Let's break down the problem step by step: ### **Given Data** - **Induction Motor:** - Real Power, \( P_1 = 275\, \text{kW} \) - Power factor, \( pf_1 = 0.5 \) lagging - **Synchronous Motor:** - Real Power, \( P_2 = 100\, \text{kW} \) - \( pf_2 \) unknown (machine is overexcited, so it supplies vars, i.e., leading or less lagging) - **Combined Power factor:** \( pf_{total} = 0.75 \) lagging We need to **find the kVA rating of the synchronous motor**. --- ## **Step 1: Find Apparent and Reactive Power of Induction Motor** ### **Apparent Power of Induction Motor (\( S_1 \))** \[ S_1 = \frac{P_1}{pf_1} = \frac{275}{0.5} = 550\, \text{kVA} \] ### **Reactive Power of Induction Motor (\( Q_1 \))** \[ Q_1 = S_1 \cdot \sin(\cos^{-1}(pf_1)) \] \[ \cos^{-1}(0.5) = 60^\circ \implies \sin(60^\circ) = 0.866 \] \[ Q_1 = 550 \times 0.866 = 476.3\, \text{kVAR} \quad (\text{lagging}) \] --- ## **Step 2: Find Total Real Power and Apparent Power with Overall Power Factor** ### **Total Real Power (\( P_{total} \))** \[ P_{total} = P_1 + P_2 = 275 + 100 = 375\, \text{kW} \] ### **Total Apparent Power (\( S_{total} \))** \[ S_{total} = \frac{P_{total}}{pf_{total}} = \frac{375}{0.75} = 500\, \text{kVA} \] ### **Total Reactive Power (\( Q_{total} \))** \[ Q_{total} = S_{total} \cdot \sin(\cos^{-1}(0.75)) \] \[ \cos^{-1}(0.75) = 41.41^\circ \implies \sin(41.41^\circ) = 0.6614 \] \[ Q_{total} = 500 \times 0.6614 = 330.7\, \text{kVAR} \quad (\text{lagging}) \] --- ## **Step 3: Find Reactive Power of Synchronous Motor** \[ Q_{total} = Q_1 + Q_2 \] But since the synchronous motor is overexcited, it **supplies vars**, i.e., **Q₂ is negative** (leading). \[ Q_2 = Q_{total} - Q_1 = 330.7 - 476.3 = -145.6\, \text{kVAR} \] So, **synchronous motor supplies 145.6 kVAR leading**. --- ## **Step 4: Find Apparent Power (\( S_2 \)) of Synchronous Motor** \[ S_2 = \sqrt{P_2^2 + Q_2^2} \] \[ S_2 = \sqrt{100^2 + (145.6)^2} = \sqrt{10000 + 21209.6} = \sqrt{31209.6} = 176.7\, \text{kVA} \] --- ## **Final Answers** ### **kVA Rating of Synchronous Motor:** \[ \boxed{176.7\ \text{kVA}} \] --- ### **Summary Table** | Motor | Real Power (kW) | Apparent Power (kVA) | Reactive Power (kVAR) | |-------------------|-----------------|----------------------|----------------------| | Induction Motor | 275 | 550 | +476.3 | | Synchronous Motor | 100 | 176.7 | -145.6 | | **Total** | 375 | 500 | +330.7 | --- ## **Explanation** - The induction motor draws a lot of lagging vars (reactive power). - The overexcited synchronous motor is used to *compensate* for some of this, by supplying leading vars. - The result is that the overall system power factor improves to 0.75 lagging. - The kVA rating of the synchronous motor is calculated based on its real and reactive power output, using the Pythagorean theorem for apparent power. --- ### **Final boxed answer:** \[ \boxed{176.7\ \text{kVA}} \] is the kVA rating of the synchronous motor.