--- title: "Sending end from A = 0.98∠3°, B = 110∠75°" subject: "Power System Analysis I" chapter: 6 key: "semester_4/power_system_analysis_i:abcd-132kv-50mva-a098-3" source: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/abcd-132kv-50mva-a098-3 site: BCT Engineering --- # Sending end from A = 0.98∠3°, B = 110∠75° Power System Analysis I · Chapter 6: Performance Analysis - Asked in: 2082 Kartik (new course), 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-a098-3 - Whole chapter: https://www.bctengineer.com/md/semester_4/power_system_analysis_i/chapter-6 ## Question A 3-phase 132 kV, 50 Hz overhead line delivers 50 MVA at pf 0.8 lagging at its receiving end. The constants of the line are A = 0.98∠3° and B = 110∠75° Ω per phase. Find (i) sending end voltage and power angle (ii) sending end active and reactive power (iii) line losses. ## Short answer Using V_S = A V_R + B I_R per phase, V_R = 76.21 kV, I_R = 218.7∠−36.87° A. Sending-end line voltage is 165.19 kV, power angle 11.35°. Sending-end powers: P_S = 62.49 MW, Q_S = 33.38 Mvar. Line losses: 22.49 MW, 3.38 Mvar. ## Answer **Given:** 132 kV, 50 MVA at 0.8 p.f. lagging at the receiving end; $A = 0.98\angle 3^\circ$, $B = 110\angle 75^\circ\ \Omega$ per phase. Symmetrical line assumed, so $D = A$. $$ \begin{aligned} V_R &= \frac{132}{\sqrt3} = 76.21\angle 0^\circ\ \text{kV} \\ I_R &= \frac{50\times10^6}{\sqrt3\times132\times10^3} = 218.7\angle -36.87^\circ = 174.95 - j131.22\ \text{A} \end{aligned} $$ $P_R = 50\times0.8 = 40$ MW, $Q_R = 30$ Mvar. #### (i) Sending-end voltage and power angle $$ \begin{aligned} AV_R &= 0.98\angle 3^\circ\times76{,}210 = 74{,}584 + j3909\ \text{V} \\ BI_R &= 110\angle 75^\circ\times218.7\angle -36.87^\circ = 24{,}056\angle 38.13^\circ = 18{,}923 + j14{,}854\ \text{V} \\ V_S &= AV_R + BI_R = 93{,}507 + j18{,}762 = 95.37\angle 11.35^\circ\ \text{kV/phase} \\ V_{S(L)} &= \sqrt3\times95.37 = 165.19\ \text{kV} \end{aligned} $$ Power angle $\delta = 11.35^\circ$. #### (ii) Sending-end active and reactive power Using the sending-end power equations with line values ($\beta = 75^\circ$, $\alpha = 3^\circ$, $D = A$): $$ \begin{aligned} P_S &= \frac{|D||V_S|^2}{|B|}\cos(\beta - \alpha) - \frac{|V_S||V_R|}{|B|}\cos(\beta + \delta) \\ Q_S &= \frac{|D||V_S|^2}{|B|}\sin(\beta - \alpha) - \frac{|V_S||V_R|}{|B|}\sin(\beta + \delta) \end{aligned} $$ $$ \begin{aligned} \frac{|D||V_S|^2}{|B|} &= \frac{0.98\times165.19^2}{110} = 243.10\ \text{MVA} \\ \frac{|V_S||V_R|}{|B|} &= \frac{165.19\times132}{110} = 198.22\ \text{MVA} \\ P_S &= 243.10\cos 72^\circ - 198.22\cos 86.35^\circ = 75.12 - 12.63 = 62.49\ \text{MW} \\ Q_S &= 243.10\sin 72^\circ - 198.22\sin 86.35^\circ = 231.20 - 197.82 = 33.38\ \text{Mvar} \end{aligned} $$ (Check: the receiving-end form of the same equations gives back $P_R = 40$ MW and $Q_R = 30$ Mvar.) #### (iii) Line losses $$ \begin{aligned} P_{loss} &= P_S - P_R = 62.49 - 40 = 22.49\ \text{MW} \\ Q_{loss} &= Q_S - Q_R = 33.38 - 30 = 3.38\ \text{Mvar} \end{aligned} $$ **Answer:** (i) $V_S = 165.19$ kV (line), $\delta = 11.35^\circ$; (ii) $P_S = 62.49$ MW, $Q_S = 33.38$ Mvar; (iii) losses $= 22.49$ MW and $3.38$ Mvar. Note: the active loss is large because the 3° angle of $A$ implies a resistive (lossy) shunt branch: from $AD - BC = 1$, $C = 9.996\times10^{-4}\angle 39.1^\circ$ S. The series resistance alone ($110\cos 75^\circ = 28.5\ \Omega$) would account for only about 4.1 MW. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the receiving-end phase current found? A. I_R = S/(sqrt3 V_L) = 218.7 A at -36.87 deg (lagging for 0.8 pf) B. I_R = S/V_L = 378.8 A C. 218.7 A at +36.87 deg D. I_R = P/(3 V_L) Answer: A. 50 MVA at 132 kV gives 218.7 A, lagging by cos^-1(0.8) = 36.87 deg. (Key point: Receiving current) 2. Which equation gives the sending-end voltage? A. V_S = V_R + B I_R B. V_S = A V_R + B I_R, with V_R = 132/sqrt3 kV as reference C. V_S = A V_R - B I_R D. V_S = C V_R + D I_R Answer: B. The first ABCD equation gives V_S = 95.37 kV per phase at 11.35 deg, i.e. 165.19 kV line. (Key point: Sending voltage) 3. What is the power angle delta? A. The angle of A, 3 deg B. The angle of B, 75 deg C. The phase angle of V_S relative to V_R, 11.35 deg D. The pf angle, 36.87 deg Answer: C. With V_R as reference, the angle of V_S is the power angle. (Key point: Power angle) 4. How are the line losses found? A. Receiving minus sending power B. 3 I^2 R with R = 110 ohm C. Only the Q difference D. Sending-end power minus receiving-end power: 62.49 - 40 = 22.49 MW and 3.38 Mvar Answer: D. Losses are P_S - P_R and Q_S - Q_R. (Key point: Line loss) ## Close variants set in other exams - [Sending end from A = 0.98∠0.3°, B = 110∠75°](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Aabcd-132kv-50mva-a098-0-3) — 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 - [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 - [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 - [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 - [Sending and receiving power, 140∠10° and 132 kV](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Apower-140kv-132kv-a099) — 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