Temporary overvoltage for SLG fault at line end

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: $r = 0.031\ \Omega$/km, $L = 1$ mH/km, $C = 10$ nF/km, length 400 km, fault at the far end ($x = 400$ km), $Z_n = 5 + j50\ \Omega$, $Z_f = 0$. Line resistance is included, as asked.

Step 1: line constants

$$\begin{aligned} z &= 0.031 + j(314.16)(1\times10^{-3}) = 0.031 + j0.31416\ \Omega/\text{km} \\ y &= j(314.16)(10\times10^{-9}) = j3.1416\times10^{-6}\ \text{S/km} \\ \gamma &= \sqrt{zy} = (0.04896 + j0.99466)\times10^{-3}\ /\text{km} \\ Z_c &= \sqrt{z/y} = 316.6 - j15.6\ \Omega \\ \beta l &= 0.9947\times10^{-3}\times400 = 0.3974\ \text{rad} = 22.77^\circ \end{aligned}$$

Step 2: prefault voltage at the open end (Ferranti)

$$E_f = \frac{1}{|\cosh\gamma l|} = 1.0845\ \text{pu} \quad (\text{lossless: } 1/\cos 22.77^\circ = 1.0845)$$

Step 3: sequence impedances up to the fault

$$\begin{aligned} Z_1 &= Z_2 = 400(0.031 + j0.31416) = 12.4 + j125.66\ \Omega \\ Z_0 &= Z_1 + 3Z_n = 27.4 + j275.66\ \Omega \\ Z_1 + Z_n &= 17.4 + j175.66\ \Omega \\ m &= \frac{Z_n}{Z_1 + Z_n} = \frac{5 + j50}{17.4 + j175.66} = 0.2847\angle{-0.05^\circ} \end{aligned}$$

Step 4: healthy-phase voltages

$$\begin{aligned} V_b/E_f &= a^2 - m = -0.7847 - j0.8658, \quad |V_b/E_f| = 1.1684 \\ V_c/E_f &= a - m = -0.7847 + j0.8663, \quad |V_c/E_f| = 1.1688 \end{aligned}$$

Step 5: temporary overvoltage

$$\begin{aligned} V_{TOV} &= 1.1688 \times 1.0845 = 1.2676\ \text{pu} \\ &= 1.2676 \times \frac{400}{\sqrt3} = 292.7\ \text{kV (rms, phase)} \end{aligned}$$

 Gen ~--[ 400 km line ]--x  SLG at far end
  |                          (phase a)
 Zn = 5+j50 ohm
  |
 gnd

Answer: the temporary overvoltage on the healthy phases ≈ 1.27 pu (≈ 293 kV rms to ground, peak ≈ 414 kV). This is the Ferranti rise (1.085) multiplied by the earth-fault factor (1.169). Since $X_0/X_1 < 3$ here, the system behaves as effectively grounded (earth-fault factor below 1.4).