--- title: "15 km line: regulation and zero-regulation capacitor" subject: "Power System Analysis I" chapter: 6 key: "semester_4/power_system_analysis_i:15km-line-zero-regulation" source: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/15km-line-zero-regulation site: BCT Engineering --- # 15 km line: regulation and zero-regulation capacitor Power System Analysis I · Chapter 6: Performance Analysis - Asked in: 2077 Chaitra, 2+2+4 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/15km-line-zero-regulation - Whole chapter: https://www.bctengineer.com/md/semester_4/power_system_analysis_i/chapter-6 ## Question A 15 km long 3 phase overhead line delivers 5 MW at 11 kV at a power factor of 0.8 lagging. Line loss is 12% of the power delivered. Line inductance is 1.1 mH per km per phase. Calculate: (i) sending end voltage and regulation, (ii) power factor of the load to make regulation zero, (iii) the value of capacitor to be connected at the receiving end to reduce regulation to zero. ## Short answer Sending end voltage is 13.72 kV, regulation 24.73%. For zero regulation, load power factor must be about 0.941 leading. A 5.54 MVAr capacitor bank (≈145.8 µF per phase, star-connected) at the receiving end reduces regulation to zero. ## Answer Short line (15 km), so shunt capacitance is neglected. #### Line constants $$ \begin{aligned} I &= \frac{5\times10^6}{\sqrt3 \times 11\,000 \times 0.8} = 328.04\ \text{A} \\ P_{loss} &= 0.12 \times 5 = 0.6\ \text{MW} = 3I^2R \\ R &= \frac{0.6\times10^6}{3\times 328.04^2} = 1.859\ \Omega \\ X &= 2\pi(50)(1.1\times10^{-3})(15) = 5.184\ \Omega \end{aligned} $$ #### (i) Sending-end voltage and regulation $$ \begin{aligned} V_R &= \frac{11\,000}{\sqrt3} = 6350.9\angle0^\circ\ \text{V} \\ I &= 328.04\angle-36.87^\circ\ \text{A} \\ V_S &= V_R + I(R + jX) \\ &= 6350.9 + 328.04\angle-36.87^\circ\,(1.859 + j5.184) \\ &= 7921.5\angle7.21^\circ\ \text{V} \end{aligned} $$ Line value $= \sqrt3 \times 7921.5 = 13.72$ kV. $$ \%VR = \frac{7921.5 - 6350.9}{6350.9}\times100 = 24.73\% $$ #### (ii) Load pf for zero regulation Using the approximate drop $V_S - V_R \approx I(R\cos\phi \pm X\sin\phi)$, regulation is zero when the load is **leading** and $$ \begin{aligned} IR\cos\phi - IX\sin\phi &= 0 \\ \tan\phi &= \frac{R}{X} = \frac{1.859}{5.184} = 0.3585 \\ \phi &= 19.73^\circ,\quad \cos\phi = 0.941\ \text{leading} \end{aligned} $$ (Solving $|V_S| = |V_R|$ exactly with $P$ = 5 MW gives 0.890 leading.) #### (iii) Capacitor at the receiving end The 5 MW load stays at 0.8 lag (needs 3.75 MVAr). For zero regulation the net receiving-end reactive power must be $$ Q_{net} = -P\tan\phi = -5 \times 0.3585 = -1.793\ \text{MVAr} $$ Capacitor rating: $$ Q_C = 3.75 + 1.793 = 5.543\ \text{MVAr (3-phase)} $$ Capacitance per phase (star connected): $$ C = \frac{Q_C}{\omega V_L^2} = \frac{5.543\times10^6}{314.16\times 11\,000^2} = 145.8\ \mu\text{F} $$ **Answer: (i) $V_S$ = 13.72 kV, regulation 24.73 %; (ii) pf ≈ 0.941 leading; (iii) 5.54 MVAr bank, ≈ 145.8 µF per phase (star).** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is R found from the given loss? A. P_loss = 0.12 x 5 MW = 3I^2R, so R = 1.859 ohm B. R = X tan phi C. R = V/I D. R = 0.12 x 15 Answer: A. I = 328 A from 5 MW at 11 kV, 0.8 pf. (Key point: Finding R) 2. How is regulation computed? A. (V_R - V_S)/V_S B. (V_S - V_R)/V_R with V_S = V_R + I(R + jX) = 7921.5 V against 6350.9 V, giving 24.73 % C. I R / V_R only D. (13.72 - 11)/13.72 Answer: B. Short line: no-load voltage equals V_S. (Key point: Regulation) 3. For zero regulation the load must be A. Lagging with tan(phi) = X/R B. Unity C. Leading, with tan(phi) = R/X so IR cos phi = IX sin phi D. Lagging with tan(phi) = R/X Answer: C. A leading current makes the reactive drop oppose the resistive drop. (Key point: Zero regulation pf) 4. How is the capacitor MVAr found? A. 3.75 - 1.793 B. 5 tan 0.3585 only C. 3.75 only D. Load Q (3.75) plus net Q to be supplied for leading pf (1.793) = 5.543 MVAr Answer: D. Net Q must be -1.793 MVAr, so the capacitor supplies 3.75 + 1.793. 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