabla}' \mathbf{u}'\end{aligned} \hspace{\stretch{1}}(4.25)

or

\begin{aligned}\frac{\partial {\mathbf{u}'}}{\partial {t'}} + (\mathbf{u}' \cdot \boldsymbol{\nabla}') \mathbf{u}' = \boldsymbol{\nabla}' p' + \frac{\nu}{U L} \boldsymbol{\nabla}' \mathbf{u}'\end{aligned} \hspace{\stretch{1}}(4.26)

Introducing the Reynold’s number

\begin{aligned}R = \frac{L U}{\nu}\end{aligned} \hspace{\stretch{1}}(4.27)

We have NS in dimensionless form

\begin{aligned}\frac{\partial {\mathbf{u}'}}{\partial {t'}} + (\mathbf{u}' \cdot \boldsymbol{\nabla}') \mathbf{u}' = \boldsymbol{\nabla}' p' + \frac{1}{{R}} \boldsymbol{\nabla}' \mathbf{u}'\end{aligned} \hspace{\stretch{1}}(4.28)

The implications of this will be discussed further in the next lecture.