 # Power Factor Correction (example problem)

## Given the following circuit: ## Obtain an expression for the complex power of load "A":

Real Power for load "A": $$P_A = 24 \; kW$$ Apparent power for load "A": $$S_A = \frac{P_A}{P_{f_A}} = \frac{24 \; kW}{0.8} = 30 \; kVA$$ Power factor angle for load "A": $$\cos(\theta_A) = 0.8$$ $$\theta_A = \cos^{-1}(0.8) = 36.87^{\circ}$$ Reactive power for load "A": $$Q_A = S_A \sin(\theta_A)$$ $$= 30,000 \sin(36.87^{\circ})$$ $$Q_A = 18 \; kVAR$$ The complex power for load "A": can now be expressed as: $$\mathbb{S}_A = P_A + jQ_A$$ $$\mathbb{S}_A = 24,000 + j 18,000 \qquad (Eqn \; 1)$$

## Obtain an expression for the complex power of load "B":

Real Power for load "B": $$P_B = 40 \; kW$$ Apparent power for load "B": $$S_B = \frac{P_B}{P_{f_B}} = \frac{40 \; kW}{0.95} = 42,105 \; VA$$ Power factor angle for load "B": $$\cos(\theta_B) = 0.95$$ $$\theta_B = \cos^{-1}(0.95) = 18.19^{\circ}$$ Reactive power for load "B": $$Q_B = S_B \sin(\theta_B)$$ $$= 42,105 \sin(18.19^{\circ})$$ $$Q_B = 13,144 \; VAR$$ The complex power for load "B": can now be expressed as: $$\mathbb{S}_B = P_B + jQ_B$$ $$\mathbb{S}_B = 40,000 + j 13,144 \qquad (Eqn \; 2)$$

## Apply the law of conservation of AC power to get the complex power of the circuit:

$$\mathbb{S} = \mathbb{S}_A + \mathbb{S}_B$$ Substituting equations 1 and to into the above expression gives us: $$\mathbb{S} = (24,000 + j 18,000) + (40,000 + j 13,144)$$ $$\mathbb{S} = 64,000 + j31,144 = 71,175 \angle 25.95^{\circ} \qquad (Eqn \; 3)$$ The real power of the circuit will be: $$P = 64,000 \; W$$ From equation #3 we see that : $$25.95^{\circ} = (\theta_v - \theta_i) = power \; factor \; angle$$ We can now determine the power factor of the original circuit: $$P_f = \cos(25.95^{\circ})$$

$$P_f = 0.899$$

### B) Determine the capacitor value required to correct power factor to unity:

In order to get a power factor of "1" we need the new power factor angle to be 0 degrees (cos=1). Recall the formula for the capacitance value required to achieve a unity power factor for an inductive load (found on the previous page) $$C = \frac{P[\tan(\theta_1) - \tan(\theta_2)]}{\omega V^2_{rms}}$$ ...where: $$\theta_2 = new \; power \; factor \; angle = 0^{\circ}$$ Recall that : $$\omega = 2 \pi f$$ and the frequency (f) of our circuit was given to us as 60 Hz. Therefore: $$\omega = 120 \pi$$ We now have all of the necessary information to determine the required capacitor value: $$C = \frac{64,000[\tan(25.95^{\circ}) - \tan(0^{\circ})]}{120 \pi (120)^2}$$

$$C = 5.74 \; \mu F$$