Air clearances for 220 kV single circuit line

The air clearances of a 220 kV single-circuit line are found from the maximum system voltage, the insulator string and its swing. Assumptions: $V_m = 245$ kV, swing 45°, 254 × 154 mm discs, tower width at cross-arm level 1.5 m.

1. Phase-to-earth (tower) clearance

$$\begin{aligned} \text{Method I:}\quad & V_{ph,peak} = \frac{245\sqrt{2}}{\sqrt{3}} = 200.0\ \text{kV} \Rightarrow a = 200 + 30 = 230\ \text{cm} \\ \text{Method II:}\quad & V_{ph,rms} = \frac{245}{\sqrt{3}} = 141.45\ \text{kV} \Rightarrow a = 6.5 \times 14.145 + 8 = 99.9\ \text{in} = 2.54\ \text{m} \end{aligned}$$

Adopt the larger value, a = 2.54 m.

2. Phase-to-phase clearance

$$a_{pp} \approx 1.15a = 2.9\ \text{m}$$

The actual spacing chosen below is much larger.

3. Insulator string length

10 discs meet the wet (395 kV) and impulse (900 kV) levels. Adding one spare gives 11 discs:

$$l = 11 \times 0.154 + 0.3 \approx 2.0\ \text{m}$$

4. Cross-arm length (clearance with swing)

$$L_{ca} = a + l\sin45° = 2.54 + 1.41 = 3.95 \approx 4.0\ \text{m}$$

5. Vertical spacing between conductors

$$V = l + a = 2.0 + 2.54 = 4.54 \approx 4.5\ \text{m}$$

6. Horizontal spacing between conductors on opposite sides

$$H = 2 \times 4.0 + 1.5 = 9.5\ \text{m}$$

7. Earth wire clearance (shielding angle 30°, one earth wire)

$$y = \frac{4.0 + 0.75}{\tan30°} = 8.23 \approx 8.3\ \text{m above the top conductor}$$

Also check the midspan clearance between the earth wire and the conductor, using the IS 5613 rule of thumb: at least $0.012L + 1$ m, about 4.6 m for a 300 m span.

8. Minimum ground clearance

$$H_g = \frac{V - 33}{33} + 17 = \frac{220 - 33}{33} + 17 = 22.67\ \text{ft} = 6.91\ \text{m}$$

This compares with 7.0 m in IS practice.