--- title: "280 km, 60 Hz line: nominal-π vs long line" subject: "Power System Analysis I" chapter: 5 key: "semester_4/power_system_analysis_i:280km-60hz-nominal-pi-long" source: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/280km-60hz-nominal-pi-long site: BCT Engineering --- # 280 km, 60 Hz line: nominal-π vs long line Power System Analysis I · Chapter 5: Transmission Line Modeling - Asked in: 2081 Chaitra (new course), 4+4 marks - Video: [PS25 Numerical on Medium Transmission Lines](https://www.youtube.com/watch?v=Dt4Y5BlylMo) — Lectures in Electrical Engineering - 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/280km-60hz-nominal-pi-long - Whole chapter: https://www.bctengineer.com/md/semester_4/power_system_analysis_i/chapter-5 ## Question A 280 km long transmission line on a 60 Hz system has line impedance of (33 + j104) Ω, a total shunt admittance of 10⁻³ mho. The receiving end feeds a load of 80 MW at 220 kV at 0.8 p.f. lagging. Find the sending-end voltage, current, power and power factor using (i) nominal pi method (ii) long line, and comment on the results. ## Short answer Per-phase: V_R=127.02 kV, I_R=262.4 A at 0.8 lag. Nominal-π: V_S=251.03 kV, I_S=201.8 A, p.f. 0.9715 lag, P_S=85.24 MW. Exact long-line: V_S=250.23 kV, I_S=202.1 A, p.f. 0.9729 lag, P_S=85.21 MW. The difference is <0.3%, confirming nominal-π is acceptable for 280 km. ## Answer **Given:** $Z = 33 + j104\ \Omega$, $Y = j10^{-3}$ S (totals, so frequency is not needed), load 80 MW at 220 kV, 0.8 p.f. lagging. Per-phase values are used. $$ \begin{aligned} V_R &= \frac{220}{\sqrt3} = 127.02\angle 0^\circ\ \text{kV} \\ I_R &= \frac{80\times10^6}{\sqrt3\times220\times10^3\times0.8} = 262.4\angle -36.87^\circ = 209.95 - j157.46\ \text{A} \end{aligned} $$ #### (i) Nominal-Π method $$ \begin{aligned} ZY &= (33 + j104)(j10^{-3}) = -0.104 + j0.033 \\ A = D &= 1 + \frac{ZY}{2} = 0.948 + j0.0165 = 0.9481\angle 0.997^\circ \\ B &= Z = 109.1\angle 72.40^\circ\ \Omega \\ C &= Y\left(1 + \frac{ZY}{4}\right) = 9.74\times10^{-4}\angle 90.49^\circ\ \text{S} \end{aligned} $$ $$ \begin{aligned} V_S &= AV_R + BI_R = (120{,}410 + j2096) + (23{,}304 + j16{,}638) \\ &= 143{,}720 + j18{,}734 = 144.93\angle 7.43^\circ\ \text{kV/phase} \\ V_{S(L)} &= \sqrt3 \times 144.93 = 251.03\ \text{kV} \\ I_S &= CV_R + DI_R = 200.58 - j22.09 = 201.8\angle -6.29^\circ\ \text{A} \end{aligned} $$ p.f. $= \cos(7.43^\circ + 6.29^\circ) = \cos 13.71^\circ = 0.9715$ lagging $P_S = 3V_SI_S\cos\phi_S = 85.24$ MW (and $Q_S = 20.80$ Mvar) #### (ii) Long-line (exact) method $$ \begin{aligned} \gamma l &= \sqrt{ZY} = 0.3303\angle 81.20^\circ = 0.05055 + j0.3264 \\ Z_c &= \sqrt{Z/Y} = 330.3\angle -8.80^\circ\ \Omega \\ A = D &= \cosh\gamma l = 0.9484 + j0.0162 = 0.9485\angle 0.98^\circ \\ B &= Z_c\sinh\gamma l = 107.2\angle 72.71^\circ\ \Omega \\ C &= \frac{\sinh\gamma l}{Z_c} = 9.828\times10^{-4}\angle 90.32^\circ\ \text{S} \end{aligned} $$ $$ \begin{aligned} V_S &= AV_R + BI_R = 143{,}270 + j18{,}538 = 144.46\angle 7.37^\circ\ \text{kV/phase} \\ V_{S(L)} &= 250.23\ \text{kV} \\ I_S &= CV_R + DI_R = 200.98 - j21.11 = 202.1\angle -6.00^\circ\ \text{A} \end{aligned} $$ p.f. $= \cos(7.37^\circ + 6.00^\circ) = 0.9729$ lagging, $P_S = 85.21$ MW ($Q_S = 20.25$ Mvar) #### Results and comment | Quantity | Nominal-Π | Long line | |---|---|---| | $V_S$ (line) | 251.03 kV | 250.23 kV | | $I_S$ | 201.8 A | 202.1 A | | Sending p.f. | 0.9715 lag | 0.9729 lag | | $P_S$ | 85.24 MW | 85.21 MW | | Efficiency | 93.86 % | 93.89 % | - The two methods differ by only about 0.3 % in voltage, so for 280 km the nominal-Π model is reasonably accurate. - The long-line method is exact. The error of the Π model grows with length, so the exact model should be used for lines well above 250 km. - In both, $I_S < I_R$ because the line's charging current partly cancels the lagging load current. **Answer:** Nominal-Π: $V_S = 251.03$ kV, $I_S = 201.8$ A, p.f. 0.9715 lag, $P_S = 85.24$ MW. Long line: $V_S = 250.23$ kV, $I_S = 202.1$ A, p.f. 0.9729 lag, $P_S = 85.21$ MW. