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

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\newcommand{\PSbox}[3]{\mbox{\rule{0in}{#3}\special{psfile=#1}\hspace{#2}}}
\begin{document}
\begin{enumerate}
\item Heat conduction and diffusion in alloy casting

  \begin{enumerate}
  \item This was a bit confusing because we used a single coefficient $h$ to
    describe both convective transfer from the mold to the environment, and
    also conduction through the mold.

    That said, for pseudo-steady-state conduction, temperature is linear across
    the solid metal, varying between (unknown) $T_s$ at the mold-metal
    interface and $T_m$ at the melt interface.  The Biot number $hY/k$
    determines the ratio of temperature differences:
    $$\frac{hY}{k} = \frac{T_m-T_s}{T_s-T_{env}}.$$
    For a small Biot number, this ratio will be small, putting $T_s$ closer to
    $T_m$; for a large Biot number, $T_s$ wil be closer to $T_{env}$.  So your
    small- and large-Biot number sketches should have looked something like:
    \begin{center}
      \PSbox{casting-a.ps}{190pt}{137pt}
      \PSbox{casting-b.ps}{190pt}{137pt}
    \end{center}

  \item Start with a heat balance at a moving melt interface (from the equation
    sheet):
    $$|q_\ell| - |q_s| = -U\rho\Delta H_f$$
    where $U=dY/dt$, the velocity of the melt interface.  Because the liquid
    temperature is uniform, there's no temperature gradient there, and no flux,
    so $q_\ell=0$.  This leaves us with the solid, where $q_s=k_sdT/dx$.  But
    since we don't know $T_m-T_s$ very well (see the left graph above), we can
    instead use the expression given in the problem, simplified since
    $T_s\simeq T_m$:
    $$q_s = h(T_s-T_{env}) \simeq h(T_m-T_{env})$$
    This gives us:
    $$\frac{dY}{dt} = U = \frac{h(T_m-T_{env})}{\rho\Delta H_f}$$
    This leads to a roughly constant front velocity, and linear growth of the
    solid layer.

  \item For long times (large Biot number), growth is limited by conduction
    through the solid metal, and $T_s\simeq T_{env}$.  In this case, we can use
    the pseudo-steady-state approximation to estimate the temperature gradient
    to give the heat flux as:
    $$q_s=-k_s\frac{\partial T}{\partial x} = -k_s\frac{T_m-T_s}{Y} \simeq
    -k_s\frac{T_m-T_{env}}{Y}.$$
    This too goes into the melt interface heat balance, giving the
    solidification rate:
    $$\frac{dY}{dt} = \frac{k_s(T_m-T_{env})}{Y\rho\Delta H_f}$$
    Solving this differential equation gives $Y\propto\sqrt{t}$.

  \item This sketch should have included combined and parabolic growth rates
    something like:
    ({\em e.g.} oxidation):
    \begin{center}
      \PSbox{growth3.ps}{172pt}{115pt}
    \end{center}

  \item This part involves solute diffusion in a moving frame of reference,
    which gives us convective mass transfer.  We start with the equation for
    diffusion in a moving body (from the equation sheet):
    $$\frac{\partial C}{\partial t} + u_x\frac{\partial C}{\partial x'} =
    D\frac{\partial^2 C}{\partial x'^2} + G.$$
    This is steady-state, so $\partial C/\partial t=0$, and in a metal alloy,
    the species are elements which aren't created or destroyed, so $G=0$.
    We're left with an ordinary differential equation:
    $$D\frac{d^2C}{dx'^2} - u_x\frac{dC}{dx'} = 0$$
    As a homogeneous equation with constant coefficients, we can solve this by
    assuming a solution of the form $C=e^{Rx'}$ and solve for $R$:
    $$DR^2e^{Rx'} - u_xRe^{Rx'} = 0$$
    $$R(DR-u_x) = 0$$
    $$R=0\ \ {\rm or}\ \ R=\frac{u_x}{D}$$
    $$C=A\exp\left(\frac{u_x}{D}x'\right) + B$$

  \item In the liquid, the concentration at the interface where $x'=0$ is
    $5C_L$; a long way from the interface (at $x'=\infty$), $C=C_L$.  Taking
    the second condition first, and noting that $u_x$ is negative (since the
    interface is moving in the positive $x$-direction):
    $$C_L = Ae^{-\infty} + B\ \Rightarrow\ B=C_L,$$
    then the first condition:
    $$5C_L = Ae^0 + C_L\ \Rightarrow\ A=4C_L;$$
    $$C = 4C_L\exp\left(\frac{u_x}{D}x'\right) + C_L.$$

  \item The order-of-magnitude thickness of the high-concentration layer ends
    where the argument of the exponential is $-1$ (for simplicity):
    $$\frac{u_x}{D}x'=-1\ \Rightarrow\ x'=-\frac{D}{u_x}.$$
    To get the thickness to a specific tolerance, {\em e.g.} the 1\% criterion
    like a boundary layer, we just multiply this by a constant such as
    $\ln(0.01)$.
  \end{enumerate}
\end{enumerate}
\end{document}