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6.2: Complex Potential- Basic examples

  • Page ID
    76231
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    Basic examples


    Uniform flow

    The complex potential

    \(\begin{eqnarray}\label{uniform}
                F(z)=Ue^{-i\alpha}z
            \end{eqnarray}\)

    corresponds to uniform flow at speed \(U\) in a direction making an angle with the \(x\)-axis.

    Here we are interested in finding the velocity field

    \(\mathbf V = \left(u(x,y), v(x,y)\right).\)

    But first we need to obtain the stream funcion \(ψ\), which is the imaginary component of (1).

    Rewriting (1) we obtain

    \(\begin{eqnarray*}
            F(z)&=& Ue^{-i\alpha}z \\
                &=& U\left( \cos \alpha - i \sin \alpha \right)\left( x+iy\right) \\
                &=& U\left( x\cos \alpha + y \sin \alpha  \right) + i U\left( y\cos \alpha - x \sin \alpha  \right).
            \end{eqnarray*}\)

    Thus

    \(\begin{eqnarray*}
        \psi =  U\left( y\cos \alpha - x \sin \alpha  \right).
        \end{eqnarray*}\)

    Finally, since \(u=\frac{\partial \psi}{\partial y}\) and \(v=-\frac{\partial \psi}{\partial x}\), we have that

    \(\begin{eqnarray*}
            u=U \cos \alpha ,\quad v=U\sin \alpha.
            \end{eqnarray*}\)

    The applet below shows a simulation of the uniform flow. Drag the sliders to change parameters. Click on Trace button to show streamlines. Click on Field button to show vector field.


    Stagnation point flow

    The complex potential

    \(\begin{eqnarray*}
                F(z)=\frac{k z^2}{2}
            \end{eqnarray*}\)

    corresponds to the stagnation point flow with strength \(k≥0\).


    Source & Sink

    A source of strength \(Q>0\) at the origin is represented by the complex potential

    \(\begin{eqnarray}\label{source-sink}
                F(z)=\frac{Q}{2\pi}\log z .
            \end{eqnarray}\)

    Note that this is a multi-valued function, with a branch point at the origin. If \(Q<0\), then the complex potential corresponds to a sink.

    It is easy to generalise (2) for an arbitrary point \((a,b)\) in the complex plane. The required complex potential is

    \(\begin{eqnarray*}
                F(z)=\frac{Q}{2\pi}\log(z-c).
                \end{eqnarray*}
    \)

    where \(c=a+ib\).


    Vortex

    A vortex of strength \(C\) at the origin is represented by the complex potential

    \(\begin{eqnarray*}
                F(z)=\frac{-iC}{2\pi}\log z .
            \end{eqnarray*}\)

    This is again a multi-valued function. For \(C>0\), rotation is anticlockwise, and for \(C<0\) rotation is clockwise.

    A vortex at an arbitrary point \(c\in \mathbb C\) is represented by the complex potential

    \(\begin{eqnarray*}
                F(z)=\frac{-iC}{2\pi}\log(z-c).
                \end{eqnarray*}\)

    where \(c=a+ib\).

    Exercise \(\PageIndex{1}\)

    Find the velocity fields of the Stagnation point, Source & Sink and Vortex flows.


    Combining complex potentials

    The basic flows presented above can be combined by simply superimposing the corresponding complex potentials.

    For example, consider a uniform flow \(Uz\), with speed \(U≥0\), and a source \(\frac{Q}{2\pi}\log z\), with \(Q≥0\). Thus we can produce the complex potential

    \(\begin{eqnarray}\label{comb}
                F(z)=Uz+\frac{Q}{2\pi}\log z .
            \end{eqnarray}\)

    The following applet shows the flow produced by (3). Drag the sliders to change parameters.

    Exercise \(\PageIndex{2}\)

    Find the velocity field of the flow produced by a source of strength \(Q\) in a uniform flow at speed \(U\) in the \(x\)-direction.


    This page titled 6.2: Complex Potential- Basic examples is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by Juan Carlos Ponce Campuzano.

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