Completely new to fluid dynamics so a bit unsure of what I'm doing here - could I have some guidance as to whether I'm doing this correctly and a little help in plotting please? I've been asked to plot the stream functions of the following complex potentials:

a) $\Omega(z) = Uz$

My understanding is that the stream function is the function associated with the imaginary part, so my solution is:

$Uz = U(x+iy)$ hence the stream function is: $ \psi(x,y) = Uy$?

If that is correct, I can see the streamlines will just be horizontal to the x-axis and the distance between the streamlines would depend on the constant $U$ (which I assume is usually velocity)?

b) $ \Omega(z) = \frac{m}{2\pi}ln(z)$

Taking the parameterisation $z=re^{i\theta}$:

$\frac{m}{2\pi}ln(z) = \frac{m}{2\pi}(ln(r) + i\theta)$ and the stream function is : $\psi(r,\theta) = \frac{m\theta}{2\pi}$

I think I've probably gone wrong here, I'm aware that it's a source flow and what it should look like. Any help appreciated.


EDIT: I think maybe I should be asking how to plot streamlines from the stream function, maybe that's where I'm getting confused?


Your answers seem correct to me - I haven't touched this stuff in a long time, but from Wikipedia it seems you're right that $\psi = \operatorname{Im} \Omega$ and that the streamlines are just the level sets of $\psi$.

You seem on top of (a) - note that $U$ is indeed the velocity since the definition of potential flow is $v = \nabla \operatorname{Re}(\Omega)$, which in this case is just $U e_x$.

For (b), your stream function seems perfect for a sink/source at the origin: the level sets of $m \theta/2 \pi$ are just the level sets of $\theta$, i.e. the radial lines from the origin.

  • $\begingroup$ Hopefully I'm on the right track then - thanks! How would I go about plotting the streamlines of the source flow from the stream function. I'm getting a bit confused there - would I just make the function equal to different constants? $\endgroup$ – Mike Miller Sep 3 '13 at 15:12
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    $\begingroup$ @BritMiller: yup, that's right - in this case you'd just space out some values $\theta_0$ between $0$ and $2 \pi$ and plot the solution curves to $\psi(r,\theta) = \theta_0$. $\endgroup$ – Anthony Carapetis Sep 3 '13 at 15:14
  • $\begingroup$ Brill - thanks for the help, I see what I'm doing now. $\endgroup$ – Mike Miller Sep 3 '13 at 15:18

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