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Pressure of Fluid

  1. Jan 9, 2006 #1
    I have noticed what appears to be a geometric theory of pressure vessels. Its probably already been proved, so has anyone heard of it? If not, I may play around to see if I can get any math out of it to work. I have noticed that if you have any closed body with internal pressure, and you make a cut anywhere along the body, and separate the body so that its a new free body diagram, that the net force in the direction perpendicular to the cut, is always going to be equal to the pressure, times the cross sectional area where the cut was made. Its simliar to the Gauss' law about net flux in a way, except that the magnitude of the electric field goes down in Gauss' law, here the pressure force always is the same, even at the cut. Thanks guyz!
     
    Last edited: Jan 9, 2006
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  3. Jan 10, 2006 #2

    Q_Goest

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    Hi Cyrus, it sounds like you're describing a statics problem wherein you're making a freebody diagram and finding the sum of the forces equals zero. Is there more to it than that?
     
  4. Jan 10, 2006 #3

    FredGarvin

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    Cyrus,
    I agree with Q in that you are essentially doing a statics problem. There are a couple of situations where you do need to be careful:

    1) Near ends. Your typical analysis is fine except for when you get close to the closed ends of the vessel. That's when things get a bit interersting.

    2) If the walls are effectively considered "thin". IIRC, the criteria was approximately r/t >= 10. If this isn't the case, then the plane stress is tough to be justified.
     
  5. Jan 10, 2006 #4
    Well, I already know the formula for radial stress of thick walled, and that of thin walled vessels. What im saying is that when you cut a vessel, in static equilibrium, they show that the pressure force in the x direction is the area of the cut plane, times the pressure force. But this is not directly obvious, because the pressure is acting normal to the surface around the sphere. But I did an easy proof to show that for a cylinder this is equal to the area of the cut plane times the pressure. What I'm saying is that this seems to be generally true no matter what general shape the walls of the cylinder are, that the pressure in the x direction will be equal to the area of the cut plane times the pressure.
     
  6. Apr 16, 2011 #5
    Hi Cyrus!

    Would you mind giving me an idea of exactly how you did this proof? I'm doing thin and thick cylinders and its bothering me too....

    Also, could anyone give the general proof that Cyrus was askking for? It'd be really helpful ...
     
  7. Apr 16, 2011 #6
    Not a proof, but an intuitive way to look at it.

    Have a flat plate supporting pressure. You know the force applied to its supports. Roughen the surface with some sandpaper. The force wouldn't change, even though the surface area may have just doubled, and the direction of most of the pressure force is no longer normal to the plate, but normal to the groove surfaces.

    To get all mathy, I'd try to define the total force as an integral of pressure and area and surface normals or something. Then I guess you can show this integral equals the cut area * pressure. Hmm, might get too messy for arbitrary geometries tho.
     
  8. Apr 17, 2011 #7
    HI Unrest,

    Here's my approach to what could be a simple mathematical proof...but there's something wrong....any idea what it is?......

    According to the picture I've attached, the Pressure acts in radial lines on the surface of the cylinder....it has two components PcosA and PsinA.....while considering half of the crossection of the cylindrical pipe, the components PcosA cancel out....then we get the total vertical pressure on the half-crossesction of the pipe as integral(PsinA) within the limits 0 to pi...that gives 2P, not P.....also, if we use this method to analyse a full crossesction of the pipe (considering the uppur half of the pipe also), then all the components PcosA aswell as PsinA will get cancelled ans there'll be no resultant pressure!!
     
  9. Apr 17, 2011 #8
    Here's the pic
     

    Attached Files:

  10. Apr 17, 2011 #9
    Pressure is a scalar, force is a vector.
    You don't resolve pressure.

    Notice the original posts were about the resultant force of the pressure.
     
  11. Apr 17, 2011 #10
    Oh gosh, small details can hit back big time!! Thanks, Studiot....
    ...Okay, so could anyone give me a hint as to how to prove this thing now?
     
  12. Apr 17, 2011 #11
    It sounds reasonable, but as studiot said, you're calling force pressure. If you call it force, then it's fine. The half pipe gets only a net downward force, and the complete pipe gets zero net force. You'd expect zero net force on the complete pipe because it doesn't get pushed sideways.

