Attn: Physicists, Engineers, Reloaders

[shadowdragon]

How would one calculate the volume of gas produced by the combustion of a chemical?

Specifically the volume created by the combustion of 1 gram of smokeless powder once the gas has expanded to where the psi inside the gas envelope is equal to about 14.7 psi? or rather ambient air pressure?

I am working through another random thought here.
I seem to get more of them the older i get, hmmmm.
should i worry?

edit:
new random thought- combustion rates of chemicals. formula or table?

 

[sgbrown]

1st question first -

If you know the products of combustion, water in the sample case below, and you know the density of the product at the given temperature and pressure - then you'd just divide the amount of product by the density (grams divided by grams/liter yiels liters of product.)

Sample Case 2 H2 + O2 => 2 H2O

So the first think you need is to get the chemical formula for the combustion of smokeless powder (Which I don't have sorry.)

I dug out one of my old aquatic chemistry texts and found out I'd forgotten a lot more than I thought on chemical reaction rates. The equations I've found have numerous variables and require calculus to solve. For a given reaction in a given medium under given conditions of temperature and pressure, the formulae can be simplyfied - but the simplified forms are NOT the same for all reactions or even for the same reactions under different conditions.

SO... if you have tables available I'd use them to get a "ball park answer"- or maybe some one else has a simplified answer for the particular issue of smokeless powder.

 

[TheEngineer]

Just a first not in order to provide an idea

Use the property of a gas, that 1 mol (a molecular weight unit) always take up 22,4 l under standard conditions.
E.g. 1 mol of CO2 corrospondends a mass of 14 g C + 2 * 16 g O = 46 g.
Those 46 g of CO2 would take up 22,4 l at 1 bar and 293 degrees Kelvin.

So, the exact values really depend on the chemicals, we are talking about.
Maybe its not too wrong for a rule over the tumb to assume that 1 g produces 1 l.

And about combustion rates, I would refer to tables found in civil architecture, plant safety or fire fighting booklets

There are of course formulars for that, buts its still getting very chemical .....

regards,

TE

 

[sgbrown]

Shadowdragon,

Ideal gas law is PV = nRT where

P is pressure in atm
V is volume in liters
n is the number of moles of the gas
R is the gas constant (0.0821 liter atm / mol deg Kelvin)
T is the absoulte temperature in degrees Kelvin (deg Kelvin above)

Solving for volume yeilds: V = nRT/P

This yeilds 22.4 liters per mole at standard temp (273 K) and pressure (1 Atm) as TE indicated in his post.

Some iterations of this would likely get answers to your first question - as long as you know the products of the reaction. I'm betting they'll include CO2, at least one oxidated form of nitrogen, and possibly some oxidated sulfer compounds as well - but since I don't know the chemical formula for smokeless powder much less the combustion reactions for it, I'm guessing at this stage.

 

[TheEngineer]

Add-on:
So far, I just gave a hint on the volume of the "pure" reaction results.

I don't know if its about really just the volume of the combustion products or about secondary smoke volumes...

AFAIR you can safely assume a distribution factor of perhaps 500 - 1000 e.g. for the line of sight blocking "feature". Thats an amount I found specified in my old plant safety stuff.
So then, 1 gram of the powder should effectivly produce a volume of 1000 l smoke (in fact 1 l pure combustion product distributed in 1000 l air).

 

[shadowdragon]

Basically what happened was i had a thought about CPR weapons, and how to "formulize" figuring accuracy, range and damage based on some simple easy measurements- bore, barrel length, powder charge, projectile weight. I have a ballistics program i can use for certain parts but i need a comparison point to be able to create game stats from the available info, and once i have that i can work out a simple formula ( i hope) to allow me to convert/create cpr weapons for any game system.
just a thought

 

[TheEngineer]

I see.
Perhaps thats an issue for the weapon geeks here ?

Could a powder charge be related with a specific energy content, so that the chemical part could be bypassed ?

 

[atpollard]

I would start with the weight and velocity of the projectile as it leaves the barrel )often available for real weapons) and skip the powder charge calculations. The volume will change with the pressure and temperature over time as the powder burns - and combustion will not be complete. Any formula to simulate such a complex reaction would be too complex to be useful and no more accurate than a formula to estimate muzzle velocity.

 

[Zparkz]

What you need is FF&S and then convert from there.

 

[shadowdragon]

Combustion of smokeless powder is pretty complete - you get a little residue but not all that much. i've fired 200 or so rounds thru my .44 and cleaning it is a breeze. its a revolver so it vents gasses all over the place and most residue comes off with a wipe. once i run it thru the dishwasher it sparkles real nice. btw arm and hammer(TM) dishwasher detergent is the best so far.....

 

[shadowdragon]

Quote:
Use the property of a gas, that 1 mol (a molecular weight unit) always take up 22,4 l under standard conditions.


understand mol, dont understand 22,4 1. is there a number missing? or is it a ratio or what?

 

[Anthony]

You're usually better off just using the total chemical energy of the propellant rather than worrying about expansion volume. However, a typical bullet is only around 30% efficient, the rest of the energy goes into noise, heating up the barrel, expanding gas, and other sorts of waste energy.

 

[Plankowner]

Originally posted by shadowdragon:
Quote:
Use the property of a gas, that 1 mol (a molecular weight unit) always take up 22,4 l under standard conditions.


understand mol, dont understand 22,4 1. is there a number missing? or is it a ratio or what?

