Weight of Compressed Air

[Thordic]

Using this formula:

Air has an average molecular weight of 29 (28 for N2*.8 + 32 for O2 * .2)
1 mole of gas at STP occupies a volume of 22.4 L, therefore, 29g {~1 oz)of air at STP occupies 22.4 L

Take you tank volume in Liters (ft3X28.3 or cu in X .0164) and correct to STP (forget the temperature, just multiply the volume by 204 or 306{3000psi/14.7psi; 4500/14.7psi}. This gives you the gas volume at STP.

Divide this volume by 22.4 and you get the weight of air in oz.

Using that, you get the following information:

3000 PSI = 204 atm
4500 PSI = 306 atm

47ci = 0.77 L
68ci = 1.11 L
88ci = 1.44 L
114ci = 1.87 L

So now, just plug it into what I said above, and you get these weights:

47ci 3000PSI tank: 7.0 ounces of air
68ci 3000PSI tank: 10.1 ounces of air
88ci 3000PSI tank: 13.1 ounces of air
114ci 3000PSI tank: 17.0 ounces of air

47ci 4500PSI tank: 10.5 ounces of air
68ci 4500PSI tank: 15.1 ounces of air
88ci 4500PSI tank: 19.7 ounces of air
114ci 4500PSI tank: 25.5 ounces of air

As you can see, the weight of compressed air is not negligible, especially in higher pressure setups.

In any case, feel free to comment on it.

[the JoKeR]

Hey, I just found a monkey wrench! Let's see how it flies!


I assume these calculations are for sea level. Do they change based on altitude, or since we are dealing with a compressed gas does it matter?

[Thordic]

Temperature and air pressure have negligible effects on these calculations. I'll try to explain :)

Temperature in these calculations is in Kelvin, where room temp. is ~300* for STP. Any temperature you play in will be less than say ~30*K. Thats a 10% max, but more likely you are looking at closer to a 5% max. Once you run that through the numbers, you'll see that the temperature of where you are playing has very little effect on the results.

As for pressure, at pressures this high, it has very little effect. Look at it like this - Even when your tank is empty, it STILL has 14.7 PSI in it. Everything around you is under a constant 14.7 PSI. So your true tank pressure is 3014.7 when you fill it. The extra atmosphere of pressure, that 14.7, gets effectively negated by atmopheric pressure. Just like taring a scale.

Now lets assume you are playing in, say, Colorado. Lets assume the atmospheric pressure is .8 ATM, or 11.76 PSI. I don't care what the pressure is really like in CO, just work with me. That means that the pressure in your tank at top pressure is 3011.76 PSI.

3014.7 - 3011.76 = 2.94

Thats one thousandth of a percent of the tank capacity. If you want to weigh your tank in milligrams or something that will detect that, feel free :)

The only time atmospheric pressure would become a problem is with low-pressure storage systems. If you are storing 100 PSI, then that 3 PSI becomes 3%, which is a statistically significant number. 0.098% isn't very significant :)

[the JoKeR]

Sorry, but I didn't really expect you go to through that much trouble to explain it. I knew that it would be minimal, if any difference based on altitude. It is nice to get a refresher in all the stuff I forgot from high school. Guess some people really do use what they learn! Then again, you could have made all that stuff up and I'd be in no position to argue.:D

[Thordic]

1 year high school physics + 1 year of studying mechanical engineering + an hour on the phone with my dad, a chemical engineer = what you just saw.

My dad was the one who gave me the formula and backed up my numbers.

Besides, this is kinda the point of Deep Blue, right? I know I find these types of discussions really interesting.

Next thing I'm diving into is gas effeciency. I may need some help with that one, though, to get data on different guns.

[the JoKeR]

Originally posted by Thordic
Next thing I'm diving into is gas effeciency

I had Mexican food for lunch, so I'm all about gas!:eek:

Yes, not real Deep Blue type stuff, but I gotta be me!

[cphilip]

Good stuff Thordic. And I was one of those who had always dismissed the weight of a tank with or without air as negligable and never thought of it as being worth bothering to add into the mix. But its surpringly heavy isn't it?

[animal]

You talk about using stuff you learned... I scared myself the other day when I actually used calculus & trig while building a subwoofer enclosure. Who knew I'd use those 5 semesters of calc.

