11.8 Power Factor Correction

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11.8 Power Factor Correction

Most domestic loads (such as w ashing machines, air conditioners, and refrigerators) and industrial loads (such as induction motors) are induc tive and operate at a lo w lagging po wer factor. Although the inducti ve nature of the load cannot be changed, we can increase its power factor.

The process of increasing the power factor without altering the voltage or current to the original load is known as power factor correction.

Since most loads are inducti ve, as shown in Fig. 11.27(a), a load’ s power factor is improved or corrected by deliberately installing a capacitor in parallel with the load, as shown in Fig. 11.27(b). The effect of adding the capacitor can be illustrated using either the power triangle or the phasor diagram of the currents in volved. Figure 11.28 sho ws the latter, where it is assumed that the circuit in Fig. 11.27(a) has a power factor of cos θ1, while the one in Fig. 11.27(b) has a power factor of cos θ2. It is evident from Fig. 11.28 that adding the capacitor has caused the phase angle between the supplied v oltage and current to reduce from θ1 to θ2, thereby increasing the power factor. We also notice from the magnitudes of the vectors in Fig. 11.28 that with the same supplied v oltage, the circuit in Fig. 11.27(a) dra ws larger current IL than the current I drawn by the circuit in Fig. 11.27(b). Power companies charge more for larger currents, because they result in increased power losses (by a squared factor, since P = IL 2 R). Therefore, it is beneficial to both the power company and the consumer that every effort is made to minimize current level or keep the power factor as close to unity as possible. By choosing a suitable size for the capacitor, the current can be made to be completely in phase with the voltage, implying unity power factor.

Alternatively, power factor correction may be viewed as the addition of a reactive element (usually a capacitor) in parallel with the load in order to make the power factor closer to unity.

An inductive load is modeled as a series combination of an inductor and a resistor.

Figure 11.27 Power factor correction: (a) original inductive load, (b) inductive load with improved power factor.

We can look at the power factor correction from another perspective. Consider the power triangle in Fig. 11.29. If the original inducti ve load has apparent power S1, then

$$P = S_1 \cos \theta_1$$

, $Q_1 = S_1 \sin \theta_1 = P \tan \theta_1$ (11.57)

Figure 11.29 Power triangle illustrating power factor correction.

If we desire to increase the po wer factor from cos θ1 to cos θ2 without altering the real power (i.e., P = S2 cos θ2), then the new reactive power is

$$Q_2 = P \tan \theta_2 \tag{11.58}$$

The reduction in the reacti ve power is caused by the shunt capacitor; that is,

$$Q_C = Q_1 - Q_2 = P(\tan \theta_1 - \tan \theta_2)$$

(11.59)

But from Eq. (11.46), QC = V² rms∕XC = ωCV ² rms. The value of the required shunt capacitance C is determined as

$$C = \frac{Q_C}{\omega V_{\text{rms}}^2} = \frac{P(\tan \theta_1 - \tan \theta_2)}{\omega V_{\text{rms}}^2}$$

(11.60)

Note that the real po wer P dissipated by the load is not af fected by the power factor correction because the a verage power due to the capaci tance is zero.

Although the most common situation in practice is that of an inductive load, it is also possible that the load is capaciti ve; that is, the load is operating at a leading power factor. In this case, an inductor should be connected across the load for po wer factor correction. The required shunt inductance L can be calculated from

$$Q_L = \frac{V_{\text{rms}}^2}{X_L} = \frac{V_{\text{rms}}^2}{\omega L} \qquad \Rightarrow \qquad L = \frac{V_{\text{rms}}^2}{\omega Q_L} \tag{11.61}$$

where QL = Q¹ − Q2, the dif ference between the ne w and old reacti ve powers.

Example 11.15 When connected to a 120-V (rms), 60-Hz po wer line, a load absorbs 4 kW at a lagging po wer factor of 0.8. Find the v alue of capacitance necessary to raise the pf to 0.95.

Solution:

If the pf = 0.8, then

cos θ1 = 0.8 ⇒ θ1 = 36.87°

where θ1 is the phase difference between voltage and current. We obtain the apparent power from the real power and the pf as

$$S_1 = \frac{P}{\cos \theta_1} = \frac{4000}{0.8} = 5000 \text{ VA}$$

The reactive power is

$$Q_1 = S_1 \sin \theta = 5000 \sin 36.87 = 3000$$

VAR

When the pf is raised to 0.95,

$$\cos \theta_2 = 0.95 \qquad \Rightarrow \qquad \theta_2 = 18.19^\circ$$

The real power P has not changed. But the apparent power has changed; its new value is

$$S_2 = \frac{P}{\cos \theta_2} = \frac{4000}{0.95} = 4210.5 \text{ VA}$$

The new reactive power is

$$Q_2 = S_2 \sin \theta_2 = 1314.4 \text{ VAR}$$

The difference between the new and old reactive powers is due to the parallel addition of the capacitor to the load. The reactive power due to the capacitor is

$$Q_C = Q_1 - Q_2 = 3000 - 1314.4 = 1685.6 \text{ VAR}$$

and

$$C = \frac{Q_C}{\omega V_{\text{rms}}^2} = \frac{1685.6}{2\pi \times 60 \times 120^2} = 310.5 \,\mu\text{F}$$

Note: Capacitors are normally purchased for voltages they expect to see. In this case, the maximum voltage this capacitor will see is about 170 V peak. We would suggest purchasing a capacitor with a voltage rating equal to, say, 200 V.

Find the value of parallel capacitance needed to correct a load of 140 kVAR at 0.85 lagging pf to unity pf. Assume that the load is sup plied by a 220-V (rms), 60-Hz line.

Answer: 7.673 mF.