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Drawing constant pressure lines in flow field

  1. Mar 23, 2017 #1

    joshmccraney

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    Hi PF!

    Can someone help me understand how to draw lines of constant pressure in an inviscid flow field, say flow around a cylinder. I am having trouble understanding how to draw these. Any help is greatly appreciated!

    I ask this question because I am preparing for the Q exam for PhD and one of my classmates asked me about a Hele-Shaw flow: imagine two circular plates very close to each other (infinitely close). Now compare the streamlines of this Hele-Shaw flow and inviscid flow around a cylinder. Also draw lines of constant pressure for both cases. Evidently the streamlines looks identical but the pressure lines are different.
     
    Last edited: Mar 23, 2017
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  3. Mar 23, 2017 #2
    What is your equation for the variations of pressure in inviscid flow?
     
  4. Mar 23, 2017 #3

    joshmccraney

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    Bernoulli's equation, will give a pressure distribution for inviscid flow regimes, so this could give us the pressure variations too, right? For flow around a cylinder this is expressed as $$p = \frac{1}{2} \rho U^2 (2 R^2 \cos(2\theta)/r^2-R^4/r^4)+p_\infty$$
     
  5. Mar 23, 2017 #4
    Yes. That's fine. How about in general, in terms of the stream function.
     
  6. Mar 23, 2017 #5

    joshmccraney

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    My thoughts are to take Bernoullis equation ##p + \rho v^2/2 =p_\infty + \rho u_\infty^2/2## (ignoring unsteady and gravity terms) and then write ##v## in terms of the streamfunction. but this will depend on the coordinate system we choose. Since you're asking for a "general" description I'm confused what coordinate system to represent ##\psi## in. Or am I missing the point?
     
  7. Mar 23, 2017 #6
    You should be able to do it irrespective of the coordinate system.
     
  8. Mar 24, 2017 #7
    In 2D, what is the gradient of the stream function dotted with itself?
     
  9. Mar 25, 2017 #8

    joshmccraney

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    You're asking for ##\nabla \psi \cdot \nabla \psi##? I am not sure. I know ##\nabla \psi## is normal to the velocity, but this is all I can say about it, and without selecting a coordinate system I am not sure how to proceed. Any help?
     
  10. Mar 25, 2017 #9
    Isn't it ##V^2##?
     
  11. Mar 27, 2017 #10

    joshmccraney

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    Is it? I've always seen the relationship with ##\psi## and the velocity as either ##\vec{V} = \nabla \times \psi \vec{e}## where ##\vec{e}## is the basis vector we do not need in order to describe the flow. How could I prove ##\nabla \psi \cdot \nabla \psi = V^2##?
     
  12. Mar 27, 2017 #11
    Just do it in cartesian coordinates.
     
  13. Mar 27, 2017 #12

    joshmccraney

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    Yea, I see it's true in cartesian, but writing this in cartesian coordinates goes against your comment in post 6. Could you direct me where to go to find the general proof that ##\nabla\psi\cdot\nabla\psi=V^2##? I've searched around but have been unable to find anything.
     
  14. Mar 27, 2017 #13
    I don't have a reference. Try cylindrical and see if it works for that.
     
  15. Mar 27, 2017 #14

    joshmccraney

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    Ok, so in cylindrical I'll assume we have flow only in ##r## and ##\theta##, so there is no ##z## component (I could do a different flow if you prefer?). Define the streamfunction ##\psi## as $$\vec{V} = \nabla \times \psi \hat{z} =-\frac{\psi_\theta}{r}\hat{r}+\psi_r\hat{\theta} \implies\\ |\vec{V}|^2 = v_r^2+v_\theta^2 = \frac{\psi_\theta^2}{r^2}+\psi_r^2$$ and notice
    $$\nabla \psi \cdot\nabla\psi = \frac{\psi_\theta^2}{r^2}+\psi_r^2$$ so this works. Bernoullis is then
    $$p + \rho v^2/2 =p_\infty + \rho u_\infty^2/2 \implies\\ p =p_\infty + \rho u_\infty^2/2 - \rho \nabla\psi\cdot\nabla\psi/2.$$
    But then how to draw ##p##; I don't see how writing it in terms of the streamfunction helps. Any ideas?
     
  16. Mar 28, 2017 #15
    You calculate p on a grid, and then use a graphics package with 2D contour plotting capability to draw the diagram showing the lines of constant p.
     
  17. Mar 30, 2017 #16

    joshmccraney

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    So there's no real way to intuitively do this sort of thing "on the fly"?
     
  18. Mar 31, 2017 #17

    Nidum

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    There are complete analytic solutions for stream function and velocity potential for a few idealised situations .

    These can sometimes be used to generate analytic solutions for more practical problems .

    They are also useful when devising efficient numerical solution methods and when just sketching flow patterns .

    http://www.freestudy.co.uk/fluid mechanics/t5203.pdf

    Ch.4-7 are most relevant .
     
  19. Mar 31, 2017 #18
    I don't know what you mean. In some cases, you can solve for ##\psi## as a function of the spatial coordinates, and then solve for the lines of constant ##\psi##.
     
  20. Apr 9, 2017 #19

    joshmccraney

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    Sorry for the ambiguity: by "intuitive" I mean, can you look at a velocity field with a known geometry and infer the pressure lines? For example, the streamlines of a flow around a cylinder are not normal to lines of constant pressure. However, if we have to cylindrical plates that are very very close, evidently the streamlines won't change, yet now the pressure is always normal to the streamlines. Can you help me understand why this is and how I would know they are normal to the streamlines?
     
  21. Apr 10, 2017 #20
    If the lines of constant pressure are perpendicular to the streamlines, then the gradient of pressure is in the same direction as the velocity vector. Using the Navier Stokes equations, what can you do to test this?
     
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