Question about the "Valve Ineffectiveness Chart"....

Question about the "Valve Ineffectiveness Chart"....

Joined: February 17th, 2010, 3:30 am

July 5th, 2012, 5:18 pm #1

Quite a while back I found Steve in NC's threads on the "Sonic Horizon Theory" and his "Valve Ineffectiveness Chart"....



It occured to me while pondering this chart that it should be possible to have the left axis of the chart read the amount of "usuable HPA" which (I think) should be double the percent of barrel length used (as a percent of bore volume).... As an example, at the 850 fps shown, the pellet is 33% of the way down the bore when the valve becomes ineffective (ie no more useful air is released by the valve).... At that point, assuming the pressure in the bore behind the pellet and the valve is equal (which would, of course, require no restrictions and 100% efficiency), the volume of air that can do useful work (either accelerating the pellet, or pushing the air that CAN accelerate the pellet out of the valve) would be twice the volume of the bore behind the pellet, ie 66% of the bore volume.... If the valve was a closed system (ie no other air could enter it during the shot) then we should be able to calculate the MINIMUM volume of the valve required, in combination with the pressure of the air inside it.... ie the MINIMUM bar-cc of air required....

In the simplest example, if the valve was 33% of the bore volume, and started at 3000 psi, then that air would have expanded to 66% bore volume (half in the barrel, half still in the valve) at 1500 psi at the moment of valve ineffectiveness.... Other combinations are possible, however, the valve could be equal to the bore volume and start at 2250 psi and still produce 66% bore volume at 1500 psi at valve ineffectiveness, as an example.... In the limiting case, the valve/reservoir volume would be infinite, and the pressure would not drop during the time the valve was "effective" (eg. 1500 psi in the example given).... In this case, the bar-cc would be 1500/14.5 x 66% of the bore volume.... By using this limiting case, the volume would be twice the value on the left axis of the chart.... while the MINIMUM pressure could be calculated with just a few variables....

The major variables are:

1. The barrel length and caliber (and hence the bore volume)
2. The muzzle velocity (and hence the position of valve ineffectiveness)
3. The pellet weight (which determines the force, ie pressure x area, required to reach the MV in the length of the barrel)

There are some minor variables which could affect the results:

A. The transfer port volume (ie wasted volume between the valve seat and pelelt base)
B. The drag of the pellet in the bore (friction losses)
C. Restrictions to airflow (and overall efficiency)
D. The mass of the air in the bore (at the point of valve ineffectiveness?)

Item "A." is usually pretty insignificant, and Item "B." is relatively small compared to the force available (it might use up ~50 psi).... Item "C." can be ignored if we just want to calculate the MINIMUM air required (more air will be needed to overcome it), but Item "D." will become quite significant as the pressure and volume increase....

I throw this idea out there just for discussion.... is there something I'm missing?....

Bob

Last edited by rsterne on July 5th, 2012, 5:30 pm, edited 1 time in total.
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Joined: November 28th, 2002, 6:26 pm

July 5th, 2012, 8:44 pm #2

1. The basic idea behind "Sonic Horizon" was, due to the fundamental limit imposed by the speed of sound on the speed of propagation of pressure change, it makes no difference to the acceleration of the pellet whether the valve closes or remains open after the pellet crosses the "Horizon," because the pressure change created by the valve closing (or not) can't reach the pellet before it departs from the muzzle. In other words, the only gas that can contribute to accelerating the pellet in the example MV = 850fps case, is the gas occupying the first 33% of the bore. That is to say, 33% of the volume of the bore.

2. When air expands in an airgun, the expansion is so rapid that no significant heat has time to transfer from the walls of the bore etc. to the air. That is to say, the expansion is (to a good approximation) adiabatic. Under that condition, pressure isn't inversely proportional to volume, but (because air is effectively a diatomic gas) to volume raised to the 1.4 power. Therefore, for example, doubling volume wouldn't reduce pressure by factor of 2, but by 21.4 = 2.64.

Steve
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Joined: February 17th, 2010, 3:30 am

July 5th, 2012, 9:03 pm #3

to your first point, I understand that the only volume that can affect the pellet is the 33% IN the bore.... however, if the pressure in the bore and the valve was equal at that instant, wouldn't the total volume that was contributing to the movement of the pellet be (at a minimum) twice that 33% - ie 66% of the bore volume?.... I understand that the half that is still in the valve doesn't have any FURTHER affect (and in fact with perfect valve timing, could be captured).... but after all, it helped push out the half that IS doing the work, didn't it?.... What I'm trying to quantify is the TOTAL amount of air (minimum bar-cc) required to get the pellet to accelerate to (in this example) 850 fps....

To your second point, that could be worked into the pressure/volume relationship calculation, couldn't it?.... Even if you want to ignore the air still in the valve, you know the volume of air in the first third of the bore (33% of the bore volume), the muzzle velocity, the distance available for acceleration, and the weight of the pellet.... Shouldn't it be possible to calculate the MINIMUM pressure required?.... and hence the MINIMUM bar-cc?.... From that, we could determine the actual efficiency (in percent)....

Bob
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Joined: March 1st, 2002, 12:22 am

July 5th, 2012, 9:23 pm #4

1. The basic idea behind "Sonic Horizon" was, due to the fundamental limit imposed by the speed of sound on the speed of propagation of pressure change, it makes no difference to the acceleration of the pellet whether the valve closes or remains open after the pellet crosses the "Horizon," because the pressure change created by the valve closing (or not) can't reach the pellet before it departs from the muzzle. In other words, the only gas that can contribute to accelerating the pellet in the example MV = 850fps case, is the gas occupying the first 33% of the bore. That is to say, 33% of the volume of the bore.

