Knowledgebase
Knowledgebase: Sterling ACR Guidance
 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