is there any way to find out how much gas is produced by certain amount of a powder?

If you burn the powder inside a sealed pressure vessel and measure the pressure before and after it is burned you could calculate the volume of gas produced. It would only be a close estimation since to know the true volume you would need to know the partial pressures of all the gasses produced.
 
i would attempt such a test by getting a bottle and putting a measured amount of said powder inside it. run some kind of electrical igniter in it and put a plastic bag over the top.
 
Warning do not do this:rolleyes:
Just think about it :eek::eek::eek::eek:

Spawn-Inc said:
i would attempt such a test by getting a bottle and putting a measured amount of said powder inside it. run some kind of electrical igniter in it and put a plastic bag over the top.
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You will not be able to do this out side of a specilized lab
 
You can do it by calculation, if you have a knowledge of combustion. Since you are asking on the internet, its fair to assume you dont... If you want to get started you need to know the elemental composition of smokeless gunpowder. Then any text on combustion will help you out.
 
cosmic said:
You can do it by calculation, if you have a knowledge of combustion. Since you are asking on the internet, its fair to assume you dont... If you want to get started you need to know the elemental composition of smokeless gunpowder. Then any text on combustion will help you out.
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The exact chemical composition of any one given powder will be different from the next and a trade secret of the manufacturer. Short of having access to a mass spectrometer it wouldn't be possible to calculate past a general value for, say, pure nitrocellulose.
 
LUTNIT said:
If you burn the powder inside a sealed pressure vessel and measure the pressure before and after it is burned you could calculate the volume of gas produced. It would only be a close estimation since to know the true volume you would need to know the partial pressures of all the gasses produced.
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I thought about this method but then I think of the following question
The amount of oxygen in your gun chamber is definitely different from a pressure vessel. Would that change the combustion result?
 
strikeraj said:
I thought about this method but then I think of the following question
The amount of oxygen in your gun chamber is definitely different from a pressure vessel. Would that change the combustion result?
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Not by any really measureable amount as the oxidizer is a part of the powder. Basically meaning that the ambient oxygen plays no part in the burning of powder. Same reason a gun can fire underwater
 
Chamber pressures could change the rate or completeness of a burn as well though. In a large-ish sealed container some powders may not burn as completely as others because the pressure doesn't meet the required amounts. The powder could be thrown around by the initial ignition and land far enough away to not ignite.
 
Your question is straightforward to get an answer that is a reasonably good estimate, however I'm not sure if you'll find it particularly satisfying or not.

At atmospheric pressure and at standard temperature, one mole of (an ideal) gas occupies 22 litres of volume. If the pressure is higher, the volume is lower; if the temperature is higher, the volume is larger.

A fact that comes in handy is that one gram of powder produces one gram of gun gas. There is very little effect from the bits of air that happen to also be in a cartridge case, the vast majority of the reactants are all contained within the powder.

You need to know the amount (weight/mass) of powder that you are burning, and the average molecular weight of the products of combustion (O2 is 32 grams per mole, N2 is 28 grams per mole, CO2 is 44 grams per mole, H2O is 18 grams per mole, CO is 28 grams per mole, etc). An approximation that's good enough for many purposes would be to use something in the low to mid 30s, let's choose 34 grams per mole here.

To tie these together, we need to know how many moles of gun gas are produced by a certain number of grains of powder. There are ~15 grains per gram, so we have:

Code: 
                       # grains of powder        22 litres of gas per mole
litres of gun gas =   ----------------------- * ----------------------------
                       15 grains per gram        34 grams per mole

Working all this out we get a simpler expression:
Code: 
litres of gun gas = # grains of powder / 23

So one grain of powder produces 1/23rd of a litre of gas. Or, 23 grains of powder produces a litre of gas. Or,46 grains of powder produces two litres of gas.
 
Not sure why you would want to know that unless you also know a way to establish the speed that the gas is generated. The two in combination might lead to a pressure curve of sorts.
 
