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Up: 3.8 - Method of Previous: 3.9.2 Forces

3.10 Lift due to circulation

Example: force on a vortex in a uniform stream.

\begin{displaymath}\phi = Ux + \frac{\Gamma }{2\pi }\theta
\end{displaymath}


\begin{figure}
\begin{center}
\epsfig{file=lfig1111.eps,height=1in,clip=}
\end{center}
\end{figure}

Consider a control surface in the form of a circle radius R centered at the point vortex. Then according to Newton's law:

\begin{displaymath}\underbrace{0}_{\begin{array}{c}\mbox{\small {For}} \\
\mbo...
...
\frac{d}{dt}M_{cv} = \left( {F_{cs} + F_v } \right) + M_{IN}
\end{displaymath}

$M_{CV}$ = Total (linear) momentum of control volume
$F_{CS}$ = Hydrodynamic force on CS by surrounding fluids
$F_{V}$ = Hydrodynamic force on CS by vortex = -(force on vortex by fluid)
$M_{IN}$ = net flux of momentum in CV through CS
\begin{figure}
\begin{center}
\epsfig{file=lfig1112.eps,height=3in,clip=}
\end{center}
\end{figure}

Therefore, force on vortex $F = -F_{V} =
F_{CS}+M_{IN}$.
Momentum flux:
\begin{align}u = & U - \frac{\Gamma }{2\pi r}\sin \theta , \notag \\
v = & \fr...
...riptscriptstyle\rightharpoonup$ }}\over
{V}} \cdot \hat {n} \notag
\end{align}

\begin{figure}
\begin{center}
\epsfig{file=lfig1113.eps,height=1.2in,clip=}
\end{center}
\end{figure}

Evaluation of $M_{IN}$.
\begin{align}(M_x )_{IN} = & - \rho \int\limits_0^{2\pi } {d\theta ruV_r = } -
...
...^{2\pi } {\cos ^2\theta d\theta = - \frac{\rho U\Gamma }{2}} \notag
\end{align}

Pressure:

\begin{displaymath}p = - \frac{1}{2}\rho \left\vert
\mathord{\buildrel{\lower3p...
...ptscriptstyle\rightharpoonup$ }}\over
{v}} \right\vert^2 + C
\end{displaymath}

Velocity vector:
\begin{align}\left\vert
\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\r...
... r}U\sin \theta + \left( {\frac{\Gamma }{2\pi r}}
\right)^2 \notag
\end{align}

Force on the control surface - $F_{CS}$:

\begin{displaymath}\begin{array}{l}
\left( {F_{CS} } \right)_x = \int\limits_0...
...2\theta } }_\pi = - \frac{1}{2}\rho U\Gamma \\
\end{array}
\end{displaymath}

Finally,

\begin{displaymath}\begin{array}{l}
F_x = \left( {F_{CS} } \right)_x + \left( ...
...left( {M_y } \right)_{IN} = - \rho
U\Gamma \\
\end{array}
\end{displaymath}


i.e. the fluid exerts a downward force $ - \rho U\Gamma
$ on the vortex

Kutta-Joukowski law:

\begin{displaymath}\begin{array}{l}
F = - \rho U\Gamma \\
3D: \mathord{\bui...
...style\rightharpoonup$ }}
\over {\Gamma }} \\
\end{array}
\end{displaymath}

Generalized Kutta-Joukowski Law:

\begin{displaymath}\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightha...
...riptstyle\rightharpoonup$ }}\over
{\Gamma }} _i } } \right),
\end{displaymath}

where $\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$ }}\over
{F}} $ is the total force on a system of n vortices in a free stream with speed $\vec{U}$.

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Up: 3.8 - Method of Previous: 3.9.2 Forces