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Water discharge and Bernoulli's equation

  1. Jul 18, 2012 #1
    Dear all,


    I have found in the literature that if there is a open tank discharging water into the atmosphere the pressure at the discharge point is considered to be equal to the atmospheric pressure.

    If one looks to the Bernoulli's equation and assuming no friction:

    Pressure + Elevation Pressure + Dynamic pressure = Total Pressure

    So, my question is:

    Why do they consider the Pressure at the discharge point to be equal to the atmospheric pressure and not the Total Pressure instead? Shouldn't it be similar to the Venturi case, where there is a pressure decrease due to an increase in velocity.

    Couldn't we be pumping water against atmospheric pressure with a Pressure< 101 325 Pa but compensated by the discharge velocity of 5 m/s, for example (no calculations done)?


    Regards
     
  2. jcsd
  3. Jul 18, 2012 #2

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    There are the pressures before the hole and after. The total pressure is equal on both sides. The elevation pressure after the hole is zero. The atmospheric pressure difference across the hole is zero. The dynamic pressure before the hole is assumed to be very small and set to zero. So you get Pa+Pe+0 = Pa+0+Pd or Pe=Pd, where Pe is the elevation pressure before the hole, Pd is the dynamic pressure after the hole.
     
  4. Jul 18, 2012 #3
    Thanks for being so swift answering me.

    The dynamic pressure before the hole is not assumed to be very small. In the literature aforementioned in my post, the authors seek to calculate the velocity for the discharge.

    I, too, thought that the total pressure before the hole and after the hole should be the same. But, the total pressure after the hole is Patm, and after the hole is Patm + Dynamic pressure. Therefore, the two don't match as they should.
     
  5. Jul 18, 2012 #4

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    Total pressure before the hole is Patm+Elevation pressure, after the hole its Patm + Dynamic pressure.
     
  6. Jul 18, 2012 #5
    The static pressure at the discharge point is equal to Patm, not the total pressure. The total pressure = static + dynamic = Patm + dynamic.
     
  7. Jul 19, 2012 #6

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    Right, just after (downstream, outside the tank) of the outlet, the static pressure is Patm and the total pressure is Patm+Pdynamic. Just before (upstream, inside the tank) the outlet, the static pressure is Patm+Pelev and the dynamic pressure is very small, because the tank is much larger in cross section than the hole, so the velocity of the water in the tank as it heads towards the hole is small compared to the velocity at which the water is ejected. So small that it is generally ignored and the dynamic pressure before the hole is set to zero. So setting the total pressure equal on both sides gives Patm+Pelev=Patm+Pdynamic or Pelev=Pdynamic.
     
  8. Jul 19, 2012 #7
    I would like to understand why do you consider the static pressure at the discharge to be equal to the atmospheric pressure. All that is required is the total pressure to be bigger than the atmospheric pressure. As long as the static pressure is compensated by the dynamic pressure, the static pressure can be lower than the atmospheric pressure.

    I have found this link that I think helps to prove my point:

    http://en.wikipedia.org/wiki/De_Laval_nozzle

    I would like to stress this paragraph:

    "the exit pressure can be significantly below ambient pressure it exhausts into, but if it is too far below ambient, then the flow will cease to be supersonic"

    So, as one can see, it is said that the discharge pressure to the atmosphere can be lower than the atmospheric pressure.
     
  9. Jul 19, 2012 #8

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    The De Laval nozzle is for compressible, supersonic flow. I don't think Bernoulli's equation is valid for supersonic flow. Anyway, water is practically incompressible, so the situation being considered here is for incompressible, very subsonic flow.

    The static pressure is atmospheric after the hole because there is no other source of pressure. The potential energy of the water due to the elevation, before the hole, has been entirely converted to kinetic energy of the flow, so there is no elevation pressure after the hole.
     
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