Low Voltage Capacitor Banks,Medium Voltage Capacitor Banks,Static Var Compensatior

An example of design the capacity of Static Var Compensatior(Capacitor banks)

Jan 21,2021

ZDDQ Engineers give an example for reference,

An Alternator is supplying a load of 650 kW at Power factor 0.65. What size of Capacitor banks (kVAR) is required to raise the Power Factor to unity (1 or >0.95)? And how many more kW can the alternator supply for the same kVA loading when P.F improved.

Supplying kW = 650 kW

Original P.F = Cosθ1 = 0.65

Final P.F = Cosθ2 = 1

θ1 = Cos-1 = (0.65) = 49°.45; Tan θ1 = Tan (41°.24) = 1.169

θ2 = Cos-1 = (1) = 0°; Tan θ2 = Tan (0°) = 0

Required Capacitor kVAR to improve P.F from 0.75 to 0.90

Required Capacitor kVAR = P (Tan θ1 – Tan θ2)

= 650kW (1.169– 0)

= 759.85 kVAR