If you picture what goes on inside your chamber/barrel when that primer pops it must be something!

I wonder how much volume or space the powder fills when fully burned...for example...is the volume something like the volume of a 55 gallon drum compressed into the dimensions of say a 7mmRemMag Case??

I couldn't guess.

Yes I am bored tonight.

Take the psi, pounds per square inch; divide by 14.7 psi per atmosphere. Multiply the volume of the case by the number of atmospheres and you have the volume you are looking for.

At least I think so.

youp50 is correct -- except that chamber pressure measurements are transient and not static. Ballisticians have been attempting for well over a century to translate the transient pressures in a rifle's chamber to something consistent with the static pressure, say, in a truck tire, but it is an illusive goal. Even today's piezo-electric "PSI" figures don't have a clear correlation with yesterday's "CUP" ("copper units of pressure", which were originally intended to have a one-to-one correlation with static pressure.)

As stated you can get close but you have to make some assumptions. Been way too many years but I believe it is Boyle's law.

Take your max pressure (60000) times the volume (assume case volume of 30-06 of 4.428cm3 volume will be greater than this since the peak is after the bullet has moved)divide by 14.7 says you have 18,073 cm3.

Or as youp50 says simply take pressure divided by 14.7 and your volume increases by that amount. In this case 4081 times.

Pressure and volume are directly proportional

So

P1V1=P2V2

You have to keep your units consistent.

Maybe I shouldn't have asked!

So for a 7mm or 30-06 sized case what would that translate to??

5 Gallon Jug or 55 Gallon Drum??

quote:

30-06 sized case

Using the 30-06 from my note above 18,073 cm3 is 4.77 gallons

Paul, -5 Gallons doesn't seem like allot, I thought that it would be much more than that

P1V1 = P2V2 assumes that T1 = T2
Temperature units are in the absolute scale like Kelvin or Rankin

quote:

Originally posted by SR4759:
Pressure and volume are directly proportional

So

P1V1=P2V2

You have to keep your units consistent.

actually studied under the inventor of that formula, back in my undergrad days, which is actually
(note the numbers below are subscripts not integers)
P1V1T1=P2V2T2

which takes into account temp changes... Only if temp is the same can it be ignored..

quote:

Paul, -5 Gallons doesn't seem like allot, I thought that it would be much more than that

Need but I think as Jeff said ignoring Temp isn't something you can do

It is a pointless exercise unless you are just exercising your brain. You physics majors are forgetting to calculate the temperature variable. The higher the pressure the higher the temperature and the higher the temperature the higher the pressure.

When the pressure and temperature is reduced to standard atmospheric conditions the volume will be lower than your calculations represent. What is the temperature of the gasses in the chamber? What is the temperature at peak pressure? What is the volume at peak pressure?

I am not motivated enough to do the math so I will just accept what you guys come up with. As I said - a pointless exercise.

quote:

Originally posted by jeffeosso:
[QUOTE]Originally posted by SR4759:

You have to keep your units consistent.

actually studied under the inventor of that formula, back in my undergrad days, which is actually
(note the numbers below are subscripts not integers)
P1V1T1=P2V2T2

Really??? roger

Yes it is pointless and has no real world meaning or importance...I just posted it for fun and out of curiosity.

Since no one here is going to measure the temperature - so what?

quote:

Originally posted by jeffeosso:

quote:

Originally posted by SR4759:
Pressure and volume are directly proportional

So

P1V1=P2V2

You have to keep your units consistent.

actually studied under the inventor of that formula, back in my undergrad days, which is actually
(note the numbers below are subscripts not integers)
P1V1T1=P2V2T2

which takes into account temp changes... Only if temp is the same can it be ignored..

quote:

Originally posted by SR4759:
Since no one here is going to measure the temperature - so what?

Since we are counting angels on a pinhead, we should also go along and count the Devils in the details.

You might be careful with that "no one" generality... I assume by no one you mean yourself, as there are some wilely characters here that just might... Or might have access to a community college level calorimetry... And they might be interested enough to test it...

Then again they might call a powder make or western labs

Several problems; You don't know the volume
that exists when the powder stops burning. And
as already pointed out you don't know the
temperatures. The actual formula is PV/T.
This formula is for Ideal Gasses under steady
state conditions not for a heavy, hot, dynamic
gas flow.
Taint simple.

Quickload makes a darn good simulation of bullet movement and percent unburned powder... Merely a matter of calculating the volume of the bore...

" Merely a matter of calculating the vo;ume of the bore". Then what?, you still don't know
the other variables.

Simple ansver to a simple question:
1gram~15,3grains of smokeless burns to 8,0-1,1liters~1.69-2.32pint of gas at atmospheric pressure.

