Firearms thermodynamic efficiency


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PercyShelley
March 19, 2012, 06:42 AM
Some years ago I ran across this article:

http://www.z-hat.com/Efficiency%20of%20the%20300%20Hawk.htm

On the thermodynamic efficiency of firearms; that is, how much of the energy of the burning gunpowder gets converted into motion of the bullet.

Does anyone know if the figures in this article are correct? They certainly pass the smell test; about 30% efficiency is close to what most internal combustion engines achieve, and a firearm is not so different from a piston engine.

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The Lone Haranguer
March 19, 2012, 08:03 AM
It doesn't seem unreasonable. Perhaps a little on the high side, particularly with "overbore," very high velocity rifle cartridges that produce a huge amount of muzzle blast and flash, or some handgun cartridges from short barrels. Once the bullet has left the barrel, all that noise isn't doing anything to add additional "push" or speed to it.

Owen
March 19, 2012, 08:24 AM
looks good to me.

Loosedhorse
March 19, 2012, 10:31 AM
What would an 100% efficient rifle be like? Silent, cold barrel, cold bullet, 0 chamber pressure on bullet exit, no recoil.

Not likely.

This is a bit of a marketing gimmic. You can make any cartridge seem more efficient if you choose a powder charge to achieve the preferred chamber pressure quickly and minimize the pressure as the bullet exits. (Heck, you can also goose it a bit using moly.)

To the extent that you mimic the "magic chamber pressure curve" that is illustrated, you will achieve similar "efficiency." I see no reason to believe that curve is uniquely available to one cartridge only.

But yes; we all know that you pay a lot for a little extra speed. That is the magnum downside.

Driftertank
March 19, 2012, 11:13 AM
Read an article a while back where a guy charted "cartridge efficiency" by measuring velocity, energy, calculating "L factor," then comparing these numbers to grains of powder in a typical load. The most efficient cartridge he tested, IIRC, was the .35 Whelen, Ackley Improved.

It was a good read, while his calculations were not all-inclusive, he pointed out that higher cartridge efficiency pays dividends in things like barrel life and cost of components when reloading.

It also sort of vindicated my love of the .308, as it was the most efficient of the .30 caliber rounds. Before reading the article, i just knew it was a popular, flexible round that offered good performance for a lot of uses with less recoil than the magnums.

brickeyee
March 19, 2012, 01:38 PM
Energy efficiency is not the goal of a firearm, so who really cares?

Owen
March 19, 2012, 02:04 PM
Thermodynamic efficiency is something that internal ballistics people pay lots of attention too.

<mod hat> If you don't care about a topic, you can feel free to not post in a thread </mod hat>

eye5600
March 19, 2012, 03:00 PM
Energy efficiency is not the goal of a firearm, so who really cares?

The gun designer cares. Better efficiency means more velocity with a given round.

Old krow
March 19, 2012, 03:56 PM
Looks close to me, even then it may be a little on the high side in some cases. In this case I believe that it was a 24" barrel used. If the exact same bullet (round) that was shot out these guns was shot out of a 20" barrel, or a 22" barrel, it would be even less efficient.

Energy efficiency is not the goal of a firearm, so who really cares?

Reloaders/handloaders use at least some of this, whether they know it or not. The Ideal Gas Law (PV = nRT) still applies, it's why we don't seat bullets too deep or change powder charges when we change bullet weights. Some even go so far as to compare the amount of powder used to gain a given velocity to determine the relative "value" of a given powder. Speer Gold Dot "short barrel" ammo is a good example of trying to find an efficient sweet spot. Barrel life is another reason.

Interesting read, thanks for posting that.

brickeyee
March 19, 2012, 04:13 PM
The gun designer cares. Better efficiency means more velocity with a given round.

No, they care about meeting a performance goal.

X ft/s at the muzzle within allowable pressure.

The thermodynamic efficiency is not even on the list.

If you even calculated the efficiency of an Otto cycle motor, you would understand that SOME things have efficiency concerns.

Firearms are not one of them.

The Ideal Gas Law (PV = nRT) still applies

Actually, since the gasses in the barrel are still undergoing chemical reactions it does NOT apply.

