--- title: "150 km line, A = 0.96∠1°: regulation, max power" subject: "Power System Analysis I" chapter: 6 key: "semester_4/power_system_analysis_i:abcd-150km-30mw-max-power" source: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/abcd-150km-30mw-max-power site: BCT Engineering --- # 150 km line, A = 0.96∠1°: regulation, max power Power System Analysis I · Chapter 6: Performance Analysis - Asked in: 2075 Bhadra, 8 marks - 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-150km-30mw-max-power - Whole chapter: https://www.bctengineer.com/md/semester_4/power_system_analysis_i/chapter-6 ## Question A 3-phase long line of about 150 km has the following parameters: A = D = 0.96∠1.0° and B = 100∠80°. For a load of 30 MW at 0.8 p.f. lag, 110 kV, find (i) sending end voltage and regulation of the line (ii) reactive power supplied by the line (iii) maximum power that can be transferred on the line if sending and receiving end voltages are the same as in the question. ## Short answer Sending end voltage = 132.87 kV, regulation = 25.82%; the line supplies about 2.28 MVAr (sending end provides 20.22 MVAr of the 22.5 MVAr load demand); maximum transferable power ≈124.0 MW. ## Answer Data: $A = D = 0.96\angle1^\circ$, $B = 100\angle80^\circ\ \Omega$, load 30 MW at 0.8 pf lag, $V_R$ = 110 kV. $C$ from $AD - BC = 1$: $C = (A^2-1)/B = 8.53\times10^{-4}\angle77.84^\circ$ S. #### (i) Sending-end voltage and regulation $$ \begin{aligned} V_R &= 110/\sqrt3 = 63.51\angle0^\circ\ \text{kV} \\ I_R &= \frac{30\times10^6}{\sqrt3\times110\times10^3\times0.8} = 196.82\angle-36.87^\circ\ \text{A} \\ AV_R &= 60.97\angle1^\circ\ \text{kV} \\ BI_R &= 19.68\angle43.13^\circ\ \text{kV} \\ V_S &= AV_R + BI_R = 76.71\angle10.91^\circ\ \text{kV} \end{aligned} $$ Line value $= \sqrt3\times76.71 = 132.87$ kV. $$ \%VR = \frac{|V_S|/|A| - |V_R|}{|V_R|}\times100 = \frac{76.71/0.96 - 63.51}{63.51}\times100 = 25.82\% $$ #### (ii) Reactive power supplied by the line $$ \begin{aligned} I_S &= CV_R + DI_R = 174.38\angle-19.35^\circ\ \text{A} \\ S_S &= 3V_SI_S^* = 34.66 + j20.22\ \text{MVA} \end{aligned} $$ Load reactive power $Q_R = 30\times0.75 = 22.5$ MVAr. $$ Q_S - Q_R = 20.22 - 22.5 = -2.28\ \text{MVAr} $$ The sending end supplies only 20.22 MVAr, so the **line itself supplies about 2.28 MVAr** net to the load (its charging MVAr exceeds its $I^2X$ consumption). #### (iii) Maximum power transfer With $|V_S|$ = 132.87 kV and $|V_R|$ = 110 kV, maximum $P_R$ occurs at $\delta = \beta$: $$ \begin{aligned} P_{R,max} &= \frac{|V_S||V_R|}{|B|} - \frac{|A||V_R|^2}{|B|}\cos(\beta-\alpha) \\ &= \frac{132.87\times110}{100} - \frac{0.96\times110^2}{100}\cos79^\circ \\ &= 146.15 - 22.16 = 123.99\ \text{MW} \end{aligned} $$ **Answer: (i) $V_S$ = 132.87 kV, regulation 25.82 %; (ii) the line supplies ≈ 2.28 MVAr (the sending end gives 20.22 MVAr of the 22.5 MVAr load); (iii) $P_{max}$ ≈ 124.0 MW.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which expressions give V_S and regulation? A. V_S = D V_R + C I_R; %VR = (V_S - V_R)/V_R B. V_S = A V_R + B I_R; %VR = (|V_S|/|A| - |V_R|)/|V_R| x 100 = 25.82 % C. V_S = V_R + B I_R; %VR = (V_S - V_R)/V_S D. V_S = A V_R - B I_R; %VR = |V_S|/|V_R| Answer: B. ABCD: V_S = A V_R + B I_R, and no-load receiving voltage is V_S/|A|. (Key point: Sending voltage and regulation) 2. How is it concluded that the line supplies about 2.28 MVAr? A. P_S is greater than P_R B. Because the pf is leading C. Q_S (20.22) is less than Q_R (22.5), so the line generates the difference D. Q_S is greater than Q_R, so the line absorbs it Answer: C. If sending end provides only 20.22 MVAr but load takes 22.5 MVAr, the line's charging exceeds its I^2X absorption by 2.28 MVAr. (Key point: Reactive power of line) 3. Maximum receiving power occurs at what angle and with what expression? A. delta = 90 degrees, P = |V_S||V_R|/|B| B. delta = beta, P_max = |V_S||V_R|/|B| - |A||V_R|^2 cos(beta-alpha)/|B| C. delta = alpha, P_max = |A||V_R|^2/|B| D. delta = 0, P_max = 30 MW Answer: B. From P_R = (V_S V_R/B) cos(beta-delta) - (A V_R^2/B) cos(beta-alpha), the first term is largest at delta = beta, giving 123.99 MW. (Key point: Maximum power) ## Related in Power System Analysis I - [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× - [Regulation and efficiency of 100 km, 110 kV line](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Aregulation-efficiency-100km-110kv) — Power System Analysis I, chapter 6, asked 2× - [Nominal-T 100 km line, 20 MW at 110 kV](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Anominal-t-100km-20mw) — Power System Analysis I, chapter 5 - [220 kV line, 200 MW: sending end from ABCD](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Aabcd-220kv-200mw) — Power System Analysis I, chapter 6 - [150 km line by T-model, 60 MW](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3A150km-t-model-60mw) — Power System Analysis I, chapter 5 - [70 km, 40 MW line: regulation and zero-regulation pf](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3A70km-40mw-regulation) — Power System Analysis I, chapter 6 ## Related in other subjects - [Conductor selection, 100 MW 70 km 132 kV](https://www.bctengineer.com/question.md?key=semester_8%2Ftransmission_and_distribution_system_design%3Aconductor-selection-100mw-70km-132kv) — Transmission and Distribution System Design, chapter 4 - [Empirical formula; voltage for 150 MW over 100 km](https://www.bctengineer.com/question.md?key=semester_8%2Ftransmission_and_distribution_system_design%3Aempirical-formula-150mw-100km) — Transmission and Distribution System Design, chapter 2 - [Conductor selection, 100 MW 80 km, no corona](https://www.bctengineer.com/question.md?key=semester_8%2Ftransmission_and_distribution_system_design%3Aconductor-selection-100mw-80km-corona) — Transmission and Distribution System Design, chapter 4 - [Conductor selection, 220 kV line for 200 MW](https://www.bctengineer.com/question.md?key=semester_8%2Ftransmission_and_distribution_system_design%3Aconductor-selection-220kv-200mw-120km) — Transmission and Distribution System Design, chapter 4