Power Factor Correction in a Workshop
The electrical wing of a DSSSB-maintained workshop records a connected load of 250 kW with a demand factor of 0.6. The maximum demand meter shows a power factor of 0.78 lagging throughout the first shift, when the load is at its peak; in the second shift the shop runs at about 30% of that peak kW. The DISCOM tariff charges ₹350 per kVA of billing demand each month in addition to the energy charge, and a leading power factor attracts a penalty. The junior engineer proposes a single fixed 75 kVAR capacitor bank switched on with the main breaker, at an installed cost of ₹500 per kVAR plus ₹15,000 for the panel. The executive engineer wants the kVAR requirement, the monthly saving, the payback, and an opinion on the fixed-bank proposal before the estimate goes to the sanctioning authority on Friday.
1. Compute the maximum demand in kW, the present kVA demand, and the capacitor rating needed to raise the power factor to 0.95 at peak load.
Read the worked answer
Maximum demand = connected load × demand factor = 250 × 0.6 = 150 kW.
At pf 0.78 lagging: kVA demand = 150 / 0.78 = 192.31 kVA. Reactive power = √(192.31² − 150²) = √(36,983 − 22,500) = √14,483 ≈ 120.3 kVAR. tan φ₁ = 120.3 / 150 = 0.802.
At the target pf 0.95: φ₂ = cos⁻¹ 0.95 = 18.19°, tan φ₂ = 0.3287, so the reactive power allowed is 150 × 0.3287 = 49.3 kVAR.
Compensation required = 120.3 − 49.3 = 71 kVAR.
A 71 kVAR step does not exist as a standard unit, so specify 75 kVAR (three 25 kVAR steps). Note that the kVAR requirement is set by the kW and the two power factors only; it does not depend on the size of the connected load, which is why the demand factor has to be applied first before any calculation is attempted.
2. Work out the monthly saving in billing demand and the simple payback of the installed cost, then state whether the estimate should be sanctioned as proposed.
Read the worked answer
New kVA demand at the same 150 kW and pf 0.95 = 150 / 0.95 = 157.89 kVA. Reduction in billing demand = 192.31 − 157.89 = 34.42 kVA. Monthly saving = 34.42 × ₹350 = ₹12,047 (about ₹1.45 lakh a year).
Installed cost = 75 × ₹500 + ₹15,000 = 37,500 + 15,000 = ₹52,500. Simple payback = 52,500 / 12,047 = 4.36 months, roughly 4.4 months.
The payback is short enough that the expenditure is justified on billing demand alone; the further savings from reduced I²R losses in the transformer and cables have deliberately not been counted, so the case is conservative. Sanction is recommended, but not in the proposed form: the estimate should be revised to three switched 25 kVAR steps with an automatic power factor control relay, which costs little more than the fixed bank and avoids the problem in the next task.
3. Evaluate the junior engineer's proposal of a single fixed 75 kVAR bank switched on with the main breaker, using numbers from the second shift.
Read the worked answer
In the second shift the shop runs at about 30% of peak kW, i.e. 0.30 × 150 = 45 kW. At the uncorrected pf 0.78 the reactive demand is 45 × 0.802 = 36.1 kVAR. If the whole 75 kVAR bank stays connected, the net reactive power becomes 36.1 − 75 = −38.9 kVAR, i.e. the installation runs at a leading power factor of about 0.76. That produces a penalty under this tariff, raises the secondary voltage, and can cause capacitor-switching transients and relay maloperation. The correct requirement at 45 kW is only 45 × (0.802 − 0.3287) ≈ 21.3 kVAR.
The fixed bank is therefore oversized for most of the working day even though it is correctly sized for the peak. The fix is three 25 kVAR steps under an APFC relay with a target of 0.95 and a suitable dead band, so that the second shift connects one step (net reactive 11.1 kVAR, pf about 0.97 lagging) and only the first shift connects all three. Steps of 25 kVAR also keep the switching transient small, and a detuned reactor should be considered if the shop has significant variable-frequency drive load.