Let's break down the problem and solve each part step-by-step. Some values will come from the provided problem, while others may require formulas or standard knowledge for synchronous machines. **We will do each part in order.** --- ### **a) Rotor mechanical and electrical speeds, and load impedance value** #### **Mechanical Speed (\( \omega_{mech} \)):** Given: - Wind speed = 12.0 m/s - Generator operates at 1.0 pu rotor speed (rated speed) - Frequency \(f = 11.25\) Hz The **synchronous speed** formula: \[ N_s = \frac{120 f}{P} \] where \(P\) = number of poles. But, let's find angular speed: \[ \omega_{mech} = 2\pi N_s / 60 \] But first, let's get \(N_s\): - \(f = 11.25\) Hz - Assume \(P\) is given in Appendix B. (If not specified, let's leave \(P\) as a variable.) \[ N_s = \frac{120 \times 11.25}{P} \quad [\text{rpm}] \] \[ \omega_{mech} = 2\pi \times \frac{N_s}{60} \] #### **Electrical Speed (\( \omega_{elec} \)):** \[ \omega_{elec} = 2\pi f = 2\pi \times 11.25 = 70.685 \text{ rad/s} \] #### **Load Impedance:** Given: - \(R_L = 0.19125\, \Omega\) - \(L_L = 1.3104\, \text{mH} = 1.3104 \times 10^{-3}\, \text{H}\) - \(f = 11.25\) Hz \[ X_L = 2\pi f L_L = 2\pi \times 11.25 \times 1.3104 \times 10^{-3} = 0.0926\, \Omega \] So, \[ Z_L = R_L + jX_L = 0.19125 + j0.0926\, \Omega \] --- ### **b) \( dq \)-axis and rms stator currents** The load is supplied at 690 V (line-to-line, rms), so per phase: \[ V_{ph} = \frac{690}{\sqrt{3}} = 398.37\, V \] \[ I_{ph} = \frac{V_{ph}}{Z_L} \] Magnitude of \(Z_L\): \[ |Z_L| = \sqrt{0.19125^2 + 0.0926^2} = 0.2128\, \Omega \] So, \[ I_{ph} = \frac{398.37}{0.2128} = 1872.2\, A \] **dq-axis currents** normally require machine parameters (from Appendix B). If only load current is needed, then above is the answer. Otherwise, use Park's transformation if needed. --- ### **c) \( dq \)-axis and rms stator voltages** Already calculated \(V_{ph, rms} = 398.37\, V\). **dq-axis voltages** need the stator resistance, reactances (from Appendix B), and current. Typically: \[ V_{d} = V_{ph, rms} \cos(\theta) \] \[ V_{q} = V_{ph, rms} \sin(\theta) \] Where \(\theta = \tan^{-1}(X_L / R_L)\). \[ \theta = \tan^{-1} \left(\frac{0.0926}{0.19125}\right) = 25.6^\circ \] \[ V_{d} = 398.37 \cos(25.6^\circ) = 359.2\, V \] \[ V_{q} = 398.37 \sin(25.6^\circ) = 172.2\, V \] --- ### **d) Mechanical Power and Torque** \[ P_{out} = \sqrt{3} V_{LL} I_{ph} \cos\phi \] Where \(\cos\phi\) is the power factor: \[ \cos\phi = \frac{R_L}{|Z_L|} = \frac{0.19125}{0.2128} = 0.899 \] \[ P_{out} = \sqrt{3} \times 690 \times 1872.2 \times 0.899 = 2,013,000\, W = 2.013\, MW \] **Torque:** \[ T = \frac{P_{mech}}{\omega_{mech}} \] Where \(\omega_{mech}\) (from above, needs number of poles). --- ### **e) Stator winding loss and power factor** **Stator winding loss:** \[ P_{loss} = 3 I_{ph}^2 R_L = 3 \times (1872.2)^2 \times 0.19125 = 2,009,000\, W \] **Power factor:** Already calculated above, \(\cos\phi = 0.899\). --- ### **f) Stator active and reactive powers, generator efficiency** **Active Power:** \[ P = \sqrt{3} V_{LL} I_{ph} \cos\phi = 2,013,000\, W \] **Reactive Power:** \[ Q = \sqrt{3} V_{LL} I_{ph} \sin\phi \] \[ \sin\phi = \sqrt{1 - \cos^2\phi} = \sqrt{1 - 0.899^2} = 0.438 \] \[ Q = \sqrt{3} \times 690 \times 1872.2 \times 0.438 = 976,000\, VAR \] **Efficiency:** \[ \eta = \frac{P_{out}}{P_{in}} \] Assuming only stator winding loss, \[ P_{in} = P_{out} + P_{loss} = 2,013,000 + 2,009,000 = 4,022,000\, W \] \[ \eta = \frac{2,013,000}{4,022,000} = 0.5006 = 50.06\% \] --- ## **Summary Table** | Quantity | Value | |-----------------------------|------------------------------| | Mech. Speed (\(N_s\)) | \( \frac{120 \times 11.25}{P} \) rpm | | Elec. Speed (\(\omega_{elec}\)) | 70.685 rad/s | | Load Impedance (\(Z_L\)) | \(0.19125 + j0.0926\, \Omega\) | | rms Stator Current (\(I_{ph}\)) | 1872.2 A | | rms Stator Voltage (\(V_{ph}\)) | 398.37 V | | \(V_d\), \(V_q\) | 359.2 V, 172.2 V | | Output Power (\(P_{out}\)) | 2,013,000 W | | Power Factor (\(\cos\phi\)) | 0.899 | | Stator Loss (\(P_{loss}\)) | 2,009,000 W | | Reactive Power (\(Q\)) | 976,000 VAR | | Efficiency (\(\eta\)) | 50.06% | --- **Note:** For exact mechanical speed and torque, the number of poles is required from Appendix B. If you provide it, I'll update the calculations. Let me know if you need a detailed calculation for a specific part, or if you have the number of poles!