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

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\begin{enumerate}
\item Radiation in facing target sputtering

  \begin{enumerate}
  \item There are two ways to use the diagrams to calculate $F_{21}$, one is by
    the paralell rectangles method:

    \begin{center}
      \PSbox{factarg1.eps hscale=25 vscale=25}{333pt}{178pt}
    \end{center}

    $$F_{21}=\frac{F_{2A}-F_{2B}}{2}$$
    For $F_{2A}$, $x=b/c=10{\rm cm}/10{\rm cm}=1$, and $y=a/c=10{\rm cm}/10{\rm
    cm}=1$, so the graph gives $F_{2A}=0.2$.  For $F_{2B}$, $x=b/c=10{\rm
    cm}/20{\rm cm}=0.5$, and $y=a/c=10{\rm cm}/20{\rm cm}=0.5$, so the graph
    gives $F_{2B}=0.07$.  These and the equation above result in
    $F_{21}=0.065$.

    The other way is by the common-edge rectangles method, using
    $\phi=90^\circ$:

    \begin{center}
      \PSbox{factarg2.eps hscale=25 vscale=25}{336pt}{178pt}
    \end{center}

    $$F_{21}=2(F_{2C}-F_{2D})$$
    For $F_{2C}$, $N=a/b=20{\rm cm}/10{\rm cm}=2$ and $L=c/b=10{\rm cm}/10{\rm
    cm}=1$, so the graph gives $F_{2C}=0.24$.  For $F_{2D}$, $N=a/b=10{\rm
    cm}/10{\rm cm}=1$, and $L=c/b=10{\rm cm}/10{\rm cm}=1$, so the graph
    gives $F_{2D}=0.20$.  These and the equation above result in
    $F_{21}=0.080$, which is, well, somewhat close to the parallel rectangles
    viewfactor.

    Now use the reciprocity rule to calculate $F_{12}$:
    $$A_1F_{12}=A_2F_{21}$$
    $$F_{12}=\frac{A_2F_{21}}{A_1}$$
    $S_1$ includes two $10\times10$ squares, so $A_1={\rm200cm^2=0.02m^2}$;
    $S_2$ is one $10\times10$ square, so $A_2={\rm100cm^2=0.01m^2}$.  These and
    the $F_{21}$s calculated by the two methods above give $F_{12}=0.032$ to
    0.040.

  \item This is the total power emitted by the targets multiplied by the
    viewfactor and by the absorptivity of the substrate:
    $$Q_{12} = \alpha_2 F_{12} A_1\epsilon_1\sigma T_1^4$$
    Using $\alpha_2=\epsilon_2$ and $F_{12}\simeq0.035$ gives us:
    $$Q_{12} = {\rm 0.7 \cdot 0.035 \cdot 0.02 m^2 \cdot 0.4 \cdot
      5.67\times10^{-8}\frac{W}{m^2K^4} \cdot (1200K)^4 = 23W}$$
    Not a whole lot of power, but for a small substrate, enough to get it
    decently hot.

  \item This is a similar situation; using $F_{21}=0.07$ gives us:
    $$Q_{21} = {\rm 0.4 \cdot 0.07 \cdot 0.01 m^2 \cdot 0.7 \cdot
      5.67\times10^{-8}\frac{W}{m^2K^4} \cdot (900K)^4 = 7.3W}$$
    Note that the ratio $Q_{12}/Q_{21}$ is simply $T_1^4/T_2^4$.

    This method misses one thing, which is reflection of a surface's radiation
    off another surface and back onto itself.  With viewfactors as small as
    these, this secondary absorption of reflected radiation will be negligible.
    But in general, this is the motivation for the total exchange viewfactors.
  \end{enumerate}
\end{enumerate}
\end{document}