I've a doubt about my professor's approach on the following apparatus:

Consider a stationary, incompressible, potential and bidimensional flow in the duct shown in the figure bellow:


A stream line is given by $y=\frac{c}{x}$ with $c$ constant. The full flow is parameterized by $c\in[1,4]$.

Some questions were then asked. One of them involved expressing mathematically the stream function $\psi$. My professor just said that $\psi=xy$. It satisfies the stream line equation because $\psi=xy=const\Rightarrow y=\frac{const}{x}$ (note that $\psi$ is constant along a stream line).

But is $\psi=xy$ the only stream function that can result from stream lines defined by $y=\frac{c}{x}$, with $c$ constant? For example, if $\psi=2xy$, we would get $\psi=2xy=const\Rightarrow y=\frac{\frac{const}{2}}{x}$. And because $\frac{const}{2}$ is a constant, then $\psi=2xy$ could be also a candidate stream function.

How could I obtain $\psi$ just through the knowledge of the stream lines equations? Wouldn't we get infinite $\psi$ candidates for such stream lines?

  • 2
    $\begingroup$ Any non-constant function of xy, say $S=f(xy)$, will yield the hyperbolic streamlines you are after, and the flow will be incompressible, but will the flow also be potential, i.e., curl-free? That requires ${{\nabla }^{2}}S=0$. $\endgroup$ – Bert Barrois Apr 11 '18 at 11:31
  • $\begingroup$ Thank you! I think that now I understand. From the additional condition $\nabla^2\psi=0$, I got $\psi=c_1xy+c_2$, where $c_1$ and $c_2$ are constants. The constant $c_2$ can be neglected, it doesn't matter for the flow. Constant $c_1$ is imposed by the volumetric flow that goes through the duct. $\endgroup$ – Élio Pereira Apr 11 '18 at 14:10
  • $\begingroup$ @BertBarrois $\nabla^2S=0$ gives a divergence-free flow, not a potential flow. $\endgroup$ – Deep Apr 12 '18 at 3:23
  • $\begingroup$ @Deep You may be confusing a stream fn with a potential fn. If S is the stream fn, then${{v}_{x}}=+\partial S/\partial y$, ${{v}_{y}}=-\partial S/\partial x$, and $curl(\mathbf{v})={{\nabla }^{2}}S$. $\endgroup$ – Bert Barrois Apr 12 '18 at 11:49
  • $\begingroup$ @BertBarrois You are right. Sorry my bad. $\endgroup$ – Deep Apr 13 '18 at 3:52

Your Answer

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

Browse other questions tagged or ask your own question.