Surge impedance and travel time of 400 km line

Data: $L = 1.26$ mH/km, $C = 0.009\ \mu$F/km, $l = 400$ km. Resistance is neglected for $Z_0$ and $v$, as asked.

(i) Surge impedance

$$\begin{aligned} Z_0 &= \sqrt{\frac{L}{C}} = \sqrt{\frac{1.26\times10^{-3}}{0.009\times10^{-6}}} \\ &= \sqrt{1.4\times10^{5}} = 374.17\ \Omega \end{aligned}$$

(ii) Velocity of propagation

$$\begin{aligned} v &= \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{1.26\times10^{-3} \times 0.009\times10^{-6}}} \\ &= \frac{1}{\sqrt{1.134\times10^{-11}}} = \frac{1}{3.3675\times10^{-6}} \\ &= 2.9696\times10^{5}\ \text{km/s} \end{aligned}$$

(Since $L$ and $C$ are per km, $v$ is in km/s.) This is very close to the speed of light, as expected for an overhead line.

Travel time over the line

The velocity does not depend on the surge magnitude (150 kV) or its tail length:

$$\begin{aligned} t &= \frac{l}{v} = \frac{400}{2.9696\times10^{5}} \\ &= 1.347\times10^{-3}\ \text{s} = 1.347\ \text{ms} \end{aligned}$$

The current wave accompanying the 150 kV surge is $I = V/Z_0 = 150/374.17 = 0.401$ kA $= 400.9$ A.

Answer: $Z_0$ = 374.2 Ω, $v$ = 2.97 × 10⁵ km/s, travel time = 1.347 ms.