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Does gas pressure change linearly as it changes volume?


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#1 dosh

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Posted 24 March 2013 - 05:01 PM

Lets say I had 800 psi sitting in a one cubic inch chamber. If I changed the volume to two cubic inches would the pressure drop to 400 psi?

#2 Latsabb

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Posted 24 March 2013 - 05:17 PM

Pressure=Force/Area

If you had a cube, with a volume of one cubic inch, that means that each side would be one inch. As a cube, it would be like a die, and therefore have 6 sides, and 6 square inches.

800 lb/sqinch=force/6sqinches

or

force=(800lb/sqinch)*6sqinches

Force equals 4800 pounds

So now remove a side of the die, and put another one on. Four sides with 2 square inches, and 2 sides with 1 square inch is 10 square inches.

Pressure=4800 pounds/10 sqinches

Pressure then equals 480 PSI

Edited by Latsabb, 24 March 2013 - 05:18 PM.

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#3 rntlee

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Posted 24 March 2013 - 07:33 PM

Yes. It's inversely proportional. Look up Boyle's Law.

Edited by rntlee, 24 March 2013 - 07:34 PM.


#4 Danny D

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Posted 25 March 2013 - 03:57 PM

PV=nRT

assuming the temperature remains constant if the volume is increased by a factor of 2 then the pressure must be equivalent to 1/2 of the starting value.


So as per your original question: Once the temperature stabalizes to room temperature again, it will be 400psi.

Edited by Danny D, 25 March 2013 - 03:57 PM.


#5 Latsabb

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Posted 25 March 2013 - 03:59 PM

PV=nRT

assuming the temperature remains constant if the volume is increased by a factor of 2 then the pressure must be equivalent to 1/2 of the starting value.


So as per your original question: Once the temperature stabalizes to room temperature again, it will be 400psi.


This is correct. You can safely ignore my fairly embarrassing post above, lol. Although my formula and delivery are correct, it is not for a fluid pressure inside a container, but for the pressure a force from an item applies upon something else. Not sure why I didnt think about it being fluid mechanics.
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#6 dosh

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Posted 25 March 2013 - 07:54 PM

Thanks. I've got this idea of being mulling on how to get a gun to shoot a ball on 150 psi using only a HPA regulator set at 800, but it involves carefully set volumes. I'm working on piecing together the formulas and conversions I need now that I know this works.

#7 cockerpunk

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Posted 25 March 2013 - 07:56 PM

yes, as long as your are reasonably close to reversible processes. then the above mentioned ideal gas law should work.

most expansions/compressions are not reversible processes, esp with air as your working fluid. air is nothing like an ideal gas.

Edited by cockerpunk, 25 March 2013 - 07:57 PM.

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#8 dosh

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Posted 27 March 2013 - 02:02 PM

Would it get me in the ballpark?

#9 btrettel

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Posted 29 March 2013 - 09:04 PM

yes, as long as your are reasonably close to reversible processes. then the above mentioned ideal gas law should work.

most expansions/compressions are not reversible processes, esp with air as your working fluid. air is nothing like an ideal gas.


I'm a little confused.

The ideal gas law has no requirements about the reversibility of the process. You can find reversible process relationships for ideal gases, but the ideal gas law itself has no such requirement.

Very fast expansions like those in a paintball gun are nearly reversible because the heat transfer (etc.) is nearly negligible over short time scales. This isn't true for other processes like filling an air tank, which usually has enough time for heat transfer to allow the temperature to equilibrate (a non-reversible process).

I wouldn't say that air is nothing like an ideal gas. Air is usually very close to an ideal gas at the temperatures and pressures in paintball guns. Look at a compressibility chart. The compressibility factor for air is pretty near 1 (within 0.05 or so) for pressures up to 70 bar or so at room temperature. If you use HPA, then non-idealities are more important (see my earlier posts here for an example), but not too much; the compressibility factor is about 1.15 at room temperature and gets closer to 1 as the tank cools down.

