6° phase shift per 100 km of lossless line

For a lossless line, the voltage at distance $x$ from the receiving end is

$$V(x) = V_R\cos\beta x + jZ_cI_R\sin\beta x$$

The phase of the voltage changes along the line through the factor $\beta x$. Here $\beta$ is the phase constant (radians per km):

$$\beta = \omega\sqrt{LC}$$

Velocity of propagation

For a lossless overhead line, with $L$ and $C$ per km:

$$L = \frac{\mu_0}{2\pi}\ln\frac{D}{r}, \qquad C = \frac{2\pi\varepsilon_0}{\ln(D/r)}$$

(taking $r' \approx r$), so

$$LC = \mu_0\varepsilon_0 \;\Rightarrow\; v = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{\mu_0\varepsilon_0}} = 3\times10^5\ \text{km/s}$$

Phase constant at 50 Hz

$$\begin{aligned} \beta &= \frac{\omega}{v} = \frac{2\pi \times 50}{3\times10^5} \\ &= 1.0472\times10^{-3}\ \text{rad/km} \end{aligned}$$

Phase shift per 100 km

$$\begin{aligned} \beta \times 100 &= 0.10472\ \text{rad} \\ &= 0.10472 \times \frac{180^\circ}{\pi} = 6^\circ \end{aligned}$$

So the phase shift between the sending-end and receiving-end voltages is 6° for every 100 km of a lossless line, at 50 Hz. Equivalently, the wavelength is $\lambda = v/f = 6000$ km, which corresponds to 360°: $360^\circ/6000\ \text{km} = 6^\circ/100$ km.

For example, a 300 km line has $\beta l = 18^\circ$.