--- title: "Compensation capacity for 15 MW, 33 kV load" subject: "Power System Analysis I" chapter: 6 key: "semester_4/power_system_analysis_i:compensation-capacity-15mw-33kv" source: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/compensation-capacity-15mw-33kv site: BCT Engineering --- # Compensation capacity for 15 MW, 33 kV load Power System Analysis I · Chapter 6: Performance Analysis - Asked in: 2079 Chaitra, 8 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/compensation-capacity-15mw-33kv - Whole chapter: https://www.bctengineer.com/md/semester_4/power_system_analysis_i/chapter-6 ## Question A three-phase overhead line has resistance and reactance per phase of 5 and 25 ohm, respectively. The load at the receiving-end is 15 MW, 33 kV, 0.8 power factor lagging. Find the capacity of the compensation equipment needed to deliver this load with a sending-end voltage of 33 kV. Calculate the extra load of 0.8 lagging power factor which can be delivered with the compensating equipment (of capacity as calculated above) installed, if the receiving-end voltage is permitted to drop to 28 kV. ## Short answer Using short-line model (Z=25.495∠78.69° Ω) and power equations, to deliver 15 MW, 33 kV at 0.8 lag with equal sending voltage, compensation of 17.39 MVAr capacitive is required. With that compensator and receiving voltage dropping to 28 kV, an extra 3.52 MW at 0.8 lag can be supplied. ## Answer Treat it as a short line: $A = 1\angle0^\circ$, $B = Z = 5 + j25 = 25.495\angle78.69^\circ\ \Omega$. With line kV and three-phase MW/MVAr, the receiving-end power circle equations are: $$ \begin{aligned} P_R &= \frac{|V_S||V_R|}{|Z|}\cos(\theta-\delta) - \frac{|V_R|^2}{|Z|}\cos\theta \\ Q_R &= \frac{|V_S||V_R|}{|Z|}\sin(\theta-\delta) - \frac{|V_R|^2}{|Z|}\sin\theta \end{aligned} $$ with $\theta = 78.69^\circ$. #### Part 1: Compensation for $|V_S| = |V_R| = 33$ kV $$ \frac{33\times33}{25.495} = 42.714,\quad \frac{33^2}{25.495}\cos\theta = 8.377,\quad \frac{33^2}{25.495}\sin\theta = 41.885 $$ Active power condition: $$ \begin{aligned} 15 &= 42.714\cos(78.69^\circ-\delta) - 8.377 \\ \cos(78.69^\circ-\delta) &= 0.5473 \Rightarrow \delta = 21.87^\circ \end{aligned} $$ Reactive power the line can deliver at this condition: $$ Q_R = 42.714\sin(56.82^\circ) - 41.885 = -6.14\ \text{MVAr} $$ The load needs $Q_L = 15\tan(\cos^{-1}0.8) = 11.25$ MVAr (lagging), and the line itself needs 6.14 MVAr fed into its receiving end. So the compensator must supply: $$ Q_C = 11.25 - (-6.14) = 17.39\ \text{MVAr} $$ #### Part 2: Extra load with $|V_R| = 28$ kV and the same $Q_C$ $$ \frac{33\times28}{25.495} = 36.242,\quad \frac{28^2}{25.495}\cos\theta = 6.031,\quad \frac{28^2}{25.495}\sin\theta = 30.154 $$ For a new load $P$ at 0.8 pf lag, the line must deliver $Q_R = 0.75P - 17.39$. So: $$ \begin{aligned} P &= 36.242\cos(78.69^\circ-\delta) - 6.031 \\ 0.75P - 17.39 &= 36.242\sin(78.69^\circ-\delta) - 30.154 \end{aligned} $$ Solving these together (eliminating $P$): $$ \delta = 31.33^\circ,\quad P = 18.52\ \text{MW},\quad Q_R = -3.49\ \text{MVAr} $$ Check: $0.75 \times 18.52 - 17.39 = -3.49$ MVAr. Extra load $= 18.52 - 15 = 3.52$ MW. **Answer: Compensator capacity ≈ 17.39 MVAr; extra load ≈ 3.52 MW at 0.8 pf lag (total 18.52 MW at 28 kV).** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. In part 1 the active power equation is used with |V_S| = |V_R| = 33 kV to find A. the load angle delta (21.87 deg) needed to deliver 15 MW B. The compensator rating directly C. The line resistance D. The sending pf Answer: A. Solve 15 = 42.714 cos(theta - delta) - 8.377 for delta. (Key point: Condition for part 1) 2. How is the compensator capacity found? A. Q_C = 15 tan(phi) = 11.25 MVAr B. Q_C = load Q (11.25 MVAr) minus Q_R of the line (-6.14 MVAr) = 17.39 MVAr C. Q_C = 11.25 - 6.14 = 5.11 MVAr D. Q_C = P/pf = 18.75 Answer: B. The line needs 6.14 MVAr fed at its receiving end in addition to the load's 11.25. (Key point: Compensator rating) 3. In part 2 with the same compensator at 28 kV and 0.8 pf load, what relation links P and Q_R? A. Q_R = 0.75P + 17.39 B. Q_R = 0.8P C. Q_R = 0.75P - 17.39 D. Q_R = 17.39 only Answer: C. Load Q is 0.75P and the compensator supplies 17.39 MVAr, so the line delivers the difference. (Key point: Part 2 relation) 4. How is the extra load found? A. 18.52 MW B. 28 - 15 = 13 MW C. 17.39 - 15 = 2.39 MW D. New total P (18.52 MW) minus 15 MW = 3.52 MW Answer: D. Extra load is the increase over the original 15 MW. 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