--- title: "Efficiency at 0.995 leading pf and zero regulation" subject: "Power System Analysis I" chapter: 6 key: "semester_4/power_system_analysis_i:leading-pf-efficiency-zero-regulation" source: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/leading-pf-efficiency-zero-regulation site: BCT Engineering --- # Efficiency at 0.995 leading pf and zero regulation Power System Analysis I · Chapter 6: Performance Analysis - Asked in: 2076 Bhadra, 4+4 marks - Video: [Effect of Load Power Factor on Regulation and Efficiency | EE301 KTU PGTP](https://www.youtube.com/watch?v=Cw78lNzS788) — Er. PRAVEESH V V - 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/leading-pf-efficiency-zero-regulation - Whole chapter: https://www.bctengineer.com/md/semester_4/power_system_analysis_i/chapter-6 ## Question A 3-phase 50 Hz line operating at 33 kV is supplying a load of 20 MW at 0.995 p.f. leading. The series impedance of the line is 4 + j8 Ω/ph. Determine the power transmission efficiency of the line. How can the line voltage regulation be made zero? ## Short answer Efficiency is about 93.09%. To make regulation zero, the net power factor must be around 0.894 leading, requiring a shunt capacitor of approximately 8 MVAr at the receiving end. ## Answer Short line, $Z = 4 + j8\ \Omega$/phase, $V_R = 33$ kV, load 20 MW at 0.995 pf leading. #### Efficiency $$ \begin{aligned} I &= \frac{20\times10^6}{\sqrt3\times33\,000\times0.995} = 351.67\ \text{A} \\ P_{loss} &= 3I^2R = 3(351.67)^2(4) = 1.484\ \text{MW} \\ \eta &= \frac{20}{20 + 1.484}\times100 = 93.09\% \end{aligned} $$ Present regulation (for reference): $\phi = 5.73^\circ$ leading, $$ V_S = 19\,052.6 + 351.67\angle5.73^\circ\,(4 + j8) = 20\,384.3\angle8.29^\circ\ \text{V} $$ i.e. 35.31 kV line, regulation $= 6.99\%$. It is still positive because the pf is only slightly leading. #### Making regulation zero For a short line, $$ \%VR \approx \frac{I(R\cos\phi - X\sin\phi)}{V_R}\times 100 \quad (\text{leading pf}) $$ This is zero when $$ \tan\phi = \frac{R}{X} = \frac{4}{8} = 0.5 \Rightarrow \cos\phi = 0.894\ \text{leading} $$ So regulation is made zero by making the load (or load plus compensator) take more leading current: - Connect a **shunt capacitor bank / synchronous condenser** at the receiving end. - Required net leading reactive power: $Q = P\tan\phi = 20\times0.5 = 10$ MVAr. - The load already supplies $20\tan(5.73^\circ) = 2.01$ MVAr leading, so the capacitor needed is $$ Q_C = 10 - 2.01 = 7.99\ \text{MVAr} $$ $$ C_{ph} = \frac{Q_C}{\omega V_L^2} = \frac{7.99\times10^6}{314.16\times33\,000^2} = 23.36\ \mu\text{F (star)} $$ (An exact solution of $|V_S| = |V_R|$ gives net pf 0.847 leading and $Q_C$ ≈ 10.55 MVAr.) **Answer: Efficiency ≈ 93.09 %. Zero regulation needs a net pf of about 0.894 leading, i.e. a shunt capacitor of about 8 MVAr at the receiving end.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the power transmission efficiency of this short line found? A. Using sending-end pf only, eta = cos(phi_S) B. Loss = 3I^2R with I = 351.67 A, then eta = P/(P + loss) = 93.09 % C. Loss = 3I^2X, then eta = P/(P + loss) D. Loss = I^2R for one phase, then eta = 100 - loss Answer: B. Only resistance dissipates active power; total loss 3I^2R = 1.484 MW gives eta = 20/21.484 = 93.09 %. (Key point: Efficiency method) 2. For a leading load on a short line, regulation is zero when: A. R sin(phi) = X cos(phi) with lagging pf B. tan(phi) = R/X, i.e. R cos(phi) - X sin(phi) = 0 with leading pf C. tan(phi) = X/R with lagging pf D. cos(phi) = 1 Answer: B. For a leading load %VR is proportional to R cos(phi) - X sin(phi); setting it to zero gives tan(phi) = R/X = 0.5, cos(phi) = 0.894 leading. (Key point: Condition for zero regulation) 3. Why is the regulation at 0.995 pf leading still positive (6.99 %)? A. The line has shunt capacitance B. The load pf is only slightly leading, much less than the 0.894 leading needed C. Leading loads always give positive regulation D. The efficiency is below 100 % Answer: B. Zero regulation needs about 0.894 leading; at 0.995 the load is too close to unity pf, so the resistive drop still dominates. (Key point: Why still positive) 4. How is zero regulation obtained here, and what rating is needed? A. A series resistor; about 2 MW B. A shunt reactor of about 10 MVAr C. A shunt capacitor / synchronous condenser of about 8 MVAr at the receiving end D. A transformer tap change of about 8 % Answer: C. Net leading Q needed is P tan(phi) = 10 MVAr; the load already gives 2.01 MVAr leading, so a capacitor of 10 - 2.01 = 7.99 MVAr is added. (Key point: Compensation) ## Related in Power System Analysis I - [33 kV, 12 MW leading load: efficiency, zero regulation](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3A33kv-12mw-leading-efficiency) — Power System Analysis I, chapter 6, asked 2× - [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 - [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× - [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 - [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 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 - [Current, pf, power and efficiency of 3.3 kV motor](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machines_ii%3Amotor-3-3kv-efficiency) — Electric Machines II, chapter 2 - [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