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Confused about water/compressed air

  1. Dec 1, 2009 #1
    This is my first post in this forum. First and foremost, big thanks to anyone who takes time out to help me with this. I work for a fire protection company; one thing i service is fire hydrants.

    I recently discovered this and do not understand why this is happening:
    A hydrant contains only air in its barrel at atmospheric pressure. When the hydrant valve is opened and the 150psi water source begins to fill the hydrant, the air in the hydrant is compressed. (Note: the hydrant is not discharging water, only filling the air occupied barrel between the valve and top section.) This is to be expected. However, when a gauge is placed on the top section to measure the pressure in the barrel I expected to find the gauge reading 150psi - the water source pressure. Instead the pressure was much higher. Anyone know why or know of a formula to calculate this? Thanks for any info in advance!
     
  2. jcsd
  3. Dec 1, 2009 #2

    stewartcs

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    Sounds like a bad gauge...unless the air pressure that exists before you open the valve is a lot higher than the 150 psi of the water...but that would mean your water pressure is also now higher...

    CS
     
  4. Dec 1, 2009 #3

    FredGarvin

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    Your gauge wouldn't be measuring in psia would it?
     
  5. Dec 1, 2009 #4
    The gauge is calibrated and measures PSI/kpa. Forgive me, I am not familiar with the difference between psig and psia. The air in the barrel is at atmospheric pressure and temperature as well. When I contacted the hydrant manufacturer I was told this was normal and would occur every time. Now I am scratching my head. The only other variable that I think may be relevant is the water source temperature is going to be lower than atmospheric air temperature.
     
  6. Dec 1, 2009 #5

    stewartcs

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    PSIG is referenced to atmosphere, so atmospheric pressure would read 0 psig whereas it would read about 14.73 psia if the gauge where absolute.

    So you are saying that you have a space in the hydrant that is at atomspheric pressure, and then suddenly it is filled with water at a pressure of 150 psi...correct?

    The only thing I can think of is that there is a hydrostatic head effect that you are not accounting for somewhere.

    CS

    EDIT: We're talking about a fire hydrant that's outside on the curb right?
     
  7. Dec 1, 2009 #6
    Yes. OK, the gauge is PSIG - not absolute. It reads zero at atmospheric pressure. Right, it's a hydrant like outside at the curb. It's what's called a traffic model. The valve is 5' below grade and above this valve the vertical barrel and top section are dry until this valve is opened.
     
  8. Dec 1, 2009 #7

    stewartcs

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    Where is the system pressure of 150 psig measured from? Is it just behind the valve before the space containing atmospheric pressure, or is it measured at some other elevation?

    CS
     
  9. Dec 1, 2009 #8
    The water source pressure is measured at a different location with a calibrated psig gauge at aproximately identical elevation. Hope this helps.
     
  10. Dec 1, 2009 #9

    stewartcs

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    Sounds like there is an elevation change that you are experiencing (e.g. you have some extra head pressure due to the hydrant being at a lower elevation than where the system pressure is being measured from).

    How high is the pressure above the 150 psig system pressure?

    CS
     
  11. Dec 1, 2009 #10
    Not exactly sure how much higher the pressure reading is. My boss recorded them isn't available right now. I greatly appreciate your time and will think about this more and report back tomorrow.
     
  12. Dec 1, 2009 #11

    stewartcs

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    No problem...Good luck.

    CS
     
  13. Dec 1, 2009 #12
    Unlike liquids, gasses are somewhat elastic. That's why your car has vacuum hoses attached to the intake manifold; as the piston is being drawn down with the intake valve open, a partial vacuum is set up, limiting the amount of air that's available for the stratified charge.
    What happens on the exhaust cycle is similar to what's happening in the fire hydrant: as the piston travels upward, exhaust gases are sent through the exhaust valve; however, a region of high-pressure combustion gases is set up that are under much more pressure than the ambient air. 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.
     
  14. Dec 1, 2009 #13
    Thanks guys. I like to think of myself as a pretty intelligent person, but physics doesn't take long to make my brain hurt. I appreciate the insight. Now, if I could just come up with a formula that takes the variables (water volume and psi, ambient air volume and the temperatures) so that I can calculate the rise in pressure that will occur in the barrel. See, the problem stems from this: The water source pressure (approx 150psi) is below the hydrant's rated pressure, but the pressure we are recording once the barrel is filled with water and the ambient air is compressed is higher than what the hydrant is rated for. This, as you may have guessed, can cause hydrant and other property damage and even injury or death as these hydrants we service are in industrial plants.
     
