--- title: "Sending and receiving power, 140∠10° and 132 kV" subject: "Power System Analysis I" chapter: 6 key: "semester_4/power_system_analysis_i:power-140kv-132kv-a099" source: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/power-140kv-132kv-a099 site: BCT Engineering --- # Sending and receiving power, 140∠10° and 132 kV 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/power-140kv-132kv-a099 - Whole chapter: https://www.bctengineer.com/md/semester_4/power_system_analysis_i/chapter-6 ## Question Compute the sending end and receiving end power of a 3-phase line if A = 0.99∠0.1° and B = 18∠68° Ohm, if VS,L = 140∠10° kV and VR,L = 132∠0° kV. ## Short answer Using ABCD parameters, receiving-end power is 183.51 MW and −17.25 MVAr; sending-end power is 192.11 MW and −5.43 MVAr. ## Answer Use the power-circle equations (taking $D = A$). Line kV give three-phase MW/MVAr. Data: $|V_S| = 140$ kV, $|V_R| = 132$ kV, $\delta = 10^\circ$, $|A| = 0.99$, $\alpha = 0.1^\circ$, $|B| = 18\ \Omega$, $\beta = 68^\circ$. #### Constants $$ \begin{aligned} \frac{|V_S||V_R|}{|B|} &= \frac{140\times132}{18} = 1026.67 \\ \frac{|A||V_R|^2}{|B|} &= \frac{0.99\times132^2}{18} = 958.32 \\ \frac{|A||V_S|^2}{|B|} &= \frac{0.99\times140^2}{18} = 1078.0 \end{aligned} $$ Angles: $\beta-\delta = 58^\circ$, $\beta-\alpha = 67.9^\circ$, $\beta+\delta = 78^\circ$. #### Receiving-end power $$ \begin{aligned} P_R &= 1026.67\cos58^\circ - 958.32\cos67.9^\circ = 544.05 - 360.54 = 183.51\ \text{MW} \\ Q_R &= 1026.67\sin58^\circ - 958.32\sin67.9^\circ = 870.66 - 887.91 = -17.25\ \text{MVAr} \end{aligned} $$ #### Sending-end power $$ \begin{aligned} P_S &= 1078.0\cos67.9^\circ - 1026.67\cos78^\circ = 405.57 - 213.46 = 192.11\ \text{MW} \\ Q_S &= 1078.0\sin67.9^\circ - 1026.67\sin78^\circ = 998.80 - 1004.23 = -5.43\ \text{MVAr} \end{aligned} $$ Line loss $= 192.11 - 183.51 = 8.60$ MW. **Answer: Receiving end: 183.51 MW − j17.25 MVAr; sending end: 192.11 MW − j5.43 MVAr.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which expression gives P_R? A. |Vs||Vr|cos(beta-delta)/|B| - |A||Vr|^2 cos(beta-alpha)/|B| B. |Vs||Vr|cos(beta+delta)/|B| - |A||Vr|^2 cos(beta-alpha)/|B| C. |A||Vs|^2 cos(beta-alpha)/|B| - |Vs||Vr|cos(beta+delta)/|B| D. |Vs||Vr|sin(delta)/|B| Answer: A. P_R = 1026.67 cos58 - 958.32 cos67.9 = 183.51 MW. (Key point: Receiving-end active power equation) 2. Which expression gives P_S, and which angle appears with the cross term? A. |A||Vr|^2 cos(beta-alpha)/|B| - |Vs||Vr| cos(beta-delta)/|B| B. |A||Vs|^2 cos(beta-alpha)/|B| - |Vs||Vr| cos(beta+delta)/|B| with beta+delta = 78 degrees C. |Vs||Vr|cos(beta+delta)/|B| only D. |A||Vs|^2 cos(beta+alpha)/|B| Answer: B. The sending-end uses the |A||Vs|^2 term and cos(beta+delta); P_S = 405.57 - 213.46 = 192.11 MW. (Key point: Sending-end power equation) 3. Q_R = -17.25 MVAr and Q_S = -5.43 MVAr. What does a negative Q mean here? A. The end is absorbing reactive power as a lagging load B. The end delivers leading VArs (leading power factor) C. Reactive power is lost as heat D. The calculation is invalid Answer: B. With S = P + jQ as delivered power, negative Q corresponds to a leading pf at that end. (Key point: Sign of reactive powers) 4. How is the line active power loss obtained? A. P_R - P_S B. P_S - P_R = 192.11 - 183.51 = 8.60 MW C. P_S + P_R D. Q_S - Q_R Answer: B. Loss is the difference between sending-end and receiving-end active power, 8.60 MW. 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