Temporary overvoltage for SLG fault at line middle

A single line-to-ground (SLG) fault raises the power-frequency voltage of the healthy phases. On an unloaded long line it is raised further by the Ferranti effect. The result is a temporary overvoltage (TOV).

Method and assumptions (data not given are taken as below and stated):

1. Generator emf $E = 1$ pu (phase), with the generator's own sequence impedances neglected (not given).
2. Prefault voltage at the fault point is found from the long-line equation with the far end open (Ferranti rise):

$$E_f = E\,\frac{\cosh\gamma(l - x)}{\cosh\gamma l}$$

where $x$ is the distance of the fault from the source. 3. Positive- and negative-sequence impedance up to the fault: $Z_1 = Z_2 = (r + j\omega l)\,x$. 4. Zero-sequence impedance $Z_0 = Z_1 + 3Z_n$. No separate zero-sequence line data are given, so the line's $Z_0$ is taken equal to $Z_1$, and $Z_n$ is the generator neutral impedance.

Healthy-phase voltage for an SLG fault on phase $a$ through $Z_f$:

$$V_b = E_f\left[a^2 - \frac{Z_0 - Z_1}{2Z_1 + Z_0 + 3Z_f}\right] = E_f\left[a^2 - \frac{Z_n}{Z_1 + Z_n + Z_f}\right]$$ $$V_c = E_f\left[a - \frac{Z_n}{Z_1 + Z_n + Z_f}\right], \qquad a = 1\angle120^\circ$$

Data: $r = 0.04\ \Omega$/km, $l = 1.01$ mH/km, $C = 11$ nF/km, length 400 km, fault at $x = 200$ km, $Z_f = j35\ \Omega$, $Z_n = 4 + j50\ \Omega$.

Step 1: line constants

$$\begin{aligned} z &= 0.04 + j(314.16)(1.01\times10^{-3}) = 0.04 + j0.3173\ \Omega/\text{km} \\ y &= j(314.16)(11\times10^{-9}) = j3.456\times10^{-6}\ \text{S/km} \\ \gamma &= \sqrt{zy} = (0.0659 + j1.0492)\times10^{-3}\ /\text{km} \\ \beta l &= 1.0492\times10^{-3}\times400 = 0.4189\ \text{rad} = 24^\circ \end{aligned}$$

Step 2: prefault voltage at the midpoint (Ferranti)

$$E_f = \frac{\cosh(200\gamma)}{\cosh(400\gamma)} \approx \frac{\cos 12^\circ}{\cos 24^\circ} = \frac{0.9781}{0.9135} = 1.0707\ \text{pu}$$

Step 3: sequence impedances

$$\begin{aligned} Z_1 &= Z_2 = 200(0.04 + j0.3173) = 8 + j63.46\ \Omega \\ Z_1 + Z_n + Z_f &= (8 + j63.46) + (4 + j50) + j35 = 12 + j148.46\ \Omega \\ m &= \frac{Z_n}{Z_1 + Z_n + Z_f} = \frac{4 + j50}{12 + j148.46} = 0.3368\angle0.05^\circ \end{aligned}$$

Step 4: healthy-phase voltages

$$\begin{aligned} V_b/E_f &= a^2 - m = (-0.5 - j0.8660) - 0.3368 = -0.8368 - j0.8663 \\ |V_b/E_f| &= 1.2044 \\ |V_c/E_f| &= |(-0.5 + j0.8660) - 0.3368| = 1.2040 \end{aligned}$$

Step 5: temporary overvoltage

$$\begin{aligned} V_{TOV} &= 1.2044 \times 1.0707 = 1.290\ \text{pu} \\ &= 1.290 \times \frac{400}{\sqrt3} = 297.8\ \text{kV (rms, phase)} \end{aligned}$$

Answer: TOV on the healthy phase ≈ 1.29 pu (≈ 298 kV rms to ground, peak ≈ 421 kV), with the stated assumptions. The ratio $Z_0/Z_1$ is high here because of the generator grounding impedance. A lower neutral impedance would reduce the TOV.