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Jimmi9999
10-24-2002, 09:34 PM
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
10-24-2002, 10:03 PM
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
10-25-2002, 01:10 AM
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
10-25-2002, 06:08 AM
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
10-25-2002, 07:37 AM
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
10-25-2002, 06:56 PM
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
10-25-2002, 09:25 PM
[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
10-26-2002, 12:58 AM
MasterYoda,

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

AGD

vamicromagger
10-26-2002, 10:35 AM
"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
10-26-2002, 08:22 PM
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
10-26-2002, 11:58 PM
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
10-27-2002, 12:41 AM
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
10-27-2002, 10:52 PM
Originally posted by AGD
MasterYoda,

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

AGD

AGD,

I have been playing paintball for a little over five years. I’ve always enjoyed tinkering with mechanical systems, such as paintball guns, cars, etc. Right now I am a junior in the School of Mechanical Engineering at Purdue University. I hope to get my B.S. in 2004 then go to grad school and get my masters. Aside from studying I am a part of Purdue’s intercollegiate paintball team. Recently I have found time to spend in our machine shop in which I have made a delrin bolt and other things. Would there be any internship opportunities that I might be able to take advantage of at AGD? I think that it would be a very interesting and exciting experience for me.

MasterYoda
10-28-2002, 12:37 AM
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
10-28-2002, 01:44 AM
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.

Schmitti
10-28-2002, 02:49 PM
No More Math! Enough Already! You are thinking to much!

Anyways... You can get 1700psi cartridges that are for our purposes the same size as a 12gr at Leland Ltd. http://www.lelandltd.com/

Just follow the link for Products then follow 12gr airgun cartridges and you will see them there.

MasterYoda you over thought the problem, believe me it could have been done easier.

Take care...

PhantomPaul17
10-28-2002, 02:54 PM
Those 13CI tanks are from Catilina. I know because a friend of mine works for ATS and sells em with their ATS markers.
But their more like 5 ounce co2 tanks....size comparision.


talk to ya later
-Paul-

MasterYoda
10-28-2002, 05:09 PM
Originally posted by Schmitti
No More Math! Enough Already! You are thinking to much!

MasterYoda you over thought the problem, believe me it could have been done easier.

Take care...

I appologize if you would have rather had the answer and not the means, but I was trying to give enough information to justify the answer to those who were interested. It is true that I could have done without the thermodynamic analysis of the CO2 cartridge if I wanted to search for the actual volume of the container, but for me it was a lot quicker and easier to just do a couple very simple calculations. Anyone can say, "Well you will get fewer shots with a N2 cartridge than a CO2 cartridge" or "you will get about 30 shots with a N2 cartridge." However, these are not close to solutions, they are merely speculation. I just tried to get numbers based on fact not guessing. I don't know how I could have simplified my analysis any more. All of my assumptions set up the problem for an ideal situation. I really should have done a fluid dynamic analysis of the ball being shot through the barrel, but I felt that the way I did it was close enough. To be more precise one would need to perform several trials to find the average number of shots delivered. In your oppinion this may be easier, but I am not going to spend the time or the money to do such an experiment. If you have a more direct way, please tell me. It may help me in future situations. However, if you don't have one, here is my answer:

Approximate maximum number of shots: 12
Which in my oppinion is not enough to be worth it. Changing a cartridge every 12 shots would be a annoying and expensive.

Wat
10-28-2002, 05:26 PM
Master Yoda,

The assumptions you make to calculate the volume of a 12 gram co2 cartridge is still invalid. You cannot figure out the volume of a container purely by knowing the mass of saturated co2 it holds.

From your equations

v = vfl + x(vg - vfl)

x = (mg)/(mf+mg)

And an an additional unlisted equation

mf + mg = 12grams

We have the following known variables

vfl, vg, mtotal

And the following unkowns

v, x, mg and mf

System of 3 equations and 4 unknowns is unsolvable. Look at it this way, I have a 12 gram co2 cartridge and a 3.5 oz co2 tank and fill them both with 12 grams of co2. Here are two containers of different volume holding the same amount of co2. How could i possibly calculate the volume given that there are clearly many different possiblities?

I think your mistake lies in assumption #2, that the co2 outputs 850psi. As a saturated gas, the pressure is purely dependant on temperature. There is no reason to assume output pressure, you can easily look it up. This is not a way to estimate the volume of a co2 cartridge at all because the estimate is completely unbounded. At best we could make the assumption that co2 is saturated and our estimates would be bounded between when mg = 12grams (pure gas) to mf = 12grams (pure fluid). Or (0.00142m^3/kg)x(12g) to (0.00544m^3/kg)*(12g) or between 17ccs to 67ccs. Thats a pretty large range.

I do not see how you calculated the quality x without knowing any parameters about the container. I don't actually have my thermo charts either, but i'm thinking you can't just read x off a table either. If i'm missing something, let me know.

And FYI, i finished my bachelors degree in mechanical engineering from MIT in 1998.

