How do changes in internal pressure affect airflow in a room?

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Changes in internal pressure significantly affect airflow in a room, as demonstrated by a model using the vent opening equation. With a vent area of 2500 mm² and specific parameters, the calculated airflow rate is approximately 0.0043 m³/s. An increase in mass entering the room raises the internal pressure to 101715 Pa, leading to a pressure difference of -475 Pa, indicating a potential negative flow from inside to outside. The discussion highlights the importance of time intervals in pressure calculations, suggesting that pressure may equalize faster than initially estimated. Understanding these dynamics is crucial for effective airflow management in enclosed spaces.
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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^3q = Cd A ((2 *delta P )/p)^0.5

so q = 0.004308549 m^3/s or 0.258512945 m^3/ minIf 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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Jimmy Boy said:
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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