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\textbf{Question}

Steady viscous incompessible flow takes place parallel to the $z$-axis
in a cylindrical pip of cross section $D$. The kinematic viscosity and
density of the fluid are both constant and are given by $\nu$ and
$\rho$ respectively and there are no body forces. By assuming a
velocity of the form $\un{q} = (0,0,w(x,y))$, show that the pressure
gradient $-P$ in the $z$-direction must be constant. Show also that
$w$ satisfies the equation
$$\nabla^2w=A,$$
where $A$ is a constant that you should specify, and give suitable
boundary conditions for $w$.

When $D$ is a circular disc of radius $a$, find $w$ and show that the
mass flus $M$ is given by the Hagen-Poiseulle discharge formula.
$$M= \frac{\pi P a^4}{8 \nu}.$$

Suppose now that viscous incompressible flow takes place in the same
circular pipe but is driven by an $unsteady$ pressure gradient
$p_z=p_0e^{-k^2t}$ where $t$ denores time and $p_0$ and $k$ are
constants. By seeking a solution of the form
$\un{q}=(0,0,R(r)e^{-k^2t})$, show that the function $R$ satisfies the
ordinary differential equation
$$R''+\frac{1}{r}R' +AR = B$$
where $A$ and $B$ are constants which you should specify. Give
suitable boundary conditions for this equation.

[You may use, without proof, the fact that in cylindrical polar
coordinates $(r,\theta, z)$
$$\nabla^2 = \frac{1}{r}\frac{\partial}{\partial r} \left ( r
\frac{\partial}{\partial r} \right ) + \frac{1}{r^2}
\frac{\partial^2}{\partial \theta^2} + \frac{\partial^2}{\partial z^2}.]$$

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\textbf{Answer}


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Assume $\underline{q}=(0,0,w(x,y))$ then $div(\underline{q})=0$

The three components of the Navier-Stokes equations become

$\begin{array}{rcl} 0 & = & -p_x/\rho + 0\\ 0 & = & -p_y/\rho +
0\\ 0 & = & -p_z/\rho + \nu (w_{xx}+w_{yy}) \end{array}$ $\left.
\begin{array}{l} \\ \\ \end{array} \right \}$ $\Rightarrow p=p(z)$
only.

Now by the standard separation of variables argument, since
$w=w(x,y)$ only we have

$\begin{array}{crcl} & -p_z/rho & = & constant
\\ \Rightarrow & -p_z/\rho & = & P/\rho \ \rm{say}. \ \ (P\
\rm{constant})\\ \Rightarrow & 0 & = & P/\rho +\nu \nabla^2 w\\
\Rightarrow & \nabla^2 w & = & -P/mu \ \ (\underline{x} \in D)
\end{array}$

Boundary conditions:- by no-slip we need $w=0$ on $\delta D$.

If D is a circle of radius $a$, then the problem becomes (assuming
symmetry so that $w=w(r)$ only)

$\begin{array}{crcl} &\displaystyle \frac{1}{r}(rw_r)_r & = &
\displaystyle -\frac{p}{\mu} \ \ (w=0 \ \rm{on} \ r=a)\\
\Rightarrow & (rw_r)_r & = & \displaystyle -\frac{Pr}{\mu},\\ &
rw_r & = & \displaystyle -\frac{Pr^2}{2\mu}+C \\ \Rightarrow & w_r
& = & \displaystyle -\frac{Pr}{2\mu} +\frac{C}{r} \end{array}$

So $\displaystyle w=-\frac{Pr^2}{4\mu}+ C\log r+D.$

Must choose $C=0$ so that $w<\infty$ at $r=0$

Now imposing $w(a)=1 \ \ \Rightarrow$ $$\displaystyle
w=\frac{P}{4\mu}(a^2-r^2)$$

\begin{eqnarray*} M & = & \displaystyle \int_{pipe} \rho w \,dA =
\int_{\theta=0}^{2\pi} \int_{r=0}^a \frac{\rho P}{4\mu}(a^2-r^2)
\,rdrd\theta\\ & = & 2\pi\left [ \frac{\rho P}{4\mu}\left (
\frac{a^2r^2}{2}-\frac{r^4}{4} \right ) \right ]_0^a =
\frac{2\pi\rho P}{4\mu}\frac{a^4}{4} \end{eqnarray*}

Thus as required $$\displaystyle M=\frac{\pi Pa^4}{8\nu}$$

Now the flow is UNSTEADY with $p_z=-p_oe^{-k^2t}$.

Again we seek $\underline{q}=(0,0,w)$ where now $w$ is a function
of $r$ and $t$. So still $div(\underline{q})=0$.

The only N/S equation that does not give $0=0$ is $$w_t=-p_x/\rho
+\nu\nabla^2 W$$

So, with $w$ as suggested, we find that

$\begin{array}{crcl} & \displaystyle -k^2Re^{k^2t} & = &
\displaystyle -\frac{p_0}{\rho}e^{-k^2t}+\nu
e^{-k^2t}\frac{1}{r}(rR')'\\ \Rightarrow & -k^2R & = &
\displaystyle -\frac{p_0}{\rho}+\frac{\nu}{r}(R'+rR'')\\
\Rightarrow & \displaystyle \nu R'' + \frac{\nu R'}{r} + k^2R & =
& \displaystyle \frac{p_0}{\rho} \\ i.e. & \displaystyle R'' +
\frac{R'}{r} +AR & = & B \end{array}$

$$\displaystyle (A=\frac{k^2}{\nu}, \ \ B=\frac{p_0}{\nu\rho})$$

Boundary conditions:- need $R(a)=0$

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