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Steam, preesure and water flow project. I Cant understand the behavior.(+Movie)

  1. Dec 10, 2011 #1
    Hello All,

    I'm working on a project and one of the main parts is this pressure vessel that is sealed air tight that has 2 tubes coming out. One is straight and the other is a U tube. I have a 2.5 KW heater coil in the water to boil the water in order to create steam to raise pressure and move the water through the tubes into a holding/cooling tank (open to atmospheric pressure).
    What my common sense and calculations should me was that the water level of should be the same throughout the compression of the water through the tubes. But- as seen in the video the water through the U tube flows quicker and makes a "tunnel" of steam allowing the pressure to drop causing an instant boil...
    Note that the tubes are connected on top before they enter the second tank. (-same pressure and water head).


    Pic of system (before i got the heater):

    **The brass parts are one way valves. The left is directed for upwards flow and the one on the right is directing from the clear hose back to the bottom tank.

    The question: Why isn't the water level descending equally through out both the tubes?
  2. jcsd
  3. Dec 11, 2011 #2


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    Actually the pressures are only equal at the center of the tee for the two paths. There is a slight pressure drop across the tee to the U-tube, another slight pressure drop across the directional change at the bottom of the U-tube and a slightly greater pressure drop due to the length of the U-tube compared to the straight tube. There is a slightly lower pressure at the opening of the U-tube compared to the straight tube due to elevation head. Finally as the velocity of the steam and heated water moving past the opening of the U-tube, it sreates a lower pressure due to the Bernoulli effect. The result is that the pressure is lower and boiling begins earlier in the U-tube.
    Last edited: Dec 11, 2011
  4. Dec 11, 2011 #3
    I understand what your saying but i don't think the water is boiling in the U tube. It is being pushed down buy the steam pressure and once it passes the lowest part if the U it rises and creates a "Tunnel" of steam resulting in a imediet pressure drop in the vessel (thus the sudden boil). For some reason the water in the U has a lower head and is easier to push down.. It might be the losses and the Bernoulli effect but that sounds a little to far away, the water at the bottom of the long hose is at higher pressure (Due to the water head) and from what I understand it should be easier to flow from there..
    I appreciate your replay and i'll give it some thought and maybe look up in my thermodynamics book to see if i can find some calculations to back it up..
  5. Dec 11, 2011 #4


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    Have it your way. In any case, if the temperature of the fluid in the U-tube is over the saturation temperature for the pressure in that region, the contents will be steam, not liquid. The effects , as I said, are slight, so the effect occurs just before the entire contents initiate bulk boiling. Install a pressure guage and see if there is any bulk pressure drop in the boiling chamber.

    Think of this another way. As the water level inside the U-tube drops it is actually showing you how the bulk conditions in the tank have reached saturation conditions. As you say the pressure in the bottom of the tank is higher than the the pressure at the top. Thus the saturation temperature is reached in the upper part of the chamber first. There is still thermal mixing so the level of boiling remains below bulk boiling conditions as the water continues to heat. Thus the level in the U-tube leg drops until it reaches the bottom. of the curve. There is bvery little distance between the bottom of your tubes and the heater. So once the level in the U-tube reaches its minimum, there is only a short time to bulk boiling. If you could increase the elevation of the tubes you would see a longer delay to bulk boiling after the U-tube reaches minimum level.

    Another possibility is to measure the rate the water level decreases and use it to predict the bulk boiling point when the level in the U-tube would theoretically drop to the heater.
    Last edited: Dec 11, 2011
  6. Dec 11, 2011 #5


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    I'll provide you one more explanation. The pressure in this system is initially based on the elevation of the level in your condensation tank and the temperature density of the water in the two tanks. As you heat the water there is some level swell due to heating and expansion of steam (decrease in density), but basically the pressure at the bottom of the tank will still be determined by the height of the water in the condensation tank until either the boiling rate exceeds the relief flow rate capacity of the tubes or the water in the condensing tank reaches boiling off.

    There is no physical way to cause a sudden pressure drop in the lower tank because your one way valves ensure that you can't "burp" the system and quench the steam/water mixture in the boiling tank.
  7. Dec 11, 2011 #6
    How about rapidly condensing the steam that flows through the U tube as hits the cold water that just flowed back through the one way valve from the top water tank.
    Even if the water is ,lets say at 90 degrees C, the steam will condense quit rapidly and cause a pressure drop, don't you think?
    I checked it out and there is a ratio of 1/1000 between water and steam (at 100 degrees 100 Kpa).
  8. Dec 11, 2011 #7
    What are you're thoughts about this:
    Maybe i will eliminate the thermal intervention. Lets say i make another hole on the lid of the bottom vessel and attach a one way valve and to a compressor. If i raise the pressure do you predict the same behavior (flow through the U tube before the other tube)?
  9. Dec 11, 2011 #8


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    There is a very small area where the steam is in contact with the liquid in the U-tube. There is condensation at this boundary but it is swamped by the heat transfer from the bulk water to the liquids in the tube. Remember that heat transfer is proportional to area and temperature difference. Basically the liquid and steam at that interface inside the tube is at local saturation temperature or you wouldn't see a stable level.

    If you pump air into the lower chamber it will accumulate at the top of the chamber and gradually force level down and up into the condensing chamber. The level in the condensing chamber will rise due to the displaced water. Once the level reaches the opening of the U-tube, it will begin to force the level down. The straight tube will remain filled with liquid until the level reaches its opening. Once the level reaches the bottom of the U-tube and the bottom of the straight tube the air pressure will be vented through the condensing chamber and pressurization and level change in the condensing tank will cease. (If I understand the physical construction of the apparatus)
  10. Dec 11, 2011 #9
    Thank you for you're responses, you've given me quit a bit of insight. I have to think about this and try to figure out how to overcome this problem.
    It seems you understood the physical construction pretty well. Maybe I will post the detailed sketch of this project just to be a bit clearer.
    In the next few days i will give this project some thought and work to see if i can achieve the desired behavior.
    If i have a major breakthrough i will post it.
    Thanks again. Good Evening.
  11. Jan 6, 2012 #10
    Hello again,

    I continued working on my project and i had quit a bit of progress. I achieved a full work cycle and was able to bring both water levels to the same hight (changed the piping to a bigger diameter).

    Movie link:

    I have a new question, there is something I'm missing in my calculation of the compression stage (when the water descends).

    According to my calculations:

    Q dot=2.5 [KW] (my heater)
    hfg=2260 [KJ/Kg]

    m dot= Q dot / hfg = 0.001 [kg/sec]
    => V dot= 0.0018 [m^3/sec] (multiplied the mass rate by steam volume at that pressure and temp).

    The area of the pressure chamber is 0.2 X 0.2 [m^2].

    => Speed of the water descending is the steam creation volume rate divided by the area of the boiler:

    V dot / A = 0.045 [m/sec]

    After measuring the time of descend on the recent run of the model i calculated an average speed of: 0.002 [m/sec]

    This is a big miscalc and I'm having a hard time figuring out the problem.

    Any suggestions?
    Last edited: Jan 6, 2012
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