---
title: "200 mile line: Zc, propagation constant, SIL"
subject: "Power System Analysis I"
chapter: 5
key: "semester_4/power_system_analysis_i:200-mile-zc-propagation-sil"
source: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/200-mile-zc-propagation-sil
site: BCT Engineering
---

# 200 mile line: Zc, propagation constant, SIL

Power System Analysis I · Chapter 5: Transmission Line Modeling

- Asked in: 2071 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/200-mile-zc-propagation-sil
- Whole chapter: https://www.bctengineer.com/md/semester_4/power_system_analysis_i/chapter-5

## Question

A 200 mile long, 3-phase, 220 kV, 60 Hz overhead transmission line has series impedance of 0.8431∠79.04° Ω/mi and shunt admittance of 5.105 × 10⁻⁶∠90° S/mi. Determine the characteristic impedance, propagation constant, phase shift constant, attenuation constant and SIL of the line.

## Short answer

Characteristic impedance Zc = 406.4 angle -5.48° Ω; propagation constant gamma = 2.075×10⁻³ angle 84.52° per mile; attenuation constant alpha = 1.981×10⁻⁴ Np/mi; phase constant beta = 2.065×10⁻³ rad/mi; surge impedance loading SIL = 119.1 MW (using |Zc| from sqrt(z/y)).

## Answer

**Given:** $l = 200$ mi, 220 kV, 60 Hz, $z = 0.8431\angle 79.04^\circ\ \Omega$/mi, $y = 5.105\times10^{-6}\angle 90^\circ$ S/mi.

#### Characteristic impedance

$$
\begin{aligned}
Z_c &= \sqrt{\frac{z}{y}} = \sqrt{\frac{0.8431\angle 79.04^\circ}{5.105\times10^{-6}\angle 90^\circ}} = \sqrt{165{,}152\angle -10.96^\circ} \\
&= 406.4\angle -5.48^\circ\ \Omega = 404.5 - j38.8\ \Omega
\end{aligned}
$$

#### Propagation constant

$$
\begin{aligned}
\gamma &= \sqrt{zy} = \sqrt{0.8431\times5.105\times10^{-6}\angle 169.04^\circ} \\
&= 2.075\times10^{-3}\angle 84.52^\circ\ \text{per mile} \\
&= 1.981\times10^{-4} + j2.065\times10^{-3}\ \text{per mile}
\end{aligned}
$$

For the whole line: $\gamma l = 0.4149\angle 84.52^\circ = 0.0396 + j0.4130$.

#### Attenuation and phase-shift constants

$$
\begin{aligned}
\alpha &= \text{Re}(\gamma) = 1.981\times10^{-4}\ \text{Np/mi}\quad(\alpha l = 0.0396\ \text{Np}) \\
\beta &= \text{Im}(\gamma) = 2.065\times10^{-3}\ \text{rad/mi}\quad(\beta l = 0.4130\ \text{rad} = 23.66^\circ)
\end{aligned}
$$

#### Surge impedance loading

$$
\text{SIL} = \frac{V_L^2}{|Z_c|} = \frac{(220)^2}{406.4} = 119.1\ \text{MW}
$$

(Strictly, SIL is defined with the lossless surge impedance $\sqrt{L/C} = \sqrt{x/b} = \sqrt{0.8277/5.105\times10^{-6}} = 402.7\ \Omega$, which gives 120.2 MW. The difference is under 1%.)

**Answer:** $Z_c = 406.4\angle -5.48^\circ\ \Omega$, $\gamma = 2.075\times10^{-3}\angle 84.52^\circ$ /mi, $\beta = 2.065\times10^{-3}$ rad/mi, $\alpha = 1.981\times10^{-4}$ Np/mi, SIL $= 119.1$ MW (120.2 MW with the lossless $Z_c$).

## Check yourself

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

1. How is the characteristic impedance computed from z and y?
   A. Zc = sqrt(z y)
   B. Zc = sqrt(z / y)
   C. Zc = z / y
   D. Zc = sqrt(y / z)
   Answer: B. Zc = sqrt(z/y); dividing magnitudes and subtracting angles gives 406.4∠-5.48° Ω. (Key point: Zc formula)

2. How is the propagation constant found and what is its form?
   A. gamma = sqrt(z/y), purely real
   B. gamma = sqrt(z y) = alpha + j beta, per mile
   C. gamma = z y, purely imaginary
   D. gamma = sqrt(z y) per whole line, without division by length
   Answer: B. gamma = sqrt(zy) = 2.075e-3∠84.52° per mile, whose real part is alpha and imaginary part beta. (Key point: Propagation constant)

3. From gamma = 1.981e-4 + j2.065e-3 per mile, which are alpha and beta?
   A. alpha = 2.065e-3 Np/mi, beta = 1.981e-4 rad/mi
   B. alpha = 1.981e-4 Np/mi, beta = 2.065e-3 rad/mi
   C. alpha = beta = 2.075e-3
   D. alpha = 0, beta = 2.075e-3
   Answer: B. Alpha is the real part (attenuation) and beta the imaginary part (phase shift). (Key point: Alpha and beta extraction)

4. How is SIL computed for the 220 kV line with |Zc| = 406.4 Ω?
   A. 220 x 406.4 = 89,408 MW
   B. 220^2 / 406.4 = 119.1 MW
   C. 220 / 406.4 = 0.54 MW
   D. sqrt3 x 220^2 / 406.4 = 206 MW
   Answer: B. SIL = VL^2/|Zc| = 48400/406.4 = 119.1 MW. (Key point: SIL calculation)

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