Leistung Factor Correction Rechner

Berechnen Sie capacitor bank size for power factor correction. Reduce reactive power charges und line losses.

kW

Required Capacitor Bank

55.3 kVAR

Current Reactive Power88.2 kVAR
New Reactive Power32.9 kVAR
Apparent Power Reduction28.1 kVA

Required Capacitor Bank vs Real Power

Formel

## Power Factor Correction Low power factor means the utility must deliver more current (and apparent power) than the load actually uses. Capacitor banks supply reactive power locally. ### Formula **kVAR needed = kW x (tan(arccos(PF_current)) - tan(arccos(PF_target)))** ### Benefits - Reduced utility power factor penalties - Lower line current (smaller cables, less loss) - Freed transformer and distribution capacity Most utilities penalize power factors below 0.90. Correcting to 0.95 is a common economic target.

Lösungsbeispiel

100 kW load at PF 0.75, correcting to PF 0.95.

  1. 01Current angle: arccos(0.75) = 41.41 degrees, tan = 0.8819
  2. 02Target angle: arccos(0.95) = 18.19 degrees, tan = 0.3287
  3. 03kVAR needed: 100 x (0.8819 - 0.3287) = 55.3 kVAR
  4. 04kVA reduction: 100/0.75 - 100/0.95 = 133.3 - 105.3 = 28.1 kVA

Häufig Gestellte Fragen

What causes low power factor?

Inductive loads like motors, transformers, and fluorescent lighting draw reactive current that lags voltage, reducing power factor.

Can I overcorrect power factor?

Yes. A leading power factor (above 1.0 correction) can cause voltage rise and resonance. Do not exceed PF 0.98-0.99.

Where should capacitors be installed?

At the load for maximum benefit, at the main bus for simplicity, or a combination. Automatic switching banks handle varying loads.

Lernen

Ohm's Law Guide

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