--- title: "450 km 400 kV line, 420 MW, 0.9 pf" subject: "Power System Analysis I" chapter: 5 key: "semester_4/power_system_analysis_i:450km-400kv-420mw-090" source: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/450km-400kv-420mw-090 site: BCT Engineering --- # 450 km 400 kV line, 420 MW, 0.9 pf Power System Analysis I · Chapter 5: Transmission Line Modeling - Asked in: 2069 Poush, 10 marks - 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/450km-400kv-420mw-090 - Whole chapter: https://www.bctengineer.com/md/semester_4/power_system_analysis_i/chapter-5 ## Question A 450 km long, 50 Hz, 400 kV overhead transmission line has series resistance of 0.033 Ohm/km, series inductance of 1.067 mH/km and shunt capacitance of 0.0109 × 10⁻⁶ F/km. Compute the ABCD parameters, sending end active power, voltage and currents, and voltage regulation if the line is supplying a load of 420 MW at 0.9 power factor lagging. Also compute the power transmission efficiency of the line. ## Short answer Using long-line model: A = D = 0.8861∠0.71°, B = 145.8∠84.60° Ω, C = 1.482×10⁻³∠90.22° S. Sending-end voltage = 467.27 kV, current = 546.3 A, active power = 436.24 MW. Voltage regulation = 31.84 %, efficiency = 96.28 %. ## Answer **Given:** 50 Hz, 400 kV, $l = 450$ km; $r = 0.033\ \Omega$/km, $L = 1.067$ mH/km, $C = 0.0109\ \mu$F/km; load 420 MW at 0.9 p.f. lagging at 400 kV. Long-line model is used. #### Step 1: Line constants $$ \begin{aligned} z &= 0.033 + j(2\pi\times50\times1.067\times10^{-3}) = 0.033 + j0.3352\ \Omega/\text{km} \\ y &= j2\pi\times50\times0.0109\times10^{-6} = j3.424\times10^{-6}\ \text{S/km} \\ Z &= zl = 14.85 + j150.84 = 151.6\angle 84.38^\circ\ \Omega, \quad Y = yl = j1.541\times10^{-3}\ \text{S} \end{aligned} $$ #### Step 2: ABCD parameters $$ \begin{aligned} \gamma l &= \sqrt{ZY} = 0.4833\angle 87.19^\circ = 0.02370 + j0.48271 \\ Z_c &= \sqrt{Z/Y} = 313.6\angle -2.81^\circ\ \Omega \\ A = D &= \cosh\gamma l = 0.88599 + j0.01100 = 0.8861\angle 0.71^\circ \\ B &= Z_c\sinh\gamma l = 145.8\angle 84.60^\circ\ \Omega \\ C &= \frac{\sinh\gamma l}{Z_c} = 1.482\times10^{-3}\angle 90.22^\circ\ \text{S} \end{aligned} $$ #### Step 3: Receiving-end values $$ \begin{aligned} V_R &= \frac{400}{\sqrt3} = 230.94\angle 0^\circ\ \text{kV} \\ I_R &= \frac{420\times10^6}{\sqrt3\times400\times10^3\times0.9} = 673.6\angle -25.84^\circ = 606.22 - j293.60\ \text{A} \end{aligned} $$ #### Step 4: Sending-end voltage and current $$ \begin{aligned} V_S &= AV_R + BI_R = (204{,}610 + j2541) + (50{,}925 + j83{,}948) \\ &= 255{,}540 + j86{,}489 = 269.78\angle 18.70^\circ\ \text{kV/phase} \\ V_{S(L)} &= \sqrt3\times269.78 = 467.27\ \text{kV} \\ I_S &= CV_R + DI_R = (-1.33 + j342.24) + (540.33 - j253.46) \\ &= 539.01 + j88.78 = 546.3\angle 9.35^\circ\ \text{A} \end{aligned} $$ #### Step 5: Sending-end power p.f. $= \cos(18.70^\circ - 9.35^\circ) = 0.9867$ lagging $$ P_S = 3|V_S||I_S|\cos\phi_S = 3\times269.78\times10^3\times546.3\times0.9867 = 436.24\ \text{MW} $$ #### Step 6: Voltage regulation $$ \begin{aligned} V_{R(NL)} &= \frac{|V_S|}{|A|} = \frac{467.27}{0.8861} = 527.35\ \text{kV (line)} \\ \%\text{VR} &= \frac{527.35 - 400}{400}\times100 = 31.84\ \% \end{aligned} $$ #### Step 7: Efficiency $$ \eta = \frac{P_R}{P_S}\times100 = \frac{420}{436.24}\times100 = 96.28\ \% $$ **Answer:** $A = D = 0.8861\angle 0.71^\circ$, $B = 145.8\angle 84.60^\circ\ \Omega$, $C = 1.482\times10^{-3}\angle 90.22^\circ$ S; $V_S = 467.27$ kV, $I_S = 546.3$ A, $P_S = 436.24$ MW, regulation $= 31.84\%$, efficiency $= 96.28\%$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is IR for 420 MW, 400 kV, 0.9 p.f. lagging? A. About 673.6 A at -25.84° B. About 638 A at -18.19° C. About 606 A at 0° D. About 673.6 A at +25.84° Answer: A. IR = 420e6/(sqrt3 x 400e3 x 0.9) = 673.6 A, lagging by cos^-1(0.9) = 25.84°. (Key point: Receiving-end current at 0.9 p.f.) 2. How is the voltage regulation computed here? A. (VS - VR)/VR B. VR(NL) = |VS|/|A|, then (VR(NL) - VR(FL))/VR(FL) x 100 C. (VR - VS)/VS D. |A| x 100 Answer: B. At no load VS = A VR, so VR(NL) = 467.27/0.8861 = 527.35 kV and regulation is (527.35 - 400)/400 = 31.84%. (Key point: Voltage regulation method) 3. How is the transmission efficiency found? A. PR/PS x 100 = 420/436.24 B. PS/PR x 100 C. (VR/VS) x 100 D. cos φ x 100 Answer: A. Efficiency is output over input power: 420/436.24 = 96.28%. (Key point: Efficiency) 4. How is PS computed from the sending-end quantities? A. PS = 3 |VS| |IS| cos(angle VS - angle IS) using phase values B. PS = |VS(L)| |IS| C. PS = sqrt3 |VS| |IS| with phase VS D. PS = 3 |VS| |IS| cos(angle of VS) Answer: A. Three-phase power is 3 times per-phase VS IS times the cosine of the angle between them (9.35°), giving 436.24 MW. 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