root/trunk/matml/transport/problems/cdmold1/cdmold1-solution.tex

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\begin{enumerate}
\item CD Injection Molding I: Navier-Stokes

  \begin{enumerate}
  \item Assumptions which could be made here included:
    \begin{itemize}
    \item Newtonian behavior with uniform viscosity (since that was given).
    \item Incompressible flow with uniform density; this and the previous
      assumption permitted use of the simpler closed-form Newtonian
      Navier-Stokes equations.
    \item Laminar flow (with small thickness and high viscosity this is
      reasonable).
    \item Steady-state (given); time derivatives are all zero.
    \item Fully-developed (given as short entrance length); flow is purely
      radial, $u_\theta=u_z=0$.  But velocity derivatives in the flow direction
      (the $r$-direction) are {\em not} zero, as conservation of mass requires
      that there be faster flow near the center and slower flow near the
      outside.
    \item Edge effects and axisymmetric are the same thing here; velocity
      derivatives in the $\theta$ direction are zero.
    \item Gravity acts in the negative $z$-direction; $F_r=F_\theta=0$,
      $F_z=-\rho g$.
    \end{itemize}

  \item Since $u_\theta=u_z=0$ (because flow is axisymmetric and
    fully-developed), the mass conservation equation reduces to:
    $$\frac{1}{r}\frac{\partial}{\partial r} \left(ru_r\right) = 0.$$
    Steady-state, fully-developed, axisymmetric, and $F_r=0$ reduce the
    $r$-momentum equation to:
    $$\rho u_r\frac{\partial u_r}{\partial r}= - \frac{\partial p}{\partial r}
    + \mu\left[\frac{\partial}{\partial r}\left(\frac{1}{r}
        \frac{\partial}{\partial r}\left(ru_r\right)\right)
      + \frac{\partial^2u_r}{\partial z^2}\right].$$
    With a bit more sophistication, one might realize that the $r$-derivative
    in the viscosity term is just a derivative of what we found to be zero in
    the mass equation, so that too could be eliminated.

    The $\theta$-momentum equation is almost completely eliminated:
    $$0 = -\frac{1}{r}\frac{\partial p}{\partial\theta},$$
    so there are no pressure gradients in the $\theta$-direction, and the
    $z$-momentum equation only has pressure gradients due to weight of the
    polymer:
    $$0 = -\frac{\partial p}{\partial z} + F_z
    \left(= -\frac{\partial p}{\partial z} - \rho g\right).$$
  \end{enumerate}
\end{enumerate}
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