Air Required to Remove Moisture

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Required air to remove moisture or water vapor production in a room can be calculated as 

L = G / (x_r - x_m)   (1)

where 

L = required air flow (kg/h)

G = moisture or vapor production in the room (kg/h)

x_r = humidity ratio room air (kg_H2O/kg dry air)

x_m = humidity ratio make-up air (kg_H2O/kg dry air)

Example - Removing Moisture from a Room

Required air flow to keep a humidity ratio of 0.010 (kg_H2O/kg dry air) in a room with input air humidity ratio of 0.001 (kg_H2O/kg dry air) and vapor production 100 (kg/h) can be calculated as 

L = 100 (kg/h) / (0.010 (kg_H2O/kg dry air) - 0.001 (kg_H2O/kg dry air))  

  = 11020 (kg/h)

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Related Documents

• Evaporation from Water Surfaces - The amount of evaporated water from a water surfaces - like swimming pools or an open tanks - depends on the temperature in the water and in the air, and the humidity and velocity of the air above the surface - online calculator
• Humidity Ratio of Air - Humidity ratio (GPP) is the ratio between actual mass of water vapor present in moist air - to the mass of dry air