An example of how to calculate the capacity of reactive power (kVAr)?

An Motor is supplying a load of 700 kW at a P.F (Power factor) of 0.65. What size of Capacitor in kVAr is required to raise the P.F (Power Factor) to unity (1)?

And how many more kW can the motor supply for the same kVA loading when P.F improved.

Supplying kW = 700 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 0.65 to 1

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

= 700kW (1.169– 0)

= 818.3 kVAr

ZDDQ Reactive Power Compensation Devices,

Low voltage Automatic Capacitor banks

Medium voltage Automatic Capacitor Banks

Low voltage Static Var Generator

Medium voltage Statcom

Low voltage hybrid power factor correction.