Power factor Improvement formula; Additional KVAR required for Power factor improvment.

Power Factor Improvement with Capacitor Bank Sizing

Why Power Factor Improvement is Needed?

• Most industries operate with inductive loads (motors, pumps, lighting ballasts, welding machines, etc.), which lower the power factor.

• Typical industry power factor: 0.80 – 0.85.

• Low PF leads to:

  • Higher current flow in the system.

  • Energy losses (15–20%).

  • Utility penalties for reactive power consumption.

• Maintaining PF near unity (1.0) reduces losses and avoids penalties.

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Formula for Capacitor Bank Sizing

To calculate the required capacitor (kVAR) for PF correction:

$$\text{Q (kVAR)} = \text{P (kW)} \times \left[\tan(\cos^{-1} \, \text{PF}_{\text{initial}}) - \tan(\cos^{-1} \, \text{PF}_{\text{desired}})\right]$$

Where:

• P (kW) = Active load

• PF_initial = Present running power factor

• PF_desired = Target power factor (close to 1)

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Example Calculation

• Load = 200 kW

• Initial PF = 0.8

• Desired PF = 1.0

$$\text{Q} = 200 \times [\tan(\cos^{-1}(0.8)) - \tan(\cos^{-1}(1))]$$ $$\text{Q} = 200 \times [\tan(36.86°) - \tan(0°)]$$ $$\text{Q} = 200 \times (0.75 - 0) = 150 \, \text{kVAR}$$

✅ Answer: An additional 150 kVAR capacitor bank is required to improve PF from 0.8 → 1.0.

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Key Notes

• In practice, PF correction is usually done up to 0.95–0.99, not exactly 1.0, to avoid leading PF issues.

• Capacitor banks can be:

  • Fixed type (constant loads)

  • Automatic / APFC panels (variable loads)

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