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Energy stored in compressed air

Post questions and info about pneumatic (compressed gas) powered cannons here. This includes discussion about valves, pipe types, compressors, alternate gas setups, and anything else relevant.
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Energy stored in compressed air

Unread postAuthor: Biopyro » Tue Dec 30, 2008 10:27 am

I am doing a physics project with my copper QEV gun, and part of it involves determining where the energy for the projectile comes from.
I need to know the equation which relates the pressure and volume of air to the energy stored so that I can work out how efficient the transfer to kinetic energy is.
Any excuse to use a spud gun in college hey!
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Unread postAuthor: btrettel » Tue Dec 30, 2008 10:46 am

The ideal gas law combined with the work-pressure-volume and polytropic process relationships will give you a good idea of the total energy stored. Rather than derive these for you I'll just give you the equations. These are for ideal gases decompressed to atmospheric temperature.

For an isothermal process (constant temperature): E = P * V * ln(P/Patm)

For an adiabatic process (no heat transfer): E = (Patm * (P/Patm)^(1/k) * V - P * V)/(1 - k)

Where
P is the absolute chamber pressure
Patm is atmospheric pressure
V is the chamber volume
k is the ratio of specific heats (for air this is about 1.4)

The adiabatic process is more realistic, but the differences probably won't matter much.

I've found difficulty getting better than 40% efficiency in general, and better than 30% efficiency when minimizing the amount of energy used in total. Don't expect any spectacular energy efficiencies.
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Last edited by btrettel on Tue Dec 30, 2008 11:48 am, edited 2 times in total.

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Unread postAuthor: Biopyro » Tue Dec 30, 2008 11:03 am

The isothermal says that only .44J are stored at 1 bar? and adiabatic says -2.91J? I entered all the values in terms of dm^3 (meant to do m^3) and bar, so I assumed the answer would come out in near enough SI units?
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Unread postAuthor: btrettel » Tue Dec 30, 2008 11:26 am

The equation I posted for an adiabatic process had an error. I've fixed it.

I just did some quick math for 1 bar gauge pressure and a 3.9e-3 m^3 volume chamber (about 50 cm of 4 inch pipe). The isothermal process said 539.1 J and the adiabatic process said 349.7 J. This is what you would expect because to keep the temperature constant heat must be transfered, so the isothermal process will have more energy available.

The difference is larger than what I thought it would be though. Maybe it was the fairly large air chamber I used that made the difference.
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Unread postAuthor: Biopyro » Tue Dec 30, 2008 11:56 am

Hmmm i am still getting very small numbers.

=(1*(1.5/1)^(1/1.4)*0.00125-1.5*0.00125)/(1-1.4),
so a 1250cm^3 (2cm*100cm) chamber at a pressure of 0.5bar above atmospheric gives a stored energy value of 5.16E-004, far lower than the kinetic energy of 0.77J.

Also, GGDT seems to think that the volume of my chamber is 320cm^3, despite giving pretty accurate readings (as far as I can tell) for velocity. I can't think why??
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Unread postAuthor: Ragnarok » Tue Dec 30, 2008 12:04 pm

You shouldn't need to do any of that, because we don't need to be concerned about any thing other than the energy stored by the system pressure.

Consider the units of Pressure and Volume: Pressure is Pascals, which is Newtons per square metre, or N/m<sup>2</sup>. Volume is cubic metres, or m<sup>3</sup>
Multiply them together and you get Nm - Newton metres, or Joules, a unit of energy.

In other words, energy stored is simply pressure times volume. Just looking at the units of a quantity can tell you a huge amount about how to multiply them to get the result you want.

In slightly more convenient units than Pascals and cubic metres, it's (pressure in Bar * Volume in CC / 10)

With regards to the GGDT point, GGDT is correct, I suspect you've used the diameter in the equations rather than the radius.
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Unread postAuthor: Biopyro » Tue Dec 30, 2008 12:20 pm

I thought I might be able to to do that, but I used cubic metres, so assumed that it was wrong. Why only /10?
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Unread postAuthor: Ragnarok » Tue Dec 30, 2008 12:27 pm

Biopyro wrote:I thought I might be able to to do that, but I used cubic metres, so assumed that it was wrong. Why only /10?

Because there are 100,000 pascals to one bar, and there are 1,000,000 cc in one cubic metre.

Subbing cubic centimetres in for cubic metres requires the result to be divided by a million, and bar for Pascals requires it to be multiplied by 100,000 - cancelling, that leaves a factor of 10 to be divided by.
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Unread postAuthor: btrettel » Tue Dec 30, 2008 12:30 pm

I'm not completely sure what you're doing wrong Biopyro. Make sure the units work is all I can suggest at the moment.

