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Old 11-24-2012, 02:17 PM   #11 (permalink)
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hmm.

If you compressed a 20,400ci piston/cylinder full of free air to 68ci, the pressure increases to 4500psi. Since the piston area remains fixed (lets say one square inch) the applied force increases exponentially with distance traveled. If you graphed this force versus distance, the area under the curve would be the work performed and therefore the energy stored in the 68/4500.

What happens when you change volume and/or pressure?
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Old 11-25-2012, 03:27 PM   #12 (permalink)
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hmm.

If you compressed a 20,400ci piston/cylinder full of free air to 68ci, the pressure increases to 4500psi. Since the piston area remains fixed (lets say one square inch) the applied force increases exponentially with distance traveled. If you graphed this force versus distance, the area under the curve would be the work performed and therefore the energy stored in the 68/4500.

What happens when you change volume and/or pressure?
Well since no one wants to play along...

Going from 2000psi to 4000psi would give you 228% more shots assuming your gun was equally efficient at all tank pressures.....

Last edited by P0E; 11-25-2012 at 04:15 PM.
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Old 11-25-2012, 07:36 PM   #13 (permalink)
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Going from 2000psi to 4000psi would give you 228% more shots assuming your gun was equally efficient at all tank pressures.....
How would that give me 228% more shots assuming my gun was equally efficient at all tank pressures?....

If a gun shoots off 576 rounds at 2000 PSI, and shoots off roughly 1,150 rounds at 4000 PSI (I got ~1,090 after I filled the tank up to 4k), that's 100% more shots.
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Old 11-25-2012, 08:54 PM   #14 (permalink)
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How would that give me 228% more shots assuming my gun was equally efficient at all tank pressures?....

If a gun shoots off 576 rounds at 2000 PSI, and shoots off roughly 1,150 rounds at 4000 PSI (I got ~1,090 after I filled the tank up to 4k), that's 100% more shots.
the extra 28% comes from having the magic box attached to an expansion chamber. you also need to have tom kaye's shoebox compressor hooked to the breech to maintain pressure while the gas is expanding in the barrel.
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Old 11-25-2012, 09:36 PM   #15 (permalink)
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You must do all pressure volume calculations in aboslute units of measurements, guage pressures are not aboslute.
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Old 11-25-2012, 09:59 PM   #16 (permalink)
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You must do all pressure volume calculations in aboslute units of measurements, guage pressures are not aboslute.
Hmm... Well. The best I could do was throw one of these on the tank.
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Old 11-26-2012, 10:40 AM   #17 (permalink)
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How would that give me 228% more shots assuming my gun was equally efficient at all tank pressures?....

If a gun shoots off 576 rounds at 2000 PSI, and shoots off roughly 1,150 rounds at 4000 PSI (I got ~1,090 after I filled the tank up to 4k), that's 100% more shots.
heinous, I don't understand how an inefficient, hastily designed to be cheaply made compressor would increase system efficiency at all.

Energy is not simply pressure times volume. If that was the case, a fully expanded tank would have the same amount of energy as when it was compressed.

To calculate energy (assuming isothermal expansion) you need to also multiply by the natural log of the pressure ratio.

E = P1*V* ln(P1/P2)

Although P1 might increase by 2x in your example, the other factor (ln (P1/P2)) increases by 14%.

It is derived from the example I started to outline several posts back. Although I tried to exclude the math bits.

oh.. P.S.
E is energy in Joules
P1 is your starting pressure in Pascals
V is volume in cubic meters
P2 is your pressure 'floor'. In this case atmospheric pressure, roughly 14.7psi.

Last edited by P0E; 11-26-2012 at 10:44 AM.
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Old 11-26-2012, 11:38 AM   #18 (permalink)
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Quote:
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heinous, I don't understand how an inefficient, hastily designed to be cheaply made compressor would increase system efficiency at all.

Energy is not simply pressure times volume. If that was the case, a fully expanded tank would have the same amount of energy as when it was compressed.

To calculate energy (assuming isothermal expansion) you need to also multiply by the natural log of the pressure ratio.

E = P1*V* ln(P1/P2)

Although P1 might increase by 2x in your example, the other factor (ln (P1/P2)) increases by 14%.

It is derived from the example I started to outline several posts back. Although I tried to exclude the math bits.

oh.. P.S.
E is energy in Joules
P1 is your starting pressure in Pascals
V is volume in cubic meters
P2 is your pressure 'floor'. In this case atmospheric pressure, roughly 14.7psi.
It is probably closer to an adiabatic expansion not isothermal, thats why tanks and barrels get cold when fired. If everything was heated than maybe it would be isothermal, but the process is faster than the heat transfer is in this situation.

Adiabatic expansion is generally less efficient than isothermal expansion.

And also, unfortunately, the compressability of HPA goes down as pressure goes up, so n(or the number of moles) is not a constant ratio of PV/RT. Higher pressures are less "efficient" at holding air, by maybe 10%.
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Old 11-26-2012, 12:13 PM   #19 (permalink)
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It is probably closer to an adiabatic expansion not isothermal, thats why tanks and barrels get cold when fired. If everything was heated than maybe it would be isothermal, but the process is faster than the heat transfer is in this situation.

Adiabatic expansion is generally less efficient than isothermal expansion.

And also, unfortunately, the compressability of HPA goes down as pressure goes up, so n(or the number of moles) is not a constant ratio of PV/RT. Higher pressures are less "efficient" at holding air, by maybe 10%.
The gas released in the breech is fast enough to be considered adiabatic, but not the gas in the tank if released slowly. In this application we are looking at the maximum energy in the tank and not considering how much less energy we would get if we emptied the tank at a faster rate.

Even with an non-ideal gas, we can roughly assume isothermal in this application since we would ideally be allowing the gas to expand over days. Temperature remains constant because the air in the tank absorbs heat from the atmosphere as it slowly expands. This heat was stored here during/after compression.

I'm eager to see how you would calculate this problem and your result.
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Old 11-26-2012, 12:27 PM   #20 (permalink)
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