--- title: "33 kV, 12 MW leading load: efficiency, zero regulation" subject: "Power System Analysis I" chapter: 6 key: "semester_4/power_system_analysis_i:33kv-12mw-leading-efficiency" source: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/33kv-12mw-leading-efficiency site: BCT Engineering --- # 33 kV, 12 MW leading load: efficiency, zero regulation Power System Analysis I · Chapter 6: Performance Analysis - Asked in: 2070 Bhadra, 6 marks - Video: [Short Transmission Line Numerical 1 - Performance of Transmission Line - Power System Engineering 1](https://www.youtube.com/watch?v=T5_SJHAmo0k) — Ekeeda - 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/33kv-12mw-leading-efficiency - Whole chapter: https://www.bctengineer.com/md/semester_4/power_system_analysis_i/chapter-6 ## Question A 50 Hz, 3-phase, 33 kV overhead line is supplying a load of 12 MW at 0.9 p.f. leading. The line has a series impedance of 6 + j16 Ω/phase. Compute the efficiency of the transmission line. How can the voltage regulation of the line be made zero? ## Short answer Efficiency is 92.45%. To make voltage regulation zero, add about 0.2 MVAr of shunt capacitance at the load, resulting in a net receiving-end power factor of approximately 0.894 leading. ## Answer Short line: $Z = 6 + j16\ \Omega$, $V_R$ = 33 kV, load 12 MW at 0.9 pf leading. #### Efficiency $$ \begin{aligned} I &= \frac{12\times10^6}{\sqrt3\times33\,000\times0.9} = 233.27\ \text{A} \\ P_{loss} &= 3I^2R = 3(233.27)^2(6) = 0.979\ \text{MW} \\ \eta &= \frac{12}{12 + 0.979}\times100 = 92.45\% \end{aligned} $$ #### Present regulation $$ \begin{aligned} V_S &= 19\,052.6 + 233.27\angle25.84^\circ\,(6 + j16) = 19\,102.3\angle11.99^\circ\ \text{V} \\ &\Rightarrow 33.09\ \text{kV (line)},\quad \%VR = 0.26\% \end{aligned} $$ The regulation is already very small, because the leading load current nearly cancels the drop. #### Making regulation zero Zero regulation needs $|V_S| = |V_R|$. With line kV, MW, MVAr: $$ (V_R^2 + RP + XQ)^2 + (XP - RQ)^2 = V_R^4 $$ $$ (1089 + 72 + 16Q)^2 + (192 - 6Q)^2 = 1089^2 $$ The root nearer zero is $$ Q = -6.01\ \text{MVAr} \quad (\text{pf} = 0.894\ \text{leading}) $$ The load supplies $-12\tan25.84^\circ = -5.81$ MVAr, so a little more leading VAr is needed: $$ Q_C = 6.01 - 5.81 = 0.20\ \text{MVAr} $$ So connect a small **shunt capacitor** (about 0.2 MVAr, $C = Q_C/\omega V_L^2 \approx 0.58\ \mu$F per phase in star) at the receiving end, or raise the leading current of a synchronous load slightly. In general, regulation is made zero by adjusting receiving-end reactive power (capacitor or reactor) until the net pf is near $\cos\phi = X/|Z|$ leading. Note: the approximate rule $\tan\phi = R/X$ gives pf 0.936 lead here, but at such small regulation the neglected quadrature term matters, so the exact value above is used. **Answer: Efficiency ≈ 92.45 %. Regulation (now 0.26 %) becomes zero with about 0.2 MVAr of extra shunt capacitance at the load (net pf ≈ 0.894 leading).** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the line current found for a 12 MW, 33 kV, 0.9 pf three-phase load? A. I = P / (3 x V_L x pf) B. I = P / (sqrt(3) x V_L x pf) C. I = P / (sqrt(3) x V_L) D. I = P x pf / (sqrt(3) x V_L) Answer: B. Three-phase power is sqrt(3) V_L I cos(phi), so I = P/(sqrt(3) V_L pf) = 233.27 A. (Key point: Line current from load data) 2. Which expression gives the efficiency of this short line? A. Received power / (received power + 3I^2R) B. Received power / (received power + 3I^2X) C. Received power x 3I^2R D. (received power - 3I^2R) / received power Answer: A. Only the resistance dissipates active power, so loss = 3I^2R and eta = P_R/(P_R + loss) = 92.45%. (Key point: Efficiency of a short line) 3. Why is the present regulation only about 0.26% although the line has a large reactance? A. The line resistance is negligible B. The line is long enough to have charging current C. The leading load current nearly cancels the voltage drop D. The sending voltage is lower than the receiving voltage Answer: C. A leading current makes the reactive drop oppose the rise, so |V_S| is almost equal to |V_R|. (Key point: Why regulation is tiny for a leading load) 4. How is the regulation made zero in this answer? A. Increase the line resistance B. Add a series inductor at the sending end C. Reduce the load to below 12 MW only D. Adjust receiving-end reactive power, here a small shunt capacitor, so that |V_S| = |V_R| Answer: D. Zero regulation means |V_S| = |V_R|; this is achieved by changing receiving-end VAr, about 0.2 MVAr more leading here. (Key point: Making regulation zero) ## Close variants set in other exams - [Zero regulation for 40 MW leading load](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Azero-regulation-40mw-leading) — Power System Analysis I, chapter 6, asked 2× ## Related in Power System Analysis I - [33 kV, 15 MW line: efficiency and regulation](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3A33kv-15mw-efficiency-regulation) — Power System Analysis I, chapter 6 - [Efficiency at 0.995 leading pf and zero regulation](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Aleading-pf-efficiency-zero-regulation) — 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 - [Capacitor for zero regulation, 70 km 220 kV line](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3A70km-220kv-zero-regulation-capacitor) — Power System Analysis I, chapter 6 - [12 km line at 32.5 kV: regulation and capacitor](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3A12km-32-5kv-regulation-capacitor) — Power System Analysis I, chapter 6 - [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× ## Related in other subjects - [220 kV double circuit efficiency and regulation](https://www.bctengineer.com/question.md?key=semester_8%2Ftransmission_and_distribution_system_design%3Adouble-circuit-220kv-efficiency-regulation) — Transmission and Distribution System Design, chapter 4 - [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 - [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× - [Currents and efficiency of 132/11 kV transformer at 5 MW](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Astar-delta-132-11kv-5mw-currents) — Electrical Machines I, chapter 2, asked 3×