Ragnarok's method would overestimate the total energy by a large amount when your chamber is below about 2.72 atma and underestimate when above that. Yes, the units are correct, but that doesn't mean the formula is correct.

If you are pumping up a gun, my equation would be the amount of work you're putting in (just negative because it is entering the system).

Edit: Corrected.
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Unread postAuthor: Biopyro » Tue Dec 30, 2008 12:34 pm

ah ha. I was using bar as the SI measurement instead of pascals. Thank you both very much for your help!

and yes, I was using the diameter. schoolboy error haha!
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Unread postAuthor: btrettel » Tue Dec 30, 2008 12:42 pm

For the area of a circle, I would highly suggest using pi*d^2/4 instead of pi*r^2. It's much more convenient, especially with mental math (for me at least).

Edit: One last thing: If you're comfortable with the adiabatic method, use that because it'd be much closer to reality than the isothermal method. It also probably would be a little more impressive.
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Unread postAuthor: Ragnarok » Tue Dec 30, 2008 1:13 pm

btrettel wrote:If you are pumping up a gun, this would be the amount of work you're putting in (just negative because it is entering the system

No, because it doesn't consider the energy that is lost due to adiabatic heating under pumping, which increases the work done during pumping, but after the heat is lost to the atmosphere isn't considered.

However, I assure you, that makes it exactly you want.

And indeed, if I take your equations and enter the stats for HEAL, I'd actually have to be getting >100% efficiencies to have achieved some of the muzzle energies I've recorded!
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Unread postAuthor: btrettel » Tue Dec 30, 2008 1:43 pm

Interesting point about adiabatic heating. The adiabatic equation does take that into account (edit: I earlier said neither equation had but I wasn't thinking). That statement was for the sake of illustration. That shouldn't change the total possible work the gas can do from whatever state it is pressurized to and starts at, which is what the derived equations I posted represent as far as I know.

I've been working slowly over the past week on a simulation tool like GGDT and I haven't seen any efficiencies over 50% using the adiabatic equation I posted as the reference, and that was when I was trying to improve efficiency ignoring everything else rather than having fixed muzzle velocities and finding the maximum efficiency for that configuration as I had done earlier. The highest efficiencies I've seen used a HEAR valve and were contrived--they wouldn't be practical. My model has agreed fairly closely with GGDT given the system doesn't accelerate anything over about Mach 0.3 and the pressures and temperatures are in the range valid for my constant specific heats, so I don't think it's a fault of the model. For this reason I'm having a hard time believing that you'd get greater than 100% efficiency. Perhaps if you posted some specific numbers I could take a look (though I could probably find them myself... let me see...).

Edit: You said the chamber was 120 cm and had a 26.3 mm ID. That makes the energy about 4.36 kJ for about 9.2% efficiency with an isothermal process.

For an isentropic/adiabatic process, the energy is about 2.07 kJ, making the efficiency about 19.4%. That's the difference the heat transfer makes and is a respectable efficiency from what I've looked at.

Again, am I missing something? From what I've seen, for a fixed volume chamber the efficiency doesn't vary greatly with pressure unless something else is changed.

Let me reiterate that when the pressure ratio between operating and atmospheric ressure is higher than 2.72 your energy equation will return energies lower than mine, so yours will underestimate the total energy. If mine would make impossible efficiencies at that state, yours would do worse.
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Unread postAuthor: Ragnarok » Tue Dec 30, 2008 2:28 pm

btrettel wrote:I've been working slowly over the past week on a simulation tool like GGDT.

I've been doing the same, but for a rather longer time.
I'm more interested in getting accurate results in the transonic region though.

For this reason I'm having a hard time believing that you'd get greater than 100% efficiency.

I think for the simple reason that it's impossible by the 2nd law of Thermodynamics, I also believe I would have a very hard time of managing it.

Okay, let's look at this from a different angle. If you're saying that the raw pressure times volume is the amount of work put in by pumping, where are you saying the energy is lost/gained compared to the results given by your own equations?

EDIT: Arseface, I've just noticed an error in my maths, I managed to hit the / key over the * key when using the numeric pad, and ended up dividing where I should have been multiplying, which explains some of what I've been seeing. Disregard the >100% efficiency stuff.
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Unread postAuthor: btrettel » Tue Dec 30, 2008 2:42 pm

I think for the simple reason that it's impossible by the 2nd law of Thermodynamics, I also believe I would have a very hard time of managing it.


Ah, yes, I was referring to your math, not any physical efficiency.

Okay, let's look at this from a different angle. If you're saying that the raw pressure times volume is the amount of work put in by pumping, where are you saying the energy is lost/gained compared to the results given by your own equations?


I'm not saying pressure times volume is the work put in by pumping, in fact, that's what I thought you were saying. All the equations I have posted do is tell you how much work an ideal process from one pressure and one volume to a lower pressure and higher volume will do.
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