The formula
kW = kVA × pf ; kVAr = √(kVA² − kW²) ; hp = kW × η / 0.7355
- kVA
- apparent power — what the supply has to carry
- kW
- real power — what does work and what the meter bills
- kVAr
- reactive power — carried but never consumed
- pf
- power factor, cos φ, between 0 and 1
- η
- motor efficiency as a fraction — needed only for horsepower
Worked example
A 10 kVA supply at 0.8 power factor, feeding a motor of 88 % efficiency.
- kW = 10 × 0.8 = 8.0 kW
- kVAr = √(10² − 8²) = √36 = 6.0 kVAr
- Shaft power = 8.0 × 0.88 = 7.04 kW
- hp = 7.04 / 0.7355
8 kW real, 6 kVAr reactive, and 9.57 metric horsepower at the shaft.
Where you'll use it
Sizing a DG set or UPS against a load stated in kW, reading a transformer nameplate, and the standard exam question on the power triangle. The 2 kVA of headroom that disappears between a 10 kVA set and an 8 kW load is not a rounding error — it is the reactive power the set still has to carry.