kW is real electrical power; kVA is apparent electrical power. For a consuming AC load, power factor connects them: kW = kVA × PF. A 10 kVA load at PF 0.8 draws 8 kW. A 10 kW load at the same PF requires 12.5 kVA.
Need a numerical result? Open the kVA to kW calculator or the kW to kVA calculator.
kW vs kVA: which number do you need?
| Quantity | Meaning | How it is used here |
|---|---|---|
| kW | Real power P | Electrical energy rate; use with hours to calculate kWh |
| kVA | Apparent power S | Relates RMS voltage and current |
| Power factor | Ratio P/S | Connects real and apparent power |
| Efficiency | Useful output/input power | Connects the input and output of equipment; it is not PF |
For a single-phase load, S = V × I ÷ 1,000 in kVA, using RMS values. In a balanced three-phase system, S = √3 × V × I ÷ 1,000, where V is line-to-line voltage and I is line current. Use the kVA to amps calculator when current is the result you need.
How to convert between kW and kVA
kW = kVA × PF; kVA = kW ÷ PF. Use the PF of the load and operating condition being studied. For these positive-load examples, 0 < PF ≤ 1. Do not use 80 where a field expects 0.80.
At PF 1, the numerical values in kW and kVA match. Below PF 1, kVA is numerically larger. If PF is unknown, the answer is not unique: 10 kVA could represent 10 kW at PF 1, 9 kW at PF 0.9 or 8 kW at PF 0.8.
Six worked conversion examples
These are mathematical scenarios with specified inputs, not verified product ratings.
| Known quantity | PF | Calculation | Result |
|---|---|---|---|
| 6 kVA | 1.00 | 6 × 1.00 | 6 kW |
| 15 kVA | 0.80 | 15 × 0.80 | 12 kW |
| 25 kVA | 0.92 | 25 × 0.92 | 23 kW |
| 9 kW | 0.75 | 9 ÷ 0.75 | 12 kVA |
| 18 kW | 0.90 | 18 ÷ 0.90 | 20 kVA |
| 30 kW | 0.95 | 30 ÷ 0.95 | 31.58 kVA |
Why equipment may have both a kW and a kVA limit
Consider a hypothetical source with separate continuous output limits of 10 kVA and 9 kW. A load of 8 kW at PF 0.8 requires 10 kVA: it reaches the apparent-power limit even though its real power is below 9 kW. A 9.5 kW load at PF 1 requires 9.5 kVA: it is below the kVA limit but exceeds the kW limit.
The arithmetic checks two limits, not the complete suitability of a UPS or generator. Actual equipment also has operating conditions, load compatibility and transient limits. Use its documentation; passing one rating does not cancel another.
Power factor and motor efficiency are different
If a motor delivers 18 kW at its shaft with an assumed efficiency of 90%, its electrical input is 18 ÷ 0.90 = 20 kW. At an assumed PF of 0.80, its input apparent power is 20 ÷ 0.80 = 25 kVA. The order is shaft output → electrical input → apparent power.
If 20 kW is already a measured electrical input, do not divide by efficiency again. The U.S. Department of Energy distinguishes input power, rated shaft power and efficiency in its guide to determining motor load and efficiency. It also shows why operating load matters when interpreting motor measurements.
Does changing PF change energy use?
Take a hypothetical load held at 20 kW. At PF 0.80 it requires 25 kVA; at PF 0.95 it requires about 21.05 kVA. In both cases, four hours at 20 kW corresponds to 80 kWh at the stated measurement point. Lower apparent power is not automatically an equal percentage reduction in energy.
For a separate energy calculation, use kW and operating time to calculate kWh.
Frequently asked questions
Is 1 kVA always 0.8 kW?
No. That result assumes PF 0.8. The power factor must be known or explicitly stated as an assumption.
Do I multiply by three when converting total three-phase kVA to kW?
No. When the kVA value already represents the whole balanced load, multiply it by the relevant overall PF. Do not count the phases again.
Is power factor the same as efficiency?
No. PF compares real and apparent electrical power; efficiency compares output and input real power.
Can I use cos φ for every load?
For sinusoidal voltage and current, PF corresponds to cos φ. With waveform distortion, the displacement angle alone may not describe total PF. Use an appropriate measured total power factor for a P/S conversion.
Can kVA alone predict runtime?
No. Runtime requires an energy source and real-power demand, with the relevant conversion losses and operating constraints.