valving/pressure Q

[PakProtector]

hey-Hey!!!,
I'd like to size a storage chamber for two shots of identical performance. First one limited by the valve time at high pressure, the second by the remaining pressure and size of the chamber. I want to be able to charge up to some particular pressure, and then make two shots w/o adding any HP air.

This is for a pumper. Primary use for hunting. Most likely a 35 FPE 25 cal. Looks like I'll have to consume 1.5 litre-Atm per shot IFF I can get .3 efficiency from the air. I can't see wanting to start much higher than 2000 psi...but we'll see how the pump design goes. I can put ~65 FPE into my RX2 in one shot. So figure the pump is 50% efficient and I'm pouring in at least 450 FPE for two shots...

Yes I still want to continue...heh-heh-heh. Looking for a modern Super Range I think.
cheers,
Douglas

[rsterne]

The simplest way is to forget a separate chamber and just size the valve to contain enough air for the two shots.... Adjust the hammer strike until the velocity of the two shots is equal.... Basically, you are building a retained air pumper.... My "Uber-Carbine" Grouse gun delivers two shots of 600 fps with 14.3 gr (11 FPE).... You just need to scale it up.... and then pump like mad !!!

Most PCPs can attain about 1 FPE/CI of air used at 1 bar pressure.... For a 35 FPE shot, you need 35 CI.... If you had a 1 CI valve that dropped 35 bar (~500 psi) in pressure during the shot.... or a 1/2 CI valve that dropped 70 bar (~1000 psi) you should be close.... Let's say an 8-10 cc valve, pumped to 2000 psi, dump maybe 40% of the air on the first shot and the rest on the second shot.... whatever was needed to get the shots the same velocity.... The numbers may not be exact, but it should get you started....

Bob

[PakProtector]

1 FPE/ CI @ 1 Bar? 1 liter-Atmosphere is 74.3 FPE....so those things are 6/7.5 efficient on air? Most impressive... I see L-W barrels are almost 2' long...should be enough. Can't help thinking grafting a pump tube onto a Marauder would be a bit of a time saver for the first prototype.
cheers,
Douglas

[rsterne]

I'm not familiar with your "litre-atmosphere" term for power available.... I do know that I have seen values as high as 1.86 FPE/CI however.... or as low as 0.5 FPE/CI.... The efficiency of a shot string is arrived at as follows:

FPE per shot times total shots = total FPE

Pressure drop (psi) divided by 14.5 = pressure drop in Bar
Volume of reservoir (cc) divided by 16.4 = volume of reservoir in CI
Pressure drop (Bar) times reservoir volume (CI) = total volume in CI (at 1 Bar)

Total FPE for string divided by total CI used for string = Efficiency (FPE/CI)

Some people use the formula the other way up and in Metric.... but a bigger number is less efficient....
Total volume used (cc) at 1 Bar divided by FPE = Barcc/FPE
I prefer the former method as the results average around 1 FPE/CI, and a bigger number is more efficient....

Bob

[PakProtector]

liter-atm is about equal to 1000cc-bar or 60 CI-Bar= 73 FPE. 1 atm is 1.01 bar...just another pressure-volume energy number. Given the conversion I am going to question your 1.86 FPE/CI-Bar number...
cheers,
Douglas

[rsterne]

How do you relate liter-atm to FPE?.... ie how is the conversion done, ie how do you determine that it is equal to 73 FPE?.... I have never seen that number before.... The way you are talking,  this is a theoretical "limit" of some kind.... Since 73 FPE from 61 CI is only 1.2 FPE/CI.... and even a stock Disco regularly exceeds that value.... we seem to have a problem.... A stock Disco returns from 1.23 to 1.32 FPE/CI depending on pellet weight (heavier is more efficient)....

I am 100% confident in the 1.86 FPE/CI number.... Here is the raw data to support that number....

75 shots averaging 620 fps with 14.3 gr. pellets on a regulated 22XX PCP.... This was attained from a 13 CI tank which lost 550 psi during the string....

Avg FPE = 12.2.... Total FPE = 915 FPE
550 psi = 37.9 Bar.... Total CI = 37.9 x 13 = 493 CI
Efficiency = 915 / 493 = 1.86 FPE/CI

That data was obtained during testing of varying hammer spring settings and was the highest efficency obtained.... Other values obtained, for reference, were:

1.03 FPE/CI @ 31 FPE
1.20 FPE/CI @ 29 FPE
1.36 FPE/CI @ 24 FPE
1.60 FPE/CI @ 18 FPE

I look forward to hearing where the 73 FPE number for 1 liter-atm. comes from....

