Air clearances for 132 kV double circuit line

The dimensions follow from the minimum phase-to-earth air clearance, the insulator string length and the insulator swing.

Assumptions: maximum system voltage 145 kV; swing 45°; disc 254 × 154 mm; tower width at cross-arm level 1.2 m.

Minimum air clearance

$$\begin{aligned} \text{I:}\quad & V_{ph,peak} = \frac{145\sqrt{2}}{\sqrt{3}} = 118.4\ \text{kV} \Rightarrow a = 118.4 + 30 = 148.4\ \text{cm} = 1.48\ \text{m} \\ \text{II:}\quad & V_{ph,rms} = \frac{145}{\sqrt{3}} = 83.72\ \text{kV} \Rightarrow a = 6.5 \times 8.372 + 8 = 62.4\ \text{in} = 1.585\ \text{m} \end{aligned}$$

Take the larger value: a ≈ 1.6 m.

ii) Insulator string length

For 145 kV, Table A-3 gives:

• wet withstand 230 kV needs 6 discs (250 kV),
• impulse 550 kV needs 6 discs (610 kV),
• dry withstand 265 kV needs 4 discs.

Adding one spare disc gives 7 discs:

$$l = 7 \times 0.154 + 0.30\ (\text{hardware}) = 1.38 \approx 1.4\ \text{m}$$

Common practice uses 9 discs (about 1.6 m) where pollution is a concern.

i) Cross-arm length

$$L_{ca} = a + l\sin45° = 1.6 + 1.4 \times 0.707 = 2.59 \approx 2.6\ \text{m}$$

This keeps clearance $a$ from the tower body when the string swings by 45°.

iii) Separation of conductors

• Vertical (between cross-arms on the same side):

$$V = l + a = 1.4 + 1.6 = 3.0\ \text{m}$$

The hanging conductor keeps clearance $a$ from the cross-arm below.

• Horizontal (between the two circuits across the tower):

$$H = 2L_{ca} + w = 2 \times 2.6 + 1.2 = 6.4\ \text{m}$$

This is well above the phase-to-phase clearance of about $1.15a = 1.8$ m.

            ew (earth wire)
             |
   o---------+---------o   top     \
             |                      | 3.0 m
   o---------+---------o   middle  /
             |
   o---------+---------o   bottom
   |<-2.6->|1.2|<-2.6->|

iv) Height of earth wire above the top conductor

One earth wire at the peak is enough for a 132 kV double-circuit vertical tower. For a 30° shielding angle, the horizontal distance to the outer conductor is $2.6 + 0.6 = 3.2$ m:

$$y = \frac{3.2}{\tan 30°} = 5.54 \approx 5.6\ \text{m}$$