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Fluid Dynamics question

  1. Mar 6, 2017 #1
    1. The problem statement, all variables and given/known data
    upload_2017-3-6_19-23-1.png

    upload_2017-3-6_19-23-31.png
    2. Relevant equations
    upload_2017-3-6_19-24-8.png

    3. The attempt at a solution
    I don't even know where to start. I don't understand the question.
     
  2. jcsd
  3. Mar 6, 2017 #2
    Have you learned that the speed of the flow is proportional to the magnitude of the gradient of the stream function?
     
  4. Mar 6, 2017 #3
    do you mean

    V^2=u^2+v^2

    where
    upload_2017-3-6_19-42-50.png
     
  5. Mar 6, 2017 #4
    Yes. But you need to express theta in terms of x and y to evaluate these derivatives.(except, of course, far upstream).
     
  6. Mar 6, 2017 #5
    So would it be better to use
    upload_2017-3-6_19-48-8.png
    and make y = rsin(theta)
     
  7. Mar 6, 2017 #6
    Wait, I see what you mean
    upload_2017-3-6_19-54-29.png
     
  8. Mar 6, 2017 #7
  9. Mar 6, 2017 #8
    upload_2017-3-6_20-13-48.png
     
  10. Mar 6, 2017 #9
    I'm not going to check your math. I leave it to you to get the math correct.
     
  11. Mar 6, 2017 #10
    That's fine. I'm not here to learn math. Since I now have V^2 what do I do from here? I still don't understand the question I need to solve.
     
  12. Mar 6, 2017 #11
    You need to show that, at the x and y corresponding to theta = 66.8 degrees and psi = 0, the speed is the same as at y = 0, x = infinity
     
    Last edited: Mar 6, 2017
  13. Mar 6, 2017 #12
    So I know at a large value of -x and y= o that my V^2 is equal to 1.

    would it be appropriate to sub in
    x=rcos(theta)
    y=rsin(theta)
    r=x^2+y^2
    and then plug in 66.8 = theta
     
  14. Mar 6, 2017 #13
    You have to evaluate it at psi = 0.
     
  15. Mar 6, 2017 #14
    Bernoulli eq?
     
  16. Mar 6, 2017 #15
    What about it?????
     
  17. Mar 6, 2017 #16
    The only thing I can think of that would relate the velocity eq and pressure would be that. Is that what I should be using?
     
  18. Mar 6, 2017 #17
    No. You should be setting psi = 0 and theta = 66.8 degrees (in radians). This gives you an equation for y. Once you know y and theta, you know x.
     
  19. Mar 6, 2017 #18
    I see exactly what you mean. I must be losing my mind thinking psi is pressure. I did exactly what you said and got x and y. I plugged into the the V^2 equation and got 1 as my answer.
     
  20. Mar 6, 2017 #19
    Thanks for helping me. I can sleep easy tonight. You are the man Chestermiller!
     
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