The Psychrometric Chart 4 – Cooling and dehumidification

In my previous post, we became familiar with the heating process and looked at how to calculate the power required to raise the air temperature from 21.0C db to 31.0C db. In this post we will look at the cooling process.

The method for calculating cooling power is the same as for calculating heating power. We determine the specific enthalpy of the starting point and the specific enthalpy of the finishing point. The difference between them is the amount of energy that needs to be put in (heating) or taken out (cooling).

When air is heated, the moisture content does not change. However, it is possible to cool air to below its due point causing water to precipitate out of the air. Due point is reached when the air humidity is 100%RH or saturated. When air is cooled below its due point, it follows the line of 100% RH. The chart below shows what happens when air at 21.0C db 50% RH is cooled to 5.0Cdb

 

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

When this air is cooled to 5.0C db it has the following properties.

Temperature db = 5.0 C
Humidity = 100% RH
Moisture Content = 0.0054 kg/kg
Specific Volume = 0.844 m³/kg
Temperature wb = 5.0 C
Specific enthalpy = 18.5 kJ/kg

The reduction in dry bulb temperature is 16.0C. The difference in specific enthalpy is 22.5 kJ/kg. The difference in moisture content is 0.0024 kg/kg. If this air were to be heated back up to 21.0C db, it would have a humidity of 34% RH. This is one method and probably the most common method of achieving dehumidification.

The next post will look at humidification.

 

See more at: http://www.gibbonsgroup.co.uk/blog/2013/06/the-psychrometric-chart-4-cooling-and-dehumidification