--- title: "Static compensation to hold sending end at 140 kV" subject: "Power System Analysis I" chapter: 6 key: "semester_4/power_system_analysis_i:static-compensation-140kv" source: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/static-compensation-140kv site: BCT Engineering --- # Static compensation to hold sending end at 140 kV Power System Analysis I · Chapter 6: Performance Analysis - Asked in: 2074 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/static-compensation-140kv - Whole chapter: https://www.bctengineer.com/md/semester_4/power_system_analysis_i/chapter-6 ## Question A 3-phase 132 kV overhead line delivers 50 MVA at 132 kV and 0.8 pf lagging. The constants of the line are A = 0.983∠3°, B = 110∠75° Ω per phase. Determine the capacity of the static compensation equipment at the receiving end to reduce the sending end voltage to 140 kV for the same load condition. ## Short answer A capacitive compensator of approximately 35 MVAr must be installed at the receiving end to reduce the sending end voltage to 140 kV for the same load. ## Answer Data: $|A| = 0.983$, $\alpha = 3^\circ$, $|B| = 110\ \Omega$, $\beta = 75^\circ$, $V_R$ = 132 kV, load 50 MVA at 0.8 lag = 40 MW + j30 MVAr. With line kV, powers are three-phase. Without compensation, $V_S = AV_R + B(P - jQ)/V_R = 165.6$ kV, which is too high. We need $|V_S| = 140$ kV with the same 40 MW load. #### Receiving-end power-circle equations $$ \begin{aligned} P_R &= \frac{|V_S||V_R|}{|B|}\cos(\beta-\delta) - \frac{|A||V_R|^2}{|B|}\cos(\beta-\alpha) \\ Q_R &= \frac{|V_S||V_R|}{|B|}\sin(\beta-\delta) - \frac{|A||V_R|^2}{|B|}\sin(\beta-\alpha) \end{aligned} $$ $$ \frac{140\times132}{110} = 168.0,\qquad \frac{0.983\times132^2}{110} = 155.71,\qquad \beta-\alpha = 72^\circ $$ #### Find $\delta$ from $P_R$ = 40 MW $$ \begin{aligned} 40 &= 168.0\cos(75^\circ-\delta) - 155.71\cos72^\circ \\ 40 &= 168.0\cos(75^\circ-\delta) - 48.12 \\ \cos(75^\circ-\delta) &= 0.5245 \Rightarrow 75^\circ-\delta = 58.37^\circ,\ \delta = 16.63^\circ \end{aligned} $$ #### Reactive power the line delivers $$ \begin{aligned} Q_R &= 168.0\sin58.37^\circ - 155.71\sin72^\circ \\ &= 143.04 - 148.09 = -5.05\ \text{MVAr} \end{aligned} $$ So the line can deliver no lagging VAr; it actually needs 5.05 MVAr from the receiving end. #### Compensator rating Load requires 30 MVAr lagging. The compensator must supply: $$ Q_C = 30 - (-5.05) = 35.05\ \text{MVAr} $$ **Answer: Static (capacitive) compensator of about 35 MVAr at the receiving end.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is held fixed and what is found to size the compensator? A. delta fixed at 0 B. |V_R| is varied C. |V_S| = 140 kV and P_R = 40 MW fixed; find delta then the Q_R the line can deliver D. Q_R fixed at 30; find P Answer: C. Using the power-circle equations with |V_S| = 140 kV, P_R = 40 MW gives delta = 16.63 deg and then Q_R. (Key point: Method) 2. What is the line's Q_R at |V_S| = 140 kV and what does it mean? A. +30 MVAr; no compensator needed B. -5.05 MVAr; line cannot supply lagging VAr, needs 5.05 from the receiving end C. +5.05 MVAr; load Q reduced D. 0 MVAr; unity pf Answer: B. Q_R = 143.04 - 148.09 = -5.05 MVAr: the line delivers no lagging VAr at that sending voltage. 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