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PF improvement of 1020 kVA factory load

Short answer

Required capacitor 569.18 kVAr. Transformer copper loss drops by 4.12 kW (12 036 kWh/yr, Rs 90 267). Maximum demand reduces by 332.84 kVA saving Rs63 240/month. Payback period 0.54 year (6.4 months).

A factory takes an average load of 1020 kVA at 0.64 p.f.(lag) from the 1600 kVA transformer. The no-load loss and full load loss of transformer are 2.4kW and 18.57 KW respectively. If the unit cost of capacitor bank per kVAr is Rs.800, maximum demand charge is Rs.190 per kVA and unit charge of energy is Rs.7.50. Calculate: i) kVAR required to improve p.f. to 0.95 ii) Reduction in loss in transformer if the loading duration is 2920 hours per year. iii) Reduction in maximum demand charge iv) Payback Period

Given: average load 1020 kVA at 0.64 pf (lag); transformer 1600 kVA; no-load loss 2.4 kW; full-load (copper) loss 18.57 kW; capacitor cost Rs 800/kVAr; maximum demand charge Rs 190/kVA (taken as per month); energy charge Rs 7.50/kWh; loading 2920 h/year.

i) kVAr required to improve pf to 0.95

P  = 1020 × 0.64          = 652.8 kW
Q₁ = 1020 × sin(cos⁻¹0.64) = 1020 × 0.7684
   = 783.74 kVAr
Q₂ = P × tan(cos⁻¹0.95)   = 652.8 × 0.3287
   = 214.56 kVAr
Qc = Q₁ − Q₂              = 569.18 kVAr
New kVA S₂ = 652.8 / 0.95 = 687.16 kVA

ii) Reduction in transformer loss

No-load loss (2.4 kW) does not change. Copper loss ∝ (S/S_rated)²:

Cu loss before = 18.57 × (1020/1600)²   = 7.547 kW
Cu loss after  = 18.57 × (687.16/1600)² = 3.425 kW
Reduction      = 4.122 kW
Energy saved   = 4.122 × 2920 = 12 035.6 kWh/yr
Cost saved     = 12 035.6 × 7.50 = Rs 90 266.8/yr

iii) Reduction in maximum demand charge

kVA reduction = 1020 − 687.16 = 332.84 kVA
Saving        = 332.84 × 190 = Rs 63 240 per month
              = Rs 758 880 per year

iv) Payback period

Capacitor cost = 569.18 × 800      = Rs 455 342
Annual saving  = 758 880 + 90 266.8 = Rs 849 147
Payback        = 455 342 / 849 147
               = 0.536 year ≈ 6.4 months

Answer: (i) 569.18 kVAr; (ii) 4.12 kW less loss, 12 036 kWh/yr = Rs 90 267/yr; (iii) Rs 63 240/month (Rs 758 880/yr); (iv) payback ≈ 0.54 year (about 6.4 months). (If the demand charge is taken as a one-time yearly figure, annual saving = 63 240 + 90 267 = Rs 153 507 and payback ≈ 2.97 years.)

✓ Answer checked by Claude Opus 5.5 (high effort)

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