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Old October 28, 2012, 09:51 PM   #1
The Comedian
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Firearm science question...

I'm not quite sure where to post this, so I decided this forum would be the most relevent. Is there any data available regarding the volume of gas produced by a gunshot? I understand that this will have extreme variances with regard to the caliber of ammunition used, but any info would be helpful.
Thanks.
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Old October 28, 2012, 10:49 PM   #2
Jimro
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Are you looking at volume in terms of molecules of gas produced by molecules of powder?

Using a simple nitrocellulose molecule and assuming as complete combustion as possible.

C6H8(NO2)2O5 => 2CO2 4H2O 2N2 CO (remainder 4 Carbons).

That gives you 9 molecules of gas for every molecule of nitrocellulose. But you can't really compare the volume of a solid with the volume of a gas, as the volume of the gas is both temperature and pressure dependent.

Or are you asking for the equivalent volume of gas produced at normal atmospheric pressure?

Because then we get into pV=nRT and right now it is too late for me to even try to do that math with any sort of fidelity.

9 molecules of gas for every molecule of nitrocellulose means I (or someone with plenty of time) would have to compute for each type of molecule separately and add the partial pressures, adjusting volume of the final product to normal pressure to one ATM.

Let us know what you really want to know, I'm sure we've covered it before.

Jimro

I pulled the following from wikipedia as it may explain detonation velocity a bit better, which might be what you are really asking for anyways.

Currently, propellants using nitrocellulose (detonation velocity 7,300 m/s) (typically an ether-alcohol colloid of nitrocellulose) as the sole explosive propellant ingredient are described as single-base powder.

Propellants mixtures containing nitrocellulose and nitroglycerin (detonation velocity 7,700 m/s) as explosive propellant ingredients are known as double-base powder.

During the 1930s triple-base propellant containing nitrocellulose, nitroglycerin, and a substantial quantity of nitroguanidine (detonation velocity 8,200 m/s) as explosive propellant ingredients was developed. These propellant mixtures have reduced flash and flame temperature without sacrificing chamber pressure compared to single and double base propellants, albeit at the cost of more smoke.
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Old October 28, 2012, 11:09 PM   #3
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Jimro,
Wow, you really went all out! I was just wondering the volume at atmospheric pressure, and its application towards suppressor design. I figure that the larger the volume of the can, the quieter the report of the shot. I understand now that this has so many variables. Thank you for your answer.
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Old October 28, 2012, 11:17 PM   #4
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You are correct, larger volume means more surface area means more interaction between the gasses and the suppressor.

But when one ounce of water when vaporized turns into 9 gallons of steam (at normal atmospheric pressure) it becomes impossible to create a suppressor that can actually contain the volume of gas produced (well you could, but no one really wants a 55 gallon drum on the end of a rifle).

Your best bet is to look at the amount of surface area you can pack into a suppressor instead of total volume (within reason, and really there are no new suppressor designs as they have been around commercially since around the turn of the last century if I recall correctly).

Jimro
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Old October 29, 2012, 10:58 AM   #5
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I wonder if someone with QuickLoad can speak to whether or not this is on one of the output pages? I have NO IDEA, but I can tell you the QuickLoad puts out (what seems like) volumes of information, 95% of which I'm too dense to make use of!
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Old October 29, 2012, 12:44 PM   #6
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No, but you can work it out from the expansion and the muzzle pressure, which are provided. Obviously it will be different for every combination of powder charge and bullet and chambering and barrel length. The other caveat is that the gas is still pretty hot when it gets to the muzzle and will reduces in volume substantially as it cools. That temperature will have to be guessed at, but the ratio of peak to muzzle pressure can probably be applied to the estimated peak temperature of 4500 to 5000 degrees to get a very rough ballpark figure. But a suppressor begins to see gas while it's still hot, anyway, and cools as it expands through the baffles, so this is a complex dynamic problem.


Jimro,

Detonation velocity is the speed of a blasting cap initiated compression wave sustained through a solid piece of a high explosive while it is detonating. Since we carefully avoid detonation in favor of much slower controlled deflagration of gun powder in firearm chambers, it isn't apparent to me what use you have in mind for that detonation velocity information. Is it something specific or just an aside?
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Old October 29, 2012, 06:56 PM   #7
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Unclenick,

Just an aside to point out the velocity of change from solid to gas changes with different types of powder. Most of us know that even powders of the same chemistry with different burn rates will produce the same amount of gas, but in a different time rate which means different pressures.

