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Two-Dimensional Irrotational Flow in Cylindrical Coordinates

As we have seen, in a two-dimensional flow pattern, we can automatically satisfy the incompressibility constraint, $\nabla\cdot{\bf v}=0$, by expressing the pattern in terms of a stream function. Suppose, however, that, in addition to being incompressible, the flow pattern is also irrotational. In this case, Equation (477) yields
\begin{displaymath}
\nabla^2\psi = 0.
\end{displaymath} (516)

In cylindrical coordinates, since $\psi=\psi(r,\theta,t)$, this expression implies that (see Section C.3)
\begin{displaymath}
\frac{1}{r}\,\frac{\partial}{\partial r}\!\left(r\,\frac{\pa...
...c{1}{r^{\,2}}\,\frac{\partial^2\psi}{\partial\theta^{\,2}}
=0.
\end{displaymath} (517)

Let us search for a separable steady-state solution of Equation (517) of the form

\begin{displaymath}
\psi(r,\theta) = R(r)\,\Theta(\theta).
\end{displaymath} (518)

It is easily seen that
\begin{displaymath}
\frac{r}{R}\,\frac{d}{dr}\!\left(r\,\frac{dR}{dr}\right)= - \frac{1}{\Theta}\,\frac{d^2\Theta}{d\theta^{\,2}},
\end{displaymath} (519)

which can only be satisfied if
$\displaystyle r\,\frac{d}{dr}\!\left(r\,\frac{dR}{dr}\right)$ $\textstyle =$ $\displaystyle m^2\,R,$ (520)
$\displaystyle \frac{d^2\Theta}{d\theta^{\,2}}$ $\textstyle =$ $\displaystyle -m^2\,\Theta,$ (521)

where $m^2$ is an arbitrary (positive) constant. The general solution of Equation (521) is a linear combination of $\exp(\,{\rm i}\,m\,\theta)$ and $\exp(-{\rm i}\,m\,\theta)$ factors. However, assuming that the flow extends over all $\theta $ values, the function $\Theta(\theta)$ must be single-valued in $\theta $, otherwise $\nabla\psi$--and, hence, ${\bf v}$--would not be be single-valued (which is unphysical). It follows that $m$ can only take integer values (and that $m^2$ must be a positive, rather than a negative, constant). Now, the general solution of Equation (520) is a linear combination of $r^m$ and $r^{-m}$ factors, except for the special case $m=0$, when it is a linear combination of $r^0$ and $\ln r$ factors. Thus, the general stream function for steady two-dimensional irrotational flow (that extends over all values of $\theta $) takes the form
\begin{displaymath}
\psi(r,\theta) = \alpha_0+ \beta_0\,\ln r + \sum_{m>0} (\alpha_m\,r^m+\beta_m\,r^{-m})\,
\sin[m\,(\theta-\theta_m)],
\end{displaymath} (522)

where $\alpha_m$, $\beta_m$, and $\theta_m$ are arbitrary constants. We can recognize the first few terms on the right-hand side of the above expression. The constant term $\alpha_0$ has zero gradient, and, therefore, does not give rise to any flow. The term $\beta_0\,\ln r$ is the flow pattern generated by a vortex filament of intensity $2\pi\,\beta_0$, coincident with the $z$-axis. (See Section 5.11.) The term $\alpha_1\,r\,\sin(\theta-\theta_1)$ corresponds to uniform flow of speed $\alpha_1$ whose direction subtends a (counter-clockwise) angle $\theta_1$ with the minus $x$-axis. (See Section 5.9.) Finally, the term $\beta_1\,\sin(\theta-\theta_1)/r$ corresponds to a dipole flow pattern. (See Section 5.10.)

The velocity potential associated with the irrotational stream function (522) satisfies [see Equations (459) and (474)]

$\displaystyle \frac{\partial\phi}{\partial r}$ $\textstyle =$ $\displaystyle \frac{1}{r}\,\frac{\partial\psi}{\partial\theta},$ (523)
$\displaystyle \frac{1}{r}\,\frac{\partial\phi}{\partial\theta}$ $\textstyle =$ $\displaystyle -\frac{\partial\psi}{\partial r}.$ (524)

It follows that
\begin{displaymath}
\phi(r,\theta) = \alpha_0-\beta_0\,\theta+\sum_{m>0}(\alpha_m\,r^m-\beta_m\,r^{-m})\,\cos[m\,(\theta-\theta_0)].
\end{displaymath} (525)


next up previous
Next: Inviscid Flow Past a Up: Incompressible Inviscid Fluid Dynamics Previous: Two-Dimensional Vortex Filaments
Richard Fitzpatrick 2012-04-27