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\textbf{Question}

An axisymmetric jet of viscous incompressible fluid of consatn density
$\rho$ and constant kinematice viscosity $\nu$ is injected in the
positive $z$-direction through a jole at $r=z=0$ into the same fluid
at rest. Cylindrical polar coordinates $(r,\theta,z)$ are used and the
flow is independent of $\theta$ so that the fluid velocity is given by
$\un{w} = u\un{e}_r+w\un{e}_z$ where $\un{e}_r$ and $\un{e}_z$ are
unit vectors in the $r-$ and $z-$ directions respectively. YOU MAY
ASSUME that in a boundary layer treatment of the flow where the radius
of the jet is assumed to be much less than its length, the
(dimensional) boundary layer equations are
\begin{eqnarray*}
uw_r+ww_z & = & \nu \left ( w_{rr} + \frac{1}{r} w_r \right )\\
(ru)_r + (rw)_z & = & 0
\end{eqnarray*}

%Question 4i
\begin{description}
\item{(i)}
Assuming that bot $ruw$ and $rw_r$ then to zero as $r\to\infty$, show
that the quantity
$$M = 2\pi\rho \int_0^{\infty} rw^2 \,dr$$
is independent of $z$ and give a physical interpretation of this
result. State two condtions that must be satisfies by the velocity on
the $z$-axis and give brief reasons why these must hold.

%Question 4ii
\item{(ii)}
By using the fact that $dM/dz = 0$ to help determine $m$ and $n$,
verify that a similarity solution exists to the problem in the form
\begin{eqnarray*}
\psi & = & z^m f(\mu)\\
\mu & = & rz^{-n}\\
(ru & = & -\phi_z, \ \ \ rw=\phi_r)
\end{eqnarray*}
provided $f$ satisfies the ordinary differential equation
$$ff'-\mu(ff''+f'^2) = \nu (f'-\mu f'' + \mu^2 f''')$$
and give suitable boundary conditions for this equation.

\end{description}

\newpage
\textbf{Answer}

\begin{center}
\begin{tabular}{cc}
\epsfig{file=311-jet.eps, width=70mm} &
$\underline{q}=u\underline{\hat{e}}_r+w\underline{\hat{e}}_z$
\end{tabular}
\end{center}

We may assume that \begin{eqnarray*} uw_r+ww_z & = &
\nu(w_{rr}+\frac{1}{r}w_{r})\\ (ru)_r +(rw)_z & = & 0
\end{eqnarray*}

\begin{description}
%Question 4i

\item{(i)} Consider $$\displaystyle M=2\pi\rho\int_0^{\infty} rw^2
\,dr, \ \ \ \frac{dM}{dz}=2\pi\rho\int_0^{\infty} 2rww_z \,dr$$

Using momentum equation

$\displaystyle \Rightarrow \frac{dM}{dz}=2\pi\rho\int_0^{\infty}
2r(\nu(w_{rr}=\frac{1}{r}w_r)-uw_r) \,dr$

Integrate by parts:-
\begin{eqnarray*} \displaystyle \frac{dM}{dz} & = & 4\pi\rho \displaystyle
\left \{ \int_0^{\infty} \nu rw_{rr}+\nu w_r-ruw_r \,dr \right
\}\\ & = & \displaystyle 4\pi\rho \left \{ \left [ \nu rw_r -ru_w
\right ]_0^{\infty} - \int_0^{\infty} \nu w_r -\nu w_r -(ru)_rw
\,dr \right \} \end{eqnarray*}

Now since $rw_r,\ ruw$ are zero at $0$ by symmetry and $\to 0$ as
$r\to\infty$ we have
\begin{eqnarray*} \displaystyle \frac{dM}{dz}& = &
4\pi\rho\int_0^{\infty} (ru)_r w_r \,dr\\ & = &
-4\pi\rho\int_0^{\infty} (rw)_z w\\ & = & -4\pi\rho\int_0^{\infty}
rww_z \,dr
\end{eqnarray*}

Thus $\frac{dM}{dz}=-\frac{dM}{dz}$ so $\frac{dM}{dz}=0$ and $M$
is independent of $z$.

