Condensate

To present an example, we shall take point 1 on diagram, temperature of 35°C and relative humidity of 50%, thus the air at this point includes only 50% of the humidity in steam form that is the maximum for this temperature.

The moisture content ,,x,, for this point is 19 grams of H2O in one kilogramme of air.

When we start to cool such air down, the temperature shall drop to the constant “x”.

When the temperature shall be lowered to the point 1' on diagram, which is located on the relative humidity curve φ = 100%, this shall mean that the air in the temperature of point 1' (approx. 23°C) includes the maximum amount of water in the form of water steam for the given temperature. Temperature of the point 3 is called the wet thermometer temperature, dew point temperature is point 4.

Cooling air below this temperature shall result in formation of small water drops in the air, this is the way that dew is formed on exchangers in air conditioning systems.

In the example, we cool the air down to the temperature of 13°C this is point 2 on diagram.

Relative humidity at this point shall amount to 100% and the humidity content shall be 10 grams of H2O per kilogramme of air.

The difference in humidity content in point 1' and point 2 is the amount of the formed condensate, that is 19-10=9 grams of H2O per kilogramme of air.

• An example of how to count the amount of condensate for a particular expenditure of cooled air:

We cool the air down from the temperature t1 = 35°C, relative humidity φ1 = 50% and humidity content x1 = 19 g/kg . Air quantity Qv = 1000 m³/h. We cool the air down to t2 = 13°C.

For the calculation let us adopt the air density p = 1.2 kg/m³.

The air mass volume equals to: 

The difference in the air’s humidity content in point “1” and point “2” is: 

Amount of formed condensate “A” is:

Therefore, during cooling of a given amount of air from temperature at point 1 to temperature at point 2 on diagram, there shall be 10.8 kilogrammes of condensate formed during an hour.