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

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\begin{enumerate}
\item Radiation in Hall-H\'{e}roult cell aluminum smelting

  \begin{enumerate}
  \item At 1273 K (1000$^\circ$ C), radiative emission is given by:
    $$e = \epsilon\sigma T^4 =
    {\rm0.75\cdot\sigma=5.67\times10^{-8}\frac{W}{m\cdot K}\cdot(1273K)^4 =
      1.1\times10^5\frac{W}{m^2}}.$$
    Given the assumptions of a cold black environment, this is also the net
    radiative heat flux from the surface.

  \item First the ``radiative heat transfer coefficient'':
    $$h=\frac{q}{T} = \epsilon\sigma T^3 = 88\frac{\rm W}{\rm m^2\cdot K}.$$
    Then the Biot number:
    $${\rm Bi}=\frac{hL}{k} =
    \frac{\rm88\frac{W}{m^2\cdot K}\cdot 1m}{\rm200\frac{W}{m\cdot K}} =
    0.44.$$
    The temperature is only somewhat non-uniform across the graphite anode.

  \item A real Hall cell is more complex in several ways:
    \begin{itemize}
    \item In terms of radiation (which was the focus of the problem), the
      environment is neither cold nor black, especially the cryolite, so net
      flux will be significantly less than predicted here.
    \item The biggest reason cryolite doesn't freeze across the bottom of the
      anode is because heat is generated by ionic current through the cryolite
      (Joule heating).
    \item 88 $\rm\frac{W}{m^2\cdot K}$ is a very small $h$, so natural
      convection in the air is likely to add significantly to this heat loss;
      even so, the air temperature in the cell enclosure is a lot closer to the
      anode temperature than the assumed 0 K.
    \item The real Hall cell {\em does} have a frozen shell of cryolite across
      the top of the bath (just barely visible in the figure), which acts as a
      low-conductivity thermal blanket, allowing Joule heating to self-heat the
      process.  It just doesn't freeze across the anode-cryolite interface,
      because of internal heating.
    \end{itemize}
  \end{enumerate}
\end{enumerate}
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