Source:
ISBN: 978-91-633-4647-7
"Allt om krut och lite till"
www.nordicballistics.com

So to ansver your question.
To raise the internal pressure of a 55gal drum to 1 bar~14½ psi you need = 55gal/2quart x 15,3grains of powder = 1683 grains (give or take a little!)

quote:

Originally posted by hawkins:
" Merely a matter of calculating the vo;ume of the bore". Then what?, you still don't know
the other variables.

Wow.. Given the barrel length you get volume and estimated unburned powder. The model it with shorter and shorter barrel mapping percent unburned. You will have a pretty good histogram model and can model to 0.001 in length. Given calorimetry data you can have the heat. Bullet friction can be computed as the difference in expansion rate, given by powder manf.

Give you heat mass volume pressure and loss. A decent model anyone that had a little software and a friend in a community college you could verify calcations to empirical data to verify model.

The tools are there and the math ain't hard...

When I saw this I just started thinking... always bad. Perfect gas law PV=NRT or pressure times volume equals number of moles times gas constant times temperature. Problem is we start with one complex mixture of solid Stuff at say room temp. and 1 atm pressure, we turn it into a complex mixture of new imperfect Stuff like CO2, H20, various oxides of nitrogen and who knows what else, possible some plasmas ??, at say 60,000 psi at maybe 4-5000 deg. F, then we let this mix cool back down to room temp. and hold it at 1 atm. pressure. So what is its volume??? Could be done (maybe), if we knew the quantity of all of the final combustion products. Maybe the powder manufacturer could give us some insight here??
C.G.B.

Re thinking this (always bad) the analytical approach would be ok for an upper division thermo final exam but there is some empirical data out there that would come pretty close without much math. I have seen 1 shot suppressors made out of 2(1?)L pop bottles (probably a youtube vid. on this). If you assume that all the gas from a 9mm can be contained in a 2L bottle without rupturing it and say the 9mm uses about 6 gr. of powder and your 7mm uses about 60 gr then it would take about 20L to contain the gas or a little over 5 gal. That would probably be a good first guess, and a lot less arm waving.
C.G.B.

For commercial explosives, the gas yield is usually given in the data sheets. It typically ranges between 700 and 950 liter per kilogram of explosive. This is very similar to the data mentioned by The Dane specifically for smokeless powder (the difference is due to slightly different proportions of hydrogen, nitrogen and carbon).

The data gives the volume at standard conditions, which is basically 1 bar and 20 degrees Celsius.

As previously mentioned, without complicated computer simulation, it would not be possible to calculate this from peak pressures, since you do not know how far the reaction has proceeded at peak pressure, what the temperature is, how far the bullet has already moved at that time etc. Also, the mixture of unburnt powder, smoke (solid reaction products) and swath (gaseous reaction products) is not really a good ideal gas.

P1V1 = P2V2
____ ____

T1 T2 ( I cannot even make this computer show the T2 as the divisor on P2V2

The absolute temperature is a denominator.

As better minds than mine stated this is a static assumption. The powder produced gas is continually expanding as the bullet speeds off into the yonder.

quote:

Originally posted by youp50:
P1V1 = P2V2
____ ____

T1 T2 ( I cannot even make this computer show the T2 as the divisor on P2V2

The absolute temperature is a denominator.

P1V1/T1 = P2V2/T2. NOT P1V1T1=P2V2T2. roger

Modern smokeless powder is a dynamic flammable solid that burns faster as pressure is added. It also burns hotter with added pressure because the heat transfer has less time to dissipate. All the combustible products are burned before the peak pressure is reached. That is why Quick Load and other computer simulations cannot be relied upon as statistical data.
Computer simulations can be very close as long as the parameters they use are within the limits of the algorithms in use. A long and short cartridge can have the same internal volume and use the same bullet but they will respond differently to the same powder because one builds pressure faster than the other. The shorter fatter cartridge will use a slower powder than the the long skinny case.
The pressure rise times are therefore different.

JV,
That is interesting and less than I would have expected. Looks like 60 gr would only yield about 3.7l at stp. Of course how fast does the gas cool once expelled?
C.G.B.

quote:

Originally posted by cgbach:
JV,
That is interesting and less than I would have expected.

To put the data in perspective, a kg of air has an approximate volume of 770 liters in standard conditions. This means that the gaseous reaction products of explosives have a similar or slightly lower density than air.

The thermal behavior after the bullet leaves the barrel is coupled with the pressure drop that happens at the same time. Within milliseconds, the pressure equalizes with that of the surrounding atmosphere. This goes along with a significant drop in temperature due to adiabatic cooling. If you were to catch all the gas at the muzzle, but allowed it to expand to atmospheric pressure, I would not expect it to be very hot anymore.

Shouldnt atmospheric pressure also be a nessecary variable? I expect the volume would be greater in space then it would be on the earths surface.. Or am I splitting hairs?

quote:

Originally posted by Wstrnhuntr:
Shouldnt atmospheric pressure also be a nessecary variable? I expect the volume would be greater in space then it would be on the earths surface.. Or am I splitting hairs?

