Ratio of vertical separation, 132 vs 220 kV

The vertical separation between conductors is fixed by the insulator string length $l$ plus the minimum air clearance $a$.

Derivation of the expressions

1. Minimum air clearance. As given with IOE papers, allow 6.5 inch per 10 kV of the maximum phase voltage (rms), plus 8 inch as a factor of safety:

$$a = \left(6.5\times\frac{V_{max}/\sqrt3}{10} + 8\right)\ \text{inch}$$

2. String length. $l = n\times 146$ mm plus about 0.3 m of hardware, where $n$ is the number of 254 × 146 mm discs.

3. Vertical spacing. Take a double-circuit tower with conductors arranged vertically. The upper conductor hangs a distance $l$ below its cross arm, and it lies directly above the tip of the next cross arm, which is $Y$ below.

  ======o  upper cross arm
        |  l (string)
        *  upper conductor
        :  >= a (air clearance)
  ======o  lower cross arm
        |  l
        *  lower conductor

The gap between the upper conductor and the lower cross arm is $Y - l$, and it must be at least $a$:

$$Y - l \ge a \;\Rightarrow\; Y = l + a$$

When the string swings by θ, the conductor rises: its vertical distance from the cross arm above becomes $l\cos\theta$. That only increases the gap, so the unswung position governs.

132 kV double circuit

• Maximum system voltage 145 kV, so the phase voltage is $145/\sqrt3 = 83.72$ kV.
• $a = 6.5\times8.372+8 = 62.42$ inch $= 1.585$ m.
• Discs: 9 discs, the standard practice for 132 kV. This covers the table check (1-min wet withstand 230 kV needs 6 discs; impulse 550 kV needs 7 discs) plus a margin for switching surges, pollution and one defective disc.
• $l = 9\times0.146+0.3 = 1.614$ m.
• $Y_{132} = 1.614+1.585 = 3.199$ m.

220 kV double circuit

• Maximum system voltage 245 kV, so the phase voltage is 141.45 kV.
• $a = 6.5\times14.145+8 = 99.94$ inch $= 2.539$ m.
• Discs: 14, the standard 220 kV suspension string.
• $l = 14\times0.146+0.3 = 2.344$ m.
• $Y_{220} = 2.344+2.539 = 4.883$ m.

Ratio

$$\frac{Y_{132}}{Y_{220}} = \frac{3.199}{4.883} = 0.655$$

Answer: the vertical separation is about 3.2 m for 132 kV and 4.9 m for 220 kV, a ratio of about 0.66 : 1. The 132 kV spacing is roughly two-thirds of the 220 kV spacing, because both the clearance and the string length grow with voltage.