Estimating Muzzle Energy

Discussion in 'General Airgun Chat' started by DeanB, Feb 7, 2019.

  1. DeanB

    DeanB Active Member

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    I posted earlier showing some calculations that support Jim Tyler’s findings on the effect of temperature on muzzle energy, and I got to thinking further.

    WARNING - DON’T CONTINUE READING THIS POST IF YOU LIKE TO SHOOT AND THINK EQUATIONS ARE A WASTE OF TIME!

    The internal energy of the air in the regulator is the only source of energy used to propel a pellet from a PCP gun. The internal energy can be calculated from:

    Internal energy for a diatomic gas = 5/2 NkT = 5/2 nRT - Eqn 1

    (Air consists mostly of nitrogen and oxygen which are both diatomic gases.)

    where N = number of molecules; k = Boltzman constant; T = gas temperature in Kelvin; n = number of molecules; R = gas constant

    The ideal gas law states that:

    nRT = PV - Eqn 2

    where P = pressure; V = volume

    Inserting equation 2 into equation 1 shows that the energy available can be calculated simply from:

    Internal Energy = 5/2 P.V

    Inserting typical values shows that

    P = Regulator pressure = 90 bar (say) = 9 x 106 Pa = 9 x 106 N m-2

    V= Volume of air in the regulator = 1.5 cm3 (approx..) = 1.5 x 10-6 m3

    Estimated Internal energy = 13.5 J = 10 ft lb

    So, the estimated value is a bit low and the calculations ignore frictional losses of the air and the pellet to barrel contact. Nonetheless, not far off.
     
  2. DeanB

    DeanB Active Member

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    Just realised....forgot to multiply by 5/2. Also seen that others estimate the used air volume as 1 cm3 at 84 bar.
    So revised calculations is
    Energy = 5/2 (8.4 x 106) (1 x 10-6) = 21 J = 28.5 ft lb
    So with a muzzle energy of 12 ft lb, we are using about 16 ft lb in air transfer, losses from the reg, and pellet to barrel contact, etc.
     
  3. DeanB

    DeanB Active Member

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    Error (J to ft lb conversion) spotted by Jim and correction made...
    Energy = 5/2 (8.4 x 106) (1 x 10-6) = 21 J = 15.5 ft lb
    So, not far out. But if the air used at 84 bar is 2.35 cm3, then the energy will be 2.35 greater than estimated which again suggests high friction losses.
     
  4. Jesim1

    Jesim1 Active Member

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    This is way over my head from a maths point of view, but due to the "guestimates" of reg pressure and internal reg volume these figures could be dramatically out as we are talking tiny specifications.

    I'm not knocking the equations, but most people will not accurately know the reg pressure (not just take it from an uncalibrated gauge) or indeed the exact internal volume of the reg - so although this may be technically accurate, I can't see it being of value in the real world to 99% of guns?:rolleyes:

    James
     
  5. cloverleaf

    cloverleaf Active Member

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    Nice work (I find the calcs for internal energy of a pressurised gas particularly useful), however FWIW I think the approach you're taking is somewhat flawed for a number of reasons:

    - It works on the assumption that the reg pressure falls to ambient with every shot, which is not the case.
    - It doesn't account for a number of external variables (which granted, most of which you've mentioned) - another being barrel volume.

    On account of the above I think the method has limited usefulness for estimating muzzle energy from reg variables as you describe, however would be much more suited to a "fixed volume, free expansion" situation - i.e. a single / multi-stroke pneumatic that dumps its full charge with every shot. In addition this could be made a lot more accurate by taking the barrel volume into account - combining with the initial plenum volume to give a final volume; from which a final pressure could be calculated and finally the amount of internal energy remaining in the expanded air charge once the pellet reaches the muzzle and the air ceases to do any more useful work on the projectile.

    In addition your calcs could be useful for working out PCP air efficiency (which can vary significantly from one design to another and is sometimes a good diagnostic tool); since it would allow you to calculate the total stored energy in the cylinder both before and after a (preferably long) shot string; from which you could calculate efficiency based on the number of shots taken and their mean muzzle energy.

    Ta for the equations - I'm off to chuck them into a spreadsheet ;)
     
  6. DeanB

    DeanB Active Member

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    Thanks Cloverleaf. I was pleased that I could explain the changes in muzzle energy due to temperature changes that Jim Tyler had found. Then today I was waiting for some data from a company and I thought "I wonder if I can predict the speed of pellet from an air rifle". Totally agree that there are improvements that could be made to the model but, as you say, much of the data needed is not known. Bob Evans (Evo ) has posted that his Airfwolf only needs a pressure drop of 0.3 bar for each shot. That seems very efficient compared to the 1 bar used by many other guns. So, as you indicated, it could be interesting to look at efficiency as you suggested.
    Think I'd better focus on work tomorrow.
     
