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Sizing of capacitor for power factor correction of system


Consider a system having following loads , we want to improve the system power factor to 0.95

EquipmentOutput
KW
EfficiencyLoad
KW
Power factorKVA
Motor A300.933.30.841.7
Motor B900.9297.80.85115
Lighting200.8523.50.9525
Heaters10 10110
Total150 164.6 191.7

Total load in KW = 164.6

And total load in KVA = 191.7

Therefore , the average power factor of the system will be = 164.6/191.7 =0.86

As we want to improve this system power factor from 0.86 to 0.95.

Existing power factor = 0.86 = Cosθ1 ; therefore tan θ1 =0.593

Required power factor = 0.95 = Cosθ2; therefore tan θ2=0.328

So, amount of capacitors to be installed to improve power factor will be

= Total Load (tan θ1 – tan θ2)

= 164.6(0.593-0.328)

= 44 KVAr


One Comment

  1. Manu Pareek

    Usefull article

    Reply

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