Skip to main content

Capacitor bank for 3-phase motor pf

Short answer

Reactive power required to improve pf from 0.8 to 0.9 lag is 15.94 kVAR total, 5.31 kVAR per phase. Star capacitor per phase: 317 μF; delta capacitor per phase: 105.7 μF.

A 400V, 50 Hz, 3 phase induction motor takes 60 KW power from supply mains at 0.8 power factor lagging. Calculate the capacitance per phase and KVAR rating per phase of capacitor in order to improve the power factor to 0.9 lagging using (i) star connected capacitor bank and (ii) Delta connected capacitor bank.

Given P = 60 kW, V_L = 400 V, f = 50 Hz, cos φ₁ = 0.8 lag, cos φ₂ = 0.9 lag.

Reactive power

  • tan φ₁ = tan(36.87°) = 0.75, so Q₁ = 60 × 0.75 = 45.0 kVAR
  • tan φ₂ = tan(25.84°) = 0.4843, so Q₂ = 60 × 0.4843 = 29.06 kVAR
  • Total capacitor kVAR: Q_c = 45.0 − 29.06 = 15.94 kVAR
  • Per phase: 15.94/3 = 5.31 kVAR (the rating per phase is the same for star or delta bank)

(i) Star-connected bank

Voltage across each capacitor = 400/√3 = 230.9 V.

C_Y = (Q_c/3) / (ω V_ph²) = 5314 / (314.16 × 230.9²) = 5314 / (1.6753 × 10⁷)

C_Y = 317 μF per phase

(ii) Delta-connected bank

Voltage across each capacitor = 400 V.

C_Δ = 5314 / (314.16 × 400²) = 5314 / (5.0265 × 10⁷)

C_Δ = 105.7 μF per phase

(C_Y = 3 C_Δ, so the delta bank needs only one third of the capacitance.)

Line current of the motor falls from 60 000/(√3 × 400 × 0.8) = 108.3 A to 96.2 A.

✓ Answer checked by Claude Opus 5.5 (high effort)

Related questions

All Basic Electrical Engineering chapter 8 questions →