1. **Problem Statement:** We have three loads supplied from a 230 V, 50 Hz single-phase AC supply: (a) Lighting load: 4 kW at unity power factor (PF = 1) (b) Motor load: 10 kVA at 0.8 lagging power factor (c) Intermittent load: 5 kVA at 0.6 lagging power factor We need to: (i) Draw the power triangle for the three loads showing real power (P), reactive power (Q), apparent power (S), and angles. (ii) Find total real power, apparent power, and overall power factor. (iii) Correct overall power factor to 0.9 lagging by adding a capacitor and find the reactive power supplied by it. (iv) Calculate the capacitance value needed for power factor correction. --- 2. **Formulas and Important Rules:** - Real power: $P = S \times \text{PF}$ - Reactive power: $Q = S \times \sin\phi$, where $\cos\phi = \text{PF}$ - Apparent power: $S = \sqrt{P^2 + Q^2}$ - Power factor angle: $\phi = \cos^{-1}(\text{PF})$ - For power factor correction, capacitor supplies reactive power $Q_c$ to reduce total reactive power. - Capacitive reactive power: $Q_c = V^2 \times 2\pi f C$ --- 3. **Step-by-step solution:** **(i) Calculate P, Q, S, and angles for each load:** - Lighting load (a): - $P_a = 4$ kW (given) - $\text{PF}_a = 1 \Rightarrow Q_a = 0$ - $S_a = P_a = 4$ kVA - Angle $\phi_a = \cos^{-1}(1) = 0^\circ$ - Motor load (b): - $S_b = 10$ kVA - $\text{PF}_b = 0.8$ lagging - $P_b = S_b \times 0.8 = 8$ kW - $\phi_b = \cos^{-1}(0.8) = 36.87^\circ$ - $Q_b = S_b \times \sin 36.87^\circ = 10 \times 0.6 = 6$ kVAR - Intermittent load (c): - $S_c = 5$ kVA - $\text{PF}_c = 0.6$ lagging - $P_c = S_c \times 0.6 = 3$ kW - $\phi_c = \cos^{-1}(0.6) = 53.13^\circ$ - $Q_c = S_c \times \sin 53.13^\circ = 5 \times 0.8 = 4$ kVAR **Power triangle for each load:** - Real power $P$ on horizontal axis - Reactive power $Q$ on vertical axis - Apparent power $S$ is hypotenuse - Angles $\phi$ between $P$ and $S$ --- **(ii) Total real power, reactive power, apparent power, and overall power factor:** - Total real power: $$P_{total} = P_a + P_b + P_c = 4 + 8 + 3 = 15 \text{ kW}$$ - Total reactive power: $$Q_{total} = Q_a + Q_b + Q_c = 0 + 6 + 4 = 10 \text{ kVAR}$$ - Total apparent power: $$S_{total} = \sqrt{P_{total}^2 + Q_{total}^2} = \sqrt{15^2 + 10^2} = \sqrt{225 + 100} = \sqrt{325} \approx 18.03 \text{ kVA}$$ - Overall power factor: $$\text{PF}_{total} = \frac{P_{total}}{S_{total}} = \frac{15}{18.03} \approx 0.832 \text{ lagging}$$ - Overall angle: $$\phi_{total} = \cos^{-1}(0.832) \approx 33.6^\circ$$ --- **(iii) Power factor correction to 0.9 lagging:** - Desired power factor: $\text{PF}_{new} = 0.9$ - Desired angle: $$\phi_{new} = \cos^{-1}(0.9) = 25.84^\circ$$ - New reactive power: $$Q_{new} = P_{total} \times \tan \phi_{new} = 15 \times \tan 25.84^\circ = 15 \times 0.484 = 7.26 \text{ kVAR}$$ - Reactive power supplied by capacitor: $$Q_c = Q_{total} - Q_{new} = 10 - 7.26 = 2.74 \text{ kVAR}$$ - This $Q_c$ is capacitive reactive power and will be shown downward on the power triangle reducing total reactive power. --- **(iv) Calculate capacitance $C$ needed:** - Supply voltage $V = 230$ V (rms) - Frequency $f = 50$ Hz - Capacitive reactive power formula: $$Q_c = V^2 \times 2 \pi f C$$ - Rearranged for $C$: $$C = \frac{Q_c}{V^2 \times 2 \pi f}$$ - Convert $Q_c$ to VAR: $$Q_c = 2.74 \times 1000 = 2740 \text{ VAR}$$ - Calculate $C$: $$C = \frac{2740}{230^2 \times 2 \pi \times 50} = \frac{2740}{52900 \times 314.16} = \frac{2740}{16622464} \approx 1.65 \times 10^{-4} \text{ F} = 165 \mu F$$ --- **Final answers:** - Total real power: 15 kW - Total reactive power: 10 kVAR - Total apparent power: 18.03 kVA - Overall power factor: 0.832 lagging - Capacitive reactive power needed: 2.74 kVAR - Capacitance required: 165 microfarads --- **Power triangle summary:** - Horizontal side: $P = 15$ kW - Vertical side before correction: $Q = 10$ kVAR - Vertical side after correction: $Q_{new} = 7.26$ kVAR - Capacitor supplies $Q_c = 2.74$ kVAR reactive power - Hypotenuse before correction: $S = 18.03$ kVA - Hypotenuse after correction: $S_{new} = \frac{P}{0.9} = \frac{15}{0.9} = 16.67$ kVA - Angles: $33.6^\circ$ before, $25.84^\circ$ after correction