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

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\begin{enumerate}
\item Freezing Lake
  \begin{enumerate}
  \item This was a straightforward application of finite difference principles:
    \begin{eqnarray}
      \label{eq:energydifference}
      \frac{H_{i,n+1}-H_{i,n}}{\Delta t} &=&
      \frac{\left[k\frac{\partial T}{\partial x}\right]_{i+\frac{1}{2},n} -
        \left[k\frac{\partial T}{\partial x}\right]_{i-\frac{1}{2},n}}{\Delta x} \\
      \nonumber &=& \frac{k_{i+\frac{1}{2},n}(T_{i+1,n}-T_{i,n}) -
        k_{i-\frac{1}{2},n}(T_{i,n}-T_{i-1,n})}{(\Delta x)^2}
    \end{eqnarray}

  \item At the bottom of the lake, the flux at $x_{i+\frac{1}{2}}$ is zero, so
    we remove it from equation \ref{eq:energydifference}:
    \begin{equation}
      \label{eq:bottombound}
      \frac{H_{i,n+1}-H_{i,n}}{\Delta t} =
      \frac{k_{i+\frac{1}{2},n}(T_{i+1,n}-T_{i,n})}{(\Delta x)^2}. 
    \end{equation}
    At the top of the lake, the flux is given by $h(T_{air}-T)$, so we replace
    the $i+\frac{1}{2}$ part of the equation.  But we have to be careful with
    the $\Delta x$es, distinguishing between the one used for the divergence of
    the flux, and that used to calculate the flux itself at
    $x_{i-\frac{1}{2}}$:
    \begin{equation}
      \label{eq:topbound}
      \frac{H_{i,n+1}-H_{i,n}}{\Delta t} = \frac{h(T_{air}-T_{i,n}) -
        \frac{k_{i-\frac{1}{2},n}(T_{i,n}-T_{i-1,n})}{\Delta x}}{\Delta x}. 
    \end{equation}

  \item \label{timestep} First, extend the properties table in the problem to
    include thermal diffusivity and stable timestep (for $\Delta x=0.01$m):
    \begin{center}
      \begin{tabular}{l|c|c|}
        Material                            & Liq. water & Sol. water \\ \hline
        Conductivity $k$, $\rm\frac{W}{m\cdot K}$     & 0.56 & 2.3 \\
        Heat capacity $c_p$, $\rm\frac{J}{kg\cdot K}$ & 4200 & 2100 \\
        Density $\rho$, $\rm\frac{kg}{m^3}$           & 1000 & 920 \\ \hline
        Thermal diffusivity $\alpha$, $\rm\frac{m^2}{s}$ & $1.33\times10^{-7}$&
        $1.19\times10^{-6}$ \\
        Stability criterion, sec& $\Delta t\leq375$& $\Delta t\leq42$ \\ \hline
      \end{tabular}
    \end{center}
    If we use the liquid water criterion, the liquid is stable but the solid
    clearly is not.  So we use the smaller timestep criterion (solid water, 42
    seconds), which corresponds to the larger thermal diffusivity.

  \item \label{timescales} The three timescales:
    \begin{enumerate}
    \item \label{condtime} For $L=0.1$m, the time required to reach
      steady-state in liquid water is given by $t\sim L^2/\alpha = 75,000$
      seconds (about 20 hours).
    \item \label{condlimit} First we solve the differential equation:
      \begin{equation}
        \label{eq:condeqn}
        \rho\Delta H_f\frac{dX}{dt} = k\frac{\Delta T}{X}
      \end{equation}
      $$X dX = \frac{k\Delta T}{\rho\Delta H_f} dt$$
      $$\frac{X^2}{2} = \frac{k\Delta T}{\rho\Delta H_f} t + const$$
      For $X=0$ at $t=0$, the integration constant is zero, so the time is
      given by
      \begin{equation}
        \label{eq:condresult}
        t = \frac{\rho\Delta H_f X^2}{2k\Delta T} =
        \frac{\rm 920\frac{kg}{m^3}\cdot334,000\frac{J}{kg}\cdot(0.1m)^2}
        {\rm 2\cdot2.3\frac{J}{s\cdot m\cdot K}\cdot10K} = 66,800{\rm s}. 
      \end{equation}
    \item \label{convlimit} The differential equation for convection-limited
      freezing is even simpler:
      \begin{equation}
        \label{eq:conveqn}
        \rho\Delta H_f\frac{dX}{dt} = h\Delta T
      \end{equation}
      $$\rho\Delta H_f X = h\Delta T t + const.$$
      Again the integration constant is zero, so using $h=10\frac{W}{m^2\cdot
        K}$ (from part \ref{fdsolution}), required time is given by:
      \begin{equation}
        \label{eq:convresult}
        t = \frac{\rho\Delta H_f X}{h\Delta T} =
        \frac{\rm 920\frac{kg}{m^3}\cdot334,000\frac{J}{kg}\cdot0.1m}
        {\rm 10\frac{W}{m^2\cdot K}\cdot10K} = 307,280{\rm s}. 
      \end{equation}
      This is clearly the dominant timescale in this problem. 
    \end{enumerate}
    We can add the above results to give a rough estimate of the total time
    required to freeze the lake, which gives us 449,000 seconds, or about 5
    days.  At 42 seconds/timestep (from part \ref{timestep}), this will require
    about ten thousand timesteps (!).

  \item \label{fdsolution} The spreadsheet provided implements the above
    equations and boundary conditions.  Two plots are provided: one giving the
    information requested in the asignment, and the other with plots at a few
    more times in between.  Times for each set of plots are as follows:

    \begin{center}
      \begin{tabular}{|lcc||lcc|}
        Time   & Plot 1 series \# & Plot 2 series \# &
        Time   & Plot 1 series \# & Plot 2 series \# \\ \hline
        0      & 1 & 1 &  158634 &   & 8 \\
        3360   &   & 2 &  204414 & 3 & 9 \\
        6846   &   & 3 &  251034 &   & 10 \\
        10080  &   & 4 &  301434 &   & 11 \\
        13760  & 2 & 5 &  351834 &   & 12 \\
        62034  &   & 6 &  402234 &   & 13 \\
        110334 &   & 7 &  448560 & 4 & 14 \\ \hline
      \end{tabular}
    \end{center}

    Note that the calculated total time of freezing is within 0.2\% of the
    estimate in part \ref{timescales}.
  \end{enumerate}
  \newpage
  \begin{center}
    \pdfimageresolution 180
    $\ $\pdfximage{plot1.png}\pdfrefximage\pdflastximage$\ $

    $\ $\pdfximage{plot2.png}\pdfrefximage\pdflastximage$\ $
  \end{center}
\end{enumerate}
\end{document}
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