Consider for example a two-dimensional potential flow: a line of sources with volumetric flow per line length, $m$, centered on the origin. For this type of flow, the associated complex potential is:


And so, the stream function corresponds to:


where $\theta=arg\{z\}$ and $k\in \mathbb{Z}$.

Normally books show the previous expression with $k$=0 (they don't even mention the periodicity of the $\arg()$ function). In that situation the lines $\psi=const$ can be drawn as shown on the figure. The volumetric flow per line length can be computed by:


where $Q$ has a conventional direction from left to the right if $\psi_1$ is in a higher position than $\psi_2$.

The configuration $k=0$ will result in a physical problem: the computed volumetric flow between some different stream lines is right, but for example, between $\psi=\frac{7m}{8}$ and $\psi=0$ it will be $Q=-\frac{7m}{8}$, which is wrong. Between those lines is expected a volumetric flow of $\frac{m}{8}$ instead. One way of solving this problem (I'm just guessing), is by considering that each streamline can have multiple simultaneous values, i.e., by considering the factor $k$ on the general expression of $\psi$ (see $(2)$) (see for illustration $\psi_2$ and $Q_2$ on the figure). One confusion that this approach can produce is that $(3)$ can't work for all $\psi_1$,$\psi_2$ (we would need to pick the correct ones), which brings more ambiguity to the problem. My question is: can a streamline be really associated to multiple simultaneous values? enter image description here


No, it cannot. First of all, the angle $\theta$ in your expression for the stream function is to be understood as $\mod(\theta,2\pi)$. Second, you need to be careful regarding the orientation of your streamlines. The flow rate "between" the two streamlines you mention is indeed $7m/8$ if you take the flow outside the acute angle enclosed by the streamlines, and it is $m/8$ inside that angle.

As for the value of $k$, it is completely irrelevant: You can add an arbitrary constant to $\psi$ without changing the flow. What you cannot do, of course, is change the values of $k$ for a single stream function. You can pick any $k$ you like, but once you have picked one, it's fixed.

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  • $\begingroup$ But why is the flow rate between the two streamlines the one outside the acute angle? If I consider $\psi_1=0$ the upper streamline and $\psi_2=\frac{7m}{8}$ the lower one, shouldn't the flow be inside? Actually I can undestand what you said. Because if the streamlines start from the same point, the flow can be outside or inside depending on view perspective. $\endgroup$ – Élio Pereira Apr 28 '17 at 22:16
  • $\begingroup$ Yes. If you take $\psi_1=0$ and $\psi_2=7m/8$, then you need to turn counter-clockwise from the $\psi_1$ streamline and keep going until you hit the $\psi_2$ streamline. The flow rate across that path is $7m/8$. $\endgroup$ – Pirx Apr 28 '17 at 22:22

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