--- title: "Nominal-π derivation, losses and UPF phasor diagram" subject: "Power System Analysis I" chapter: 5 key: "semester_4/power_system_analysis_i:nominal-pi-200km-derivation-losses" source: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/nominal-pi-200km-derivation-losses site: BCT Engineering --- # Nominal-π derivation, losses and UPF phasor diagram Power System Analysis I · Chapter 5: Transmission Line Modeling - Asked in: 2078 Chaitra, 4+1+2 marks - Video: [5. Phasor diagram of medium transmission line nominal π method](https://www.youtube.com/watch?v=wpxhHJraBaM) — Bookforum - 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/nominal-pi-200km-derivation-losses - Whole chapter: https://www.bctengineer.com/md/semester_4/power_system_analysis_i/chapter-5 ## Question For a 3-ph, 200 km long transmission line, derive an expression relating sending end voltage and current to receiving end voltage and current using the nominal-π model. Also write down the expressions for determining active and reactive power losses in the line. Construct a phasor diagram of currents and voltages for a unity power factor load at the receiving end. ## Short answer For a 200 km line (medium), nominal-π ABCD: A=D=1+ZY/2, B=Z, C=Y(1+ZY/4). Active loss P_loss=3|I_L|²R, reactive loss Q_loss=3|I_L|²X−3(ωC/2)(|V_S|²+|V_R|²). Phasor diagram for unity pf: V_R reference, I_R in phase; I_C2 leads V_R 90°; I_L=I_R+I_C2; V_S=V_R+I_LR+jI_LX; I_C1 leads V_S90°; I_S=I_L+I_C1. ## Answer A 200 km line is a **medium line**, so its shunt capacitance cannot be ignored. In the nominal-Π model the total series impedance $Z = R + jX$ is lumped in the middle and $Y/2$ is placed at each end (all quantities per phase). ```text IS IL Z = R + jX IR o-->--+-----[ R ]--[ jX ]-----+-->--o | | VS [Y/2] Ic1 Ic2 [Y/2] VR | | o-----+-----------------------+-----o ``` #### Relation between sending and receiving quantities Current through the receiving-end shunt: $I_{C2} = \frac{Y}{2}V_R$ Series current: $I_L = I_R + \frac{Y}{2}V_R$ $$ \begin{aligned} V_S &= V_R + ZI_L = \left(1 + \frac{ZY}{2}\right)V_R + ZI_R \\ I_S &= I_L + \frac{Y}{2}V_S = Y\left(1 + \frac{ZY}{4}\right)V_R + \left(1 + \frac{ZY}{2}\right)I_R \end{aligned} $$ In matrix form: $$ \begin{bmatrix} V_S \\ I_S \end{bmatrix} = \begin{bmatrix} 1 + \frac{ZY}{2} & Z \\ Y\left(1 + \frac{ZY}{4}\right) & 1 + \frac{ZY}{2} \end{bmatrix}\begin{bmatrix} V_R \\ I_R \end{bmatrix} $$ #### Active and reactive power losses (3-phase) With $Y/2 = j\omega C/2$ at each end: $$ \begin{aligned} P_{loss} &= 3|I_L|^2 R = P_S - P_R \\ Q_{loss} &= 3|I_L|^2X - 3\frac{\omega C}{2}\left(|V_S|^2 + |V_R|^2\right) = Q_S - Q_R \end{aligned} $$ The first term of $Q_{loss}$ is the reactive power absorbed by the series inductance. The second is the reactive power generated by the two shunt capacitances. A negative $Q_{loss}$ means the line supplies net reactive power, which is normal at light load. #### Phasor diagram for unity p.f. load Take $V_R$ as reference; $I_R$ is in phase with $V_R$. 1. $I_{C2}$ leads $V_R$ by 90°. 2. $I_L = I_R + I_{C2}$ leads $V_R$ slightly. 3. $V_S = V_R + I_LR + jI_LX$: $I_LR$ is parallel to $I_L$, $I_LX$ is perpendicular to $I_L$. 4. $I_{C1}$ leads $V_S$ by 90°; $I_S = I_L + I_{C1}$. ```text IS IC1 ^ ^ VS \ | IL / | \ | / / | IL.X IC2 ^ \|/ / | | / d /____| |/ / IL.R O------------------------> VR , IR ``` At unity p.f. the $I_LX$ drop is almost at right angles to $V_R$, so $|V_S|$ is only slightly greater than $|V_R|$ while the load angle $\delta$ is noticeable. $I_S$ leads $I_R$ because of the two charging currents. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is a nominal-Π model used for a 200 km line? A. It is a medium line, so shunt capacitance cannot be ignored B. It is a short line so capacitance is zero C. It is a long line needing distributed parameters D. Only Π models can handle 3-phase lines Answer: A. 200 km is in the 80-250 km medium range where lumped shunt capacitance is needed. (Key point: Why nominal-Π for 200 km) 2. The current through the series impedance in the nominal-Π model is: A. IL = IR - (Y/2)VR B. IL = IR C. IL = IS + (Y/2)VS D. IL = IR + (Y/2)VR Answer: D. The receiving-end shunt current (Y/2)VR adds to IR before flowing in the series arm. (Key point: Series current IL) 3. The active power loss in the Π line is: A. 3|IL|^2 R B. 3|IR|^2 X C. 3|VS|^2 / R D. 3 wC |VR|^2 Answer: A. Only R dissipates power, carrying IL, so Ploss = 3|IL|^2 R = PS - PR. (Key point: Active power loss) 4. Qloss = 3|IL|^2 X - 3(wC/2)(|VS|^2 + |VR|^2). A negative Qloss means: A. The line absorbs net reactive power B. The line has zero resistance C. The line supplies net reactive power, as at light load D. The load is purely resistive Answer: C. The shunt capacitances generate more reactive power than the series inductance absorbs. (Key point: Reactive loss and sign) 5. In the unity p.f. phasor diagram, IS compared with IR: A. IS leads IR because of the two charging currents B. IS lags IR C. IS is in phase with IR D. IS equals IR in magnitude Answer: A. IC1 and IC2 lead the voltages by 90°, so IS leads IR. 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