Worked problem

Power Factor Correction: How Many kvar of Capacitors?

Find the capacitor kvar needed to raise a 500 kW load from 0.80 to 0.95 power factor, and the drop in apparent power and line current.

Electrical & Power · Last reviewed

Given

A 480 V, 3-phase facility draws 500 kW at a power factor of 0.80 lagging. The utility asks for 0.95 or better.

Find

The capacitor kvar to add, and the change in apparent power and line current.

Solution

kvar = P × (tan φ1 − tan φ2), where φ = arccos(PF)
  1. tan φ1 = tan(arccos 0.80) = 0.750. tan φ2 = tan(arccos 0.95) = 0.329.
  2. kvar = 500 × (0.750 − 0.329) = 210.7 kvar. A standard bank of about 200 to 225 kvar is the practical choice.
  3. Apparent power before: 500 ÷ 0.80 = 625 kVA. After: 500 ÷ 0.95 = 526.3 kVA, a drop of about 16%.
  4. Line current before: 625,000 ÷ (√3 × 480) = 752 A. After: 526,300 ÷ (√3 × 480) = 633 A.

Answer

about 211 kvar | 625 → 526 kVA | 752 → 633 A

Common mistakes

  • Using the apparent power in place of the real power in the kvar formula.
  • Overcorrecting. A leading power factor can cause overvoltage and resonance problems.
  • Ignoring harmonics. Where drives or rectifiers are present, capacitors can resonate with the system and need detuning reactors.
  • Sizing for full load when the load varies. Switched or automatic stages follow the load.

Try two variations

  1. A 300 kW load from 0.75 to 0.95: tan φ1 = 0.882, tan φ2 = 0.329, so kvar = 300 × 0.553 = 166 kvar. Apparent power falls from 400 kVA to 316 kVA.
  2. A 500 kW load from 0.85 to 0.95: tan φ1 = 0.620, so kvar = 500 × (0.620 − 0.329) = 145.5 kvar.

The multiplier shortcut

Tables often give a multiplier for each pair of power factors, so that kvar = kW × multiplier. From 0.80 to 0.95 the multiplier is 0.421, which gives the 210.7 kvar in the main problem. The multiplier is just tan φ1 − tan φ2.

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Questions

Does power factor correction reduce real power use?

It reduces the current and the losses in the supply conductors and transformer, which is a small saving. It mainly cuts the apparent power, which is what many utilities bill against.

Does it lower the current inside the plant?

Only upstream of the capacitors. Loads on the load side of the bank still draw their original current.

Where should the capacitors be installed?

At the individual load, at a motor control center, or at the service. Capacitors at the load cut current in the most conductors but cost more per kvar. A central bank at the service is simpler and can switch in stages.

What is a detuned or filtered capacitor bank?

A bank with series reactors that moves its resonant frequency away from the harmonics present, which prevents resonance in systems with drives and other nonlinear loads.

What does a power factor penalty look like?

Utility rules differ. Many bill on peak kVA or add a charge when the power factor falls below a threshold such as 0.90 or 0.95. Check the tariff.

Educational material. Examples use simplified, hypothetical inputs and are not a design, a construction document or professional engineering advice. Confirm requirements with the codes and standards adopted where the work is built and with a licensed engineer.

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