Enter Values
Formula
Capacitor kVAR = kW × [tan(cos-1 PF1) - tan(cos-1 PF2)]
PF1 is the existing power factor and PF2 is the higher target power factor.
Examples
Example 1
100 kW, 0.75 PF to 0.95 PF
kVAR = 100 × [tan(acos(0.75)) - tan(acos(0.95))]
kVAR = 55.30
Example 2
250 kW, 0.80 PF to 0.95 PF
kVAR = 250 × [tan(acos(0.80)) - tan(acos(0.95))]
kVAR = 105.33
Example 3
75 kW, 0.70 PF to 0.90 PF
kVAR = 75 × [tan(acos(0.70)) - tan(acos(0.90))]
kVAR = 40.19
What Correction Changes
Correction capacitors supply part of the reactive power locally. This can reduce source current, feeder losses, voltage drop, and utility power-factor penalties without reducing the load's real kW.
Engineering Considerations
Actual capacitor-bank selection depends on available standard steps, system voltage, frequency, switching method, load variation, harmonics, resonance, and utility requirements.
Do not correct beyond the actual operating requirement. Overcorrection can produce leading power factor and unacceptable voltage or resonance conditions.
Common Mistakes
- Correcting to 1.0 PF without checking utility requirements or load variation.
- Ignoring harmonics, resonance, capacitor switching steps, and voltage rating.
- Using a nameplate kW value when measured operating kW is needed.
- Assuming correction reduces real energy used by the load.
Reference Notes
The formula uses the standard displacement power-factor correction relationship based on real power and phase-angle tangent difference. Practical capacitor-bank design should use measured load data, harmonic review, utility requirements, manufacturer data, and applicable standards.