Sonic Choke? Maybe not. 2162 FPS

[Scotchmo]

Bob,

My thoughts on adiabatic vs isothermal model:

The reservoir air in a multi shot gun is undergoing primarily an isothermal expansion between shots. Since it usually has seconds to recover. If it were an insulated air tube, and a semi-auto, then it might be better treated as adiabatic.

Though not many processes are purely adiabatic, the shot charge expansion is primarily adiabatic. With an infinite dump chamber, there is no expansion, so it does not matter. With the large dump chamber used in the test, it is not a major factor. Though I do factor it in.

I am talking about overall affect.

There is also localized expansion, localized variations in pressure along the barrel and radially from the bore center. If there is any super low temp regions, they would be localized to a very small area. I ignore them.

[rsterne]

I agree that in Lloyd's most recent shot, which is a dump shot with a reservoir larger than barrel volume.... the difference between Isothermal and Adiabatic is small, on the order of about 11 fps.... However, if you look at a gun like my regulated QB79, and apply 100% Adiabatic expansion, it is difficult to come up with numbers that match the measured efficiency without having the gun end up 100% efficient, by the time you include the mass of air being accelerated.... and we know that kind of efficiency just isn't possible....

Bob

[Scotchmo]

Bob,


On the QB79, are you talking about the overall efficiency of the fill? That will definitely have an isothermal contribution to the apparent air use efficiency.

The isothermal energy contribution is happening between shots, not during the shot. At least the vast majority of it.

Unless you are able to shoot it fast enough to deplete the main tank before the temperature restabilizes. Than your efficiency would not appear to be so high.

Insulate the gun and repeat shots as fast as you can until depleted. The apparent air use efficiency will be lower than otherwise.

[rsterne]

Nope, I'm talking about taking the air for each shot, starting at the regulator output of only 1200 psi, and calculating the available energy in that air.... By the time you add up the FPE of the shot, plus the FPE taken to accelerate that air.... it comes out pretty darn close to the total theoretical FPE available from that amount of air if it expands Adiabatically.... too close, IMO....

I get 90 shots at 26 FPE on a fill.... That is from 3000 psi down to 1100 psi (a 1900 psi drop) on a 13 CI (213 cc) tank, running through a 1200 psi regulator.... That works out to 1900/14.5 x 213 = 27,910 bar.cc, divided by 90 = 310 bar.cc per shot.... At 1200 psi, that is 310 / (1200/14.5) = 3.75 cc per shot.... Run that through the calculator I posted, and you even if you expand it to 1 bar (which it doesn't, or it wouldn't have any report left), you get 41 FPE, and the air exiting the barrel would be only 83*K (starting to condense).... There is 26 FPE energy in the pellet, and the mass of air in the barrel is 27% that of the pellet, so now you are up to 33 FPE required to produce the shot.... Then you have to take into account that we still have a report, as the air is not fully expanded at the muzzle.... If the muzzle pressure is only 10 bar (145 psi), and we go back to the calculator, the maximum energy in the 3.75 cc of 1200 psi air is only 26 FPE if the expansion is Adiabatic.... It would be impossible to get that much energy in the pellet, and still accelerate the air in the barrel at the same time.... If you use Isothermal expansion, there is plenty of energy to go around, however.... and the efficiencies come back to what we are used to....

Going back to Lloyd's calculator, and setting it up to emulate the performance of my QB79, using Isothermal expansion, requires a efficiency factor of 72%.... If I use Adiabatic expansion, it can only work if the efficiency is set to 100%, and then only barely....  I'm not saying the expansion is 100% Isothermal.... but I'm willing to bet it's not 100% Adiabatic, either....

Bob

[Bill G]

WOW!  I go away for a few weeks and you get all kinds of crazy velocity.  Hope to catch up over this long weekend.  Lloyd that was nicely done. 

Bill G

[Scotchmo]

....  I'm not saying the expansion is 100% Isothermal.... but I'm willing to bet it's not 100% Adiabatic, either....

Bob

It is somewhere in between. As long as it's not over 100% efficiency, I'm happy. 

