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PF improvement of 490 kW load at 11 kV

Short answer

Capacitor bank rating 377.09 kVAr total (125.7 kVAr/phase in delta), capacitance 3.31 µF per phase. Cost NRs 7.54 lakh; kVA demand falls from 700 to 505.15 kVA.

The power factor on an industrial 3φ load 490 kW is to be improved from 0.7 lagging to 0.97 lagging by connecting loss free delta connected capacitor across 11 kV, 50 Hz supply. The cost of Var controlling device with switching is Nrs.2000 per kVAR.

Given: 3-phase load P = 490 kW, pf improved from 0.7 lag to 0.97 lag, delta-connected loss-free capacitors on 11 kV, 50 Hz; cost of var control with switching NRs 2000/kVAr.

kVAr rating of capacitor bank

tan φ₁ = tan(cos⁻¹0.7)  = 1.0202
tan φ₂ = tan(cos⁻¹0.97) = 0.2506
Q₁ = 490 × 1.0202 = 499.90 kVAr
Q₂ = 490 × 0.2506 = 122.81 kVAr
Qc = Q₁ − Q₂      = 377.09 kVAr  (total, 3-phase)

Capacitance per phase (delta)

In delta, each capacitor has full line voltage 11 kV across it.

Qc per phase = 377.09 / 3 = 125.70 kVAr
C = Q_ph / (ω V²)
  = 125 698 / (2π × 50 × 11 000²)
  = 125 698 / (314.16 × 1.21 × 10⁸)
  = 3.307 × 10⁻⁶ F = 3.307 µF
Current per capacitor = 125.70 kVAr / 11 kV = 11.43 A

Cost

Cost = 377.09 × 2000 = NRs 7,54,189

Effect

kVA before = 490 / 0.7  = 700.0 kVA
kVA after  = 490 / 0.97 = 505.15 kVA
Reduction  = 194.85 kVA

Answer: capacitor bank of 377.09 kVAr (125.7 kVAr per phase), C ≈ 3.31 µF per phase in delta, cost ≈ NRs 7.54 lakh; kVA demand falls from 700 to 505.2 kVA.

✓ Answer checked by Claude Opus 5.5 (high effort)

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