Temporary overvoltage at no load and its remedy

Treat the line as lossless and find the no-load receiving-end voltage from the long-line equations. This rise is the Ferranti effect.

Data: $l = 400$ km, $L = 1$ mH/km, $C = 12$ nF/km, $f = 50$ Hz ($r$ neglected as stated).

Line constants

$$\begin{aligned} \beta &= \omega\sqrt{LC} = 2\pi(50)\sqrt{1\times10^{-3} \times 12\times10^{-9}} \\ &= 314.16 \times 3.464\times10^{-6} = 1.0883\times10^{-3}\ \text{rad/km} \\ \beta l &= 1.0883\times10^{-3} \times 400 = 0.4353\ \text{rad} = 24.94^\circ \\ Z_c &= \sqrt{L/C} = \sqrt{\frac{1\times10^{-3}}{12\times10^{-9}}} = 288.68\ \Omega \end{aligned}$$

Temporary overvoltage at no load

With $I_r = 0$, $V_s = V_r\cos\beta l$:

$$\frac{V_r}{V_s} = \frac{1}{\cos 24.94^\circ} = \frac{1}{0.9067} = 1.1029\ \text{pu}$$

So the receiving end rises to about 1.103 pu (about 441 kV for 400 kV at the sending end).

Countermeasure: shunt reactor at the receiving end

Connect a shunt reactor $X_L$ at the receiving end. Then $I_r = V_r/(jX_L)$ and

$$\frac{V_s}{V_r} = \cos\beta l + \frac{Z_c}{X_L}\sin\beta l$$

For $V_r/V_s = 1.04$:

$$\begin{aligned} \cos\beta l + \frac{Z_c}{X_L}\sin\beta l &= \frac{1}{1.04} = 0.96154 \\ \frac{Z_c}{X_L}\sin\beta l &= 0.96154 - 0.90674 = 0.05480 \\ X_L &= \frac{288.68 \times \sin 24.94^\circ}{0.05480} = \frac{288.68 \times 0.42173}{0.05480} \\ &= 2221.4\ \Omega\ \text{per phase} \end{aligned}$$

Three-phase rating at rated voltage:

$$Q = \frac{V^2}{X_L} = \frac{(400)^2}{2221.4} = 72.0\ \text{MVAr}$$

(about 77.9 MVAr at the 1.04 pu operating voltage).

Answer: no-load TOV = 1.103 pu. Install a shunt reactor of about 2221 Ω per phase (≈ 72 MVAr, 3-phase) at the receiving end to limit the voltage to 1.04 pu.