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Calculating Distance fluid will travel in small tube with Piston on other side

  1. Jan 18, 2009 #1
    I want to learn how to calculate what distance a fluid will travel up a small tube if a large piston is used on one side, and no piston on the other side.

    Example Diagram
    (Consider everything is square instead of round, for easier measurements.)

    [​IMG]

    Assume there is a 10kg mass placed on a square piston.

    Under the piston is 10L of water, 1.0 density, 10KG. The 10L water is spread out very wide. The piston (barrel) is 50cm wide, 20cm long, 10cm high. Assume the piston is near zero weight and the 10kg (100N) is all that is acting.

    Gravity is in effect.

    A 1cm by 1cm tube comes out at the very bottom of the water chamber. The tube heads up 1 meter high (100cm, 100mL volume).

    If the 10kg mass is let go right there on the piston, how far up would the water go.

    Forget atmosphere pressure if it makes it much simpler.

    Assume the tube is open to air at the top, not closed (sorry for the crumby drawing).

    A picture is attached to make it easier to visualize.

    This is a problem I made up myself, just to understand how to calculate this sort of situation. It is not homework (someone moved it to the homework section for some reason). It is just a brain exercise to learn more about pistons when they push fluid in an open system.

    I figure that if there is 0-100ml of water on one side maximum at any point in time, then the formula to calculate that side of the system may be p=pgh. Whereas on the piston side the formula is p=f/a.

    Since this is an open system, the pressure is dynamic rather than static though.

    Gravity is working on both sides of the system.

    If it were a column of water on each side, which is not the case in this example, the water in the tube would simply level out per p=pgh. However in this case, shown above, I am unsure what will happen. Help me understand?
     

    Attached Files:

    Last edited: Jan 18, 2009
  2. jcsd
  3. Jan 19, 2009 #2
    When your system settles down the pressure on the left will equal the pressure on the right.
    The pressure on the left is LeftLiquidHeight*Density
    The pressure on the right is RightLiquidHeight*Density + PressureDueTo10kgWeight
     
  4. Jan 19, 2009 #3
    What about including the "g", force of gravity. Did it get canceled out somehow.
     
  5. Jan 20, 2009 #4
    Pressure at depth h in a fluid of density d is equal to hd.

    Let us check the dimensions...

    Pressure = Force/Area
    Pressure = Mass*GravitationalAcceleration/Area
    Pressure = (Mass*Length)/(Time^2*Length^2)
    Pressure = Mass/(Time^2*Length)

    Pressure = Height*Density
    Pressure = Length*Weight/Volume
    Pressure = Length*Mass*GravitationalAcceleration/Volume
    Pressure = (Length*Mass*Length)/(Time^2*Length^3)
    Pressure = Mass/(Time^2*Length)


    Where you need g is in calculating
    (a) the density of the liquid (but you gave it already(without units!))
    (b) the force due to your 10kg mass
     
  6. Jan 20, 2009 #5
    Ahh, the density I actually use in calculations is 1000kg/m^3, I said "1.0" referencing 1 gram per mL, and apologize.

    The force I tend to use is newtons (100N) but in diagram I show 10KG static mass to make sure people know it is a simple constant mass and not some finger or foot pressing down with variable force or such.
     
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