Sonic Choke? Maybe not. 2162 FPS

[match]

there are also micro state changes in the electron shells and quantum effects which store and release energy in diatomic gasses but I digress....

[I_like_Irons]

LLoyd,

I've reached the point of (tentatively) selecting an appropriate ADC (analog/digital converter).  It is the LTC2351-12.  I think it is fast enough, has simultaneous multi-channel capabilities, and its12 bit resolution should be adequate for this.  I am now at the point of selecting the strain gauges and appropriate op-amps. Choices, and choices, and still more choices.  I'm still  quite fuzzy about software and programming, however.   But, as with any worthwhile endeavor, it is a significant learning experience. 


Now back to the thread:

A thought has occurred to me about how to treat the column of gas.  Like Tim McMurray stated in another thread, it is like a compressed spring. 

I am wont to think the column of gas like a very long coil spring that is fed into the barrel at up to Mach 1 for the pressure and temperature of the gas.  The pellet, initially, has enough resistance to keep the spring at "full" compression.  The mass of the spring itself prevents it from being fed faster than that rate.

However, when the pellet has reached that speed, the front part of the spring (also traveling at that rate) is now free to expand, and release stored energy. But, the coils behind the Mach 1 limit cannot yet expand.  Those coils that wind up in front of this Mach 1 "wave" get progressively more numerous and thus add more energy to the system.  They get in front due to the expansion of the coils in front of them.

The upper limit will be when the friction and compressed air force in front of the pellet (in a very long barrel) equal the expanding air behind the pellet.  There will be a barrel length  (for a given pellet mass and initial pressure) that will allow a maximum velocity of the pellet inside of it, while a longer barrel will not allow this velocity to occur, as there will be too much air in front of it. The optimum barrel will release just enough air to the atmosphere.

Now if one were to make a vented barrel whose vents would close just as the pellet reached them . . . . 

[rsterne]

David, I like your analogy of the compressed air spring.... The second part, about the limit coming when the air in front of the pellet compresses to the point of balancing the force behind would take an incredibly long barrel.... and could be completely eliminated by applying a vacuum in front of the pellet.... Therefore, such a gun in space, should have no limit, right?.... Somehow I think that the compressed air behind the pellet will eventually lose it's ability to continue to accelerate the pellet.... I still think that will occur at twice the molecular velocity for air at the temperature involved.... ie at room temp about 3300 fps.... but it really is just a hunch, I can't prove it.... The three things that point me in that direction are my "thought experiment" with packets of air, the fact that the maximum velocity of flow in a tube is twice the average velocity, and Steve's equation for the "point of valve ineffectiveness", but using the molecular velocity instead of the speed of sound.... That equation goes to zero effect at 3300 fps for room temperature air....

Bob

[rsterne]

Further to my post # 517, if we apply the idea of using 1/3 of the air mass to the equation in post # 386, we get the following results for pellet mass equals zero....

V = sqrt { 2 x P / 12 / (D / 890575 / 3) } = sqrt (2 x 890575 x 3 / 12 ) x sqrt ( P / D ) = sqrt (445288 ) x sqrt ( P / D) = V = 667.3 sqrt ( P / D ) with P in psi, D in kg/m^3, and V in fps.... Here are the results for pressures up to 10,000 psi....



These results for zero pellet mass are interesting, because they show how much air diverges from an Ideal Gas at very high pressures.... the pressure (force) increasing faster than the density (mass).... Remember, the prediction for maximum velocity using the same equation (and 1/3 of the mass), but assuming air was an Ideal Gas, resulted in a velocity of 2257 fps, regardless of pressure.... If nothing else, I think the above numbers may be high enough that they will prove difficult to reach....

It occurred to me that I should be consistent in the units, and provide a Metric and Imperial version of the above equation.... In Imperial units, using Pressure P in psi and Density D in lb./in^3, and calculating the maximum Velocity V in fps, for zero pellet mass, and using 1/3 of the air mass, we have the following....

V = sqrt { 2 x P / 12 / (D / 32.174 / 3) } = sqrt ( 2 x 3 x 32.174 / 12 ) x sqrt ( P / D ) = sqrt ( 16.087) x sqrt ( P / D ) = V = 4.011 sqrt ( P / D )

In Metric units, using pressure P in bar (1 bar = 100,000 N/m^2), Density D in kg/m^3, and calculating the Velocity V in m/s, for zero pellet mass, we have the following....

