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\textbf{Question}

A viscous incompressible fluid (having constant density $\rho$ and
constant kinematic viscosity $\nu$) flows steadily in a pipe of
constant cros-sectional are which has its generators parallel to the
$z$-axis. There are no body forces. The cross-section of the pipe is
denoted by $D$, so that the region occupied by the fluid is
$$\{(x,y,z):(x,y) \in D, -\infty < z < \infty \}$$.

\begin{description}
%Question 3a
\item{(a)}
By assuming a flow velocity $\un{q}$ of the form $\un{q}=(0,0,w(x,y))
$, show that $w$ satisfies
$$\nabla^2 w = - \frac{P}{\mu}$$
where $P$ is a constant that should be identified. Give the boundary
consdtion that is satisfied by $w$ on $\delta D$, the boundary of $D$.

%Question 3b
\item{(b)}
Verify that the function
$$\phi(x,y) = K(x^2-y^2/3)(y-a\sqrt{3})$$
satisfies $\nabla^2\phi = -P/\mu$ for a particular choice of $K$,
which you should determine. By finding where $\phi=0$ in the
$(x,y)$-plane, identify the shape of the pipe through which the flow
velocity is given by $\un{q}=(0,0,\phi(x,y))$.

%Question 3c
\item{(c)}
Determine the mass flow rate along the pipe.
\end{description}

\newpage
\textbf{Answer}
\begin{description}
\item{(a)}
$$ \ $$

\begin{center}
\epsfig{file=311-tubeflow2.eps, width=70mm} $\ \ \ $
\epsfig{file=311-crossect.eps, width=30mm}
\end{center}


Assume that $\underline{q}=(0,0,w(x,y))$. Then
$$div(\underline{q})=\frac{\partial}{\partial z}(w(x,y))\equiv 0$$

Navier-Stokes give (note $\underline{q}_t=0$ and
$(\underline{q}.\nabla)\underline{q}=0)$

$\left. \begin{array}{ll} 0 = &\displaystyle \frac{-p_x}{\rho}+\nu
(0)\\ 0 = &\displaystyle \frac{-p_y}{\rho}+\nu (0)\\ 0 =
&\displaystyle \frac{-p_z}{\rho}+\nu (w_{xx}+w_{yy}) \end{array}
\right \}$ (no body forces)

So $p_x=p_y=0$ and $p=p(z)$ only. But from the 3rd equation, if
$p=p(z)$ only and $w=w(x,y)$ only we must have $p_z=constant=-P$
say (thus identifying P).

$\Rightarrow$ $$\displaystyle
0=\frac{P}{\rho}+\frac{\mu}{\rho}(w_{xx}+w_{yy})$$ $$\displaystyle
\Rightarrow \nabla^2w=\frac{-P}{\mu} \ \ \ ((x,y)\in D)$$

On $\partial D$ we must have no-slip, so $w=0$ on $\partial D$.

%Question 3b
\item{(b)}
Consider $\displaystyle\phi
(x,y)=K(x^2-\frac{y^2}{3})(y-a\sqrt{3})$

\begin{eqnarray*}
\phi_x & = & K2x(y-a\sqrt{3})\\
\phi_y & = & -(\frac{2Ky}{3})(y-a\sqrt{3})+K(x^2-\frac{y^2}{3})\\
\Rightarrow \phi_y & = & -Ky^2+Kx^2+2Ka\sqrt{3}/3\\
\phi_{xx} & = & K2(y-a\sqrt{3}) 3\\
\phi_{yy} & = & -2Ky+2Ka/\sqrt{3}
\end{eqnarray*}

$\Rightarrow$
\begin{eqnarray*} \nabla^2\phi-\phi_{xx}+\phi{yy} &
= & 2Ky-2Ka\sqrt{3}-2Ky+2Ka/\sqrt{3}\\ & = & -4Ka/\sqrt{3}
\end{eqnarray*}

Thus $\displaystyle \nabla^2\phi=\frac{-P}{\mu}$ if $\displaystyle
\frac{-P}{\mu}=\frac{-4Ka}{\sqrt{3}}$ $$\displaystyle \Rightarrow
K= \frac{\sqrt{3}P}{4\mu a}$$

Now let us consider where $\phi =0$ in the xy plane.

We have $\phi =0$ when

$\left. \begin{array}{rl} (i) & \ \ y=a\sqrt{3}\\ (ii) & \ \
x=y/\sqrt{3}\\ (iii) & \ \ x=-y/\sqrt{3} \end{array} \right \}$
i.e. on 3 lines.

\begin{center}
$\begin{array}{c} \epsfig{file=311-triangletube.eps, width=40mm}
\end{array}
\begin{array}{c}
\rm{The\ lines\ intersect\ at:}\\
O=(0,0)\\
A=(a,\sqrt{3}a)\\
B=(-a,\sqrt{3}a)
\end{array}$
\end{center}

So the pipe shape is a triangle. Moreover,

$|AB|=2a$,
$|OA|=\sqrt{a^2+(a\sqrt{3})^2}=2A=|OB|$.

Thus the pipe is an EQUILATERAL TRIANGLE.

%Question 3c
\item{(c)}
To find the mass flow, need $\int_D \rho w\,dA.$

Since obviously symmetric about $x=0$

\begin{eqnarray*} M & = & \displaystyle 2\rho \int_{y=0}^{\sqrt{3}a}
\int_{x=0}^{y/\sqrt{3}} \frac{\sqrt{3}P}{4\mu
a}(x^2-y^2/3)(y-a\sqrt3) \,dx \,dy\\ & = & \displaystyle
\frac{2\rho\sqrt3P}{(3\sqrt3)4\mu a}\int_0^a\sqrt3
(\frac{y^3}{3}-y^3)(y-a\sqrt3) \,dy\\ & = & \frac{2\rho P}{12\mu
a} \int_0^{a\sqrt3} \frac{-2y^3}{3}(y-a\sqrt3) \,dy \\ & = &
\frac{-2P\rho}{18\mu a}\left [ \begin{array}{l} \displaystyle
\frac{a^5(\sqrt3)^5}{5}-\frac{\sqrt3a^5(\sqrt3)^4}{4}
\end{array} \right ]\\ & = & \frac{Pa^4\sqrt3 \rho}{20\mu}
\end{eqnarray*}

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