Not at all. (splitting hairs) One would use standard atmospheric data in the calculations. 14.7 psi, x degrees, altitude, etc...

The free volume of a pressurized gas would be nearly infinite in a vacuum.

To a physicist "Temperature" and "pressure" are essentially the same thing, an expression of energy.

The fact is that the propellants we use deflagrate into a cloud of very hot nitrogen gas that wants to expand and equalize with atmospheric pressure.... but there's something in the way...

What is being forgotten is that the propellant has not completely burned at the point where that peak pressure is being recorded, and it likely ends up producing three (or more) times as much gas as that 18liter figure posted above.

Also remember, as gasses expand they cool.
How much of that expansion can be used to transfer energy is
a separate issue.

Water converted to "Steam" expands some 1800 times when it changes phase from liquid to gas, and look how much power can be transmitted through that media...

The arguments I always love are ones on OTHER forums where people suggest that firearms wouldn't function in outer-space (I.E. in a Vacuum) because there is no air for the powder to burn (Smokeless powder functions quite well in a vacuum)

OR that the bullets will tumble in a vacuum (Why? it is air resistance that causes them to tumble in the first place)

Great discussion.

Question, If there were a chamber strong enough to contain all the released gasses of x amount of y powder, with none escaping, what would that pressure/volume be?

quote:

Originally posted by mike_elmer:
Great discussion.

Question, If there were a chamber strong enough to contain all the released gasses of x amount of y powder, with none escaping, what would that pressure/volume be?

Mike,
That is exactly how the powder burn rate is determined. It is called a "closed bomb" test. A certain amount of powder is placed in a vessel fitted with electrical ignition and a battery of transducers. Then the powder is ignited and the pressure/time curve is recorded. I don't know specifics about the amount of powder or the volume of the enclosure but it must be a lot of pressure that is contained and it must take a while to cool back down.

quote:

Originally posted by PaulS:

quote:

Originally posted by mike_elmer:
Great discussion.

Question, If there were a chamber strong enough to contain all the released gasses of x amount of y powder, with none escaping, what would that pressure/volume be?

Mike,
That is exactly how the powder burn rate is determined. It is called a "closed bomb" test. A certain amount of powder is placed in a vessel fitted with electrical ignition and a battery of transducers. Then the powder is ignited and the pressure/time curve is recorded. I don't know specifics about the amount of powder or the volume of the enclosure but it must be a lot of pressure that is contained and it must take a while to cool back down.

Ok, Now that leads to my next question... After the temps inside the "closed bomb" cool, is there still pressure, and how much?

I am certain that high temps equal high pressure, and that is the pressure we concern ourselves with when reloading.

But, how much pressure is actually left after cooling?... like a net amount of created pressure/volume.

Interesting thread. How fitting that the following adage was shared with me today:

"If you can't explain it well enough for your grandmother to understand, you probably don't understand it yourself."

Hence, I will not try.

Sam

quote:

Originally posted by mike_elmer:
Ok, Now that leads to my next question... After the temps inside the "closed bomb" cool, is there still pressure, and how much?

I am certain that high temps equal high pressure, and that is the pressure we concern ourselves with when reloading.

But, how much pressure is actually left after cooling?... like a net amount of created pressure/volume.

I have never seen any data or participated in any discussions on that information. I have no idea and wonder if that data is even recorded anyplace. I do know that when I had questions about the process (amount of powder and volume of the container) I was told it was proprietary information.
Can it really be that top secret? I guess it can and is. If you find the information I would ask that you share your source so I can look it over too.

Paul

quote:

Originally posted by PaulS:

quote:

Originally posted by mike_elmer:
Ok, Now that leads to my next question... After the temps inside the "closed bomb" cool, is there still pressure, and how much?

I am certain that high temps equal high pressure, and that is the pressure we concern ourselves with when reloading.

But, how much pressure is actually left after cooling?... like a net amount of created pressure/volume.

I have never seen any data or participated in any discussions on that information. I have no idea and wonder if that data is even recorded anyplace. I do know that when I had questions about the process (amount of powder and volume of the container) I was told it was proprietary information.
Can it really be that top secret? I guess it can and is. If you find the information I would ask that you share your source so I can look it over too.

Paul

Well, Paul, I would have to assume that the volume and weight would correlate to the mass of the propellant before ignition... but I am not a physicist, or even a scientist... I just have a lot of questions... and I would not be surprised at all, if the answer is proprietary. That would just mean that it was a really good question!

quote:

Originally posted by Allan DeGroot:
To a physicist "Temperature" and "pressure" are essentially the same thing, an expression of energy.

not really, no.. as that would include every other expression of energy, and while inter dependent, they arent, except in the extreme, dependent variables...

in a CLOSED system, adding more energy (then its not really closed) can change temp and pressure.. as volume would be inelastic .. unless we are talking about cooking off 2g of h335 in a 55gallon drum at 0 psi absolute..