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Required Air Flow to Remove Moisture

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The required air flow to remove moisture or water vapor production in a room can be calculated as 

L = G / (xr - xm)                                    (1)


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)

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Related Topics


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.

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