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Pressure drop across valves

  1. Jan 1, 2010 #1
    Hi guys, i want to ask why is there pressure drop across valves? Because if you look at bernoulli's equation, the head will be the same in 3 points, before the valve, at the valve and after the valve. The only thing changes when a fluid flow through a valve is its pressure and velocity at the valve, due to the change of area. But before and after the valve, pressure should be the same, because the area of the pipe goes back to the orginal size.
    So why is there a pressure drop across a valve?

    The best i can come up with is maybe because some of the energy in the fluid flow has been converted to vibration energy (i think there is vibrations in the valve disks but i am not sure how/why the convertion takes place). Is this assumption true?

    And is there a way to predict how much pressure loss you will have across a valve? Because without some kind of predictions, how would you design/select a pump for a pipe line with valves (since your pressure drop across a valve, you might not have enough NPSHr for the pump)? What do people in the industry do? Do they size the pump with NPSHr value lower than the pressure after the valve?

    Thank you for the helps!
  2. jcsd
  3. Jan 1, 2010 #2


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    Hi Ask1122! :smile:

    Yes, it's because of "unrecoverable" energy loss.

    See the whole of https://www.physicsforums.com/showthread.php?t=67757" :approve: for fascinating details. Here's some extracts …
    Last edited by a moderator: Apr 24, 2017
  4. Jan 3, 2010 #3
    Thanks tiny-tim ;)
    It is a very informative thread, but some parts are above my head though, i think i will need to get my thermodynamics up to scratch before i go back to that thread ;P
  5. Jan 3, 2010 #4
    Bernoulli's equation does not apply to a viscous fluid, or a fluid that experiences nonconservative body forces.
  6. Jan 5, 2010 #5
    well which equation is used to calculate pressure drop in viscous flows??
  7. Jan 5, 2010 #6


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