To the extent the gases are doing work and increasing their volume as the bullet is driven down the barrel, Charles's law can be applied, but as the gases continue to react even it becomes less than perfect.

Improvements in thermodynamic efficiency are not going to help a firearm very much.
It is far more profitable to improve the powder and its burn curve through chemistry, and things like granule shape (not exactly chemistry, but related to how the powder burns).

Shadow 7D
March 19, 2012, 04:13 PM
Kinda, but according to Newton, the best you can hope to achieve is 50% (equal and opposite)
So losing 20% to heat and friction isn't bad.

It gets complicated, the gas is produced by a pressure dependent reaction (for speed) that is also influenced by the rapid increase in volume as the bullet moves down the barrel.

Old krow
March 19, 2012, 05:27 PM
No, they care about meeting a performance goal.
The thermodynamic efficiency is not even on the list.

This will easily get off topic fast, so, I'll just say this, read some on Julian Hatcher, specifically "Hatcher's Notebook." They do indeed care about thermal efficiency and it is measured. The Army researched it quite a bit. And, Charles's law is a special case of the Ideal Gas Law. It does apply.

On the OP. Hatcher measured the thermal efficiency of the Browning Machine Rifle and found a thermal efficiency of 29%, so I'd say that the article is on target. Other small arms ranged from 17% - 37%.

So losing 20% to heat and friction isn't bad.

Heat to cartridge case 4%
Kinetic energy to bullet 29%
Kinetic energy to gases 19%
Heat to barrel 22%
Heat to gases 19%
Heat to bullet friction 7%

That was the Browning Machine Rifle though and will vary.

Shear_stress
March 19, 2012, 06:24 PM
I guess that's why they call guns "heaters."

Still, they'd work even better if they didn't squander a 1/3 of their energy trying to fling a small chunk of lead down a narrow tube. Just like those stupid ceiling heaters that waste 10% of their energy making light for some reason.

Loosedhorse
March 19, 2012, 11:10 PM
Kinda, but according to Newton, the best you can hope to achieve is 50% (equal and opposite)Not for energy. If the rifle is unimpeded, it will achieve a momentum equal to the bullet's momentum (and in the opposite direction), but the energy of the bullet will be much greater (in fact, by the same ratio that the rifle mass is greater than the bullet mass).

And of course, if you impede the rifle (let's say bolt it rigidly into bedrock), then you effectively increase the mass of the rifle tremendously, so essentially all of the kinetic energy goes to the bullet.

PercyShelley
March 21, 2012, 01:53 AM
Thanks all.

If I had to guess, pistol caliber carbines are the most thermodynamically efficient firearms, since they have an enormously long barrel for the amount of powder they're burning. Magnum rifles, especially with shorter barrels, are probably the least efficient.

Shadow 7D
March 21, 2012, 02:40 AM
no, actually the barrel length doesn't add that much to it after a certain portion
pistol powders burn fast and are designed for tossing lead out of 3" barrels.

Rifle powders are for longer barrels
a .22 will SLOW down after 22? inches

PercyShelley
March 21, 2012, 04:26 AM
How fortunate are we in the twenty first century to have experimental data concerning just this matter, and available for free on the world wide intertubes!

http://www.ballisticsbytheinch.com/


Looks like even very high expansion ratio, low pressure cartridges like .45 ACP are still accelerating most loads past the 10" mark, albeit pretty marginally.

The pattern appears to be that the light bullet weights see the biggest gains in the long barrels. It would be interesting to know the powder charges that they're sitting on.

denton
March 21, 2012, 10:07 PM
PV=nRT does not apply. This equation only works for ideal gasses, which are monatomic. Propellant gasses do not meet this requirement.

Efficiency is always (effectively employed energy)/(total energy).

Energy thankfully does not split evenly between the bullet and the rifle. Otherwise, none of us would fire more than one shot per lifetime. The accepted formulation is that the mass of the rifle + cartridge system remains at rest unless acted on by an outside force, if the system is freely recoiling, which a shoulder fired rifle is not. So for a rifle free to slide on a frictionless surface, the center of mass would not move.

