root/trunk/matml/transport/problems/zincdimens/zincdimens-solution.tex

\documentclass{article}
\usepackage{fullpage}
\newcommand{\PSbox}[3]{\mbox{\rule{0in}{#3}\special{psfile=#1}\hspace{#2}}}
\begin{document}
\begin{enumerate}
\item Dimensional analysis: diffusion of a zinc coating

  The ``Shrinking Gaussian'' solution to the time-dependent diffusion equation:
  $$C = \frac{\beta}{\sqrt{\pi Dt}}\exp\left(-\frac{x^2}{4Dt}\right)$$

  \begin{enumerate}
  \item Dimensions and units:
    \begin{center}
      \begin{tabular}[h]{c|c}
        Dimension & Units \\ \hline
        $C$     & $\rm\frac{mol}{m^3}$ \\
        $\beta$ & $\rm\frac{mol}{m^2}$ \\
        $D$     & $\rm\frac{m^2}{s}$ \\
        $t$     & seconds \\
        $x$     & meters \\ \hline
      \end{tabular}
    \end{center}

    Note: substitution of g or kg for mol works just as well.

  \item As we can see above, there are five dimensions, and three base units
    (mol or g or kg, m, s), therefore, there are just two dimensionless
    parameters.

  \item We want to construct $\pi_C$ and $\pi_x$, eliminating $\beta$, $D$ and
    $t$.

    \begin{center}
      \begin{tabular}[h]{c|c|c|c|}
        Dimension    & mol  & m    & s \\ \hline
        $C$          & 1    & $-3$ & 0 \\
        $\beta^{-1}$ & $-1$ & 2    & 0 \\
        $D^{1/2}$    & 0    & 1    & $-\frac{1}{2}$ \\
        $t^{1/2}$    & 0    & 0    & $\frac{1}{2}$ \\ \hline
        Total        & 0    & 0    & 0 \\ \hline
      \end{tabular}
      \hspace{0.5in}
      \begin{tabular}[h]{c|c|c|c|}
        Dimension  & mol & m    & s \\ \hline
        $x$        & 0   & 1    & 0 \\
        $\beta^0$  & 0   & 0    & 0 \\
        $D^{-1/2}$ & 0   & $-1$ & $\frac{1}{2}$ \\
        $t^{-1/2}$ & 0   & 0    & $-\frac{1}{2}$ \\ \hline
        Total      & 0   & 0    & 0 \\ \hline
      \end{tabular}
    \end{center}

    So we have:
    $$\pi_C = \frac{C\sqrt{Dt}}{\beta}$$
    $$\pi_x = \frac{x}{\sqrt{Dt}}$$

  \item Start with the solution itself:
    $$C = \frac{\beta}{\sqrt{\pi Dt}}\exp\left(-\frac{x^2}{4Dt}\right)$$
    One rearrangement should do it:
    $$\frac{C\sqrt{Dt}}{\beta} =
    \frac{1}{\sqrt{\pi}}\exp\left(-\frac{x^2}{4Dt}\right)$$
    $$\pi_C = \frac{1}{\sqrt{\pi}}\exp\left(-\frac{\pi_x^2}{4}\right)$$

  \item When made dimensionless like this, the ``Shrinking Gaussian'' becomes a
    simple Gaussian with a maximum value of $1/\sqrt{\pi}$:

    \begin{center}
      \PSbox{gauss.ps}{190pt}{135pt}
    \end{center}

    As for the width, you could define any measure of it, and give the
    corresponding value.  One such measure is where the Gaussian reaches $1/e$
    of its maximum; this is relatively easy because you just set the exponent
    to $-1$, so $-\pi_x^2/4 = -1$ and $\pi_x=2$.

    Another popular measure for Gaussian distributions is the ``full width at
    half maximum'' (FWHM).  For this, you solve:
    $$\exp\left(-\frac{\pi_x^2}{4}\right) = \frac{1}{2}$$
    $$\pi_x = 2\sqrt{\ln 2} \simeq 1.39$$
    Either of these measures was fine.
  \end{enumerate}
\end{enumerate}
\end{document}