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\newcommand{\ds}{\displaystyle}
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{\bf Question}

A pipe has a variable circular cross section; the radius $r$
depends on the distance along the pipe, $x,$ as $r = a - be^{-x},$
with $a>b.$  If the speed of the water in the pipe is $v_0$ at the
inlet $(x = 0)$ find the speed of the water in the pipe as a
function of $x.$

\vspace{.25in}

{\bf Answer}

\begin{center}
$ \begin{array}{l}
\textrm{Mass Flux:}\ \  {\bf j} = \rho v(x) {\bf i}
\end{array}
\ \ \
\begin{array}{c}
\epsfig{file=215-3-1.eps, width=45mm}
\end{array} $
\end{center}

Mass is conserved, therefore

 $$\frac{dm}{dt}  =  - \int_{S+} {\bf j} \cdot{\bf
n} \, ds + \int_{S-} {\bf j} \cdot{\bf n} \, ds$$

Now $\ds\frac{dm}{dt} = 0.$ Thus $\ds\int_{x=0} {\bf j} \cdot{\bf
n} \, ds = \int_{\alpha} {\bf j} \cdot{\bf n} \, ds$ \ as $\rho$
is constant.

Therefore $\ds\rho \, 2 \pi (a - b) v_0 = \rho 2 \pi (a -
be^{-x})v(x) \Rightarrow v(x) = \frac{a-b}{a-be^{-x}}$



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