Required Air Flow to Remove Moisture
The required air flow to remove moisture or water vapor production in a room can be calculated as
L = G / (xr - xm) (1)
where
L = required air flow (kg/h)
G = moisture or vapor production in the room (kg/h)
xr = humidity ratio in room air (kgH2O/kg dry air)
xm = humidity ratio in make-up air (kgH2O/kg dry air)
Example - Removing Moisture from a Room
Required air flow to keep a humidity ratio of 0.010 (kgH2O/kg dry air) in a room with input air humidity ratio of 0.001 (kgH2O/kg dry air) and vapor production 100 (kg/h) can be calculated as
L = (100 kg/h) / ((0.010 kgH2O/kg dry air) - (0.001 kgH2O/kg dry air))
= 11020 (kg/h)
Related Topics
• Ventilation Systems
Design of systems for ventilation and air handling - air change rates, ducts and pressure drops, charts and diagrams and more.
Related Documents
Air - Humidity Ratio
The mass of water vapor present in moist air - to the mass of dry air.
Air - Maximum Moisture Carrying Capacity
Maximum water content in humid air vs. temperature.
Evaporation from a Water Surface
Evaporation of water from a water surface - like a swimming pool or an open tank - depends on water temperature, air temperature, air humidity and air velocity above the water surface - online calculator.