How much air is used in say a .25 cal gun per shot lets say pushing a 25 grain pellet at 900 FPS producing 50 FPE. What I am looking for is the pressure and volume of air used per shot.
Thanks Mike
Thanks Mike
Technical question about air volume?
- Thread starter Gundog
- Start date
Thats a good question, but, there are same things to consider, the gun is regulated? What is the work pressure?
The EDgun Matador shoots almost 70 shots from 210 to +-120 bar, and the regulator is set to 130bar, so, it's a small volume at 130 to do it, but probably you dont waste all the 130 during the valve stays open.
Vinny
The EDgun Matador shoots almost 70 shots from 210 to +-120 bar, and the regulator is set to 130bar, so, it's a small volume at 130 to do it, but probably you dont waste all the 130 during the valve stays open.
Vinny
Upvote 0
My question is not related to one gun. I have an idea I want to explore for a new airgun technology and I need a number to plug in for more calculations. This idea if it is feasible would not require a regulator as it will only store air for one shot at a time. What size tank is that? I can get my figure I think by knowing the volume of the tank?
Thanks Mike
Thanks Mike
Upvote 0
My Vulcan in .25 is shooting 50 ft lbs maybe a pound or two more. One magazine which is nine shots takes twenty Bar which is 290 psi. So 290psi divided by 9 shots = 32 psi per shot. my barrel length 17.7 in long. And I don't know how to would convert that out to volume.
Upvote 0
The piece I need is volume used per shot it does not need to be exact but a range I know there are many variables and that is where tuning would come in but I need a base number for volume the pressure I pretty much have figured using my Wildcat I had I filed it to 220 Bar and ran it down to 150 bar after 40 shots. The air tank was 300CC but I am not sure how to get that volume from those numbers I will ask my neighbor the engineer if he can calculate the volume from those numbers.
Thanks Mike
Thanks Mike
Upvote 0
I think I found my answer it is right at 1CC per shot volume @ 150 bar using the numbers from my old Wildcat. I found this online calculator. http://www.endmemo.com/chem/boyle.php
Upvote 0
The volume of air in cc or cubic inches has to be associated with a pressure...normally standard pressure of 14.7 psia.
Using the Vulcan example, the volume of the air cylinder or reservoir is 290 cc. Using the ideal gas law of P1V1=P2V2; the initial volume of air in the reservoir at standard atmospheric conditions (3625 psi*290 cc)=(14.7 psi*?). Solving....71,514 cc
The volume of air left in the cylinder after 9 shots is (3335 psi*290 cc)=(14.7*?). Solving....65,793 cc
So 9 shots used 71514-65793 = 5,721 cc of air or 636 cc of air per shot.
Or very simply as KY calculated, 32 psi per shot...use ideal gas law to calculated volume at standard atmospheric conditions of 14.7 psi
(32 psi)*(290 cc)=(14.7 psi)*(? cc). Solving...631 cc of air per shot.
Not sure about the 1 cc calculation...I calculate much more than that at an average air reservoir pressure of say 170 bar or 2500 psi. I calculate 3.7 cc per shot at 2500 psi but this is using the Vulcan example.
Using the Vulcan example, the volume of the air cylinder or reservoir is 290 cc. Using the ideal gas law of P1V1=P2V2; the initial volume of air in the reservoir at standard atmospheric conditions (3625 psi*290 cc)=(14.7 psi*?). Solving....71,514 cc
The volume of air left in the cylinder after 9 shots is (3335 psi*290 cc)=(14.7*?). Solving....65,793 cc
So 9 shots used 71514-65793 = 5,721 cc of air or 636 cc of air per shot.
Or very simply as KY calculated, 32 psi per shot...use ideal gas law to calculated volume at standard atmospheric conditions of 14.7 psi
(32 psi)*(290 cc)=(14.7 psi)*(? cc). Solving...631 cc of air per shot.
Not sure about the 1 cc calculation...I calculate much more than that at an average air reservoir pressure of say 170 bar or 2500 psi. I calculate 3.7 cc per shot at 2500 psi but this is using the Vulcan example.
Upvote 0
It's been a while since I did this math for dive cylinders at different depths, but AJ's math looks right to me. Good ole PVT!
Upvote 0
My calculation is 1CC @ 150 Bar not atmospheric pressure. Maybe I am wrong I won;t say I am right but how are you getting more CC out than what the bottle holds? Unless you are talking after the air is released to the atmosphere? The number I am looking for is the volume of air to be released and stored at 150Bar to fire the round at a FPS speed the same as a PCP. I have an idea on how to compress that air for one shot unlike anything that I have heard done before.