He means that 1 mol = 22.4 liters

 

[shadowdragon]

Perhaps thats an issue for the weapon geeks here ?

no offense to the weapon geeks here but thats how greatswords wound up 15lbs in dnd, and one suppliment i read printed for third ed dnd stated unequivically that the greatsword was used by holding the blade and bludgeoning your opponent with the handle. another showed a drawing of a daiyku (japanese longbow used by samurai) and completely ignored the fact the hand position is offset so it can be more easily used from horsback (thereby making it look like an oriental influenced english longbow instead of the unique item it actually is) and then stated it could ONLY be used from horesback. so i decided to do something radical- try to learn a little about the physics involved before trying to break something down to dice rolls and paper sheets. many other gamer geeks see these issues, many dont, and from what i have seen a larger percentage just dont care either way.

i may be onto something, it may be that reality is just to frackin big for distillation, or i may be a wingnut wasting my time, but im havin fun so....

ok - when chemicals react, frequently gasses are part of the result. i am interested in the chemical reaction between O2, smokeless (anyone know the chemical formula for it, i think it is a derivitive of nitroglycerine). the reaction is started by heat and increased by pressure and additional heat. got that part. but instead of producing another clump of solid matter, the solids are residual, and the rest are hot gasses.
what am thinking is the "perfect" cpr weapon has a barrel length such that the projectile exits the barrel at the moment the gas pressure reaches ambient atm pressure, thereby imparting as close to 100% of the energy to the projectile as possible. (this is an "ideal case" situation)
the volume of the bore will determine the barrel length based upon the amount of propellant burned, and the weight of the projectile. and i just realized i am way over thinking what i want.

thanks for the help anyway guys

hee....

 

[atpollard]

If it helps, the old rule of thumb for canon is barrel length = 50 times projectile diameter.

Good luck and have fun.

 

[Anthony]

Without criticizing the basic goal -- solving the performance of guns based on applied physics, while not impossible, is not trivial either. If you don't already know gas dynamics by heart, the math is almost certainly over your head.

 

[sgbrown]

Originally posted by Anthony:
Without criticizing the basic goal -- solving the performance of guns based on applied physics, while not impossible, is not trivial either. If you don't already know gas dynamics by heart, the math is almost certainly over your head.

Maybe a set of tables based on current balistic reports for various rounds would work. If your wanting to predict down to 4 or 5 significant digits - the dyanamics are probably more than most of us want to get into just for a hobby. But if all your after is a rule of thumb based on physics principals to get within 1 or 2 significant digits, then it may be doable.

For a game I really don't care if the actual muzzle velocity would be 1558 fps or 1621 fps - calling it roughly 1500 to 1700 fps is more than close enough. However if you have a system that in reality would produce a muzzle velocity of 1500 fps and you insist that the velocity is only 500 fps - that's a big difference. Trending several sets of ballistic numbers may be the answer to some generic but close to accurate charts.

WARNING Beginning of old military story / rant: Quess what I'm saying is I'd like to see a gaming system without obvious errors in the stuff we know - such as ACRs with an effective range of only 100 meters

- Don't they believe in marksmanship training in far future armies? M-16's (off the armory shelf - not match grade weapons) are notoriously inaccurate and I'm far from the best shot in the world, but I rarely if ever missed the 200 meter target on the qualifying range when I was in the service. (250 and 300 meter targets were a different story for me - but I knew lots of soldiers who rarely if ever missed those.) End Rant.

Okay - all better now ...

SGB

 

[shadowdragon]

nothing is "over my head"- its all in how much effort i feel up to putting into learning it. math is math- its always "doable". besides, do gas dynamicists spring from the earth fully knowing gas dynamics by heart? i don't think so.

so if we are done with the condensation.....

basically i am looking for a starting point to make a formula that i can use which may not be accurate to 4,3, 2, or even 1 significant digit, EXCEPT at the starting point- we are reducing the world to the random rolls of dice here. its not supposed to BE real just easy to see the REAL origins and say "hey this s&^* aint too far off" if i can get three or five data points and plot a curve even better.

so i figure i'll get an "ideal" rifle, the barrel length provides the exact bore volume needed for the powder charge to expand from initial ? to 14.7 psi, at which point the projectile has left the barrel with all the energy the expanding gas can impart without losing energy due to extra barrel length. from this calculate muzzle velocity, drop moa, sight circle etc... do this for 5 or 10 bore sizes and see if we get a predictable curve - if so i can devise a formula to work it out. if not i find another way to formulize it.

or not- i might get bored after all........

 

[TheEngineer]

My humble engineering sense tells me, that maybe the formula approach might point to some direction, but the overlay of side effects like real reaction dynamics, heat transfer, friction and what the hell more might be more significant.

As I am said, I'm no weapon geek at all, but I guess "gun" design is a highly empirical thing, too.

For your modelling I really would rely on the energetic approach, as the actual dynamics of the gas expansion are perhaps hard to grasp.

BTW, I found some additional info about smokeless powder. Major component is nitrocellulose, with a molaric mass of 297 g/mol.
Combustion (needs no oxygen) results are N2, H2O, CO and CO2, all which produce a considerable volume expansion. So 1 g nc are transformed into 0,83 l gasous products.


Regards,

TE

 

[shadowdragon]

quote:
So 1 g nc are transformed into 0,83 l gasous products.

ok i followed everything up to the bold. i have never seen it written out that way- what does 0,83 1 mean?

sorry but everything above a high school education i've taught myself so i have some significant gaps to deal with. my own fault but we cope as we can.

edit:
somehow missed plankowners post 6 or 8 times. now i understand. that last bit isn't a one its an ell. doh! *opens eyes to actually see what he is looking at*

tanks all...