Much of the lost energy disappears when the bullet uncorks the barrel. All the propellant gas that is still in the barrel is vented into the air, and the energy stored in it is lost to propulsion.

Shadow 7D
March 23, 2012, 06:31 PM
Energy thankfully does not split evenly between the bullet and the rifle. Otherwise, none of us would fire more than one shot per lifetime.
Wrong
the way energy works in ballistics is by pressure
see a PUNCH has more Energy (joules) than a bullet does, yet a bullet may go right through you due to exerting it's force over a SMALLER surface, but a punch can knock you down by exerting a MUCH STRONGER force over a much larger area.

(and we have proven time and time again a bullet lacks the ability to knock you down short of causing you to collapse due to mechanical structural or neurological damage)

Equal force is exerted, newtons law is correct, but your mass added to the gun is effected by it orders less than a bullet measured in GRAINS, now there are 7000 grains to the pound, want to do the math, and find out how many times MORE massive you, and your gun are than a bullet?

denton
March 23, 2012, 08:03 PM
Shadow 7D, you're debating physics with a physicist. Energy does not split evenly between the bullet and the rifle... not even close.

Here's a recoil calculator for you: http://www.handloads.com/calc/recoil.asp

Enter a 180 grain bullet, and 2800 FPS as the muzzle velocity. Choose 57 grains of powder, and a 9 pound rifle and scope. Hit CALCULATE.

The 18.87 foot lbs of energy shown in the results is the kinetic energy of the rifle if it is allowed to recoil freely (sliding on a frictionless surface). Yet the KE of the 180 grain bullet is in the neighborhood of 3,000 foot lbs. That's nowhere near an even split. It just doesn't work that way.

Conservation of Momentum is a wonderful thing. That is the accepted method for analyzing firearm recoil.

Loosedhorse
March 23, 2012, 10:00 PM
Shadow 7D, you're debating physics with a physicist.:D

As I made exactly the same point about energy in post #14--perhaps Shadow has me on his "Ignore" list?--I appreciate your post, denton.

Shadow 7D
March 24, 2012, 02:15 AM
Yield, I'm more of a chemistry type of person, if I understand the principle I can usually see it in my head.
And Loosed I took you off my list, I was just pretty annoyed that I was trying to agree with you for a page, and just didn't seem to get it, but bygones are bygones.
As for the other guy, I'd have to go look at my list, he's still on it.

Blackstone
March 24, 2012, 08:30 AM
Very interesting read, will pass this on to some friends of mine

Loosedhorse
March 24, 2012, 09:17 AM
bygones are bygonesSincerely appreciated; no sarcasm. And no hard feelings here at all.

Chemistry Guy
March 24, 2012, 11:36 AM
I don't know exactly how the 70%+ of the energy that doesn't go towards driving the bullet is dissipated, but my gut tells me that the vast majority of it is heat. Acoustic energy and PV work done by the expanding gas on the atmosphere are probably peanuts compared to the heat released both inside and outside the barrel. I would imagine that this type of research is also very important in the design of heat shielding, suppressors, barrels, and the like.

denton
March 24, 2012, 01:27 PM
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
March 24, 2012, 02:06 PM
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
March 24, 2012, 06:03 PM
Yeah and the m240 saw still gets red hot after a full mag dump LOL.

exavid
March 24, 2012, 07:10 PM
I think I'll just wait until handheld versions of rail guns become available.;) So much less cleaning and maintenance.:)

Shadow 7D
March 24, 2012, 07:11 PM
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
March 24, 2012, 07:38 PM
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
March 24, 2012, 07:44 PM
its true voltia, so so true.

Jim K
March 24, 2012, 07:52 PM
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
March 24, 2012, 07:55 PM
I won't remember half the thermodynamics I'm learning right now in uni

MX26
March 24, 2012, 08:09 PM
I used to believe that only a cynical attitude was logical. Then I learned to reason.

