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Tower dimensions for 220 kV double circuit line

Short answer

Cross arm length: 4.20 m; insulator string length: 2.34 m (14 discs); vertical separation: 4.88 m; horizontal separation: 9.89 m; one earth wire; earth wire height above top conductor: 8.57 m.

For a 220 kV double circuit transmission line, suggest the dimensions for following: i) Cross arm length ii) Insulator string length iii) Horizontal and vertical separation of conductors iv) Number of earth wires v) Height of earth wire from the top most conductor. (Take max. swing of insulator as: 45°)

The tower dimensions follow from the minimum air clearance, the insulator string length, the 45° swing of the string, and the shielding requirement for lightning.

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=6.5×141.4510+8=99.94 inch=2.54 ma = 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=14×0.146+0.3 (hardware)=2.344≈2.34 ml = 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 aa from the tower body:

X=a+lsin⁡θ=2.54+2.344sin⁡45∘=4.20 mX = 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 aa:

Y=l+a=2.344+2.539=4.88 mY = 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:

Sh=2X+w=2(4.196)+1.5=9.89 mS_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=4.946X + w/2 = 4.946 m, so

he=X+w/2tan⁡30∘=4.9460.577=8.57 mh_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

Number of earth wires

One earth wire is used. Lines up to 220 kV, which have a modest cross-arm spread, are adequately shielded by a single peak earth wire within a 30° shielding angle. Two earth wires are normally needed only for 400 kV and above, where horizontal configurations are wide. In very high-lightning hilly areas, two earth wires may also be used for a 220 kV double-circuit line to reduce the shielding angle.

ItemValue
Cross arm length4.20 m
Insulator string length2.34 m (14 discs)
Vertical separation4.88 m
Horizontal separation9.89 m
Number of earth wires1
Earth wire above top conductor8.57 m

Answer: X ≈ 4.2 m, l ≈ 2.34 m, Y ≈ 4.9 m, horizontal spacing ≈ 9.9 m, one earth wire about 8.6 m above the top conductor.

✓ Answer checked by Claude Opus 5.5 (high effort)

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