--- title: "Capacitor for zero regulation, 70 km 220 kV line" subject: "Power System Analysis I" chapter: 6 key: "semester_4/power_system_analysis_i:70km-220kv-zero-regulation-capacitor" source: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/70km-220kv-zero-regulation-capacitor site: BCT Engineering --- # Capacitor for zero regulation, 70 km 220 kV line Power System Analysis I · Chapter 6: Performance Analysis - Asked in: 2069 Bhadra, 6 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/70km-220kv-zero-regulation-capacitor - Whole chapter: https://www.bctengineer.com/md/semester_4/power_system_analysis_i/chapter-6 ## Question A 70 km long, 50 Hz, 3-phase overhead transmission line is operating at 220 kV. The series impedance of the line is 8 + j22 Ohm/phase. The line is supplying a load of 100 MW at 0.9 p.f. lagging. Compute the capacitance per phase to be connected across the load so as to make the voltage regulation of the line zero. ## Short answer To achieve zero regulation, a three-phase shunt capacitor of 87.8 MVAr is needed, corresponding to 5.77 µF per phase in star connection. ## Answer Short line: $Z = 8 + j22\ \Omega$/phase, $V_R$ = 220 kV, load 100 MW at 0.9 pf lag ($Q_L = 100\tan25.84^\circ = 48.43$ MVAr). Without compensation the regulation is 3.92 %. #### Condition for zero regulation With line kV and three-phase MW/MVAr, the net receiving-end $P + jQ$ must satisfy $|V_S| = |V_R|$: $$ \begin{aligned} V_S &= V_R + \frac{(R + jX)(P - jQ)}{V_R} \\ |V_R|^4 &= (V_R^2 + RP + XQ)^2 + (XP - RQ)^2 \end{aligned} $$ Substituting $V_R = 220$, $P = 100$, $R = 8$, $X = 22$: $$ 48\,400^2 = (48\,400 + 800 + 22Q)^2 + (2200 - 8Q)^2 $$ $$ 548Q^2 + 2\,129\,600Q + 82\,920\,000 = 0 $$ Taking the root nearer zero: $$ Q = -39.34\ \text{MVAr}\quad(\text{net pf} = 0.931\ \text{leading}) $$ #### Capacitor rating $$ Q_C = Q_L - Q = 48.43 + 39.34 = 87.77\ \text{MVAr (3-phase)} $$ #### Capacitance per phase (star connected) $$ C = \frac{Q_C}{\omega V_L^2} = \frac{87.77\times10^6}{314.16\times(220\times10^3)^2} = 5.77\ \mu\text{F} $$ (For a delta-connected bank, $C_\Delta = C/3 = 1.92\ \mu$F.) Using the approximate rule $\tan\phi = R/X$ leading gives $Q_C = 48.43 + 100(8/22) = 84.80$ MVAr and $C$ ≈ 5.58 µF, close to the exact value. **Answer: About 87.8 MVAr of shunt capacitors, i.e. C ≈ 5.77 µF per phase (star).** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What condition defines zero voltage regulation? A. |V_S| = |V_R| at full load B. V_S = 0 C. Load current is zero D. The pf is exactly unity Answer: A. Regulation is zero when the sending and receiving voltage magnitudes are equal with the load connected. (Key point: Zero regulation condition) 2. Net required Q is -39.34 MVAr and the load Q is +48.43 MVAr (lagging). What is the capacitor rating? A. 48.43 - 39.34 = 9.09 MVAr B. 39.34 MVAr C. 48.43 + 39.34 = 87.77 MVAr D. 100 MVAr Answer: C. The capacitor must cancel the lagging 48.43 MVAr and then supply a further 39.34 MVAr leading: Q_C = Q_L - Q = 87.77 MVAr. (Key point: Capacitor rating) 3. For a star-connected bank, how is C per phase found? A. C = Q_C / (omega V_L^2) B. C = Q_C / (omega V_ph) C. C = Q_C x omega / V_L^2 D. C = Q_C / (3 omega V_L^2) Answer: A. Q_C = omega C V_L^2 for a star bank (3 x omega C V_ph^2), giving 5.77 microfarad. (Key point: Capacitance per phase) 4. How does the delta-connected capacitance per phase compare with the star value? A. Three times larger B. Equal C. One-third, 1.92 microfarad D. Square root of 3 times larger Answer: C. For the same MVAr, a delta bank sees line voltage across each unit, so C_delta = C_star/3. 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