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flumech2

  • 3 years ago

Are velocity components Vr=5rcos(theta), Vtheta = -5rsin(theta) irrotational?

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  1. wio
    • 3 years ago
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    Umm, irrotational? What do you mean?

  2. flumech2
    • 3 years ago
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    Irrotational flow

  3. wio
    • 3 years ago
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    LIke the curl is 0?

  4. flumech2
    • 3 years ago
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    Ya....is that all I have to do?

  5. flumech2
    • 3 years ago
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    gradient X V....?

  6. wio
    • 3 years ago
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    Hold on, I'm not sure what irrotational is. Can you give me a definition?

  7. flumech2
    • 3 years ago
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    Well my textbook it as flows for which no particle rotation occurs

  8. wio
    • 3 years ago
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    Okay so curl is 0... meaning we need to find if it has a potential function.

  9. flumech2
    • 3 years ago
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    This is impossible in reality because all fluids have viscosity, but flows can be assumed irrotational in certain cases

  10. wio
    • 3 years ago
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    \[ \frac{\partial f}{\partial r} =5r\cos(\theta) \implies f = \frac{5}{2}r^2\cos(\theta) +g(\theta) \]

  11. wio
    • 3 years ago
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    \[ \frac{\partial f}{\partial \theta} = -\frac{5}{2}r^2\sin(\theta) +g'(\theta)=-5r\sin(\theta) \]This let's us solve for \(g(\theta)\)

  12. wio
    • 3 years ago
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    there might be another method, I'm not completely sure right now.

  13. flumech2
    • 3 years ago
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    Okay. Well, I think it has to do with the gradient (del) cross velocity function (V)

  14. flumech2
    • 3 years ago
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    The only thing I am confused about is whether there is a different between the irrotational flow and the incompressible flow. It seems like if I find that there is an incompressible flow, it's automatically going to give me an irrotational flow

  15. flumech2
    • 3 years ago
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    My textbook defines the curl as (1/2)(del) X V.....but in rectangular coordinates it's simply defined as (1/2) (dv/dx - du/dx)

  16. wio
    • 3 years ago
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    What's confusing me is if we take into consideration that it is in polar coordinates, and just treat it like Cartesian coordinates.

  17. flumech2
    • 3 years ago
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    Ya. I mean every example I see in my book seems to look different ha....for example, for Vr = 0 and Vtheta = f(r) (1/r)* d/dr(r*Vtheta) = 0 for irrotational flow

  18. flumech2
    • 3 years ago
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    I feel like I kind of understand what is going on, but just not quite able to piece it together.

  19. wio
    • 3 years ago
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    \[ \frac{\partial f}{\partial \theta} = -\frac{5}{2}r^2\sin(\theta) +g'(\theta)=-5r\sin(\theta)\\ g'(\theta) =(5r^2/2+r) (-\sin\theta)\implies g = (5r^2/2+r) (\cos\theta)+C \]

  20. wio
    • 3 years ago
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    \[ \frac{\partial f}{\partial r} =5r\cos(\theta) \implies f = \frac{5}{2}r^2\cos(\theta) +g(\theta)\\ f = 5r^2\cos(\theta)+r\cos(\theta) \]

  21. wio
    • 3 years ago
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    It has a potential function, that means it's curl is 0, it's conservative, irrotational, etc.

  22. flumech2
    • 3 years ago
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    Another example my prof. gave in class was this: Given: velocity field V= \[\frac{ -q }{2 \Pi r } e _{r} + \frac{ K }{ 2 \Pi r } e _{}\] is it irrotational

  23. flumech2
    • 3 years ago
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    and what he did was gradient x V

  24. flumech2
    • 3 years ago
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    and got 0.....so I guess that is the same thing as you are saying essentially, right?

  25. wio
    • 3 years ago
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    Whoops, did a bit of algebra wrong there.

  26. wio
    • 3 years ago
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    \[ \frac{\partial f}{\partial \theta} = -\frac{5}{2}r^2\sin(\theta) +g'(\theta)=-5r\sin(\theta)\\ g'(\theta) =(5r^2/2-5r) (\sin\theta)\implies g(\theta) = -(5r^2/2+r) (\cos\theta)+C \]

  27. wio
    • 3 years ago
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    I keep messing up trying to find the damn potential function. Maybe it doesn't have one.

  28. flumech2
    • 3 years ago
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    I think it is \[del = e _{_{r}} \frac{\partial}{\partial r} + e _{_{\theta}} \frac{ 1 }{ r } \frac{\partial}{\partial \theta} \]

  29. flumech2
    • 3 years ago
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    and then that cross the the velocity vector of vr and vtheta

  30. flumech2
    • 3 years ago
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    \[e _{_{r}} \frac{\partial}{\partial r} + e _{_{\theta}} \frac{ 1 }{ r } \frac{\partial}{\partial \theta} X [ e _{_{r}} 5\cos(\theta) + e _{_{\theta}} -5\sin(\theta)]\]

  31. flumech2
    • 3 years ago
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    = 0

  32. wio
    • 3 years ago
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    yeah, I don't know those formula unfortunately. The fact you bring them up makes me think we're dealing with polar coords though.

  33. flumech2
    • 3 years ago
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    We are definitely dealing with polar coordinates, but I think it's still the curl

  34. wio
    • 3 years ago
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    \[ \frac{dx}{dt} = \frac{dx}{dr}\frac{dr}{dt} \]

  35. wio
    • 3 years ago
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    anyway I'm going to bed.

  36. flumech2
    • 3 years ago
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    Okay, thanks for your help

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