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In studying the stream function of a doublet in a uniform flow, I obtained the following stream function: $$\Psi = Uy - \frac{m}{2\pi}\arctan{\frac{2x_0 y}{x^2 + y^2 - x_0^2}} $$ where $x_0$ is half the distance between the source and sink and m is the strength of the source and sink. When considering the streamline which corresponds to the equation $\Psi = 0$, I get the equation $$y = \frac{m}{2\pi U}\arctan{\frac{2x_0 y}{x^2 + y^2 - x_0^2}}$$ Now this streamline is called the $\textbf{Rankine Oval}$.

$\textbf{My question is:}$ How can I prove that it is indeed an oval from the above equation? I am not familiar with any 'oval equations'. My intuition says that it is not always an ellipse either. Is this intuition right?

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  • $\begingroup$ The classification of shapes is mathematics, not physics. $\endgroup$ – sammy gerbil Jan 27 '17 at 6:19
  • $\begingroup$ But since this has ties to a physical problem, it seems to me it is acceptable here as well. That said, have you tried plotting the function? $\endgroup$ – Kyle Kanos Jan 27 '17 at 11:03
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The stream function you give is for a parallel flow plus a source-sink pair located at $x=\pm x_0$. This is not the same thing as a doublet. A doublet is defined as the limit of $x_0\rightarrow0$, $m\rightarrow\infty$ of this source-sink pair such that $m\,x_0=\mbox{const.}=K$

Second, there is no simple closed-form expression of the curve $\Psi=0$ for the case of the Rankine oval that I know of. Numerical computation of that contour will give you the Rankine oval, however, (plus the line $y=0$, of course).

Yes, you are correct, this oval is not an ellipse for any finite choice of the parameters. However, in the above limit of $x_0\rightarrow0$, $m\,x_0=K$ the oval becomes a circle.

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