Is the post reg volume exactly 3.75cc?

13ci, 3000psi, 70F=1.80096moles

13ci, 1100psi, 70F=0.66035moles

1.80096-0.66035=1.14061moles

3.75cc, 1200psi, 70F=0.01268moles/shot

1.14061/0.01268=90 shots

I guess you are right!

[K.O.]

...
It will be interesting to see which way applying the VanDerWaals corrections takes us.... ie which is more important the lower density at high pressure, or greater pressure loss during the shot, compared to an Ideal gas.... or if they pretty much cancel out?....

Bob

Bob,

Isn't that backwards once you get over about 2000psi?

Looking at your earlier graph:


It will result in the model showing a higher than expected air mass at pressures over 3500psi. So that reduces the velocity estimate.

However, with the correction, air at 2000psi is wanting to take up a larger volume than an ideal gas. As the 4500psi dump chamber depletes, the pressure will not be falling off as fast as we originally predicted. So that increases the velocity estimate.

As you suggested, maybe it will still be a wash. It could end up balancing out to be the same with or without VanDerWaals correction.

Guessing what happens with an infinite or very large dump chamber should be easy:

An infinite dump of 2000psi could use an air mass multiplier of about 93%. - higher velocity than ideal gas.

An infinite dump chamber of 3500psi air needs no correction. - velocity as predicted

An infinite dump of 4500psi might need an air mass multiplier of about 109%. - velocity lower than ideal gas

An infinite dump of 6000psi might need an air mass multiplier of about 125%. - not good for performance

So am I wrong in seeing a heavy powerful (low repulsive) charge tapering to a faster lighter(higher repulsive) charge leading back to a slower an lighter charge (low repulsive) as the pressure drops...

 which sets up for supersonic flow thru over scavenging right as the pressure start dropping to about 1500-1000 psi..? which may set up conditions for a shockwave once super sonic flow at the throat collapses..?

But even if that is not happening with the break away is now closer to a rupture disk; enough that the pellet is much closer to riding a shock wav..?

[K.O.]

I think I will ask my daughter to take a look at this paper and see what she thinks?

http://www.sciencedirect.com/science/article/pii/S0895717709003367

[rsterne]

The post reg. volume is about 15 cc, the 3.75 cc is the amount of 1200 psi air used per shot....

Sorry, K.O., I don't know what a high or low "repulsive force" is.... nor do I understand how a pellet can "ride a shockwave".... You did see that graph was drawn for a specific reason, to show how air density at higher pressures compares to 3000 psi, right.... and that is was mis-labelled.... The correct graph is in reply # 291....

Bob

[Scotchmo]

You guys have got me laying awake half the night trying to wrap my head around this stuff. The following is a part of a follow up post that I put over on the yellow. There are some smart people over there as well. Their discussions don't usually have the same businesslike "air" , as I see on the GTA. And those discussions fall off fairly quickly on the yellow.
----------------------------------

v = 172* sqrt(Q/(m+Z*Q))

I looked into the origins of that equation. It balances. It makes sense as far as the math goes and as far as the underlying formulas. Very clean/elegant with a maximum possible velocity as a result. Presently, I can't find anything wrong with it. But, the implications are difficult to accept. In the image below, the top view represents what I imagine would need to occur.

1) The equation has no time or distance required for acceleration. All air in the tank is at V(0). Any acceleration would be in the barrel side of the transition where is must already be at 1640fps.

2) when pellet mass goes to zero, pressure and barrel length become irrelevant. The air will always reach 1640fps, even with no barrel length, and just a tiny pressure variance.

3) The three blue/red dots represents small masses of the air. They have to transition instantly from 0fps to 1640fps. How can that be possible?



The bottom view in the prior image is my attempt at a visual representation of the tank/barrel as modeled in the spreadsheet. The air undergoes acceleration in the tank, just before entering the barrel. So we take everything else into account as best we can. Mostly driven by the same formulas as the 1640fps equation. Though using numerical integration methods. Using small divisions of the barrel in my case, and small divisions of time in other peoples models. Each (time or distance) gives approximately the same results, neither showing a hard limit as to maximum velocity. But this method also has some things with implications that don't sit well. At any moment in time there are pressure and velocity gradients throughout the system, both radially and linearly in both the tank and the barrel. To correctly model it, maybe it needs a finite element model of the fluid dynamics that are involved in the process. Way beyond me.

