Power Triangle and Power Factor Correction: Sizing the Capacitor Bank
Induction motors, welding transformers and fluorescent ballasts draw magnetising current that lags the voltage. This lagging reactive current does no useful work but still flows through every cable, switch and transformer between the load and the generator. The result is higher current, more copper loss and, for industrial consumers, higher electricity bills, because many Indian distribution companies either penalise low power factor or bill on kVAh instead of kWh.
Capacitors supply leading reactive power locally and cancel much of the lagging demand. In this article you will use the power triangle to calculate the kVAR required and then convert that into actual capacitance values for single-phase and three-phase installations.
The Power Triangle
Real power P (kW), reactive power Q (kVAR) and apparent power S (kVA) form a right-angled triangle. P is the base, Q is the vertical side and S is the hypotenuse. The angle between P and S is the power factor angle φ.
S (kVA)
/|
/ |
/ | Q (kVAR)
/φ |
/____|
P (kW)
S = √(P² + Q²)
cos φ = P/S
Q = P × tan φTo improve power factor from cos φ1 to cos φ2 without changing the real power, the reactive power must fall from P tan φ1 to P tan φ2. The capacitor bank must supply the difference.
Qc = P × (tan φ1 − tan φ2) (kVAR) Capacitance for a known Qc: Single-phase: C = Qc / (2πf V²) Three-phase delta bank, per phase: C = (Qc/3) / (2πf VL²) Three-phase star bank, per phase: C = (Qc/3) / (2πf Vph²)
| Power factor | Angle φ | tan φ |
|---|---|---|
| 0.70 | 45.57° | 1.0202 |
| 0.75 | 41.41° | 0.8819 |
| 0.80 | 36.87° | 0.7500 |
| 0.85 | 31.79° | 0.6197 |
| 0.90 | 25.84° | 0.4843 |
| 0.95 | 18.19° | 0.3287 |
| 0.99 | 8.11° | 0.1425 |
| 1.00 | 0° | 0 |
Worked Example 1: Single-Phase Pump Load
A 230 V, 50 Hz single-phase load takes 2 kW at 0.70 lagging. Find the capacitor needed to raise power factor to 0.95, and the current before and after.
- Qc = 2 × (1.0202 − 0.3287) = 2 × 0.6915 = 1.383 kVAR.
- C = 1383 / (2π × 50 × 230²) = 1383 / (314.16 × 52900) = 83.2 µF.
- Current before: I1 = 2000 / (230 × 0.70) = 12.42 A.
- Current after: I2 = 2000 / (230 × 0.95) = 9.15 A.
A standard 80 µF or two 40 µF motor-run capacitors rated for at least 440 V AC would be a practical choice, giving a final power factor slightly below 0.95. The supply current drops by about 26%, so cable copper loss, which is proportional to I², drops by about 46%.
Worked Example 2: Three-Phase Workshop at 415 V
A workshop has a total load of 50 kW at 0.75 lagging on a 415 V, 50 Hz supply. Size a delta-connected capacitor bank to raise power factor to 0.95.
- Qc = 50 × (0.8819 − 0.3287) = 50 × 0.5532 = 27.66 kVAR.
- Per phase: 27.66/3 = 9.22 kVAR.
- Each delta capacitor sees the full 415 V, so C = 9220 / (314.16 × 415²) = 9220 / (314.16 × 172225) = 170 µF per phase.
- kVA before: 50/0.75 = 66.67 kVA. Line current = 66670 / (√3 × 415) = 92.75 A.
- kVA after: 50/0.95 = 52.63 kVA. Line current = 52630 / (√3 × 415) = 73.22 A.
| Quantity | Before (0.75) | After (0.95) |
|---|---|---|
| Real power | 50 kW | 50 kW |
| Reactive power | 44.10 kVAR | 16.43 kVAR |
| Apparent power | 66.67 kVA | 52.63 kVA |
| Line current | 92.75 A | 73.22 A |
In practice you would buy a standard 25 or 30 kVAR three-phase capacitor unit rated at 440 V or higher rather than individual µF values. Note that a capacitor rated at 440 V delivers less kVAR at 415 V, because kVAR is proportional to V². A 30 kVAR, 440 V unit gives 30 × (415/440)² = 26.7 kVAR at 415 V, which is close to the target.
Practical Installation Points
- Fixed vs automatic: A fixed bank suits a constant load such as a single motor. Varying loads need an automatic power factor control (APFC) panel that switches capacitor steps through contactors based on measured power factor.
- Avoid over-correction: A leading power factor at light load raises voltage and can also attract penalties under kVAh billing. Size for the normal load, not the peak.
- Individual motor correction: A capacitor connected at the motor terminals should not exceed the motor no-load magnetising kVAR; otherwise self-excitation can produce dangerous voltages when the motor is switched off while still spinning.
- Harmonics: VFDs, UPS units and LED drivers produce harmonic currents. Plain capacitors can resonate with the transformer inductance. Use detuned reactors in series with capacitors where harmonic loads are significant.
- Use capacitor-duty contactors: They include pre-charge resistors to limit the very high inrush current when a capacitor is switched on.
Summary
The power triangle links kW, kVAR and kVA through the power factor angle. The capacitor kVAR needed is P (tan φ1 − tan φ2), and capacitance follows from Q = 2πfCV². Correcting a 50 kW load from 0.75 to 0.95 needs about 28 kVAR and cuts line current from about 93 A to 73 A. Always choose a capacitor voltage rating above the supply voltage, allow for the reduced kVAR at lower voltage, and watch for harmonics.