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C/B ratios and diminishing returns

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: clide » Thu Jan 24, 2008 6:01 pm

It took me a while to get my head around the whole "If P*V is constant, how does compressing it add energy?". Thing is, it doesn't add energy to the air. The air still has the same amount of energy in it, but the thing is, you've created a "pressure gradient" for it to flow across, and when you talk about the "energy in the chamber", you talk about the amount of energy that can flow across that gradient.

Arg, it does add energy to the air. Energy is the capacity to do work. The gas in that piston cylinder arrangement has the capacity to do (approximately) the same amount of work that you put into it when you are compressing it. After you push in the cylinder it can apply a force over a distance and get back to its original state, so it MUST have more energy in its compressed state because it can do work and return to its original state.

Ragnarok wrote:The distance the piston needs to move will be the initial gas volume / piston area, and the force the piston needs to exert will be the integral of the pressure * piston area.

Because those are multiplied together, the area terms cancel.
As pressure * volume is a constant value here (assuming no leaks, and no heating), the integration can also be omitted.

And that simplifies down to pressure * volume again.

The work done is the integral of pressure times dV since you are summing the force times distance over tiny divisions of distance. So to pull a constant PV out of the integral you need to first multiply it by V/V. Then you can pull the PV out and you are left with PV*integral of dV/V which will give you an equation for work that is P1*V1*ln(V2/V1)

That is what you will find for the equation for isothermal compression work in a thermodynamics textbook. Of course that equation is the work done by the gas so it will be negative for compression.
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Last edited by clide on Thu Jan 24, 2008 6:10 pm, edited 1 time in total.
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Unread postAuthor: DYI » Thu Jan 24, 2008 6:08 pm

Dewey, good job finding my old C:B ratio topic. At the point I made it, there wasn't really anything quite like it that I could find, so I posted what I had found. It was probably unnecessary, and I got flamed a bit for it. Probably moreso than you for starting this topic, actually.

But it seems to have amounted to something useful, so no harm done.
And I had never known that the efficiency box in the GGDT was broken, and had never gone to the trouble of doing the math to check it.
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Unread postAuthor: Hotwired » Thu Jan 24, 2008 8:02 pm

Ah springs!

I thought of posting the spring analogy before but I was derailed by some other formulas and forgot, clides already done one but dammit I'm not wasting a good analogy after so long just staring at 100cc @ 2bar apparently equalling 10cc @ 20 bar and wondering how the devil...

A gas piston arranged to act like a spring by that equation has the same energy either compressed or relaxed but you know that when compressed it can output a force before returning to its relaxed state.

But still there are the same moles of gas and so by E=PV there is still no energy difference between either state.

I think we need to go back to clides first post, do what D_Hall did and find energy put into compressing the gas, which as has been said should pretty much equal the energy coming back out again. Minus a load of things on the way.
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