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Calculating pressure

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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Calculating pressure

Unread postAuthor: shud_b_rite » Fri Feb 02, 2007 3:31 am

Im not sure if i posted this in the right section but moderators are welcome to move it.

Say I have a bike pump that holds 100cc (cubic centimeters) of air. If I were to press the handle down halfway so that the volume becomes 50cc and the pressure inside becomes 10psi (example only). Then I press the handle down even further so that the volume becomes 25cc. What would the pressure now be?

My guess is that it would be 20psi because you are taking 50cc of air at 10psi, and packing it into a space that is half the size as it was before so you are doubling the psi. Is this correct? If anyone knows could they please explain it?

Dont say that I havnt researched this because I have. I am trying to figure out how much psi my bike pump goes up to because I somehow broke my pressure guage.

Any help would be great
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Unread postAuthor: MrCrowley » Fri Feb 02, 2007 4:42 am

I remember reading somewhere that isn't the case....it was a long time back though, Someone was saying if they pumped 5.7psi(example) per pump it would take them 10 pumps to get to 57psi. Which isn't the case as the pressure inside the chamber grows and it becomes harder to pump as you go. There is no set pressure per pump.
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Unread postAuthor: Hotwired » Fri Feb 02, 2007 5:18 am

It won't work trying to guesstimate the pressure like that, your guesses could be way off (probably under but still). The safest way to get to a high pressure without a gauge is to have known how many full strokes it took to get there with a gauge....
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Unread postAuthor: joannaardway » Fri Feb 02, 2007 5:54 am

There is a set pressure per pump (assuming no leakage), but I wouldn't use that to calculate how many pumps it would take.

It gets harder to pump, but the pressure still increases by the same amount per pump.

Take the ideal gas equation:

pV = nRt

P is pressure, v is volume, n is number of moles of gas (proportional to the number of molecules of gas), R is the gas constant, and t is the absolute temperature.

V,R and t are considered constant for this. (In truth, t will rise slightly, but the effect is small).

To increase the amount of gas in the chamber by a set amount, you add n moles of gas with a pump. This quantity is the same for each pump, as the pump is always the same volume, and at the same pressure (14.7 psi from the atmosphere) - so the same number of molecules are present.

Given that v, r, and t aren't changing, the only value that changes is the pressure. As n is changing by the same amount each time, P must do the same in exact proportion.

So there is a set pressure per pump, but taking into account leakage, then your actual pressure will be different to your calculated pressure.
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Unread postAuthor: MrCrowley » Fri Feb 02, 2007 5:57 am

Yeah,it was something along those lines, i'll try and dig up that old topic.

Edit: Found it: http://www.spudfiles.com/forums/viewtopic.php?t=2495&
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Unread postAuthor: shud_b_rite » Sat Feb 03, 2007 4:50 pm

Oh thanks a lot for that link. I tried searching this forum using the search form but it didnt return anything useful like that.
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Unread postAuthor: boilingleadbath » Sat Feb 03, 2007 8:52 pm

Not quite true, joanna; the pump has a miminal volume, so the efficiency (and therefore introduced quantity) decreases as the pressure increases.

Anyway, pressure (absolute) goes as:
14.7/(fractional volume)
The gauge pressure (what a gauge will read) will be 14.7 psi lower. (at sea level)

So, if you compressed it to 1/2 of the piston tube, it'd be at 14.7/(1/2)
...or, if you compressed it to 1/4 of the piston tube's volume, it'd be 14.7/(1/4)

Respectively, these give you roughly 30 PSIA and 60 PSIA - but note that you are probably interested in PSIG (what a gauge will read).
In which case the answers would be ~15 psig and ~45 psig.

I don't know how useful this methology is, but I figured I'd explain it anyways.
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