Considering a temperature T_A (starting temperature) and a temperature T_B (end temperature), the energy required to cool down humid air, at constant absolute humidity is then :

H_B-H_A = (1005+1884*ω)*T_B-(1005+1884*ω)*T_A

Example of humid air cooling calculation : Step by Step calculation

Air a 35c and 30% RH is to be cooled down until 25c, what is the energy to be absorbed ?

STEP 1 : determine the absolute humidity of the air

It can be determined on a Mollier diagram or psychrometric chart : the absolute humidity is ~0.010 kg water / kg of dry air

STEP 2 : calculate the difference in enthalpy in between states

H_B-H_A = (1005+1884*0.010)*(273.15+25)-(1005+1884*0.010)*(273.15+35) = -10238 J/kg of dry air

Graphically

It is possible to calculate the energy required to cool down humid air by using a psychrometric chart. Knowing the starting conditions, it is possible to determine the final conditions by moving along the lines of constant specific humidity. The difference of enthalpy read on the graph between the starting and final conditions allows to calculate the energy required to heat up the humid air considered.

2. Cooling humid air with condensation : dehumidification

When cooling humid air, if the dew point is reached, water will start to condense and the absolute humidity of the air will decrease. It is possible to calculate the energy to be absorbed to perform this condensation thanks to a psychrometric diagram.

Compared to the example above, if we consider we cool down at 10c instead of 25c, starting from point A, the 1st transformation step will be to reach the saturation (RH=100%) then move along the saturation curve until reaching 10c. This would require a large amount of energy as the latent heat of the water in the air would have to be removed.

This transformation is actually ideal, in a air handling unit, if the cooling coil is at 10c, the resulting air will not be at 10c or at 100% RH% but in between. The air will indeed be cooled down, some water will condensate at the contact of the coil thus the absolute humidity will decrease, but without reaching the conditions that would bring the whole volume of air to saturation.

Example of humid air cooling (with condensation) calculation : Step by Step Calculation Guide

A air handling unit is conditioning outside air at 35c and 30% RH to 25c by using a coil at 10c. Calculate the energy required for cooling the air.

Step 1 : assess if there is condensation

There will be condensation if the temperature of the cooling coil is lower than the dew point of the air. In our case, the dew point is around 14c, this means that water will condense on the coil which is at 10c.

Step 2 : Assess the conditions on the coil

On the coil at 10c, the air will be at saturation, the point C can thus be positioned on the diagram at 10c and 100% RH

Step 3 : Assess the conditions of the air at the outlet of the AHU

The air leaves the AHU at 25c, the point B can then be determined by drawing a line in between points A and C and placing point B where the line crosses 25c.

Step 4 : calculate the energy required for cooling

As point A and B are now on the diagram it is possible to determine the specific enthalpy of both conditions. The actual heat to remove can be calculated by multiplying the difference of specific enthalpy by the mass flow.

It is also possible to calculate the amount of water condensed thanks to the difference in absolute humidity in between points A and B.