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Pressure forces from a liquid in rotation

  1. Nov 21, 2012 #1

    A liquid in rotation give two kind of forces:

    1/ centripetal
    2/ pressure forces (Fa, Fb in the drawing)

    I'm ok that centripetal forces can't give energy (we lost kinetics energy) but Fa/Fb seems in direction to external circle, why it's not possible to recover energy from these forces ?

    Liquid in the drawing don't move in the center, it's put at same place (same velocity) but don't return to the center.

    Attached Files:

  2. jcsd
  3. Nov 22, 2012 #2
    I add additional view for show zoom of forces (or like I understand them), maybe Fa and Fb don't exist ?
    Fc=centripetal force depend only from velocity, Fc=mw²r
    Fa+Fb=additional forces from pressure

    Attached Files:

  4. Nov 22, 2012 #3
    If water is rotating at W, we increase radius of water but don't change the number of molecules for each "layer" with a triangular shape (see drawing). Like that, water move at external and can give energy. Why water lost kinetics energy in this case because force Fa+Fb help molecule to go at external circle ?

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  5. Nov 22, 2012 #4


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    Staff: Mentor

    Because force isn't energy. Why isn't it possible to recover wood from metal?
  6. Nov 22, 2012 #5
    If you watch my last drawing, water move when I increase radius of water, each molecule of water move in the direction of force Fa+Fb. The triangular shape allow to keep constant the number of molecule in each "layer" of molecules. Triangular shape turn only. If water give energy the rotational speed increase or decrease ?
  7. Nov 25, 2012 #6
    1/ If I take an elliptical section of a screw with liquid inside. I place it under gravity, at bottom the pressure is high compare at top. Torque from gravity pressure is zero, ok. But these forces Fa+Fb are on each molecule of liquid, the torque at top is not the same that bottom.

    2/ It's the same reflexion with an asymetrical object put in a liquid (liquid under gravity). Watch second drawing for that. In right line there is no less force from outside but in part of curve there is less pressure from liquid due to the forces Fa+Fb. For this I think temperature cancel the effect, but imagine balls without temperature effect (big balls), the pressure add force one time and another force adding by Fa+Fb.

    I think when a fixed pressure is put on a curve surface there is more than fixed pressure, there is a pressure from the geometry too.

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    Last edited: Nov 25, 2012
  8. Nov 26, 2012 #7
    I understood the problem, in fact molecules of liquid don't see a border like we see, the pressure is all electrostatic pressure, from each molecule to each molecule, that from a liquid to a liquid or liquid to solid, nothing move from Fa+Fb because these forces don't exist with molecules. The shape is not liquid and a curve solid, it's a lot of molecules, that's all. It's not the problem of temperature, only how pressure is apply.

    But with macroscopic balls for replace molecules it's possible to have a border. Like this Fa+Fb exist, how these forces cancel themselves ? Sure there is friction but it's strange these force are adding in pressure system, I never heard about that.
  9. Nov 29, 2012 #8
    For now, I don't understand where is my error.

    W3.png : if I take a yellow solid, I attach on it compressible balls, between balls and solid there is very small pressure, outside there is 1 bar for example. Like that I understood the solid+balls don't have force because dx forces cancel Fa+Fb forces (zoomb.png)

    W2.png: if I take a black solid with right line at left and curve line at right. I place it in a lot of compressible balls. Fa+Fb forces exist, but here dx forces are not put on solid but on others balls. Like that I have a force on black solid and a torque witch can produce energy when the solid rotate (even there is friction). Like balls are compressible and because there are spheres, there is space between balls and this space can be use by balls for move.

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  10. Nov 30, 2012 #9
    Nobody can help me ?
  11. Nov 30, 2012 #10


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    Staff: Mentor

    The problem hasn't changed. Your diagrams and descriptions are incomprehensible. Until you learn how to communicate effectively, no, I don't think we can help you. Some of that is your use of the English language, but most is not: You need to make your descriptions clearer and more rigorous. More math in the descriptions would be a big help. RMPs, flow rates, sizes, etc.

    In addition, it seems like you are looking to design a perpetual motion machine. If your designs were more comprehensible and that were clearer, these discussions would be closed anyway.
  12. Dec 1, 2012 #11
    Oh, sorry if I'm not clear. I will make efforts. I'm not fluently in english.

    I'm interesting about energy, efficiency but I don't try to create perpetual motion, I know it's not possible. I like to understand how things works in physics especially with energy.

    Forget my last messages, I try to explain my problem:

    I take a curved shape solid and I put in a lot of small compressible balls. For this, watch the first drawing E1, blue color is a lot of balls. It's a top view. I'm interesting about forces on black solid and torque on it. For me, F1=-F2. But there is another force Fa+Fb, it's possible to watch on E2 drawing (only half solid on it), I exagerate the size of balls for see details and forces. The sum of Fa+Fb would give a torque on solid. Dx1 and Dx2 compensate sum of Fa+Fb but with the shape I give for solid, Dx1 and Dx2 can't cancel torque (sure, for me), I think the error is easy to find but not for me. The pressure in balls is put with compressed one side to another watch third drawing, it's a side view.

    thanks for your help

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  13. Dec 1, 2012 #12


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    Except that by definition, your Fa and Fb are acting tangent to the surface, so they provide exactly zero force to that surface.
  14. Dec 1, 2012 #13
    For me, solid has F1 and F2 forces, these forces cancel themselves. Fa+Fb reduce pressure from balls, solid don't see F2, but F2-sum(Fa+Fb). For me, each layer of ball give a "curve" of balls (like E4 showing), this give Fa+Fb and Dx1/Dx2 for each layer. The shape of the recipient give curve too but it's possible to change the recipient to a square shape. Can you explain a little more your method ?

    Attached Files:

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