Zero length barrels in the spreadsheet model just causes a divide by zero error.

At this point, I'm watching test results with great interest. What happens if we eliminate all possibility of any outside energy contribution and still get air way over 1640fps? I would look for what could be wrong with the equation. My first thoughts is that we can't simply divide by 2 for the average mass of the air column. It seems the correct thing to do, but if the pressure is not homogeneous throughout the entire barrel, that can't work. Right? Who knows. I don't.

What if after taking all possible precautions, the tests stay under 1640fps? Then, I would need to resolve the fact of an air particle having instantaneous velocity of both zero and 16404fps. Maybe it really is something at the atomic/molecular level. Again - way beyond me.

[rsterne]

I saw Steve_in_NC's answer.... that the molecules are already moving at 1640 fps, and hence don't accelerate.... You are not the only one who doesn't understand this, as the logical extension of that is that if there is only 1 bar in the reservoir, when you open the valve, the molecules coming out are going at the very same speed as if there is 100 bar inside it, and don't accelerate further.... Interestingly, if you apply his formula to Lloyd's tests, they predict velocities much lower than observed.... not only for the 2031 fps test, but also for the .22 cal tests with the 23.3" barrel at over 3000 psi....

EDIT:  I had a couple of graphs posted here, and have removed them as somewhat irrelevant.... I found a thread on the Yellow that proposed Steve's equation was out by a factor of PI/4 in one term, which raises the maximum velocity to 2088 fps.... In fairness, I have therefore deleted the graphs which were based on his original equation, even though he did not choose to change it.... Here is the pertinent thread, quite brilliantly written by Herb1836....

http://www.network54.com/Forum/79537/message/1236402420/Formula+for+maximum+theoretical+muzzle+velocity+for+an+airgun-

Steve's answer follows, and Herb1836 points out the flaw, where PI/4 is omitted from the "Q" in the numerator of the equation (but is included in the constant of 172).... thereby changing the maximum velocity to 2088 fps.... That was pointed out to Steve, and he reverted to the basic idea that 1640 fps was the absolute limit, so there could be a problem in the calculation of the average air density which he took to be Z = D/2.... The matter was left unsettled.... Therefore, I saw no point in using a flawed equation, and removed the graphs based on it.... Actually, when I work through it, I get yet another answer, of 1854 fps.... 

Interestingly, for that zero mass projectile, as you state, you can set the barrel length at 0.001", and the pressure to 1 psi, and you still get the same 1644 fps maximum velocity....

Bob

[K.O.]

http://research.chem.ox.ac.uk/claire-vallance.aspx


and this says its her lecture notes...  oops pdf to large to attach.

http://vallance.chem.ox.ac.uk/pdfs/PropertiesOfGasesLectureNotes.pdf

 Heading 5 subset compression factor
as the temp drops the pull/attractive force  gets stronger helping pull the charge out of the chamber: helps achieve supersonic flow at some point, maybe?

about riding the shock wave... and this being close to using a Rupture disk thus giving a much higher initial  acceleration... so now the chamber and pellet are acting much more like a shock tube than before...

Might have some one who reads this that understands shock tube simulators... might be close enough to do some modeling in that direction..?

[K.O.]

and no did not read any but your  post(mislabeled) will go back and pay better attention...  

[rsterne]

Another version of the thought experiment.... Start with Lloyd's test setup.... 4000 psi pushing against the back of the bullet, but assume an infinite reservoir volume.... Therefore, no matter where the bullet is in the barrel, the pressure of the reservoir remains at 4000 psi.... We will assume the temperature is a constant (initially at least)....

OK, so now put a membrane, with a mass of zero, against the back of the bullet.... release enough air to move the bullet 1% of the barrel length, and insert another mass-less membrane.... repeat this until the bullet is at the muzzle, and you have 100 packets of air, each separated by a membrane.... got the picture?....

