Confused about water/compressed air

  • #26
S_Happens
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There's no need to do a thought experiement to determine if it's a static condition or not, the OP stated that it is. The only questions are left are the details of the rest of the system.

When he says "industrial plant" I'm assuming it has it's own firewater system. The system should be designed/required to maintain a minimum pressure while supplying a certain number number of hydrants. We'd simply have to find out the control scheme to determine what the supply pressure is supposed to be.

Is 150 psig at the pump discharge? If the system is designed to control at 150 psig, then where is the sensing point in relation to this hydrant?

If we don't get any more info, I'll see what I can dig up as far as standards/requirements (chemical plant).
 
  • #27
Thanks, I will get on that asap.
 
  • #28
170
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It would seem that you are wrong (if I have understood what you are saying - in particular, I assume what you mean by "the middle of the water" means just below the surface).
The pressure just above and just below the surface of the water must be precisely the same. How could it not be? The respective moduli of the two fluids is irrelevant. If there were a difference in pressure then there would be a resultant force and things would move to change the situation.

stewartcs seems to be right
You guys, this isn't rocket science for cryin' out loud!

Look. Consider a "J"-shaped cavity. Consider that the vertical part of the "J" is quite long in relation to the curved part of the "J". The curved part of the "J" has a 1 foot vertical section after the bend. The entire apparatus is, say, 20 feet long and 12 inches in diameter. It's closed on the curved end and has an opening at the top--or vertical--end.
Now add 20 gallons of water to the "J". What will happen? Yep! That's right--the water will almost fill the whole "J", but will leave a pocket of air at the end of the curved part of the "J" (remember, there's a 1-foot vertical section right after the bend). What's the air pressure in this little region? Far greater than the pressure that's being exerted by the water at the top of the vertical part of the "J"!!!

Thus we have unequal pressures between one end and the other end, which is the precise discrepancy the OP was bringing to the thread body.
 
  • #29
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I'll see what I can dig up as far as standards/requirements (chemical plant).
Don't bother. We all should know by now that the unequal air psig measurements are due to the effect of gravity on the water in the watermain, and that the sheer weight of the water on the air pressure that's being measured is the reason for the PSI discrepancy.
 
  • #30
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0
Every ~2.3 feet in elevation difference equals 1 psi. You might have a slope of 1-2%, 1-2 feet per 100 feet of horizontal difference, and it would appear level. Or etc.

But the air pressure at the hydrant is exactly equal to the water pressure at the hydrant. It's the other location and hydrant that are likely different. Or the gauge is giving a bad reading.


Now, evacuate the gas from the cavity. The gauge that's measuring the gas should read 0 psi, and the gauge that's reading the water should read something other than 0 psi. In fact, the more water that's being measured, the higher the reading of the water gauge.
Impossible. Water or water vapor will fill the vacuum until the pressures are equalize. Just like when you suck on a straw you reduce the pressure in your mouth. You can't have two different pressures next to eachother without causing some flow. This really is a simple situation. Once the water stops moving the air and water are at the same pressure no matter what.
 
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  • #31
stewartcs
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You guys, this isn't rocket science for cryin' out loud!

Look. Consider a "J"-shaped cavity. Consider that the vertical part of the "J" is quite long in relation to the curved part of the "J". The curved part of the "J" has a 1 foot vertical section after the bend. The entire apparatus is, say, 20 feet long and 12 inches in diameter. It's closed on the curved end and has an opening at the top--or vertical--end.
Now add 20 gallons of water to the "J". What will happen? Yep! That's right--the water will almost fill the whole "J", but will leave a pocket of air at the end of the curved part of the "J" (remember, there's a 1-foot vertical section right after the bend). What's the air pressure in this little region? Far greater than the pressure that's being exerted by the water at the top of the vertical part of the "J"!!!