Honestly I though the original question was going to be about replicating firearm ballistics with a pressurized gas source.

If I remember correctly, the deflagration max velocity is 5,200 fps for the expanding gas (which is why they were trying to reach 5,000 fps with the Eargenshplitten Loudenboomer...). Anyways, the only firearm I know of that actually gets close is the smoothbore main gun of an M1A1 (or A2) Abrams main battle tank.

I was pretty tired when I wrote that original reply, so I probably didn't make a point so much as a ramble.

Jimro
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Old October 29, 2012, 07:19 PM   #8
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I believe the OP was looking for the volume of the gas after the phase change from the solid to a gas. IIRC, the powder type has little to do with it, and the constant is 70,000x.
If this is incorrect, I'd like to know. I've been using that for years.
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Old October 29, 2012, 08:16 PM   #9
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JohnMoses,

Different powders have different energies because of additions to nitrocellulose (double base has nitroglycerin added, triple base has nitroguanidine in addition to that), but they are all going to be very similar in the end.

pV=nRt

If we calculate the molecular weight of a simple molecule of nitrocellulose, C6H8(NO2)2O5, we come up with a molecular weight of 251.14 grams per mol.

Let us go with a 41.5 grain charge and convert to grams and we get 2.689155 grams.

That gives us 0.010707 mols of nitrocellulose. C6H8(NO2)2O5 => 2CO2 4H2O 2N2 CO (remainder 4 Carbons). That means we multiply 0.010707 by 9 to get the number of mols of gas (gross oversimplification, each product will have a slightly different R value).

In SI units, P is measured in pascals, V is measured in cubic metres, n is measured in moles, and T in kelvin (273.15 Kelvin = 0 degrees Celsius). R has the value 8.314 J·K−1·mol−1 or 0.08206 L·atm·mol−1·K−1 if using pressure in standard atmospheres (atm) instead of pascals, and volume in litres instead of cubic meters.

so we set P to 1 atm, T to 293 K, and calculate for V.

V = 0.963701325*.08206*293 = 2.31708 liters.

Now we have another assumption to make, that we can use a volumetric density as a legitimate proxy for actual density. So 0.0653 cc/gr is the densest I can find on a chart, so our 41.5 grains becomes 2.70995 cubic centimeters. Since one cubic centimeter equals one milliliter we are looking at 2.70995/2317.08 or an a solid volume of 0.001169, or a thousand fold expansion in volume from solid to gas.

Considering that 1,000 and 70,000 are only an order of magnitude apart I'd say my math is close enough to demonstrate the process. In reality, one would go through and do this for every molecule in the gunpowder they were interested in (accounting for binders, stabilizers, etc) and then adding up all the partial pressures to get a total. I didn't even calculate for complete combustion (a remainder of 4 carbons, from the simplest nitrocellulose formula I could find).

My chemistry professors would be horrified by my math and assumptions here, but my engineering professors would be happy with a "ball park" first shot at looking at the problem, so don't take any of this as gospel.

Jimro
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Old October 29, 2012, 08:53 PM   #10
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Wow, thanks for the responses everyone. Now that I have the rough idea of the equation, I can plug in the values.
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Old October 29, 2012, 09:58 PM   #11
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Unclenick, can you crunch some numbers and see if any of my calculations are remotely valid?

I was going over my math and assuming that we 2.7 liters of gas is the max produced, we divide back down by 2's until we reach "approximately" our starting point for the solid.

2.3 liters/(2 to the 8th) = 0.07, so we are looking at 32 atmospheres of compression (inversely related, volume and pressure).

1 atmosphere is roughly 15 psi (slightly less, I'm rounding for easy math). 15x32 is 480 psi. So how do we turn 480 psi into 50,000 psi? Heat.

P1/T1 = P2/T2, so 0.0027m3/293k = 0.00008m3/X solve for x...

0.00008(293)/0.0027 = X = 8.61 degrees Kelvin. Which means to maintain 1 atmosphere of pressure there is a minimum difference of 285 degrees kelvin (or Celcius) between the solid and gaseous states, that's a 545 degree Fahrenheit difference, which seems about right to me based on metal spalling of projectiles.

