PF improvement of 490 kW load with demand charge

Loss-free capacitors connected in delta across the 6.6 kV supply provide leading kVAR. The real power (490 kW) and kWh stay the same; only the kVA demand falls, so the saving is in the kVA demand charge.

a) Total kVAR required

tan(cos⁻¹0.7)  = 1.0202
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

b) Capacitance required

In a delta connection, each phase capacitor is across the line voltage of 6.6 kV.

Qc/phase = 377.09 / 3 = 125.70 kVAR
C = Qph / (ω V²)
  = 125 698 / (2π × 50 × 6600²)
  = 125 698 / (1.3685 × 10¹⁰)
  = 9.19 µF per phase

(Line current of the capacitor bank = 377 090 / (√3 × 6600) = 33.0 A.)

c) Payback period

kVA1 = 490 / 0.70 = 700.00 kVA
kVA2 = 490 / 0.97 = 505.15 kVA
ΔkVA = 194.85 kVA

Annual saving in demand charge
  = 194.85 × 200 × 12 = Rs 467 629 /year

Cost of capacitor bank with switching
  = 377.09 × 1200 = Rs 452 513

Payback = 452 513 / 467 629
        = 0.968 year ≈ 11.6 months

Answer: (a) 377.1 kVAR, (b) 9.19 µF per phase in delta, (c) payback ≈ 0.97 year (≈ 11.6 months).