Power Factor Correction and Improvement

Power factor (PF) is a critical metric in electrical power systems that measures how effectively electrical power is being converted into useful work output. Understanding and correcting low power factor is essential for reducing utility penalties and improving system capacity.

What is Power Factor?

Power factor is the ratio of real power ($P$, measured in Kilowatts, kW) to apparent power ($S$, measured in Kilovolt-Amperes, kVA):

$$PF = \frac{\text{Real Power } (P)}{\text{Apparent Power } (S)}$$

In terms of AC circuit trigonometry, it is represented as:

$$PF = \cos(\theta)$$

Where $\theta$ is the phase angle between voltage and current.

There are three components of power in an AC system, forming the Power Triangle:

1. Real Power ($P$, kW): The actual work-producing power (e.g., shaft rotation, heat, light).
2. Reactive Power ($Q$, kVAR): The non-working power required to sustain magnetic fields in inductive loads like motors and transformers.
3. Apparent Power ($S$, kVA): The vector sum of real and reactive power: $$S = \sqrt{P^2 + Q^2}$$

The Cost of Low Power Factor

A low power factor (typically under 0.90) indicates that a facility is drawing more current than necessary to do the same amount of work. This leads to:

• Utility Penalties: Most industrial power utilities charge extra fees for low power factor.
• Overloaded Infrastructure: Higher current causes thermal stress on transformers, switchgear, and cabling, reducing their operating lifespan.
• Increased Line Losses: Voltage drops and $I^2R$ power losses increase as current rises.

Sizing Capacitor Banks for Correction

To correct a low power factor, we install capacitor banks that supply leading reactive power (kVAR), offsetting the lagging inductive kVAR from motors and transformers.

Formula for Capacitor Rating ($Q_c$)

To determine the required capacitor size in kVAR ($Q_c$) to improve the power factor from $PF_{initial}$ ($\cos\theta_1$) to $PF_{target}$ ($\cos\theta_2$):

$$Q_c = P \times [\tan(\theta_1) - \tan(\theta_2)]$$

Where:

• $P$ = Active load power (kW)
• $\theta_1 = \arccos(PF_{initial})$
• $\theta_2 = \arccos(PF_{target})$

Example Calculation

An industrial plant has a constant load of 500 kW at an initial power factor of 0.75. The target is 0.95.

1. Find the angles: $$\theta_1 = \arccos(0.75) \approx 41.41^\circ$$ $$\theta_2 = \arccos(0.95) \approx 18.19^\circ$$
2. Find the tangents: $$\tan(\theta_1) \approx 0.882$$ $$\tan(\theta_2) \approx 0.329$$
3. Calculate required capacitor rating ($Q_c$): $$Q_c = 500 \times [0.882 - 0.329] = 500 \times 0.553 = 276.5\text{ kVAR}$$

Installing a 275 kVAR capacitor bank will successfully correct the power factor, lowering currents and eliminating utility penalties.