So we start with a solid at normal temp, it starts turning to gas, we get 32 atmospheres of pressure from gas alone, but also a 285 degree C increase in temperature, which would need to account for the remaining pressure, and once again I'm too tired to go back to the equation and figure out what an increase of 285 K would do, especially since I calculated for the volume of powder, and not the volume of a cartridge case.

And this is assuming that there isn't "waste heat" somewhere in the system.
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Old October 30, 2012, 04:47 PM   #12
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This is actually quite a complicated subject. Powders include deterrents and stabilizers. As a result, not only are the generated gas species different for different powders, the total potential energy content varies substantially. QuickLOAD's database has the black powder substitutes as the lowest energy powders, at about 3050 J/g. The lowest energy rifle powder, no doubt owing to massive quantities of deterrent, is Accurate 8700 at 3460 J/g. At the other extreme are Alliant Power Pistol and Bullseye, at 5150 J/g and 5158 J/g, respectively. So the ratio is as much as 1.7:1. Obviously, a higher energy powder with the same burn rate as a lower energy one can produce higher average temperature and therefore a higher pressure with less total gas (you would need less of the higher energy powder to get the same velocity). The average for all the QL database powders is about 3950 J/g. Further, due to ash and particle suspension and heat transfer to the bullet and bore and kinetic energy translation into the bullet’s kinetic energy, and the constantly changing temperature, this stuff won’t behave exactly like an ideal gas.

Since QuickLOAD has the complicating factors taken into account, I can use its output to make an estimate of gas volume. At 3910 J/g, IMR 4198 is reasonably close to average in energy content. If I put a 21 grain charge of it behind a 53 grain SMK bullet in a .223 Remington with a 24" barrel with the bullet seated 1 caliber into a case that has 28.8 grains fired water overflow capacity (lower than usual, but the program default), the program tells me the volume under the bullet is 1.725 cm³. The expansion ratio of that barrel length is then 9.28 relative to that bullet base position and case capacity, so the volume behind the bullet at the moment the bullet reaches the muzzle in this case is 16 cm³. The muzzle pressure at that moment will be 6075 psi, while the breech pressure will be 7520 psi (there's a pressure gradient), for an average of 6798 psi. Divided by one atmosphere (14.7 psi), you get an expansion of 462 times after the gas exits the muzzle.

462 times 16 cm³ = 7.4 liters. However, that's at whatever temperature the gas has at the muzzle. If the peak temperature is 5,000°F (5,460°R) at the peak pressure of 49,500 psi, at which point the expansion ratio has progressed only to about 1.2, volume behind the bullet is at that moment 2.07 cm³ and about 52% of the powder has burned. Almost 100% of the powder is burned by the time the bullet gets to the muzzle, so then, based on volume alone, you would expect muzzle temperature taken as a proportion of gas and pressure to be about 5,460°R×2.07 cm³/16 cm³/0.52 or about 1358°R, which is 899°F. If your ambient temperature is 75°F or 535°R, then you’d expect 7.4 liters of gas at 899°F to decrease in volume by 535°R/1358°R or a factor of 0.39, becoming 2.9 liters when it finally got all the way down to air pressure and temperature.

Anyway, that’s for 21 grains of a fairly average energy single-base powder burned in a 24” barrel. It comes out at 7.4 liters and drops rapidly toward 2.9 liters at standard temperature and pressure. So your estimate wasn't far off if you pick the right temperatures. Double that charge as in, say, a .308 Winchester and you’ll be at roughly double those volumes as long as the powder choice burns fast enough to burn completely in the barrel. Real volumes will obviously be affected by that percent of burn which, in most instances, won’t be that high. But for a worst case it’s best to pretend it will.
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Old October 30, 2012, 04:51 PM   #13
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Thanks Unclenick, I appreciate the confirmation. I don't have access to Quickload, but I probably should buy a copy as it seems to be a necessity when you are "going off the manual" so to speak in terms of components.

Jimro
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Old October 30, 2012, 04:57 PM   #14
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I don't think you'll regret it. It's got so many default arguments you can change that it becomes an education all by itself just seeing how much effect each tweak has. The free QuickTARGET ballistics program that comes with it is very useful as it imports the QuickLOAD results directly. The newer QuickTARGET Unlimited 3DOF ballistics program that also comes free with it has proven to be educational, too.
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