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The psychrometric chart 6 - energy saving with humidification

In my previous post, we became familiar with the humidification process for both isothermal (steam) and adiabatic (cold water) humidifiers. In this post, we will look at examples of how it is possible to save energy by changing a humidifier type or re-instating a turned-off humidifier. It is not sufficient to just consider the energy usage of the humidifier, the total energy input to the air system must be taken into account to determine if energy savings have been achieved.

In the following discussions, we'll be looking at the winter condition. This is because cold air, even if totally saturated, has a relatively low moisture content. When this air is warmed up to room temperature, the relative humidity will be low and the air will require humidifying.

There are three main types of air system used to supply conditioned air to a building. These are full fresh air (also known as total loss), fresh air with heat recovery and mixed-air systems.

Full fresh air

With full fresh-air systems, outside air is taken in in one condition and leaves the AHU (air handling unit) in a different state. We have seen in the previous post that increasing the air humidity, raising the specific enthalpy from one value to a higher value requires the same amount of energy no matter which type of humidification is used. Energy saving by changing the humidifier type or re-instating a humidifier is not possible. In most cases, where a system is designed to operate with steam humidifiers, there will be insufficient capacity in the heating coils to allow the humidifiers to be changed to adiabatic systems. Where an old type of adiabatic system has been removed, perhaps over concerns about legionella risks, it would be possible to install a modern adiabatic humidifier which is safe.

Full fresh air with heat recovery

With these air systems, it is often possible to save energy by changing the humidifiers from steam systems to adiabatic systems. These air systems recover energy from the air being exhausted from the building and use it to preheat air coming into the building. This is achieved by means of static plate heat exchangers, run-around coils or thermal wheels. In moderate outside air conditions, the energy recovered from the exhaust air is sufficient to provide part of the energy required to offset the adiabatic cooling. There is generally an increase in the heating energy but not as much energy as would be required to make steam. We have found that when air is humidified in office buildings, the temperature on the floors can be reduced, so it is generally better to humidify these air systems.

We've created a calculator to work out how much energy can be saved using MET office data for Kew in west London to give the outside air conditions. You can get the calculator free from the downloads section of our website:

Mixed-air systems

Mixed-air systems offer a very good opportunity for energy savings, especially if they are currently fitted with steam humidifiers. Mixed-air systems use an arrangement of dampers to mix the return air (air that has already been through the building) with fresh air. Control over the damper positions is normally a function of the BMS (Building Management System). The most common type of control is enthalpy control where by the mix of fresh and recirculated air is adjusted to achieve the desired enthalpy of the supply air. The damper position can be calculated using the expression:

DP = 100 x (Es – Er)/(Ef – Er)


  • DP = damper position (% fresh air)
  • Es = specific enthalpy of the supply air
  • Er = specific enthalpy of the return air
  • Ef = specific enthalpy of the fresh air

The expression is valid between 100% fresh air and 0% fresh air and not usable when Ef = Er.

To represent the mixed air condition on the psychrometric chart, a line is drawn from the fresh air condition (normally after the frost coil) to the return air condition. The mixed air condition can be found by marking on the line the percentage of fresh air as a percentage of the line. See the chart below showing outside air at -5.0°C saturated (100%RH) heated by a frost coil to +5.0°C, mixed with return air at 23.0°C, 44.0%RH.

If our building has a supply air temperature into the work space of 17.0°C and a work space condition of 21.0°C, 50%RH, we can draw this onto the chart and determine the amount of fresh air that would be required using enthalpy control. Careful measurement of the chart reveals that the amount of fresh air required is just over 19%.

It should be noted that the mixed air condition has the same enthalpy as the supply air condition, 36.9 kJ/kg. If the mixed air is humidified with an adiabatic humidifier (like the Gibbons rotary atomiser system) to get to the required supply air condition, , the amount of energy required is virtually zero (no change in enthalpy). If an isothermal system (steam) is used, the air has to be both humidified and cooled. See the chart below for details.


To calculate the energy needed for steam humidification the air has to be cooled to 17.0°C and the air will have a specific enthalpy of 34.25 kJ/kg. this is a reduction in energy of 34.25  – 36.9 = -2.65 kJ/kg. In practice, a typical cooling system will have a COP (coefficient of performance) of 3, so the energy required by the cooling system will be 2.65 / 3 = 0.88 kJ/kg.

The steam humidifier will increase the specific enthalpy of the air back up to 36.9 kJ/kg i.e. energy input of 2.65 kJ/kg. The total amount of energy required will be:

2.65 + 0.88 = 3.53 kJ/kg.

If the air flow rate were, say, 10.0 kg/s, the power required using steam humidification would be 35.3 kJ/s or kW. The equivalent power for a Gibbons humidifier for the same duty would be 0.2 kW giving a saving of 35.1 kW. If the air system were to run for 10 hours the energy saving would be 351.0 kWh.

The energy savings, cost savings and CO2 savings can be very large in this type of application. We frequently replace steam humidifiers with our system and normally achieve a payback in less than two years.

We've made a calculator to work out how much energy can be saved using MET office data for Kew in west London to give the outside air conditions. You can access the calculator free from the downloads section of our website to see how much it would be possible to save in your application.

In my next post, we will take a look at other energy-saving opportunities. If you would like to discuss any points raised in these blogs, please feel free to post a comment, question or observation.

For all humidification enquiries, call Steve Rix on 07966423165, email or click here.

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  • Date: 6 April 2014 Anas
    i want to ask that how you have calculated 17degC and 34.25KJ/Kg
  • Date: 8 April 2014 Steve Rix
    This is an example so we have made up some conditions that we would expect to see in practice. In the example, We have assumed that the desired room condition is 21.0C db (dry bulb), 50%RH, which is fairly common. There will be some heat gain in the room (due to lights, computers, people and machinery) so the air being supplied into the room has to be at a lower temperature. Therefore we have used, for the sake of this example, a typical supply air temperature (air being supplied into the room) of 17.0C db.

    The most common mode of control with mixed air systems, for both steam humidifiers and adiabatic humidifiers is enthalpy control, i.e. the damper positions are modulated to give a mixed air condition which has the same enthalpy as the supply air. The mixed air condition in this example has a temperature of 19.5C db. The steam humidification process is isothermal (it doesn’t change the air temperature). It is therefore necessary to reduce the mixed air temperature so that it is equal to the required supply air temperature (17.0C db). The cooling process does not change the moisture content of the air in this example (it has not dropped below due-point) so it remains the same as the mixed air condition 0.00675 kg/kg. The air condition before the steam humidifier (17.0C db, 0.00675 kg/kg) is plotted on the chart. All we do now is draw a line of constant enthalpy that goes through this point. This works out at an enthalpy of 34.25 kJ/kg.

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