Worked Example: Sizing a Power Factor Correction Capacitor Bank for an Industrial Site

An industrial manufacturing facility is experiencing power factor penalties from their electricity retailer. The measured power factor is 0.72 lagging, well below the target of 0.95 lagging typically required to avoid penalty charges. This worked example sizes a capacitor bank to correct the power factor, then determines the cable and fuse ratings for the capacitor bank connection.

Poor power factor increases apparent power (kVA) drawn from the supply, resulting in higher current for the same real power output. This wastes transformer and cable capacity, increases losses, and triggers utility penalty charges — typically 1–3% of the electricity bill per 0.01 below the target power factor.

The real (active) power drawn by the facility:

P = S × cosφ1  — (Eq. 1)

Where:
  S = 500 kVA (apparent power)
  cosφ1 = 0.72 (existing power factor)

P = 500 × 0.72
P = 360 kW

This real power remains constant — the loads still consume 360 kW regardless of power factor correction. Only the reactive component changes.

Calculate the reactive power at the existing and target power factors:

Existing reactive power:
  Q1 = P × tanφ1  — (Eq. 2)

  φ1 = arccos(0.72) = 43.95°
  tanφ1 = tan(43.95°) = 0.9640

  Q1 = 360 × 0.9640
  Q1 = 347.0 kVAr

Target reactive power:
  Q2 = P × tanφ2  — (Eq. 3)

  φ2 = arccos(0.95) = 18.19°
  tanφ2 = tan(18.19°) = 0.3287

  Q2 = 360 × 0.3287
  Q2 = 118.3 kVAr

The required capacitor bank must supply the difference between the existing and target reactive power:

Qc = Q1 − Q2  — (Eq. 4)

Qc = 347.0 − 118.3
Qc = 228.7 kVAr

Select the next standard capacitor bank size. Standard ratings are typically available in 25, 50, 75, 100, 125, 150, 200, 250, 300 kVAr increments. For automatic stepped banks, common step sizes are 12.5, 25, or 50 kVAr per step.

Selected capacitor bank: 225 kVAr (automatic, 9 × 25 kVAr steps)

Verify achieved power factor with 225 kVAr:
  Qcorrected = Q1 − Qc(actual) = 347.0 − 225 = 122.0 kVAr

  Scorrected = √(P² + Qcorrected²)
  Scorrected = √(360² + 122²)
  Scorrected = √(129,600 + 14,884)
  Scorrected = √144,484
  Scorrected = 380.1 kVA

  PFcorrected = P / Scorrected = 360 / 380.1 = 0.947

  0.947 ≈ 0.95 ✓ (close enough to target)

The full-load current drawn by the capacitor bank at rated output:

Ic = Qc / (√3 × V)  — (Eq. 5)

Ic = 225,000 / (√3 × 400)
Ic = 225,000 / 692.8
Ic = 324.8 A

This is the steady-state capacitor current. However, capacitors experience transient inrush currents during switching that can reach 20–200 times the rated current for a few milliseconds. The cable and fuse must be rated for the steady-state current with an overrating factor.

Per IEC 61439-1 and capacitor manufacturer recommendations, the protective device for a capacitor bank must be rated at 1.36 to 1.50 times the capacitor rated current to account for:

• Capacitor tolerance (+5% to +15% capacitance per IEC 60831)
• Harmonic currents (even with low THD)
• Voltage variations (+10%)

Minimum fuse rating:
  Ifuse = 1.43 × Ic  — (Eq. 6, using 1.43 as mid-range factor)

  Ifuse = 1.43 × 324.8
  Ifuse = 464.5 A

Select next standard fuse rating: 500 A gG (general purpose HRC fuse)

Verify: 500 A ≥ 464.5 A  ✓

The cable must carry the overrated current continuously. Per AS/NZS 3008.1.1:2017 and IEC 60364-5-52, the cable current-carrying capacity after derating must be at least equal to the fuse rating:

Required cable capacity:
  Iz ≥ Ifuse / (k1 × k2)  — (Eq. 7)

Derating factors:
  k1 (ambient 35°C, 90°C XLPE cable):
    Reference ambient = 30°C (IEC 60364)
    From IEC 60364-5-52 Table B.52.14: k1 = 0.96

  k2 (3 grouped circuits on tray):
    From IEC 60364-5-52 Table B.52.17: k2 = 0.82

  ktotal = 0.96 × 0.82 = 0.787

  Iz ≥ 500 / 0.787
  Iz ≥ 635.3 A

From IEC 60364-5-52, Table B.52.2 for multicore XLPE copper on perforated tray (Installation Method E):

Selected cable: 185 mm² XLPE copper, rated 679 A

Verify: 679 A × 0.787 = 534.4 A > 500 A (fuse rating)  ✓

Wait — 534.4 A > 500 A, so the cable is protected by the fuse  ✓

Also verify the derated capacity exceeds the actual capacitor current:
  534.4 A > 324.8 A (capacitor rated current)  ✓

For completeness, check the voltage drop on the 15 m cable run. For capacitive loads the reactive voltage drop component actually raises the voltage, but we conservatively check the resistive component:

From cable data, 185 mm² XLPE copper:
  mV/A·m (resistive) = 0.210

ΔV = mV/A·m × Ic × L / 1000  — (Eq. 8)
ΔV = 0.210 × 324.8 × 15 / 1000
ΔV = 1.023 V

ΔV% = 1.023 / 400 × 100 = 0.26%

The voltage drop of 0.26% is negligible for a 15 m run. This is not a governing factor.

The 225 kVAr automatic capacitor bank corrects the power factor from 0.72 to 0.947, freeing 120 kVA of transformer capacity and reducing supply current by 24%. The annual savings from avoided penalty charges and reduced losses typically pay back the capacitor bank installation cost within 12–18 months.

• IEC 61439-1 — Low-voltage switchgear and controlgear assemblies, general rules
• IEC 60831-1 — Shunt power capacitors for AC power systems ≤ 1 kV, tolerances and ratings
• AS/NZS 3000:2018, Clause 4.7 — Requirements for capacitor installations
• IEC 60364-5-52, Tables B.52.2, B.52.14, B.52.17 — Cable current ratings and derating factors
• AS/NZS 3008.1.1:2017 — Cable sizing and current-carrying capacities
• Schneider Electric Guide, Chapter K — Power factor correction design methodology

Use the ECalPro Cable Sizing Calculator to verify cable and fuse ratings for your power factor correction installation. Enter the capacitor bank current with the 1.43× overrating factor applied, and the calculator will determine the correct cable size with all derating factors and voltage drop verification.