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Technical question about air volume?

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
 
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.
 
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
 
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!
 
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
 
 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
 
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.
 
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