Knowledgebase
Specific Fan Power
Posted by Mike Gordon on 22 July 2015 02:42 PM

The following basic calculation should be supplied as part of your audit pack

Example

Supply volume: 5 m3·s–1 (5000 litre/s)

Extract volume: 4 m3·s–1 (4000 litre·s–1)

Supply motor capacity: 7.5 kW

Extract motor capacity: 4 kW

Step 1

Calculate the sum of the installed supply and extract motor capacities:

7.5 + 4.0 = 11.5 kW

Step 2

The sum of the motor capacities is then multiplied by 0.7 to

give 70% of the installed motor capacity, which allows for

typical design tolerances:

11.5 x 0.7 = 8.05 kW

The 0.7 multiplier is to reflect the typical difference between

nominal motor power and actual energy absorbed. For an

ideally-matched motor and load, the factor would be 1. The

absorbed power would be slightly above the nominal power

because motors are not 100% efficient. The SFP could

therefore be higher than this approximate figure.

Note: it may be possible to determine power input more

directly by using a non-contact meter. Where a frequency

inverter is in use, absorbed fan powers may be indicated on

the inverter’s display. Where this is not available, commissioning

data should be consulted (if available) to establish

the designer’s intended operational frequency. These data

can then be used in conjunction with installed motor

capacity to establish the anticipated absorbed power of the

system. When the operational motor frequency differs from

the commissioned value, air flow rates shall be recalculated

accordingly.

Step 3

Express the installed motor capacity in watts:

8.05 kW = 8050 W

Step 4

This figure is then divided by the greater of either the supply

or extract volumes:

8050 ÷ 5000 = 1.61 W/(litre·s–1)

 

Where specific airflow rates are unavailable, they can be

estimated by multiplying the cross-sectional area of the

cooling coil by its design face velocity (typically 2.5 m·s–1).

For example, an AHU cooling coil having width of 2.0 m and

a height of 1.5 m would have an estimated airflow rate as

follows:

Cross sectional area = 2.0 m x 1.5 m = 3 m2

Estimated flow rate = 3 m2 x 2.5 m·s–1 = 7.5 m3·s–1