Sonic Choke? Maybe not. 2162 FPS

[rsterne]

Let's look at this another way.... Let's assume that all the air in the barrel moves as a solid rod, all at the same velocity, with no friction with the barrel.... The accelerating force acts on the end of this "rod" of air....

We know that the KE of that air, once accelerated to velocity V (fps) is 1/2 MV^2.... We can calculate the mass by multiplying the volume by the density, and we need the result in slugs.... Let's use 2000 psi air, which has a density of 164.4 kg/m^3, which is 0.0001846 slugs/in^3 (from an online calculator).... Let's use a barrel with an area of 1 in^2 and a length of 100 in.... It has a volume of 100 in^3, so the mass of that rod of 2000 psi air is 0.01846 slugs....

Now let's look at the maximum possible KE such a gun can produce.... We know that is equal to the force F (lbs.) times the distance (ft.).... We have 2000 psi pushing on 1 in^2, so that is a force of 2000 lbs.... The barrel is 100", which is 8.333 ft, so the maximum possible KE the gun can produce is 2000 x 8.333 = 16,666 FPE.... Let's combine the two equations and solve for the Velocity V....

KE = F x D = 2000 x 8.333 = 16,666 FPE

KE = 1/2 MV^2 = 0.01846/2 x V^2 = 0.00923 x V^2 = 16,666

V^2 = 16,666 / 0.00923 = 1,805,634

V = sqrt 1,805,634 = 1344 fps

This seems like a very low number to me, but I welcome anyone (and particularly Lloyd) to doublecheck my math.... Since in a "real" barrel, the center of the airstream moves much faster, it makes the value calculated from my formula of 2327 fps a lot more believable....

Bob

[lloyd-ss]

=====I was typing this as Bob was making another response, so the posts might seem out of sequence.  Lloyd
========================
Bob, Yes, 1/3, not 1/6th.  Ok, I see what you are saying and the explanation sounds good. I'll go with it. 

I'd like to take a slightly different approach to this (at least I think its different), at a fairly elementary level.
Going back to the original Hi Vel shot of 1745 fps, 7.5 gn , 26 cc, 4500 psi, 23.3" barrel, dump shot.
This shot yielded 50.7 FPE and was 70% efficient, meaning that 70% 0f the energy that was in the air entering the barrel was actually transferred tot he projectile.  If the shot had been 100% efficient, the velocity would have been 2090 fps and 72.9 FPE.  Therefore, the difference in energy between the actual shot and the theoretical shot, was 22.2 FPE.  That 22.2 FPE was lost, so where did it go?  It had to be converted to something, and heat is the only possibility, correct?

If the energy was converted to heat in the air in the barrel, final air mass is .0069 lbs, specific heat of air at that final average pressure is about .32 BTU/lb deg F.  So, the temperature change in the air would have been:

delta T (deg F) of the air in barrel = (22.2 x .0013 BTU/fpe) / (.32 BTU/lb air x .0069 lbs) = 13 deg F temperature rise of the air.

If all the energy were converted to heat in the barrel (.73 lbs) the calculation is this:

delta T (deg F) of the barrel itself = (22.2 x .0013 ) / ( .115 BTU/ lb steel x .73 lbs ) = 0.34 deg F temperature rise of the barrel.

Seems logical that all the energy loss is in heat.  But maybe that is already a given.
Lloyd

[wilsonj1018]

hey Lloyd , could you do me a favor and contact my buddy David Goldstein, he has been trying to reach you and isnt getting replies or returned calls.its concerning his gun that he still doesnt have back yet.
thanks
Josh

[lloyd-ss]

hey Lloyd , could you do me a favor and contact my buddy David Goldstein, he has been trying to reach you and isnt getting replies or returned calls.its concerning his gun that he still doesnt have back yet.
thanks
Josh

Josh,
Yes, I will try and contact him again.  I emailed him on April 7th and May 2nd with data and information and have not heard back from him.
Lloyd

[wilsonj1018]

thanks

[rsterne]

Lloyd, the 70% efficiency isn't 70% of the energy in the air ending up in the projectile, as I see it.... It means that the FPE is 70% of what COULD be in the projectile if ALL the input FPE ended up in the projectile AND the air in the barrel, right?.... but it's done incrementally, not all at once, so a lot of the acceleration of the projectile occurs while there is little air mass being accelerated.... It's pretty complex, which is why doing the calculations by intervals, like your spreadsheet, is the only proper way to do it....

