# Gas compression and Interaction with Fluids

1. Aug 31, 2009

### FishBlackpool

I have been trying to find out how to calculate a number of things in a system involving liquid and gas.

In the attached diagram there is a U-bend with gravity acting down on it. If one side is sealed with a plunger and the other side is filled with a fluid (L) we would end up with the trapped gas (G). Assuming the plunger remained stationary (held in place). I would like to calculate:-

1) The Pressure of G and also the volume by which it would be compressed. (I presume this will involve the densities of L and G and the mass of X?)

2) The force required to push the plunger down and move the gas and liquid around the tube. (would this just be the force required to lift L?)

My degree is in Zoology!! and I really have trouble remembering my GCSE Physics so any help would be FANTASTIC, even if its just directed reading material.

Thanks

PS. Not sure if this post belongs in the mechanics section

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2. Aug 31, 2009

### Q_Goest

Hi fish'. Welcome to the board. Let's assume the gas density is small compared to the liquid. The pressure in a column of any fluid (gas or liquid) is equal to the density times gravity times the height. See Bernoulli's equation.

For your system, there's a column of liquid with a height of X. So the difference in pressure between the two levels is:
dP = rho * g * h
where rho = liquid density
g = acceleration due to gravity
h = height of liquid (X)

The absolute pressure of the gas is the ambient pressure plus the additional pressure dP.

The force on the plug is equal to the pressure times the sealed area.

3. Sep 2, 2009

### FishBlackpool

Wow thanks for the quick resonse Q_Goest!

So if I could just run this past you and make sure I'm doing this right?

If we use the same example U-Bend, make it at sea level
liquid L = pure water at 20°C
Gas G = standard air
X = 1.5 metres
Plug Surface Area = 0.01m3

using dp= rho x g x h

dp = 998.2 kg/m3 (pure water density at 20°C) x 9.81m/s2 x 1.5m

dp = 14.685 KPa

and if I add this to the standard atmospheric air pressure at sea level (101.325KPa)
I should get the absolute pressure
=116.01KPa

Also can I ask, I think the pressure used to calculate the force on the plug is the dp not the absolute pressure, is that right? so Force = 0.01m3 x 14.685KPa

I hope this is how to do it!! thanks for the help

PS. Please can you help me calculate the compression of the air as it changes from atmospheric air pressure at sea level to the new pressure?

4. Sep 2, 2009

### Q_Goest

hi fish'.
Yes, you did all that correctly, though I didn't check the math.

Not sure what you mean by that. You've already calculated the pressure. Perhaps you mean the volume? You can determine the volume (or density) using the http://en.wikipedia.org/wiki/Ideal_gas_law" [Broken]. If the compression is very rapid, the air could be heated slightly, so it may be at an elevated temperature. Otherwise, I'd just assume the temperature is ambient.

Last edited by a moderator: May 4, 2017
5. Sep 20, 2009

### FishBlackpool

Hi

Thanks for the response, yes I did mean volume...as in, calculating the volume that the gas will decrease to as the pressure increases!

I'm going to try and test these calculations in a real experiment so instead of messing about with monatomic gas in a closed system (which I believe the ideal gas equation requires) I need a formula that can handle normal air, so I'll post on the chemistry bit of this site asking for that formula. I'll call the post 'Un-ideal Gas Equation' just incase your knowledge can astound me again!!

Thanks for all the help
Fish

6. Sep 20, 2009

### Q_Goest

Hi Fish,
The ideal gas law doesn't require a monatomic gas. It can handle air to a very high degree of accuracy. The ideal gas law only starts to break down when gas density becomes very high such as at very high pressure or very low temperature. Your situation won't come close to seeing density that high.

You might try an online calculator like this one:
http://www.ajdesigner.com/idealgas/index.php

This one looks pretty decent because you can also change the variable you're solving for and even put in a density instead of a number of moles, which should make it easier.

If you have trouble understanding how to use the calculator, or if you just don't understand how to apply the ideal gas law to this, just shout.

7. Sep 20, 2009

### Q_Goest

I see Wikipedia says the ideal gas law is most accurate for a monotomic gas, but that's a sweeping gneralization, if not downright incorrect.

From a fluids properties database I use: At 70 F, air remains within 1% of being ideal up to 450 psi (Z=0.99). Argon only to about 200 psi (Z=0.99) and helium to about 300 psi (Z=1.01). Nitrogen on the other hand, seems to be ideal up to about 1800 psi (Z=1.01). Oxygen isn't as good, at about 200 psi Z=0.99

Anyway, your pressure is no doubt much lower than that so I wouldn't worry about it.

8. Sep 20, 2009

### alxm

It's almost correct. What it really should say is 'noble gas', since no other monoatomic gases are likely to stay that way in ordinary conditions. Noble gases have near-ideal behavior because the ideal gas law neglects intermolecular forces and in the case of noble gases those are at a minimum, being limited to van der Waals forces.

A nit-pick for the sake of completeness: This all assumes that the gas is immiscible in the fluid, which is a fair enough approximation but worth knowing about. Increasing the air pressure above water will increase the amount of air dissolved in the water.