Bob

[PakProtector]

Google will find you an appropriate energy conveter. Ask it to convert 1 l-A to foot-pound force; it will be very close to 74.3 FPE per. Exchange Atmosphere for Bar( at an almost 1:1 ratio ), CI-Bar is energy. Per the converter I found earlier, 1 FPE=.013381 l-A, 1 Joule= .0098692 l-A. Pressure x volume=energy.

I can't wait to see what I've missed that gives you the numbers you got.
cheers,
Douglas

[rsterne]

I found this calculator for the energy stored by compressed air.... http://www.tribology-abc.com/abc/thermodynamics.htm

Using 1450 psi = 100 bar = 10 MPA.... expanding to 1 bar = 0.1 MPA
1 litre = 0.001 cubic metres (therefore enter 1 into the volume box in the calculator)
The result is 46.1 KJ = 34002 FPE

Using 1 CI = 0.0164 L.... The result is 46.1 x 0.0164 = 0.756 KJ = 558 FPE.... which is the amount of energy released when 1 CI of air at 1450 psi expands to a pressure of 14.5 psi....

Pressure differential is 100-1 = 99 bar times 1 CI = 99 CI at 1 bar
558 FPE / 99 CI = 5.64 FPE/CI.... That is the maximum potential efficiency of 1 CI of air at 1450 psi....
Therefore, an output of 1.86 FPE/CI is an efficiency of only 1.86 / 5.64 = 33%.... I think that's pretty reasonable....

The calculator shows that it's not a linear function in relation to the stored pressure.... For a 1 litre tank, expanded to 1 bar (0.1 MPA) you get the following energy output for various storage pressures....

1 MPA (145 psi) = 2.3 KJ (1,696 FPE)
2 MPA (290 psi) = 6 KJ (4,425 FPE)
4 MPA (580 psi) = 14.8 KJ (10,916 FPE)
8 MPA (1160 psi) = 35.1 KJ (25,888 FPE)
16 MPA (2320 psi) = 81.2 KJ (59,890 FPE)

It looks to me like the function is logarithmic.... I used the values for isothermal expansion.... I realize that the value will be somewhere between isothermal and adiabatic.... but since the temperature of the air coming out of the barrel certainly is much higher than what the calculator gives for adiabatic I didn't know how to estimate the "blend" between isothermal and adibatic that is occurring....

The actual work being done by the expanding air is greater than the FPE given to the pellet.... The air itself has mass, and has a measureable FPE (or at least one that can be calculated).... In one example I calculated, using data from a gun that shot at 20 FPE at 1.2 FPE/CI with a mechanical efficiency of 50% including the air.... would have only had a mechanical efficiency of about 42% with the energy of the air ignored.... The harder you push a given design, the more air wasted.... and the weight of the air discharged becomes more and more significant....

Bob

[PakProtector]

hey Bob,
Thanks for the period of instruction. Seems what I missed is not realizing how much atrophy has effected my computing/skill set... Back to the books...I kept them for just such an eventuality.

Douglas

[Bill G]

Bob very nice job of explaining that bit of info.  Your knowledge is second to none.

[rsterne]

HEY Doug!.... Not to worry, it's easy to apply stuff the wrong way, I do it all the time.... 
Bill G.... He had me going for a while, so I had to start digging to find a theory that fit the facts.... 
I learned something VERY valuable from this that I already knew intuitively.... that higher pressures are more efficient.... Now I know how to quantify it.... Don't ya love Math ?!?!?

EDIT: I decided to use that calculator to figure out the maximum possible efficiency for 0-5000 psi and create a graph....



I converted psi to MPA, used the calculator to get KJ, converted that to FPE.... Using the pressure drop (expanding to 1 bar) I then calculated the FPE/CI and graphed it using Excel.... The easy way to remember it is that the maximum efficiency of a PCP is about 5 FPE/CI at 800 psi.... 6 FPE/CI at 2000 psi.... and 7 FPE/CI at 4500 psi.... You can calculate the percentage efficiency of your gun by dividing the maximum (from the graph) by the FPE/CI your gun produces.... For example, a Disco shooting 1.3 FPE/CI at 1500 psi is about 1.3 / 5.7 = 23% efficient....

Bob

[lloyd-ss]

I have been away for a few days doing hard physical labor and you guys have been having fun doing all this  geeky math stuff. Jeez!  A great job of bringing Doug's original energy conversion formula into a real-world application by adding all the caveats that need to be applied. 
I have not studied the thread fully, but Bob, thanks for doing all the research and presentation. And Doug, thanks for starting this discussion with that conversion.
You have mentioned the additional weight of the air column being accelerated along with the pellet.  That not worthy up some of the energy. Pellet friction in the barrel?  Losses due to the inherent flow characteristics of the valve (shape of the pulse)? The original conversion had to do with expanding the volume from the stated pressure all the way down to atmospheric.  But in reality, the final pressure in the barrel might anywhere from 10 or 20% of the original pressure to 70% or more of the original pressure.  So Bob, would you just be working with a particular window of your graph for each shot, with the starting and ending pressure, or the delta?   This is good stuff!
My head hurts.  Time to stop. LOL.
Lloyd

