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Homework Help: Stream function for double sink / source flow

  1. Apr 5, 2010 #1
    1. The problem statement, all variables and given/known data
    Consider a source/sink in the origin of a coordinate system with a strength equal to 2m and two sources/sinks, each with a strength of −m, i.e. with a opposite sign with respect to the source/sink in the origin. The two sources/sinks are positioned along the x-axis at x = δx and x = −δx, respectively. Consider the limit for δx → 0 under the condition that

    [tex]\lim_{\delta x \rightarrow 0} \delta x^2 m = \mu_2[/tex]

    with [tex]\mu_2[/tex] finite. Consider this problem in three dimensions and sketch the resulting streamline patterns. Is there a closed streamline, which does not pass through the origin.

    2. Relevant equations
    The potential [tex]\Phi_e[/tex] for a single source/sink of strength m in 3D is:
    [tex]\Phi_e=-\frac{m}{4\pi r}[/tex]
    [tex]\bar{u}_e=\nabla\Phi_e=-\frac{m \bar{r}}{4\pi r^3}[/tex]
    with [tex]\bar{r}=\bar{x}-\bar{x}_0 [/tex] where x = (x, y, z) and [tex] r = |\bar{x}-\bar{x_0}| = \sqrt{x^2 + y^2 + z^2}[/tex]

    3. The attempt at a solution
    Since the poisson equation which leads to the potential solution above is linear, I can add the 2 sinks and 1 source together to get 1 potential:
    [tex]\Phi_e=-\frac{m}{4 \pi}\left(\frac{-1}{|\bar{x}-\delta\bar{x}|}+\frac{2}{|\bar{x}|}+\frac{-1}{|\bar{x}+\delta\bar{x}|}\right)[/tex]

    I have to take the limit to 0 for δx and I want to introduce [tex]\mu_2[/tex] so I take:
    [tex]\Phi_e=\lim_{\delta x \rightarrow 0} -\frac{m \delta x^2}{4 \pi}\frac{1}{\delta x^2}\left(\frac{-1}{|\bar{x}-\delta\bar{x}|}+\frac{2}{|\bar{x}|}+\frac{-1}{|\bar{x}+\delta\bar{x}|}\right)[/tex]

    From this point on I'm highly unsure about my method.

    Taking the limit it follows that [tex]\lim_{\delta x \rightarrow 0} \delta x^2 m = \mu_2[/tex] and [tex]\frac{1}{\delta x^2}\left(\frac{-1}{|\bar{x}-\delta\bar{x}|}+\frac{2}{|\bar{x}|}+\frac{-1}{|\bar{x}+\delta\bar{x}|}\right)=\frac{\partial}{\partial x^2}\left(\frac{3}{|\bar{x}|}\right)[/tex]

    So my potential becomes:
    [tex]\Phi_e=\frac{\mu_2}{4\pi}\frac{\partial}{\partial x^2}\left(\frac{3}{|\bar{x}|}\right)[/tex]

    From here I could proceed to calculate the velocity, but honestly I don't trust my answer enough to take the effort so I first wanted to ask whether anybody could run through my calculation and comment on any mistakes/errors!

    Thanks in advance!
     
    Last edited: Apr 5, 2010
  2. jcsd
  3. Apr 5, 2010 #2
    I do not know what streamline patterns are, but I can help with the math some. I have no idea where you came up with the solution you got. But I played with it some and got a nice solution for the potential in the end, so I can only guess I did it right.

    You will first want to write out [tex]|\vec{x}-\vec{\delta x}|[/tex] out in square roots.

    Then you are going to want to expand your potential in a taylor series since you know [tex]\delta x[/tex] is small.

    I also recommend expanding it out to 2nd order since you know you want [tex]\delta x^2[/tex] (because they give you a limit for it). That, and you will see something happen to the first order term.
     
