Receiving-end voltage infinite at 1500 km

For a lossless line the ABCD constants are

$$\begin{bmatrix} V_s \\ I_s \end{bmatrix} = \begin{bmatrix} \cos\beta l & jZ_c\sin\beta l \\ \dfrac{j\sin\beta l}{Z_c} & \cos\beta l \end{bmatrix}\begin{bmatrix} V_R \\ I_R \end{bmatrix}$$

where $\beta = \omega\sqrt{LC}$ is the phase constant and $Z_c = \sqrt{L/C}$ is the surge impedance.

No-load condition

At no load $I_R = 0$, so

$$V_s = A V_R = V_R\cos\beta l \;\Rightarrow\; V_R = \frac{V_s}{\cos\beta l}$$

Phase constant for an overhead line

For a lossless overhead line the wave velocity is $v = 1/\sqrt{LC} \approx c = 3\times10^5$ km/s, so

$$\begin{aligned} \beta &= \omega\sqrt{LC} = \frac{\omega}{v} = \frac{2\pi \times 50}{3\times10^5} \\ &= 1.0472\times10^{-3}\ \text{rad/km} \;(= 0.06^\circ/\text{km} = 6^\circ \text{ per 100 km}) \end{aligned}$$

Wavelength: $\lambda = v/f = 3\times10^5/50 = 6000$ km.

Line of 1500 km

$$\begin{aligned} \beta l &= 1.0472\times10^{-3} \times 1500 = 1.5708\ \text{rad} = 90^\circ \\ \cos\beta l &= \cos 90^\circ = 0 \\ V_R &= \frac{V_s}{0} \to \infty \end{aligned}$$

So a 1500 km line is a quarter-wavelength line ($l = \lambda/4 = 6000/4$). At no load it is in resonance: the line inductance and capacitance resonate at 50 Hz. In theory the receiving-end voltage is infinite, for any finite sending-end voltage.

In practice line resistance and corona losses limit the voltage, but it would still be dangerously high. Very long lines are therefore compensated (shunt reactors, series capacitors) or split into sections by intermediate substations.