Changes in internal pressure

  • Thread starter Jimmy Boy
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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?
 

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  • #2
jack action
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So the next pressure difference is 475
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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