One gram of powder will not produce one gram of gas. The residue in the bore, the smoke and other bits that get in your eye have mass, which needs to be subtracted. Also, different powders have different density at atmospheric pressure, -think Trail Boss versus Bullseye- so measuring pressure will only get you within a close estimate. Don't bother with the air in your math. The saltpeter in bp has 3000 times more oxygen by weight (or volume?) than air, that's why it's an ingredient. The air you're taking in only has 20.9% oxygen. Not the best oxidizer going, good for campfires and running the car. When the guys on our old diesel submarines couldn't light a cigarette they knew it was time to get the machine running again. The rest is 79% nitrogen, and 0.1% "other," of which a large part is Argon. Nitogen is inert, but the other stuff isn't so much, and that makes any calculations a royal pain if you want to go past the decimal point. Easier to consult an interior ballistician, or e-mail a powder company.
 
rnbra-shooter said:
Your question is straightforward to get an answer that is a reasonably good estimate, however I'm not sure if you'll find it particularly satisfying or not.

At atmospheric pressure and at standard temperature, one mole of (an ideal) gas occupies 22 litres of volume. If the pressure is higher, the volume is lower; if the temperature is higher, the volume is larger.

A fact that comes in handy is that one gram of powder produces one gram of gun gas. There is very little effect from the bits of air that happen to also be in a cartridge case, the vast majority of the reactants are all contained within the powder.

You need to know the amount (weight/mass) of powder that you are burning, and the average molecular weight of the products of combustion (O2 is 32 grams per mole, N2 is 28 grams per mole, CO2 is 44 grams per mole, H2O is 18 grams per mole, CO is 28 grams per mole, etc). An approximation that's good enough for many purposes would be to use something in the low to mid 30s, let's choose 34 grams per mole here.

To tie these together, we need to know how many moles of gun gas are produced by a certain number of grains of powder. There are ~15 grains per gram, so we have:

Code: 
                       # grains of powder        22 litres of gas per mole
litres of gun gas =   ----------------------- * ----------------------------
                       15 grains per gram        34 grams per mole

Working all this out we get a simpler expression:
Code: 
litres of gun gas = # grains of powder / 23

So one grain of powder produces 1/23rd of a litre of gas. Or, 23 grains of powder produces a litre of gas. Or,46 grains of powder produces two litres of gas.
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you know what
how did i forget conservation of mass LOL
even tho it is not the most accurate but definitely a good estimate
 
Headcase said:
One gram of powder will not produce one gram of gas. The residue in the bore, the smoke and other bits that get in your eye have mass, which needs to be subtracted. Also, different powders have different density at atmospheric pressure, -think Trail Boss versus Bullseye- so measuring pressure will only get you within a close estimate. Don't bother with the air in your math. The saltpeter in bp has 3000 times more oxygen by weight (or volume?) than air, that's why it's an ingredient. The air you're taking in only has 20.9% oxygen. Not the best oxidizer going, good for campfires and running the car. When the guys on our old diesel submarines couldn't light a cigarette they knew it was time to get the machine running again. The rest is 79% nitrogen, and 0.1% "other," of which a large part is Argon. Nitogen is inert, but the other stuff isn't so much, and that makes any calculations a royal pain if you want to go past the decimal point. Easier to consult an interior ballistician, or e-mail a powder company.
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Thanks for your advice
i might try to email and see if they consider this as a trade secret or not lol
 
stocker said:
Not sure why you would want to know that unless you also know a way to establish the speed that the gas is generated. The two in combination might lead to a pressure curve of sorts.
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I am thinking to do a CFD simulation to help designing a more efficient comp.....
just a project idea in mind, not sure how it is gonna turn out yet
 
Strikeraj, sounds to me like you are interested in interior ballistics. Do a search of "interior ballistics" and follow where that leads. There is a 1950s book "Interior Ballistics of Guns" by Corning that is one of the classics. You'll probably find a good bibliography on "JBM"s web site.

If you want to simulate the burning of powder in a chamber, you start with a burn model for the powder. Typically you get an empircal equation along these lines:

b = A * p ^ r

b = burn rate (typically in units of depth of exposed surface burnt per second)
A = constant
p = pressure
r = burn rate exponent, typically in the range 1.5-4 IIRC (but it has been a *long* time)
 
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