# Power Factor Correction

The design of any power transmission system is very sensitive to the magnitude of the current in the lines as determined by the applied loads. Increased currents result in increased power losses (by a squared factor since $P = I^2R$) in the transmission lines due to the resistance of the lines. Heavier currents also require larger conductors, increasing the amount of copper needed for the system, and, quite obviously, they require increased generating capacities by the utility company.
Every effort must therefore be made to keep current levels at a minimum. Since the line voltage of a transmission system is fixed, the apparent power is directly related to the current level. In turn, the smaller the net apparent power, the smaller the current drawn from the supply. Minimum current is therefore drawn from a supply when $S = P$ and $Q_T = 0$. Note the effect of decreasing levels of $Q_T$ on the length (and magnitude) of S in Fig. 1 for the same real power.
Fig. 1: Demonstrating the impact of power-factor correction on the power triangle of a network.
Note also that the power-factor angle approaches zero degrees and $F_p$ approaches 1, revealing that the network is appearing more and more resistive at the input terminals. The process of introducing reactive elements to bring the power factor closer to unity is called power-factor correction. Since most loads are inductive, the process normally involves introducing elements with capacitive terminal characteristics having the sole purpose of improving the power factor.
Fig. 2: Demonstrating the impact of a capacitive element on the power factor of a network.
In Fig. 2(a), for instance, an inductive load is drawing a current IL that has a real and an imaginary component. In Fig. 2(b), a capacitive load was added in parallel with the original load to raise the power factor of the total system to the unity power-factor level. Note that by placing all the elements in parallel, the load still receives the same terminal voltage and draws the same current IL. In other words, the load is unaware of and unconcerned about whether it is hooked up as shown in Fig. 2(a) or Fig. 2(b). Solving for the source current in Fig. 2(b): $$\begin{split} I_s &= I_C + I_L\\ &=j I_C(I_{mag}) + I_L(R_e) + j I_L(I_{mag})\\ &= I_L(R_e) + j [I_L(I_{mag}) + I_C(I_{mag})] \end{split}$$ If $X_C$ is chosen such that $I_C(I_{mag}) = I_L(I_{mag})$, then $$I_s= I_L(R_e) + j [0] = I_L(R_e) \angle 0^\circ$$ The result is a source current whose magnitude is simply equal to the real part of the load current, which can be considerably less than the magnitude of the load current of Fig. 2(a). In addition, since the phase angle associated with both the applied voltage and the source current is the same, the system appears "resistive" at the input terminals, and all of the power supplied is absorbed, creating maximum efficiency for a generating utility.
Example 1: A 5-hp motor with a 0.6 lagging power factor and an efficiency of $92%$ is connected to a 208-V, 60-Hz supply.
a. Establish the power triangle for the load.
b. Determine the power-factor capacitor that must be placed in parallel with the load to raise the power factor to unity.
c. Determine the change in supply current from the uncompensated to the compensated system.
d. Find the network equivalent of the above, and verify the conclusions.
Solution:
a. Since 1 hp = 746 W, $$P_o = 5 hp = 5 (746W) = 3730 W$$ and $$P_i (drawn from the line) = { P_o \over \eta}\\ = { 3730 W \over 0.92}= 4054.35 W$$ Also $$F_p = \cos \theta = 0.6$$ Applying $$\tan \theta = {Q_L \over P_i}$$ we obtain $$Q_L = Pi \tan \theta = (4054.35 W) \tan 53.13^\circ$$ $$Q_L = 5405.8 VAR (L)$$ and $$S = \sqrt{P_1^2 + Q_L^2}\\ = \sqrt{(4054.35W)^2 + (5405.8VAR)^2}\\ = 6757.25 VA$$ The power triangle appears in Fig. 3.
Fig. 3: Initial power triangle for the load of Example 1.
b. A net unity power-factor level is established by introducing a capacitive reactive power level of $5405.8 VAR$ to balance $Q_L$. Since $$Q_C = { V^2 \over X_C}$$ then $$X_C = { V^2 \over Q_C} = { (208 V)^2 \over 5405.8 VAR (C)} = 8 Ω$$ and
$$C = { 1\over 2\pi f X_C} = { 1 \over (2\pi)(60 Hz)(8 Ω)} = 331.6 \muF$$
c. At $0.6 \,F_p$, $$S = VI = 6757.25 VA$$ and $$I = {S \over V} = {6757.25 VA \over 208 V} = 32.49 A$$ At unity $F_p$, $$S = VI = 4054.35 VA$$ and $$I = {S \over V} = {4054.35 VA \over 208 V} = 19.49 A$$ producing a $40%$ reduction in supply current. d. For the motor, the angle by which the applied voltage leads the current is $$v\theta = \cos^{-1} 0.6 = 53.13^\circ$$ and $P = EI_m \cos \theta = 4054.35 W$, from above, so that $$I_m = {P \over E \cos \theta}\\ ={4054.35 W \over (208V) (0.6)} = 32.49 \, \text{(as above)}$$ resulting in $$I_m = 32.49 A \angle 53.13^\circ$$ Therefore, $$Z_m = { E \over I_m} = { 208 V \theta 0^\circ \over 32.49 A \angle -53.13^\circ}\\ 6.4Ω \angle 53.13^\circ \\ = 3.84 Ω + j 5.12 Ω$$ as shown in Fig. 4(a).
Fig. 4: Demonstrating the impact of power-factor corrections on the source current.
The equivalent parallel load is determined from $$\begin{split} Y &= { 1 \over Z} = { 1 \over 6.4 Ω\angle 53.13^\circ}\\ &= 0.156 S \angle -53.13^\circ\\ &= 0.094S - j 0.125 S\\ &= {1 \over 10.64 Ω} + { 1 \over j 8Ω}\\ \end{split}$$ as shown in Fig. 3(b),
It is now clear that the effect of the $8 Ω$ inductive reactance can be compensated for by a parallel capacitive reactance of $8 Ω$ using a power-factor correction capacitor of $332 \mu F$.
Since $$Y_T = { 1\over -j X_C} + {1 \over R} + {1 \over jX_L} = {1 \over R}$$ $$I_s = EY_T = E {1 \over R}\\ = (208 V){1 \over 10.64 Ω} = 19.54 A \, \text{as above}$$ In addition, the magnitude of the capacitive current can be determined as follows: $$I_C = {E \over X_C} = { 208 V \over 8Ω} = 26 A$$