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Exit velocity of a water pipe.

  1. Apr 20, 2005 #1
    In my application I have a very long horizontal water pipe moving water at very high pressure. I want to keep the velocity low so as to reduce the viscosity/frictional losses to the pressure. But when I apply Bernoulli's
    principle to the find the exit velocity, I seem to get the same high speed irregardless of the pipes diameter. Is this correct?
    Here's the calculation:

    Pi + 1/2(density)(Vi)^2 = Pe + 1/2(density)(Ve)^2 , where P and V are the pressure and velocity and the i and e subscripts indicate initial and exit values.
    There is no height term since the pipe is horizontal.

    For my application, I have Pi = 6000psi = 4*10^7Pa, Vi = 1m/s, and density of water = 1000kg/m^3. Pe is the ambient pressure of the air, Pe = 1bar = 100,000Pa . Solving this for Ve I get 282m/s . This is too high for me as a speed moving through the pipe because of the frictional losses in the pressure that would result over the long distance involved. But because water is incompressible if it is leaving the pipe at this speed it must be moving through the pipe at this speed. Note that the diameter of the pipe is immaterial.
    Is this correct?

    Bob Clark
  2. jcsd
  3. Apr 20, 2005 #2


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    By using Bernoulli in this fashion you are assuming non viscous flow. There is no head loss proportional to the L/d of the pipe or the velocity. You really need to calculate the head loss, which is a function of Reynolds number, which is a function of the pipe diameter. It is also going to be dependent on the surface roughness of the pipe.

    This will add another term to the right side of your equation:
    [tex]...+ f (\frac{L}{d})(\frac{V^2}{2g}) [/tex]
    You will need to find a Moody friction factor chart to find out what [tex]f[/tex] will be as a function of Reynolds number.

    That is only the dynamic head loss for this pipe. It does not take into account minor losses for items such as bends, reductions, flanges, etc...which can be appreciable.
  4. Apr 20, 2005 #3


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    Um....high pressure = high velocity, there's not really any way around it. You can always make the pipe bigger which will decrease the velocity, but off the top of my head, that's the only way I can think of to decrease the velocity while keeping pressure constant (or somewhat at least).

    Adding to what Fred said, get yourself a Moody Diagram and calculate the Reynolds number. Then calculate you're roughness factor. Using those two parameters, you will find f, your friction coefficient, which will plug into the equation he gave.

    Note: If you do make the pipe bigger, remember that minimum required thickness is a function of pressure and ID. At 6000 psi, it would be very dangerous to have pipe below tmin
    tmin = (P*d)/(40,000+(Py)) where y = 0.4 for ferric steels.
  5. May 3, 2005 #4
    Bernoulli's equation reads:

    Pi + 1/2(density)(Vi)^2 = Pe + 1/2(density)(Ve)^2 , where P and V are
    the pressure and velocity and the i and e subscripts indicate initial
    and exit values. There is no height term since the pipe is horizontal.

    Let's say the inital pressure Pi is very high say thousands of bars. I
    want to find the velocity on exit from the pipe when it empties to the
    ambient pressure, Pe = 1 bar.
    In addition to the Bernoulli principle an equation used to calculate
    the velocity of a fluid stream is the continuity equation AeVe = AiVi,
    where A is the cross-sectional area of the pipe, with e and i again
    meaning initial and exit values. This equation holds since assuming an
    incompressible liquid the total volume flowing in must equal the volume
    flowing out.
    Therefore if the pipe is at constant diameter the velocity should stay
    the same. But that means in the Bernoulli equation above the velocity
    cancels out so the equation becomes Pi = Pe = 1 bar, which is
    incorrect. What's the resolution of this?
    Secondly let's say the pipe is of smaller diameter on exit. Then Vi =
    (Ae/Ai)Ve. Substituting this into Bernoulli gives:

    Pi + 1/2(density)(AeVe/Ai)^2 = Pe + 1/2(density)(Ve)^2.

    Then Pi-Pe = 1/2(density)(1-(Ae/Ai)^2)Ve^2. And solving for Ve this

    Ve = [2(Pi-Pe)/(density)(1-(Ae/Ai)^2)]^.5

    The curious thing about this equation is if we make the exit
    cross-section arbitrarily close to the initial cross-section then we
    can make the exit velocity arbitrarily large.
    Is this correct?

    Bob Clark
  6. May 3, 2005 #5
    Fred is correct in saying that the velocity is so high because you are neglected the effects of viscous flow, which in real life application is quite significant. However if we want to use this concept you have to know several things about the pipe such as its roughness coefficeint [tex]\epsilon[/tex]. Then you can apply the equation for headloss that Fred described.
    The reasoning is that you need a value for [tex]\epsilon/D[/tex], Reynolds number, and the friction factor of f to effectively use the Moody Chart.
    If in your application however, you are not provided the roughness of the pipe or do not have access to a Moody chart or it is simply not required to be used by some sort of guidline, the only way I can see to decrease your velocity is to increase the diameter of the pipe after the inlet.
    By just the use of Bernoulli's equation without losses and Q1=Q2 the only real way to decrease your velocity with that high of a pressure difference would be to drasticaly increase the diameter of the pipe.
  7. May 4, 2005 #6


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    You don't assume that pressure at the start of pipe is your 1000 bar. You assume that flow is coming from some source where velocity can be approximated as 0. Instead of thinking of it as a a pipe where pressure at the beginning is 1000bar, think of it as an infinately large resevoir where the pipe inlet can be approximated as V=0. Now you have a difference in pressure creating a force driving the water out of the pipe into P=1bar.

    Hope that makes sense.
  8. May 6, 2005 #7


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    Just some thoughts to let you know:

    In the problem stated above, water is being discharged to an atmosphere of pressure [tex]P_e[/tex]. If there is no variation on diameter and height, then [tex]P_i=P_e[/tex] is absolutely right. The whole pipe is equally pressurized no matter which is the pressure of the reservoir. Imagine such reservoir pressure is [tex]P_o[/tex]. This pressure will be a total pressure, so that by Bernoulli as the flow is accelerated towards the exit pipe there is a transformation of static pressure into dynamic pressure: [tex]P_o=P_i+1/2\rho U^2=P_e+1/2\rho U^2[/tex].

    Your apparent contradiction in the formula Ve = [2(Pi-Pe)/(density)(1-(Ae/Ai)^2)]^.5 is not real. If [tex] P_i\rightarrow P_e [/tex] and [tex] A_e\rightarrow A_i[/tex] then Ve is indetermined (0/0) which yields a finite value.
  9. Sep 4, 2008 #8
    It seems like everyone is headed in opposite directions with this.

    If Pi=Pe then there is no flow.
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