Mike
Mike
Upvote 0
Hey Mike, Glad to hear you are thinking outside the box. Can't wait to see your idea come to light! As far as the calculations, you are correct, AJ is bringing the volumes back to 1 atmophere (14.7psi) absolute. Hense the (psia) designation. Most pressure cylinders are quoted as to their volume at full pressure, so when you have different volume cylinders at different pressures the only way to compare apples to apples is to bring everything back to the equivalent pressure and volume of 1 atm. Hope that helped? Don't mean to hijack AJ's thread (answer) but just wanted to pipe in and help keep your idea moving. Good job AJ!
Upvote 0
Hi gundog.
I don't know if the calculation of Aj3 that's right, but i'm sure that 1cc is phisically impossible. Fx bobcat for example have 41cc prechamber for 160bar (+2 -2 bars)
Kind regards
I don't know if the calculation of Aj3 that's right, but i'm sure that 1cc is phisically impossible. Fx bobcat for example have 41cc prechamber for 160bar (+2 -2 bars)
Kind regards
Upvote 0
This is all good information and what I have in mind most likely won't work but I need to see it through either way I will let you know what I was thinking at some point. I need to keep the idea to myself at this point because if it does work I may patent it. I have 2 US patents now so I am not just a dreamer I actually do make things but like many ideas sometimes they just don't pan out.
Mike
Mike
Upvote 0
The +/- 630 cc of air per shot for the Vulcan example is the volume of air that would expand from the pressurized air reservoir to the atmosphere with each shot.
If you want to do the calculation for your own PCP, the equations can be simplified:
cc of air used per shot @ atmospheric conditions= (Bar used per shot)*(actual air reservoir volume cc)
**bar used per shot=(initial air reservoir fill pressure-ending pressure)/# shots fired
to convert volume from atmospheric conditions to air reservoir pressure of 150 bar:
cc of air used per shot @ 150 bar=(cc of air per shot @ atmospheric)/150
If you want to do the calculation for your own PCP, the equations can be simplified:
cc of air used per shot @ atmospheric conditions= (Bar used per shot)*(actual air reservoir volume cc)
**bar used per shot=(initial air reservoir fill pressure-ending pressure)/# shots fired
to convert volume from atmospheric conditions to air reservoir pressure of 150 bar:
cc of air used per shot @ 150 bar=(cc of air per shot @ atmospheric)/150
Upvote 0
The smallest one shot tank for airguns that I can think of is The Brocock Air Cartridge System.
http://www.airgunmagazine.co.uk/features/10-years-after/
You should be able to work out the volume from the calibre and a guess at the length of the cartridges.
I think James Bond should have one for his next mission.
http://www.airgunmagazine.co.uk/features/10-years-after/
You should be able to work out the volume from the calibre and a guess at the length of the cartridges.
I think James Bond should have one for his next mission.
Upvote 0
Hi gundog
If you give us more speciffic information maybe someone can give exact caculation of what you need.
If you give us more speciffic information maybe someone can give exact caculation of what you need.
Upvote 0
I just checked my 22 Evanix and used 50 bar in 22 shots or 2.27 bar/shot. The Evanix has a 290 cc air cylinder so the cc used per shot is 2.27*290=658 cc per shot @ atmospheric pressure
Converting 658 cc to 150 bar, 4.4 cc per shot. The Evanix shoots 18.1 grain at 980 fps. Pretty sure these numbers are in the ball park for our PCP's
Gundog, using the numbers from your Wildcat I calculate 525 cc per shot atmospheric or 3.5 cc per shot at 150 bar.
Converting 658 cc to 150 bar, 4.4 cc per shot. The Evanix shoots 18.1 grain at 980 fps. Pretty sure these numbers are in the ball park for our PCP's
Gundog, using the numbers from your Wildcat I calculate 525 cc per shot atmospheric or 3.5 cc per shot at 150 bar.
Upvote 0
I have the information with a couple engineers right now the initial conversation is they don't think what I have in mind is going to work but they need to do some calculations thank you to everyone who helped with information.
Mike
Mike
Upvote 0
I ran through the calculation that you suggest. Doing so, discovered the equation can be simplified to
((Initial pressure bar - final pressure bar)/# shots) X (air reservoir volume cc)= cc of air used per shot @ 1 bar or atmospheric pressure
This volume can be converted to any pressure in bar by simply dividing by the desired pressure.
would be interested to see if you agree with the math...it's been a while for me even though are an Engineer
((Initial pressure bar - final pressure bar)/# shots) X (air reservoir volume cc)= cc of air used per shot @ 1 bar or atmospheric pressure
This volume can be converted to any pressure in bar by simply dividing by the desired pressure.
would be interested to see if you agree with the math...it's been a while for me even though are an Engineer
Upvote 0