1
$\begingroup$

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.

Thanks.

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

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ 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
  • 1
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.