# Humidifying Air by Adding Steam or Water

## Air can be humidified by adding water or steam

Sponsored Links

Air can be humidified by

- adding water
- adding water vapor - steam

### Humidifying Air by adding Water

If water is added to air without any heat supply the state of air changes **adiabatic** along a constant enthalpy line - *h** - in the Mollier or psychrometric chart*. The dry temperature of the air decreases as indicated in the Mollier diagram above.

The amount of 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 air flow (m^{3}/s)

ρ= density of air - varies with temperature - 1.293 kg/m^{3}at 20^{o}C (kg/m^{3})

x= specific humidity of air (kg/kg)

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

#### Example - Humidifying Air by adding Water

An airflow of *3000 m ^{3}/h* at

*25*and

^{o}C*10%*relative humidity (A) is humidified to

*60%*relative humidity (C) by adding water through spray nozzles.

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

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 m^{3}/h) / (3600 s/h)) (1.184 kg/m^{3}) ( (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 state will change along a constant *dh/dx* line for steam as indicated in the figure above.

When adding saturated steam at atmospheric pressure the constant line *dh/dx = 2502 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 ^{o}C*. 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 estimated by using the Mollier diagram or a psyhrometric chart.

### Example - Humidifying Air by adding Steam

An airflow of *3000 m ^{3}/h* at

*25*and

^{o}C*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 ^{o}C* to

*60% relative humidity*and approximately

*25.5*(a temperature increase less than

^{o}C*1*).

^{o}CThe 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 m^{3}/h) / (3600 s/h)) (1.184 kg/m^{3}) ( (0.012 kg/kg) - (0.002 kg/kg) )

= 0.01 kg/s

= 35.5 kg/h

The enthalpy change can be estimated from 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 m^{3}/h) / (3600 s/h)) (1.184 kg/m^{3}) (55 kJ/kg - 30 kJ/kg)

= 24.7 (kJ/s, kW)

Sponsored Links

Sponsored Links

## Related Topics

## Related Documents

## Tag Search

- en: humid air steam water
- es: agua vapor aire hÃºmedo
- de: feuchte Luft Wasserdampf