--- title: "33 kV, 15 MW line: efficiency and regulation" subject: "Power System Analysis I" chapter: 6 key: "semester_4/power_system_analysis_i:33kv-15mw-efficiency-regulation" source: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/33kv-15mw-efficiency-regulation site: BCT Engineering --- # 33 kV, 15 MW line: efficiency and regulation Power System Analysis I · Chapter 6: Performance Analysis - Asked in: 2072 Asoj, 6 marks - Video: [Transmission line: efficiency and regulation](https://www.youtube.com/watch?v=CmoRX8gfGDs) — SHAHRIN BIN MD AYOB FKE - 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/33kv-15mw-efficiency-regulation - Whole chapter: https://www.bctengineer.com/md/semester_4/power_system_analysis_i/chapter-6 ## Question A 3-phase, 50 Hz, 33 kV transmission line is delivering a load of 15 MW at 0.9 pf lagging. The line series impedance per phase is 5 + j18 Ω. Determine the power transmission efficiency and voltage regulation of the line. Construct the phasor diagram of currents and voltages in the line. ## Short answer The transmission efficiency is 92.16% and voltage regulation is 20.82% (sending-end line voltage 39.87 kV). The phasor diagram shows VR as reference, I lagging by 25.84°, and VS leading VR by 10.23° with IR and IX drops added. ## Answer Short line: $Z = 5 + j18\ \Omega$/phase, shunt capacitance neglected. #### Current $$ \begin{aligned} V_R &= 33\,000/\sqrt3 = 19\,052.6\angle0^\circ\ \text{V} \\ I &= \frac{15\times10^6}{\sqrt3\times33\,000\times0.9} = 291.59\ \text{A},\quad \phi_R = 25.84^\circ\ \text{lag} \end{aligned} $$ #### Efficiency $$ \begin{aligned} P_{loss} &= 3I^2R = 3(291.59)^2(5) = 1.275\ \text{MW} \\ \eta &= \frac{15}{15 + 1.275}\times100 = 92.16\% \end{aligned} $$ #### Voltage regulation $$ \begin{aligned} IR &= 1458.0\angle-25.84^\circ\ \text{V},\quad IX = 5248.6\angle64.16^\circ\ \text{V} \\ V_S &= V_R + I(R + jX) = 19\,052.6 + 5447.4\angle48.63^\circ \\ &= 23\,018.5\angle10.23^\circ\ \text{V} \end{aligned} $$ Line value $= \sqrt3\times23.019 = 39.87$ kV. $$ \%VR = \frac{23\,018.5 - 19\,052.6}{19\,052.6}\times100 = 20.82\% $$ #### Phasor diagram (lagging load) ```text Vs (23.02 kV, 10.23 deg) /| / | I*X (5.25 kV, leads I by 90) / | / | O-------------------+--> I*R (1.46 kV, along I) \ Vr (19.05 kV, reference) \ \ phi_R = 25.84 deg v I (291.6 A) ``` - $V_R$ is the reference. - $I$ lags $V_R$ by $\phi_R = 25.84^\circ$. - $IR$ is drawn from the tip of $V_R$ parallel to $I$; $IX$ is drawn from its tip, perpendicular to $I$ (leading). - $V_S$ joins $O$ to the end of $IX$ and leads $V_R$ by $\delta = 10.23^\circ$; $I$ lags $V_S$ by $36.07^\circ$ (sending pf 0.808 lag). **Answer: Efficiency ≈ 92.16 %, regulation ≈ 20.82 % ($V_S$ = 39.87 kV).** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the load current found? A. I = P/(3 V_L) B. I = P/V_L C. I = P cos(phi)/V_L D. I = P/(sqrt3 V_L cos phi) = 291.59 A Answer: D. 15e6/(sqrt3 x 33000 x 0.9) = 291.59 A. (Key point: Current) 2. What gives the efficiency? A. 15/33 B. 3I^2R = 1.275 MW and eta = 15/(15 + 1.275) = 92.16 % C. 3I^2X D. I^2R single phase Answer: B. Only R dissipates: 3 x 291.59^2 x 5 = 1.275 MW. (Key point: Efficiency) 3. In the phasor diagram for the lagging load, how are drops drawn? A. Both along V_R B. IX lags I C. IR along I, IX perpendicular leading I by 90 degrees, added to V_R D. IR perpendicular, IX along I Answer: C. V_S = V_R + IR (parallel to I) + jIX (leading I by 90 degrees). (Key point: Phasor diagram) 4. What is the regulation? A. (39.87 - 33)/39.87 B. (33 - 39.87)/33 C. 6.9 % D. (23018.5 - 19052.6)/19052.6 = 20.82 % Answer: D. Phase values: V_S = 23.02 kV, V_R = 19.05 kV. 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