Capacitor Bank Calculator
Size capacitor banks for power factor correction from measured kW, kVA, and target PF. Enter values for instant results with step-by-step formulas.
Formula
kVAR = kW x [tan(arccos(PF1)) - tan(arccos(PF2))]
Where kW is the real power, PF1 is the current (lower) power factor, PF2 is the target (higher) power factor, and arccos converts power factor to phase angle. The difference of tangents gives the reactive power that must be supplied by the capacitor bank to achieve the desired power factor improvement.
Worked Examples
Example 1: Industrial Plant Power Factor Correction
Problem: A 500 kW industrial plant has a power factor of 0.75 lagging. Calculate the capacitor bank size to correct to 0.95 PF at 480V, 3-phase, 60 Hz.
Solution: theta1 = arccos(0.75) = 41.41 degrees\ntheta2 = arccos(0.95) = 18.19 degrees\nkVAR1 = 500 x tan(41.41) = 500 x 0.882 = 441.0 kVAR\nkVAR2 = 500 x tan(18.19) = 500 x 0.329 = 164.4 kVAR\nRequired kVAR = 441.0 - 164.4 = 276.6 kVAR\nkVA before: 500/0.75 = 666.7 kVA\nkVA after: 500/0.95 = 526.3 kVA\nCurrent reduction: 666.7 to 526.3 = 21.1%
Result: Install 280 kVAR capacitor bank | kVA reduced by 140.4 (21.1%) | Current drops from 802A to 633A
Example 2: Motor Power Factor Correction
Problem: A 75 kW motor operates at 0.82 PF. Size individual capacitors to correct to 0.97 PF at 400V, 3-phase, 50 Hz.
Solution: kVAR1 = 75 x tan(arccos(0.82)) = 75 x 0.698 = 52.4 kVAR\nkVAR2 = 75 x tan(arccos(0.97)) = 75 x 0.251 = 18.8 kVAR\nRequired = 52.4 - 18.8 = 33.6 kVAR\nCheck: 33.6 / 75 = 44.8% of motor kW (slightly high)\nUse 30 kVAR standard size (40% of motor kW)\nActual PF with 30 kVAR: improved to ~0.955
Result: Use 30 kVAR capacitor (nearest standard below 40% of motor kW) | PF improves from 0.82 to 0.955
Frequently Asked Questions
How do capacitor banks improve power factor?
Capacitor banks supply reactive power locally at the load, reducing the reactive power that must be transmitted from the utility. Inductive loads like motors draw reactive current that lags voltage by 90 degrees. Capacitors supply reactive current that leads voltage by 90 degrees. When connected in parallel, the capacitive current cancels the inductive current, reducing the total current flowing from the source. The real power consumed by the load remains unchanged because capacitors supply only reactive power. By sizing the capacitor bank to cancel the required amount of reactive power, the power factor improves from its original value to the desired target, reducing the apparent power and line current proportionally.
What is the difference between fixed and automatic capacitor banks?
Fixed capacitor banks provide a constant amount of reactive power compensation regardless of load conditions. They are suitable when the load is relatively constant, such as a single large motor running continuously. Automatic capacitor banks use a power factor controller that monitors the power factor in real time and switches capacitor steps on and off as needed. They are essential when loads vary throughout the day, such as in manufacturing plants where different machines run at different times. Automatic banks typically have 4 to 12 steps of capacitors controlled by contactors. The controller measures the reactive power demand and adds or removes capacitor steps to maintain the target power factor.
What are the dangers of capacitor banks and safety precautions?
Capacitor banks store electrical energy and can deliver a dangerous shock even after power is disconnected. Discharge resistors must be installed to reduce the stored voltage to 50 volts or less within one minute of disconnection per NEC requirements. Capacitors can also cause harmonic resonance with the system inductance, amplifying harmonic currents from variable frequency drives and other nonlinear loads. This can damage equipment and cause capacitor failure. Proper fusing of each capacitor unit is essential because a short-circuited capacitor can cause cascading failures. Over-voltage due to leading power factor can damage sensitive equipment. Always consult a power systems engineer before installing large capacitor banks.
How do harmonics affect capacitor bank sizing and design?
Harmonics, particularly the 5th and 7th produced by variable frequency drives and rectifiers, can be amplified by capacitor banks due to parallel resonance with the system inductance. The resonant frequency equals 1 / (2 x pi x sqrt(L x C)), and if this frequency coincides with a harmonic, the harmonic current flowing through the capacitors can exceed their thermal rating and cause premature failure. De-tuned capacitor banks include series reactors (typically 5.67 or 7 percent) that shift the resonant point below the lowest harmonic frequency, typically to the 4.3rd or 3.8th harmonic. In systems with more than 20 percent nonlinear load, de-tuned or active harmonic filter systems should be used instead of plain capacitor banks.
Where should capacitor banks be installed in the electrical system?
Capacitor banks can be installed at three levels: at the utility service entrance, at distribution switchboards, or at individual motor terminals. Service entrance installation is simplest and cheapest, reducing utility penalties but not reducing internal cable and transformer losses. Distribution-level installation reduces losses in the feeder cables and transformers serving that section. Individual motor correction provides the greatest loss reduction and is best for large motors that run continuously. A common approach is to install fixed capacitors at each large motor (sized at about 90 percent of motor no-load kVAR to avoid self-excitation) and an automatic bank at the main switchboard to handle the varying remainder.
Is my data stored or sent to a server?
No. All calculations run entirely in your browser using JavaScript. No data you enter is ever transmitted to any server or stored anywhere. Your inputs remain completely private.