Tower top potential and discs against back-flashover

When lightning strikes the tower top, the stroke current $I_s$ divides between all the paths that meet at the tower top. These paths are in parallel: the stroke channel itself (surge impedance $Z_s$), the tower footing resistance $R$, and the ground wires (surge impedance $Z_g$ each). So the tower-top potential is

$$V_t = I_s Z_{eq}, \qquad \frac{1}{Z_{eq}} = \frac{1}{Z_s} + \frac{1}{R} + \frac{n}{Z_g}$$

where $n$ is the number of ground wires. (Tower inductance and travel time in the tower are neglected.)

Data: $R = 40\ \Omega$, $Z_g = 500\ \Omega$ (each), $Z_s = 400\ \Omega$, $I_s = 50$ kA.

(i)(a) Considering all impedances (two ground wires)

$$\begin{aligned} \frac{1}{Z_{eq}} &= \frac{1}{400} + \frac{1}{40} + \frac{2}{500} \\ &= 0.0025 + 0.025 + 0.004 = 0.0315\ \text{S} \\ Z_{eq} &= 31.75\ \Omega \\ V_t &= 50 \times 31.75 = 1587.3\ \text{kV} \end{aligned}$$

(i)(b) Neglecting ground wires and stroke impedance

Only the footing resistance carries the current:

$$V_t = I_s R = 50 \times 40 = 2000\ \text{kV}$$

(i)(c) One ground wire and stroke impedance

$$\begin{aligned} \frac{1}{Z_{eq}} &= \frac{1}{400} + \frac{1}{40} + \frac{1}{500} = 0.0295\ \text{S} \\ Z_{eq} &= 33.90\ \Omega \\ V_t &= 50 \times 33.90 = 1694.9\ \text{kV} \end{aligned}$$

The ground wires and the stroke channel take part of the current, so they lower the tower-top potential. The footing resistance is the largest single factor.

(ii) Voltage across the insulator string, case (a)

The surge on the ground wire induces a voltage $K V_t$ on the phase conductor through the coupling factor $K$. The conductor rises with the tower, so the insulator string sees only the difference:

$$\begin{aligned} V_{ins} &= V_t - K V_t = (1 - K) V_t \\ &= (1 - 0.2) \times 1587.3 \\ &= 1269.8\ \text{kV} \approx 1270\ \text{kV} \end{aligned}$$

If this exceeds the flashover voltage of the string, a back-flashover occurs from the tower (at high potential) to the conductor.

Number of discs to prevent back-flashover

Lightning strokes normally come with rain, so the wet (rain) flashover voltage of 80 kV per disc is used (the conservative choice):

$$\begin{aligned} n &= \frac{V_{ins}}{V_{rain}} = \frac{1269.8}{80} = 15.87 \\ n &= 16\ \text{discs (rounded up)} \end{aligned}$$

If dry conditions are assumed: $1269.8/125 = 10.16$, so 11 discs. Since a thunderstorm is wet, the higher number is chosen.

Answer: $V_t$ = 1587.3 kV, 2000 kV and 1694.9 kV for (a), (b), (c); insulator voltage = 1269.8 kV; minimum 16 standard discs (rain FOV basis).

Note: the textbook method ignores the power-frequency voltage of the conductor and assumes the per-disc flashover values add linearly. In practice the string would be checked against its impulse flashover level and the footing resistance would be reduced.