Optimum pf under two-part tariff (200 kW)

Data: P = 200 kW at cos φ1 = 0.707; energy charge Rs 10/kWh; demand charge Rs 200/kVA/month; correcting equipment Rs 1200/kVAR.

Annual kVA charge:

A = 200 × 12 = Rs 2400 /kVA/year
B = Rs 1200 /kVAR

(No interest/depreciation rate is given, so the full equipment cost is set against one year's kVA charge.)

a) Optimum power factor

Total annual cost = A·P/cos φ2 + B·P(tan φ1 − tan φ2) is minimum when

sin φ2 = B / A = 1200 / 2400 = 0.5
cos φ2 = √(1 − 0.5²) = 0.866 lagging

b) Capacity of pf correcting equipment

Q1 = 200 × tan(cos⁻¹0.707) = 200 × 1.0003 = 200.06 kVAR
Q2 = 200 × tan(cos⁻¹0.866) = 200 × 0.5774 = 115.47 kVAR
Qc = 200.06 − 115.47 = 84.59 kVAR
Cost = 84.59 × 1200 ≈ Rs 101 508

c) Annual bill (16 h/day, 350 days)

Energy = 200 × 16 × 350        = 1 120 000 kWh
Energy charge = 1 120 000 × 10 = Rs 11 200 000
kVA demand = 200 / 0.866       = 230.94 kVA
Demand charge = 230.94 × 2400  = Rs 554 256
Annual bill = 11 200 000 + 554 256 = Rs 11 754 256

For comparison, without correction the demand is 200/0.707 = 282.89 kVA and the demand charge is Rs 678 925, so the bill would be Rs 11 878 925; correction saves Rs 124 669 per year in demand charge.

Note: if a 10 % annual interest + depreciation on the equipment is assumed instead (B = Rs 120/kVAR/year), sin φ2 = 0.05 and the optimum pf becomes about 0.999.

Answer: (a) optimum pf = 0.866 lagging, (b) ≈ 84.6 kVAR, (c) annual bill ≈ Rs 1.175 crore (Rs 11 754 256).