--- title: "Nominal-T relations, losses and Ferranti condition" subject: "Power System Analysis I" chapter: 5 key: "semester_4/power_system_analysis_i:nominal-t-derivation-losses-ferranti" source: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/nominal-t-derivation-losses-ferranti site: BCT Engineering --- # Nominal-T relations, losses and Ferranti condition Power System Analysis I · Chapter 5: Transmission Line Modeling - Asked in: 2068 Bhadra, 8 marks - Video: [ABCD parameters of medium transmission line by Nominal T method explained by Irfan Mujawar](https://www.youtube.com/watch?v=HzHGlzKCBPI) — Dr. Irfan Mujawar - 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-t-derivation-losses-ferranti - Whole chapter: https://www.bctengineer.com/md/semester_4/power_system_analysis_i/chapter-5 ## Question Derive the expressions relating sending end current and voltage to the receiving end voltage and current in a medium length transmission line by nominal-T model. Express the active and reactive losses in the line and the condition for voltage at the receiving end being higher than at the sending end. ## Short answer In the nominal-T model, A = D = 1 + ZY/2, B = Z(1 + ZY/4), C = Y. Active loss = 3(|I_S|² + |I_R|²)R/2, reactive loss includes line inductive and capacitive parts. Ferranti effect occurs at light load when |1 + ZY/2| < 1, making receiving voltage exceed sending voltage due to capacitance. ## Answer In the **nominal-T model**, the total shunt admittance $Y = j\omega C$ is at the middle of the line and $Z/2 = (R + jX)/2$ is in each arm (per phase). ```text IS Z/2 Z/2 IR o-->--[R/2 jX/2]--+--[R/2 jX/2]-->--o | VS [ Y ] Ic (VC) VR | o-----------------+-----------------o ``` #### Sending-end voltage and current $$ \begin{aligned} V_C &= V_R + \frac{Z}{2}I_R \\ I_S &= I_R + YV_C = YV_R + \left(1 + \frac{ZY}{2}\right)I_R \\ V_S &= V_C + \frac{Z}{2}I_S = \left(1 + \frac{ZY}{2}\right)V_R + Z\left(1 + \frac{ZY}{4}\right)I_R \end{aligned} $$ So $A = D = 1 + \frac{ZY}{2}$, $B = Z\left(1 + \frac{ZY}{4}\right)$, $C = Y$. #### Active and reactive losses (3-phase) Current $I_S$ flows in the sending half and $I_R$ in the receiving half: $$ \begin{aligned} P_{loss} &= 3\left(|I_S|^2 + |I_R|^2\right)\frac{R}{2} \\ Q_{loss} &= 3\left(|I_S|^2 + |I_R|^2\right)\frac{X}{2} - 3\omega C|V_C|^2 \end{aligned} $$ The last term is the reactive power generated by the line capacitance. Equivalently, $P_{loss} = P_S - P_R$ and $Q_{loss} = Q_S - Q_R$. #### Condition for receiving voltage higher than sending voltage At no load or light load ($I_R \approx 0$): $$ V_S = \left(1 + \frac{ZY}{2}\right)V_R \;\Rightarrow\; V_R = \frac{V_S}{1 + \frac{ZY}{2}} $$ With $Z = R + jX$ and $Y = j\omega C$: $$ 1 + \frac{ZY}{2} = \left(1 - \frac{\omega CX}{2}\right) + j\frac{\omega CR}{2} $$ Neglecting the small $R$ term, $|V_R| > |V_S|$ when $$ \left|1 + \frac{ZY}{2}\right| < 1 \quad\Leftrightarrow\quad 0 < \frac{\omega CX}{2} < 2 $$ which is always true for a practical line (and $R$ small). So at light load the receiving voltage rises above the sending voltage; this is the **Ferranti effect**. The rise is $$ |V_R| - |V_S| \approx \frac{\omega CX}{2}|V_S| = \frac{\omega^2LC\,l^2}{2}|V_S| $$ so it grows with the square of line length. With load, $V_R > V_S$ happens when the capacitive charging current outweighs the drop of the load current, i.e. when the load is below about SIL or is leading. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which is correct for the nominal-T model? A. VC = VR + (Z/2)IR; IS = IR + Y VC; VS = VC + (Z/2)IS B. VC = VR + Z IR; IS = IR + Y VR C. VC = VS - Z IS D. IS = IR Answer: A. The shunt branch is at the mid-point, so VC includes the receiving half of the impedance, and VS adds the sending half. (Key point: Nominal-T relations) 2. Why is the active loss 3(|IS|^2 + |IR|^2)R/2? A. IS flows in the sending arm and IR in the receiving arm, each with resistance R/2 B. The shunt branch dissipates power C. Only IR flows in the line D. R is split in three Answer: A. Each half of the line has resistance R/2 and carries its own current (IS or IR). (Key point: Active loss) 3. In the reactive loss expression, what does the term 3ωC|VC|^2 represent? A. Reactive power generated by the line capacitance, subtracted from that absorbed by the inductance B. Reactive power absorbed by the load C. Active power loss D. Reactive power absorbed by the resistance Answer: A. The series reactance absorbs Q while the shunt capacitance generates it, so net Q loss subtracts the capacitive term. (Key point: Reactive loss) 4. At no load, VR > VS when: A. |1 + ZY/2| < 1, i.e. the capacitive term outweighs the voltage drop (0 < ωCX/2 < 2) B. |1 + ZY/2| > 1 C. ZY = 0 D. R is very large Answer: A. At IR = 0, VR = VS/(1 + ZY/2); for X dominant, 1 + ZY/2 ≈ 1 - ωCX/2 < 1, so VR exceeds VS (Ferranti effect), and the rise grows with the square of length. 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