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Bernoulli's equation confuses me

  1. Aug 3, 2008 #1
    I have a simple pipe (with water) that reduces in diameter. I've been trying to calculate the pressure after the reduction using the first version of the Bernouli equation shown here;

    http://en.wikipedia.org/wiki/Bernoulli's_principle

    Now to calculate the velocity I've replaced the v with flow/area in the equation. I was expecting to see an increase in pressure after the reduction in pipe size, but the equation tells me it has actually dropped...? I'm just relating to the fact that if you squeze a hose the pressure increases, doesn't it? And I'm thinking that the flow needs to be the same before and after the reduction?
     
  2. jcsd
  3. Aug 3, 2008 #2

    russ_watters

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    Yes, the pressure drops. The static pressure. Bernoulli's principle states that the total pressure is constant along a streamline. If the velocity increases, then the velocity pressure increases. So to keep the total pressure constant, the static pressure must decrease.

    What you are thinking is that if you suddenly add an obstruction to a pipe that didn't have one before, you have to increase the pressure to keep the flow rate the same. That's a different problem.
     
  4. Aug 3, 2008 #3

    FredGarvin

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    The increase in pressure that you feel is not the pressure in the constriction, but the pressure slightly upstream. The pressure actually where the flow area reduces does indeed drop. That also highlights a common misconception about clogged arteries in the body and high blood pressure. The pressure isn't increased in the constriction. The pressure is higher upstream of the obstruction.
     
  5. Aug 3, 2008 #4

    russ_watters

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    Again, though, that is a situation where the pump (your heart) attempts to push the same flow through a new obstruction. The new total pressure will indeed be higher than the old total pressure.
     
  6. Aug 4, 2008 #5
    Thx for answering guys :)
    When I squeze the opening of a rubber hose that barely delievers water, then the water comes out much faster and would hit you with more force. Standing at the recieving end I wouldn't hesitate saying that the pressure has increased...? You're saying that's not correct?
     
  7. Aug 4, 2008 #6

    russ_watters

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    A hose is a tough situation because there are dynamic losses in the hose that make it different from the ideal. Generally, you lose a lot at the valve before the water even gets to the hose. Nozzles are only typically around 50% efficient, so an awful lot of the pressure just plain goes away (converted to heat). Then by putting your thumb over the end of the hose, you are just making that the place where the most pressure is lost instead of the valve. Plus, once the water gets out of the hose, Bernoulli's principle no longer applies. It is quite possible to have a more energetic thick stream of water dropping two inches from the hose vs a thin stream squirting ten feet.

    Regardless, though, you need to specify what pressure you are talking about and differentiate between pressure, force, and energy. A thin stream may hurt when it hits someone, but that doesn't necessarily mean there is more force or energy in it than a thicker stream of lower velocity.
     
  8. Aug 4, 2008 #7

    minger

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    As far as Bernoulli's, I always think of it in terms of energy. Pressure is a type of "energy" that can be converted. Kinetic energy and potential energy are also converatble between each other. Bernoulli's says that the total of these energies along a streamline stays constant. So, if the kinetic energy increases, then it can only do so by taking it from potential (i.e. going downhill) or by taking it from pressure.
     
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