PVT Pneumatics question

This somewhat relates to pisa and psig. Suppose I have a 100 in^3 air cylinder full of air with a psig of 0. Now I push the rod in halfway, the cylinder now has 50 in^3 of air and assuming the temperature doesn't change the pressure should be double, but 2 * 0 is still 0. Obviously the atmospheric pressure (14.7 pis) comes in to play here. So the psia should be twice the normal atmospheric pressure or 29.4 psi but the gage will read what the atmospheric pressure is 14.7 right? What about if we start with a pisg of 100 then psia is 114.7 so when we double the psia is 229.4 but the psig is 214.7? And it is psig that is what determines how much force is required to hold an air cylinder at a certain point right?

So am I right here or am I completely screwed up?

Reply to
Chris W
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I have always assumed that any meter always read lbs above atmosphere, or Pascal's for that matter, but since you asked the question doubt begins to creep slowly in!!

Also looking for the old text books.

-- Cheers Adrian.

Reply to
Adrian Hodgson

PSIG is the pressure above atmosphere. PSIA is the absolute pressure, ( 0 at full vacume )

For working out pressure changes with change in volume you must work in absolute pressure if you want to use the formula

P1 x V1 = P2 x V2

whare P1 and V1 are the inital pressure and volume and P2 and V2 the final value ( no leaks, tempreture changes... )

For gage pressure it's ( P1 + 14.7 ) x V1 = ( P2 + 14.7 ) x V2.


As to the force a piston generates, remember every piston has two sides..

If it's area A, and it's open to atmosphere at the back, then

Force = A x PSIG.

Force = A x ( PSIA - 14.7 )

With a double acting piston it's

Force = A x ( PSI 1 - PSI 2 )

Just make sure you don't mix gage and absolute pressures, but the formula works for either.

If there is a rod on one side of a double acting cylinder you may need to correct for it's area.

Hope this helps

-- Jonathan

Barnes's theorem; for every foolproof device there is a fool greater than the proof.

To reply remove AT

Reply to
Jonathan Barnes

Always convert known gauge pressures to absolute and then work everything in absolute pressure AND temperature and convert to gauge pressure at the end when you have finished calculations . Temperature WILL change formula is P1xV1/T1 = P2xV2/T2 ChrisR

Reply to

I had this confusion just last week. I had to do a hydro-test on a tank and the drawing had the test pressure at 13.7 psia. I went to the engineering dept. and asked what the @#$%^ was this all about and found out that it was the external design pressure and that the actual test pressure was 3.5 psig.

Reply to

Thanks for all the replies. Some one mentioned pressure so let me as a question regarding that. I have been ignoring temperature so far hoping that it won't play a significant role in my application. So here it is. I will have about 33.58 In^3 of nitrogen gas at 595.6 psi and 80 degrees Fahrenheit will then compress the gas to about 7.07 In^3 ignoring temperature that gives me 2829.4 psi. This will happen in about 1 second. How much will the temperature go up and will this have a significant effect on the PV value. The volume of gas will always be fluctuating between these 2 values, most of the time it will probably be more like in the middle 50% of that range.

Chris W

Chris W wrote:

Reply to
Chris W


An easy way to look at it is, you added one atmosphere to Gauge pressure to get PSIA so after you compress it no matter how many times, 2, 4, 6, 10, you must subtract one atmosphere to get back to PSIG. No magic, pure math.

Reply to
Bud Trinkel

Your formula assumes PV = MT = constant.

this is true only if no work is done... ( constant energy ).

If your talking about compressing a gas, then energy is required, and things get complicated. ( entropy and enthalpy need to be considered ).

Reply to
Jonathan Barnes

Ouch.... your going to find about half the Greek alphabet in the formulas you need for this one :-(

There will be a lot of heat in this process. if it's a repeat cycle the whole lot will rapidly heat up, so your start point for each cycle will change....

These are the sort of figures that might relate to the gas shock absorbers on a large lorry, what are you doing ???


Reply to
Jonathan Barnes

Funny you should mention shook absorbers. That is sort of what I am doing. I am using a hydraulic cylinder that will be plumed into an accumulator to act as both a spring and shock. The reason I am doing it this way is to have more control over the suspension for rock climbing and other wild extreme road situations. Just as an example one of the things I intend on doing is having a mode where if one wheel is going over a large obstacle what you would like to happen is for that wheel to go up and the opposite wheel to go down. By opening a valve I can link the piston end of both left and right cylinders together so when one goes up the other will be pushed down. They will also be hooked to the accumulator for spring action. There are a few other things I plan to do with this setup, like a hydraulic anti sway system, ride height, and being able to independently change the ride height of each wheel.

Reply to
Chris W

Reply to
Cliff Ray

In message , Chris W writes

There is little new under the Sun. Substitute 'following' for 'opposite' and refer to the coupled wheels of a steam locomotive, now link the spring suspension points mechanically and you have what was called 'compensated springing'!

Reply to
Mike Hopkins

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