Owen
March 24, 2012, 08:38 PM
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
March 24, 2012, 10:32 PM
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
March 25, 2012, 08:32 AM
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%20Notes%5CTech%20Note%2048,%20Barrel%20Design,%20Heat,%20and%20Reliability,%20030824%E2%80%A6.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
March 25, 2012, 08:50 AM
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
March 25, 2012, 06:54 PM
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
March 25, 2012, 10:00 PM
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
March 25, 2012, 10:44 PM
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%20Notes%5CTech%20Note%2048,%20Barrel%20Design,%20Heat,%20and%20Reliability,%20030824%E2%80%A6.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
March 26, 2012, 06:36 AM
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 (http://en.wikipedia.org/wiki/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
March 26, 2012, 03:28 PM
"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
March 27, 2012, 02:24 PM
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
March 27, 2012, 04:38 PM
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
March 27, 2012, 09:32 PM
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
March 29, 2012, 01:42 AM
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
March 29, 2012, 09:05 AM
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
March 29, 2012, 09:15 PM
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

Loosedhorse
March 29, 2012, 09:18 PM
Don't look now, but check out post #12. :o

denton
March 29, 2012, 10:03 PM
It's one of the joys of being forgetful. Everything old is new again. :)

PercyShelley
March 30, 2012, 02:32 PM
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.


OK, I see it now. Yes, if you measure the thermodynamic efficiency of the rocket in terms of increase in velocity of the rocket from a stationary point of reference, yes you do get wonky efficiency numbers like that.

Which is why it makes more sense to measure efficiency as percent of thermal energy converted to kinetic energy of the propellant.

You'd get similar weirdness (although to a lesser degree because of, as you point out, the more skewed mass ratio) if you measured the efficiency of a firearm in space as a function of propellant energy converted to projectile motion relative to a stationary point of reference. As the gun floated back with each shot, it would get less and less efficient.

The BAR tests are interesting. I would not have guessed that you lose almost as much energy to heating the case up as you do to barrel friction!

At SHOT there was a new plastic-cased ammo company that mentioned that they were getting a slight velocity boost with the same loads as regular ammo just from switching the case material. I wonder if it's because less energy was going into heating up the case, or because the case is insulative enough that it's preventing some of the energy from leaking into the firing chamber. It could also be because plastic is bulkier and the case has less volume, which would lead to higher efficiency as well. Perhaps it's all three.

brickeyee
March 30, 2012, 04:01 PM
P1*V1=P2*V2. Those laws do not require an ideal gas.

Cahrle's law,like therdieal gas law does indded requoire assumptions about reactivity.

2H + 1O yields 1 H2O.

The volume law just failed.
The ideal gas law also fails since tt encompasses Charle's law.

If the slightest bit of condensation occurs, the ideal gas law fails horribly.

It cannot work in a state change, or if liquid gas is present.

you can keep taking gas out with liquid preset above the boiling point with zero change in pressure until all the liquid vaporizes.

That is why pressure gauges cannot tell the contents of a propane tank or Dewar with any liquid nitrogen oxygen, or refrigerant. The liquid is above boiling, and any removal f gas immediately cases vaporization of the liquid.

The laws can be used carefully for short times compared to any further reaction rates, but temperature changes and composition changes can easily make the gas so far from ideal things come apart.
Even the ideal gas constant is slightly different among ideal gases.

The pressure will not decrease til the liquid is evaporated.

The military is interested in thermodynamic performance, especially in machine guns and other weapons that can fire rapidly compered to their ability to shed heat.

It is not the efficiency they care so much about, but the heating produced from deflagration.

Hot steel wears faster than cooler steel from friction.

SimplyChad
March 30, 2012, 04:18 PM
Well, our tax dollars at work, circa 1929:
Everytime I think no we wouldnt burn money like that. Well im proved otherwise.

denton
March 30, 2012, 06:31 PM
Cahrle's law,like therdieal gas law does indded requoire assumptions about reactivity.

But they work for polyatomic gasses, which are not ideal.

AntiSpin
March 30, 2012, 06:46 PM
if physics terms could mean the same thing each time they're used, and no matter who uses them. Too often such terms are used indiscriminately, or even interchangeably, when in fact they have specific definitions and produce nonproductive conclusions when not used correctly.

