--- title: "Receiving and sending end powers from ABCD, 132/134 kV" subject: "Power System Analysis I" chapter: 6 key: "semester_4/power_system_analysis_i:abcd-power-132kv-134kv" source: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/abcd-power-132kv-134kv site: BCT Engineering --- # Receiving and sending end powers from ABCD, 132/134 kV Power System Analysis I · Chapter 6: Performance Analysis - Asked in: 2080 Chaitra, 4+6 marks - Video: [Power system II _Unit01_Steps to draw receiving end circle diagram of transmission line](https://www.youtube.com/watch?v=qt0qW_Iahlg) — Elite Academy - 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/abcd-power-132kv-134kv - Whole chapter: https://www.bctengineer.com/md/semester_4/power_system_analysis_i/chapter-6 ## Question A 50 Hz, three-phase overhead line has the following parameters: A = 0.98∠0.8°; B = 45∠74° Ω; VSL = 132∠10° kV, VRL = 134∠0° kV. Determine the receiving end active and reactive power, power factor and line losses. Also find out the sending end active and reactive power, power factor and voltage regulation of the line. ## Short answer From ABCD parameters and voltages, using power-circle equations: receiving power 59.29 MW, −21.07 MVAr, pf 0.942 leading; sending power 68.59 MW, −27.65 MVAr, pf 0.927 leading; loss 9.30 MW; regulation 0.52 % (no-load voltage 134.69 kV). ## Answer Use the power-circle equations (taking $D = A$). Let $A = |A|\angle\alpha$, $B = |B|\angle\beta$, $V_S$ leading $V_R$ by $\delta$. With line voltages in kV, powers come out directly in three-phase MW/MVAr. Data: $|A| = 0.98$, $\alpha = 0.8^\circ$, $|B| = 45\ \Omega$, $\beta = 74^\circ$, $|V_S| = 132$ kV, $|V_R| = 134$ kV, $\delta = 10^\circ$. Useful terms: $$ \begin{aligned} \frac{|V_S||V_R|}{|B|} &= \frac{132\times134}{45} = 393.07 \\ \frac{|A||V_R|^2}{|B|} &= \frac{0.98\times134^2}{45} = 391.04 \\ \frac{|A||V_S|^2}{|B|} &= \frac{0.98\times132^2}{45} = 379.46 \end{aligned} $$ Angles: $\beta-\delta = 64^\circ$, $\beta-\alpha = 73.2^\circ$, $\beta+\delta = 84^\circ$. #### Receiving end $$ \begin{aligned} P_R &= 393.07\cos64^\circ - 391.04\cos73.2^\circ = 59.29\ \text{MW} \\ Q_R &= 393.07\sin64^\circ - 391.04\sin73.2^\circ = -21.07\ \text{MVAr} \\ \text{pf}_R &= \frac{59.29}{\sqrt{59.29^2 + 21.07^2}} = 0.942 \end{aligned} $$ $Q_R$ is negative, so reactive power flows from the receiving end into the line: the receiving-end pf is **0.942 leading**. This is expected because $|V_R| > |V_S|$. #### Sending end $$ \begin{aligned} P_S &= 379.46\cos73.2^\circ - 393.07\cos84^\circ = 68.59\ \text{MW} \\ Q_S &= 379.46\sin73.2^\circ - 393.07\sin84^\circ = -27.65\ \text{MVAr} \\ \text{pf}_S &= \frac{68.59}{\sqrt{68.59^2 + 27.65^2}} = 0.927\ \text{(leading)} \end{aligned} $$ #### Line losses $$ \begin{aligned} P_{loss} &= P_S - P_R = 68.59 - 59.29 = 9.30\ \text{MW} \\ Q_S - Q_R &= -27.65 - (-21.07) = -6.59\ \text{MVAr} \end{aligned} $$ The negative reactive difference means the line, on net, generates about 6.59 MVAr (its charging exceeds its $I^2X$ absorption). #### Voltage regulation No-load receiving voltage $= |V_S|/|A| = 132/0.98 = 134.69$ kV. $$ \%VR = \frac{134.69 - 134}{134}\times100 = 0.52\% $$ **Answer: $P_R$ = 59.29 MW, $Q_R$ = −21.07 MVAr, pf 0.942 lead; $P_S$ = 68.59 MW, $Q_S$ = −27.65 MVAr, pf 0.927 lead; loss 9.30 MW; regulation 0.52 %.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which equation set gives receiving power for given V_S, V_R, delta? A. P_R = |VsVr/B| cos(beta - delta) - |A Vr^2/B| cos(beta - alpha), and the sine form for Q_R B. P_R = |VsVr/B| cos(beta + delta) - ... C. P_R = |Vs||Vr| sin(delta)/X only D. P_R = 3 V I cos(phi) with I from load Answer: A. These are the power-circle equations with line voltages giving three-phase MW. (Key point: Equations used) 2. Q_R = -21.07 MVAr. What does it mean? A. Load is lagging at 0.942 B. Reactive power flows from the receiving end into the line; pf is leading (0.942) C. Line absorbs 21 MVAr D. Calculation is wrong Answer: B. Negative Q_R means leading load, consistent with |V_R| > |V_S|. (Key point: Negative Q_R) 3. How is regulation found here? A. (132 - 134)/132 B. |V_S| x |A| C. (|V_S|/|A| - |V_R|)/|V_R| x 100 = 0.52 % D. (P_S - P_R)/P_S Answer: C. No-load voltage is 132/0.98 = 134.69 kV. (Key point: Regulation) 4. How are active losses calculated? A. P_R - P_S B. Q_S - Q_R C. |V_S| - |V_R| D. P_S - P_R = 68.59 - 59.29 = 9.30 MW Answer: D. Loss = sending-end minus receiving-end active power. 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