HVAC| Energy & Environment| Electrical| General

# Do you know how much heat is getting added up to the air stream through AHU fan ?

In any large air conditioning system, AHU - Air handling unit is considered as the major equipment which is used to distribute the conditioned air into the spaces. We all know that there will be a heat gain to the air because of normal heat loads in the space along with ceiling lighting, human presence etcetera. But the interesting thing to notice is even the AHU fan itself adds up the heat to the air and raises the temperature of conditioned air to some extent. That's one of the reasons why you observe a return air temperature little higher than that of the room air temperature. Other heat gains include that from lights, air stratification, duct wall and all.

In this section, I will be explaining how to calculate the heat gain from that of an AHU fan.Â

When air passes over a fan mounted in supply or return air duct, there will be some heat gain from the work done by the fan to pressurise the air-stream, as well as heat gain from the fan motor if this is mounted in the line of airflow. A fan in the supply air-stream will require a lower off-coil temperature from the heating or cooling coil in order to maintain the desired supply air temperature. A fan in the extract air-stream will raise the temperature in the return air duct coming onto the cooling coil.

Calculation of heat gains due to AHU fan can be assessed with the help of the energy balance equation.
The electrical energy supplied to the fan is equated to the heat gain of the air-stream. This approach is valid as all the energy supplied to the fan eventually ends up as a heat gain to the air-stream.

1. Fan power = fan total pressure Ã— volumetric flow rate Ã· fan total efficiency
2. Heat gain by air-stream = volumetric flow rate x density x specific heat capacity Ã—Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â temperature rise (Î”t).
Equating these two cancels out the volumetric flow rate and gives:

Fan total pressure Ã· fan total efficiency = density Ã— specific heat capacity x temperature rise

Î”t(k) = fan pressure Ã· (fan total efficiency Ã— density Ã— specific heat capacity)

Standard figures are used for density (1Â·2 kg/m3) and specific heat capacity (1Â·026 kJ/kg K). So for a nominal 1 kPa fan pressure, the heat gain to the air-stream is given by:

Î”t(k) = 0Â·812 Ã· fan total efficiency K.

For a specific fan, multiply this nominal temperature difference by the fan total pressure in kilopascals.

Case 1: Motor outside air-stream

The fan total efficiency is defined as the ratio of impeller output mechanical power to shaft electrical input. A typical figure is 70% ( widely varies with manufacturer and size). This gives a temperature difference of around 1Â·2 K/kPa.

Case 2: Motor and fan bearings inside air-stream

Here, all the heat generated by the fan, the motor and the transmission goes into the air-stream, so the efficiency of both the fan and the motor are taken into account. Typical motor efficiency is 90 %. This gives a temperature difference of 1Â·3 K/kPa.

It is evident that utmost care must be taken in calculating fan gains, as these depend on fan pressure and efficiency, which can vary according to the type of system.