So -- here:

ENERGY: the ability to perform work

WORK: a force acting through a distance; the product of force times distance

KINETIC ENERGY: the energy which an object possesses due to its motion; the work needed to accelerate a body of a given mass from rest to a specific velocity.

FORCE: an influence that causes an object to undergo a change in speed, a change in direction, or a change in shape

POWER: A measurement of the rate at which energy is dissipated; i.e. transferred, used, or transformed

MOMENTUM: This is a difficult concept to narrow down to a short definition. One of the best descriptions that I have seen applied to the term “momentum” is “the amount of force required to stop a moving object” It’s not perfect, but for our purposes here that description will suffice.

And never forget, the formula for kinetic energy is one-half the mass times the square of the velocity.

For momentum, just plain mass times velocity.

Momentum and energy are not the same thing.

Have fun!

Certaindeaf
March 31, 2012, 01:59 AM
I wonder what Clark would say. lol

Loosedhorse
March 31, 2012, 08:49 AM
Anyone game for talking about the efficiency of railguns (http://en.wikipedia.org/wiki/Railgun)? Maybe we can dust off some Maxwell?

:evil:

Airborne Falcon
March 31, 2012, 10:03 PM
Loosedhorse ... I lol'd.

Yoda
April 1, 2012, 06:52 PM
So, a guy throwing a rock, or flinging one in a sling, is more efficient that a rifle, because there is no barrel to heat up?

Airborne Falcon
April 1, 2012, 11:54 PM
So, a guy throwing a rock, or flinging one in a sling, is more efficient that a rifle, because there is no barrel to heat up?
That's over simplifying it just a bit, is it not?

Loosedhorse
April 2, 2012, 10:14 AM
So, a guy throwing a rock, or flinging one in a sling, is more efficient that a rifle, because there is no barrel to heat up? Nope (http://www.coachesinfo.com/index.php?option=com_content&view=article&id=392:efficiency-article&catid=107:rowing-general-articles&Itemid=207).The internal or muscle efficiency is determined mainly by the effectiveness of muscle contraction and is estimated for rowing to be in the range of 14-27%By the way, the rowers tested were VERY efficient.Loosedhorse ... I lol'd. ;)

brickeyee
April 2, 2012, 12:06 PM
But they work for polyatomic gasses, which are not ideal.

No one said otherwise.

O2, N2 are diatomic and behave reasonably well.
The are relatively non-reactive in a pure mixture.


Even better if you use the gas ideal gas constants for THE CORRECT GAS.

Even the 'ideal' constant varies with gases and non-reacting ,mixtures of gases.

If you mix H2 and O2 the mixture behaves 'nicely' right up until the reaction starts.
And the resulting hot water vapor behaves reasonably well until any of it condenses

Not all gases behave under all conditions as ideal gases, and state changes and reactions are two if the big poisoners.

Try using it with any liquified gas present in the container and see how far you get.
Propane, carbon dioxide, even a Dewar with liquid oxygen or nitrogen.

The ideal gas law fails with state changes or chemical reactions in the gases.

PercyShelley
April 3, 2012, 07:53 AM
Anyone game for talking about the efficiency of railguns? Maybe we can dust off some Maxwell?

My vague recollection is that railguns have pretty good efficiency, it's coilguns that have absolutely terrible, single digit efficiency.

Hmmm...

Is it a mistake to assume that the pressure detected by a strain gauge mounted on the barrel somewhere is representative of the pressure behind the projectile? Things are happening fast enough in a gun barrel that I'm not sure Pascal's Law applies. I don't think this (http://www.dtic.mil/ndia/2010armament/WednesdayLandmarkAMikeBixler.pdf) would work if the changes in pressure were being applied instantaneously across the entire bore area.

I also recall reading a small arms design manual which mentioned that propellant gasses have net velocity, and that this has practical consequences. Erosion tends to be more aggressive on the borewards side of gas ports, for example, because the gas is moving in the same direction as the bullet and tends to smack into that side of the port and wear it down faster.

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