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Next: Circulation and point vortices Up: Vorticity Previous: Evolution of vorticity

Vorticity in 2D and the Streamfunction

Let

\begin{displaymath}
{\bf v} = {\bf i} v_{x}(x,y,t) + {\bf j} v_{y}(x,y,t)
\end{displaymath} (6.9)

The constraint $\mbox{\boldmath$\nabla$}\cdot{\bf v} = 0$ can be satisfied by using a vector potential for ${\bf v}$ (c.f., ${\bf B} = \mbox{\boldmath$\nabla$}\times{\bf A})$.


\begin{displaymath}
{\bf v} = \nabla \times{\bf G}\qquad(\Rightarrow\nabla\cdot{\bf v} = 0)
\end{displaymath} (6.10)

For two dimensional flow, ${\bf G}$ need only have a non-zero ${\bf\hat{z}}$-component, e.g., ${\bf G} = -\Psi(x,y,t){\bf\hat{z}}$.


\begin{displaymath}
{\bf v} = \mbox{\boldmath$\nabla$}\times(-\Psi{\bf\hat{z}})...
...\Psi)\equiv
{\bf\hat{z}}\times(\mbox{\boldmath$\nabla$}\Psi)
\end{displaymath} (6.11)

Note that $\mbox{\boldmath$\nabla$}\times {\bf\hat{z}}=0$. Thus ${\bf v}$ and has the components


\begin{displaymath}
v_{x} = -\partial\Psi/\partial y \quad,\quad v_{y} =
\partial\Psi/\partial x
\end{displaymath} (6.12)

[check that ${\bf v}$ is incompressible: $\mbox{\boldmath$\nabla$}\cdot{\bf v} = \partial
v_{x}/\partial x + \partial
v...
...al^{2}\Psi/\partial x \partial y +
\partial^{2}\Psi/\partial y \partial x = 0.$]



The scalar $\Psi $ is called streamfunction since a contour $\Psi =$ const. is aligned with the velocity - i.e. it is a streamline. See Figure (6.1).

Figure 6.1: Contours of the streamfunction in two dimensions: Streamlines follow $\Psi $ contours.
\includegraphics [scale=0.5]{fundfig6_1.eps}

In 2D, $\displaystyle {\bf\omega} = \mbox{\boldmath$\nabla$}\times{\bf v} =
{\bf\hat{z...
...+
\frac{\partial^{2}\Psi}{\partial y^{2}}\right) =
{\bf\hat{z}}\nabla^{2}\Psi$.

Thus ${\bf\omega} = \omega(x,y,t){\bf\hat{z}}$, and


\begin{displaymath}
\omega = \nabla^{2}\Psi
\end{displaymath} (6.13)

given ${\bf\omega}$, we can solve (6.11) for $\Psi $, then use (6.9) to find ${\bf v}$.

Consequently, equation (6.8) reduces, in 2D $(\partial/\partial z)$ to


\begin{displaymath}
\frac{D{\bf\omega}}{Dt} = \left(\omega \frac{\partial}{\partial
z}\right)(v_{x},v_{y},0) = 0
\end{displaymath} (6.14)

$\Rightarrow {\bf\omega}$ in a given fluid element remains constant, since the convective derivative is zero, as illustrated in Figure (6.2).

Figure 6.2: In two dimensional incompressible flow the vorticity $\omega $ remains constant in any given fluid element throughout the subsequent evolution.
\includegraphics [scale=0.5]{fundfig6_2.eps}


\begin{displaymath}{\bf\omega}_{a}(t_{2}) = {\bf\omega}_{a}(t_{1})\quad,\quad {\bf\omega}_{b}(t_{2}) = {\bf\omega}_{b}(t_{1})\end{displaymath} (6.15)



Example 6.1 Rankine Vortex.

