12 gram nitro?

[Schmitti]

These can be found here...

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]

Actually

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]

Re: These can be found here...

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]

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]

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]

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.

[Vegeta]

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]

Fancy Shmancy Math... psh!

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]

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]

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]

Re: Fancy Shmancy Math... psh!

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]

Re: Fancy Shmancy Math... psh!

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]

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]

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.

[wes]

There is a 13ci/3000 pis tank! heres a microphantom with one



found it on www.pumpplayers.com