Power-factor correction on a workshop bus that also feeds a VFD
A small engineering workshop is supplied at 415 V, 3-phase, 50 Hz through a 63 A LT connection. The loads running together at full load during the shift are:
- 15 kW cage induction motor at power factor 0.82
- 5 kW compressor motor at power factor 0.78
- 8 kW lighting and resistance heaters at unity power factor
The tariff charges Rs 350 per kVA of billing demand and penalises a monthly average power factor below 0.90. The owner wants capacitors to reach 0.95.
On the same LT bus there is a 22 kW pump driven by a 6-pulse VFD. The electrician proposes a single fixed 10 kVAr capacitor bank at the LT panel, sized by a thumb rule of one kVAr per kW of motor rating, connected without switching or a reactor.
1. Compute the total active power, reactive power, apparent power, power factor and line current of the workshop.
Read the worked answer
Convert each load to kW and kVAr using Q = P tan(phi):
- Motor: cos = 0.82, tan = 0.698, Q = 15 x 0.698 = 10.47 kVAr
- Compressor: cos = 0.78, tan = 0.802, Q = 5 x 0.802 = 4.01 kVAr
- Lighting and heaters: Q = 0
P = 15 + 5 + 8 = 28 kW Q = 14.48 kVAr S = sqrt(28^2 + 14.48^2) = sqrt(784 + 209.7) = 31.52 kVA pf = 28/31.52 = 0.888
Line current I = S/(sqrt3 x V) = 31,520/(1.732 x 415) = 31,520/718.8 = 43.9 A.
Interpretation: the 63 A connection is adequate at 43.9 A, but the power factor of 0.888 sits just below the 0.90 penalty threshold, so the penalty applies and the correction needed to reach 0.95 is modest. Note that the unity-pf lighting load dilutes the effect of the motors, which is why the bus pf is not as poor as either motor.
2. Size the capacitor bank needed for 0.95, find the new line current, and quantify the demand-charge saving and the change in cable losses.
Read the worked answer
At pf 0.95, tan(acos 0.95) = 0.3287, so Q_target = 28 x 0.3287 = 9.20 kVAr.
Capacitor rating = 14.48 - 9.20 = 5.28 kVAr. Specify 5 kVAr (which gives pf = 0.947) or a switched 2.5 + 2.5 kVAr bank.
New S = 28/0.95 = 29.47 kVA and I = 29,470/718.8 = 41.0 A, a reduction of 2.9 A, about 6.5%.
Demand saving = (31.52 - 29.47) x 350 = Rs 718 per month, plus the removal of the low-pf penalty. Cable and transformer I2R loss falls in the ratio (41.0/43.9)^2 = 0.872, that is about 13% less loss, which is proportionally larger than the current reduction because loss goes as the square.
Judgement: only 2.1 kVA of capacity is released, so the business case rests mainly on the penalty and on the loss saving, not on a large capacity release. Any quotation assuming a dramatic current reduction is overselling the scheme.
3. Assess the electrician's proposal of a single fixed 10 kVAr bank at the LT panel. State the risks and what you would specify instead.
Read the worked answer
Three problems, in order of severity.
Size. Ten kVAr is far too large for a 5.3 kVAr requirement. Full-load pf would rise to about 0.987, but at night with only the 8 kW lighting load running, Qc = 10 kVAr against P = 8 kW gives tan(phi) = -1.25, a leading power factor near 0.62. Leading pf at light load causes voltage rise, can trip drives and can itself attract a reactive-energy penalty. The one kVAr per kW rule applies to a motor's rating at its own terminals, not to a whole mixed bus.
Location and control. A fixed bank cannot follow the load. Specify either a contactor-switched bank with a pf controller in 2.5 kVAr steps, or compensation at each motor terminal at about 30% of motor kW, never exceeding the motor's no-load reactive draw.
Harmonics. The 6-pulse VFD injects 5th and 7th harmonic currents. A fixed capacitor resonates with the source inductance at h = sqrt(Ssc/Qc); if that order lands near 5 or 7, harmonic voltage is amplified and capacitors fail early. Specify a detuned reactor, typically 7% impedance tuned to about 189 Hz, or a tuned filter, and check the network short-circuit level before energising.
Recommendation: a switched, detuned 5 kVAr scheme, not a fixed 10 kVAr one.