Firearms thermodynamic efficiency

[denton]

When the bullet uncorks the barrel, the pressure is pretty often in the 15 KPSI neighborhood. With a peak pressure of around 60 KPSI, there is quite a bit of energy stored in compressed gas that is lost as the gas vents to the atmosphere.

Bullet friction in the barrel is around 150 pounds, so in a 2 foot barrel, the work associated with friction is about 300 foot pounds or so.

Even though propellant gas is in contact with rifle steel for only a millisecond or so, there has to be a lot of heat loss there. It would be easy enough to estimate I suppose.

In the 5.56 at least, enough heat is transferred to the bullet to bring it almost to the melting point of lead.

That's not all pretty qualitative, but maybe it helps.

 

[MX26]

In response to brickeeye,

Charle's Law and the Ideal Gas Law

The observations behind this relationship were first observed by Jacques Alexander Cesar Charles in 1787. However, the relationship was first published in 1802 by Joseph Luis Gay-Lussac (hence, Charles--Gay-Lussac's Law).

Combining Charles--Gay-Lussac with Boyle-Mariotte (pV = constant at fixed T) provides the foundation of the Ideal Gas Law: pV/T = constant.

Notice Charles--Gay-Lussac is based on the premise of FIXED pressure. This is entirely unhelpful for an equation of state within a firearm (outside the bounds of deriving the ideal gas law) since pressure is hardly fixed.

Source:
Engines, Energy, and Entropy by John B. Fenn
Pgs. 39-40

In fact, if we were stuck in 1660's when Boyle first formulated his observations, Boyle-Mariotte would be far more useful in understanding and modeling propellant combustion in a PERFECT world where no heat was generated (no temperature change).

Yes, the Ideal Gas Law does break down quite often - notably under high pressure and gases with a more complex molecular structure. However, it is still very useful in an analysis of work in a thermodynamic event. There are also other models and tables which give far better state data in place of the Ideal Gas Law.


Improvements in thermodynamic efficiency will increase muzzle energy.
(given good timing, since the bullet only sees the chemical reaction when it is within the muzzle)

The First Law of Thermodynamics

The First Law is stated:

dE = Q - W

Where dE is "change in energy". This would be the energy released by the combustion of the gunpowder within the cartridge case - a release of chemical energy.
Q is heat energy (energy released by the reaction as heat)
W is mechanical work (energy transferred to the bullet - resulting in movement or "work")

Fenn, pg. 159

This is the key to the First Law. If that "dE" term produces large "Q" and small "W" we have low efficiency. This reaction would result in little muzzle energy and a hot gun.

Instead, if the "dE" term mainly produced a small "Q" and large "W" we have high efficiency. This reaction would result in high muzzle energy with minimal heat. I've never designed a firearm before but I can safely assume that this is preferable to designers looking for reliable operation and maximum effectiveness in a firearm.

Now, this analysis is simplified quite a bit, but I hope the fundamentals still show through.

Please don't read this as some kind of "prove you wrong" statement. I only wish to clarify the thermodynamics discussed here. It's quite a neat subject.

 

[xfyrfiter]

Yeah and the m240 saw still gets red hot after a full mag dump LOL.

 

[exavid]

I think I'll just wait until handheld versions of rail guns become available. So much less cleaning and maintenance.

 

[Shadow 7D]

Oh, I'm sure they could make a handheld one now,
It's just the few tonnes of capacitor banks needed to feed it that's hard to lug around.

 

[Voltia]

I used to think that, when I got my degrees in engineering, I would be able to make everyone better informed on the internet. Now that I have them, I realize that people make so many assumptions, quote so much half-remembered high school science, most of which doesn't apply anyway, that is isn't worth the bother even starting.

 

[Owen]

its true voltia, so so true.

 

[Jim K]

That kinetic energy in the movement of the bullet in the barrel can be converted back to heat if the bullet suddenly stops, as with a barrel obstruction. It is that heat that softens the barrel steel and allows the pressure to bulge or burst the barrel.

Jim

 

[Blackstone]

I won't remember half the thermodynamics I'm learning right now in uni

 

[MX26]

I used to believe that only a cynical attitude was logical. Then I learned to reason.

 

[Owen]

I think the bulge/rupture in a bore obstruction scenario happens too fast for it to be caused by heating. Wish I remembered how to figure that out...

 

[brickeyee]

Charle's Law and the Ideal Gas Law

Good description.

To bad you missed all the assumptions that went in to the law being valid.

Like non-reacting gases.

try that with hydrogen and oxygen.

Everything will work just fine until they react.

you can then use the ideal gas law (and not Charles's law) but you need to be adiabatic.

As soon as you lose heat to the enclosure things start to vary.

While the ideal gas law will work on the final temperature of the products, no condensation can occur or it fails.

Look up what an ideal gas is.

Non-reactive is right up there on the list.

 

[PercyShelley]

I had an idea, but it's not baked all the way yet:

So, could you take one of these time/pressure curves:

http://www.armalite.com/images/Tech...el Design, Heat, and Reliability, 030824….pdf

Somehow convert it to a time/volume curve (this would require extremely precise knowledge of the position of the bullet), and compare the pressure curve then to an ideal adiabatic curve, integrate the difference of the curves and figure out how much of the energy is being wasted?