    But it doesn't quite count as a proof because it only works for a cylindrical pipe. You might as well use a square pipe and it'd be even easier!
     
  13. Apr 17, 2011 #12
    Hmm, okay, but then what is the proof?
     
  14. Apr 17, 2011 #13
    I want to say:

    S is the surface of the half pipe, or whatever pressure vessel
    [tex]\textbf{n}[/tex] is the normal vector
    p is pressure

    Total force = p [tex]\int_{S} \textbf{n}dS[/tex] = independent of the shape of S, as long as the edge is fixed.

    I'm sure there must be a theorum (Something like Stokes?) that says this is true.

    If the surface is closed then there must also be a theorum saying that this surface integral over a closed surface is zero. In that case it would be a special case of the proof of Archimede's principle for bouyancy force. Here the pressure is constant, but with bouyancy the pressure varies with depth.
     
    Last edited: Apr 17, 2011
  15. Apr 17, 2011 #14
    OKay...but we were trying to prove that the net effect of pressure in any direction will be equal to the area of the cut plane times the pressure... like what Cyrus said....
     
  16. Apr 17, 2011 #15
    Yea I understand that. I just don't quite have my 'proof' complete. The integral I showed should be the net force caused by pressure on an arbitrarily shaped surface S (such as your half pipe). So if you can prove that it's equal to the integral over any other surface with the same boundary, then you can choose to use the flat surface directly across the cut.

    Writing it this way takes the details of the physics out of the picture. It's just the integral of the normal vectors of a surface. It seems intuitive that that should be zero for a closed surface, and for an open surface (like our case), it should be independent of the shape.

    I suspect you can use the same kind of theorum as for proving the Archimedes principle, as well as Gauss's law as Cyrus said.
     
  17. Apr 17, 2011 #16
    I think I'm just about there.

    This other thread says

    [tex]
    \mathbf{F} = \int_{surface}p(\mathbf{-n})\,dA = 0
    [/tex]

    That means the total force caused by pressure on a closed surface is zero. That itself needs to be proved, but should be a simpler problem.

    Imagine the pressure vessel is a coke bottle with the cap on and pressure inside.

    Assuming the equation above is true, the total force on the inside surface is zero. But we know the force on the cap is Fcap = p*Acap. That means the total force on every other part of the bottle must be Fbottle = 0 - Fcap = -p*Acap

    So that's it! The magnitude of the total force on the bottle, excluding the cap, is p*Acap.
     
  18. Apr 17, 2011 #17

    AlephZero

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    You don't need to tie yourself in knots with surface integrals.

    Just cut the vessel and the fluid inside it into two parts with a plane. Draw a free body diagram for one part.

    The vessel was not moving, so the total force on the cut surface = 0.

    The total force on the cut surface = the force caused by pressure in the fluid + the force caused by stress in the vessel.

    Force caused by pressure = pressure x area
    Force cause by stress = mean stress x area
    Cut area of vessel = circumference x wall thickness, approximately.

    If the cut surface is a circle

    [tex]\pi r^2 P = 2 \pi r t \sigma[/tex]

    [tex]\sigma = P r / 2t[/tex]
     
  19. Apr 17, 2011 #18
    I guess you're considering something similar to a thin walled sphere filled with liquid...but the question is why you take

    [tex]\pi r^2 [/tex]...which is the projected area...in case of a cylindrical vessel, it'll be 2rLP,where 2rL is the projected area of one half of the cylinder.... you might want to go back to the OP and other posts made by Cyrus to get an idea of what the problem is...in short, the question is why we take the projected area and not the total area, which would be 2(pi) r^2 in case of one half of a sphere and 2(pi)rL in case of a half-cylinder..
     
  20. Apr 17, 2011 #19

    AlephZero

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    I am not taking the projected area. I am taking the area where I cut the body and the fluid into two parts. Sure, that area looks the same as the "projected area", but I don't care whether it really is the projected area or something different.

    When you make a free body diagram, you can cut up the body any way you like, to make the math simple.
     
  21. Apr 18, 2011 #20
    Okay, but then we do in many cases take the projected area...and a derivation of the same formula as done in my textbook aswell as on all related webpages use and specifically mention it as projected area....could you give me an explanation of this(this was the original question too)?
    I'll give you an example of such a website if I can find one...
     
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