Thus we have unequal pressures between one end and the other end, which is the precise discrepancy the OP was bringing to the thread body.
Of course the pressure at the top of the J will be less, it's open to atmosphere and the hydrostatic head from the elevation change is causing the air pressure to be high since it's trapped at the bottom...that's what's been said all along...i.e. the increase in pressure is probably attributed an elevation difference. You need to read the other posts more thoroughly. The elevation change was the first thing I proposed as the cause for the increased pressure. Hence me asking where the 150 psig was measured from.

The air pressure at the air/water interface is exactly the same...which is the crux of the problem. So, if he is reading 150 psig right at the valve, and then the valve is opened, the air pressure in the chamber will equalize to 150 psig. Period.

If the pressure read from another location at a higher elevation, then he'll have to add in the elevation head.

You're right, it's not rocket science, so stop trying to make it to be.

CS
 
  • #32
stewartcs
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...Likewise, the water in the hydrant is exerting a tremendous force on the air that you're measuring, and due to air's elastic nature, will be under a pressure that's significantly higher than just 150 psig.
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
 
  • #33
sophiecentaur
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You guys manage to get so cross with each other! Why?
I wouldn't mind betting that most of you would actually agree with each other about the basic Physics - it's just that you're not communicating the problem i.e writing what you actually mean and reading what's actually written. TX and RX need to be in sync.
Getting ratty doesn't help the Science one jot nor tittle.
 
  • #34
S_Happens
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Don't bother. We all should know by now that the unequal air psig measurements are due to the effect of gravity on the water in the watermain, and that the sheer weight of the water on the air pressure that's being measured is the reason for the PSI discrepancy.
I'm well aware of "head" pressure (so is pretty much everyone else in this thread, but you haven't seemed to notice). I'm not talking about flipping through some basic intro to physics book to figure out the magic of manometers. I'm talking about checking OSHA standards and anything else applicable to firewater systems to get any details, because we've been supplied almost nothing.

I asked where the 150 psig supply pressure was being read as well, and how it's being maintained. Even ignoring the fact that the OP doesn't know for certain that the supply is 150 psig, we can be almost certain that there isn't a gauge to read the pressure at the isolation for the hydrant. The systems (dry barrel like the OP is talking about) are setup where the hydrant has an underground isolation valve (two isolations really in case the hydrant needs to be worked on) connecting it to an underground header. The only question is to find out how the header pressure is maintained. The norm for industrial plants is to have their own firewater pumps that pull from a canal/river/large body of water. If this is the case then it might not be elevation change that we need to be concerned with.

I'm sure everyone participating already knew all this as they plow hatred first into this discussion.

It's obviously a simple situation that can be handled without thought experiments, confusion, and frustration.
 
  • #35
sophiecentaur
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It seems obvious to me that your query is, actually, totally specific. The quoted pressure can only apply at the level where it is measured and you'd have to find out which level this is done on the supply. Unless you are drawing a significant flow of water then the head, locally, will differ just by the height difference. The presence of air is pretty much a red herring unless there is a significant height of trapped air. If there is, then the 'dry' pressure will be higher than the 'wet' pressure because air is much less dense. This is unlikely except when the system is first installed because who is going to bother to pump out water from a rising main?

If the gauge is giving a reading which does not agree with the above calculations then it has to be faulty. But this isn't a Physics question; it's surely one for the installer of the system or equipment or the recommendations of the local fire authority / insurance company. Also, if it's a safety of life consideration, I wouldn't rely on the potential BS you can read on a Forum (Mine included)!
 
  • #36
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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.
 
  • #37
sophiecentaur
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A gas IS a fluid. So is a liquid.
They will both FLOW - i.e. they are fluid.
 
  • #38
stewartcs
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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
 
  • #39
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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!
 
  • #40
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?
 
  • #41
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?
 
  • #42
sophiecentaur
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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.
 
  • #43
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.
 
  • #44
sophiecentaur
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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 lets air through but not water and gradually lets the pressure return to the normal water pressure.
 
  • #45
stewartcs
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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
 
  • #46
stewartcs
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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
 

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