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Theoretical Fluid Mechanics - Ideal Fluid Flow

  1. Oct 5, 2014 #1
    1. The problem statement, all variables and given/known data
    Write down the complex potential for a source of strength m located at [itex]z=ih[/itex] and a source of strength m located at [itex]z=-ih[/itex]. Show that the real axis is a streamline in the resulting flow field, and so deduce that the complex potential for the two sources is also the complex potential for a flat plate located along the line [itex]y=0[/itex] with a source of strength m located a distance h above it.

    Obtain the pressure on the upper surface of the plate mentioned above from Bernoulli's equation. Integrate the pressure difference over the entire surface of the plate, and so show that the force acting on the plate due to the presence of the source is [itex]pm^2/(4\pi h)[/itex]. Take the pressure along the lower surface of the plate to be equal to the stagnation pressure in the fluid.

    2. Relevant equations
    [itex]z=x+iy[/itex]
    Complex potential general form: [itex]F(z)=\phi +i\psi[/itex]
    Complex potential for a source of strength m : [itex]F(z)=\frac{m}{2\pi}\log(z-z_0)[/itex]
    Bernoulli's equation for ideal irrotational flow: [itex]\frac{\partial\phi}{\partial t}+\frac{p}{\rho}+\frac{1}{2}\nabla\phi\cdot\nabla\phi-G=F(t)[/itex] where [itex]F(t)[/itex] is the unsteady Bernoulli constant

    3. The attempt at a solution
    I have set up the problem in [itex] ({x,iy}) [/itex] space with two sources along the [itex] iy [/itex] axis of strength m, one at +h and one at -h. I then came up with the complex potential of the flow due to these two sources as [itex]F(z)=\frac{m}{2\pi}\log(z+h)+\frac{m}{2\pi}\log(z-h)[/itex]. This is about as far as I get before starting to become lost.

    The next step is that I need to show that [itex]y=0[/itex] is a streamline in the flow field. To do so, I believe I need to separate real and imaginary parts of the complex potential so that I can isolate [itex]\psi[/itex] and show that [itex]\psi=0[/itex] at [itex]y=0[/itex]. When I attempt to tease out [itex]\psi[/itex] from the complex potential I show that [itex]\phi + i\psi=\frac{m}{2\pi}\log(z^2-h^2)[/itex], I can take the exponential of each side to get rid of the log function, but then my [itex]i\psi[/itex] term is tied up in an exponential [itex]e^{\phi+i\psi} =e^{\frac{m}{2\pi}}(z^2-h^2)[/itex]. I have tried using Euler's formula, however, I still end up with a mess in which I can't seem to get equations for [itex]\phi[/itex] and [itex]\psi[/itex] in terms of x and y.

    I am also somewhat uncomfortable with the form of Bernoulli's equation the book gives. I assume that since this flow is ideal, irrotational, and steady that Bernoulii's equation would simplify down to [itex]\frac{p}{\rho}+\frac{1}{2}\nabla\phi\cdot\nabla\phi-G=0[/itex], is this correct?


    Thank you all in advance for your time and guidance. I'm way out of my comfort zone in this course.

    Also, would this topic be better of in one of the other subforums?
     
  2. jcsd
  3. Oct 6, 2014 #2

    BvU

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    Hello Bio, and welcome to PF :)

    This is not my expertise, but if you want to write a complex potential in this way, don't you want something like ##{\bf F}({\bf z}) =\frac{m}{2\pi}\log({\bf z}+ih)+\frac{m}{2\pi}\log({\bf z}-ih)## (bold face F and z are complex, but h is a real number), which indeed also is a little healthier at z=h ? And then you rewrite it to ## {\bf F}({\bf z}) = \phi + i\psi=\frac{m}{\pi}\log({\bf z}^2+h^2)## to easily see ##\psi = 0## if z is real.
     
    Last edited: Oct 6, 2014
  4. Oct 6, 2014 #3
    Thank you for your input BvU. You are correct that it is easy to see that [itex]\psi=0[/itex] when [itex]y=0[/itex] (i.e. when z is real), however, I still need to isolate [itex]\phi[/itex] for use in Bernoulli's equation. I believe I need to do this for all arbitrary y values, and not just for [itex]y=0[/itex]...
     
  5. Oct 6, 2014 #4

    BvU

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    to answer which question ?
    Otherwise, I would simplify the Bernoulli eqn (no time dependence, no gravity) and see if I can find ##\nabla \phi## for y = 0 ...
     
    Last edited: Oct 6, 2014
  6. Oct 7, 2014 #5
    For whatever reason I was hesitant to find [itex]\phi[/itex] for y=0 and felt as though it needed to be a general equation for [itex]\phi[/itex] applicable to any value of y. Why is it okay to use [itex]\phi[/itex] at a specific value of y for Bernoulli's equation? Is it because it's asking for the pressure on the upper surface of a flat plate, therefore y=+0? ... how obvious!
     
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