PF improvement of 490 kW load at 6.6 kV

Capacitors connected in delta across the 6.6 kV supply provide the leading kVAR needed to raise the pf from 0.7 to 0.97.

(i) Total kVAR required

Q1 = P tan φ1 = 490 × tan(cos⁻¹0.7)  = 490 × 1.0202 = 499.90 kVAR
Q2 = P tan φ2 = 490 × tan(cos⁻¹0.97) = 490 × 0.2506 = 122.81 kVAR
Qc = Q1 − Q2  = 377.09 kVAR

(ii) Capacitance (delta connection)

In delta, each capacitor sees the line voltage of 6.6 kV and supplies one-third of Qc.

Qc per phase = 377.09 / 3 = 125.70 kVAR
Qph = V² ω C   →   C = Qph / (2π f V²)
C = 125 698 / (2π × 50 × 6600²)
  = 125 698 / (314.16 × 43 560 000)
  = 9.19 × 10⁻⁶ F = 9.19 µF per phase

(iii) Payback period

The question does not give the demand tariff, so assume a kVA demand charge of Rs 200/kVA/month (as given in the companion version of this problem). Since the capacitors are loss-free, the kWh consumption does not change; the saving is only in kVA demand.

kVA before = 490 / 0.7  = 700.00 kVA
kVA after  = 490 / 0.97 = 505.15 kVA
Reduction  = 194.85 kVA

Annual saving = 194.85 × 200 × 12 = Rs 467 629
Capital cost  = 377.09 × 1200     = Rs 452 513
Payback = 452 513 / 467 629 = 0.968 year ≈ 11.6 months

Answer: Qc ≈ 377.1 kVAR; C ≈ 9.19 µF per phase (delta); payback ≈ 0.97 year (about 11.6 months) with an assumed demand charge of Rs 200/kVA/month.