As the bullet moves down the barrel, each packet of air, when released, will be at 4000 psi, it will be 1% long, and have the same mass.... The molecules within that packet of air will be rattling around at 1650 fps, relative to its center of mass, pushing on the membranes at 4000 psi.... However, the bullet is accelerating, and the lighter it is, the faster it accelerates.... The packet of air immediately behind it must be travelling at (nearly) the velocity of the bullet, while the one at the breech is barely moving (while it is filling).... One of two things must happen, or a combination of the two....

1. Each successive packet (which remains 1% of the barrel length long), will be filled identically, but in a shorter period of time (the flow rate out of the valve will increase)....
OR
2. If that doesn't happen then the packet of air next to the bullet must expand, and the pressure decrease (and so on, progressively, towards the breech)....
 
I would suggest that since each packet of air produced is at 4000 psi, and it is pushing on the one ahead of it (and ultimately the bullet) with that pressure, there is no reason (or mechanism) for the air in each packet to expand (at least initially).... This is condition 1.... Each packet is filled, from the reservoir, with molecules moving at 1650 fps.... As the packet in front of it accelerates away, progressively more and more of the molecules in each packet are travelling in the direction of the muzzle.... Eventually, when the bullet reaches 1650 fps, all the packets of air behind it, all still at 4000 psi, are all travelling at 1650 fps....

If the bullet has not yet reached the muzzle, now things change.... The air coming out of the valve cannot exceed the molecular velocity of 1650 fps.... In other words, the minimum time it takes to fill that 1% long packet has been reached.... However, there is still 4000 psi pushing on the bullet, accelerating it.... so as the bullet continues to accelerate, the pressure in that first packet starts to drop.... Actually, the pressure in ALL the packets starts to drop.... This is condition 2.... It's not like the valve is closed, it is still feeding air into the packet next to the breech at a constant rate.... However, the pressure between the breech and the bullet starts to decline.... It doesn't matter (for the purpose of our thought experiment) if that drop in pressure is continuous along the barrel or progressive.... The point is, that the air entering from the reservoir is NOT exceeding 1650 fps, while the force on the bullet is enough to keep accelerating it.... The air in the packet immediately behind the bullet is expanding, still being accelerated by the expanding packet behind it, and so on back to the breech.... The average pressure in the barrel is dropping, the air is entering from the reservoir at 1650 fps, and the bullet continues to accelerate well beyond that velocity, because of the expanding air behind it.... Each packet is now getting longer as the air inside it expands and the pressure drops.... The membranes are getting further apart.... The one closest to the breech is still moving 1650 fps.... so each one forward of that is moving faster than 1650 fps.... and so is the bullet, which is in front of them all, getting a push from each packet in turn.... The result is it easily surpasses 1650 fps by the time it reaches the muzzle....

If we look at the temperature of the air, since we have an infinite reservoir, the air entering the barrel stays at constant temperature.... The air in each packet, as long as it is not expanding, also stays at that temperature.... so while we are in condition 1, we have an Isothermal situation.... Once the velocity of the air exiting the valve reaches 1650 fps, we enter condition 2.... The air exiting the valve, filling the next packet, is still Isothermal.... However, the air in all the other packets is now expanding, and that is likely to be Adiabatic.... Using Lloyd's shot as the model, for the bullet to be travelling 2031 fps at the muzzle, it will be going 1650 about 28% of the way down the barrel, about 0.001 sec after launch, and will exit the muzzle at 0.0025 sec.... I would suggest only the latter part of that cycle is Adiabatic....

