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Fluid Dynamics

  1. May 10, 2014 #1
    Hey guys, I've been wracking my brain for a while, and thus I have landed here yet again.

    My problem is the following:

    Say you have a pipeline with a venturi injector, and intermittently a large tank of water with a portion of air on top -- I.e. a closed circuit communicating between the venturi and the tank. Let us of course neglect friction. In this setup the water is added and taken from the bottom of the tank, so air that is in the tank is trapped on top and incoming air becomes trapped as well.

    Now, at first we have everything at rest, I.e. Nothing moving and the pressure of the air inside the tank is 1atm plus whatever the weight of the water in the pipes relatively on top of it has added. If we capped off the venturi the system is closed (let's let the pipes/tank/venturi be made of a perfect insulator so that the system is truly closed). Then let us add some kinetic energy to the water so that it travels around the circuit: nice, now we have a circuit of water and the air in the tank is still 1atm+whatever. This circling flow of water will create a depression in the venturi pipe, and if we open it up to the atmosphere then air will be injected through the pipe and travel to the tank where it will be trapped, thus increasing the pressure in the tank. This will go on until the pressure inside the system has built up sufficiently that the relative depression in the venturi tube will nevertheless be 1atm and thus no more air will be injected from the atmosphere. Neglecting friction/turbulence/whatever other forms of energy loss that are non-ideal we of course know that the water in the system will then keep circulating with its final speed until the end of time, just as it would have circulated with its initial speed until the end of time had we not opened the venturi to the atmosphere.

    All fine and dandy, here is my question: is the final water speed less than the initial water speed, I.e. Has the water lost energy in order to compress the gas? Intuition leads me to think the atmosphere is doing the pushing since it is forcing itself into the depression, which necessitates an immediate compression of of the air already trapped in the tank in order to make room (water incompressible on this scale). But this seems to contradict thermodynamics, specifically the impossibility of creating free energy I.e. Creating a pressure gradient from otherwise uniform air without ourselves doing work (second law as I recall). On the other hand I have no idea how to take entropy into consideration for this process, although superficially it seems as though entropy is decreasing, which again contradicts the intuitive idea of the atmosphere spontaneously forcing itself into the depression.

    Any insight you guys may have will be very much appreciated -- I hope perhaps one of you may even point me to an appropriate text (or perhaps compose it yourself if it is a matter of a couple sentences) that would educate me as to how much energy the water loses, why, at what stage etc. So far I've only come across theory dealing with fluids on their own (I.e. Incompressible or compressible flow) but never two different fluids interacting with eachother along a pipeline.

  2. jcsd
  3. May 11, 2014 #2


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    I got dizzy trying to follow your description. A simple sketch of the problem would go a long way to clarifying your question.
  4. May 11, 2014 #3
    Yes, sorry bout that ^^'

    Attached is a simple (mostly thanks to my paint skills :/) sketch of the system at start with water velocity V1 (note we don't know the relationship between the pressure in the tank, where the water is practically stationary -- the tank should be rather large -- to that of the water in the venturi since we haven't specified the dimensions of the pipes and such. I think this is irrelevant for the discussion for now, since we are only looking at things qualitatively in terms of energy)

    Now as air is sucked in through the venturi tube, eventually the pressure of the water in the injection port of the venturi will reach 1 atm and no more air will be sucked in -- equilibrium. Neglecting friction etc. what can we say about the velocity of the water at this equilibrium stage?

    1. If V2<V1 then what is the physical explanation -- how is the water doing work on the atmosphere? (To meet it seems the atmosphere must be doing the work to get in the Venturi, as it is the one "pushing" -- just as it is the atmosphere doing work on the gas in a piston if the gas in the piston is at lower pressure.)

    2. V2=V1 would be nice, but seems (at least to me) to contradict the 2nd law of thermo (created pressure differential from homogenous air without doing work).

    3. V2>V1 doesn't make any sense so we can leave it out.

    Thanks again, and let me know if there is any way to make this clearer for the reader!

    Attached Files:

  5. May 15, 2014 #4
    Are bumps allowed on this forum? I hope so ^^
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