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 October 28, 2012, 09:51 PM #1 The Comedian Junior Member   Join Date: October 28, 2012 Posts: 14 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.
 October 28, 2012, 10:49 PM #2 Jimro Senior Member   Join Date: October 18, 2006 Posts: 6,594 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. __________________ Machine guns are awesome until you have to carry one.
 October 28, 2012, 11:09 PM #3 The Comedian Junior Member   Join Date: October 28, 2012 Posts: 14 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.
 October 28, 2012, 11:17 PM #4 Jimro Senior Member   Join Date: October 18, 2006 Posts: 6,594 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 __________________ Machine guns are awesome until you have to carry one.
 October 29, 2012, 10:58 AM #5 Sevens Senior Member   Join Date: July 28, 2007 Location: Central Ohio Posts: 10,435 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! __________________ Attention Brass rats and other reloaders: I really need .327 Federal Magnum brass, no lot size too small. Tell me what caliber you need and I'll see what I have to swap. PM me and we'll discuss.
 October 29, 2012, 12:44 PM #6 Unclenick Staff   Join Date: March 4, 2005 Location: Ohio Posts: 12,125 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? __________________ Gunsite Orange Hat Family Member CMP Certified GSM Master Instructor NRA Certified Rifle Instructor NRA Benefactor Member
 October 29, 2012, 06:56 PM #7 Jimro Senior Member   Join Date: October 18, 2006 Posts: 6,594 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 __________________ Machine guns are awesome until you have to carry one.
 October 29, 2012, 07:19 PM #8 JohnMoses Senior Member   Join Date: April 11, 2008 Posts: 148 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.
 October 29, 2012, 08:16 PM #9 Jimro Senior Member   Join Date: October 18, 2006 Posts: 6,594 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 __________________ Machine guns are awesome until you have to carry one.
 October 29, 2012, 08:53 PM #10 The Comedian Junior Member   Join Date: October 28, 2012 Posts: 14 Wow, thanks for the responses everyone. Now that I have the rough idea of the equation, I can plug in the values.
 October 29, 2012, 09:58 PM #11 Jimro Senior Member   Join Date: October 18, 2006 Posts: 6,594 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. __________________ Machine guns are awesome until you have to carry one.
 October 30, 2012, 04:51 PM #13 Jimro Senior Member   Join Date: October 18, 2006 Posts: 6,594 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 __________________ Machine guns are awesome until you have to carry one.
 October 30, 2012, 04:57 PM #14 Unclenick Staff   Join Date: March 4, 2005 Location: Ohio Posts: 12,125 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. __________________ Gunsite Orange Hat Family Member CMP Certified GSM Master Instructor NRA Certified Rifle Instructor NRA Benefactor Member

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