  4. Apr 6, 2010 #3
    Ok, I know that the source/sink are positioned on the x-axis so [tex]\delta\vec{x}=\delta x[/tex]. Then I get:

    [tex]
    |\vec{x}-\vec{\delta x}|=\sqrt{x^2+y^2+z^2+\delta x^2-2x\delta x}
    [/tex]
    and
    [tex]
    |\vec{x}+\vec{\delta x}|=\sqrt{x^2+y^2+z^2+\delta x^2+2x\delta x}
    [/tex]

    Taylor expanding the [tex]\frac{-1}{|\vec{x}+\vec{\delta x}|}[/tex] and [tex]\frac{-1}{|\vec{x}-\vec{\delta x}|}[/tex] terms yields:

    [tex]
    \frac{-1}{|\vec{x}-\vec{\delta x}|}=\frac{-1}{\sqrt{x^2+y^2+z^2}}-\frac{x}{(x^2+y^2+z^2)^{3/2}}\delta x+0.5\left(\frac{1}{(x^2+y^2+z^2)^{3/2}}-\frac{3x^2}{(x^2+y^2+z^2)^{5/2}}\right)\delta x^2
    [/tex]

    and

    [tex]
    \frac{-1}{|\vec{x}+\vec{\delta x}|}=\frac{-1}{\sqrt{x^2+y^2+z^2}}+\frac{x}{(x^2+y^2+z^2)^{3/2}}\delta x+0.5\left(\frac{1}{(x^2+y^2+z^2)^{3/2}}-\frac{3x^2}{(x^2+y^2+z^2)^{5/2}}\right)\delta x^2
    [/tex]

    So the [tex]\delta x[/tex] term drops out of the potential and the 0-order terms cancel out against the [tex]\frac{2}{|\vec{x}|}[/tex] term so I get:

    [tex]
    \Phi_e=\frac{m\delta x^2}{4\pi}\left[\frac{1}{(x^2+y^2+z^2)^{3/2}}-\frac{3x^2}{(x^2+y^2+z^2)^{5/2}}\right]
    [/tex]

    Taking the limit then turns [tex]m\delta x^2[/tex] into [tex]\mu_2[/tex].

    The problem I have with this equation is that I expect a derivative to pop up somewhere, because that's how the solution for a 'normal' sink/source doublet(single source, single sink with infinitesimally small distance to each other) is. Take a look at this: http://web.mit.edu/fluids-modules/www/potential_flows/LecturesHTML/lec1011/node26.html"
     
    Last edited by a moderator: Apr 25, 2017
  5. Apr 6, 2010 #4
    That last answer looks like the right answer except you lost a negative sign somewhere. Also thanks for the link, I know what you want now. You need to use the trick:

    [tex]\frac{\partial^2 f(x,y,z)}{\partial x^2}=\lim_{h\to 0}\frac{f(x-h,y,z)-2f(x,y,z)+f(x+h,y,z)}{h^2}[/tex]

    Just treat your:

    [tex]f(x,y,z)=\frac{1}{|\vec{x}|}=\frac{1}{\sqrt{x^2+y^2+z^2}}[/tex]

    That will give you a very similar expression to what you had originally, except you were originally off by a factor. Hope this helps. It will also give you the same answer you got in the last post (besides a negative sign) if you take the derivatives of your expression, just to check your work.
     
  6. Apr 6, 2010 #5
    Then I get:
    [tex]
    \Phi_e=\frac{\mu_2}{4\pi}\frac{\partial^2}{\partial x^2}\left(\frac{1}{|\bar{x}|}\right)=\frac{\mu_2}{4\pi}\frac{\partial^2}{\partial x^2}\left(\frac{1}{\sqrt{x^2+y^2+z^2}}\right)
    [/tex]

    Solving the differential part yields for my potential:
    [tex]

    \Phi_e=-\frac{\mu_2}{4\pi}\left[\frac{1}{(x^2+y^2+z^2)^{3/2}}-\frac{3x^2}{(x^2+y^2+z^2)^{5/2}}\right]

    [/tex]

    Just like you said I would. Thanks! That was really helpful!!
     
  7. Apr 6, 2010 #6
    No problem. Hah, that method was easier than a taylor expansion anyways. Didn't know about that method until you showed me, so we both learned something.
     
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