Use $(R, \phi)$ coordinates, rather than $(x,y)$ : ${\bf\omega} = \omega {\bf e}_{z}$, still. The geometry of the Rankine Vortex is illustrated in Figure (6.3) and is defined to be

$\omega = \left\{\begin{array}{cl} \omega_{0} &,\quad R<a\\ 0
&,\quad R>a
\end{array}\right.$

Figure 6.3: Rankine Vortex: Vorticity is constant inside $R=a$, and zero outside. The velocity only has a $\phi $ component.
\includegraphics [scale=0.5]{fundfig6_3.eps}

$\displaystyle \nabla^{2}\Psi = \frac{1}{R}\frac{\partial}{\partial
R} \left(R\...
...partial R}\right) +
\frac{1}{R^{2}}\frac{\partial^{2} \Psi}{\partial \phi^{2}}$

since $\omega = \omega (R)$ is axisymmetric we look for solutions of the form $\Psi = \Psi(R)$


\begin{displaymath}
{\bf v} = {\bf e}_{z}\times \mbox{\boldmath$\nabla$}\Psi = {...
...si}{dR};\Rightarrow v_{R}\equiv 0, v_{\phi} = \frac{d\Psi}{dR}
\end{displaymath} (6.16)

$\Psi = \left\{\begin{array}{cl} \Psi_{1}(R) &,\quad R<a\\
\Psi_{2}(R) &,\quad R>a
\end{array}\right.$

$ \underline{R<a}: {\displaystyle\frac{1}{R}\frac{d}{dR}\left(R\frac{d\Psi_{1}}{...
...
\omega_{0}\Rightarrow R\frac{d\Psi_{1}}{dR}} =
\frac{1}{2}\omega_{0}R^{2} + A$ ($A =$ const)

Dividing by $R$, followed by integration we find


\begin{displaymath}\frac{d\Psi_{1}}{dR} = \frac{1}{2} \omega_{0}R + \frac{A}{R} ...
... \frac{1}{4}\omega_{0}R^{2} + A\log R + B
(B = \textrm{const})\end{displaymath}

since $\Psi_{1}$ must be regular at $R=0,\textrm{we require}A=0;\Psi_{1}=\frac{1}{4}\omega_{0}R^{2} + B$

Now find ${\bf v}$ using (6.16):

$v_{R1} = 0, v_{\phi 1} = d\Psi_{1}/dR = \frac{1}{2}\omega_{0}R$ (This is similar to ``solid body'' rotation with angular velocity $\Omega
{\bf e}_{z}; \Omega = \frac{1}{2} \omega_{0}$.)



$\underline{R>a}:$ (6.13) $\Rightarrow \mbox{\boldmath$\nabla$}^{2}\Psi = 0$


\begin{displaymath}
\frac{1}{R} \frac{d}{dR}\left(R\frac{d\Psi_{2}}{dR}\right) ...
...
R\frac{d\Psi_{2}}{dR} = C \Rightarrow \Psi_{2} = C\log R +D.
\end{displaymath}

(6.16) $\Rightarrow v_{R2}=0, v_{\phi 2} = d\Psi_{2}/dR=C/R$.

${\bf v}$ is continuous at $R=a$, i.e., $v_{\phi 1}(a) = v_{\phi
2}(a)$

$ \frac{1}{2}{\displaystyle\omega_{0}a = \frac{C}{a}, \Rightarrow C = }\frac{1}{2}
\omega_{0}a^{2}$.

Hence the full solution for ${\bf v}$ is $v_{r} = 0$ and (see Figure (6.4))


\begin{displaymath}
V_{\phi} = \left\{\begin{array}{cl} \frac{1}{2}\omega_{0}R &...
...
\frac{1}{2}\omega_{0}a^{2}/R &,\quad R>a
\end{array}\right.
\end{displaymath} (6.17)

Figure 6.4: The variation of $V_\phi $ with $R$ in a Rankine Vortex.
\includegraphics [scale=0.5]{fundfig6_4.eps}


next up previous
Next: Circulation and point vortices Up: Vorticity Previous: Evolution of vorticity
Andrew Wright
2002-09-16