12 gram nitro?

[Jimmi9999]

what would it take to make a 12 gram filled with nitro instead of co2? i personally would like to see this because i would rather use a 12gr than a big tank.

[vamicromagger]

I think you would only get about 2 shots out of a N2 12 gram. Co2 has alot more "potential" because it is a liquid and expands into a gas.

[MasterYoda]

Assuming air is an ideal gas, one can use the ideal gas equation (p*V=m*R*T)where R= 287J/(kg*K). Using the values m=.012kg p=31.02MPa (4500psi) and T = 293.15K (room temp.). You get a volume of about 2in^3. An average tank gets around 8 shots per cubic inch. So in the most ideal circumstance you would get about 16 shots then you would be done. This cartrige would also be larger than the standard co2 cartrige, depending on the pressure. The smaller the cartrige, the higher the pressure inside the 12g air cartrige. For instance if you didn't want to have to regulate the air, then the volume would be around 12.5ci for 800psi. A 12g air cartrige would also be heavier, and/ or more expensive than a 12g co2 cartrige, due to the tremendous pressure difference. It would probably be so expensive that no one would consider producing such a system, let alone a disposable unit such as the standard 12g co2. With current technology and materials, it is not feasible to produce such a disposible compressed air cartridge.

[Top Secret]

Actually I believe they do exist. Some of the old school guys were telling me about the 12g HPA cartridge that gave less shots, but shot alot faster than a standard C02.

[oldsoldier]

the military used to (may still do) use compressed nitrogen, I think, for their decontamination apparatus. I have used them; they are dark green, used for the decon sprayer thingy for a vehicle.

[Wat]

MasterYoda,

you can't use the ideal gas law because a 12 gram Co2 cartridge hold CO2 gas in a saturated state, ie it is part gas part liquid. It hold gas at a much higher density than HPA would.

What you want to do is compare internal volume of a 12 gram tank which is probably on the order of 1 cubic inch. I can't imagine you could get more than a half dozen shots out of that.

[MasterYoda]

[QUOTE]Originally posted by Wat
[B]MasterYoda,

you can't use the ideal gas law because a 12 gram Co2 cartridge hold CO2 gas in a saturated state, ie it is part gas part liquid. It hold gas at a much higher density than HPA would.

My application of the ideal gas law is correct if you assume that air is indeed an ideal gas. The calculations that I did were using air as the working fluid, not carbon dioxide. I pretty much set up my hypothetical situation to get a ratio of output pressure to volume. The original question was for a 12g HPA system. Since the mass of 12g was specified, that is what I used for the mass of the air. I felt no need to perform an analysis of the current carbon dioxide cartridge as most of us are familiar with them. Even though it wouldn't be a difficult analysis, I also didn't feel like dragging out my thermodynamic tables. I appologize if I was unclear in specifying that air was the working fluid, and not CO2 in my analysis.

[AGD]

MasterYoda,

Welcome to the forums! You seem to have extensive knowledge of physics. How about telling us about yourself?

AGD

[vamicromagger]

"You seem to have extensive knowledge of physics. How about telling us about yourself?"


I was thinking that to after I read his post and my head stopped spinning!

[Wat]

the original poster listed using a 12 gram as in 12gram co2 cartridge filled with nitro instead of co2. He did not ask for a system holding 12 grams of N2.

Thus your application of using M = 12grams in the ideal gas law is incorrect.

[Jimmi9999]

Actually i was asking about a 12 gram nitro not co2 because my mag doesn't really like the 12 gram co2 cartidges. i understand that nitro probably won't work in the regular co2 cartidges they have now but something along the idea of disposable cartidges for nitro.

[SeeK]

Trying to get 12 grams of N2 would be interesting to see. How much volume would that be?

The user is asking about N2 carts. in a 12 gram form factor for lack of a better word. You could also call them chargers like NO for the whipped cream machines. I saw the term on a fire extinguisher mfg. page. The N2 version does exist in this form. I think this person got it from a Military supply place. If I remember correctly it was 1800 psi so you wouldn't get many shots but you should be able to calculate that if you knew the volume of the valve. I've seen that number on one of the threads if you search.

It would be more practical to use one of those 13ci HPA tanks. It's the size of a 9oz Aluminium tank. They were selling at Shatnerball for a little over $100.

[MasterYoda]

Originally posted by AGD
MasterYoda,

Welcome to the forums! You seem to have extensive knowledge of physics. How about telling us about yourself?

AGD

[MasterYoda]

Originally posted by Wat
the original poster listed using a 12 gram as in 12gram co2 cartridge filled with nitro instead of co2. He did not ask for a system holding 12 grams of N2.

Thus your application of using M = 12grams in the ideal gas law is incorrect.

Since you insist that my previous analysis is incorrect, I will perform one based on using a set volume for the cartridge in order to figure out the output pressure. Since I don't have the volume of a standard 12g right off the top of my head, I will perform a simple analysis in order to get a rough estimate. First of all I will list the assumptions that I will make:
1. STP (standard atmospheric conditions)
2. 850psi output for CO2 cartridge at STP
3. C02 and Air cartridges are both closed systems.
4. CO2 is in the two-phase region on a T-v diagram.
5. N2 is an ideal gas

First, to find the volume of a "12g" cartrige, we will have to use the equation V=v*m where V is the volume of the control volume, v is the specific volume of carbon dioxide at 850psi (5.86MPa) and 25C (298.15K) and m is the mass of the fluid in the system. Well, I guess I have to drag those thermo tables after all.
m = 12g (co2)
To find v one must use the equation for a substance in the two phase region. v= vfl + x(vg - vfl) and the quality of the mixture x = (mg)/(mf+mg). Using thermodynamic tables, one obtains these values.
vfl=.00142m^3/kg
vg= .00544m^3/kg
solving for x gives the value .207(--)
thus v=.0022m^3/kg
and V=.0022m^3/kg*.012kg= 2.701(10)^-5 m^3= 1.648in^3

Next we can evaluate a system for compressed air in the same volume. Note that the working fluid is no longer CO2. For this analysis I will use N2 such that the assumption that the ideal gas assumption is minutely more valid. It really makes no difference at stp.
Since we have two free parameters (p and m) in the equation form pV=mRT, a relationship can be set up between the density (rho) and the pressure. rho = p/RT. Which tell us very little. Basically, the more you pressurize the container, the denser the gas will be. Therefore the conclusion can be drawn that the shot capacity of an N2 system with a volume of 1.648in^3 is dependant on the pressure. This parameter will be dependant on the characteristics of the materials used for the container.

[MasterYoda]

Using the value that I found for the volume, and 1800psi pressure, one can set up a situation to roughly calculate the number of shots. I assumed that a ball was accellerated from 0-285fps in a distance 8in. The bore of the barrel I assumed to be .689in. I assumed that while being accellerated in the barrel, the ball experienced constant accelleration without friction. One can then find the accelleration multiply that by the mass to find the force. I assumed that a paintball was roughly .01lbf or 4.97(10)^-4 lbm. I then found the pressure, and used the ideal gas equation to figure out how much mass of N2 is required to accelerate the ball. I used the ideal gas equation to find how much mass is inside the cartrige. Dividing the mass in the cartridge by the mass used accelerating the ball gives a ratio of 12.25. Meaning in the most ideal circumstance with no gas required to operate pnuematics, you could get 12 shots out of such a system. I am guessing that a more realistic value would fall between 6 and 9 shots depending on the efficiency of the gun.