kVAr and techno-economic analysis for 1000 kVA transformer

Given: transformer 1000 kVA, average load 800 kVA at 0.6 pf (lag), no-load loss 1.6 kW, full-load loss 13.5 kW, capacitor cost NRs 1500/kVAr, demand charge NRs 210/kVA/month.

kVAr required to improve pf to 0.95

P  = 800 × 0.6               = 480 kW
Q₁ = 800 × 0.8               = 640 kVAr
Q₂ = 480 × tan(cos⁻¹0.95) = 480 × 0.3287
   = 157.77 kVAr
Qc = 640 − 157.77            = 482.23 kVAr

(In practice a standard 500 kVAr APFC panel would be used; the calculation uses 482.23 kVAr.)

Reduction of kVA demand

S₂ = 480 / 0.95   = 505.26 kVA
ΔS = 800 − 505.26 = 294.74 kVA

Reduction in transformer loss

No-load loss (1.6 kW) stays the same; copper loss ∝ (load/rating)²:

Cu loss before = 13.5 × (0.8)²     = 8.640 kW
Cu loss after  = 13.5 × (0.5053)²  = 3.446 kW
Reduction      = 5.194 kW

If the transformer runs all year (8760 h): 5.194 × 8760 = 45 496 kWh saved per year (extra benefit; energy rate not given).

Techno-economic analysis

Cost of capacitor bank = 482.23 × 1500   = NRs 723 347
Demand charge saving   = 294.74 × 210    = NRs 61 895 / month
                       = NRs 742 737 / year
Simple payback         = 723 347 / 61 895
                       = 11.7 months ≈ 0.97 year

The investment is recovered in less than one year from demand charge alone; the loss reduction, lower current, better voltage and released transformer capacity (about 295 kVA) are extra gains. So the pf improvement is technically and economically justified.