We all know that when gunpowder burns it creates gas. My question is, "How much?" If you burn a grain of IMR (type) powder how many cubic inches of gas will you get at sea level pressure? Anybody ever seen that number?

Thanks,
Ed

No. And the reason is simple.

Smokeless powder burning rate (how fast it burns and how much pressure it produces) is determined by a whole bunch of factors: type, single or double base, deterrent coatings, number and size of perforations, size of granules, size of cartridge case, caliber, etc.

If you identify a single, specific powder type, say, an IMR number, you could make a good guess based on caliber and bullet weight, of the RANGE of pressures that cartridge could produce for a given powder charge. Change any of the variables, and the pressure range can change greatly.

There are slide rule type ballistic calculators that can closely approximate chamber pressures if you input the caliber, powder charge, and cartridge case. These chamber pressures are determined by measuring in interior ballastics laboratories, and doing a LOT of loading and firing of actual ammunition.

PV=nRT

Would make a good chemistry experiment if you had the right equipment.

I'm sure the volume of gas is very consistent with volume of powder burned. The reaction may be more complete under higher pressures though? I wonder.

Its a good question, I doubt very much gas is produced. Maybe a few cubic feet? The trick is its produced in way less than a second in a really small reaction chamber with only one direction to go.

I don't know any figures, it would be an interesting question for a ballistics lab to answer. Really though gas volume is just one factor. Case volume and overall volume in the system is a big one too. Think of how much velocity and energy is created by a tiny little 22lr case, with about 3.5gr of powder in it. Or in the case of subsonic, no powder at all, just primer compound.

Another good comparison is 308 vs. 30-06. The 308 cartridge can produce nearly the same velocities with lower weight bullets, maybe only short by 200 fps. But an average 308 handload takes around 41-45 grains of powder. A similar load for 30-06 might require 50 grains of powder. All that extra gas created is merely filling up empty space in that big 30-06 case to build similar pressure.

Additionally, some powders are touted as producing more gas, for the purposes of feeding compensators and such. Winchester AutoComp is an example. I dunno how these work as I've never used one, probably just a little slower.

The amount of gas produced by combustion is directly proportional to the mass of the combustion by products in moles. As smokeless powder is a combination of several chemical elements, usually nitroglycerin, nitrocellulose, and other various compounds, the calculation would be extremely complex.

An experimental method would be the best way to determine average gas volume production. This was probably done by the chemists when developing the propellant but finding the information would be difficult.

Here's what prompted my question. Yesterday I was looking through my old tattered copy of "Hatcher's Notebook" and in the back is the hand written passage stating that, "1 gr. of IMR produces 20.54 ci of gas." That note was written by me perhaps 35 years ago and I just can't remember where I got it. I suspect very much that it was deduced from the pressure/bullet travel curves in that book and Boyle's law. I had hoped you guys world save me from plowing through the arithmetic again.

In response to the assertion that all the gas/grain ratios would be different I can only offer the following. In the the 10th edition of "Handloader's Digest" Edward Yard wrote an article on the energy content of gunpowder. He found that most single based powders (liker the IMR's) produced 1800 to 1900 btu's per pound. Double base items like Bullseye and Unique were a bit stronger at 2460 btu's per pound. Now I could be wrong but it seems to me that if two powders produce the same energy when burned, they should produce the same volume of gas.

Thanks,

Ed

This is false on its face, unless the powders have the exact same chemical composition. The energy produced by oxidation per mole depends on the element being oxidized.

When starting with a weight of powder, there is also the problem of the ratio of elements being oxidized. A mole of carbon produces the same volume of gas as a mole of hydrogen when oxidized, but a mole of carbon is about 14 times as heavy as a mole of hydrogen.

Internal ballistics is incredibly comlicated, and made more so by the fact that combustion is still in progress when the bullet exits, and the pressure is still high. Look at pressure curves from actual gun firings. Internal ballistics as a science relies on a series of differential equations (calculus), and attempts to reduce it to arithmetic or algebra are crude approximations of a limited set of conditions. Calculus is the math required to deal with processes undergoing rapid change, while algebra requires everything to hold still or remain constant.

Fortunately for us handloaders, we don't need to understand the internal ballistics theory to use the proven results.

__________________
Science plus Art

Why couldn't two completely different chemical compounds produce exactly the same amount of gas in a reaction?

Correct me if I'm wrong, but I thought all IMR type powders did have the same chemical composition. (And yes, I understand that the ballistics part of this discussion would devolve into integral calculus. I'm asking a much simpler question.)

They can (or not, depending) but that wasn't the question.

Two reactions that produce the same energy may or may not produce the same amount of gas after the reaction. It depends on the ratio of the elements that react.

Gasoline and diesel fuel can both produce energy to propel vehicles of the same type and weight, but the diesel produces more carbon dioxide and less water vapor compared to gasoline.

__________________
Science plus Art

Good point, depending on how approximate an answer you're willing to accept, and for what purpose.
Roughly 15% of the weight of an IMR powder is not the actual powder but coatings, graphite, etc that participate in the combustion, and their amounts and types vary. Also the pressure during combustion will affect how much of the powder actually reacts, and how much unburned remains.

So if "about X cc after complete combustion, at standard temp and pressure" is good enough purely as a matter of interest, then sure, why not? I would tend to change the 20.54 ci to "about 20" and leave it at that.
Obviously the volume of the gas when the bullet leaves the barrel is going to be the volume of the barrel and case, at a much higher pressure and temp.

__________________
Science plus Art

OKFC05,

That's close enough for me. I don't intend to do anything with the number. My curiosity would be satisfied with the belief that I had the first digit right and the decimal in the correct place. Are you saying that "around 20" sounds right or simply that 20 is a more legitimate way of posting the answer than 20.54?

Thanks,
Ed

Sorry, I can't verify your number off the top of my head.
I recall 1 kg black gunpowder produces approx. 350 liter gas and smoke.
Nitrocellulose (single base smokeless powder) is about 6 times as effective as blackpowder by weight.
So your number at least sounds the right order of magnitude from a quick units conversion, if that's any encouragement.

I was talking more about the uncertainty in the expected empirical results from different powders, and the inability to predict them accurately due to content and condition variability. Like when somebody asks "How fast can a pickup go?" and I answer "typically less than 200 kph."

__________________
Science plus Art

I like the second answer better.

Seems like I read somewhere a long time ago that 40 or 50 times as much gas is produced by the same volume of powder.

I'm getting a little tired, hope you can decipher that...