V = sqrt { 2 x P x 100000 / (D / 3) } = sqrt ( 2 x 3 x 100000 ) x sqrt ( P / D ) = sqrt ( 600000 ) x sqrt ( P / D ) = V = 774.6 sqrt ( P / D )

All three equations give the same results, of course....

Bob

[I_like_Irons]

Therefore, such a gun in space, should have no limit, right?.... Somehow I think that the compressed air behind the pellet will eventually lose it's ability to continue to accelerate the pellet....

When the pressure x area equals friction, that will be the limit in a vacuum.  If frictionless, then the limit will likely have to do with relativity.    Although, the acceleration after some point will be quite small well before the relativity limit.

[Scotchmo]

Therefore, such a gun in space, should have no limit, right?.... Somehow I think that the compressed air behind the pellet will eventually lose it's ability to continue to accelerate the pellet....

When the pressure x area equals friction, that will be the limit...

That's the simple view, and I like it.

It would be nice if there were a single equation to tell us that limit. I'm not so sure that there is.

[Scotchmo]

there are also micro state changes in the electron shells and quantum effects which store and release energy in diatomic gasses but I digress....

I think I'll skip that part. For now at least. 

[phoebeisis]

 
How hard would it be to test how quickly the muzzle blast/pressure change-no pellet- is detectable?
Might tell you how fast the fastest detectable  energy wave is going

[Bill G]

Sorry, but could someone define Cp and Cv?.  I seem to have missed this.

[rsterne]

They are the Specific Heat Constants for a gas.... Cp is at constant pressure and Cv is at constant volume.... The constant gamma, (k =Cp/Cv) is most used as the exponent in the Adiabatic expansion process.... Beyond that, I have little understanding of them.... other than the oft-quoted value of k = 1.4 for air varies with pressure and temperature.... There is another gas constant, R = Cp-Cv, which I know nothing about....

Bob

[rsterne]

Charles, I think as soon as you try and measure the velocity of the muzzle gas, with no pellet in place, you will alter the reading....

Bob

[phoebeisis]

Bob
Thanks

[Scotchmo]

Further to my post # 517, if we apply the idea of using 1/3 of the air mass to the equation in post # 386, we get the following results for pellet mass equals zero....

v = sqrt { 2 x P / 12 / (Da / 890575 / 3) } = sqrt (2 x 890575 x 3 / 12 ) x sqrt ( P / Da ) = sqrt (445288 ) x sqrt ( P / Da) = v = 667.3 sqrt ( P / Da ) with P in psi, Da in kg/m^3, and v in fps.... Here are the results for pressures up to 10,000 psi....



These results for zero pellet mass are interesting, because they show how much air diverges from an Ideal Gas at very high pressures.... the pressure (force) increasing faster than the density (mass).... Remember, the prediction for maximum velocity using the same equation (and 1/3 of the mass), but assuming air was an Ideal Gas, resulted in a velocity of 2257 fps, regardless of pressure.... If nothing else, I think the above numbers may be high enough that they will prove difficult to reach....

Bob

How about some higher numbers:

After a recent discussion on the yellow about de Laval nozzles and how the associated equations might apply to airgun, I decided to look at the Fanno flow equations a little more closely. When compared to the de Laval, the Fanno model looks a lot more like an airgun with a dump chamber. I'm looking for something that we might be able to use. There is no easy way to include pellet mass, but maybe it will give a maximum possible velocity of the air.

Here is the wiki link to some of the equations:
https://en.wikipedia.org/wiki/Fanno_flow



I used the velocity ratio equation V/V*, substituting SOS for V*, and used the k (gamma) values for air at different pressures. I used some very high Mach numbers in order to maximize the velocity ratios:



The unrealistic, very high mach numbers (>M10) indicate where the air had already been fully expanded, as velocity had mostly maxed out already.

Maybe these are the max velocities for air at 70F? They may just be a theoretical limit rather than what can actually be achieved.

Edit: for the equation used, as k approaches 1, max possible velocity approaches infinity.

[rsterne]

If you look over the Siegel equations, you will find that the maximum velocity is Mach 5.... This applies even to light gas guns, providing you are working at the correct temperature....

I think if you look at the equation again, when k = 1, the value becomes M, does it not?....

Bob

[Scotchmo]

If you look over the Siegel equations, you will find that the maximum velocity is Mach 5.... This applies even to light gas guns, providing you are working at the correct temperature....

