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I have the following streamline function,

$$(r^2 - \frac{a^3}{r})sin^2\theta = b$$

how do I get the speed of this fluid at each point on the streamfunction?

Incompressible, this is the streamline for very large r.

It is derived from the streamfunction in spherical polars for a fluid perturbed by a sphere,

$$\Psi(r,\theta) = -\frac{1}{2}U(r^2 - \frac{a^3}{r})sin^2\theta$$

user57142
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  • You better specify your boundary condition and what sort of fluid this is (e.g. incompressible, irotational ). Otherwise, you will never have a unique solution. – achille hui Feb 25 '13 at 12:46
  • have edited to specify – user57142 Feb 25 '13 at 13:00
  • I apologise for misreading the question, now I look closer I am a little confused. The first expression you give is not a general definition for a streamline of the streamfunction below. A streamline is an equation describing a curve on which $\psi =$ constant. What information are you given and what have you derived? – Tom Oldfield Feb 25 '13 at 13:58

1 Answers1

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The velocity of a fluid in terms of the streamfunction is $u = \nabla\wedge(0,0,\psi)$. The speed is then the magnitude of this vector. In terms of spherical polar co-ordinates:

$$u =\bigg(\frac1r\frac{\partial\psi}{\partial\theta},-\frac{\partial\psi}{\partial r},0\bigg)$$

Note that in general, the streamfunction contains a lot more information than the streamline equation. The streamline equation will give the curves in the plane that the streamlines flow across, but will not have information about the speed of the flow on an particular streamline.

Tom Oldfield
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  • Yes I agree, but I want to find the velocity of the fluid on this streamline, the $u = \nabla \psi$ wont give the velocity on the unique streamline path. – user57142 Feb 25 '13 at 13:22
  • I'm confused. Isn't the $\psi$ in $u = \nabla \psi$ is usually called the velocity potential while the stream function is the name of those $\phi$ where $(u_x, u_y) = (\frac{\partial \phi}{\partial y}, -\frac{\partial \phi}{\partial x})$? – achille hui Feb 25 '13 at 13:25
  • $\Psi$ is the stream function with $u=\nabla \times \Psi$. http://en.wikipedia.org/wiki/Stream_function – Thomas Feb 25 '13 at 13:48
  • @achillehui Indeed I was assuming that $\psi$ was the velocity potential, rather than the streamfunction. I saw something that looked familiar and didn't read the question fully, I apologise! I also realise now that what you give in the original question is not a streamline for the given streamfunction, although if you replace the $C$ with an $-a^3$ then it should be. – Tom Oldfield Feb 25 '13 at 13:50
  • @tomoldfield, no problem. I know the math but not the exact naming conventions of these stuff. I just want to make sure. – achille hui Feb 25 '13 at 14:01
  • @user57142 a streamline is a contour of $\psi$. That is, a curve along which $\psi$ is constant. The equation here is given by $\psi = \Psi(r,\theta) = -\frac{1}{2}U(r^2 - \frac{a^3}{r})\sin^2\theta =$ some constant. Where have you found this question? It might help if you posted it in it's entirety. – Tom Oldfield Feb 25 '13 at 14:09