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Pressure drop across an orifice (orific drop in meters?)

  1. Mar 23, 2014 #1
    1. The problem statement, all variables and given/known data

    ΔP = 1000 x 9.81 x (Orifice pressure drop in m)

    Pressure drop across orifice = 470.72 Pa


    2. Relevant equations



    3. The attempt at a solution

    I am not sure how this works. How can pressure be converted to meters? It does not make sense to me.

    Any help would be much appreciated

    Thanks
    Scott
     
    Last edited: Mar 23, 2014
  2. jcsd
  3. Mar 23, 2014 #2

    rude man

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    Might help if you stated the problem clearly.
     
  4. Mar 23, 2014 #3
    Question

    Orifice pressure drop in meters = (Pa (N/m^2 ))/(ρg (N/m^3 )) This gives answer in meters


    Question 1

    Given:
    T = 312 K
    ρ = 1.1333 kg/m^3 (From Air properties table)
    g = 9.81 m/s^2

    Orifice pressure drop = 470.72 Pa

    Calculation:

    Orifice pressure drop in meters = 470.72 / (9.81x1.1333) = 42.34 meters

    This answer does not seem right, looks way too large.
     
  5. Mar 23, 2014 #4

    SteamKing

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    Pressure drops can be expressed as a head of a fluid, often water, but others like mercury can be used as well. The head would be measured in meters of fluid, or some other unit of length. That's what the reading on a barometer is, after all. The reading of 760 mm is the height of a column of mercury supported by the difference in pressure between a vacuum and atmospheric pressure. In working with modest pressure drops, water is used in place of mercury. A pressure drop of 1 meter of water is equivalent to 1000 kg/m^3 * 9.81 m/s^2 = 9810 pascals.
     
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