I think if you look at the equation again, when k = 1, the value becomes M, does it not?....

Bob

The equation is not giving a Mach number, it gives a velocity ratio. When k approaches 1, the ratio becomes 1x. At M1, the ratio is always 1/1, for any value of k.

If I change k to 1, the air can expand indefinitely and continue to accelerate itself. At normal k values, the downstream Mach velocity falls faster than the Mach number grows. It runs out of "gas" after M10:

[rsterne]

OK, but using the equation, with k = 1, I get the following....

V/V* = M x { 1 / sqrt  [ ( 2 / k + 1 ) x ( 1 + [ ( k-1 ) / 2 ] x M^2 ) ] } = M x { 1 / sqrt [ ( 2 / 2 ) x ( 1 + [ 0 / 2 ] x M^2 ) ] } = M x { 1 / sqrt [ ( 1 ) x ( 1 ) ] }  = M x { sqrt 1 } = M ....

If you say the trend is to infinity, please show what I did wrong?....

BTW, here is the reference to Siegel's equation showing Mach 5 as the maximum for air (gamma = 1.4).... http://dspace.dsto.defence.gov.au/dspace/bitstream/1947/4048/1/DSTO-TR-1092%20PR.pdf .... the equation appears on page 12.... As gamma increases, the maximum Mach number decreases....

Bob

[rsterne]

I think I get it.... V/V* equals Mach if gamma = 1 (which is the proof I did above).... However, gamma can never drop to 1 unless you are adding energy to the system (Isothermal = 1).... and Mach can never exceed the value given by the Siegel equation, based on the gamma from column C in your table (k = 1.4 at STP).... With air, the lowest gamma is 1.4 (see below), which in Siegel's equation gives you a maximum of Mach 5....



In fact, if we want the output of Siegel's equation in Mach, it becomes even simpler.... 

Vmax = 2/(k-1) .... which yields Mach 5 when gamma = 1.4 (STP).... and at the maximum gamma in your table (@ 5000 psi, k = 1.745) it gives....

Vmax =  2 / ( k-1 ) = 2 / ( 1.745 - 1 ) = 2 / 0.745 = Mach 2.68 (at 5000 psi and 70*F)

Using Mach 2.68 as the limit in your equation gives....

V/V* = 2.68 x ( 1 / sqrt ( 2 / 2.745 ) x ( 1 + ( 0.745 / 2 ) x 2.68^2 ) ) = 2.68 x ( 1 / sqrt ( (0.729) x ( 1 + (0.373 x 7.18) ) ) = 2.68 x ( 1 / sqrt (0.729 x 3.68) ) = 2.68 x 1 / 1.64 = 1.63

which for V* = 1615 fps yields 2632 fps.... It would seem that if you first apply Siegel's equation (which is based on gamma, which is in turn based on pressure) to calculate the maximum Mach number, and then use that Mach number in your Fanno Flow equation, the result should be the absolute maximum velocity for that pressure, at 70*F....

Until we come up with a better idea.... *LOL*....

Bob

[Scotchmo]

Bob,
That's it.

If the equation for maximum Mach# given by Siegel is correct, you could use that with the Fanno flow equation and get the max velocity.

At 5000psi, 70F:

With no limit on Mach#, we get a 3100fps velocity limit.

With the Siegel Mach limit of M2.68, we get a 2632fps velocity limit.

[rsterne]

OK, I'm not saying if the method is sound or not, but just for discussion and visualization purposes, here are the results.... I used the raw data you sent me for the values of gamma, for every 500 psi increment, to calculate Siegel's maximum Mach number.... and then from that, using your Fanno Flow equation I calculated the maximum V/V* ratio (ie maximum Mach number) for Fanno Flow.... assuming Siegel is correct....



I then used the raw data I have for the Speed of Sound in air, at the same pressure increments, to convert the Fanno Mach numbers into the velocity in fps....



It might be interesting to extend these curves out to 10,000 psi, just out of curiosity, but I don't have the gamma values for that.... I note that the maximum velocity at 2000 psi is not all that fast, at just over 2200 fps.... I don't know if that is a problem or not.... but that's the results I got, anyway....

Bob

[Scotchmo]

Bob,
That last graph is the velocity allowed by the Siegel Mach limits.

The max Fanno velocity is higher at each pressure. The M100 column from my chart is high enough to flat line the velocity for each pressure.