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Changes in internal pressure

  1. Jun 8, 2016 #1
    I am trying to develop a basic model for calculating internal room pressure. I am using the vent opening equation. q = Cd A ((2 *delta P )/p)^0.5



    If I have a set the vent 2500 mm2; A = 2.5e-3 m^2
    Cd = 0.61
    Internal pressure to be 102,135 and the external pressure to be 102,140;
    p = 1.21205 (at 20 C)
    delta P = 5 Pa.
    volume = 55 m^3


    q = Cd A ((2 *delta P )/p)^0.5

    so q = 0.004308549 m^3/s or 0.258512945 m^3/ min


    If I wanted to calculate the next internal pressure,

    the mass entering the room is 0.313331 kg,

    Converting this back into a pressure value,
    P = m * R *T / V

    P = 480 Pa. So the new internal pressure is 102135 + 480 = 101715 Pa

    So the next pressure difference is 475, which causes a larger air flow and a larger difference pressure the next time. It starts to spiral out of control.

    Any ideas?
     
  2. jcsd
  3. Jun 8, 2016 #2

    jack action

    User Avatar
    Science Advisor
    Gold Member

    Actually, the next pressure difference is -475, which would give a negative flow (from inside to outside).

    The pressure will probably equalize before the 1 minute time difference you estimated. Reduce your time difference.
     
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