  15. Dec 2, 2009 #14

    sophiecentaur

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    If you are talking about the static condition, the pressure in the air cannot be different from the pressure in the water (at the same level) or the water level would change accordingly. The fact that the air is springy is not relevant. Think of a mechanical analogy: the tension in a (rigid) wire will be exactly the same as in a spring that it pulls against.

    There is a relevant difference, however, and that is in the enormous amount of energy stored in compressed air compared with the very small amount of energy stored in an almost incompressible fluid like water. This is because you do much more work in compressing a large volume of air into a small volume in order to raise the pressure significantly - the work done is pressure change times volume. Your air assisted brakes use this energy. There is virtually no change in the water volume so very little energy involved.
     
  16. Dec 2, 2009 #15

    stewartcs

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    This is not what is happening here. If the air is exposed to the fluid and it is in static equilibrium, the air pressure must equal the water pressure. Nothing more to it than that.

    CS
     
  17. Dec 2, 2009 #16

    stewartcs

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    Since this is a static system, you only need to know the pressure at one elevation (the 150 psig at whatever elevation), and the elevation change to the fire hydrant. The new pressure will be 150 +/- head pressure due to the elevation change. If the hydrant is lower than the referenced elevation (the point where the 150 psig is measured from) the head added will be positive. I suspect this is the case.

    It would be helpful if you knew could find out how high the pressure is at the hydrant to get some idea of the elevation changed.

    CS
     
  18. Dec 2, 2009 #17

    Dale

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    I think this is the key point.

    To the OP: is the increased pressure a transient thing where it spikes up and then settles down to the 150 psi, or is it static even after you wait for a while.
     
  19. Dec 2, 2009 #18
    I'd expect a post like this from someone with just a couple of posts, but a Science Advisor getting it wrong?

    Consider a closed, gaseous cavity that's halfway filled with water. Insert a pressure gauge in the middle of the cavity that's taking a reading of the gas, and another pressure gauge in the middle of the water that's taking a reading of that water.
    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. The different reading is due to the gravitational field being exerted on the water which has the effect of increasing the water pressure. Thus, in the static condition, we have the two gauges reading different results.

    Apply the same anology to the situation where air is being pumped into the cavity: After the air is pumped in, the gauge situated in the gas will read a very high reading (the gas, after all, is now far more pressurized than the air outside the cavity), and will exert an increase of force on the water that's in the cavity. Since the water does not compress (whereas the gas does compress), the gauge that's in the middle of the water does not change in its reading. Sure, the pressure that's being exerted on the cavity walls itself will be constant (even at the water-cavity interface), but in the center of the water the pressure will not change.

    Please correct me if I'm wrong.
     
    Last edited: Dec 2, 2009
  20. Dec 2, 2009 #19

    stewartcs

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    The part you are missing is that the water has a constant pressure of 150 psig sitting behind a valve. On the other side of the valve there is a cavity with nothing but air in it at atmospheric pressure (which is 0 psig). As soon as the valve is opened, the air will be compressed by the water until it is in equilibrium with the water. In order for it to be in equilibrium with the water, the gas pressure MUST equal the water pressure. Hence, it can only be a maximum of whatever the water pressure is right behind the valve (assumed to be 150 psig based on the OP).

    If you still have trouble understanding that, try to imagine a cylinder separated by a piston with water on one side and air on the other with the piston initially locked mechanically (both at 0 psig initially). Now pressurize the water to 150 psig and unlock the piston. What will happen to the air side?

    Well, it will be compressed due to the unbalanced force on the piston by the water (F = PA). Since the area of the piston is the same on either side, the force is a function of the pressure only. Since the air pressure is initially 0, and the water is now 150 psig, a greater force is applied to the water side...hence the piston will move such that it compresses the air until the air pressure is equal to that of the water pressure, which will be 150 psig in this example.

    CS
     
  21. Dec 2, 2009 #20

    sophiecentaur

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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
     
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