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Pressure drop across an orifice (Orifice pressure 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
     
  2. jcsd
  3. Mar 23, 2014 #2
    To convert a pressure (Pa) to a pressure head (m) divide by ρg (N/m^3).
     
  4. Mar 23, 2014 #3
    Hi,

    Thank you for the reply. Still slightly confused.

    It is ΔP that I need to find in order to calculate the ideal mass flowrate of air.

    Do I already have the answer to the question, ie ΔP = 470.72 Pa ?
     
  5. Mar 23, 2014 #4
    or do I need to use an Air properties table to find ρ at the temperature I am given?
     
  6. Mar 23, 2014 #5
    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.
     
  7. Mar 23, 2014 #6
    Sorry, I thought you were only looking for the conversion and I assumed you were talking about an incompressible fluid not a compressible gas. So I can't help you there.
     
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