 

[JimBoIHN]

What's this bs about ignore lists? Sounds pretty immature to me. There is usually something to be learned from both the idiot and the genius. Hey, I can show you how to be an idiot, if nothing else. I ain't blocking nobody...........................................yet.

 

[MX26]

brickeyee,

Yes, I understand that the ideal gas model breaks down readily. The point of the Charles/ideal gas discussion was to demonstrate the relationship between the two. They are both equations of state which define ideal gases. In fact, an ideal gas is literally defined as "a gas that obeys the equation of state pV=nRT". So by continually arguing that the ideal gas law does not hold in this case (I am aware that it does not), you are further disproving your original statement that Charles's law holds in its place. This is why I went into the discussion regarding the ideal gas law in the first place. You cannot declare that the ideal gas law does not apply, then go on to say that Charles's law will be more useful in its place when both are based on the fundamental premise of modeling an ideal gas. That is all I intended to prove in the discussion you are questioning - nothing more, nothing less.

I would like to further point out something you said:
"As soon as you lose heat to the enclosure things start to vary."

Yes, this eludes to the first law: the topic of my second discussion. As heat (Q) generation increases (this heat generated will subsequently be lost to the materials constructing the gun and the atmosphere), work (W) decreases. Efficiency subsequently decreases by definition. This thermodynamic interaction governs the entire process. To say that efficiency is of no concern completely disregards this fact.

If thermodynamic efficiency does not apply in firearm design, I question what would happen if one touched off a 30-06 rifle designed for combustion demonstrating 30% efficiency, then touched off another round in the same rifle possessing 90% efficiency - all other variables held constant. (This setup is likely impossible - but demonstrates the concepts discussed)

I am enjoying our discussion quite a bit. Thank you!

 

[Yoda]

Aren't rockets more efficient?

Wouldn't a rocket "gun" be more efficient? I seem to recall that the was a rocket pistol, maybe back in the 1950s or 60s.

Now, recoiless rifles don't seem to be very efficient at all. They'd certainly be harder to make into a handgun.

- - - Yoda

 

[MX26]

I had an idea, but it's not baked all the way yet:

So, could you take one of these time/pressure curves:

http://www.armalite.com/images/Tech...el Design, Heat, and Reliability, 030824….pdf

Somehow convert it to a time/volume curve (this would require extremely precise knowledge of the position of the bullet), and compare the pressure curve then to an ideal adiabatic curve, integrate the difference of the curves and figure out how much of the energy is being wasted?

I think you're on the right track.

What we need is a P-v diagram. If we knew the position of the bullet at some discrete points in time we could easily convert the "time" axis on the graph you provided to a "volume" axis. If we integrated the area under this curve we would have work.

Compare this to the area under the curve of an adiabatic P-v diagram and we would have our answer.

 

[PercyShelley]

Wouldn't a rocket "gun" be more efficient? I seem to recall that the was a rocket pistol, maybe back in the 1950s or 60s.

Now, recoiless rifles don't seem to be very efficient at all. They'd certainly be harder to make into a handgun.

I think so; rockets are supposed to be stupid efficient (~60% so sayeth wikipedia). That said, I usually see rocket "efficiency" quoted as specific impulse, or how much velocity the rocket can pick up per unit mass of propellant, rather than as thermodynamic efficiency. It's perfectly possible to calculate rocket thermodynamic efficiency, it's just not the figure of merit that's quoted most often by convention.

There might be issues in miniaturizing rockets though. Solid fuel rockets, which would be the only practical handgun-sized projectiles tend to have low specific impulse compared to liquid fuel rockets. Another issue would be the back-pressure on the rocket nozzle caused by being fired off in a barrel, which would drop off rapidly after it un-corked. Atmospheric pressure is an important variable in nozzle design, and it's hard to design a nozzle that has good performance over a wide range of atmospheric pressures.

You could conceivably have a rocket handgun that doesn't have a barrel, but then you'd get a larger amount of rocket exhaust in your face. Not pleasant.

That rocket handgun you're thinking of was the gyrojet.

I think you're on the right track.

What we need is a P-v diagram. If we knew the position of the bullet at some discrete points in time we could easily convert the "time" axis on the graph you provided to a "volume" axis. If we integrated the area under this curve we would have work.

Compare this to the area under the curve of an adiabatic P-v diagram and we would have our answer.

Thanks. The math seemed like it would work, but I just don't wanna do it. I think you would need to know the bullet position at a given time with some degree of precision for this to work, as well.

Oh well, it was easier than my first idea of surrounding the barrel in an enormous insulated water tank to act like a big bomb calorimeter.

 

[brickeyee]

"You cannot declare that the ideal gas law does not apply, then go on to say that Charles's law will be more useful in its place when both are based on the fundamental premise of modeling an ideal gas."

Within a less restricted environment Charle's law has slightly greater applicability.

Not by a lot, but by some.

It to breaks down quickly in reactive unstable gas mixtures.

