--- title: "Nominal-T 100 km line, 20 MW at 110 kV" subject: "Power System Analysis I" chapter: 5 key: "semester_4/power_system_analysis_i:nominal-t-100km-20mw" source: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/nominal-t-100km-20mw site: BCT Engineering --- # Nominal-T 100 km line, 20 MW at 110 kV Power System Analysis I · Chapter 5: Transmission Line Modeling - Asked in: 2071 Bhadra, 6 marks - Video: [Medium Transmission Line: T eq. ckt: 1st problem on Sending end voltage, current & power factor](https://www.youtube.com/watch?v=El7seSAY75M) — Gargi Basu - 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-100km-20mw - Whole chapter: https://www.bctengineer.com/md/semester_4/power_system_analysis_i/chapter-5 ## Question A 100 km long 3-phase 50 Hz transmission line is nominal-T modelled. The resistance and reactance are 0.2 Ω per phase per km and j0.4 per phase per km respectively. The shunt admittance is j2.5 × 10⁻⁶ S/phase per km. Calculate current and voltage at the sending end if the line delivers 20 MW at 110 kV and at 0.9 p.f. lagging. ## Short answer Total Z=20+j40 Ω, Y=j2.5×10⁻⁴ S. Receiving V_R=63.51 kV, I_R=116.6∠-25.84° A. Nominal-T: mid-point V_C=65.59∠1.39° kV, I_S=110.1∠-18.23° A, V_S=67.39∠2.84° kV/phase (116.73 kV line). ## Answer **Given:** $l = 100$ km, $z = 0.2 + j0.4\ \Omega$/km, $y = j2.5\times10^{-6}$ S/km; load 20 MW at 110 kV, 0.9 p.f. lagging; nominal-T model. #### Step 1: Total parameters $$ \begin{aligned} Z &= (0.2 + j0.4)\times100 = 20 + j40\ \Omega, \quad \frac{Z}{2} = 10 + j20\ \Omega \\ Y &= j2.5\times10^{-6}\times100 = j2.5\times10^{-4}\ \text{S} \end{aligned} $$ #### Step 2: Receiving-end values $$ \begin{aligned} V_R &= \frac{110}{\sqrt3} = 63.51\angle 0^\circ\ \text{kV} \\ I_R &= \frac{20\times10^6}{\sqrt3\times110\times10^3\times0.9} = 116.6\angle -25.84^\circ = 104.97 - j50.84\ \text{A} \end{aligned} $$ ```text IS Z/2 Z/2 IR o-->--[R/2 jX/2]--+--[R/2 jX/2]-->--o | VS [ Y ] Ic (VC) VR | o-----------------+-----------------o ``` #### Step 3: Mid-point voltage $$ \begin{aligned} V_C &= V_R + I_R\frac{Z}{2} = 63{,}509 + (104.97 - j50.84)(10 + j20) \\ &= 65{,}575 + j1591 = 65.59\angle 1.39^\circ\ \text{kV} \end{aligned} $$ #### Step 4: Shunt current and sending-end current $$ \begin{aligned} I_C &= YV_C = j2.5\times10^{-4}(65{,}575 + j1591) = -0.40 + j16.39\ \text{A} \\ I_S &= I_R + I_C = 104.58 - j34.45 = 110.1\angle -18.23^\circ\ \text{A} \end{aligned} $$ #### Step 5: Sending-end voltage $$ \begin{aligned} V_S &= V_C + I_S\frac{Z}{2} = 65{,}575 + j1591 + (104.58 - j34.45)(10 + j20) \\ &= 67{,}310 + j3338 = 67.39\angle 2.84^\circ\ \text{kV/phase} \\ V_{S(L)} &= \sqrt3\times67.39 = 116.73\ \text{kV} \end{aligned} $$ Check with ABCD: $A = 1 + ZY/2 = 0.995\angle 0.14^\circ$, $B = Z(1 + ZY/4) = 44.61\angle 63.51^\circ\ \Omega$, $C = Y$; $AV_R + BI_R$ gives the same $V_S$. Sending-end p.f. $= \cos(2.84^\circ + 18.23^\circ) = 0.9331$ lagging; $P_S = 20.77$ MW. **Answer:** $I_S = 110.1\angle -18.23^\circ$ A, $V_S = 116.73$ kV (line) $= 67.39\angle 2.84^\circ$ kV per phase. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For the 100 km line with z = 0.2 + j0.4 Ω/km and y = j2.5e-6 S/km, what are Z/2 and Y? A. Z/2 = 10 + j20 Ω, Y = j2.5e-4 S B. Z/2 = 20 + j40 Ω, Y = j2.5e-4 S C. Z/2 = 10 + j20 Ω, Y = j1.25e-4 S D. Z/2 = 0.1 + j0.2 Ω, Y = j2.5e-6 S Answer: A. Z = 100z = 20 + j40 Ω so Z/2 = 10 + j20 Ω; Y = 100y = j2.5e-4 S (the whole Y at the middle in the T model). (Key point: Total line parameters) 2. What is IR for 20 MW at 110 kV and 0.9 p.f. lagging? A. About 105 A at 0° B. About 116.6 A at -25.84° C. About 202 A at -25.84° D. About 116.6 A at +25.84° Answer: B. IR = 20e6/(sqrt3 x 110e3 x 0.9) = 116.6 A, lagging by cos^-1(0.9) = 25.84°. (Key point: Receiving-end current) 3. What is the correct sequence of calculation for the nominal-T circuit? A. VS = VR + IR Z, then IC, then IS B. VC = VR + IR Z/2, IC = Y VC, IS = IR + IC, VS = VC + IS Z/2 C. IC = Y VR, IS = IR + IC, VS = VR + IS Z D. VC = VR - IR Z/2, then VS = VC - IS Z/2 Answer: B. First find the mid-point voltage VC, then the shunt current, then IS, and finally VS = VC + IS(Z/2). (Key point: Order of calculation in the T model) 4. With VS = 67.39 kV per phase, what is the sending-end line voltage? A. 67.39 kV B. About 116.7 kV C. About 39 kV D. About 202 kV Answer: B. VS(L) = sqrt3 x 67.39 = 116.73 kV. 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