Air clearances for 220 kV double circuit line

Air clearances fix the size of the tower. They come from the minimum air clearance $a$, the insulator string length $l$, and the swing of the string.

Data and assumptions: maximum system voltage 245 kV (phase rms 141.45 kV); 254 × 146 mm disc insulators; maximum swing of the suspension string 45°; minimum air clearance 6.5 inch per 10 kV (phase rms) plus 8 inch safety, as given with IOE papers; tower body width at cross-arm level about 1.5 m.

Minimum air clearance (a)

$$a = 6.5\times\frac{141.45}{10} + 8 = 99.94\ \text{inch} = 2.54\ \text{m}$$

Insulator string length (l)

• Number of discs: the standard 220 kV suspension string has 14 discs.
• Check against the withstand and flashover tables: the 1-min wet withstand of 395 kV and the impulse withstand of 900 kV each need 10 discs. Extra discs are added for switching surges, pollution and one defective disc, giving 13–14.
• Length:

$$l = 14\times0.146 + 0.3\ (\text{hardware}) = 2.344 \approx 2.34\ \text{m}$$

Cross arm length (X)

The swung string must keep the conductor at least $a$ from the tower body:

$$X = a + l\sin\theta = 2.54 + 2.344\sin45^\circ = 4.20\ \text{m}$$

measured from the tower face to the point of suspension.

Vertical separation (Y)

The conductor hanging under one cross arm must clear the cross arm below by $a$:

$$Y = l + a = 2.344 + 2.539 = 4.88\ \text{m}$$

Horizontal separation

In a double-circuit tower, each circuit is arranged vertically on one side of the tower. The horizontal distance between the conductors of the two circuits at the same level is:

$$S_h = 2X + w = 2(4.196) + 1.5 = 9.89\ \text{m}$$

Height of earth wire above the top conductor

Use one earth wire at the tower peak with a shielding angle of at most 30° to the outer top conductors. The horizontal offset is $X + w/2 = 4.946$ m, so

$$h_e = \frac{X + w/2}{\tan30^\circ} = \frac{4.946}{0.577} = 8.57\ \text{m}$$

This is about 6.2 m above the top cross arm, since the conductor hangs 2.34 m below that cross arm.

              E  earth wire
             /|\     30 deg shielding
            / | \
    *------o--+--o------*   top arm, conductors
           |     |          hang l below arm
    *------o-----o------*   Y = 4.88 m
           |     |
    *------o-----o------*
     X=4.2m  w    X=4.2m

Answer: cross arm about 4.2 m, string 2.34 m, vertical spacing about 4.9 m, horizontal spacing about 9.9 m, and earth wire about 8.6 m above the top conductor.