When does condensation occur during humid air cooling?

Condensation occurs whenever the surface temperature of the cooling coil (or cooling media) drops below the dew point temperature of the incoming humid air.

What is the cooling coil Bypass Factor (BF)?

The Bypass Factor (BF) represents the fraction of air that passes through the cooling coil without contacting the cold fins or tubes. It depends on coil depth, fin spacing, and air face velocity.

How is the total cooling load calculated for humid air with dehumidification?

Total cooling load is the mass flow rate of dry air multiplied by the difference between inlet specific enthalpy and outlet specific enthalpy: Q_total = m_dry * (H_in - H_out).

Cooling Humid Air

Follow us on Twitter
Question, remark ? Contact us at contact@myengineeringtools.com

1. Cooling humid air without condensation
2. Cooling humid air with condensation : dehumidification
3. Interactive Cooling Humid Air Calculator

Air handling units require the cooling of humid air to condition it. It is then important to be able to calculate the energy required to bring air from a temperature T₁ to a temperature T₂ especially as the cooling of humid air may involve the condensation of some water.

⚡ Interactive Humid Air Cooling & Dehumidification Calculator

⚠️ ENGINEERING NOTICE & EDUCATIONAL DISCLAIMER: This interactive calculator is provided exclusively for preliminary estimation and educational purposes. It is not intended for detailed design or equipment procurement without certified vendor rating. No warranty, expressed or implied, is provided, and no liability is assumed.

Unit System: Metric SI (°C, kg/h, kW, g/kg) US Customary (°F, lb/h, BTU/h, gr/lb)

Inlet Air Temp (°C):

Inlet Relative Humidity (%):

Target Outlet Temp (°C):

Cooling Coil Surface Temp (°C):

Air Mass Flow Rate (kg/h):

Atmospheric Pressure (kPa):

Calculation Results

Inlet Dew Point

14.8 °C

Inlet Abs. Humidity

10.53 g/kg

Outlet Abs. Humidity

9.37 g/kg

Coil Bypass Factor (BF)

0.60

Sensible Cooling Duty

10.24 kW

Latent Condensation Duty

2.91 kW

Total Cooling Power

13.15 kW

Condensate Flow Rate

4.18 kg/h

1. Cooling humid air without condensation

Humid air is cooled down without change of state of the water, at constant specific humidity, but not at constant relative humidity, as the relative humidity increases when the temperature decreases. This case is observed when the temperature of the cooling media is not below the dew point of the humid air.

1.1 Calculation of humid air heating energy requirements

By calculation

The 1st step is to define the enthalpy of humid air :

\[ H' = C_{p}' \cdot (T - T_0) + \omega \cdot \Lambda_0 \]

With :

\(H'\) = enthalpy of humid air (\(\text{J/kg}_{\text{dry air}}\))
\(C_{p}'\) = specific heat of humid air (\(\text{J/K/kg}_{\text{dry air}}\))
\(T\) = temperature of calculation (\(\text{K}\))
\(T_0\) = temperature of reference (\(\text{K}\))
\(\omega\) = Absolute humidity of air (\(\text{kg}_{\text{water}}/\text{kg}_{\text{dry air}}\))
\(\Lambda_0\) = vaporization enthalpy at \(T_0\) (\(\text{J/kg}\))

In practice, taking reference at 273.15 K (0°C):

\[ H' = (1005 + 1884 \cdot \omega) \cdot (T - 273.15) + 2.5023 \times 10^6 \cdot \omega \]

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 :

\[ H_B - H_A = (1005 + 1884 \cdot \omega) \cdot T_B - (1005 + 1884 \cdot \omega) \cdot T_A \]

Example of humid air cooling calculation : Step by Step calculation

Air at 35°C and 30% RH is to be cooled down until 25°C, 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 \(\omega \approx 0.010 \text{ kg water / kg of dry air}\).

STEP 2 : calculate the difference in enthalpy in between states

\[ H_B - H_A = (1005 + 1884 \times 0.010) \times (273.15 + 25) - (1005 + 1884 \times 0.010) \times (273.15 + 35) = -10238 \text{ J/kg of dry air} \]

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

Psychrometric chart : cooling humid air with condensation

This transformation is actually ideal. In an air handling unit, if the cooling coil surface is at 10°C, the resulting air will not be at 10°C 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.

\[ BF = \frac{T_{B} - T_{C}}{T_{A} - T_{C}} = \frac{\omega_B - \omega_C}{\omega_A - \omega_C} \]

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

An air handling unit is conditioning outside air at 35°C and 30% RH to 25°C by using a coil at 10°C. 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 14.8°C, which means that water will condense on the coil surface at 10°C.

Step 2 : Assess the conditions on the coil

On the coil at 10°C, the air film in contact is at saturation. Point C can thus be positioned on the diagram at 10°C and 100% RH.

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

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

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 rate:

\[ \dot{Q}_{total} = \dot{m}_{dry} \cdot (H_A - H_B) \] \[ \dot{m}_{cond} = \dot{m}_{dry} \cdot (\omega_A - \omega_B) \]

Psychrometric chart : cooling humid air with condensation in AHU

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

💡 Industrial Rules of Thumb & Design Engineering Limits

Coil Air Face Velocity: Maintain cooling coil face velocity between 2.0 m/s and 2.5 m/s (400 - 500 fpm). Exceeding 2.5 m/s risks stripping condensed water droplets off the fins and carrying water down the ductwork.
Cooling Coil Bypass Factor (BF): Typical commercial AHU coils have BF values ranging from 0.05 to 0.20 (5% to 20%), depending on row depth (4 to 8 rows) and fin pitch (8 to 14 fins/inch). High BF (>0.30) indicates shallow coils or poor thermal contact.
Frosting Risk: If the coil refrigerant/chilled water temperature drops below 0°C (32°F) while handling humid air, ice frosting will occur, clogging fin passages and reducing airflow rate.
Condensate Drain Traps: Always design condensate trap seals with a minimum depth equal to the maximum static fan pressure plus 25 mm (1 in. w.g.) to prevent negative pressure blowout.