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\begin{enumerate}
\item Czochralski crystal growth

  \begin{enumerate}
  \item \label{czoch1} Cartesian coordinates clearly don't make sense here, and
    the complexity of the spherical equations only justify their use if the
    boundary conditions would be greatly simplified, e.g. if the crucible is
    exactly hemispherical and the top surface of the liquid is flat.  So we use
    cylindrical coordinates.

    The liquid silicon is incompressible, and its homogeneity (the silicon is
    very pure) suggests constant density.  As a liquid metalloid, it is most
    likely Newtonian, and unless there are very strong temperature gradients,
    it has constant viscosity.

    Under these conditions with laminar flow, we can use the Newtonian
    incompressible form of the Navier-Stokes equations.

    Now let's look at the other assumptions:
    \begin{itemize}
    \item Steady-state: the crystal and crucible are rotating at constant
      speeds, so the only source of time-dependence is removal of the liquid as
      the crystal grows and pulls out.  Because this pullout is orders of
      magnitude slower than the rotations, we can assume a
      ``quasi-steady-state'' where at any given moment the flow is at
      steady-state, but that steady-state is changing {\em very} slowly with
      time as there is progressively less liquid in the crucible.  So we can
      cancel the time derivatives.

    \item Fully-developed flow: because this is rotating, it's hard to define
      an fully-developed, since ``entrance length'' is essentially zero and the
      ``overall length'' essentially infinite.

      Instead, it's better to think of it as having axial symmetry, that is, we
      can lose the velocity derivatives in the $\theta$-direction.

      However, the centrifugal force varies greatly with both $r$ and $z$, so
      there is likely to be significant $r$-velocity variation, and the
      continuity equation with the time and theta derivatives cancelled tells
      us that this will produce flow in the $z$-direction.

    \item Edge-effects: it's unclear whether the ``thickness'' or ``width''
      direction is the $r$- or $z$-direction, because the sizes in those two
      directions are so similar, so we really can not neglect edge effects.
    \end{itemize}

    The resulting equations, starting with continuity:
    $$\frac{1}{r}\frac{\partial}{\partial r}\left(rv_r\right) +
    \frac{\partial v_z}{\partial z} = 0$$
    Motion $r$-component:
    $$\rho\left(v_r\frac {\partial v_r}{\partial r} - \frac{v_\theta^2}{r} +
      v_z\frac{\partial v_r}{\partial z}\right) =
    -\frac{\partial p}{\partial r} +
    \eta\left[\frac{\partial}{\partial r}\left(\frac{1}{r}
        \frac{\partial}{\partial r}\left(rv_r\right)\right)+
      \frac{\partial^2v_r}{\partial z^2}\right]$$
    Motion $\theta$-component:
    $$\rho\left(v_r\frac{\partial v_\theta}{\partial r} +
      \frac{v_rv_\theta}{r} + v_z\frac{\partial v_\theta}{\partial z}\right) =
    \eta\left[\frac{\partial}{\partial r}\left(\frac{1}{r}
        \frac{\partial}{\partial r}\left(rv_\theta\right)\right)+
      \frac{\partial^2v_\theta}{\partial z^2}\right]$$
    Motion $z$-component:
    $$\rho\left(v_r\frac{\partial v_z}{\partial r} +
      v_z\frac{\partial v_z}{\partial z}\right)=
    -\frac{\partial p}{\partial z} +
    \eta\left[\frac{1}{r}\frac{\partial}{\partial r}
      \left(r\frac{\partial v_z}{\partial r}\right)+
      \frac{\partial^2v_z}{\partial z^2}\right]+\rho g_z$$
    Isn't that much better?  We still need a computer to solve the system
    though.

  \item This is a bit of a trick question, as there are nonzero $r$-, $\theta$-
    and $z$-components to flow, making it seem three-dimensional.  However,
    because velocity and pressure are not functions of $\theta$ but only of $r$
    and $z$, by the definition given in the problem, the flow is considered
    two-dimensional.  (Likewise, all of the problems we actually solve in this
    course can be considered one-dimensional, even if the flow direction is
    different from the direction in which it's varying, {\em e.g.} $x$-velocity
    as a function of $y$.)

    Note that because of the non-zero $\theta$ velocity, some people use the
    expression ``2.5-dimensional''...
  \end{enumerate}
\end{enumerate}
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