I agree, that missing 30% (22.2 FPE) ends up as heat.... but there is also less KE in the air as well, as it is only reaching 1745 fps instead of 2090 as well.... so that missing KE must be going somewhere (heat) as well.... I think the total energy ending up in heat is the input KE minus the final KE in the pellet and air mass.... Since the starting pressure is 4500 psi, and some of the losses occur early, when the air is likely conducting heat to the barrel better.... it is likely REALLY complicated....

Bob

[Scotchmo]

=====I was typing this as Bob was making another response, so the posts might seem out of sequence.  Lloyd
========================
Bob, Yes, 1/3, not 1/6th.  Ok, I see what you are saying and the explanation sounds good. I'll go with it. 

I'd like to take a slightly different approach to this (at least I think its different), at a fairly elementary level.
Going back to the original Hi Vel shot of 1745 fps, 7.5 gn , 26 cc, 4500 psi, 23.3" barrel, dump shot.
This shot yielded 50.7 FPE and was 70% efficient, meaning that 70% 0f the energy that was in the air entering the barrel was actually transferred tot he projectile.  If the shot had been 100% efficient, the velocity would have been 2090 fps and 72.9 FPE.  Therefore, the difference in energy between the actual shot and the theoretical shot, was 22.2 FPE.  That 22.2 FPE was lost, so where did it go?  It had to be converted to something, and heat is the only possibility, correct?

If the energy was converted to heat in the air in the barrel, final air mass is .0069 lbs, specific heat of air at that final average pressure is about .32 BTU/lb deg F.  So, the temperature change in the air would have been:

delta T (deg F) of the air in barrel = (22.2 x .0013 BTU/fpe) / (.32 BTU/lb air x .0069 lbs) = 13 deg F temperature rise of the air.

If all the energy were converted to heat in the barrel (.73 lbs) the calculation is this:

delta T (deg F) of the barrel itself = (22.2 x .0013 ) / ( .115 BTU/ lb steel x .73 lbs ) = 0.34 deg F temperature rise of the barrel.

Seems logical that all the energy loss is in heat.  But maybe that is already a given.
Lloyd

Yes, all losses go to heat gain.

The heat gain is not evenly distributed. Most of the heat gain is in the air that travels the farthest. That would be the air right behind the pellet.

Knowing the air is heated some, and if we simplify and assume pressure to be constant, density will be reduced. So you are accelerating and heating less air than you think. Since the mass and therefore the KE of the air is less than you think, the losses are therefore higher than you think.

That may balance out with the fact that you are not inputting as much energy into the system as you think you are. So the losses are not as great as you think. Confused? I am a little.

You can probably assume a solid rod of air up until inlet reaches Mach 1. With a zero pellet mass, that happens instantly.

------------------------------------

Imagine it like this:
Once you reach supersonic, you are pushing a "solid" rod of air into the barrel at a fixed velocity of Mach 1. There is no more acceleration at the inlet. Once it enters the barrel, all additional acceleration comes from the expansion of the "air rod" inside the barrel, at the front of the "rod".