#10 Pyrate Jim

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Posted 30 March 2013 - 02:46 PM

The answer is Yes.
I was taught that Boyles' Law is volume over pressure is equal to pressure over volume.
IE: One Cubic inch at 500 PSI has equal energy to Five Cubic inches at 100 PSI.
I believe Charles' Law deals with the relationship of temperature to pressure for a given volume, a whole different set of parameters.

However, those laws only apply to volume/area/pressure as set values.
How you use the available energy is a whole 'nother discussion.

Open expansion or Linear (closed tube) expansion have different aspects.
A "Linear expansion" would describe how the gas loses pressure within a confined area, say a Nelson power tube opening down a larger tube like the barrel.
An "Open expansion" would describe the gas leaving the barrel into the atmosphere.

It's the "linear" rules that also govern constrictions, hydraulic damming and Venturi effects.
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#11 cockerpunk

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Posted 01 April 2013 - 08:43 AM


yes, as long as your are reasonably close to reversible processes. then the above mentioned ideal gas law should work.

most expansions/compressions are not reversible processes, esp with air as your working fluid. air is nothing like an ideal gas.


I'm a little confused.

The ideal gas law has no requirements about the reversibility of the process. You can find reversible process relationships for ideal gases, but the ideal gas law itself has no such requirement.

Very fast expansions like those in a paintball gun are nearly reversible because the heat transfer (etc.) is nearly negligible over short time scales. This isn't true for other processes like filling an air tank, which usually has enough time for heat transfer to allow the temperature to equilibrate (a non-reversible process).

I wouldn't say that air is nothing like an ideal gas. Air is usually very close to an ideal gas at the temperatures and pressures in paintball guns. Look at a compressibility chart. The compressibility factor for air is pretty near 1 (within 0.05 or so) for pressures up to 70 bar or so at room temperature. If you use HPA, then non-idealities are more important (see my earlier posts here for an example), but not too much; the compressibility factor is about 1.15 at room temperature and gets closer to 1 as the tank cools down.


expansion in a paintball gun is not a very reversible process, because there is serious cooling in the process. as a quick calc, there is something like 2700 shots worth of energy in a 68/45, i have yet to actually see a gun demonstrate even 2000 from a tank (i know, many people claim ...), that's almost a 1/3rd of the energy from the tank lost because of irreversibility.

Edited by cockerpunk, 01 April 2013 - 08:43 AM.

The ultimate truth in paintball is that the interaction between the gun and the player is far and away the largest factor in accuracy, consistency, and reliability.

And yes, Gordon is the sexiest manifestation of "to the front."


#12 Egomaniacal

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Posted 02 April 2013 - 08:44 PM

Lets say I had 800 psi sitting in a one cubic inch chamber. If I changed the volume to two cubic inches would the pressure drop to 400 psi?


Yes. And with NkT at the beginning = NkT at the end, it doesn't matter how you change it.

Edited by Egomaniacal, 02 April 2013 - 08:45 PM.

eiπ = − 1

#13 btrettel

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Posted 03 April 2013 - 05:25 PM

expansion in a paintball gun is not a very reversible process, because there is serious cooling in the process. as a quick calc, there is something like 2700 shots worth of energy in a 68/45, i have yet to actually see a gun demonstrate even 2000 from a tank (i know, many people claim ...), that's almost a 1/3rd of the energy from the tank lost because of irreversibility.


Reversible adiabatic processes cool down during expansion. What makes a process irreversible is heat transfer, friction, turbulence, etc. The expansion in a paintball gun is nearly reversible because it occurs so quickly, which makes heat transfer during the shot negligible. Dynamic friction is not usually a big effect (static friction is, however, but that is not an irreversible effect). Neglecting turbulence and all other effects in a computer simulation seems to produce pretty good results, so I'm led to believe other irreversible effects are negligible.

