---
title: "400 km line at UPF: Zc, propagation, regulation"
subject: "Power System Analysis I"
chapter: 5
key: "semester_4/power_system_analysis_i:400km-upf-zc-regulation"
source: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/400km-upf-zc-regulation
site: BCT Engineering
---

# 400 km line at UPF: Zc, propagation, regulation

Power System Analysis I · Chapter 5: Transmission Line Modeling

- Asked in: 2068 Magh, 6 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/400km-upf-zc-regulation
- Whole chapter: https://www.bctengineer.com/md/semester_4/power_system_analysis_i/chapter-5

## Question

A 50 Hz, 400 km long, 400 kV, 3-phase overhead transmission line delivering a power of 400 MW at unity power factor has series resistance of 3.2 × 10⁻⁵ Ohm/m, inductance of 1.066 × 10⁻⁶ H/m and shunt capacitance of 1.09 × 10⁻¹¹ F/m per phase. Compute the characteristic impedance, propagation constant and phase shift constant, and voltage regulation of the line.

## Short answer

Characteristic impedance Z_c = 313.4∠−2.73° Ω, propagation constant γ = 1.073×10⁻³∠87.27° per km, phase constant β = 1.072×10⁻³ rad/km. Sending voltage computed via ABCD parameters gives voltage regulation = 9.61 %.

## Answer

**Given:** 50 Hz, $l = 400$ km, 400 kV, 400 MW at unity p.f.; $r = 0.032\ \Omega$/km, $L = 1.066$ mH/km, $C = 0.0109\ \mu$F/km (converted from per-metre values). Receiving voltage 400 kV.

#### Step 1: Series impedance and shunt admittance per km

$$
\begin{aligned}
z &= 0.032 + j(2\pi\times50\times1.066\times10^{-3}) = 0.032 + j0.3349 = 0.3364\angle 84.54^\circ\ \Omega/\text{km} \\
y &= j2\pi\times50\times1.09\times10^{-8} = j3.424\times10^{-6}\ \text{S/km}
\end{aligned}
$$

#### Step 2: Characteristic impedance

$$
Z_c = \sqrt{\frac{z}{y}} = \sqrt{\frac{0.3364\angle 84.54^\circ}{3.424\times10^{-6}\angle 90^\circ}} = 313.4\angle -2.73^\circ\ \Omega
$$

#### Step 3: Propagation and phase-shift constants

$$
\begin{aligned}
\gamma &= \sqrt{zy} = 1.073\times10^{-3}\angle 87.27^\circ = 5.11\times10^{-5} + j1.0721\times10^{-3}\ /\text{km} \\
\beta &= 1.0721\times10^{-3}\ \text{rad/km}, \quad \alpha = 5.11\times10^{-5}\ \text{Np/km} \\
\gamma l &= 0.02044 + j0.42884
\end{aligned}
$$

#### Step 4: ABCD constants

$$
\begin{aligned}
A &= \cosh\gamma l = 0.90964 + j0.00850 = 0.9097\angle 0.54^\circ \\
B &= Z_c\sinh\gamma l = 130.5\angle 84.71^\circ\ \Omega
\end{aligned}
$$

#### Step 5: Sending-end voltage

$$
\begin{aligned}
V_R &= 230.94\angle 0^\circ\ \text{kV}, \quad I_R = \frac{400\times10^6}{\sqrt3\times400\times10^3} = 577.35\angle 0^\circ\ \text{A} \\
V_S &= AV_R + BI_R = 230.3\angle 19.53^\circ\ \text{kV/phase} = 398.83\ \text{kV (line)}
\end{aligned}
$$

#### Step 6: Voltage regulation

$$
\begin{aligned}
V_{R(NL)} &= \frac{|V_S|}{|A|} = \frac{398.83}{0.9097} = 438.43\ \text{kV} \\
\%\text{VR} &= \frac{438.43 - 400}{400}\times100 = 9.61\ \%
\end{aligned}
$$

The load (400 MW) is below the SIL ($400^2/313.4 \approx 511$ MW), so the line generates net reactive power and $V_S$ is slightly below $V_R$ at full load.

**Answer:** $Z_c = 313.4\angle -2.73^\circ\ \Omega$, $\gamma = 1.073\times10^{-3}\angle 87.27^\circ$ per km, $\beta = 1.072\times10^{-3}$ rad/km, voltage regulation $= 9.61\%$.

## Check yourself

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

1. How is the characteristic impedance obtained from the per-km line data?
   A. Zc = sqrt(z/y), with z the series impedance and y the shunt admittance per km
   B. Zc = sqrt(z*y)
   C. Zc = z*y
   D. Zc = sqrt(y/z)
   Answer: A. Zc = sqrt(z/y); the product sqrt(zy) is the propagation constant instead. (Key point: Characteristic impedance)

2. gamma = sqrt(zy) = alpha + j beta. Which statement is correct for this line?
   A. alpha is the phase-shift constant and beta is the attenuation constant
   B. beta (about 1.072e-3 rad/km) is the phase-shift constant and alpha (about 5.11e-5 Np/km) is the attenuation constant
   C. Both alpha and beta equal 1.073e-3
   D. beta is about 5.11e-5 rad/km
   Answer: B. The real part of gamma is attenuation alpha, the imaginary part is the phase-shift constant beta. (Key point: Propagation constant)

3. With V_R = 230.94 kV per phase and I_R = 577.35 A at unity pf, how is V_S found?
   A. V_S = V_R + I_R Zc
   B. V_S = V_R / A
   C. V_S = A V_R + B I_R using A = cosh(gamma l) and B = Zc sinh(gamma l)
   D. V_S = C V_R + D I_R
   Answer: C. The sending voltage comes from the first ABCD equation with the long-line constants. (Key point: Sending end voltage)

4. How is voltage regulation computed here?
   A. (|V_S| - |V_R|)/|V_S| x 100
   B. (|V_S| - |V_R|)/|V_R| x 100 always
   C. |A| x |V_S| / |V_R| x 100
   D. (|V_S|/|A| - |V_R|)/|V_R| x 100, since V_S/|A| is the receiving voltage at no load
   Answer: D. At no load I_R = 0, so V_R(NL) = V_S/|A|; regulation compares it with full-load V_R. This gives 9.61 %. (Key point: Regulation method)

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