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Pressure inside pneumatic cylinder

  1. Mar 8, 2016 #1
    Hello,

    I'm having issues with pressure calculations that I'm using for a double acting pneumatic cylinder. In my application, I start the piston near one end of the cylinder with both sides closed, and then move the piston towards the other end. At each point, I update the pressure, volume, and net force which is used to provide torque to the rest of my system. For some reason, the theoretical pressure values on the side that is being compressed are much larger compared to the experimental values that I've recorded, and as a result, my system is not producing as much torque as I had expected.

    I am assuming an adiabatic process, so I am using the following equation:

    PBVBγ = PAVAγ

    where B and A stand for before and after the process, respectively. I then calculate PA according to the above equation since the other three variables are known.

    Any ideas what could be causing the discrepancy between my calculations and the actual values that I'm recording in the cylinder?

    Thanks
     
  2. jcsd
  3. Mar 8, 2016 #2

    billy_joule

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    You have assumed the system is 100% efficient, which is not true. What losses do you think exist in a real pneumatic system?
     
  4. Mar 8, 2016 #3
    I assume there will be some losses due to friction, but I am concerned since the max pressure I am recording is about 1/2 of what I calculated using the equation above, so the torque output of my device is not as high as what I designed it for.
     
  5. Mar 9, 2016 #4

    JBA

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    If this is a standard cylinder with a single rod on one end, are you accounting for the loss of piston area on the rod end side of the piston in your calculation?
     
  6. Mar 10, 2016 #5

    Baluncore

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    Maybe you have a leaking piston seal, or a gauge calibration problem.
    How have you allowed for the vacuum being drawn on one side of the piston ?
    You write that both sides are closed, but have you allowed for the volumes of the air lines between the cylinder and the closed valves ?
     
  7. Mar 10, 2016 #6
    I have included the difference in area on both sides of the cylinder. Perhaps it is not an adiabatic process, and should be modeled as Boyle's Law?

    PBVB = PAVA
     
  8. Mar 10, 2016 #7
    Thanks so much for your reply.

    I don't think that I have a leaking piston, since the pressure always returns to the same value when the piston returns to its original position.

    In the equation that I am using for force, I do include the vacuum being drawn on one side of the piston. I should just be subtracting the pressure multiplied by the cross-sectional area of that side, correct?

    I have not included the volume of the air between the cylinder and the valve, but I can include that. I imagine that it may lower the overall pressures that I have calculated, but I don't think that it will account for the discrepancy that I currently have.
     
  9. Mar 10, 2016 #8

    Baluncore

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    Yes, but you must get the signs correct. The vector forces on the piston faces are acting in the same direction, so they must be summed.
    Measure or compute the stagnant line volume. Wishful thinking that it may not be a problem, may be the problem.
     
  10. Mar 16, 2016 #9

    JBA

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    Perhaps you post a copy of your calculations for examination. Sometimes a simple error can exist that the originator of the calculation just can't spot.
     
  11. Apr 6, 2016 #10
    Yea definitely, here is a copy of my Matlab code that I'm using to update the pressure and volume as I move the cylinder from one end to the other

    P1(i) = P_eq*(Vb1^(gamma) / ( Vb1-(a-as)*dLp(i))^(gamma) );
    P2(i) = (P_eq*(Vb2^(gamma)) ) / ( (Vb2 + a*dLp(i))^(gamma) );

    V1(i) = Vb1-(a-as)*dLp(i);
    V2(i) = Vb2 + a*dLp(i);

    -- dLp is the change in length that ranges from 0 to the stroke length of the cylinder, and gamma is either 1 or 1.4, depending upon if the process is modeled as isothermal or adiabatic.
    Let me know if anything catches your eye. Thanks for the help!
     
  12. Apr 8, 2016 #11

    JBA

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    I am having trouble evaluating the combined actions of those formulas, if it were my process I would utilize an Excel spreadsheet to calculate the forces independently to allow me to evaluate how each is contributing to their combined effect.

    One comment regarding an early post of yours is that I would definitely, as you considered, use Boyle's Law of PV = PV for this analysis.

    One item I see is that concerns me is that there is no factor in your analysis addressing piston seal frictional drag; and, at high differential pressures this can be a significant load addition and would tend to increase proportionally with the increasing cylinder pressure as the stroke approaches the end of its travel.

    Additionally, If your cylinder pressures are very high (> 2000 psig) near the end of the stroke you also might investigate utilizing a Z compressibility PV/Z = PV/Z equation for your calculation.

    Beyond those comments, without seeing any detailed information on the cylinder size stroke , etc. it is not possible for me to do any comparative analysis on your results.
     
  13. Apr 8, 2016 #12

    Baluncore

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    If you assume that volume is proportional to rod displacement, then you are also assuming a zero volume of dead-space. That assumption is certainly false as the piston must be attached somehow to the rod. The way that is done mechanically will probably result in a significant dead-space, which will lower the compression ratio at each end of the cylinder. That dead-space must be added to the dead-space of the airlines that connect to the closed valves. There will also be some dead-space in the necessary clearance between the piston and the cylinder wall.

    You can measure the total dead-space by operating the cylinder in a like way to removing air from a hypodermic syringe.
    1. Stroke to minimum volume. 2. Suck in sufficient liquid to more than fill the dead-space. 3. Invert the cylinder and let any air bubbles rise to the open valve. 4. Push the rod to eject the air and all excess liquid until the piston is at the mechanical end of stroke. The volume of liquid contained is then equal to the dead-space. 5. Draw back the piston to suck sufficient air into the cylinder end with the liquid. 6. Invert and wait for the liquid to settle. 7. Eject the liquid into a measuring cylinder. That is a measure of the dead-space at that end of the cylinder. Repeat the process for the other end.

    You must allow for those dead-space volumes in your computation.
     
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