Confused about water/compressed air

[Neo_Anderson]

This is ridiculous. The "air's elastic nature" (i.e. compressibility) has nothing to do with the air pressure being higher in a static fluid. If two fluids are in static equilibrium together, their pressures are exactly the same at the interface (i.e. same elevation). I refer you again to the example given above about the cylinder/piston.

CS

We're not talking about two [different?] fluids, we're talking about a fluid and a gas in a solid, hollow structure.
The only way the OP could have gotten a very high psig reaading is if the gravitational pull on the liquid caused the high psig reading (every time he measures these things).
It's impossible to proffer an alternate explination unless there are water pumps or other hidden variables involved.

 

[sophiecentaur]

A gas IS a fluid. So is a liquid.
They will both FLOW - i.e. they are fluid.

 

[stewartcs]

We're not talking about two [different?] fluids, we're talking about a fluid and a gas in a solid, hollow structure.
The only way the OP could have gotten a very high psig reaading is if the gravitational pull on the liquid caused the high psig reading (every time he measures these things).
It's impossible to proffer an alternate explination unless there are water pumps or other hidden variables involved.

I'm well aware of what we are talking about.

Yes there are two different fluids. As Sophie has already pointed out, air is a fluid that is in a gas phase.

The gravitational pull that you are referring to is, again, the hydrostatic pressure due to an elevation change. This explanation (elevation change) has been stated over and over and over again in this post as the probable cause.

CS

 

[JohnEJ]

This is simple. As has been pointed out, a system at equilibrium must have the same pressure at all points regardless of the media, i.e. air or water, and discounting pressure changes due to elevation. When the valve is first opened, the pressure of the air (at ~0 PSIG, presumably) will increase rapidly up to the pressure at the inlet of the valve. The gauge is at the TOP of the hydrant. Therefore, the highest pressure (due to velocity of the air) is seen at this point, resulting in an artificially high reading. The same would be seen if the hydrant was open to the atmosphere, because the air (or water) has to make a 90⁰ bend at the location of the gauge, transferring kinetic energy to the gauge. This is also one way in which "water hammer" occurs due to the difference in flow rate of compressed air relative to water. (No, water hammer doesn't depend only on a bend, or on air in the pipes.) The inertial energy of the water rushing into the cavity results in a momentary pressure spike before the system reaches equilibrium (that is, stops flowing).

I'm sure that's what you're seeing here. After the sysem reaches equilibrium, the pressure will stabilize. But, it may be less than your source pressure due to pressure drop through the pipes and valves, assuming you have continuous flow. Of course, elevation does affect the pressure and you need to consider that as well.

Sorry it took so long to respond, I'm new here and just saw this thread!

 

[matt81turner]

OK guys. I think I have finally gotten close to figuring this out with the combined gas law.
(P1V1)/T1 = (P2V2)/T2
P is pressure measured in atmospheres
V is volume measured in liters (does this have to be in liters?)
T is temperature in Kelvins

1 variables are initial state (when hydrant is in closed position)
2 variables are end state (after hydrant is opened and filled with water)

It's my understanding that if I can determine the values for P1, V1, T1, T2, and V2 I will be able to solve for P2.
I appreciate all the theories and feedback. Anyone think I am on the right track?

 

[matt81turner]

Of course, when using this formula we are assuming that any other variables such as elevation change are not a factor. With that in mind, anyone see a problem with using this formula?

 

[sophiecentaur]

Having read all of this stuff again I think we are probably just discussing a transient effect.
Unless there is some detail that has been left out of the original description. What is the timescale for all this to occur? If it's brief, then JohnEJ's idea of 'overshoot' of the mass of water sounds a good one. It would be a damped resonance effect of mass against spring, where the loss is so high than you only get less than one half cycle. If the pressure excess lasts for longer, then it could be explained by initial heating of the body of air as the water rushes in - if there is a non-return valve 'upstream' then the pressure could remain high, until the air cools down to ambient again.

 

[matt81turner]

The elapsed time of this happening is a couple of minutes max. We are going to conduct this test again with a higher gauge since the previous one read 0-300psi and pegged when the water filled the hydrant. There is a check valve upstream at the fire pump hundreds of feet away. (the fire pump is not running during our testing) I didn't originally mention this because we do not believe it to be a factor.

 

[sophiecentaur]

Hundreds of feet wouldn't make a lot of difference to an incompressible fluid in rigid pipes so JohnEJ's resonance mechanism could well explain the initial rise in pressure being maintained. A couple of minutes timescale makes me think in terms of some thermal effect - or a leak which let's air through but not water and gradually let's the pressure return to the normal water pressure.

 

[stewartcs]

OK guys. I think I have finally gotten close to figuring this out with the combined gas law.
(P1V1)/T1 = (P2V2)/T2
P is pressure measured in atmospheres
V is volume measured in liters (does this have to be in liters?)
T is temperature in Kelvins

1 variables are initial state (when hydrant is in closed position)
2 variables are end state (after hydrant is opened and filled with water)

It's my understanding that if I can determine the values for P1, V1, T1, T2, and V2 I will be able to solve for P2.
I appreciate all the theories and feedback. Anyone think I am on the right track?

The equation should be dimensional correct. You could use cubic meters for volume (m^3), but you'll have to use Pascals (N/m^2) for pressure.

I don't think you're on the right track with this though.

CS

 

[stewartcs]

The elapsed time of this happening is a couple of minutes max. We are going to conduct this test again with a higher gauge since the previous one read 0-300psi and pegged when the water filled the hydrant. There is a check valve upstream at the fire pump hundreds of feet away. (the fire pump is not running during our testing) I didn't originally mention this because we do not believe it to be a factor.

The timing is critical to answering you question. This is one of possibilities that was presented initially (i.e. water hammer/transient effect).

If the pressure reading is stabilized and you still have a high reading, it's not transient...

If it last only a few seconds (depending on the way the piping is set up with the check valve upstream) then it is most likely transient...However, like I said before, you're probably not going to see the transient on an analogue gauge.

CS