Air cooling : Contact factor and Bypass factor of air cooling coils

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1. Cooling humid air without condensation
2. Cooling humid air with condensation : dehumidification

Air handling units require the cooling of humid air to condition it thanks to a cooling coil. The cooling coil is never 100% efficient. The actual efficiency of a cooling coil can then be expressed with a contact factor or a bypass factor (both are related).

The case considered here is a cooling coil that allows to cool down air from a condition A to a condition B, while the cooling coil is at the temperature of point C (part of the water in the air is condensed here).

Psychrometric chart : cooling humid air with condensation in AHU

More information on humid air cooling with condensation can be found at this page.

1. Contact factor

The efficiency can be represented with a contact factor.

Calculation of the contact factor of a cooling coil

The contact factor can be calculated with the following formula :

β = (x_A-x_B) / (x_A-x_C) = (h_A-h_B)/(h_A-h_C)

With :

β = contact factor
h_A = enthalpy of humid air at condition A, inlet of the cooling coil, (J/kg_dry_air)
h_B = enthalpy of humid air at condition B, outlet of the cooling coil, (J/kg_dry_air)
h_C = enthalpy of humid air at condition C, in contact with the cooling coil, (J/kg_dry_air)
x_A = humidity ratio (=Absolute humidity of air) at condition A (-)
x_B = humidity ratio (=Absolute humidity of air) at condition B (-)
x_C = humidity ratio (=Absolute humidity of air) at condition C (-)

In practice, the contact factor can be approximated thanks to a ratio of the temperatures of points A, B and C :

β ≈ (T_A-T_B) / (T_A-T_C)

With :

T_A = temperature of humid air at condition A, inlet of the cooling coil, (K)
T_B = temperature of humid air at condition B, outlet of the cooling coil, (K)
T_C = temperature of humid air at condition C, in contact with the cooling coil, (K)

Example of contact factor calculation : Step by Step calculation

Air a 35c and 30% RH is to be cooled down until 25c with a cooling coil at 10c, what is the contact factor value ?

STEP 1 : approximate the bypass factor with temperatures

The contact factor can be approximated as β ≈ (T_A-T_B) / (T_A-T_C) = (35-25)/(35-10)=10/25=0.4

STEP 2 : calculate with a psychrometric chart

The psychrometric chart given at the beginning of the page can be used to calculate the contact factor with the humidity ratio for example : β = (x_A-x_B) / (x_A-x_C) = (0.010-0.008)/(0.010-0.005)=0.002/0.005 = 0.4

2. Bypass factor

The efficiency can be represented with a bypass factor.

Calculation of the bypass factor of a cooling coil

The bypass factor can be calculated with the following formula :

BPF = (x_B-x_C) / (x_A-x_C) = (h_B-h_C)/(h_A-h_C) = (T_B-T_C) / (T_A-T_C) = 1 - β

With :

BPF = bypass factor
β = contact factor
h_A = enthalpy of humid air at condition A, inlet of the cooling coil, (J/kg_dry_air)
h_B = enthalpy of humid air at condition B, outlet of the cooling coil, (J/kg_dry_air)
h_C = enthalpy of humid air at condition C, in contact with the cooling coil, (J/kg_dry_air)
x_A = humidity ratio (=Absolute humidity of air) at condition A (-)
x_B = humidity ratio (=Absolute humidity of air) at condition B (-)
x_C = humidity ratio (=Absolute humidity of air) at condition C (-)
T_A = temperature of humid air at condition A, inlet of the cooling coil, (K)
T_B = temperature of humid air at condition B, outlet of the cooling coil, (K)
T_C = temperature of humid air at condition C, in contact with the cooling coil, (K)

3. Cooling requirements

It is possible to calculate the heat flow required in the cooling coil thanks to the flow of humid air to cool down. The heat flow can then be expressed as :

Q_cooling
= m_air.(h_A-h_B)
= energy used to cool down from T_A to T_B + energy used to condensate the water in between A and B
= m_air.Cp_air.(T_A-T_B)+m_air.h_vap.(x_A-x_B)

(Q_cooling is expressed here as positive, please adjust the sign according to the convention used in your calculations)

With :

Q_cooling = heat flow rate or power to bring the air from condition A to B (kW)
m_air = mass flow rate of air (kg/s)
h_A = enthalpy of humid air at condition A, inlet of the cooling coil, (kJ/kg_dry_air)
h_B = enthalpy of humid air at condition B, outlet of the cooling coil, (kJ/kg_dry_air)
Cp_air = specific heat capacity of air (kJ/kg.c) = 1.01 kJ/kg.c
h_vap = enthalpy of vaporization of water (kJ/kg) = 2502 kJ/kg