Below I present a simple and clear way to read Chart IX.

Chart IX makes life easier for every engineer by relieving them of the need to know complex equations that describe air parameters, while allowing them to graphically present the results of their own thoughts and analyses. It's even easier in the software version, as everything is editable. A mistake doesn't require a new graph; it simply requires shifting a point, which automatically recalculates all the other parameters and transformations.

Below I present a short tutorial on constructing air transformations:

1. Heating

The transformation is characterized by the fact that the absolute moisture content "X" remains constant during the transformation. The temperature and its derived parameters such as relative humidity, air density, and air enthalpy change. The dew point temperature "tr" also remains constant.

2. Wet cooling

The transformation is characterized by a decrease in both the absolute humidity content "X" and temperature "t" and their derived parameters such as relative humidity, air density, and air enthalpy. For wet cooling to occur, the temperature of the cooling coil surface must be lower than the dew point temperature "tr".

3. Dry cooling

The transformation is characterized by a decrease only in temperature "t" and enthalpy "i," while the other derived parameters such as relative humidity and air density increase, and the absolute humidity content "X" remains constant. For dry cooling to occur, the temperature of the cooling coil surface must be higher than the dew point temperature "tr".

4. Adiabatic humidification/adiabatic cooling.

Adiabatic humidification is a process of humidifying air through direct contact with water. It is realized in spray chambers or special chambers where a nozzle system sprays water into the air in the form of subatomic droplets, which then evaporate. The characteristic feature of this process is that it proceeds along a line of constant enthalpy. However, this does not mean that the temperature remains constant. Air is a mixture composed of dry air and water vapor, whose enthalpy is given by i=ips+ipw, where i=cp*t+X*(2501+1.84*tp). Since the humidity increases during humidification, and therefore the enthalpy of water vapor "X*(2501+1.84*t)" also increases, in order for the total enthalpy to remain constant, the enthalpy of dry air "cp*t" must decrease, and since the specific heat "cp" is constant, the temperature "t" must decrease as well. That is why the adiabatic humidification process is also called adiabatic cooling, as it is associated with the decrease of air temperature.

Adiabatic humidification is only one particular case of the family of polytropic transformations realized in a spray chamber. The course of the polytropic process is closely related to the temperature of water, and air can be heated, cooled, humidified, and even dehumidified. In adiabatic humidification, the temperature of the water is equal to the temperature of the air for a wet bulb thermometer.

4. Steam humidification

Air can also be humidified in a different way by directly introducing water vapor into the air. The transformation is carried out in steam humidifiers, which "boil" water and introduce it into the ventilation air using a steam lance system. The temperature of such steam is high, at least 100°C, and can even be superheated. Due to the high temperature of the water vapor, the temperature of the air increases, but fortunately, the increase is not as high as it might seem. The heating of the air will be greater, the higher the temperature of the water vapor. We know that air is a mixture composed of dry air and water vapor, whose enthalpy is i=ips+ipw, i.e., i=cp*t+X*(2501+1.84*tp). The additional enthalpy of the water vapor introduced into the air will be equal to dipw=dX*(2501+1.84*tp), where dX=X2-X1. Therefore, to determine the temperature at point 2, it is necessary to solve the following equation and determine the temperature t2 from it. t2=[dX*(2501+1.84*tp)+X1*(2501+1.84*t1)+1.005*t1-X2*2501]/[1.005+X2*1.84]

For example, let's assume that we want to humidify 1000 m3/h of air from parameters t1=+20°C and RH1=10%, which corresponds to X1=1.44 g/kg, to parameters X2=5.79 g/kg, which corresponds to a relative humidity of RH2=~40%. We use water vapor at a temperature of 100°C for this purpose. By calculating t2 using the above equation, we obtain the result t2=20.6°C, so the temperature has increased by 0.6°C.

And that's about it. If the article has brought something positive to your development, and you want the website to continue to grow, where I share my knowledge, give it a thumbs up, leave a comment or share it. And of course, you can help me and yourself by purchasing a license or You can just buy me a coffee :)

## Comentários