Humidifying Air by Adding Steam or Water

Air can be humidified by adding water or steam

Air can be humidified by either

• adding water, or
• adding water vapor - steam

Humidifying Air by adding Water

If water is added to air without any heat supply, the air condition will change adiabatic along a constant enthalpy line - h. The dry temperature of the air will decrease as shown in the Mollier diagram above.

The added water can be expressed as:

m_w = v ρ (x_C - x_A) (1)

where

m_w = mass of added water (kg/s)

v = volume flow of air (m³/s)

ρ = density of air - vary with temperature, 1.293 kg/m³ at 20^oC (kg/m³)

x = specific humidity of air (kg/kg)

If the water added is at the same temperature as the air, the change in enthalpy is zero.

Example - Humidifying Air by adding Water

An airflow of 3000 m³/h at 25^oC and 10% relative humidity (A) is humidified to 60% relative humidity (C) by adding water through spray nozzles.

Using the Mollier diagram and following the constant enthalpy line 30 kJ/kg from A to 60% relative humidity, the state at C can be found at 14.7^oC.

The specific humidity at (A) is 0.002 kg/kg and at (C) 0.0062 kg/kg.

The amount of water added can be calculated as:

m_w = ((3000 m³/h) / (3600 s/h)) (1.184 kg/m³) ( (0.0062 kg/kg) - (0.002 kg/kg) )

= 0.0041 kg/s

= 14.9 kg/h

Humidifying Air by adding Steam

If steam is added to the air, the air condition will change along a constant dh/dx line for steam as shown above.

When adding saturated steam at atmospheric pressure the constant line dh/dx = 2,502 kJ/kg (the evaporation heat of water at atmospheric pressure). When adding saturated steam at atmospheric pressure the temperature rise is very small - in general less than 1^oC. For practical purposes the process of adding saturated steam at atmospheric pressure approximates the horizontal temperature line.

The water vapor added can be calculated using (1).

The enthalpy added can be found using the Mollier diagram.

Example - Humidifying Air by adding Steam

An airflow of 3000 m³/h at 25^oC and 10% relative humidity (A) is humidified to 60% relative humidity (B) by adding saturated steam at atmospheric pressure.

Using the Mollier diagram the process from (A) to (B) can be approximated by following the constant temperature line 25^oC to 60% relative humidity and approximately 25.5^oC (a temperature increase less than 1^oC).

The specific humidity at (A) is 0.002 kg/kg and at (B) 0.012 kg/kg.

The amount of water added can be calculated as:

m_w = ((3000 m³/h) / (3600 s/h)) (1.184 kg/m³) ( (0.012 kg/kg) - (0.002 kg/kg) )

= 0.01 kg/s

= 35.5 kg/h

The enthalpy change can be found using the Mollier diagram. The enthaply at (A) is 30 kJ/kg and at (B) 55 kJ/kg. The enthalpy difference is

dh = 55 kJ/kg - 30 kJ/kg

= 25 kJ/kg

The total heat added by the steam can be calculated as:

q = ((3000 m³/h) / (3600 s/h)) (1.184 kg/m³) (55 kJ/kg - 30 kJ/kg)

= 24.7 (kJ/s, kW)

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