[cphilip]

Hey thordic, Let do something else with this. Calculate the difference when using pure compressed nitrogen verses using compressed "air". See what we get then? I am wondering if we can save an ounce of weight going with compressed air or nitrogen. I could guess but I would probably be way off.

[bjjb99]

If you can find a tank that will maintain a decent seal to contain it, why not go to compressed helium (cost)? A 68/3K fill would weigh a whopping 1.4 ounces. Everyone coming off the field after a heavy firefight would have squeaky voices.

For that matter, fill your opponents tanks with compressed radon gas and watch their arms fall off as they lug around 78 ounces (nearly five pounds!) per 68/3K fill.

BJJB

[clockworkmiller]

Very nice work. I like the work. The numbers look right.

Helium would be interesting, seeing as how it would be the lightest of the safe gases you can use. I would love to see someone try that.

[Donnyluv]

Helium, hmmm.
Helium is the only liquid that can not be soilidified by lowering the temperature. It remains liquid down to absolute zero at ordinary pressures, but it can readily be solidified by increasing the pressure. Solid 3He and 4He are unusual in that both can be changed in volume by more than 30% by applying pressure.

30% increase in volume,...very interesting!

[clockworkmiller]

I could be wrong, but wouldnt increasing the pressure cause it to change from a liquid to a gas?

If you increase pressure, you also icnrease heat.

[Donnyluv]

Helium is weird though. Kinda like water expanding when it freezes; nothing else does.

[Donnyluv]

Normally increasing pressure would cause a gas to change to a liquid, not the other way around. Like CO2 (even though CO2 actually goes straight from a solid to a gas).

[AGD]

Our friend Ed Head did try helium. Didn't hardly shoot the ball but make big boom! Not enough gram moles to get the ball going and it leaked out of all the orings (really small molecules). Great discussion here guys.

AGD

[Thordic]

Here are numbers to chew on for Compressed Air when compared to pure Nitrogen.

29 Grams equals 1.02294 Ounces (Molar weight of air)
28 Grams equals 0.987671 Ounces (Molar weight of nitrogen)

When I did the original calculations, I rounded 29 grams off to 1 ounce. It made things a lot easier, and due to the small amount I rounded off, didn't have a significant impact on the numbers.

Its under a 5% difference (3.44780730052593504995405400122 if you wanna get real picky :) ) between compressed air and nitrogen.

So in a 68/3K tank, you are looking at around .3oz lighter with pure nitrogen. You won't even shave off a full ounce in a 114/4.5K tank.

[CHK6]

Thordic,

Thanks for turning us in the right direction when calculating weight of compressed gasses. However the gears in my brain is always turning (that doesn't mean the gears always catch thou), so I was wondering if there is a difference in weight of compressible gases due to temperature? Here are my findings.... it also a validation of everything you said earlier.

Ideal gas equation: PV=nRT
P = pressure (atm)
V = volume (liters)
n = number of molecules in the container (mol)
R = universal gas constant (l-atm/mol-K) [0.082057]
T = temperature (K)

Lets plug and play.....

For a 80 degeee day (in Farhienhiet (sp?)), with a 47 ci tank, at a pressure of 3000 PSIG, and using compressed air.

T = 80 F = 299.816666 K
V = 47 ci = 0.770189 l
P = 3000 PSIG = 204.137883 atm

n = (204.137883 * 0.770189) / (0.082057 * 299.816666)
n = 6.390715 (mol)

n * Mw = 6.390725 * 28.96 (molecular weight of air) = 185.08 grams. Convert the grams to troy ounces and we get 5.95 ounces.

Now in Texas most summer mornings start out at 80 degrees, but by 11:00 it's usually 100. Using the same formula, but instead of 80 I use 100 degrees and we get 5.73 ounces.

So after a couple of games in the heat and your arm is getting tired, don't worry mother nature tries to help you out....just a tad of 3% difference for 20+ degrees. I think I lose more than that percentage in sweat alone. ( C :"

Helium would be an awesome break through for paintball.

[Thordic]

If you look up, I already went into the effect of temperature on the weight.

As for helium, as Tom just said, it isn't dense enough to propel the paintball.

I believe the best gas for paintball in theory is Argon, but its just too expensive to be practical.

