Most economical power factor for 50 kW factory

Maximum demand: load factor = average demand / maximum demand, so

Pmax = 50 / 0.5 = 100 kW

The kVA charge applies to this maximum demand.

Most economical pf (loss-free capacitors, kWh unchanged): Let A = annual kVA charge per kVA, B = cost per kVAR of capacitor. The net saving is maximum when

sin φ2 = B / A

No interest/depreciation rate is given, so the full capacitor cost is compared with one year's kVA charge (i.e. capital recovered in one year):

A = 190 × 12 = Rs 2280 /kVA/year
B = Rs 600 /kVAR
sin φ2 = 600 / 2280 = 0.2632
cos φ2 = √(1 − 0.2632²) = 0.9648 lagging

Capacitor size:

Q1 = 100 × tan(cos⁻¹0.75)   = 100 × 0.8819 = 88.19 kVAR
Q2 = 100 × tan(cos⁻¹0.9648) = 100 × 0.2728 = 27.28 kVAR
Qc = 88.19 − 27.28          = 60.91 kVAR

Saving:

kVA1 = 100 / 0.75   = 133.33 kVA
kVA2 = 100 / 0.9648 = 103.65 kVA
ΔkVA = 29.68 kVA
Annual saving in kVA charge = 29.68 × 2280 = Rs 67 670
Capacitor cost = 60.91 × 600 = Rs 36 549
Net saving in first year = 67 670 − 36 549 = Rs 31 121

The energy charge (Rs 7.50/kWh) is the same before and after, since capacitors are loss-free and kW is unchanged.

Payback period:

= 36 549 / 67 670 = 0.54 year ≈ 6.5 months

Note: if instead an annual interest + depreciation of 10 % on the capacitor were assumed (B = Rs 60/kVAR/year), sin φ2 = 60/2280 and the economic pf would be about 0.9997, i.e. practically unity.

Answer: Most economical pf ≈ 0.965 lagging; capacitor ≈ 60.9 kVAR; annual saving ≈ Rs 67 670 (net Rs 31 121 in the first year); payback ≈ 0.54 year.