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Differential pressures balancing a valve armature

  1. Jun 1, 2013 #1
    I have 2 questions on a two-port valve that I would like to ask for assistance with. There are three pairs of drawings below that need explanations:

    The first pair shows a one-piece armature with two o-rings providing sealing service on the bottoms of the upper and lower plates. The o-rings are shown with non-circular, flattened profiles representing their compressed state. The 5 relevant measurements are:
    ID_UpperSeal = 26mm
    OD_UpperSeal = 30mm
    D_Shaft = 14mm
    ID_LowerSeal = 16mm
    OD_LowerSeal = 20mm

    The second pair of drawings shows the valve body. There are two inlet orifices, an upper and a lower, and one discharge pipe that is assumed to be so large that it is not a restriction. The 2 relevant measurements are shown below, however, in my analysis below I do not use either of these orifice sizes. I include these because it seems likely that others may want to refer to them in their own solutions:
    D_UpperOrifice = 22mm
    D_LowerOrifice = 12mm

    The third pair of drawings shows the entire valve: the armature and the valve body together. The armature has an actuator moving it up and down that is not shown in these drawings. In the closed position, the gas around the outside of the valve has pressure Phigh, while inside the valve the pressure is Plow. These 2 gas pressures both act on the two armature plates to produce a single combined force Fgas on the armature. The gas pressures are quite constant, unless of course the valve is opened at which time Plow rises to be very nearly equal to Phigh.

    My thought was that in a situation with irregular surfaces, the force on each of the two armature plates would be the sum of the pressures times the areas incident to the pressures, projected onto a plane normal to the direction of the force being calculated. My first question is: can anyone state this rule of selecting boundaries in a more rigorous and general way?

    The armature is constrained to vertical motion, and so the force Fgas of interest is also vertical. The 2 o-rings are already (radially) within a plane normal to Fgas, so the force Fgas(lower) on the lower armature plate is the high-side pressure Phigh times the internal area of the lower seal, minus the low-side pressure Plow times the external area of the lower seal:

    Fgas(lower) = Phigh*[itex]\pi[/itex]*(ID_LowerSeal/2)2 - Plow*[itex]\pi[/itex]*(OD_LowerSeal/2)2

    The force Fgas(upper) on the upper armature plate is very similar: the pressures are reversed, the seal areas are from the upper seal instead of the lower seal, and the armature shaft area is subtracted from the internal seal area calculation:

    Fgas(upper) = Plow*[itex]\pi[/itex]*((ID_UpperSeal/2)2 - (D_Shaft/2)2) - Phigh*[itex]\pi[/itex]*(OD_UpperSeal / 2)2

    The total Force due to differential gas pressure on the armature is then just:

    Fgas = Fgas(lower) + Fgas(upper)

    I would like to design the valve so that in the closed position it requires very little effort for the actuator to open, which would be the case if Fgas = 0. My second question is can anyone verify my analysis of Fgas, or can anyone criticize it? The issue I'm most worried about is my somewhat shaky method of choosing the 4 area boundaries, described above in my first question.

    Thank you very much for reading my problem, and any light you can shed on it.
    Dave
     

    Attached Files:

    Last edited: Jun 1, 2013
  2. jcsd
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