[rsterne]

I think a better assumption might be that density is constant and the pressure is increased.... but it could easily be a balance between the two, less density AND more pressure as the temperature increases.... Regardless, the net effect is that the air, being hotter, accelerates faster (easier).... Interesting that this effect is exactly the opposite of Adiabatic cooling, which some like to assume is happening.... I read in an online college paper not long ago (can't remember the source) that the expansion in an airgun inside the barrel is essentially Isothermal, possibly because of these effects cancelling each other?.... Once the air exits the muzzle I have NO doubt that Abiabatic expansion takes place.... This explains the "shoot over the thermometer and see the drop" effect, and also the "cloud of condensing water vapour at the muzzle" reported as well.... Inside the barrel, I'll bet Isothermal may be a pretty close model, especially while the valve is open....

Bob

[Scotchmo]

I think a better assumption might be that density is constant and the pressure is increased.... but it could easily be a balance between the two, less density AND more pressure as the temperature increases.... Regardless, the net effect is that the air, being hotter, accelerates faster (easier).... Interesting that this effect is exactly the opposite of Adiabatic cooling, which some like to assume is happening.... I read in an online college paper not long ago (can't remember the source) that the expansion in an airgun inside the barrel is essentially Isothermal, possibly because of these effects cancelling each other?.... Once the air exits the muzzle I have NO doubt that Abiabatic expansion takes place.... This explains the "shoot over the thermometer and see the drop" effect, and also the "cloud of condensing water vapour at the muzzle" reported as well.... Inside the barrel, I'll bet Isothermal may be a pretty close model, especially while the valve is open....

Bob

"I think a better assumption might be that density is constant and the pressure is increased..."
Maybe.

"I'll bet Isothermal may be a pretty close model..."
Isothermal is constant temperature, but implies heat flow in from an outside source. Usually a slow process. But in Fanno flow, the heat is generated internally.

Constant temperature could be a way to model it. As long as all heat came from energy within the system.

[phoebeisis]

Bob
I cut out your post-below
But that approach-which you aren't suggesting be used of course
Is more or less what Yellow Forum Steve used to get the MAXIMUM VELOCITY 1640FPS
The treat barrel  air like a rod or like "water"-with some slight differences-this is what he did-
insisted the air-all the air was matching the pellets velocity-and suggesting that ALL the  barrel
air molecules  were being accelerated-in the "speeding up sense" not the changing direction sense
to me the momentum is transferred in a wave like motion-a "bunch" of tiny independent "waves"-some are moving at close to the peak speed shown in a air molecule distribution graph
Not wanting a fight here-
and I know you aren't suggesting adopting your arch enemies approach-old approach-
he has found another "lotta math to obscure what he is saying  approach" so he has more "up to date numbers-
yes joking about the arch enemy-
I feel for him-hard to give up an idea that you pushed for so long-
but his adherents seem forgiving ,so...

Let's look at this another way.... Let's assume that all the air in the barrel moves as a solid rod, all at the same velocity, with no friction with the barrel.... The accelerating force acts on the end of this "rod" of air....

We know that the KE of that air, once accelerated to velocity V (fps) is 1/2 MV^2.... We can calculate the mass by multiplying the volume by the density, and we need the result in slugs.... Let's use 2000 psi air, which has a density of 164.4 kg/m^3, which is 0.0001846 slugs/in^3 (from an online calculator).... Let's use a barrel with an area of 1 in^2 and a length of 100 in.... It has a volume of 100 in^3, so the mass of that rod of 2000 psi air is 0.01846 slugs....

Now let's look at the maximum possible KE such a gun can produce.... We know that is equal to the force F (lbs.) times the distance (ft.).... We have 2000 psi pushing on 1 in^2, so that is a force of 2000 lbs.... The barrel is 100", which is 8.333 ft, so the maximum possible KE the gun can produce is 2000 x 8.333 = 16,666 FPE.... Let's combine the two equations and solve for the Velocity V....

KE = F x D = 2000 x 8.333 = 16,666 FPE

KE = 1/2 MV^2 = 0.01846/2 x V^2 = 0.00923 x V^2 = 16,666

V^2 = 16,666 / 0.00923 = 1,805,634

V = sqrt 1,805,634 = 1344 fps

This seems like a very low number to me, but I welcome anyone (and particularly Lloyd) to doublecheck my math.... Since in a "real" barrel, the center of the airstream moves much faster, it makes the value calculated from my formula of 2327 fps a lot more believable....

