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Mis-using Bernoulli's equation

  1. Oct 4, 2015 #1
    1. The problem statement, all variables and given/known data

    An incompressible heated gas of constant temperature and pressure flows along an infinitely long tube at an unspecified velocity v1; a pressure of P1; and a density of p1 into an unheated open area of infinite volume containing the same gas at a lower pressure of P2; a density of p2,; and an effective velocity v2 of 0 m/s. The pipe is horizontal. Find the velocity of the gas inside the tube, ignoring friction and head losses.

    2. Relevant equations
    Bernoulli's equation, maybe

    3. The attempt at a solution
    Since the pipe is horizontal, h1 and h2 are treated as 0, cancelling out the pgh terms on each side of the equation. Since v2 is also 0, the .5p2v22 is eliminated. That leaves us with:

    P1 + .5p1v12 = P2

    Since P1 > P2 in this situation, the answer is always going to be the square root of a negative number.

    Common sense tells me that the gas will flow from a region of high pressure to a region of low pressure, so it should flow out of the tube; sadly, it seems that I am applying Bernoulli's equation incorrectly in attempting to form a basic model of that effect. That, or something else is terribly wrong with the way the scenario is laid out (this is my own thought experiment; it is not homework, and no teacher is to blame for this problem).

    Just looking at the reduced equation, it's all wrong. There's no way the high pressure value plus PLUS the squared velocity is going to equal the low pressure value. But the flow velocity in the tube is going to be non-zero since gas will be constantly leaving the tube ad infinitum, while the velocity outside of the tube is going to be zero since it is effectively a section of tube with a cross-sectional area approaching infinity.
     
  2. jcsd
  3. Oct 5, 2015 #2

    BvU

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    Bernoulli equation is conservation of energy. At tube inlet you have pressure energy and kinetic energy. At tube outlet you have lower pressure energy, so if all else can be ignored, you should have higher kinetic energy. I.e. v2 > v1, in contradiction with your assumption that v2 = 0.
    The change from v2 > v1, to v'2 = 0 must occur somewhere in or around the tube outlet area, apparently.

    Look at Bernoulli examples for flow through a small orifice (most have Δp > 0 from tank to jet) and compare with your case (Δp > 0 from pipe to 'tank')
     
  4. Oct 5, 2015 #3
    Your answer makes sense. I have erred in assuming that the equation covers the quiescent gas in the open area rather than the gas ejected from the tube in the immediate vicinity of the tube's outlet.
     
  5. Oct 6, 2015 #4
    Followup question: Given that the above-mentioned gas in the tube assumes a velocity of v2 immediately after exiting the tube where v2 > v1, is it safe to assume that the pressure P2 and density p2 of said gas will be the same as that of the quiescent gas in the open area? Or will that only pertain once the velocity of the exiting gas settles out after some arbitrary amount of time and effectively equals 0?
     
  6. Oct 6, 2015 #5

    BvU

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    Bernoulli to the rescue again: when the gas is slowing down, the pressure must necessarily increase ! So I would say no. There will be some contraction effects and a volume where p is even lower than P2. I think I saw them described in one of the orifice flow examples.
     
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