Once actual deflagration has been completed, and the initial stages of oxidation the remaining gas mixture, while still reactive, can be treated as adiabatic for the time scales of interest (a few milliseconds of barrel dwell) enough to allow Charle's law to at least function as the gas volume behind the bullet increases with bullet travel in the barrel.

The major volume changing relations are mostly completed.
Further oxidation will have to wait for the gasses to exit the muzzle and have access to atmospheric oxygen.

The ideal gas law still cannot be applied any better than Charle's law since some heat loss is present, and some reactions are still occurring.

Think of Charle's law at this point as a simplified gas law (PV only), that can be used.
I am well aware that holding the correct parameters constant in the ideal gas law devolves it to Charle's law.

The ideal gas laws multiple parameters make it no more effective in this case.
While you have also violated the assumption for Charle's law, the ideal gas law has more assumptions that are being violated, making the results of applying it (without devolving it to Charle's law) rather specious.

 

[denton]

Once the powder has burned, the linear relationships between pressure, temperature, and volume do work, i.e., P1*V1=P2*V2. Those laws do not require an ideal gas.

 

[Chemistry Guy]

Rockets are not at all thermodynamically efficient when compared to traditional pressure driven projectiles, as their efficiency is near zero when the projectile is at rest and they are not efficient until the velocity of the projectile approaches the velocity of the exiting gases. A 100% efficient rocket would theoretically be possible if the projectile was moving so fast that the gases, when pushed out the back, immediately stopped because their backwards velocity equaled the velocity of the projectile. This isn't likely to be approximated unless we are talking about a hypervelocity rocket in outer space.

On the other hand, an air rifle can have a very high thermodynamic efficiency. Maybe some day some advances in materials technology will give us air rifles with firearm like performance. In the meantime, roughly 30% efficiency is good enough as long as we find ways to deal with the heat.

 

[MistWolf]

I think I'll just wait until handheld versions of rail guns become available. So much less cleaning and maintenance.

Not if they use the direct impingement gauss system

 

[PercyShelley]

Rockets are not at all thermodynamically efficient when compared to traditional pressure driven projectiles, as their efficiency is near zero when the projectile is at rest and they are not efficient until the velocity of the projectile approaches the velocity of the exiting gases. A 100% efficient rocket would theoretically be possible if the projectile was moving so fast that the gases, when pushed out the back, immediately stopped because their backwards velocity equaled the velocity of the projectile. This isn't likely to be approximated unless we are talking about a hypervelocity rocket in outer space.

That doesn't seem right. As far as the exhaust gases are concerned, the nozzle is a stationary frame of reference. Why would the rocket's motion relative to its surroundings affect how effectively the rocket nozzle collimated the hot gas into a coherent, high-velocity exhaust stream? I'm pretty sure this is one of those "there are not privileged frames of reference" things.

 

[Chemistry Guy]

That doesn't seem right. As far as the exhaust gases are concerned, the nozzle is a stationary frame of reference. Why would the rocket's motion relative to its surroundings affect how effectively the rocket nozzle collimated the hot gas into a coherent, high-velocity exhaust stream? I'm pretty sure this is one of those "there are not privileged frames of reference" things.

When a rocket is sitting still, all of the burned propellant shooting out the back is just bleeding kinetic energy. Energy must be conserved, and the projectile is gaining kinetic energy much more slowly than the energy is being released from the propellant. The only time a rocket is 100% efficient is if the exhaust gases have 0 kinetic energy, which is the case when the rocket velocity = exhaust velocity. 100% efficiency is not possible for a variety of reasons, but a fast moving rocket in outer space is the closest we can get.

It is the same principle as recoil in a rifle. When a bullet is fired, both momentum and energy are conserved. The fact that the rifle is much more massive than the bullet means that the bullet gets the majority of the kinetic energy. If the rifle is bolted to something massive, the amount of kinetic energy given to the rifle is negligible.

 

[denton]

Well, our tax dollars at work, circa 1929:

Hatcher's Notebook, P399
13. Distribution of the Heath Energy of the Powder
Each pound of modern single base smokeless powder has a potential energy of about 1,250,000 foot pounds. If this powder is fired in a .30 caliber rifle, it will supply charges for about 140 cartridges, and each will fire a 150 grain bullet at 2800 feet per second muzzle velocity, with a muzzle energy of 2612 foot-pounds.
The amount of this powder potential that has appeared in the form of muzzle energy of the bullets is therefore 140 x 2612, or 365,680 foot-pounds or only about 29 1/4 percent. Where did the other 70 3/4 percent go to?
In 1929 the Ordnance Department set up a Technical staff Test Program to determine this. The firing was done in a Browning Machine Rifle, with the results as follows:
Heat distribution of one round in a Browning Machine Rifle.
Heat to Cartridge Case................. 131.0 BTU
To Kinetic Energy of the Bullet........885.3 BTU
To Kinetic Energy of the Gases.......569.1 BTU
Heat to Barrel..............................679.9 BTU
Heat in Gases..............................598.6 BTU

Total........................................2864.0 BTU
Heat Generated by Friction.............212.0 BTU