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{\bf Question}

Consider a channel with a rectangular cross-section.  If the depth
of the water is $d(x,t).$ where $x$ measures distance along the
channel and $t$ is time.
\begin{description}
\item[(a)] Show that $d$ satisfies $$\frac{\partial d}{\partial t} +
\frac{\partial (vd)}{\partial x} = 0,$$ where $v(x,t)$ is the
speed of the water in the channel.

\item[(b)] Find the analogous equation if the water flows in a
circular pipe of radius $R$.
\end{description}

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{\bf Answer}

\begin{description}
\item[(a)]
Consider an element of the channel of length $dx.$

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Conservation of  mass for this element gives

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

$\ds m = \rho A dx \Rightarrow {\bf j} = \rho v(x) {\bf i};
\hspace{.1in} {\bf n} = {\bf i}$

Therefore $\ds \rho dx \frac{\partial}{\partial t}(d \omega) = -
\frac{\partial}{\partial x}(\rho v \omega d) dx$

So $\ds \frac{\partial d}{\partial t} + \frac{\partial}{\partial
x}(vd) = 0$

\item[(b)]
Cross section:

PICTURE



\begin{eqnarray*}{\rm Area}\ A  & = & R^2 \theta - (R - d) R \sin \theta \\ & = & R^2 \cos^{-1}
\frac{R - d}{R} - R ( R - d) \sqrt{1 - \left( \frac{R -
d}{R}\right)}\end{eqnarray*}

The same procedure as above gives: $\ds \frac{\partial}{\partial
t}A(d) + \frac{\partial}{\partial x}(VA(d)) = 0$

(Substitution for A gives a messy equation)
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