--- title: "8 km line receiving 15000 kVA at 33 kV" subject: "Power System Analysis I" chapter: 6 key: "semester_4/power_system_analysis_i:8km-line-15000kva" source: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/8km-line-15000kva site: BCT Engineering --- # 8 km line receiving 15000 kVA at 33 kV Power System Analysis I · Chapter 6: Performance Analysis - Asked in: 2073 Bhadra, 8 marks - Video: [Calculate sending end voltage & efficiency](https://www.youtube.com/watch?v=kV6ZJxx5OHA) — Deepak Narkhede - 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/8km-line-15000kva - Whole chapter: https://www.bctengineer.com/md/semester_4/power_system_analysis_i/chapter-6 ## Question 15000 kVA is received at 33 kV at 0.85 power factor lagging over an 8 km three-phase overhead transmission line. Each line has R = 0.29 ohm/km and X = 0.65 ohm/km. Calculate: (i) voltage at the sending end (ii) power factor at the sending end (iii) voltage regulation of the line (iv) efficiency of the transmission line. ## Short answer Sending end voltage = 35.17 kV, sending end power factor = 0.828 lagging, voltage regulation = 6.58%, efficiency = 96.38%. ## Answer Short line (8 km): shunt capacitance neglected. #### Data $$ \begin{aligned} R &= 0.29\times8 = 2.32\ \Omega,\quad X = 0.65\times8 = 5.2\ \Omega \\ V_R &= 33\,000/\sqrt3 = 19\,052.6\ \text{V} \\ I &= \frac{15\,000\times10^3}{\sqrt3\times33\,000} = 262.43\ \text{A} \\ \cos\phi_R &= 0.85,\ \sin\phi_R = 0.527 \end{aligned} $$ #### (i) Sending-end voltage $$ \begin{aligned} V_S &= V_R + I\angle-31.79^\circ\,(2.32 + j5.2) \\ &= 19\,052.6 + 1494.3\angle34.17^\circ \\ &= 20\,288.6 + j839.3 = 20\,306.3\angle2.37^\circ\ \text{V} \end{aligned} $$ Line value $= \sqrt3\times20\,306.3 = 35.17$ kV. #### (ii) Sending-end pf Angle between $V_S$ and $I$ $= 2.37^\circ + 31.79^\circ = 34.16^\circ$. $$ \cos\phi_S = \cos34.16^\circ = 0.828\ \text{lagging} $$ #### (iii) Voltage regulation $$ \%VR = \frac{35.17 - 33}{33}\times100 = 6.58\% $$ #### (iv) Efficiency $$ \begin{aligned} P_R &= 15\,000\times0.85 = 12\,750\ \text{kW} \\ P_{loss} &= 3I^2R = 3(262.43)^2(2.32) = 479.3\ \text{kW} \\ \eta &= \frac{12\,750}{12\,750 + 479.3}\times100 = 96.38\% \end{aligned} $$ **Answer: (i) 35.17 kV, (ii) 0.828 lagging, (iii) 6.58 %, (iv) 96.38 %.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is V_S obtained for this 8 km line? A. V_S = V_R - IZ B. V_S = V_R + IR only C. V_S = I Z D. V_S = V_R + I(R + jX) with I at -31.79 degrees (lagging) Answer: D. Short line: V_S = V_R + I Z with cos(phi) = 0.85 lagging, giving 35.17 kV line. (Key point: Sending voltage) 2. How is the sending-end pf found? A. cos of angle of V_S only B. cos of (delta + phi_R) = cos(2.37 + 31.79) = 0.828 lagging C. cos(31.79) = 0.85 D. cos(delta) = 0.999 Answer: B. The angle between V_S and I is the power angle plus the receiving pf angle, since I lags V_R by phi_R and V_S leads V_R by delta. (Key point: Sending pf) 3. What is the regulation? A. I R/V_S x 100 B. (35.17 - 33)/33 = 6.58 % C. (35.17 - 33)/35.17 D. (33 - 35.17)/33 Answer: B. Short line: (V_S - V_R)/V_R using line values. (Key point: Regulation) 4. How is efficiency calculated? A. 12750/15000 B. loss = 3I^2X C. 12750/(12750 + 479.3) = 96.38 % with loss = 3I^2R D. 15000/(15000 + 479) Answer: C. P_R = 15000 x 0.85 kW; copper loss = 3(262.43)^2(2.32) = 479.3 kW. (Key point: Efficiency) ## Related in Power System Analysis I - [Power output with 46.85 kV sending, 33 kV receiving](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Apower-output-46-85kv-33kv) — 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× - [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 - [Voltage 150 km along a 400 km line](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3A400km-voltage-at-150km) — Power System Analysis I, chapter 5 - [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 - [12 km 33 kV line: losses, zero-regulation capacitor](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3A12km-33kv-losses-zero-regulation) — Power System Analysis I, chapter 6 ## Related in other subjects - [Peterson coil for 33 kV 60 km line](https://www.bctengineer.com/question.md?key=semester_6%2Fswitchgear_and_protection%3Apeterson-coil-33kv-60km) — Switchgear and Protection, chapter 5, asked 6× - [Initial line design for 200 MW over 120 km](https://www.bctengineer.com/question.md?key=semester_8%2Ftransmission_and_distribution_system_design%3Aline-design-200mw-120km) — Transmission and Distribution System Design, chapter 2 - [Received power: 100 MHz, 0.75 km](https://www.bctengineer.com/question.md?key=semester_6%2Fpropagation_and_antenna%3Areceived-power-100mhz-0-75km) — Propagation and Antenna, chapter 5 - [Received power: 15.5 km, 3.4 GHz, dish receiver](https://www.bctengineer.com/question.md?key=semester_6%2Fpropagation_and_antenna%3Amax-received-power-15-5km) — Propagation and Antenna, chapter 5