I'm not too familiar with this forum, but if you are referring to these calculations, be aware that they implicitly assume the tank expansion process occurs at constant volume. The actual process is much closer to adiabatic. That calculation also doesn't account for the energy efficiency of a paintball gun. I'm not sure it is accurate, for these reasons. I'm not sure you can attribute the difference to irreversibility. When I get the time I'll do an adiabatic calculation with some realistic energy efficiencies to see what a better estimate would be.

Edited by btrettel, 03 April 2013 - 05:34 PM.


#14 Danny D

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Posted 03 April 2013 - 05:52 PM


expansion in a paintball gun is not a very reversible process, because there is serious cooling in the process. as a quick calc, there is something like 2700 shots worth of energy in a 68/45, i have yet to actually see a gun demonstrate even 2000 from a tank (i know, many people claim ...), that's almost a 1/3rd of the energy from the tank lost because of irreversibility.


Reversible adiabatic processes cool down during expansion. What makes a process irreversible is heat transfer, friction, turbulence, etc. The expansion in a paintball gun is nearly reversible because it occurs so quickly, which makes heat transfer during the shot negligible. Dynamic friction is not usually a big effect (static friction is, however, but that is not an irreversible effect). Neglecting turbulence and all other effects in a computer simulation seems to produce pretty good results, so I'm led to believe other irreversible effects are negligible.

I'm not too familiar with this forum, but if you are referring to these calculations, be aware that they implicitly assume the tank expansion process occurs at constant volume. The actual process is much closer to adiabatic. That calculation also doesn't account for the energy efficiency of a paintball gun. I'm not sure it is accurate, for these reasons. I'm not sure you can attribute the difference to irreversibility. When I get the time I'll do an adiabatic calculation with some realistic energy efficiencies to see what a better estimate would be.


Distortion of the round itself may dissapate a lot of energy. As an aside, I know with firearms the most efficient load is one in which consistent pressure can be maintained on the bullet in the breach and down the barrel for the longest duration. This is rarely attainable as pressures tend to spike and not plateau. Some have tried mixing fast and slow burning powders to achieve this (some with disastrous consequences). It has been done before, but only with certain setups. I bring this up, as it might be the case with paintball markers. I imagine, due to the design of the various valves, behind the paintball pressures tend to spike rather than remain constant for a legnthy duration. This might be another source of inefficiency. Just speculating at this point.

#15 btrettel

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Posted 03 April 2013 - 06:13 PM

Distortion of the round itself may dissapate a lot of energy. As an aside, I know with firearms the most efficient load is one in which consistent pressure can be maintained on the bullet in the breach and down the barrel for the longest duration. This is rarely attainable as pressures tend to spike and not plateau. Some have tried mixing fast and slow burning powders to achieve this (some with disastrous consequences). It has been done before, but only with certain setups. I bring this up, as it might be the case with paintball markers. I imagine, due to the design of the various valves, behind the paintball pressures tend to spike rather than remain constant for a legnthy duration. This might be another source of inefficiency. Just speculating at this point.


Plastic deformation is another source of irreversibility, yes. No idea how important that is in paintball.

I don't know much about the ballistics of actual firearms, but I doubt constant base pressure is optimal or even achievable. In pneumatic guns like paintball guns and firearms with "infinitely fast" chemistry, the optimal time-pressure curve follows the adiabatic process relationships (by definition --- a reversible adiabatic process with the right barrel length would have 100% efficiency). That sort of curve is very difficult to obtain due to the limited flow rate of most valves. With a computer simulation (here's one) you can see all sorts of realistic curves. Most of the time the barrel pressure increases to a peak and then decreases until the projectile leaves.

#16 Danny D

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Posted 03 April 2013 - 09:32 PM

Plastic deformation is another source of irreversibility, yes. No idea how important that is in paintball.



I imagine very important. Transfer of energy is usually the most inefficient process in a system. The deformation of a ball is a huge sink of energy in getting it to reach exit velocity.

