---
title: "450 km 400 kV line, 420 MW, 0.95 pf"
subject: "Power System Analysis I"
chapter: 5
key: "semester_4/power_system_analysis_i:450km-400kv-420mw-095"
source: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/450km-400kv-420mw-095
site: BCT Engineering
---

# 450 km 400 kV line, 420 MW, 0.95 pf

Power System Analysis I · Chapter 5: Transmission Line Modeling

- Asked in: 2072 Magh, 9 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-095
- Whole chapter: https://www.bctengineer.com/md/semester_4/power_system_analysis_i/chapter-5

## Question

A 50 Hz, 400 kV, 450 km long transmission line has the following parameters: r = 0.033 Ω/km, L = 1.067 mH/km, C = 0.0109 μF/km. It is delivering 420 MW at 0.95 power factor lagging. Determine voltage and current at the sending end, sending end power and power factor.

## Short answer

Long-line: Z=151.6∠84.38° Ω, Y=j1.541×10⁻³ S; A=D=0.8861∠0.71°, B=145.8∠84.60° Ω, C=1.482×10⁻³∠90.22° S. V_S=445.63 kV (line), I_S=564.9 A, P_S=435.71 MW at 0.9993 p.f. lagging.

## 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 0.95 p.f. lagging. Receiving-end voltage taken as 400 kV. Long-line model is used.

#### Step 1: Line constants

$$
\begin{aligned}
z &= 0.033 + j2\pi\times50\times1.067\times10^{-3} = 0.033 + j0.3352\ \Omega/\text{km} \\
y &= j2\pi\times50\times0.0109\times10^{-6} = j3.424\times10^{-6}\ \text{S/km} \\
Z &= 14.85 + j150.84 = 151.6\angle 84.38^\circ\ \Omega, \quad Y = j1.541\times10^{-3}\ \text{S}
\end{aligned}
$$

#### Step 2: $\gamma l$, $Z_c$ and ABCD

$$
\begin{aligned}
\gamma l &= \sqrt{ZY} = 0.4833\angle 87.19^\circ = 0.02370 + j0.48271 \\
Z_c &= \sqrt{Z/Y} = 313.6\angle -2.81^\circ\ \Omega \\
A = D &= \cosh\gamma l = 0.88599 + j0.01100 = 0.8861\angle 0.71^\circ \\
B &= Z_c\sinh\gamma l = 145.8\angle 84.60^\circ\ \Omega \\
C &= \frac{\sinh\gamma l}{Z_c} = 1.482\times10^{-3}\angle 90.22^\circ\ \text{S}
\end{aligned}
$$

#### Step 3: Receiving-end values

$$
\begin{aligned}
V_R &= \frac{400}{\sqrt3} = 230.94\angle 0^\circ\ \text{kV} \\
I_R &= \frac{420\times10^6}{\sqrt3\times400\times10^3\times0.95} = 638.1\angle -18.19^\circ = 606.22 - j199.25\ \text{A}
\end{aligned}
$$

#### Step 4: Sending-end voltage and current

$$
\begin{aligned}
V_S &= AV_R + BI_R = (204{,}610 + j2541) + (37{,}233 + j85{,}242) \\
&= 241{,}840 + j87{,}783 = 257.28\angle 19.95^\circ\ \text{kV/phase} \\
V_{S(L)} &= \sqrt3\times257.28 = 445.63\ \text{kV} \\
I_S &= CV_R + DI_R = (-1.33 + j342.24) + (539.30 - j169.87) \\
&= 537.97 + j172.37 = 564.9\angle 17.77^\circ\ \text{A}
\end{aligned}
$$

#### Step 5: Sending-end power and power factor
Angle between $V_S$ and $I_S$ $= 19.95^\circ - 17.77^\circ = 2.18^\circ$ (current lags).

$$
\begin{aligned}
\text{p.f.} &= \cos 2.18^\circ = 0.9993\ \text{lagging} \\
P_S &= 3|V_S||I_S|\cos\phi_S = 435.71\ \text{MW}, \quad Q_S = 16.61\ \text{Mvar}
\end{aligned}
$$

Line loss $= 435.71 - 420 = 15.71$ MW, efficiency $= 96.39\%$.

**Answer:** $V_S = 445.63$ kV (line) $\angle 19.95^\circ$, $I_S = 564.9\angle 17.77^\circ$ A, $P_S = 435.71$ MW, sending p.f. $= 0.9993$ lagging.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. How is the receiving-end current found for 420 MW at 0.95 lagging and 400 kV?
   A. IR = P/(sqrt3 VL cos φ) at angle -18.19°, about 638 A
   B. IR = P/(VL cos φ), about 1105 A
   C. IR = P/(sqrt3 VL), about 606 A at 0° angle
   D. IR = P/(3 VL cos φ), about 368 A
   Answer: A. IR = 420e6/(sqrt3 x 400e3 x 0.95) = 638.1 A, lagging VR by cos^-1(0.95) = 18.19°. (Key point: Receiving-end current at 0.95 p.f.)

2. Which relation gives the sending-end phase voltage and current?
   A. VS = D VR + B IR and IS = C VR + A IR
   B. VS = A VR + B IR and IS = C VR + D IR
   C. VS = A VR + C IR and IS = B VR + D IR
   D. VS = VR + Z IR and IS = IR
   Answer: B. The ABCD relations are VS = A VR + B IR and IS = C VR + D IR with the long-line constants. (Key point: Use of ABCD for sending end)

3. How is the sending-end power factor obtained?
   A. Cosine of the angle between VS and IS (19.95° - 17.77° = 2.18°)
   B. Same as the receiving-end power factor 0.95
   C. Cosine of the angle of VS alone
   D. Ratio of |VR| to |VS|
   Answer: A. p.f. = cos(angle of VS - angle of IS) = cos 2.18° = 0.9993 lagging. (Key point: Sending-end power factor)

4. The sending-end p.f. (0.9993) is much better than the load p.f. (0.95). Why?
   A. Line resistance supplies reactive power
   B. Line capacitance generates reactive power that offsets part of the load's lagging Q
   C. The sending voltage is lower than the receiving voltage
   D. The power factor is always raised by series inductance
   Answer: B. The shunt capacitance of the long line supplies reactive power, so the sending end supplies less Q and the p.f. approaches unity. (Key point: Why sending p.f. is higher than load p.f.)

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