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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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Unread postAuthor: Ragnarok » Tue Dec 30, 2008 3:00 pm

btrettel wrote:I'm not saying pressure times volume is the work put in by pumping, in fact, that's what I thought you were saying.

Hang on... in your own words: "If you are pumping up a gun, this would be the amount of work you're putting in". I said it was the actual "pressure energy".

I'm not saying that all that energy can be put to use, indeed, few cannons, even under ideal conditions could put more than about half that into a projectile, and it would be completely impossible to get 100% of it out. However, it is the energy actually stored by the pressure.
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Unread postAuthor: btrettel » Tue Dec 30, 2008 3:04 pm

What I meant was not what I said. I believe I had deleted a paragraph but left the last sentence by accident. Reverse the limits on the integral to figure out pump work. More accuracy can be gained with an isentropic efficiency, directly including friction, etc.

Efficiency is desired output divided by required input. The required input here is what energy is put into the system minus whatever is saved. The only systems that save energy here are those with HEAR type valves, but if the gun is pulling a vacuum, those valves can actually be less efficient.

The energy put in the system is definitely not P*V and neither is some sort of "absolute energy". The work-energy relationship says the work (negative when going in) is the integral of P*dV, so perhaps this is where you are confused. The relationship between volume and pressure can be described with a polytropic process as I mentioned earlier. Integrate that and you get the equations I posted.

If you want a more accurate measure of the energy of a pressure chamber, use thermodynamic tables or thermodynamic polynomials to figure out the difference in internal energy between the two states. It'll all be approximately the same depending on what assumptions you make.
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Last edited by btrettel on Tue Dec 30, 2008 4:02 pm, edited 1 time in total.

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Unread postAuthor: jimmy101 » Tue Dec 30, 2008 3:56 pm

You might want to download and fiddle with GasEq. It'll calculate all the thermodynamic values for compressions and expansions.
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Unread postAuthor: Biopyro » Wed Feb 25, 2009 4:55 am

I thought I had it, but looks like I didnt :'(. I was still using the wrong volume and 1bar, can't remember why because it was a while ago now. I fixed that and now the adiabatic method gives me approximately 10x the kinetic energy, and the simple method give values which are about 3x higher (34% efficiency). It's the adiabatic method I'm not sure about as the numbers are so large (and positive?).

This is the spreadsheet on google documents http://spreadsheets.google.com/ccc?key=pHtQQ83ZWFP57VeyuzfecHQ

Also, I tried GasEq and while it looked useful I couldn't in anyway find out how to make it work any of these eqautions. It seems more geared towards chemical reactions.
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Unread postAuthor: btrettel » Wed Feb 25, 2009 7:37 am

If by the simple method you mean Rag's method, don't use that because it doesn't actually represent anything. It returns the correct units but it's meaningless because it's not derived from anything and how the gas is compressed matters.

Your adiabatic equation is incorrect. You neglected a set of parenthesis.
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Unread postAuthor: Biopyro » Wed Feb 25, 2009 11:30 am

AHA! thankyou! It wasn't missing a set so much as one was misplaced. now the values are much more sensible, and almost identical to rag's method.
By it's not derived from anything, do you mean that there are no fundamental equations which you can derive it from? Just trying to understand the advantage of that so that I can put it in words, because if I understand that eqaution it will definately show that I know what I'm doing for the marks.
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Unread postAuthor: Ragnarok » Wed Feb 25, 2009 11:45 am

Hang on, "Not derived from anything?"

It can be derived from the ideal gas equation and the steady flow energy equation.
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Unread postAuthor: btrettel » Wed Feb 25, 2009 1:20 pm

I'd be interested in seeing that derivation Rag. How is using a flow energy equation when you're just measuring energy required to go from one state to another under certain conditions (adiabatic, isothermal, etc.) appropriate? Maybe I'm missing something.

