1. **Problem Statement:** We have two alternators A and B operating in parallel supplying a total load of 10,000 kW at 0.8 power factor lagging. Alternator A is adjusted to supply 6,000 kW at 0.92 lagging power factor. We need to find the power factor of alternator B. 2. **Given Data:** - Total load power, $P_{total} = 10,000$ kW - Total power factor, $\cos \phi = 0.8$ lagging - Power factor angle, $\phi = \cos^{-1}(0.8) = 36.9^\circ$ - Alternator A power, $P_A = 6,000$ kW - Alternator A power factor, $\cos \phi_A = 0.92$ lagging - Alternator A power factor angle, $\phi_A = \cos^{-1}(0.92) = 23^\circ$ 3. **Formulas and Important Rules:** - Reactive power $Q = P \tan \phi$ - Total reactive power $Q_{total} = P_{total} \tan \phi$ - For lagging power factor, reactive power is positive (inductive load) - Power supplied by B: $P_B = P_{total} - P_A$ - Reactive power supplied by B: $Q_B = Q_{total} - Q_A$ - Power factor of B: $\cos \phi_B = \frac{P_B}{\sqrt{P_B^2 + Q_B^2}}$ 4. **Calculations:** - Calculate total reactive power: $$Q_{total} = 10,000 \times \tan 36.9^\circ = 10,000 \times 0.7508 = 7,508 \text{ kVAR}$$ - Calculate reactive power of A: $$Q_A = 6,000 \times \tan 23^\circ = 6,000 \times 0.4245 = 2,547 \text{ kVAR}$$ - Calculate power of B: $$P_B = 10,000 - 6,000 = 4,000 \text{ kW}$$ - Calculate reactive power of B: $$Q_B = 7,508 - 2,547 = 4,961 \text{ kVAR}$$ - Calculate apparent power of B: $$S_B = \sqrt{P_B^2 + Q_B^2} = \sqrt{4,000^2 + 4,961^2} = 6,373 \text{ kVA}$$ - Calculate power factor of B: $$\cos \phi_B = \frac{P_B}{S_B} = \frac{4,000}{6,373} = 0.628$$ 5. **Answer for (a):** The power factor of alternator B is approximately **0.628 lagging**. --- 6. **Problem (b) Statement:** If steam supply remains unchanged but excitation of B is reduced so that its power factor becomes 0.92 leading, find the new power factor of A. 7. **Given:** - Power of B unchanged: $P_B = 4,000$ kW - New power factor of B: $\cos \phi_B = 0.92$ leading - $\phi_B = \cos^{-1}(0.92) = 23^\circ$ - Leading power factor means reactive power is negative (capacitive) 8. **Calculations:** - New reactive power of B: $$Q_B = 4,000 \times (-\tan 23^\circ) = 4,000 \times (-0.4245) = -1,698 \text{ kVAR}$$ - Total reactive power remains the same: $Q_{total} = 7,508$ kVAR lagging - New reactive power of A: $$Q_A = Q_{total} - Q_B = 7,508 - (-1,698) = 7,508 + 1,698 = 9,206 \text{ kVAR}$$ - Power of A unchanged: $P_A = 6,000$ kW - Apparent power of A: $$S_A = \sqrt{6,000^2 + 9,206^2} = 10,988 \text{ kVA}$$ - New power factor of A: $$\cos \phi_A = \frac{6,000}{10,988} = 0.546$$ - Since reactive power is lagging, power factor is lagging. 9. **Answer for (b):** The new power factor of alternator A is approximately **0.546 lagging**. **Summary:** - (a) Power factor of B = 0.628 lagging - (b) New power factor of A = 0.546 lagging