Let's break down the problem step by step. --- ## **Given Data** - **Generator:** \( S_n = 25\,\text{MVA} \), \( V_n = 7\,\text{kV} \), \( x_d = .6\,pu \), \( x_q = .7\,pu \) - **Load:** \( S_L = 15\,\text{MVA}, \, pf = .8 \) lagging Supplied at \( V_L = 20\,\text{kV} \) - **Transformer:** \( S_n = 35\,\text{MVA} \), \( V_{n1}/V_{n2} = 7\,\text{kV}/22\,\text{kV} \), \( V_{s/c} = 10\% \) (short-circuit voltage) Negligible copper losses - **Transmission Line:** Length = 3 km, \( Z_{ln} = (2.5 + j12)\,\Omega/\text{km} \) Connected to 22kV side --- ## **Step 1: Find Load Current** ### **Load Apparent Power** \[ S_L = 15\,\text{MVA} \] \[ \text{Power factor (lagging)} = .8 \] \[ P = S_L \times pf = 15 \times .8 = 12\,\text{MW} \] \[ Q = S_L \times \sin(\cos^{-1}(.8)) = 15 \times .6 = 9\,\text{MVAr} \] ### **Line-to-Line Voltage at Load** \[ V_L = 20\,\text{kV} \] ### **Load Current** \[ I_L = \frac{S_L}{\sqrt{3} V_L} = \frac{15 \times 10^6}{\sqrt{3} \times 20 \times 10^3} = \frac{15 \times 10^6}{34,641.016} \approx 432.8\,\text{A} \] --- ## **Step 2: Calculate Total Impedance (Transformer + Line)** ### **Transformer:** \[ \text{Short-circuit voltage} = 10\% \rightarrow Z_{tr} (\text{base 22kV}) = .1 \times \frac{(22,000)^2}{35 \times 10^6} \] \[ = .1 \times \frac{484,000,000}{35,000,000} = .1 \times 13.8286 = 1.38286\,\Omega \] ### **Line Impedance (3 km):** \[ Z_{line} = 3 \times (2.5 + j12) = (7.5 + j36)\,\Omega \] ### **Total Impedance (on 22 kV side):** \[ Z_{tot,22} = Z_{tr} + Z_{line} = 1.38286 + (7.5 + j36) = (8.88286 + j36)\,\Omega \] --- ## **Step 3: Refer Everything to Generator Side (7kV Side)** ### **Impedance Transformation Ratio:** \[ a = \frac{22}{7} = 3.142857 \] \[ Z_{tot,7} = \frac{Z_{tot,22}}{a^2} = \frac{(8.88286 + j36)}{9.8776} \approx (.899 + j3.645)\,\Omega \] --- ## **Step 4: Find the Generator Terminal Voltage** - The voltage at the generator terminal is at 7kV side. ### **Refer Load to Generator Side** \[ V_{gen,LL} = V_{load,LL} \div a = 20\,\text{kV} \div 3.142857 = 6.367\,\text{kV} \] --- ## **Step 5: Find Generator Current** Since load is 15 MVA at 20kV, and current is the same on both sides (for a transformer), so: \[ I_{gen} = I_L = 432.8\,\text{A} \] --- ## **Step 6: Find Generator Internal Voltage** ### **Per-Phase Analysis** - Per-phase voltage at generator terminals: \( V_{t,ph} = \frac{V_{gen,LL}}{\sqrt{3}} = \frac{6367}{\sqrt{3}} = 3675\,\text{V} \) - Per-phase current: \( I_{ph} = \frac{I_{gen}}{1} = 432.8\,\text{A} \) (since it's a Y-connection) #### **Generator Reactance (on own base):** \[ x_{d,actual} = x_d \times \frac{V_{n}^2}{S_n} = .6 \times \frac{(7,000)^2}{25 \times 10^6} = .6 \times 1.96 = 1.176\,\Omega \] #### **Total Reactance (Add to Step 3 Impedance):** \[ Z_{total,ph} = Z_{tot,7} + j x_{d,actual} = (.899 + j3.645) + j1.176 = .899 + j4.821\,\Omega \] #### **Generator Internal Voltage:** \[ E = V_{t,ph} + I_{ph} Z_{total,ph} \] \[ I_{ph} = 432.8\,\angle -36.87^\circ \,\text{A} \] (Power factor angle \( \cos^{-1}(.8) = 36.87^\circ \), lagging, so current lags voltage) Now, \[ I_{ph} Z_{total,ph} = 432.8 \angle -36.87^\circ \times (.899 + j4.821) \] First, calculate impedance magnitude and angle: \[ |Z_{total,ph}| = \sqrt{.899^2 + 4.821^2} = \sqrt{.808 + 23.26} = \sqrt{24.068} = 4.906\,\Omega \] \[ \theta_{Z} = \arctan\left(\frac{4.821}{.899}\right) = \arctan(5.364) = 79.42^\circ \] So, \[ I_{ph} Z_{total,ph} = 432.8 \times 4.906 \angle (79.42^\circ - 36.87^\circ) = 2124.9 \angle 42.55^\circ\,\text{V} \] - Resolve to rectangular: - \( 2124.9 \cos(42.55^\circ) = 1564.6\,\text{V} \) - \( 2124.9 \sin(42.55^\circ) = 1437.9\,\text{V} \) So, \[ I_{ph} Z_{total,ph} = 1564.6 + j1437.9\,\text{V} \] Add to \( V_{t,ph} \) (which is reference, so \( 3675 + j \)): \[ E = (3675 + 1564.6) + j1437.9 = 5239.6 + j1437.9\,\text{V} \] \[ |E| = \sqrt{5239.6^2 + 1437.9^2} = \sqrt{27,457,025 + 2,068,570} = \sqrt{29,525,595} \approx 5434.6\,\text{V} \] --- ## **Step 7: Calculate the Loading Angle (\( \delta \))** \[ \delta = \arctan\left(\frac{1437.9}{5239.6}\right) = \arctan(.2745) = 15.4^\circ \] --- ## **Final Answers** ### **a. Internal Synchronous Generator Voltage** \[ \boxed{5435\,\text{V (per phase, rms)}} \] ### **b. Loading Angle of the Synchronous Generator** \[ \boxed{15.4^\circ} \] --- ### **Summary Table** | Quantity | Value | |------------------------------------------|--------------| | Internal generator voltage (per phase) | 5435 V | | Loading angle (\(\delta\)) | 15.4° | --- **Let me know if you need further clarification or want the answer in per-unit!**