#17 cockerpunk

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Posted 04 April 2013 - 10:18 AM


expansion in a paintball gun is not a very reversible process, because there is serious cooling in the process. as a quick calc, there is something like 2700 shots worth of energy in a 68/45, i have yet to actually see a gun demonstrate even 2000 from a tank (i know, many people claim ...), that's almost a 1/3rd of the energy from the tank lost because of irreversibility.


Reversible adiabatic processes cool down during expansion. What makes a process irreversible is heat transfer, friction, turbulence, etc. The expansion in a paintball gun is nearly reversible because it occurs so quickly, which makes heat transfer during the shot negligible. Dynamic friction is not usually a big effect (static friction is, however, but that is not an irreversible effect). Neglecting turbulence and all other effects in a computer simulation seems to produce pretty good results, so I'm led to believe other irreversible effects are negligible.

I'm not too familiar with this forum, but if you are referring to these calculations, be aware that they implicitly assume the tank expansion process occurs at constant volume. The actual process is much closer to adiabatic. That calculation also doesn't account for the energy efficiency of a paintball gun. I'm not sure it is accurate, for these reasons. I'm not sure you can attribute the difference to irreversibility. When I get the time I'll do an adiabatic calculation with some realistic energy efficiencies to see what a better estimate would be.


many folks have done similar calculations, and the shot count depends on if you use 300 fps, 280 etc etc. all we are doing is computing how many shots worth of energy are in a tank. we are then comparing them to actual shot counts, and then making claims as to the guns overall efficiency of expansion.

this simply shows that expansion in a paintball gun is not a reversible process. if it was, then those two numbers would be pretty similar. they are not, so its no reversible.

this is because the expansion process in a paintball gun is NOT adiabatic AT ALL. there is a massive heat flux out of the working fluid, that cools the tank/gun/barrel down. heck, even the muzzle condensation that you see whenever a paintball gun is fired, is frutehr evidence that the expansion is NOT adiabatic.
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#18 btrettel

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Posted 04 April 2013 - 05:52 PM

I imagine very important. Transfer of energy is usually the most inefficient process in a system. The deformation of a ball is a huge sink of energy in getting it to reach exit velocity.


Can you quantify how much energy is lost due to plastic deformation?

many folks have done similar calculations, and the shot count depends on if you use 300 fps, 280 etc etc. all we are doing is computing how many shots worth of energy are in a tank. we are then comparing them to actual shot counts, and then making claims as to the guns overall efficiency of expansion.

this simply shows that expansion in a paintball gun is not a reversible process. if it was, then those two numbers would be pretty similar. they are not, so its no reversible.

this is because the expansion process in a paintball gun is NOT adiabatic AT ALL. there is a massive heat flux out of the working fluid, that cools the tank/gun/barrel down. heck, even the muzzle condensation that you see whenever a paintball gun is fired, is frutehr evidence that the expansion is NOT adiabatic.


If other calcs are like the one I linked to, then they do not test whether the process is close to reversible. As I said, those calcs. implicitly assume the process occurs at constant volume (i.e., not adiabatic/reversible). They also do not account for the energy efficiency of the paintball gun. The actual process is somewhere between adiabatic and isothermal, depending on how long a time is taken between shots. So the simple thermodynamic processes don't really apply if you want the best accuracy. You'd need to use some sort of computer simulation or lots of hand-calcs.

I'm afraid you aren't understanding me correctly. I agree that the expansion process (i.e., the internal ballistic process) is not exactly adiabatic, but computer simulations that neglect heat transfer and other irreversible effects make accurate predictions of muzzle velocity. That indicates the heat transfer, etc., effects are not important. The cooling down of the barrel occurs after the shot leaves the barrel, yes. I'm also aware that expanding gas in a tank will cool down the tank. It's just that during the actual shot, the process is nearly adiabatic.

The condensation is a phase change of a component the working fluid. Phase changes can occur in adiabatic processes.

