Surge impedance and velocity of 220 kV line

Given: $L = 1.3$ mH/km $= 1.3 \times 10^{-3}$ H/km, $C = 8.855 \times 10^{-9}$ F/km, length $l = 200$ km, $f = 50$ Hz. The line is treated as lossless.

(i) Surge impedance

$$Z_c = \sqrt{\frac{L}{C}} = \sqrt{\frac{1.3 \times 10^{-3}}{8.855 \times 10^{-9}}} = \sqrt{146810} = 383.2\ \Omega$$

(ii) Velocity of propagation

$$v = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{1.3\times10^{-3} \times 8.855\times10^{-9}}} = \frac{1}{3.393\times10^{-6}} = 2.947 \times 10^{5}\ \text{km/s}$$

(about 98% of the speed of light).

(iii) Travel time and period of transient

One-way travel time:

$$\tau = \frac{l}{v} = \frac{200}{2.947\times10^5} = 6.786\times10^{-4}\ \text{s} = 0.679\ \text{ms}$$

For an open-ended line energised from a source, the voltage wave reflects back and forth and the transient oscillation has a period of four travel times:

$$T = 4\tau = 2.714\ \text{ms}, \qquad f_{osc} = \frac{1}{T} = 368.4\ \text{Hz}$$

(iv) Propagation constant

For a lossless line $\gamma = \alpha + j\beta$ with $\alpha = 0$ and

$$\beta = \omega\sqrt{LC} = 2\pi \times 50 \times 3.393\times10^{-6} = 1.066\times10^{-3}\ \text{rad/km}$$ $$\gamma = j\,1.066\times10^{-3}\ \text{rad/km}$$

For the whole line, $\beta l = 0.2132$ rad $= 12.2^\circ$. (Wavelength $\lambda = v/f = 5895$ km.)

Answer: $Z_c = 383.2\ \Omega$; $v = 2.947\times10^5$ km/s; $\tau = 0.679$ ms (period $4\tau = 2.71$ ms); $\gamma = j1.066\times10^{-3}$ rad/km.