---
title: "Air clearances for 220 kV double circuit line"
subject: "Transmission and Distribution System Design"
chapter: 4
key: "semester_8/transmission_and_distribution_system_design:air-clearance-220kv-double-circuit"
source: https://www.bctengineer.com/q/semester_8/transmission_and_distribution_system_design/air-clearance-220kv-double-circuit
site: BCT Engineering
---

# Air clearances for 220 kV double circuit line

Transmission and Distribution System Design · Chapter 4: Conductor and Support Selection

- Asked in: 2071 Bhadra, 8 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_8/transmission_and_distribution_system_design/air-clearance-220kv-double-circuit
- Whole chapter: https://www.bctengineer.com/md/semester_8/transmission_and_distribution_system_design/chapter-4

## Question

For a double circuit, 220 kV transmission line, find the following air clearance with justifications. i) Cross arm length ii) Insulator string length iii) Horizontal and vertical separation of conductor iv) Height of earth wire from top most conductor

## Short answer

Cross arm length 4.20 m, insulator string length 2.34 m (14 discs), vertical conductor separation 4.88 m, horizontal separation between circuits 9.89 m, and earth wire height above top conductor 8.57 m, based on 45° swing and 2.54 m minimum air clearance.

## Answer

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.

```text
              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
```

| Item | Value |
|---|---|
| Minimum air clearance | 2.54 m |
| Insulator string length | 2.34 m (14 discs) |
| Cross arm length | 4.20 m |
| Vertical separation | 4.88 m |
| Horizontal separation (circuit to circuit) | 9.89 m |
| Earth wire above top conductor | 8.57 m |

**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.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. What is the minimum air clearance a for 220 kV in this answer?
   A. About 1.59 m
   B. About 4.2 m
   C. About 2.54 m (99.94 inch)
   D. About 0.99 m
   Answer: C. a = 6.5 x 14.145 + 8 = 99.94 inch = 2.54 m. (Key point: Air clearance)

2. Why is a 14-disc string used though tables need only 10?
   A. Because each disc withstands only 10 kV
   B. Because the swing angle needs it
   C. Because the cross arm length requires it
   D. Extra discs for switching surges, pollution and one defective disc
   Answer: D. Withstand tables need 10 discs; margins give 13-14. l = 14 x 0.146 + 0.3 = 2.34 m. (Key point: String length)

3. How is the cross arm length found?
   A. X = a + l sin(theta) so the swung conductor keeps clearance from the tower body
   B. X = a + l cos(theta)
   C. X = l + a
   D. X = a only
   Answer: A. At 45 deg swing the conductor moves out by l sin45, so X = 2.54 + 1.66 = 4.20 m. (Key point: Cross arm)

4. How is the earth wire height above the top conductor determined?
   A. he = l + a
   B. From the 30 deg shielding angle: he = (X + w/2)/tan30
   C. he = (X + w/2) x tan30
   D. he = Y x 2
   Answer: B. Horizontal offset 4.946 m divided by tan30 gives 8.57 m. (Key point: Earth wire height)

## Related in Transmission and Distribution System Design

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- [Air clearances, 220 kV line for 200 MW](https://www.bctengineer.com/question.md?key=semester_8%2Ftransmission_and_distribution_system_design%3Aair-clearances-220kv-200mw-120km) — Transmission and Distribution System Design, chapter 4
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## Related in other subjects

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