Edited by btrettel, 04 April 2013 - 05:52 PM.


#19 cockerpunk

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Posted 04 April 2013 - 06:07 PM

If other calcs are like the one I linked to, then they do not test whether the process is close to reversible. As I said, those calcs. implicitly assume the process occurs at constant volume (i.e., not adiabatic/reversible). They also do not account for the energy efficiency of the paintball gun. The actual process is somewhere between adiabatic and isothermal, depending on how long a time is taken between shots. So the simple thermodynamic processes don't really apply if you want the best accuracy. You'd need to use some sort of computer simulation or lots of hand-calcs.

I'm afraid you aren't understanding me correctly. I agree that the expansion process (i.e., the internal ballistic process) is not exactly adiabatic, but computer simulations that neglect heat transfer and other irreversible effects make accurate predictions of muzzle velocity. That indicates the heat transfer, etc., effects are not important. The cooling down of the barrel occurs after the shot leaves the barrel, yes. I'm also aware that expanding gas in a tank will cool down the tank. It's just that during the actual shot, the process is nearly adiabatic.

The condensation is a phase change of a component the working fluid. Phase changes can occur in adiabatic processes.


1. they don't imply anything, they are merely a calculation of the total energy in a typical 68/45 tank, compared to the energy required to fire a paintball. there is no thermodynamic cycles or changes calculated.
2. this does tell us how reversible a process is. because the amount of useful energy that is harvested (ie shot count), compared to the total energy, is so different, we know the process is not reversible. it would be reversible if those two numbers were at all similar.
3. yes, they don't take into account the efficiency of the gun ....BECAUSE THAT IS WHAT WE ARE TRYING TO CALCULATE. stored energy in the tank - useful energy that propels the paintball = wasted energy in the expansion process
4. link to said computer simulations that model a paintball gun's expansion process?
5. you admit there is cooling, but the process is adiabatic. so where does the cooling come from if the process is adiabatic?
The ultimate truth in paintball is that the interaction between the gun and the player is far and away the largest factor in accuracy, consistency, and reliability.

And yes, Gordon is the sexiest manifestation of "to the front."


#20 Danny D

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Posted 04 April 2013 - 07:00 PM


I imagine very important. Transfer of energy is usually the most inefficient process in a system. The deformation of a ball is a huge sink of energy in getting it to reach exit velocity.


Can you quantify how much energy is lost due to plastic deformation?



One can quantify this value, yes. Have I yet? No, hence the use of the word "Imagine". I am simply hypothesising.

I believe there is slowmo footage of a paintball firing in breach. You can most likely measure the inittial deformation from that footage (or obtain new footage), then make a rig with newton scale to measure equal deformation on another paintball. This will give you a somewhat close estimate on what kind of energy is lost in deformation.

I do have access to newton scales, but I am fairly busy right now with work. If you would like to take this on, that would be a great contribution to the community!

#21 Eskimo

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Posted 04 April 2013 - 08:51 PM

Not to rain on anyone's science parade for the question:

Does gas pressure change linearly as it changes volume?

your answer is: ONLY if the gas is a ideal, monoatomic, gas under constant conditions with the temperature held fixed. at that point, Boyle's law, ye olde PV=NRT

long answer
Spoiler


BUT for all intents and purposes, yea in your head just half the pressure and enjoy

Edited by Eskimo, 04 April 2013 - 08:59 PM.

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#22 brycelarson

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Posted 05 April 2013 - 08:05 AM

I believe there is slowmo footage of a paintball firing in breach. You can most likely measure the inittial deformation from that footage (or obtain new footage), then make a rig with newton scale to measure equal deformation on another paintball. This will give you a somewhat close estimate on what kind of energy is lost in deformation.


If you're talking about the Tom Kaye footage - there wasn't any visible deformation. The resolution wasn't super - but it appears to be minimal.

The stuff from PE is better - but I don't know if the frame rate is high enough to see deformation.






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