[rgb1]

Guys, the reference to   http://hyperphysics.phy-astr.gsu.edu/hbase/pfric2.html   is incorrect.
Hagen-Poisseuille is the solution to very low Reynolds number (laminar) flow. After going
through an old college text (Boundary Layer Theory, Schlichting) I can offer the following info.
For pipe flow, the transition Rn lies somewhere between 2e3 and 4e4. In Lloyds setup, that
is greatly exceeded in less than 1 inch of barrel travel and the turbulent boundary layer
profile (u/U vs radius) will stabilize in ~ 30 diameters (about 8 inches), changing very
little beyond that point.
Sorry for the late reply.

                                                                                                         Ron

[rsterne]

HI Charles.... You are absolutely correct, I am not suggesting that the 1344 fps number is any kind of a limit.... I was merely using the "rod" model to point out that IF you do that, the mass of the air quickly puts a limit on the velocity to something at or below that number up to 2000 psi (and then increasing above that as the pressure/density ratio increases).... This is a COMPLETELY different reasoning than using the speed of the air molecules as a limit (be that 1650 which has already been disproved, or the new 2259 fps).... Pretty much comparing apples to oranges.... The argument was presented to show (dare I say prove) that the air in the barrel is NOT moving as a single unit, or that math would apply (creating a ridiculously low "limit").... and that something more along the lines of using the 1/3 factor in the Hyperphysics equation would be more realistic.... BTW, I am still waiting for somebody to check the math on the "rod" model to make sure I got it right....

Ron, I do realize, of course, that the model used in the Hyperphysics article referenced is that for Laminar Flow.... However, it also represents (as I understand it) the lowest fraction than could reasonably be used in the equation (and therefore the highest central velocity).... If the flow is Turbulent (which I believe to be the case, as you do) then the velocity profile of the air in the tube (barrel) is shortened (truncated, or flattened) and no longer a parabola.... As such, the fraction of mass used would be close to 1/2 than 1/3, which of course produces an even lower velocity for the center of the airflow.... because the "Input FPE" is still limited by the pressure, bore area, and barrel length..... In fact, using a factor of 1/2, it is so low, on the order of 1900 fps at 2000 psi (and just over 2000 fps at 4000 psi), that it cannot apply, making the whole process a non-starter, as we have already seen velocities that exceed that....

I don't think we have a very good understanding of what is occurring inside a barrel in high pressure, high density, high speed conditions.... Yes, I am grasping at straws trying to come up with suggestions about what might be occurring.... IF the flow was laminar (which I understand would produce the highest central velocity), then the 1/3 factor would apply, and the velocities could, in theory reach the values in the graph in my post # 567.... As I understand it, if the flow is Turbulent (and it should be), then an even lower limit would apply, with the bottom limit being if you treat the air as a "rod", which we know is a non-starter (~1344 fps for an Ideal Gas)....

Scott, I perhaps used the term Isothermal incorrectly, but I meant it in the truest sense.... ie constant temperature.... I agree that the mechanism, if that is occurring, is that the energy losses which are occurring, do so in the form of heat, and cancel (and possibly reverse) and Adiabatic cooling which could in theory take place inside the barrel.... at least while the valve is open....

Please let's not loose sight of the fact that all of these theories are just that.... THEORY.... They are only valid proposals until they are proven or disproven by experiment.... I have stated that several times, a quite different attitude that I have seen elsewhere.... One other thing I would like to comment on.... the reference to an "arch-rival".... I would suggest that you read Post #545, which was a different way of looking at the Sonic Horizon Theory, incorporating the increased Speed of Sound at high pressures.... a proposal that just might let "everyone" be correct.... I don't think I need to comment further on my motivations or actions.... I just want the truth, and reality, to prevail.... I also feel for him, BTW.... 

