Power-Factor Correction for a 415 V Workshop Motor
A coir-making unit runs a 415 V, 50 Hz, three-phase induction motor at a line current of 12 A and a power factor of 0.80 lagging. The utility applies a penalty when the power factor falls below 0.95. The owner has a box of three 20 µF, 440 V AC capacitors and wants to connect them in delta at the motor terminals. The motor is not fed through a variable-frequency drive. The existing main switch and 25 m cable are rated at 16 A. The owner asks whether the capacitors will remove the penalty, what the line current will become, and whether the capacitor voltage rating is adequate. The installation has no power-factor meter, only a clamp meter and a three-phase energy meter. The electrician must also warn about switching the capacitors on when the motor is unloaded or already disconnected.
1. Calculate the motor’s active power, the reactive power needed to raise the power factor to 0.95, and the per-phase capacitance for a delta connection at 415 V.
Read the worked answer
Active power P = √3 × V_L × I_L × cos φ = 1.732 × 415 × 12 × 0.80 = 6900 W, approximately 6.9 kW. Initial reactive power Q₁ = P tan(cos⁻¹ 0.80) = 6900 × 0.75 = 5175 var. At 0.95 power factor, tan(cos⁻¹ 0.95) = 0.329, so Q₂ = 6900 × 0.329 = 2268 var. The capacitor must supply ΔQ = 5175 − 2268 = 2907 var, about 2.91 kvar. For a delta connection, V_phase = 415 V and Q_total = 3 × 2πf × C × V². Therefore C = 2907 / (3 × 314.16 × 415²) = 17.9 × 10⁻⁶ F, about 18 µF per phase. A 20 µF per phase unit will give slightly more than required. The corrected line current at 0.95 would be I = P/(√3 × V × pf) = 6900/(1.732 × 415 × 0.95) = 10.1 A.
2. Are the three 20 µF, 440 V capacitors suitable? Calculate their reactive power and the resulting power factor, and state any derating or type precautions.
Read the worked answer
Three 20 µF capacitors in delta give Q_c = 3 × 2π × 50 × 20 × 10⁻⁶ × 415² = 3246 var, or 3.25 kvar. The remaining reactive power is 5175 − 3246 = 1929 var, so the new apparent power is √(6900² + 1929²) = 7175 VA and the line current is 7175/(√3 × 415) = 9.98 A. The power factor becomes 6900/7175 = 0.96, which is acceptable but slightly overcorrected. The 440 V rating is only about 6% above the 415 V system voltage. That is marginal but commonly accepted on a clean supply; if the voltage can rise or harmonics are present, use 480 V or 525 V AC capacitors. They must be continuous-duty AC power-factor-correction capacitors, not motor-start or electrolytic capacitors. Check that each unit has a discharge resistor and is protected against overcurrent.
3. Explain the switching hazard if the capacitors are connected upstream of the motor contactor or left on when the motor is unloaded.
Read the worked answer
If the capacitors are connected directly across the supply upstream of the motor contactor, they remain energized when the motor is switched off. The motor can then be driven as a self-excited induction generator by the capacitors and residual magnetism, producing a voltage that may rise above normal. If the capacitors are switched on with the motor unloaded, the motor draws mostly magnetizing current and the capacitors can overcorrect the power factor, causing a leading power factor and voltage rise. The safest arrangement is to switch the capacitor contactor with the motor contactor so the capacitors are connected only while the motor runs, and to provide a discharge path. Never use the capacitors as a substitute for a proper motor protection scheme. With no VFD, resonance risk is lower, but it is still worth checking for unusual heating or voltage rise after connection.