2. When air expands in an airgun, the expansion is so rapid that no significant heat has time to transfer from the walls of the bore etc. to the air. That is to say, the expansion is (to a good approximation) adiabatic. Under that condition, pressure isn't inversely proportional to volume, but (because air is effectively a diatomic gas) to volume raised to the 1.4 power. Therefore, for example, doubling volume wouldn't reduce pressure by factor of 2, but by 21.4 = 2.64.

Steve
are you saying that only the energy contained in the compresed air is used, not the extra input that say a springer gets from the dieseling process?

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Joined: November 28th, 2002, 6:26 pm

July 5th, 2012, 9:52 pm #5

to your first point, I understand that the only volume that can affect the pellet is the 33% IN the bore.... however, if the pressure in the bore and the valve was equal at that instant, wouldn't the total volume that was contributing to the movement of the pellet be (at a minimum) twice that 33% - ie 66% of the bore volume?.... I understand that the half that is still in the valve doesn't have any FURTHER affect (and in fact with perfect valve timing, could be captured).... but after all, it helped push out the half that IS doing the work, didn't it?.... What I'm trying to quantify is the TOTAL amount of air (minimum bar-cc) required to get the pellet to accelerate to (in this example) 850 fps....

To your second point, that could be worked into the pressure/volume relationship calculation, couldn't it?.... Even if you want to ignore the air still in the valve, you know the volume of air in the first third of the bore (33% of the bore volume), the muzzle velocity, the distance available for acceleration, and the weight of the pellet.... Shouldn't it be possible to calculate the MINIMUM pressure required?.... and hence the MINIMUM bar-cc?.... From that, we could determine the actual efficiency (in percent)....

Bob
For example, consider your example where the reservoir volume behind the valve is infinite. Is it your reasoning that, in that case, since an infinite amount of air "contributes" to maintaining the pressure and accelerating the pellet until the pellet crosses the Horizon, that an infinite amount of air is "required?"

In any case, it's that ambiguity that accounts for my preference - which is to account only the air actually expended with the shot.

As to your second point, I agree.

Steve
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Joined: November 28th, 2002, 6:26 pm

July 5th, 2012, 10:04 pm #6

are you saying that only the energy contained in the compresed air is used, not the extra input that say a springer gets from the dieseling process?

dr_subsonic's pneumatic research lab

the Lunatic Fringe of American Airgunning
Southwest Montana's headquarters for Airgunning Supremacy
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1. The number of molecules in the volume of the air in the charge.
2. Its initial temperature.

...there's a popular number that's totally absent and irrelevant: Pressure.

In other words, although initial pressure (of course) influences how efficiently energy can be extracted from the charge and translated to muzzle energy, it has no influence on the total energy available, which sets an absolute limit on the energy that can theoretically extracted by adiabatic expansion from each bar-cc of air. About 0.18 fpe.

No matter how high the pressure.

Steve
Last edited by pneuguy on July 5th, 2012, 10:23 pm, edited 1 time in total.
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Joined: March 1st, 2002, 12:22 am

July 5th, 2012, 10:21 pm #7

comes from the density of charge wanted to return to "normal" atmospheric levels?

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Joined: February 17th, 2010, 3:30 am

July 5th, 2012, 10:27 pm #8

For example, consider your example where the reservoir volume behind the valve is infinite. Is it your reasoning that, in that case, since an infinite amount of air "contributes" to maintaining the pressure and accelerating the pellet until the pellet crosses the Horizon, that an infinite amount of air is "required?"

In any case, it's that ambiguity that accounts for my preference - which is to account only the air actually expended with the shot.

As to your second point, I agree.

Steve
and it's likely because I'm not explaining it well.... In the case of the infinite reservoir, there would be no pressure change, with the reservoir providing the air to push out the 33% at 1500 psi needed to do the work of accelerating the pellet.... In the case of the 33% valve at 3000 psi, half the air forces out the other half, with a 50% pressure change.... In both cases, the result is the same.... 33% at 1500 psi to drive the pellet.... You could also start with a 3% valve at 18,000 psi, and end up with 36% (3% in the valve + 33% in the bore) at 1500 psi.... Or, you could start with a 300% valve at 1665 psi, and end up with 333% (300% valve + 33% bore) at 1500 psi....

It appears you aren't interested (or think I'm wayyyyyyyyy off base), so I'll drop it....

Bob
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Joined: November 28th, 2002, 6:26 pm

July 5th, 2012, 10:30 pm #9

comes from the density of charge wanted to return to "normal" atmospheric levels?

dr_subsonic's pneumatic research lab

the Lunatic Fringe of American Airgunning
Southwest Montana's headquarters for Airgunning Supremacy
Proud Sponsor of team_subsonic
...(i.e., the number of molecules = mass) and its temperature.

For example, for 1 bar-cc of air at room temperature (25oC) = 2.4 x 1019 molecules = 1.2mg, its 0.18fpe.

Steve
Last edited by pneuguy on July 5th, 2012, 10:31 pm, edited 1 time in total.
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Joined: March 1st, 2002, 12:22 am

July 5th, 2012, 10:38 pm #10


I'd need 10 1bar-cc's to get it (in reality more, probably do to system loss/drag and other inefficiencies)?


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Southwest Montana's headquarters for Airgunning Supremacy
Proud Sponsor of team_subsonic
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