Now, how does this relate to reality?.... First of all, the transition from condition 1 to condition 2 will not be instantaneous but somewhat gradual.... Secondly, the velocity that the air can exit the valve may not be 1650 fps, but could be the local speed of sound (at 4000 psi about 1480 fps).... In the case of Lloyd's shot, the reservoir was not infinite, but 55 cc, so the pressure would be dropping as the bullet (and each packet of air) moves down the barrel.... The point where the pressure and speed of sound cross using Lloyd's spreadsheet is about 3500 psi, and 1400 fps, which the bullet crosses about 8" of the way down the barrel.... We could enter condition 2 at that point, and from then on, air continues to enter the barrel at the (local) speed of sound, pushing all the packets down the barrel at that velocity, and the air in each packet expands, dropping in pressure, but still accelerating the bullet to it's eventual 2013 fps.... Of course there could be other factors, such as barrel friction, fluid velocity profiles, etc.etc.... but the point of this thought experiment was to provide a mechanism whereby the velocity in each packet of air, relative to that packet, and for the first packet relative to the valve, does not exceed 1650 fps.... I think it does that rather elegantly....

Bob

[Scotchmo]

I currently have three versions of the spreadsheet:

spreadsheet 1) - no friction losses - it generally predicts too high of a velocity

spreadsheet 2) - modest friction losses, but I know it's not quite right. It works fair for the range of values that we are working in. It is more like an "efficiency factor". - it predicts pretty close to test values

spreadsheet 3) - complete progressive friction losses - I have not been able to get this to work well. As soon as the barrel gets over about 10", it break down. I've played with constants and exponents but is always falls off too fast with longer barrels. I may have found a simultaneous solution to this and the 70F 1640fps conundrum:

Bob,

About the constant temperature air - it's not.

I have been subtracting out the fluid flow losses by reducing force on the pellet. That's fine, but where did the loss actually go? It is still in the system. It should show up as entropy production (at least the adiabatic portion). I have not modeled it yet, but I did a quick check at 3" of barrel in spreadsheet number two. The "Joke" formula predicts 1640fps at the barrel inlet, but now predicts 1687fps at three inches down the barrel. When I get time, I will attempt a new variation on the spreadsheet, with full fluid friction, and entropy production. I'll show the ever increasing 1640fps limit column which I expect will always register higher than the f=ma calcs and higher than the test results. As we move down the barrel, it's possible the 1640fps limit will grow so fast, that it will always run way ahead of the test values. So it will never be a factor.

I'm thinking that the 70F "Joke" formula is only valid at a barrel length of zero. Or at longer length, only if we assume no entropy production (i.e. inviscid flow?). Factoring entropy production allows the "Joke" formula to predict higher velocities, even with a 70F initial air source.

Note: I always wondered why my PCP guns heat up when filled from a SCUBA tank. Intuitively, I would have expected the gun to cool down from the isothermal expansion of air. But the gun heats up during filling.

I said, "I might be on to something" - or maybe not. Any thoughts?

[rsterne]

The reservoir heats up because you are compressing (at least partially Adiabatically) the air that is already inside it.... while the tank you are filling from cools down from the air in it expanding (at least partically Adiabatically).... That's what I learned, anyway....

You may be onto something about the gas heating as it moves along the barrel.... I take it you don't like my "thought experiment".... *pout*.... I realize the temperature of the air is not a constant, but using that, at least in the beginning, in the above post made it (at least partially) understandable.... or at least I had hoped it made sense....

Bob

[match]

The scuba tank cools and the air leaves it expanding - the air rifle's tank heats up as the air enters it and compresses.

The problem is this is NOT a net zero energy transfer - there are other losses and entropy to account for a total loss in energy during the transfer.

[match]

Bob - I was editing as you were posting...

[rsterne]

*LOL*

Bob

[Scotchmo]

The scuba tank cools and the air leaves it expanding - the air rifle's tank heats up as the air enters it and compresses.

The problem is this is NOT a net zero energy transfer - there are other losses and entropy to account for a total loss in energy during the transfer.

The gun's tank is near empty. Not much air to compress. That cannot be much of a factor.

The air flowing through the fill hose is generating entropy. Enough to heat up the gun to a temperature that is much higher than the 70F SCUBA tank.

The air going into the gun's tank is expanding. But carrying lots of heat that it picked up in the process.

(SCUBA tank and gun tank) . Does the net loss end up as a net heat loss? Rather than a net heat intake as would be expected. That seems right to me.