Physically the result means that the momentum flux of the jet in
conserved along the jet (i.e. is the same $\forall z$).

By symmetry we require $u=w_r=0$ at $r=0$


%Question 4ii
\item{(ii)} With $ru=-\psi_z$, $rw=\psi_r$ the continuity equation
is automatically satisfied.



$\begin{array}{rcl} \rm{With} \ \psi & = & z^mf(\eta),\
\eta=rz^{-n}\ \rm{we\ have}\\ u & = & -\psi_z/r =
-mr^{-1}z^{m-1}f+nz^{m-n-1}f'\\ w & = & \psi_r/r =
r^{-1}z^{m-n}f'\\ w_z & = &
r^{-1}(m-n)z^{m-n-1}f'-nz6{m-2n-1}f''\\ w_r & = &
-r^{-2}z^{m-n}f'+r^{-1}z^{m-2n}f''\\ w_{rr} & = &
2r^{-3}z^{m-n}f'-2r^{-2}z^{m-2n}f''+r^{-1}z^{m-3n}f'''
\end{array}$

$\Rightarrow
(-mr^{-1}z^{m-1}f+nz^{m-n-1}f')(-r^{-2}z^{m-n}f'+r^{-1}z^{m-2n}f'')$
$$+ r^{-1}z^{m-n}f'(r^{-1}(m-n)z^{m-n-1}f'-nz^{m-n}f'')$$ $=\nu
(2r^{-3}z^{m-n}f'-2r^{-2}z^{m-2n}f''+ r^{-1}z^{m-3n}f'''+
r^{-2}z^{m-2n}f''$ $$- r^{-3}z^{m-n}f')$$

At this stage consider the fact that $M$ is independent of $z$.

Thus $\displaystyle \int_0^{\infty} \frac{r}{r^2}
z^{2m-2n}(f'(\eta))^2 \frac{dr}{d\eta} \,d\eta$  must not depend
upon $z$.

\begin{eqnarray*} \displaystyle \Rightarrow \int_0^{\infty}
r^{-1}\frac{z^{2m-2n}}{z^{-n}}(f'(\eta))^2 \,d\eta & = &
\displaystyle \int_0^{\infty} \frac{z^{2m-2n}}{\eta}(f'(\eta))^2
\,d\eta\\ & = & 0\\ & \Rightarrow & m=n
\end{eqnarray*}

Thus
$$(-mr^{-1}z^{m-1}f+mz^{-1}f')(-r^{-2}f'+r^{-1}z^{-m}f'')$$
$$+r^{-1}f'(-mz^{-m-1}f'')$$
$$= \mu (r^{-3}f' -r^{-2}z^{-m}f''+r^{-1}z^{-2m}f''')$$

Thence \begin{eqnarray*}
& & mr^{-3}z^{m-1}ff'-mr^{-2}z^{-1}ff''-mr^{-2}z^{-1}f'^2\\
 & = & \nu (r^{-3}f'-r^{-2}z^{-m}f''+r^{-1}z^{-2m}f''')\\
& & mz^{m-1}ff'-mrz^{-1}ff''-mrz^{-1}f'^2\\
& = & \nu (f'-rz^{-m}f''+r^2z^{-2m}f''')
\end{eqnarray*}

Comparing the first terms on each side $\Rightarrow m=1$

$\begin{array}{crcl} \Rightarrow & ff'-(r/z)ff''-(r/z)f'^2 &  &\\
& = \nu (f' - (r/z)f'' +(r^2/z^2)f'') & & \\ & ff'-\eta ff''-\eta f'^2 & =
& \nu (f' - \eta f'' + \eta^2f''')\\ i.e. & ff'-\eta (ff''+f'^2) &
= & \nu (f'-\eta f'' +\eta^2f''')
\end{array}$

B/C's:- $u=w_r=0$ at $r=0$ $\Rightarrow \ \ f(0)=f'(0)=0$

Flux condition $\displaystyle \Rightarrow \int_0^{\infty}
\frac{(f'(\eta))^2}{\eta} \,d\eta =$ given constant.

\end{description}

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