[rsterne]

True, Lloyd, the actual FPE available for a shot would depend on the pressure drop to the instant the pellet left the barrel.... The larger the residual pressure at the muzzle, the less FPE available from the expansion of the air.... The fact remains, however, that for each CI you release, had you used all of it (in a long enough barrel) then you are back to the values on the graph as the maximum FPE/CI that the air could have produced (under ideal conditions).... It's just that the design you chose wasted some of the air with no hope of utilizing it.... That energy showed up at the muzzle as muzzle blast and noise....

The calculator I linked to allows you to input the "ambient" pressure as well as the pressure in the containment vessel.... I would assume that if you input the air pressure remaining at the muzzle into the "ambient" box you could calculate the possible FPE available to accelerate the pellet (and air).... I'm not sure, however, if that is the proper use of the calculator....

Bob

[PakProtector]

hey-Hey!!!,
I can always appreciate a good instructor. I can also appreciate the use of students. And now for some History... When all the top physicists got cordoned off in the Institute for Advanced Study, no more good stuff was developed by them. OTOH, R.P. Feynman was still saddled with his grad students and had yet to earn his Nobel.

So far the modern Super Range is still on track. There is much left to learn...stand by y'all if you'd be so kind.
cheers,
Douglas

[huklbery]

Its all good and gentlemen are enriching all our knowledge as a simple derp I like the exchange.

[rsterne]

I can also appreciate the use of students.

Absolutely.... There is no better way to understand something than to have to explain it to somebody else.... Listening to what they have to say doesn't hurt either.... 

Bob

[TimmyMac1]

I can also appreciate the use of students.

Absolutely.... There is no better way to understand something than to have to explain it to somebody else.... Listening to what they have to say doesn't hurt either.... 

Bob

Well done Bob,
I used to be such a math nerd but you have it worse than me. You're the guy in math club that would sit quietly in the back till everyone else had exhausted themselves over stepping their skill set. Then you would show the facts, correct way to approach problem, give the appropriate formula, graph it and then explain it in verbage for those that could not comprehend the math/graph.
In my math club guys like that were always the last man standing as the goal was to ask questions the standing guy could not answer. Once the questions stopped you ruled as long as the teacher saw it your way or was buried her/himself.
The point of the exercise was to get everyone up to their highest comprehension level by showing the relationships between the math, the graph and the mechanics of the relationship between the parameters.
For brains like mine it was well worth the lost play time to play brainiac and to tutor hot chicks.
For guys like you Math is cake and it is really nice to have someone here with the gift to explain the stuff and catch the foul balls without insulting the players, coach or Ump. Nice Work and thanks for being here. I learn a lot from you and I already know a lot.
I have spent my entire life facinated by subsonic flow dynamics and transonic air flow charachteristics. Internal and External Aero work is all over air gun technology and the more I study it the more I realize I will never completely understand it. The more you learn the more you realize you are just scraping the surface. I wish I had Adrian Neweys Brain(Head Designer at Red Bull Racing) to apply to airgun technology. Too bad the market doesn't support the R&D cost as we could see some cool stuff.
All I have is determination and a decent math skill set that lets me see it in graph form and where the data supports the direction I went.
It would be so cool to have a 5 million dollar lab to do all the testing but we don't even have computational fluid dynamics. Just seat of the pants knowledge and info from the disciplines that push the air flow threshholds. Formula one is one of those places where 20,000 psi @ 18,000 RPM is common stuff. The onboard pumps to actuate air return rather than conventional spring return for valves.

In math, aerodynamics and fluid dynamics of airguns we stand on the shoulders of giants. Geniuses are nice to have around! If you do your homework they can show you a lot. Giffard had some cool designs he never produced as far as I know. His drawings showed he had a fruitful mind with a limited budget. I know someone like that.

[QVTom]

I'm my business, embedded systems, I have several of these "geniuses" that I regularly go to when I hopelessly lost.  I have learned that the most important skill is to ASK THE CORRECT QUESTION.  On may occasions I have not asked the correct question and only got the correct answer much later and in a unrelated conversation.  Then I would say, "remember the such and such project when I ask about this problem", the answer is always the same.  "That's not what you asked"

Tom

[rsterne]

the answer is always the same.  "That's not what you asked"

YUP!.... ain't it the truth....

Tim, I'm also an F1 fan.... my racing days go back to when Hi-Tech was a Cosworth SCA (140 HP from a 1000 cc SOHC Anglia conversion).... I used to race a Morgan, which is about as far from F1 as you can get....

Bob

[william71743]

You guys are AWESOME......