--- title: "Nominal-π phasor, sending end, regulation, efficiency" subject: "Power System Analysis I" chapter: 5 key: "semester_4/power_system_analysis_i:nominal-pi-phasor-regulation-efficiency" source: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/nominal-pi-phasor-regulation-efficiency site: BCT Engineering --- # Nominal-π phasor, sending end, regulation, efficiency Power System Analysis I · Chapter 5: Transmission Line Modeling - Asked in: 2079 Chaitra, 8 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-phasor-regulation-efficiency - Whole chapter: https://www.bctengineer.com/md/semester_4/power_system_analysis_i/chapter-5 ## Question Draw the phasor diagram for a nominal π circuit of a medium transmission line. Derive expressions for sending end voltage, current, voltage regulation and efficiency. ## Short answer In nominal-π model, half shunt admittance Y/2 at each end of series impedance Z. Sending voltage: V_S=(1+ZY/2)V_R+Z I_R; current: I_S=Y(1+ZY/4)V_R+(1+ZY/2)I_R. Voltage regulation: %VR=(|V_S|/|1+ZY/2|−|V_R|)/|V_R|×100. Efficiency: η=P_R/(P_R+3|I_L|²R) where I_L=I_R+Y V_R/2. ## Answer In the **nominal-Π model** of a medium line, the total series impedance $Z = R + jX$ is in the middle and half the shunt admittance, $Y/2 = j\omega C/2$, is connected at each end. ```text IS IL Z = R + jX IR o-->--+-----[ R ]--[ jX ]-----+-->--o | | VS [Y/2] Ic1 Ic2 [Y/2] VR | | o-----+-----------------------+-----o ``` #### Phasor diagram (lagging load, $V_R$ as reference) Construction: 1. Draw $V_R$ along the reference axis and $I_R$ lagging by $\phi_R$. 2. $I_{C2} = j\frac{\omega C}{2}V_R$ leads $V_R$ by 90°. 3. $I_L = I_R + I_{C2}$. 4. From the tip of $V_R$ add $I_L R$ (parallel to $I_L$) and $I_L X$ (90° ahead of $I_L$) to get $V_S$. 5. $I_{C1} = j\frac{\omega C}{2}V_S$ leads $V_S$ by 90°; $I_S = I_L + I_{C1}$. ```text IC2 ^ IL.X /VS | | / | |/ IL.R O-----------------------> VR \ \ phiR \ \ IL (IS = IL + IC1, v v IC1 is 90 deg ahead of VS) IR ``` #### Sending-end voltage $$ \begin{aligned} I_L &= I_R + \frac{Y}{2}V_R \\ V_S &= V_R + Z I_L = V_R + Z\left(I_R + \frac{Y}{2}V_R\right) \\ V_S &= \left(1 + \frac{ZY}{2}\right)V_R + Z I_R \end{aligned} $$ #### Sending-end current $$ \begin{aligned} I_S &= I_L + \frac{Y}{2}V_S \\ &= I_R + \frac{Y}{2}V_R + \frac{Y}{2}\left[\left(1 + \frac{ZY}{2}\right)V_R + ZI_R\right] \\ I_S &= Y\left(1 + \frac{ZY}{4}\right)V_R + \left(1 + \frac{ZY}{2}\right)I_R \end{aligned} $$ So $A = D = 1 + ZY/2$, $B = Z$, $C = Y(1 + ZY/4)$. #### Voltage regulation At no load ($I_R = 0$), $V_{R(NL)} = V_S/A$. Hence $$ \%\,\text{VR} = \frac{|V_S|/|A| - |V_R|}{|V_R|}\times 100 = \frac{\left|V_S\right| - \left|1 + \frac{ZY}{2}\right|\left|V_R\right|}{\left|1 + \frac{ZY}{2}\right|\left|V_R\right|}\times100 $$ Note: $|V_S|/|A|$ is used, not $|V_S|$, because the shunt capacitance raises the no-load receiving voltage (Ferranti effect). #### Efficiency Only the series resistance carries loss, and it carries current $I_L$: $$ \begin{aligned} P_{loss} &= 3|I_L|^2 R \\ \eta &= \frac{P_R}{P_R + 3|I_L|^2R}\times100 = \frac{3|V_R||I_R|\cos\phi_R}{3|V_S||I_S|\cos\phi_S}\times100 \end{aligned} $$ ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. In the nominal-Π model, where is the shunt admittance placed? A. All of Y at the middle B. All of Y at the receiving end C. Y/2 at each end, with Z in the middle D. Y/2 in each arm with Z at the sending end Answer: C. Nominal-Π lumps Z in the middle and puts Y/2 at each end. (Key point: Nominal-Π structure) 2. Which is the sending-end voltage in the nominal-Π model? A. VS = (1 + ZY/2) VR + Z IR B. VS = VR + Z IR only C. VS = (1 + ZY) VR + Z IR/2 D. VS = VR + Y IR Answer: A. IL = IR + (Y/2)VR and VS = VR + Z IL give VS = (1 + ZY/2)VR + Z IR. (Key point: Sending-end voltage) 3. In the regulation formula, the no-load receiving voltage is |VS|/|A| because: A. A equals the load power factor B. At no load IR = 0 so VS = A VR, and shunt capacitance raises VR (Ferranti effect) C. A is always 1 D. Regulation is defined with VS as the base Answer: B. With IR = 0, VS = A VR, so VR(NL) = VS/A; capacitance causes a rise at no load. (Key point: No-load voltage in regulation) 4. In the nominal-Π efficiency, copper loss is 3|IL|^2 R. IL is: A. The receiving-end load current IR B. The sending-end current IS C. The current through the series impedance, IR + (Y/2)VR D. The current in the sending-end capacitor Answer: C. Only the series resistance has loss and it carries IL = IR + IC2. (Key point: Loss current in Π) 5. The constant C of the nominal-Π model is: A. Z(1 + ZY/4) B. Y C. 1 + ZY/2 D. Y(1 + ZY/4) Answer: D. IS = Y(1 + ZY/4)VR + (1 + ZY/2)IR. 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