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Thread: Help!! Pressure loss of air poppets

  1. #1
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    Help!! Pressure loss of air poppets

    I have need to figure out how much pressure is lost in a system when each air poppet is cracked. Can I make four poppets crack open at the same time using the same air circuit?

    I am applying 80 PSI to a .3437 dia drilled line 5 inches long. Each air poppet ties into this line (4 times evenly spaced). I need to then push a shoulder bolt (.3125 dia head). The shoulder bolt has a spring that needs to overcome 6 lbs of force from the air pressure to crack the air poppet .015. I have all this displayed in a pdf attached to this post.

    Any help is much appreciated.

    Thank you
    Attached Files Attached Files

  2. #2
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    You have too many restriction points in your poppet branch(s) design to allow a reasonable calculation of the actual flow rate and resulting inlet pressure loss for your assembly because there are too many restricting elements in the flow stream which includes a critical (sonic)flow determination point and multiple flow channel flow rate pressure drops including the length of the .344 dia. bore feeding your poppets, the flow restriction through the poppet spring chamber and the .015 in annular channel between the poppet stem and its surrounding bore.

    However, ignoring all inlet bore, spring chamber and poppet stem .015 annulus flow losses, if you can provide me with the poppet disc lift height (is it .015 in. ?), the spring chamber bore (which is the effective poppet seat I.D.), and the O.D. of the poppet seat disc, I have access to a propriatary program that will determine the critical flow restriction point from its three possible locations and the resulting maximum air flow rate for each poppet assembly at an 80 psig inlet pressure. Also, please clarify if the .015 poppet stem to bore clearance in your drawing is radial or diametrical.

    (Revision on my initally posted available analyses capability: After sending my initial post I realized that, once I determine the orfice restricted maximum flow rate, I have a another rarely used program that can calculate the pressure drop in the .344 dia x 5 in long supply port; and, therefore the maximum pressure drop you can expect in that port. However, even with this added information these calculations will not determine your actual flow through a poppet(s) because there will still be added reductions due to the potential flowing losses through the spring chamber and stem annulus regions.)

    With regard to your question about lifting all four poppets at the same time, I have to ask why you are not using a single larger poppet since you specify that you are venting all of the poppets to atmosphere.
    Last edited by JAlberts; 12-04-2013 at 07:02 PM. Reason: Additional assistance capability

  3. #3
    Administrator Kelly Bramble's Avatar
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    Quote Originally Posted by JAlberts View Post
    With regard to your question about lifting all four poppets at the same time, I have to ask why you are not using a single larger poppet since you specify that you are venting all of the poppets to atmosphere.

    What they said.....

  4. #4
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    Well the reason for the air poppets is they are being used in a silicone injection mold. It is a eight cavity with two lines of four poppets.

    Poppet lift height is .015 inch.....can be more or less I was just keeping the .015 the same as the clearance per side of the shoulder bolt clearance. I am using a standard 3/16 shoulder bolt and the hole it rides in is .405 dia.
    Spring chamber bore is .375.
    O.D. of the poppet seat disk is .625 dia.

  5. #5
    Technical Fellow jboggs's Avatar
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    That didn't really answer Kelly's question. If all four poppets are connected to a single common volume, why not just have one larger poppet?

  6. #6
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    I am using the poppets to eject the silicone parts. I need one poppet per part.

  7. #7
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    I guess i'm not understanding your drawing. is this using the same air that holds the poppets closed to use at the other end to blow parts out of a mold? So you have to drop the pressure below 6lbs for the poppets to open and then you only have 6lbs or less to do the work at the end of the nozzles? Usually you see this kind of thing opening another orifice to allow full supply pressure in to do the work then you can use a lower pressure in the control circuit.

    By the same time do you really mean exactly the same time or is it ok if each opens a few milliseconds apart?

