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

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\begin{enumerate}
\item Freezing by radiation and convection

  \begin{enumerate}
  \item The thermal conductivity is given by the Wiedmann-Franz law:
    $$k_{el}=L\sigma_{el}T = 2.45\times10^{-8} \frac{\rm W\Omega}{\rm K^2}\cdot
    5\times10^{5} (\Omega\cdot {\rm m})^{-1}\cdot 1800{\rm K} =
    22\frac{\rm W}{\rm m\cdot K}.$$

  \item Total flux away from the top surface is radiative plus convective
    ($T_s$ is top surface temperature):
    $$q_{total}=q_{rad}+q_{conv} = \epsilon\alpha_{env}\sigma (T_s^4-T_{env}^4)
    + h (T_s-T_{env}).$$

  \item \label{htotal} If the environment is much colder, then $T_s\gg T_{env}$
    so $T_{env}\simeq 0$.  If it is ``black'', then its absorbtivity is one.
    So the above expression simplifies to
    $$q_{total}=\epsilon\sigma T_s^4 + hT_s = h_{total}T_s,$$
    $$h_{total}=\epsilon\sigma T_s^3 + h = 200+100\frac{\rm W}{\rm m^2\cdot K}
    = 300\frac{\rm W}{\rm m^2\cdot K}.$$

  \item Set the Biot number to 0.1 and solve for $Y$ using $h_{total}$:
    $${\rm Bi}=\frac{h_{total}Y}{k}=0.1 \Rightarrow Y=\frac{0.1k}{h_{total}}=
    \frac{\rm0.1\cdot 22\frac{W}{m\cdot K}}{\rm300\frac{W}{m^2\cdot K}}=
    0.0073{\rm m(7.3mm)}.$$

  \item \label{rate} You're given an equation relating flux to solidification
    rate, so solve it for $dY/dt$:
    $$q_L-q_S=\rho\Delta H_f\frac{dY}{dt}\Rightarrow
    \frac{dY}{dt}=\frac{q_L-q_S}{\rho\Delta H_f}.$$
    Since liquid metal temperature is uniform, $q_L=0$.  At quasi-steady-state,
    the flux through the solid is equal to the flux leaving its top surface,
    which is $h_{total}T_s$, $T_s$ being the surface temperature.  Since the
    metal temperature is roughly uniform, $T_s\simeq T_m$ so we can use that:
    $$\frac{dY}{dt}=\frac{h_{total}T_m}{\rho\Delta H_f}=
    \frac{\rm300\frac{W}{m^2\cdot K}\cdot1800K}
    {\rm7500\frac{kg}{m^3}\cdot2.67\times10^5\frac{J}{kg}}=
    2.7\times10^{-4}\frac{\rm m}{\rm s}.$$
    The time required to reach the thickness calculated above of 7.3mm is that
    divided by the rate calculated here, or about 27 seconds.

  \item We have resistances in series, due to conduction through the solid and
    radiation/convection from the surface.  So the heat flux in the solid
    goes like:
    $$q_s=\frac{T_m-T_{env}}{\frac{Y}{k_s}+\frac{1}{h_{total}}}.$$
    The trouble is, $h_{total}$ is a function of the surface temperature $T_s$,
    which is somewhere between $T_{env}$ and $T_m$.  We can relate $T_s$ to the
    conductive flux, which is the same as the flux $q$ above:
    $$q_s=k_s\frac{T_m-T_s}{Y}\Rightarrow T_s=T_m-\frac{qY}{k_s}.$$
    We plug this into the $h_{total}$ expression from part \ref{htotal}:
    $$h_{total} = \epsilon\sigma T_s^3 + h =
    \epsilon\sigma\left(T_m-\frac{qY}{k_s}\right)^3 + h,$$
    then plug that into the heat flux:
    $$q_s=\frac{T_m-T_{env}}{\frac{Y}{k_s}+
      \frac{1}{\epsilon\sigma\left(T_m-\frac{q_sY}{k_s}\right)^3 + h}}.$$
    To complete this analysis, one would need to solve this for $q_s$ (probably
    numerically), and use that in the solidification rate equation from part
    \ref{rate}.
  \end{enumerate}
\end{enumerate}
\end{document}
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