Power Factor Correction: Capacitor Sizing & Energy Savings
Why Correct Power Factor?
Low power factor means high reactive current that does no useful work but still flows through your wiring, transformers, and utility lines. This causes:
- Utility penalties: Many utilities charge $0.50–$2.00/kVAR per month for low PF
- Higher demand charges: kVA demand may exceed kW demand, increasing monthly bills
- Voltage drop: Reactive current causes additional voltage drop in conductors
- Transformer loading: Transformers rated in KVA are derated by low PF
Power Factor Triangle
| PF | cos(θ) | tan(θ) | Reactive/Real | Rating |
|---|---|---|---|---|
| 1.00 | 1.000 | 0.000 | 0% | Perfect |
| 0.95 | 0.950 | 0.329 | 33% | Excellent |
| 0.90 | 0.900 | 0.484 | 48% | Good |
| 0.85 | 0.850 | 0.620 | 62% | Acceptable |
| 0.80 | 0.800 | 0.750 | 75% | Below average |
| 0.70 | 0.700 | 1.020 | 102% | Poor |
| 0.60 | 0.600 | 1.333 | 133% | Very poor |
Capacitor Sizing Formula
kVAR = kW × (tan(θold) − tan(θnew))
Or equivalently:
kVAR = kW × (√(1/PFold² − 1) − √(1/PFnew² − 1))
Quick Reference: kVAR per kW
| From PF | To 0.90 | To 0.95 | To 0.98 | To 1.00 |
|---|---|---|---|---|
| 0.60 | 0.849 | 1.005 | 1.108 | 1.333 |
| 0.65 | 0.692 | 0.848 | 0.951 | 1.169 |
| 0.70 | 0.536 | 0.691 | 0.794 | 1.020 |
| 0.75 | 0.398 | 0.553 | 0.656 | 0.882 |
| 0.80 | 0.266 | 0.421 | 0.524 | 0.750 |
| 0.85 | 0.141 | 0.296 | 0.399 | 0.620 |
| 0.90 | 0.000 | 0.156 | 0.258 | 0.484 |
Example: 100 kW load at PF 0.75, correct to 0.95: kVAR = 100 × 0.553 = 55.3 kVAR → install 60 kVAR bank.
Worked Example
Facility: 200 kW average demand, PF = 0.72, utility charges $1.50/kVAR for PF below 0.85.
Step 1: Current kVAR = 200 × √(1/0.72² − 1) = 200 × 0.964 = 192.8 kVAR
Step 2: Target kVAR at 0.95 PF = 200 × 0.329 = 65.8 kVAR
Step 3: kVAR to add = 192.8 − 65.8 = 127 kVAR
Step 4: Install 130 kVAR capacitor bank (next standard size)
Step 5: Penalty avoided: (192.8 − 65.8) × $1.50 = $190.50/month = $2,286/year
Step 6: Capacitor bank cost: ~$5,000–$8,000 installed
Step 7: Payback = $6,500 / $2,286 = 2.8 years
Common Power Factor Correction Mistakes
- Over-correction: Correcting above 0.98 can cause leading PF (capacitive), which is just as problematic. Target 0.95–0.98.
- Not considering harmonics: VFDs and non-linear loads create harmonics that can cause capacitor resonance and failure. Use detuned reactors with capacitors.
- Fixed vs automatic correction: Variable loads need automatic capacitor banks with PF controllers. Fixed caps only work for constant loads.
- Installing on VFD output: Never install PF correction capacitors on the output of a VFD — the VFD sees them as a short circuit.
Standards Reference
- IEEE 519 — Harmonic Limits
- NEC Article 460 — Capacitors
- NEMA CP-1 — Shunt Power Capacitors
- IEEE 18 — Shunt Power Capacitors
Frequently Asked Questions
How do I calculate power factor correction?
Use kVAR = kW × (tan(θ_old) − tan(θ_new)). Find tan(θ) from your current PF: tan(θ) = √(1/PF² − 1). Example: 100 kW at PF 0.75 to 0.95: kVAR = 100 × (0.882 − 0.329) = 55.3 kVAR. Install a 60 kVAR capacitor bank.
What size capacitor for power factor correction?
Divide your kW load by the kVAR/kW factor from the table. For PF 0.75 → 0.95: multiply kW by 0.553. A 200 kW load needs 110.6 kVAR → install 120 kVAR. Always select the next standard size up and avoid over-correction above 0.98.
Is power factor correction worth it?
For commercial/industrial facilities with utility PF penalties: absolutely. Payback is typically 1–3 years. For residential customers: rarely worth it — most residential utilities don't charge PF penalties. The ROI depends on your utility rate structure and current PF.
Where do I install power factor capacitors?
Install at the main switchboard for facility-wide correction, or at individual motors for local correction. Main switchboard correction is simpler but doesn't reduce feeder losses. Motor-level correction reduces losses throughout the facility but costs more per kVAR.
Can capacitors cause harmonics?
Capacitors don't generate harmonics, but they can amplify existing harmonics by creating a resonance circuit with the system inductance. If total harmonic distortion (THD) is above 5%, use detuned reactors (typically 7% or 14% detuning) with the capacitors.