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Delta capacitor per phase for 50 kW motor

Short answer

Reactive power reduction from 48.19 to 24.21 kVAR, requiring 23.98 kVAR total. Delta-connected capacitors see 415 V; capacitance per phase is 147.7 μF.

A 415 V, 3 phase, 50 HZ induction motor takes 50 KW power from supply mains at 0.72 power factor lagging. A bank of capacitors is connected in delta across the line to improve the overall power factor. Calculate the capacitance per phase in order to raise the power factor to 0.9 lagging.

Given P = 50 kW, V_L = 415 V, f = 50 Hz, cos φ₁ = 0.72 lagging, cos φ₂ = 0.9 lagging. The capacitors are delta connected.

Reactive power

  • tan φ₁ = tan(cos⁻¹ 0.72) = 0.9638, so Q₁ = 50 × 0.9638 = 48.19 kVAR
  • tan φ₂ = tan(cos⁻¹ 0.9) = 0.4843, so Q₂ = 50 × 0.4843 = 24.21 kVAR
  • Capacitor reactive power: Q_c = 48.19 − 24.21 = 23.98 kVAR

Per phase: Q_c/3 = 7.99 kVAR.

Capacitance per phase (delta)

Voltage across each capacitor = line voltage = 415 V.

Q_phase = ω C V² ⟹ C = 7992 / (314.16 × 415²) = 7992 / (5.411 × 10⁷)

C = 147.7 μF per phase

(Check of line current: 50 000/(√3 × 415 × 0.72) = 96.6 A before and 77.3 A after correction.)

✓ Answer checked by Claude Opus 5.5 (high effort)

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