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Complex Potential Flow - Two Vortices offset from the Origin

  1. Feb 9, 2013 #1
    1. The problem statement, all variables and given/known data

    a. Determine the complex potential for two equal counter-rotating vortices with strength [itex]\Gamma[/itex], the positive one located at z=-a and the negative one at z=a.

    b/ Show the shape of the streamlines for this case.

    2. Relevant equations

    z = x + iy = r*e^(i[itex]\theta[/itex])

    W(z) = [itex]\Phi[/itex] + i[itex]\Psi[/itex]
    where [itex]\Phi[/itex] is the potential function and [itex]\Psi[/itex] is the stream function

    3. The attempt at a solution

    a. This part is relatively easy. I know that the complex potential for a vortex at the origin is

    i[itex]\Gamma[/itex]/(2*[itex]\pi[/itex]) * ln(z)

    Therefore, for the two vortices, we will have:

    W(z) = i[itex]\Gamma[/itex]/(2*[itex]\pi[/itex]) * ln((z+a)/(z-a))


    b. This is where I start to have trouble. I need to separate W(z) into the real and imaginary parts, and then I know that the imaginary part is the stream function. However, I don't know how to do this. It is easy for a vortex at the origin, because then I would have

    ln(z) = ln(r*e^(i[itex]\theta[/itex])) = ln(r) + i[itex]\theta[/itex]

    However, the addition and subtraction of 'a' inside of the log is giving me a lot of trouble. I've tried to separate it every way that I can think of but haven't had any success. Any help would be much appreciated. Thanks!
     
  2. jcsd
  3. Feb 9, 2013 #2

    mfb

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    To separate real and imaginary parts, it is easier to write the log as ln(z+a)-ln(z-a). In both cases, this is just a shift by +-a.
     
  4. Feb 9, 2013 #3
    I tried that, but still did not make any progress.

    ln(z+a) - ln(z-a)
    ln(r*e^iθ + a) - ln(r*e^iθ - a)

    Now what? I can't split it up into ln(r)+iθ anymore.
     
  5. Feb 9, 2013 #4

    mfb

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    What about ##r^+##,##\theta^+## corresponding to the magnitude and phase of z+a?
     
  6. Feb 9, 2013 #5
    I'm sorry I don't quite follow you, could you explain in more detail? I'm unfamiliar with that notation. Thanks.
     
  7. Feb 9, 2013 #6

    mfb

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    That is a notation I invented for your specific problem.

    z+a can be written as ##z+a=r^+ e^{i \theta^+}## with some real values ##\theta^+## and ##r^+## - there are formulas how to convert an arbitrary complex number to that shape.
    In the same way, z-a can be written as ##z-a=r^- e^{i \theta^-}## with some real values ##\theta^-## and ##r^-##.
     
  8. Feb 9, 2013 #7
    Would I do something along these lines?

    ln(z+a)
    ln(r*e^iθ + a)
    ln(r*cos(θ)+i*r*sin(θ) + a)
    ln((r*cos(θ)+a) + i*r*sin(θ))

    And then try to find a new r and θ in order to put this back into the form of r*e^iθ ? That doesn't seem to me like it will be possible, and I can't find the formulas that you mentioned. Are they available online somewhere?
     
  9. Feb 9, 2013 #8

    mfb

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    That is possible.
    The standard formulas how you get r and θ if you have a complex number a+ib? They are everywhere.
     
  10. Feb 9, 2013 #9
    I think I figured it out, thanks so much!
     
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