[AGD]

Argon does have the best molecular weight numbers and doesn't go liquid. We were discussing this in the early 90's and again my buddy Ed Head shot several tanks of argon to try it out. Unfortunately didn't write down the numbers but he reported that it didn't seem to make that much of a difference. Someone should try it again for grins.

AGD

[clockworkmiller]

Does anybody know the cost to fill a 88cuft 3kpsi scuba tank with argon as opposed to compressed air? Unless the numbers are very similar, then Argon would need extrodinarily better in order for it to be feasible. Also, what would the NPPL or NCPA or other paintball associations say about using argon in tournaments.

[Dubstar112]

Tom I could try Argon... Im good friends with my local welding shop and I bet I could get one fill for a little bit of dough ;).. But would it be safe? especially for the Flatline regulator?)(i have a 68/3k)

Either way, Im not sure because I dont think they have the proper regulated fill station.. but I could try.)

[Doc Nickel]

Argon fills will be considerably more expensive than Nitrogen.

I can't recall specific costs off the top of my head at the moment, but I do know that refilling a helium or argon cylinder for the TIG is quite a bit more than refilling my nitrogen tank for paintball fills. Keep in mind those are Alaska prices, though. Your mileage may vary.

In any case, for the purposes of the discussion, the additional cost of Argon probably far outweighs the miniscule performace benefit of the gas.

Hey Tom: Next time you have a chance, see if there's any difference in regulator recharge rate between HPA, Nitrogen, Argon or Helium. And I mean on some of the aftermarket regs you've been trying, not just the RT and 'Mag valves.

That's something I'd like to see.

Doc.

[bjjb99]

Doc Nickel wrote:
>
>Argon fills will be considerably more expensive than Nitrogen.
>

Absolutely. Considering that the air we breathe is 78% nitrogen and only 0.9% argon, it's easy to see why nitrogen costs less... it's just plain easier to extract a given volume of pure nitrogen from the air than it is to get the same volume of pure argon.

Of course, a fill of compressed air should be less expensive than either nitrogen or argon, since no extraction equipment is required at all.

BJJB

[AGD]

Argon here in the big city is more expensive but not rediculous. I think its about 20-30 bucks for a tank. This would be important for tourney players looking to maximize the air capacity while minimizing the weight and going with a smaller tank. Someone should really try it just so we know.

Doc, it's good to see you hanging around again. I am deeep in the throws of paint breakage testing and haven't done any more with the regs. When I get back to it I'll try the different gasses, could be interesting.

AGD

[Redkey]

Are you measuring hard numbers or is it a guesstimation type of test?

Are you doing your tests at impact velocities or are you just performing a crush test and measuring max load? Are you generating load - displacement curves for the tests?

Are the tests being performed at various temperatures or are you just doing them at room temp?

What about preconditioning the paintballs to determine the affects of humidity? If so, are you measuring the weight gain due to absorbed water? What about the paintball dimensions?

Are you keeping track of the paintball orientation?

Shell thickness?

My concerns would be with the batch to batch variability. Do balls produced one week have similar characteristics to balls produced weeks or months later? If the company cannot hold tight tolerances you'll never know what you'll get.

thanks
Jack

[etnie002]

Hey thordic great work. One quesiton for you though. I'm doing a project dealing with the differences between CO2 and compressed air and thought this thread would defintely help. However what do you mean by
Air has an average molecular weight of 29 (28 for N2*.8 + 32 for O2 * .2) . Wouldn't the molecular weight for N2O2 be 60 considering ((14.0 x 2) + (16.0 x 2)=60), meaning the avg. weight would be 60g. I was wondering where the .8 and .2 came from because it really confused me.

[etnie002]

Ahh sorry I think I just figured out what I did wrong. By N2 and O2 i thought u meant the formula for compressed air is N202 (i didn't bother to look it up) but as i re-read your post i realized you were merely displaying nitrogen and oxygen as diatomic (which totally slipped out of my mind for some reason). So by the .8 and .2 im guessing u meant that compressed air is 80% nitrogen and 20% oxygen, meaning its formula is N4O. But then i sitll dont see why its only 29g considering N40 would weigh 72g. SO im sitll kinda' lost.

[bjjb99]

You're basically doing a weighted average of the molecular weights of the two primary gases in the atmosphere.

