# Power Factor Correction

Whatsapp
Power factor correction is the method of compensating for lagging current by connecting capacitors to the power supply to generate leading current. A sufficient capacitance is connected, allowing the power factor to be as close to unity as possible.
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 $P_f$ 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 $I_L$ 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.
View Solution

## Do you want to say or ask something?

Only 250 characters are allowed. Remaining: 250