  8. #8
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    There is a mismatch between your drawing dimension of .500 in. dia. on the "air channel" and the "O.D. of the poppet seat disk is .625 dia." in your above post. Obviously that disc size will not fit in the flow channel. Another mismatch I found is that your stated .405 bore diameter around the .375 bolt shank is greater than the .375 bore diameter given for your spring chamber in your post. If you will correct these dimension conflicts and carefully review your other values for conflicts I will run the calculation.

    On the other hand, I would like to know why you are not simply using standard electric solenoid valves, either one or four connnected to a manifold, for this application. They are readily available, well designed and reliable.
    Last edited by JAlberts; 12-05-2013 at 01:08 PM. Reason: Added dimensioning questions

  9. #9
    Administrator Kelly Bramble's Avatar
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    Checkout the following equations....

    Pressure and Flow Drop Equations

  10. #10
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    All,

    I am looking for direction on the subject and am open minded to answers. I am in the beginning stages of mechanical engineering so I don't have the knowledge. All I can do is state what I am trying to do. That is to create an air blast to eject eight silicone parts with air pressure. By inputting 80 psi of air to 8 chambers. I think 120 psi is OSHA regulation if possible I can go that high but 80 is what I would like to use. I will change the design to whatever it takes to create an air poppet that will crack 8 poppets at the same time to release 8 silicone parts from the mold. I don't know how to figure air loss at each air poppet through a system. If anything comes from this post will be we all learn something from it, especially me. The only dimensions that are set in stone are the outside dia. of the core/air poppet and the off the shelf spring (if possible)and shoulder bolt off the shelf. It needs to be a custom mechanical component. I love our discussion here and hope to learn from it along with others. So everything is open season except Th O.D of the core and air poppet, spring pocket (.375), .1875 dia shoulder bolt. Everything else is can be changed to build a system.

  11. #11
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    You haven't answered any of the questions. So far I'm thinking the same as JAlberts, Just use one or 4 (if you have to) solenoids and use your 80# of air to eject the parts,

  12. #12
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    Let me break down what I am trying to find out... It is basically just like a flute with the exhaust end capped off.
    I have a .3437 dia. hole in a pipe. I now drill four .3437 dia. holes vertical evenly spaced into the original .3437 dia. line out to atmosphere along the pipe. I now place a spring loaded plug into each vertical hole which only allows a .015/side gap along it's channel and out to atmosphere when opened. I am inputting 80 psi of air pressure.

    Where can I find an equation or formula to figure out how much each .030 dia. hole looses for pressure? Is this a basic volume and cross-sections?

  13. #13
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    OK, I understand what you want but unfortunately, to my knowledge, there are no flow loss calculations or programs that will determine the flow loss in an annular flow port. The effect of the combined boundry layers in an annular port from the surface drag of both the outer and inner port walls is simply too complex to analyze.

    At the same time, the equivalent port diameter of each of your annular port areas is,in excel format: D equiv = sqrt(.3437^2-.314^2) = .140 in., and if you give me a length of your port then I can determine the flowrate for this equivalent diameter port. What this will give you is a conservatively maximum flow that you can expect from each of your ports; and, I can use my programs to determine what, if any, pressure drop will result in your .4347 in. diameter inlet pipe at this maximum expected flow rate.
    Last edited by JAlberts; 12-08-2013 at 11:02 PM. Reason: corrected typo in formula

  14. #14
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    JAlberts,

    The length of the main .4347 dia. line is 5.00 inches. Each flow port from that line to the atmosphere is 2.00 inches.

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    I have run the required calculation using an equivalent .140 in. diameter port for each of the 4 exhaust ports and with all 4 ports open and flowing to atmosphere the conservative maximum total flow through the system calculates to be 4 x 26.34 = 105.36 scfm which will result in a flowing pressure drop of 0.82 psi in your .4347 x 5 in. inlet line.

    Just as a note, the maximum flow through each of the the 4 annular ports is sonic flow limited and not determined by the piping friction drop in the ports' 2 in. lengths.
    Last edited by JAlberts; 12-09-2013 at 11:10 PM.

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