If I want to calculate the energy use of an AHU fan, knowing the maximum airflow, fan efficiency, and static pressure, what would be the most accurate way?
The aeraulic power of a fan is the product of the mass flow rate (kg/s) by the pressure differential (in Pa) accross the fan. Therefore, the maximum airflow that occurs at zero pressure drop and and static pressure at which there is no airflow rate are of no help, since at these ends of the fan characteristics, the aeraulic power is zero. The actual aeraulic power delivered by the fan depends on the circuit to which the air is delivered or more preciseley to its araulic resistance.
The electric power used is the aeraulic power multiplied by the fan efficiency, which itself also depends on the actual airflow rate.
The most accurate way is to use the fan characteristic curve (pressure versus flow) delivered by the factory of the fan, and the pressure drop characteristic of the araulic circuit (pressure drop versus flow rate) . The intersection of these two curves provides the actual airflow rate and pressure drop.
There is a simple formula to relate the airflow and total fan pressure to the electrical power consumption of the motor and it is this:
P (kw electrical consumption) = [Qv ( airflow rate in m3/s) x Δp ( total pressure difference across fan in Pa)]/ ( 1000 x ηe ( total system static fan efficiency)
P = (Qv x Δp)/ 1000 x ηe = kW of absorbed electrical fan power
an example;
You have 10 m3/s airflow rate, 1000 Pa total fan pressure and a fan static +motor efficiency of 0.72 then you have (10 x1000)/ (0.72 x 1000) = 13.89 kW.
Note that the fan efficiency should also include the motor efficiency, with an IE3 rated fan motor you could expect an efficiency of around 0.88 to 0.92 and you must multiply this by the static fan efficiency which could be anything from 0.7 to 0.9 depending on the fan type and rating. The fan and motor manufacturer/es should give you this information.
the above equation is 100% accurate once you use the correct data,