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

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\begin{document}
\begin{enumerate}
\item Crystal-free zone in a glass-ceramic dish

  \begin{enumerate}
  \item The Biot number is:
    $${\rm Bi}=\frac{hL}{k}=
    \frac{\rm3000\frac{W}{m^2\cdot K}\cdot 0.01m}
    {0.4\frac{\rm W}{\rm m\cdot K}}=75.$$
    Since this is very large (even using half the thickness for the
    lengthscale), assume the surface cools to $T_{fl}$ very rapidly---and
    assume this material somehow survives the resulting thermal stresses.
    Taking $x$ to be the distance from one side, temperature sketches should
    look like:

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

  \item With a uniform initial condition, and a constant $T$ boundary
    condition, at short time scales the error function is the appropriate
    solution to the thermal diffusion equation.  Since the surface temperature
    (equal to the fluid temperature) is lower than the initial temperature, the
    erf is easier to use than the erfc:
    $$\frac{T-T_{fl}}{T_i-T_{fl}}=
    {\rm erf}\left(\frac{x}{2\sqrt{\alpha t}}\right).$$

  \item We want to solve for $x$ where $T=T_{nose}$ at time $t=t_{nose}$.  The
    relative temperature at $T_{nose}$ is:
    $$\frac{T_{nose}-T_{fl}}{T_i-T_{fl}}=\frac{720-300}{1000-300}=0.6.$$
    From an error function table, erf$^{-1}(0.6)\simeq0.6$, so we need to set
    the argument of the erf to 0.6:
    $$\frac{x}{2\sqrt{\alpha t}} = 0.6 \Rightarrow x = 1.2\sqrt{\alpha t} =
    1.2\sqrt{\rm\frac{0.4\frac{W}{m\cdot K}}
      {2400\frac{kg}{m^3}\cdot900\frac{J}{kg\cdot K}}4seconds}=
    0.00103{\rm m}.$$
    [Using the erfc solution, the dimensionless temperature is 0.4, and
    erfc$^{-1}(0.4)\simeq0.6$, resulting in the same answer.]

    So the all-glassy crystal-free zone is about a millimeter thick on each
    side.

  \item The criterion for validity of the error function is:
    $$\frac{L}{2\sqrt{\alpha t}}\geq2,$$
    $$\frac{L^2}{16\alpha}\geq t.$$
    Here $L$ should be {\em half of the thickness} since when they meet in the
    middle, the error function is no longer valid.  This gives $t\leq8.73$
    seconds, and since 4 seconds is less than this, the error function is still
    valid.

    Alternatively, at four seconds, this gives $L\geq0.0034$m, and since the
    half-thickness is 0.005m, the validity criterion is satisfied.
    Alternatively, at $L=0.005$m and $t=4$ seconds, $L/2\sqrt{\alpha t}=2.9$
    which is more than two, with the same conclusion.
  \end{enumerate}
\end{enumerate}
\end{document}