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

\documentclass{article}
\usepackage{fullpage}
\begin{document}
\begin{enumerate}
\item Radiation in electrostatic levitation

  \begin{enumerate}
  \item For a spherical droplet (for which $F_{11}=0$ because it is convex),
    exactly in the middle of a cube, the viewfactors from the droplet to each
    side of the cube are equal.  Because they form an enclosure, those
    viewfactors from the droplet to each side sum to 1, so each is 1/6.
    Therefore, the viewfactor to two sides is $F_{12}=2/6=1/3$.

  \item Here we just use the identity:
    $$A_1 F_{12} = A_2 F_{21}$$
    $$F_{21} = \frac{A_1 F_{12}}{A_2} = \frac{\rm\pi(0.5cm)^2\cdot\frac{1}{3}}
    {\rm2\times(6cm)^2} = 0.0036$$

  \item A viewfactor graph gives us the viewfactor from one plate to the other.
    Since these are sides of a cube, the ratio of side length to distance
    between plates is 1, so the viewfactor from the graph should be 0.2.

    The definition of the viewfactor $F_{22}$ is the fraction of power radiated
    from surface two which arrives at surface two.  We have two plates
    radiating energy, and 20\% of each arrives at the other plate.  So 20\% of
    the total power radiated by surface 2 reaches surface 2, and $F_{22}=0.2$.

  \item This was straightforward application of radiation emissivity, flux and
    absorptivity:
    $$Q_{12} = \epsilon_1 \sigma T_1^4 A_1 F_{12} \alpha_2.$$
    \begin{center}
      \begin{tabular}{l|c|c|c|c|c|c|}
        & $\epsilon$ & $T$ & $A$ & $F$ & $\alpha(=\epsilon)$ & Calculated $Q$
        \\ \hline
        $Q_{12}$: & 0.5 & 800 K & $\rm\pi(0.005m)^2$ & $\frac{1}{3}$ & 0.8 &
        0.243 W \\
        $Q_{21}$: & 0.8 & 1000 K & $\rm2\times(0.06m)^2$ & 0.0036 & 0.5 &
        0.594 W \\ \hline
      \end{tabular}
    \end{center}
    Typically in these arrangements, droplet heating is by radiation, but not
    necessarily by the charged plates which suspend it in place.  In any case,
    these numbers indicate that the plates provide a good bit more heat to the
    droplet than vice-versa, so the rest of the heat is likely radiated to the
    environment.

  \item Heat conduction mechanisms include electronic, radiative and phonon
    conduction, so the latter two are accepted here.  (Gases can also conduct
    heat by motion of atoms, especially at low pressure with large mean free
    path.)  Phonons, however, are quantized vibrations in a lattice, and this
    is a liquid, so instead we would have simple atomic collisions, not
    quantized particles as such.  Radiation is not likely to play a major role
    in metals, since the penetration depth of photons is so small (but the
    question didn't ask for magnitude, just mechanism).

    Convection or heat transfer in the surrounding gas might play a role, but
    laser flash and other related techniques are fast enough to capture just
    the conduction in the droplet, and the conductivity of gases is extremely
    small.
  \end{enumerate}
\end{enumerate}
\end{document}