Bob

[lloyd-ss]

Bob,
Your post # 591 was a great summary to help me see where you were going with this.  I was floundering a little trying to follow your discussion, but now it makes sense.  The discussion from everyone else is extremely helpful too.

But, trying to get things ready for the Fun Shoot now, and it is raining outside, so perfect shop work time.
Lloyd

[rsterne]

Just a quick summary of some of my ideas at the present time.... The graph below shows some of the possibilities for things that may limit the velocity at 20*C, and also shows the existing test shots to date....



The purple line is the 1647 fps RMS speed of air molecules at 20*C, originally proposed as the "speed limit" for PCPs....
The black dots are shots taken by Lloyd, Nick, and myself as of May 5, 2016 that exceeded that velocity....
The blue line is from the idea that the KE is calculated using 1/3 of the total mass of air in the barrel, as per the Laminar Flow Model from Hyperphysics which I used....
The red line uses the Sonic Horizon equation with the RMS velocity in place of the Speed of Sound, which places the limit at 2 x 1647 = 3294 fps....
The green line uses the Sonic Horizon equation with the Speed of Sound being the value at various pressures, so the limit is Mach 2 at that pressure....
The dotted black line uses the Sonic Horizon equation with the Mach 2 (green line) or RMS x 2 (red line), whichever is less....
Note I omitted a recent suggestion that the limit is 2259 fps from the DeLaval equations, because it doesn't take into account that air is not an Ideal Gas....
I have not graphed Scott's ideas because I don't fully understand them, other than that they are not at a steady temperature of 20*C....

This is just a snapshot of some of the current ideas.... They are all THEORY only, and as such only valid until experimental evidence proves or disproves them....

Bob

[Scotchmo]

Bob,

Here is max Fanno velocity (as best I can determine) vs pressure for pressures up to 10000psi:

PSI    SOS fps@70F     k        V max fps
500        1142.0   1.461   2638.7
1000   1169.9   1.526   2563.8
1500   1204.1   1.585   2531.0
2000   1247.0   1.637   2537.2
2500   1297.6   1.677   2580.2
3000   1354.6   1.707   2650.7
3500   1416.3   1.727   2743.1
4000     1481.0   1.739   2851.1
4500   1547.6   1.744   2972.0
5000   1615.1   1.745   3100.3
5500   1682.7   1.742   3234.8
6000   1749.7   1.737   3371.7
7000   1881.9   1.716   3665.2
8000   2008.2   1.700   3944.0
9000   2129.2   1.683   4220.1
10000   2244.4   1.666   4490.5

It looks close to the Mach 2 curve.

At the higher pressures, air is well inside the super-critical region.

[rsterne]

Yeah, that is very close to the Mach 2 curve....According to the Wolfram Widget, air becomes a Supercritical Fluid at about 550 psi at 20*C....

Bob

[MichaelM]

I should post the video of Lloyds super gun blowing my sky screens off at the fun shoot

[Privateer]

I should post the video of Lloyds super gun blowing my sky screens off at the fun shoot

Yes you should! LOL!!
Does that chrony still work by the way?

[lloyd-ss]

I should post the video of Lloyds super gun blowing my sky screens off at the fun shoot

Michael,
Your poor chrony, and it was a very early version, too.  I think it had a heart attack.  Can you clean the video up so that it is "suitable for all audiences"?
Lloyd

[MichaelM]

I should post the video of Lloyds super gun blowing my sky screens off at the fun shoot

Michael,
Your poor chrony, and it was a very early version, too.  I think it had a heart attack.  Can you clean the video up so that it is "suitable for all audiences"?
Lloyd

lol I think it did too.... still never seen a chrony sit there and just spin numbers like it did after the second shot..... and yea I was planning on probably editing it down pretty short just because we where all being goofy and laughing so much... either that or setting it to the Benny Hill theme music.... that might be even better