--- title: "Sending end from A = 0.98∠0.3°, B = 110∠75°" subject: "Power System Analysis I" chapter: 6 key: "semester_4/power_system_analysis_i:abcd-132kv-50mva-a098-0-3" source: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/abcd-132kv-50mva-a098-0-3 site: BCT Engineering --- # Sending end from A = 0.98∠0.3°, B = 110∠75° Power System Analysis I · Chapter 6: Performance Analysis - Asked in: 2072 Magh, 6 marks - Video: [PS25 Numerical on Medium Transmission Lines](https://www.youtube.com/watch?v=Dt4Y5BlylMo) — Lectures in Electrical Engineering - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/abcd-132kv-50mva-a098-0-3 - Whole chapter: https://www.bctengineer.com/md/semester_4/power_system_analysis_i/chapter-6 ## Question A three phase 132 kV overhead line delivers 50 MVA at 132 kV and power factor 0.8 lagging at its receiving end. The constants of the line are A = 0.98∠0.3° and B = 110∠75° ohms per phase. Find (a) sending end voltage and power angle (b) sending end active and reactive power (c) line losses and VARs absorbed by the line. ## Short answer The sending-end line voltage is 164.27 kV with power angle 9.25°. Sending-end powers are 43.69 MW and 35.76 MVAr. Line losses are 3.69 MW and the line absorbs 5.76 MVAr. ## Answer Data: $A = 0.98\angle0.3^\circ$, $B = 110\angle75^\circ\ \Omega$, load 50 MVA at 0.8 lag at 132 kV, i.e. $S_R = 40 + j30$ MVA. $D = A$ is assumed (symmetrical line). #### (a) Sending-end voltage and power angle $$ \begin{aligned} V_R &= 132/\sqrt3 = 76.21\angle0^\circ\ \text{kV} \\ I_R &= \frac{50\times10^6}{\sqrt3\times132\times10^3} = 218.69\angle-36.87^\circ\ \text{A} \\ AV_R &= 74.69\angle0.3^\circ\ \text{kV} \\ BI_R &= 110\angle75^\circ\times0.21869\angle-36.87^\circ = 24.06\angle38.13^\circ\ \text{kV} \\ V_S &= AV_R + BI_R = 94.84\angle9.25^\circ\ \text{kV (phase)} \end{aligned} $$ $|V_S|$ (line) $= \sqrt3\times94.84 = 164.27$ kV, power angle $\delta = 9.25^\circ$. #### (b) Sending-end active and reactive power $$ \begin{aligned} P_S &= \frac{|A||V_S|^2}{|B|}\cos(\beta-\alpha) - \frac{|V_S||V_R|}{|B|}\cos(\beta+\delta) \\ Q_S &= \frac{|A||V_S|^2}{|B|}\sin(\beta-\alpha) - \frac{|V_S||V_R|}{|B|}\sin(\beta+\delta) \end{aligned} $$ With $\frac{0.98\times164.27^2}{110} = 240.41$, $\frac{164.27\times132}{110} = 197.12$, $\beta-\alpha = 74.7^\circ$, $\beta+\delta = 84.25^\circ$: $$ \begin{aligned} P_S &= 240.41\cos74.7^\circ - 197.12\cos84.25^\circ = 43.69\ \text{MW} \\ Q_S &= 240.41\sin74.7^\circ - 197.12\sin84.25^\circ = 35.76\ \text{MVAr} \end{aligned} $$ (Check with the receiving-end equations at $\delta = 9.25^\circ$: $P_R = 40.00$ MW, $Q_R = 30.00$ MVAr.) #### (c) Line losses and VARs absorbed $$ \begin{aligned} P_{loss} &= 43.69 - 40 = 3.69\ \text{MW} \\ Q_{absorbed} &= 35.76 - 30 = 5.76\ \text{MVAr} \end{aligned} $$ **Answer: (a) $V_S$ = 164.27 kV, δ = 9.25°; (b) $P_S$ = 43.69 MW, $Q_S$ = 35.76 MVAr; (c) loss 3.69 MW, line absorbs 5.76 MVAr.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which gives V_S and power angle? A. V_S = V_R/A; delta = alpha B. V_S = A V_R only C. V_S = A V_R + B I_R; delta = angle of V_S (9.25 degrees) D. V_S = V_R + B; delta = 0 Answer: C. V_S = 74.69 angle 0.3 deg + 24.06 angle 38.13 deg = 94.84 angle 9.25 deg kV phase; delta is its angle since V_R is reference. (Key point: Sending voltage) 2. Which pair of terms gives P_S? A. |A||V_R|^2 cos(beta+alpha)/|B| B. |V_S||V_R| cos(delta)/|B| C. |V_S|^2 sin(beta)/|B| D. |A||V_S|^2 cos(beta-alpha)/|B| minus |V_S||V_R| cos(beta+delta)/|B| Answer: D. By symmetry with the receiving-end equations the sending-end power has the A|V_S|^2 term and the (beta + delta) cross term. (Key point: Sending power formula) 3. How are line losses obtained? A. P_S/P_R B. P_loss = P_S - P_R = 43.69 - 40 = 3.69 MW C. P_S + P_R D. P_R - P_S Answer: B. Loss is sending minus receiving power. (Key point: Line loss) 4. What is the VAr absorbed by the line? A. Q_S + Q_R B. 40 - 35.76 C. Q_S - Q_R = 35.76 - 30 = 5.76 MVAr D. Q_R - Q_S Answer: C. Positive difference means net absorption. (Key point: Reactive absorbed) ## Close variants set in other exams - [Sending end from A = 0.98∠3°, B = 110∠75°](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Aabcd-132kv-50mva-a098-3) — Power System Analysis I, chapter 6, asked 2× ## Related in Power System Analysis I - [Receiving and sending end powers from ABCD, 132/134 kV](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Aabcd-power-132kv-134kv) — Power System Analysis I, chapter 6 - [Sending end active power from ABCD, 132/130.5 kV](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Asending-power-132kv-130-5kv) — Power System Analysis I, chapter 6 - [110 km, 110 kV line: sending end, regulation, efficiency](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3A110km-line-performance) — Power System Analysis I, chapter 6, asked 2× - [Nominal-T 100 km line, 50 MVA at 132 kV](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Anominal-t-100km-50mva) — Power System Analysis I, chapter 6 - [Static compensation to hold sending end at 140 kV](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Astatic-compensation-140kv) — Power System Analysis I, chapter 6 - [Receiving end power, 140.2∠18.2° kV and 132 kV](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Areceiving-power-140-2kv) — Power System Analysis I, chapter 6 ## Related in other subjects - [Fault at receiving end of 132 kV lines](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Afault-132kv-receiving-end-2071) — Power System Analysis II, chapter 3 - [Fault at receiving end of 220 kV lines](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Afault-220kv-two-generators) — Power System Analysis II, chapter 3 - [Fault at load end of overhead line](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Afault-load-end-33kv-line) — Power System Analysis II, chapter 3