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How does pressure outside of a wall affect the fluid within?

  1. Dec 19, 2015 #1
    Is it similar to the Newton's cradle principle?
    This is fluid mechanics/control volumes. There is fluid flowing through a duct and it's being analysed using a control volume surrounding the duct

    For example in this question the atmospheric pressure is having an affect on the fluid within the duct even though there is a wall in between the atmosphere and the fluid
     

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  2. jcsd
  3. Dec 19, 2015 #2

    DaveC426913

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    Gold Member

    Surely this is analogous to the pressure on a submarine. They don't make submarines square/rectangular in cross-section because it would result in uneven distribution of pressure. The longest straight side forms the weakest surface, and would buckle easier.

    Unfortunately I'm only guessing because
    a] there's no context to the question, and
    b] I'm no engineer.
     
  4. Dec 19, 2015 #3
    If this is supposed to be a momentum balance to determine the axial force F exerted by the diverging wall on the fluid, and, by Newton's 3rd law, the axial force exerted by the fluid on the diverging wall, then there is a reason you are confused. It's because the analysis is incorrect. Obviously, the pressure of the air outside the duct cannot affect the fluid inside the duct. The correct momentum balance on the fluid passing through the duct is:
    $$F+p_1A_1-p_2A_2=\dot{m}(U_2-U_1)$$
    Chet
     
  5. Dec 19, 2015 #4
    The axial force F is the force on the duct/nozzle I believe, as a result of the fluid flowing through it. I've uploaded another image below. Your answer is what I thought also, but I'm fairly sure we are both wrong, or I just explained it badly to you :D, because this is what we have always been taught in lectures, and I study at Imperial College London so I would hope they would know. What did you think of my Newton's cradle theory? I.e. the momentum of the outer particles being transferred through the wall and inside, with the wall being stationary.
     

    Attached Files:

  6. Dec 19, 2015 #5
    If I described the objective of the calculation correctly, then I am confident of my answer (and you should be confident of yours)...with all due respect to Imperial College.

    I'm not familiar with Newton's cradle problem and I was unable to see the attachment well enough because everything was too small.

    Chet
     
  7. Dec 19, 2015 #6
    Capture.PNG Capture2.PNG capture3.jpg
    okay. I may have to talk with some of my lecturers then :D hopefully you can now see the images now though if you want to look at their full derivation. and I've also uploaded an image of newtons cradle. I was thinking the three balls in the middle may act like the wall and the two particle either side transfer momentum across the wall. here's a video https://en.wikipedia.org/wiki/Newton's_cradle Thanks for all the replies by the way
     
  8. Dec 19, 2015 #7
    Aside from the Newton's cradle thing, which I didn't look at, the derivation of the force F exerted by the nozzle on the fluid and the force -F exerted by the fluid on the nozzle is, in my judgment, done incorrectly. You were correct to observe that the outside air pressure cannot affect the force balance on the fluid whatsoever (except at the exit of the nozzle). So, I stand by what I said.

    This is not the first time a problem virtually identical to this has appeared on Physics Forums, and the correct solution has never involved the outside air pressure.

    Chet
     
  9. Dec 19, 2015 #8
    Here is my solution:

    Bernoulli:$$P_1+\frac{1}{2}ρU_1^2=P_a+\frac{1}{2}ρU_2^2$$

    Momentum Balance: $$F+P_1A_1-P_aA_2=ρA_2U_2^2-ρA_1U_1^2$$

    My result for the force F is the same as theirs, except for an extra term:

    $$F=ρ\left(A_2U_2^2-\frac{A_1U_2^2}{2}-\frac{A_1U_1^2}{2}\right)-p_a(A_1-A_2)$$

    Note that the two results would match if gauge pressures were being used (i.e., ##P_2=0##) rather than absolute pressures (i.e., ##P_2=p_a##), but, it is obvious that, in their development, absolute pressures are being used. And that's what I used in my analysis. If I calculate the force f that the fluid exerts on the duct, it would be the negative of my F value:
    $$f=-F=-ρ\left(A_2U_2^2-\frac{A_1U_2^2}{2}-\frac{A_1U_1^2}{2}\right)+p_a(A_1-A_2)$$
    If I include the force that the outside air exerts on the duct, I get the net force acting on the duct fnet:
    $$f_{net}=-ρ\left(A_2U_2^2-\frac{A_1U_2^2}{2}-\frac{A_1U_1^2}{2}\right)$$
    My point of all this is that, as you had initially correctly concluded, and, as I had later confirmed, they did the analysis incorrectly, both for the original duct and for the nozzle.

    I'm amazed that this is being taught this way at a prominent university, but I guess everyone is human. If they had used gauge pressures, their answer would have been correct.

    Chet
     
  10. Dec 20, 2015 #9
    Well I can't argue with your logic because it's the same as my own, that is what I would have done. I will have to go and ask my lecturers about the notes. Thanks very much for your time and help.

    Oliver
     
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