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

\documentclass{article}
\usepackage{fullpage}
\newcommand{\PSbox}[3]{\mbox{\rule{0in}{#3}\special{psfile=#1}\hspace{#2}}}
\begin{document}
\begin{enumerate}
\item Extrusion of a tube

  \begin{enumerate}
  \item In the annulus between the plug and the tube, we can assume:
    \begin{itemize}
    \item Incompressible laminar Newtonian flow with constant viscosity,
      because the problem says so for this part.  This lets us use the simpler
      form of the equations.
    \item Steady-state, so time derivatives are zero.
    \item Axisymmetric flow, so the velocity derivatives with respect to
      $\theta$ are zero (though not necessarily pressure; if gravity were
      significant, then $\partial p/\partial\theta$ and $\partial p/\partial r$
      would not be zero).
    \item Fully-developed flow, so velocity derivatives in the $z$-direction
      are zero, and flow is only in the $z$-direction ({\em i.e.}
      $u_r=u_\theta=0$).
    \end{itemize}

  \item When all of the terms are canceled, you should be left with:
    \begin{eqnarray}
      {\rm continuity:} & & 0=0 \\
      r{\rm -momentum:} & & 0 = -\frac{\partial p}{\partial r} \\
      \theta{\rm -momentum}: & &
      0 = -\frac{1}{r}\frac{\partial p}{\partial\theta} \\
      z{\rm -momentum:} & & 0 = -\frac{\partial p}{\partial z} +
      \frac{\mu}{r}\frac{\partial}{\partial r}\left(r
        \frac{\partial u_z}{\partial r}\right)
    \end{eqnarray}

  \item For all three of these, we have the same stress profile, which is the
    same as the flux profile for cylindrical diffusion with generation:
    $$\tau_{rz} = -\frac{1}{2}\frac{\partial p}{\partial z} r + \frac{A}{r}$$
    (where $\partial p/\partial z$ is negative).  In the limit of a very thin
    annulus (so the plug nearly fills the tube), this is roughly linear like
    the stress distribution between parallel plates; we can use this simplified
    case to do the graphs.
    \begin{enumerate}
    \item For a Newtonian fluid, this gives a parabolic flow profile.
    \item For a pseudoplastic (shear-thinning) fluid, the magnitude of the
      shear stress is greatest near the walls, so the apparent viscosity is
      lowest there and highest in the center.  This makes the velocity
      distribution flatter in the center, and more curved near the walls.  (For
      example, for a power law pseudoplastic fluid with $n=0.5$, the velocity
      distribution goes as $1-y^3$ instead of the $1-y^2$ parabolic Newtonian
      profile.)
    \item For a Bingham plastic, the lower shear stress near the center will
      lead to a region with $\partial u_z/\partial r=0$, so the velocity
      profile is flat there.  If the maximum shear stress is above the yield
      stress $\tau_0$, then there will be a velocity gradient between this
      central solid region and the walls; otherwise the solid won't budge
      (unless there's some slip against the walls).
    \end{enumerate}
    You were asked to graph these with the same maximum velocity, which should
    have looked like:
    \begin{center}
      \PSbox{extruder.ps}{336pt}{135pt}
    \end{center}

  \item Solid aluminum deforms significantly as it flows past the plug; the
    resulting strain is likely to strengthen it quite a bit.  Stronger means
    the yield stress will be high, so the center will likely flow as a solid
    (uniform velocity) as in the Bingham plastic model.

    Needless to say, the simple Bingham plastic model does not capture the
    change in yield stress as the material flows through the extruder.

  \item During extrusion, grains will be elongated in the direction of flow,
    and dislocation density will rise considerably, possibly resulting in
    formation of new low-angle grain boundaries.  This corresponds to stage II
    or III deformation, where both the higher dislocation density and narrower
    grains result in higher yield stress.

    If the temperature is high enough, recrystallization could dynamically form
    new grains with low defect densities, which would then be sheared along
    with the rest of the material.  This would reduce the force required to
    drive material through the extruder, and likely produce less strengthening
    than at lower temperatures.
  \end{enumerate}
\end{enumerate}
\end{document}