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For a 220 kV (line) receiving end, which per-phase voltage is used as reference? A. 220 kV B. 220 x sqrt(3) = 381 kV C. 220/3 = 73.3 kV D. 220/sqrt(3) = 127.02 kV Answer: D. Per-phase analysis uses VR = VL/sqrt(3) = 127.02 kV. (Key point: Per-phase receiving voltage) 2. In the nominal-Π method, which expression gives the constant A? A. A = 1 + ZY/2 B. A = 1 + ZY C. A = Y(1 + ZY/4) D. A = Z Answer: A. A = D = 1 + ZY/2, B = Z and C = Y(1 + ZY/4). (Key point: Nominal-Π constants) 3. In the exact long-line method, B is calculated as: A. B = cosh(γl) B. B = sinh(γl)/Zc C. B = Z/2 D. B = Zc sinh(γl) Answer: D. A = D = cosh(γl), B = Zc sinh(γl), C = sinh(γl)/Zc. (Key point: Long line exact constants) 4. For this 280 km line, how do the nominal-Π and long-line results compare? A. Nominal-Π is wildly wrong B. They differ by only about 0.3 percent in voltage, so Π is reasonably accurate C. Long-line gives much higher losses D. They give opposite power factor signs Answer: B. VS is 251.03 kV vs 250.23 kV; error of Π grows with length. (Key point: Comparing the two methods) 5. Why is IS (201.8 A) less than IR (262.4 A)? A. Series resistance reduces current B. The load is leading C. Line charging current partly cancels the lagging load current D. The frequency is 60 Hz Answer: C. Capacitive charging current is leading and subtracts from the lagging load current. (Key point: Sending current less than receiving current) ## Close variants set in other exams - [300 km line: short line vs nominal-π](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3A300km-short-vs-nominal-pi) — Power System Analysis I, chapter 5, asked 2× ## Related in Power System Analysis I - [8 km line receiving 15000 kVA at 33 kV](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3A8km-line-15000kva) — Power System Analysis I, chapter 6 - [150 mile, 60 Hz line: ABCD and sending end](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3A150-mile-60hz-line-abcd) — Power System Analysis I, chapter 5 - [Nominal-T 100 km line, 20 MW at 110 kV](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Anominal-t-100km-20mw) — Power System Analysis I, chapter 5 - [Nominal-T 100 km line, 50 MVA at 132 kV](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Anominal-t-100km-50mva) — Power System Analysis I, chapter 6 - [400 km long line, 60 MVA at 220 kV](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Along-line-400km-60mva) — Power System Analysis I, chapter 5 - [150 km line with mid-point shunt capacitor](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3A150km-mid-shunt-capacitor) — Power System Analysis I, chapter 6 ## Related in other subjects - [Peterson coil for 33 kV 60 km line](https://www.bctengineer.com/question.md?key=semester_6%2Fswitchgear_and_protection%3Apeterson-coil-33kv-60km) — Switchgear and Protection, chapter 5, asked 6× - [Receiving-end voltage infinite at 1500 km](https://www.bctengineer.com/question.md?key=semester_8%2Fhigh_voltage_engineering%3Alossless-line-1500km-infinite-voltage) — High Voltage Engineering, chapter 3 - [Expression for receiving-end shunt compensation](https://www.bctengineer.com/question.md?key=semester_8%2Fhigh_voltage_engineering%3Ashunt-compensation-xp-expression) — High Voltage Engineering, chapter 3 - [Mid-line reactor for Vr = Vs at no load](https://www.bctengineer.com/question.md?key=semester_8%2Fhigh_voltage_engineering%3Amidpoint-reactor-400kv-500km) — High Voltage Engineering, chapter 3