Compute the molecular weight for nitrogen gas (N2). It's right around 28. Do the same for oxygen gas (O2). It's right around 32.

Air is about 80 percent nitrogen and 20 percent oxygen by volume.

Take 80 percent of nitrogen's molecular weight (0.8 * 28 = 22.4). This is the weight contribution that nitrogen will provide to air's molecular weight.

Take 20 percent of oxygen's molecular weight (0.2 * 32 = 6.4). This is the oxygen's weight contribution to air.

Add the two results together to get the molecular weight of air (28.8).

BJJB

[etnie002]

yeah i had my chemistry teacher explain this to me she was like "oh yeah i forgot to teach you this part". So i understand now, thanks.

[TRIAD]

It is unclear to me how a gas with a higher molecular weight would benefit the propulsion of a paintball. At any given temperature, all substances have the same average KE (temperature being a relative measure of the average kinetic energy of a substance). Thus, while one gas molecule might be more massive (let's compare a molecule of Ar to a molecule of N2), the one that is more massive will also be traveling at a lower velocity. (KE=(1/2)M Vsquared).
So, a mole of Ar gas and a mole of N2 gas will have (essentially) the same kinetic energy (considering they have the same average kinetic energy, and you have a specific amount of each). When you fill a tank wwith Ar, since Ar has a molecular mass of 18, the molecules have, on average, a higher velocity than molecules of molecular nitrogen (molecular mass of 28.0134). Molecular weight is therefore a moot point when dealing with the propulsion of a paintball (as far as I can see), because the average KE of each molecule of gas is still the same.

oh yeah, and for temperature to pressure comparisons, you can use the formula:

P/T=P/T

T is in K
P units must be the same on both sides

*Finally, since this is Deep Blue and we're getting all technical here, K is an absolute measurement, and thus has no units. It is not degrees K, it is just K. There is no unit used with K. Also, you want to use the term mass and not weight.

[athomas]

Its not molecular weight, its molecular density that is key to which gases can produce more propulsion.

The density dictates how many of the molecules actually affect an area at a given instant.

That's why helium does not work well as a propellant. It displaces a lot of area and produces a high pressure with a lot less molecules.

[TRIAD]

I do not believe there is a such term as "molecular density". Perhaps you meant "particle density", which I do not know if that's a real term either, but we'll use it.

The "particle density" of all gases is essentially the same. This means that no matter what gas you're dealing with, given the same size container and the same temperature and pressure, you will have the same number of particles of gas. This is why a conversion factor of 22.4 liters of gas to one mole of gas (assuming room temperature and atmospheric pressure) can be employed. The distance between the molecules of gas is so great relative to the size of the individual particles that particle size becomes negligible. So, whether I have a liter of N2 or a liter of F or a liter of Ar, the number of particles in the same size container at the same temperature and pressure will ALWAYS be the same.

Helium should, in theory, work just as well as any other propellant. However, I am ignorant of the physics involved in paintball propulsion, and am analyzing this from a strictly chemistry-oriented view. Helium is a relatively small molecule, so that may be why it does not work very well as a propellant.

[athomas]

Originally posted by TRIAD
I do not believe there is a such term as "molecular density". Perhaps you meant "particle density", which I do not know if that's a real term either, but we'll use it.



I did indeed explain this wrong.

Yes, I believe I used the wrong term here. The correct terms I believe are gas density and molar mass.

I'm gonna stop because I'm just messing myself up here. Once I get it totally straight I'll try again.:confused:

[bjjb99]

Originally posted by TRIAD
It is unclear to me how a gas with a higher molecular weight would benefit the propulsion of a paintball. At any given temperature, all substances have the same average KE (temperature being a relative measure of the average kinetic energy of a substance). Thus, while one gas molecule might be more massive (let's compare a molecule of Ar to a molecule of N2), the one that is more massive will also be traveling at a lower velocity. (KE=(1/2)M Vsquared).


The mean kinetic energy of a gas molecule is related to its temperature by the following formula:

KE = (3/2) * k * T, where
k is Boltzmann's Constant (1.38x10^-23 J/K), and
T is temperature.

The mean kinetic energy of a gas molecule is also related to its velocity by the following formula:

KE = (1/2) * m * v^2, where
m is the mass of the molecule, and
v is the molecule's velocity.