A quasistatic energy balance would be more appropriate if you were to take that angle. But you'd have to use internal energy, which was absent in any form (including specific heats) from your equation. And I already mentioned you could look at the difference between internal energies earlier, which is the procedure you'd have to take for this angle.

Using the polytropic process relationships with the work-pressure-volume relationship make the most sense because the gas is going from one state to another and you can specify how it goes from one state to another.

Biopyro, I'd suggest reading the links I gave when I first posted the equation. They'll explain precisely what you need to know. To summarize, you solve the adiabatic polytropic equation for pressure and integrate it from one volume to another. It's not that difficult and if you want a more detailed explanation I'll post something tonight (I'm busy at the moment).
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Last edited by btrettel on Wed Feb 25, 2009 2:18 pm, edited 1 time in total.
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Unread postAuthor: Biopyro » Wed Feb 25, 2009 1:36 pm

I think that might be a little above the level I need, but thanks anyway, your help has been really valuable.
Thanks rag also, I should be using both equations, but it's not exactly a doctorate thesis so often emphasis is more on essay skills than actual content :(
thanks again :D
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Unread postAuthor: jimmy101 » Wed Feb 25, 2009 1:45 pm

Ragnarok wrote: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.

Based on that line of reasoning the kinetic energy of a projectile would be mv<sup>2</sup>. :?

The units must work for the formula to be correct. However, the correct units do not prove the formula is correct.
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Unread postAuthor: Biopyro » Wed Feb 25, 2009 4:26 pm

Also, can anyone shed any light on the very low efficiency at low pressures (5%)?
This is what I wrote
The second (and third) refer to the very low energy of the tests for 1 and 0.5bars. It seems to be a result of the low pressure as all the repeats are the same, however I don't think it can be attributed to suck-back, as the volume of air expelled from the reservoir is larger than the barrel volume, even at 0.5 bars. It could be that the lower velocity is causing it to roll through the barrel, increasing friction, or it could be that at higher pressures, air streaming past the bearing creates a cushion in the small amount of space between the barrel and the bearing meaning that friction is greatly reduced.
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Unread postAuthor: Ragnarok » Wed Feb 25, 2009 4:49 pm

jimmy101 wrote:The units must work for the formula to be correct. However, the correct units do not prove the formula is correct.

You are mistaking my simplified explanation for being my method.

My complaint is still that the stored energy in a gas should not have a relation to the atmospheric pressure like in Ben's equations. For one thing, it causes ridiculous amounts of energy to appear out of nowhere should the atmosphere be a low pressure vacuum instead of the normal.
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Unread postAuthor: Biopyro » Wed Feb 25, 2009 4:54 pm

Surely it's because the atmospheric pressure is what's being compressed in the first place. I think Ben's equation takes into account heat given to the atmosphere in some way, but my understanding here isn't exactly total.
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Unread postAuthor: Ragnarok » Wed Feb 25, 2009 5:06 pm

Biopyro wrote:Surely it's because the atmospheric pressure is what's being compressed in the first place.

That shouldn't affect the stored energy.

It might have a bearing on the energy used to compress said air, but that is not the same thing as the stored energy - which is the relevant factor here.

I have textbooks that state the stored energy in pressure to be P * V. I would very much hope the money spent on buying them has not gone to such a waste as that they would get something like that wrong.
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Unread postAuthor: btrettel » Wed Feb 25, 2009 5:35 pm

I believe I said earlier that P * V might be some sort of "absolute" energy, but it's definitely not what we're looking for. That's the energy required to go from one state to another. It's also the only energy that makes sense if you are looking at efficiency (as Biopyro is) because efficiency is energy out/energy in. Atmospheric pressure is the equilibrium pressure here so things are based on that.

You might be confused about the P * V term in the equation H = U + P * V. Enthalpy and internal energy differ in that term.

P * deltaV is the work done by an isobaric process. So it makes sense with that assumption, but pumps are not isobaric. I believe the enthalpy term comes from this assumption, though I'm not sure.

Anyway, I'd be interested in seeing a copy of what your textbook states.
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