  7. Darron

    Darron Dwarf Slayer

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    Take no notice of anything Evo has to say, firstly he’s the only person who uses a take down scope and secondly he’s an Everton fan so he knows ****.
     
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  8. RobF

    RobF Administrator Staff Member

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    I think my maths had it about 2.35cc at 85...

    We have a rig somwhere with a digital gauge on the bottle.
     
  9. Jesim1

    Jesim1 Active Member

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    Surely this is based on the fact the Airwolf has a 480cc bottle and most cylinder guns have more like 180/200cc - so it has around three times the available volume but uses roughly the same as a regged gun?

    James
     
  10. RobF

    RobF Administrator Staff Member

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    Yep.
     
  11. Nomads HFT

    Nomads HFT Well-Known Member

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    There are frictional and probably other losses, Dean, but my gut feeling is that most of the 'lost' (to the pellet) energy at the instant before pellet exit will stay with the air; the potential energy associated with the remaining pressure, and the kinetic energy associated with the air's lowered temperature.
     
  12. JerryD

    JerryD Active Member

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    This is postulation, so it stands to have bricks thrown at it......

    The only system that this would hold true for would be a dump valve, where the regulated volume is completely voided as the shot is taken. I.e., a GC2 - and they are relatively air-hungry.

    As most regulated systems rely on a KO valve, which then uses the remaining air behind it to close the valve, the actual amount of air consumed will be a function of the regulated pressure and the length of time that the KO valve is open?

    While this would be "somewhat difficult" to measure, and estimate of how mush is used could be made by the residual pressure left in the downstream volume after the shot has been taken. However, this in itself would be difficult to measure as the regulator will start to operate as soon as the downstream pressure drops.

    Perhaps a fast acting pressure sensor coupled to a scope to measure the static and dynamic pressure variations in the regulator volume as the shot is taken - or simply measure the amount of air released and expanded to atmospheric after the shot has been taken, back-calculated to the volume at regulated pressure?




    .
     
  13. DeanB

    DeanB Active Member

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    I thought that the 2.35 cm3 was the volume of air used at the regulator pressure to fire the pellet. So the amount of air left in the reg, and its pressure, do not have to be considered. Did I misunderstand?
     
  14. RobF

    RobF Administrator Staff Member

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    That's exactly how I see it.

    200cc give or take at atmospheric.

    It's not a great deal, 2/3 of a coke can.
     
  15. JerryD

    JerryD Active Member

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    Wow! A springer uses about 50cc of air, based on a 90mm stroke and 26mm diameter. Does that make springers 4x more efficient than PCPs? :D:p
     
  16. C.Eaton

    C.Eaton Confirmed Anschutz Nut...

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    Stick a partially deflated bag of known volume over the muzzle with a helper ready to tighten some string over the neck to hold the volume. Fire a shot with no pellet and see what the bag inflates by?
     
  17. RobF

    RobF Administrator Staff Member

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    Yes but we know it's not about volume, because if that was the case we wouldn't need springs, we could just push the piston manually. :D

    Volume of air is no measure of efficiency. It's a term misused and further bandied around. Efficiency = energy in vs energy out.

    The spring has stored energy which you put into it by cocking it. The PCP has stored energy by using compressed air which you put into it. The energy used is what creates that compressed air, not how much air is used.

    Air isn't the 'fuel'. It's the energy stored in the air that is. It's not like a car where as a rough rule there's the same amount of energy in every litre of it so MPG is a rough measure of efficiency of the car.

    I've no doubt one will be inherintly more efficient than the other, but I'm not going to bother with the maths. Someone might, and that would be interesting...
     
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  18. JerryD

    JerryD Active Member

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    The comments was tongue-in-cheek Rob: in a springer the air is simply the energy transfer medium from spring to pellet. It's much more complicated than that, but that's fundamentally what it does.

    In terms of PCPs there's a lot more going on than just the volume of air and the stored energy: the speed of the mass air flow must have some influence on this as well.

    Me: I'm happy just shooting the damn things. A perfect understanding of the skience behind it doesn't knock down any more targets.....





    .
     
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  19. bootneckbob

    bootneckbob Active Member

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    I haven't read any of this post as instructed in the opening sentence. I just have one question, did you hit the target??:confused:
     
  20. RobF

    RobF Administrator Staff Member

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    No. Because of temp shift.

    That's what is behind the post, to measure the amount of energy in the air that does the work for the pellet and see if that's a contributary factor... to do that you need to know the amount of air behind it.

    You did ask :D
     

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