We can combine these two equations and solve for velocity as a function of temperature and molecular mass:

v = sqrt(3 * k * T / m)

At room temperature (300K), the average velocity for a molecule of N2 is right around 517 meters per second. For CO2, it is around 412 meters per second.

We cannot talk solely in terms of energy transfer when considering the acceleration of a paintball. Since the molecule-paintball collisions are not elastic, kinetic energy is not necessarily conserved. Momentum, however, is conserved. If you look at things from a momentum standpoint, you can see how a heavier gas could accelerate a paintball more easily than a lighter gas at the same temperature.

Momentum is simply the product of an object's mass and velocity.

P = m * v

A molecule of N2 at 300K has a momentum of approximately 2.4x10^-23 kg-m/s.
A molecule of CO2 at the same temperature has a momentum of approximately 3.0x10^-23 kg-m/s.

The heavier CO2 molecule has more momentum to transfer when colliding with the paintball. Since we're apparently dealing with the same number of molecules in each case, a paintball fired using a heavier gas should accelerate faster than a paintball fired using the same number of molecules of a lighter gas.

Just my guess on how things work.

BJJB

[athomas]

Well said.

[TRIAD]

a=f/m

where a is acceleration
f is force
and m is mass (of the thing being accelerated, in this case the paintball)
m will not change, because the mass of the paintball is a constant

P=F/A

A is area (of the thing being accelerated, in this case the paintball)
A will not change because the area of the paintball is constant

Therefore, the force applied to paintball is directly proportion to the pressure of the propellant. The type of propellant is a moot point.

[bjjb99]

Originally posted by TRIAD
a=f/m

where a is acceleration
f is force
and m is mass (of the thing being accelerated, in this case the paintball)
m will not change, because the mass of the paintball is a constant

P=F/A

A is area (of the thing being accelerated, in this case the paintball)
A will not change because the area of the paintball is constant

Therefore, the force applied to paintball is directly proportion to the pressure of the propellant. The type of propellant is a moot point.

It's moot until you start considering that no gas is truly ideal, which is now where I think the CO2-vs-N2 effects will be seen. For example, a cubic meter of CO2 at 300K and 1 atmosphere (101325 Pa) contains 40.796 moles of gas. The same volume, pressure, and temperature for N2 contains 40.651 moles. The rates at which pressure changes as a function of expanding volume are different for these two gases once Van der Waals effects are taken into account.

Start with a fixed initial volume V of gas at a temperature of 300K and a pressure P. Compute the number of moles of CO2 and N2 that would be contained in that volume under those conditions, taking Van der Waals effects into account. Then use the same Van der Waals equation to compute P as a function of V at 300K for the CO2 and N2 mole amounts you obtained. The P-vs-V curves are not identical.

While this is not necessarily where my original post was headed, your reply made me think a bit more and I stumbled onto the possibility that these gases were not as ideal as I originally presumed them to be.

BJJB

[TRIAD]

Most gases behave close to ideal. The only time they do not behave in an ideal fashion is at extremely low temperatures (the average velocity of the gases is such that they exhibit a lower pressure than would normally be seen. The reason for this is that the IMFA of the gases is effective enough that when they interact with each other, the molecules of gas travel in slightly curved paths, and thus they do not hit the sides of the container as often as they would at higher temperatures, or if the gases were ideal.) They also don't behave in an ideal fashion at extremely high pressures. The reason for this is that the molecules of gas actually take up space, as opposed to ideal behavior. However, we are not dealing with extremely low temperatures (near O K). I will find the equation for calculating the effect of 4500 psi on the volume of space taken up by the gas, and I should be able to calculate it.

However, our subject has changed to not whether a certain gas provides better propulsion (after talking to my physics teacher, I now know that the type of gas does not matter) but whether there is MORE of a gas in a certain volume. I do not believe that 4500 psi is as high enough pressure to affect the volume, though I am not sure, and will check. (I also believe my physics teacher would've mentioned that, but I'll talk to my chem teacher to get the final word).

"It's moot until you start considering that no gas is truly ideal, which is now where I think the CO2-vs-N2 effects will be seen. For example, a cubic meter of CO2 at 300K and 1 atmosphere (101325 Pa) contains 40.796 moles of gas. The same volume, pressure, and temperature for N2 contains 40.651 moles. The rates at which pressure changes as a function of expanding volume are different for these two gases once Van der Waals effects are taken into account."

