--- title: "Nominal-T derivation, phasor, losses, regulation" subject: "Power System Analysis I" chapter: 5 key: "semester_4/power_system_analysis_i:nominal-t-derivation-losses" source: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/nominal-t-derivation-losses site: BCT Engineering --- # Nominal-T derivation, phasor, losses, regulation Power System Analysis I · Chapter 5: Transmission Line Modeling - Asked in: 2073 Magh, 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 - Whole chapter: https://www.bctengineer.com/md/semester_4/power_system_analysis_i/chapter-5 ## Question Starting from a suitable point, derive expressions relating sending end voltage and current for a medium length transmission line using the nominal-T model. Also construct the phasor diagram and expressions to compute power loss in the line, voltage regulation and efficiency. ## Short answer Nominal-T: half series impedance Z/2 in each arm, shunt Y at centre. VS = (1+ZY/2)VR + Z(1+ZY/4)IR, IS = Y VR + (1+ZY/2)IR. Power loss Ploss = 3(|IR|²+|IS|²)R/2. Regulation uses no-load VR = |VS|/|A|. Efficiency = PR/(PR+Ploss). Phasor diagram builds VC then IS and VS. ## Answer In the **nominal-T model** of a medium line, the total shunt admittance $Y = j\omega C$ is placed at the middle of the line, and half the series impedance ($Z/2$) is placed in each arm. Values are 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 **Step 1:** Voltage across the shunt (mid-point): $$ V_C = V_R + \frac{Z}{2}I_R $$ **Step 2:** Shunt current and sending current: $$ \begin{aligned} I_C &= YV_C = YV_R + \frac{ZY}{2}I_R \\ I_S &= I_R + I_C = YV_R + \left(1 + \frac{ZY}{2}\right)I_R \end{aligned} $$ **Step 3:** Sending-end voltage: $$ \begin{aligned} V_S &= V_C + \frac{Z}{2}I_S = V_R + \frac{Z}{2}I_R + \frac{Z}{2}\left[YV_R + \left(1 + \frac{ZY}{2}\right)I_R\right] \\ V_S &= \left(1 + \frac{ZY}{2}\right)V_R + Z\left(1 + \frac{ZY}{4}\right)I_R \end{aligned} $$ So $A = D = 1 + \frac{ZY}{2}$, $B = Z\left(1 + \frac{ZY}{4}\right)$, $C = Y$. #### Phasor diagram (lagging load, $V_R$ reference) 1. $I_R$ lags $V_R$ by $\phi_R$. 2. $V_C = V_R + I_R\frac{R}{2} + jI_R\frac{X}{2}$. 3. $I_C$ leads $V_C$ by 90°; $I_S = I_R + I_C$. 4. $V_S = V_C + I_S\frac{R}{2} + jI_S\frac{X}{2}$. ```text IC ^ /| VS | VC / IS.X/2 | /| / | / |IR.X/2 | / /IR.R/2 O-----------------------> VR \ \ phiR \ v IS (IS = IR + IC) v IR ``` #### Power loss $$ \begin{aligned} P_{loss} &= 3\left(|I_R|^2 + |I_S|^2\right)\frac{R}{2} \\ Q_{loss} &= 3\left(|I_R|^2 + |I_S|^2\right)\frac{X}{2} - 3|V_C|^2\omega C \end{aligned} $$ #### Voltage regulation No-load receiving voltage $= |V_S|/|A|$: $$ \%\text{VR} = \frac{|V_S|/|A| - |V_R|}{|V_R|}\times100, \quad |A| = \left|1 + \frac{ZY}{2}\right| $$ #### Efficiency $$ \eta = \frac{P_R}{P_R + P_{loss}}\times100 = \frac{3|V_R||I_R|\cos\phi_R}{3|V_R||I_R|\cos\phi_R + 3\left(|I_R|^2 + |I_S|^2\right)\frac{R}{2}}\times100 $$ ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. In the nominal-T model: A. Y/2 is at each end B. Y is at the middle and Z/2 is in each arm C. All of Z is at the middle D. Y is at the receiving end Answer: B. Z/2 in each arm and the full shunt Y at the midpoint. (Key point: T-model structure) 2. ABCD of the nominal T are: A. A = D = 1 + ZY/2, B = Z, C = Y(1 + ZY/4) B. A = D = 1, B = Z, C = 0 C. A = D = 1 + ZY/2, B = Z(1 + ZY/4), C = Y D. A = 1 + ZY/4, B = Y, C = Z Answer: C. The T has B = Z(1 + ZY/4) and C = Y; the Π has the reverse. (Key point: T constants) 3. In the T phasor diagram, IC leads which voltage by 90°? A. VR B. VS C. IR D. VC, the voltage across the shunt Answer: D. IC = Y VC leads VC by 90°; then IS = IR + IC. (Key point: Phasor diagram step) 4. The active power loss of the nominal T is: A. 3(|IR|^2 + |IS|^2)(R/2) B. 3|IR|^2 R C. 3|IS|^2 R D. 3|VC|^2 / R Answer: A. Two arms of resistance R/2 carry IR and IS. (Key point: Power loss) 5. Percent regulation is: A. (|VS| - |VR|)/|VR| x 100 B. (|VS|/|A| - |VR|)/|VR| x 100 C. (|VR| - |VS|)/|VS| x 100 D. |A| / |VS| x 100 Answer: B. No-load receiving voltage is |VS|/|A|, with |A| = |1 + ZY/2|. 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