MasterYoda
10-28-2002, 06:03 PM
I forgot to mention that I used the value of quality from a similar experiment done in my sophmore thermo class. I should have listed this in my assumptions. The equation for the quality is really unessesary in my analysis. I discovered this when, as you said, I ended up with 3 equations and 4 unknowns. Would you have assumed the quality to be 1 or 0? You are right that assumption #2 is probably not the most accurate assumption, but like the other assumptions, it is probably close enough. I was just trying to get a rough estimate of the number of shots in an N2 cartridge. I would be interested in any other input you would have about approaching such a problem. If you don't feel like filling this forum up with more numbers and equations, feel free to pm me.

hitech
10-28-2002, 06:39 PM
Originally posted by MasterYoda
If you don't feel like filling this forum up with more numbers and equations, feel free to pm me.

That's what this forum is for! Keep on keeping on. :D

Vegeta
10-28-2002, 08:20 PM
If you don't wanna go through the numbers, don't bother posting/reading in deep blue.


Welcome to the forums MasterYoda. I am very thankful for another person who seems to know his share in physics. We need em' here.

As for compressed air in a 12 gram sized cartridge (it would not be called a 12 gram anyways since '2 grams' is referring to 12g of liquid co2.).

Normal compressed air tanks are rated for either 3000psi or 4500psi. 12 gram co2 cartdiges hold liquid co2 at no more than 900psi... and not even 900, since the co2 is in liquid form inside to 12 gram cartridge itself and therefore not epanded into a gas. The walls of a 12g cylinder are rather thin... no more than 1/8". so lets do some math here. Now. assuming that the 12g-sized cartridge of the same internal volume was to be filled with 3000psi of compressed air, simple algebra tells us that the walls of the casing of the cartridge would need to be almost 1/2" thick (approx. 5/12") to contain a volume of air with a pressure of 3Kpsi (assuming the walls of a normal 12 gram are 1/8" thick, estimated)

900/(1/8) = 3000/X ,... X = 5/12.

And that's not taking in the fact that the possibility that hte container might not hold the gas probably increases per unit of psi more that waht I have stated, depending of hte tolerance of the material the cartridge is made out of. (My thesis here assumes that the materials are the same fore the CA as the 12g).

nippinout
10-28-2002, 10:46 PM
I've always been taught to simplify.

I'm studying for my construction materials midterm, so I'll be brief.

Solve it using potential.

Find the energy needed for one shot. For each successive shot, see if there is enough potential left in the 12gram for one more. Solve until you do not have enough potential for the next shot.

The only things you need to find are volume of a 12gram, and energy for one shot.

What has been calculated from previos posts is assuming ALL energy will be used.

Why solve for the volume of a 12 gram? Measure the volume of water it takes to fill it up. Jebus, that was easy.

SeeK
10-28-2002, 11:03 PM
I got beat but I was going to say just take an empty 12 gram, fill it with a syringe and measure the water you take out.

You could also use these figures:

http://www.lelandltd.com/
click on Small High Pressure Cylinders, Page 2, top chart
Their item number 42221 is a standard 12 gram.

3.240L, .745 Diameter 14 ml water capacity, 85% fill density.

The N2 version has slightly more capacity at 14.1 ml and a pressure of 1700psi.

Wat
10-28-2002, 11:33 PM
Master Yoda,

If you want more math fun, a little problem we were discussing here and on the tinkerer's guild forum a few months back.

What % of a 20oz co2 cylinder is liquid versus vapor when filled with 20oz of co2. You will need to know the volume of the cylinder for this which i believe was 720ml. You can do a search to find out for sure.

MasterYoda
10-28-2002, 11:49 PM
Originally posted by nippinout
I've always been taught to simplify.

Solve it using potential.

Find the energy needed for one shot. For each successive shot, see if there is enough potential left in the 12gram for one more. Solve until you do not have enough potential for the next shot.

The only things you need to find are volume of a 12gram, and energy for one shot.



You would still need to use at least the 1st law relations of thermodynamics, would you not?

Wat
10-29-2002, 12:03 PM
Originally posted by nippinout
I've always been taught to simplify.
Solve it using potential.


You actually can't do it this way either, you'd be off by at least a factor of 3 if not more. You have to deal with a wonderful pest known as entropy. Thanks to the second law of thermodynamics, we could never ever get 100% mechanical work out of a thermal system. And saturated co2 is chock full of entropic flow when its near phase change.

Its just like you can't say calculate how much thermal energy is in a cup of coffee and equate it to potential mechanical energy. You could never ever get 100% extraction, not because of loss in design, but because of entropy.

nippinout
10-29-2002, 01:53 PM
I probably wasn't clear enough.

My method was for estimating a shot count for a hypothetical 12gram with N2 at 900psi.

Compare that to real world results of 12gram of C02.

I never liked thermodynamics. Got a C in the class. Took me like three weeks to get enthalpy and entropy all straightened out in my head without looking at the text. lol

It is a cool and interesting class, but Holy Jebus was it tough.

Repoman-gene
11-16-2002, 09:05 PM
Originally posted by 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.

I have a crap load of these,and I am sure I can get more...if anybody wants a few to experiment with shoot me a address...none of my 12gram quick-changers seemed to be able to pierce them...i think the face is larger than co2 12grammer.:cool:

wes
11-22-2002, 06:06 PM
There is a 13ci/3000 pis tank! heres a microphantom with one

http://www.farmlandpaintballclub.com/phantom2.jpg

found it on www.pumpplayers.com

Trench
11-24-2002, 08:27 PM
Hum... that is a really great idea... Now if we could only find a tank that would fill to 10,000 psi that way we get a good amount of shots out of a fill...