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The psychrometric chart 5 - humidification

In my previous post, we became familiar with the cooling and dehumidification processes and looked at how to calculate the power required to reduce the air temperature and dehumidify the air. In this post we will look at the humidification process.

There are two main types of humidification: isothermal and adiabatic. The isothermal humidification is achieved with steam humidifiers and the adiabatic humidification process is achieved by evaporating cold water. The two processes look very different on the psychrometric chart so we will start off with steam humidification.

Steam humidifiers

Steam humidifiers work by turning water directly into water vapour. This is done by heating the water until it boils. The process of introducing water vapour into the air increases the air’s moisture content but does not change its temperature. It should be noted that the specific enthalpy of the air increases, reflecting the energy that has to be put into the water to turn it into a vapour. We will take the example of air that starts off at 21.0°C, 30% RH and humidify it until it reaches 21.0°C, 50% RH

The starting point has the following properties:

  • Temperature db        = 21.0°C
  • Humidity                    = 30% RH
  • Moisture content      = 0.00465 kg/kg
  • Specific enthalpy     = 33.0 kJ/kg

Air with a temperature of 21.0°C db and a humidity of 50% RH, has the following properties.

  • Temperature db        = 21.0°C
  • Humidity                    = 50% RH
  • Moisture Content     = 0.0078 kg/kg
  • Specific Volume       = 0.844 m3/kg
  • Temperature wb       = 14.7°C
  • Specific enthalpy     = 41.0 kJ/kg

The difference in specific enthalpy is 41.0 – 33.0 = 8.0 kJ/kg.

The difference in moisture content is 0.0078 – 0.00465 = 0.00315 kg/kg.

Using our previous example having an air flow of 4.0 m3/s at 21.0°C, 50% RH, we can now work out the required humidifier capacity.

We know that the specific volume of the air at 21.0°C, 50%RH is 0.844 m3/kg. The mass of air is 4.0/0.844 = 4.74 kg/s.

The humidification rate is 4.74 x 0.00315 = 0.0149 kg/s.

This is normally expressed in kg/h so 0.0149 x 3600 = 53.75 kg/h.

Adiabatic humidifiers

The energy required to generate a water mist or to wet a membrane, is very low and in general the energy consumption of adiabatic humidifiers is negligible. The energy required to evaporate the cold water comes from the air. The air becomes cooler as it becomes more humid (often referred to as adiabatic cooling).

There is no energy input into the air stream or energy extract from the air stream so the adiabatic humidification process follows the lines of specific enthalpy. The chart below shows the adiabatic humidification process.

Relating to our previous example, in order to reach a condition of 21.0°C, 50%RH, it is necessary to first heat the air to 29.0°C so that when it is humidified, the required condition is met. It is interesting and actually quite logical that moving from our starting point to the finishing point in our example requires the same energy. When using steam humidifiers this energy is put into the humidifier and with adiabatic humidifiers the energy is put in to the air via the heating coil. In both cases the amount of energy is the same (8.0 kJ/kg). It should be noted that the efficiency of the heating system and steam humidifiers can be different making one solution marginally better than the other from an overall energy point of view.

In this example, there is no clear advantage of one system over the other. However, in some humidification applications, very large energy savings are possible from using the most appropriate humidification type.

We will look at saving energy with humidifiers in the next post, but if you have any enquiries in the meantime, please contact Steve Rix on 07966423165 or email steve.rix@gibbonsgroup.co.uk. All our humidification products and services can be found here.

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Humidification

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