I do not know where you are getting these mole amounts. I'll have to check the deviations as I said.

[bjjb99]

Yes, most gases behave close to ideal at normal temperatures and pressures. However, we're looking at small differences in paintball velocities, and the devil's in those details. A first-order ideal gas treatment may not be sufficient to cover the phenomenon in question.

And yes, we are looking at whether there is more of a given gas contained in a given volume at a given pressure and temperature... when that different amount of gas is released, it's going to provide a different pressure curve behind the paintball and thus different paintball muzzle velocities.

Let's take a look at some gas behavior if we include Van der Waals effects.

The Van der Waals equation:

[P + a * (n/V)^2 ] * (V - n * b) = n * R * T

P in pascals, V in m^3, T in K, n in moles, R in J/mol-K, a and b in appropriate units (listed further down).

I'm going to use a hypothetical paintgun based (loosely) off an automag.

The dump chamber volume is 0.55 cubic inches (9.01x10^-6 cubic meters).

The dump chamber pressure is 400 psig (2.76x10^6 Pascals)

I'm gonig to assume a constant temperature of 300K throughout the decompression process to keep things simple.

A bit of algebra transforms the Van der Waals equation into the following form:

(a * b / V^2) * n^3 - (a / V) n^2 + (P * b - R * T) * n - (P * V) = 0.

The constants a and b are as follows for CO2 and N2:

CO2: a = 0.3643 J-m^3/mole
CO2: b = 4.27x10^-5 m^3/mole

N2: a = 0.1361 J-m^3/mole
N2: b = 3.85x10^-5 m^3/mole

[ constants obtained from http://www.ac.wwu.edu/~vawter/PhysicsNet/Topics/Thermal/vdWaalEquatOfState.html ]

Plugging in these values along with P, V, and T yields the following results for n:

CO2: n = 0.01142 moles
N2: n = 0.01013 moles

At the same 400 psig pressure, 0.55 ci volume, and 300K temperature, there are almost 13% more molecules of CO2 than there are of N2. An ideal gas at these same conditions should amount to 0.00996 moles.

Now let's start with those values for n and see how the pressure varies as a function of increasing volume. Starting at 0.55 ci and increasing the volume to encompass a 12 inch long, 0.68 inch diameter barrel.

V = 0.55 ci --> 4.91 ci

Solving the Van der Waals equation for P gives us:

P = [ (n*R*T) / (V - n*b) - (a * (n/V)^2) ]

Evaluating this equation at V = 0.55, 1, 2, 3, 4, and 4.91 ci for both CO2 and N2 resulted in the following:

CO2:
V = 0.55 ci, P = 400 psig
V = 1.00 ci, P = 234 psig
V = 2.00 ci, P = 122 psig
V = 3.00 ci, P = 82 psig
V = 4.00 ci, P = 62 psig
V = 5.00 ci, P = 51 psig

N2:
V = 0.55 ci, P = 400 psig
V = 1.00 ci, P = 222 psig
V = 2.00 ci, P = 111 psig
V = 3.00 ci, P = 74 psig
V = 4.00 ci, P = 56 psig
V = 5.00 ci, P = 45 psig

The pressure of N2 drops off faster. This shouldn't be surprising since there were fewer N2 molecules in the dump chamber to begin with, even though both dump chambers had the same initial 400 psig pressure.

If we assume constant temperature, the CO2 maintains a higher pressure as the gas expands. This would make a difference in how the paintball is accelerated.

Of course, temperature changes will also factor into the rapidly expanding gas as it propels the paintball out of the barrel. A full numerical treatment of P, V, and T for CO2 and N2 as they expand to push a paintball along is a bit more than I want to do.

To first-order, gas type does not matter. Once you start to factor in second order effects, you begin to see non-neglegible differences in gas behavior (and subsequently, paintball behavior).

BJJB

[RobAGD]

I was noodling around after a customer asked whould be get more shots from a 4500 91 or a 88 5000.

I didnt quite know off hand, did some quick math and said that the 5k would be the better deal for him.

So I started screwing around and made this.

http://www.robagd.com/misc/air-graph2.gif

the argon numbers are not 100% and are not indicitive of a good comparsion between Air and Argon.

-Robert