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Bernoulli equation - pitot tubes

  1. Oct 28, 2014 #1
    1. The problem statement, all variables and given/known data
    未命名.png

    2. Relevant equations
    P1+pV12/2+pgh1=P2+pV22/2+pgh2

    3. The attempt at a solution
    My thinking: since the pitot tubes measure the stagnation pressure (static + dynamics pressure) and the height of the tubes are the same. By Bernoulli's equation, the total pressure along a streamline is unchanged, therefore there is no pressure change between two tubes which gives h=0.

    I just want to check is my thinking correct or not.
    Thank you!
     
  2. jcsd
  3. Oct 28, 2014 #2

    andrevdh

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    The speed of the airflow at 2 will be higher than at 1 so that P2 < P1
    so that h will be as indicated in the diagram - the water level higher
    on the left than on the right.
     
  4. Oct 28, 2014 #3
    but isn't that P only the static pressure? But in this case we also have to include the dynamics pressure as well, so the total pressure
    P+(p/2)V2 stay the same everywhere on the streamline?
     
  5. Oct 29, 2014 #4

    andrevdh

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    Bernoulli's equation describes fluid in motion so that P is the pressure in the fluid.
    The pressure is transferred to the air and water in the manometer.
    The 1/2 ρ v2 term is the kinetic energy term of the equation
    while the P term is the energy of the pressure (per unit volume of the fluid) and the
    ρgh is the potential energy term of the equation.
     
    Last edited: Oct 29, 2014
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