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

\documentclass{article}
\usepackage{fullpage}
\newcommand{\PSbox}[3]{\mbox{\rule{0in}{#3}\special{psfile=#1}\hspace{#2}}}
\begin{document}
\begin{enumerate}
\item Non-Newtonian polymer flow in a channel

  \begin{enumerate}
  \item For pseudoplastic fluids, the effective viscosity $-\tau/\dot{\gamma}$
    decreases with increasing shear, so $n$ must be less than one.  (If one,
    it's constant; greater than one, it increases with increasing shear.)

  \item Let's put $y=0$ halfway between the plates, so the boundary conditions
    are $y=\pm \delta/2\rightarrow u_x=0$.

    Because we have symmetric boundary conditions and constant (and therefore
    symmetric) driving force and uniform properties, we can assume the velocity
    is symmetric about $y=0$.  For this reason, we can say that at $y=0$,
    $\frac{du_x}{dy}=0$, and see how this helps us.  We can get
    $\frac{du_x}{dy}$ from above:
    $$\frac{du_x}{dy} =
    \left(-\frac{\Delta P}{\mu_0\cdot L}y + C_1\right)^{\frac{1}{n}}$$
    and set it to zero at $y=0$:
    $$0 = \left(-\frac{\Delta P}{\mu_0\cdot L}\cdot0 + C_1
    \right)^{\frac{1}{n}}$$
    $$0 = C_1^{\frac{1}{n}}$$
    So, $C_1=0$.  What a relief!  This simplifies the general solution to:
    $$u_x = -\frac{\mu_0\cdot L}{\Delta P}\frac{n}{n+1}
    \left(-\frac{\Delta P}{\mu_0\cdot L}y\right)^{\frac{n+1}{n}} + C_2$$
    and we can go a step further to:
    $$u_x = \frac{n}{n+1}\left(-\frac{\Delta P}{\mu_0\cdot L}
    \right)^{\frac{1}{n}}y^{\frac{n+1}{n}} + C_2$$
    This is only strictly valid for non-negative
    $-\frac{\Delta P}{\mu_0\cdot L}y$ and therefore for non-positive $y$.
    We can get the positive $y$ velocities by symmetry.

    Now use the boundary condition $y=-\frac{\delta}{2}\Rightarrow u_x=0$ and
    solve for $C_2$:
    $$0= \frac{n}{n+1}\left(-\frac{\Delta P}{\mu_0\cdot L}\right)^{\frac{1}{n}}
    \left(-\frac{\delta}{2}\right)^{\frac{n+1}{n}} + C_2$$
    $$C_2 = -\frac{n}{n+1}
    \left(-\frac{\Delta P}{\mu_0\cdot L}\right)^{\frac{1}{n}}
    \left(-\frac{\delta}{2}\right)^{\frac{n+1}{n}}$$
    which can also be written
    $$C_2 = \frac{n}{n+1}\frac{\delta}{2}
    \left(\frac{\Delta P}{\mu_0\cdot L}\frac{\delta}{2}\right)^{\frac{1}{n}}$$
    So the velocity profile for non-positive $y$ is given by:
    $$u_x = -\frac{n}{n+1}
    \left(-\frac{\Delta P}{\mu_0\cdot L}\right)^{\frac{1}{n}}
    \left[\left(-\frac{\delta}{2}\right)^{\frac{n+1}{n}} -
      y^{\frac{n+1}{n}}\right]$$

    Note that the maximum velocity at $y=0$, which is given by $C_2$, is
    proportional to the thickness to the $1+\frac{1}{n}$ power.  For a
    Newtonian fluid, this increases quadratically in thickness ({\em i.e.}
    proportional to thickness squared); for dilatant fluids ($n>1$), between
    quadratic and linear; for pseudoplastic fluids ($n<1$), faster than
    quadratic.  Maximum velocity and flow rate are also nonlinear functions of
    pressure difference, pseudo-viscosity, and length!

  \item For $n=0.5$, $u_x\sim y^3$, so this looks like:
    \begin{center}
      \PSbox{pseudo3.ps}{220pt}{128pt}
    \end{center}

  \item The velocity profile is more uniform around the maximum, so the average
    will be more than 2/3 of the maximum.  In fact, for $n=0.5$, it will be 3/4
    of the maximum.
  \end{enumerate}
\end{enumerate}
\end{document}