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Evgenii Konkin
Evgenii Konkin

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The Engineering Math Behind Power Factor Correction: Capacitor Bank Sizing

#capacitorbankcalculation #powerfactorcorrection #kvarcalculation #reactivepowercompensation

A single industrial motor running at 0.75 power factor can waste over 25% of the current it draws as reactive power—current that heats conductors, overloads transformers, and inflates utility bills. That wasted current isn't imaginary; it's real Amperes that could be eliminated with the right capacitor bank.

The Formula

The capacitor bank size in kVAR is given by:

Q_c = P × (tan(arccos(PF_initial)) - tan(arccos(PF_target)))
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Where:

  • P = real power in kW. This is the actual work-producing power consumed by the load. It sets the scale of compensation—a 500 kW load needs five times the kVAR of a 100 kW load for the same PF improvement.
  • PF_initial = starting power factor (0.01–0.9999). A lower PF means more reactive power per kW, so the tangent difference is larger.
  • PF_target = target power factor (0.01–0.9999). Approaching unity requires increasingly large kVAR because the tangent function grows rapidly as the angle approaches 90° (PF=0).

The term tan(arccos(PF)) converts power factor to the ratio of reactive power to real power (kVAR/kW). The difference between initial and target ratios gives the required compensation per kW.

Worked Example 1

A facility draws 350 kW at 0.72 PF and wants to correct to 0.95 PF.

Step 1: Compute initial angle and tangent.

φ1 = arccos(0.72) = 0.766 rad
tan φ1 = tan(0.766) = 0.963
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Step 2: Compute target angle and tangent.

φ2 = arccos(0.95) = 0.318 rad
tan φ2 = tan(0.318) = 0.329
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Step 3: Calculate kVAR.

Q_c = 350 × (0.963 - 0.329) = 350 × 0.634 = 221.9 kVAR
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Result: A 222 kVAR capacitor bank is needed. This is classified as NORMAL by the Result Intelligence System.

Worked Example 2

A small workshop runs 50 kW of machinery at 0.60 PF and aims for 0.92 PF.

Step 1: Initial.

φ1 = arccos(0.60) = 0.927 rad
tan φ1 = tan(0.927) = 1.333
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Step 2: Target.

φ2 = arccos(0.92) = 0.402 rad
tan φ2 = tan(0.402) = 0.426
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Step 3: kVAR.

Q_c = 50 × (1.333 - 0.426) = 50 × 0.907 = 45.35 kVAR
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A 45 kVAR bank is required. Despite the smaller load, the low starting PF demands proportionally more compensation per kW.

What Engineers Often Miss

  1. Harmonic resonance: Adding capacitors shifts the system's resonant frequency. If harmonic sources (e.g., VFDs) are present, resonance can amplify harmonics, damaging equipment. Always conduct a harmonic study before installing fixed banks.

  2. Load variation: Fixed banks are designed for constant loads. If the load varies widely, the bank may overcorrect during light loads, leading to overvoltage and utility penalties. Use switched banks or automatic controllers for variable loads.

  3. Target PF vs. unity: Utilities typically charge penalties below 0.95 PF. Correcting to 1.0 is often uneconomical because the kVAR requirement skyrockets near unity. Always check the utility tariff before selecting the target.

Try the Calculator

Use the Capacitor Bank Calculator to quickly size banks for your projects and see the intelligence classification.

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