Number of insulator discs for 220 kV line

Data: 220 kV, single circuit, 50 Hz. From Table A-2 (220 kV class row): maximum system voltage 255 kV, 1-min dry withstand 435 kV, 1-min wet withstand 395 kV, impulse withstand 900 kV. Usual design factors assumed: flashover withstand ratio (FWR) = 1.15, non-atmospheric condition factor (NACF) = 1.1, factor of safety (FOS) = 1.1, so

$$k = 1.15 \times 1.1 \times 1.1 = 1.3915$$

1. External over-voltage: lightning

$$V_{req} = 900 \times 1.3915 = 1252.35\ \text{kV}$$

Table A-3 (impulse FOV): 13 discs = 1185 kV < 1252.35; 14 discs = 1265 kV. → 14 discs

2. Internal over-voltage: switching surge

Assume switching surge ratio SSR = 2.8 (usual for 220 kV).

$$\begin{aligned} V_{ss} &= SSR \times \frac{\sqrt{2}\,V_{max}}{\sqrt{3}} = 2.8 \times \frac{\sqrt{2} \times 255}{\sqrt{3}} = 582.98\ \text{kV (peak)} \\ V_{req} &= 582.98 \times 1.3915 = 811.21\ \text{kV (peak)} \\ &= 811.21/\sqrt{2} = 573.62\ \text{kV (rms)} \end{aligned}$$

Compared with 1-min wet FOV: 14 discs = 555 kV < 573.62, 15 discs = 590 kV. → 15 discs

3. Power-frequency over-voltage

Wet: $395 \times 1.3915 = 549.64$ kV → wet FOV 14 discs = 555 kV → 14 discs Dry: $435 \times 1.3915 = 605.30$ kV → dry FOV 11 discs = 635 kV → 11 discs

Result

The largest requirement governs: 15 discs (254 × 154 mm) per suspension string (switching surge governs). String length $\approx 15 \times 0.154 = 2.31$ m. (Tension strings commonly use one extra disc.)

(If the IEC value $V_{max} = 245$ kV were used instead, the switching requirement would be 551.12 kV rms and 14 discs would just suffice.)

Check of air clearance: $a = 6.5 \times \frac{255/\sqrt{3}}{10} + 8 = 103.70$ in $\approx 2.63$ m.