---
title: "Air clearances, 220 kV line for 200 MW"
subject: "Transmission and Distribution System Design"
chapter: 4
key: "semester_8/transmission_and_distribution_system_design:air-clearances-220kv-200mw-120km"
source: https://www.bctengineer.com/q/semester_8/transmission_and_distribution_system_design/air-clearances-220kv-200mw-120km
site: BCT Engineering
---

# Air clearances, 220 kV line for 200 MW

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

- Asked in: 2067 Chaitra (old course), 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-clearances-220kv-200mw-120km
- Whole chapter: https://www.bctengineer.com/md/semester_8/transmission_and_distribution_system_design/chapter-4

## Question

For a transmission line to deliver 200 MW of power over 120 km, it has been decided to design a 220 kV transmission line of single circuit. Find the following air clearances with justifications. i) Cross arm length ii) Insulator string length iii) Horizontal and vertical separation of conductor iv) Height of earth wire from the top most conductor.

## Short answer

Cross arm length 4.20 m, insulator string 2.34 m (14 discs), vertical separation 4.88 m, horizontal separation 9.89 m, height of earth wire above top conductor 8.57 m. Based on 220 kV and a minimum air clearance of 2.54 m.

## Answer

The air clearances fix the tower head dimensions. They depend on the voltage, not on the power or the length; those decide the conductor. The 200 MW, 120 km data are used later for conductor selection.

**Data and assumptions:** 220 kV nominal, maximum system voltage 245 kV (phase rms 141.45 kV); 254 × 146 mm discs; maximum string swing 45°; minimum air clearance 6.5 inch per 10 kV (phase rms) plus 8 inch safety; tower body width 1.5 m; single circuit in a triangular (offset vertical) layout, with two phases on one side and one on the other.

#### Minimum air clearance
$$
a = 6.5\times\frac{141.45}{10} + 8 = 99.94\ \text{in} = 2.54\ \text{m}
$$

#### ii) Insulator string length
- **14 discs**, the standard 220 kV suspension string. The table check (1-min wet withstand 395 kV and impulse 900 kV) needs 10 discs; extra discs are added for switching surges, pollution and one defective disc.
- Length:
$$
l = 14\times0.146 + 0.3 = 2.34\ \text{m}
$$

#### i) Cross arm length
With the string swung 45°, the conductor must still be at least $a$ from the tower body:

$$
X = a + l\sin45^\circ = 2.54 + 2.344\times0.707 = 4.20\ \text{m}
$$

#### iii) Separation of conductors
- **Vertical:** the conductor below one cross arm must clear the next cross arm by $a$:
$$
Y = l + a = 2.344 + 2.539 = 4.88\ \text{m}
$$
- **Horizontal** (between the phases on opposite sides of the tower):
$$
S_h = 2X + w = 2(4.196) + 1.5 = 9.89\ \text{m}
$$
- The resulting phase distances are $\sqrt{9.89^2 + 4.88^2} = 11.03$ m between adjacent phases and $2Y = 9.77$ m between the top and bottom phases. The GMD is 10.59 m.

#### iv) Height of earth wire above the top conductor
Use one earth wire at the peak with a shielding angle of at most 30°:

$$
h_e = \frac{X + w/2}{\tan30^\circ} = \frac{4.946}{0.5774} = 8.57\ \text{m}
$$

This is about 6.2 m above the top cross arm.

```text
            E   (1 earth wire)
            |\  30 deg
            | \
       o----+--*  top phase (A)
  (B)  *----+     Y = 4.88 m
            +--*  bottom phase (C)
            |  X = 4.2 m
   <------ 9.89 m ------>
```

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

## Check yourself

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

1. What determines the air clearances at the tower head?
   A. Power and line length
   B. Conductor weight
   C. The voltage level
   D. Soil type
   Answer: C. Clearances depend on the voltage; power and length are used for conductor selection. (Key point: What fixes clearances)

2. How is the string length found for 220 kV?
   A. l = 10 x 0.146 m
   B. l = 14 x 0.146 + 0.3 = 2.34 m, using 14 discs
   C. l = 2 x air clearance
   D. l = 14 x 0.254
   Answer: B. The standard 220 kV string has 14 discs of 146 mm spacing plus hardware 0.3 m. (Key point: Insulator string)

3. How is the cross arm length found?
   A. X = a + l sin45° with the string swung 45°
   B. X = l + a
   C. X = 2a
   D. X = a - l sin45°
   Answer: A. With the string swung 45°, the conductor must still be at least a from the tower body. (Key point: Cross arm length)

4. What gives the vertical separation Y of conductors?
   A. Y = X + w
   B. Y = l + a, so the lower conductor clears the next cross arm by a
   C. Y = a only
   D. Y = l/2
   Answer: B. Y = l + a = 4.88 m. (Key point: Vertical separation)

## Related in Transmission and Distribution System Design

- [Initial line design for 200 MW over 120 km](https://www.bctengineer.com/question.md?key=semester_8%2Ftransmission_and_distribution_system_design%3Aline-design-200mw-120km) — Transmission and Distribution System Design, chapter 2
- [Conductor selection, 220 kV line for 200 MW](https://www.bctengineer.com/question.md?key=semester_8%2Ftransmission_and_distribution_system_design%3Aconductor-selection-220kv-200mw-120km) — Transmission and Distribution System Design, chapter 4
- [Air clearances for 220 kV single circuit line](https://www.bctengineer.com/question.md?key=semester_8%2Ftransmission_and_distribution_system_design%3Aair-clearances-220kv-single-circuit) — Transmission and Distribution System Design, chapter 4
- [Air clearances for 220 kV double circuit line](https://www.bctengineer.com/question.md?key=semester_8%2Ftransmission_and_distribution_system_design%3Aair-clearance-220kv-double-circuit) — Transmission and Distribution System Design, chapter 4
- [Air clearances for 132 kV double circuit line](https://www.bctengineer.com/question.md?key=semester_8%2Ftransmission_and_distribution_system_design%3Aair-clearance-132kv-double-circuit) — Transmission and Distribution System Design, chapter 4
- [Tower dimensions for 220 kV single circuit line](https://www.bctengineer.com/question.md?key=semester_8%2Ftransmission_and_distribution_system_design%3Atower-dimensions-220kv) — Transmission and Distribution System Design, chapter 4, asked 2×

## Related in other subjects

- [220 kV line, 200 MW: sending end from ABCD](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Aabcd-220kv-200mw) — Power System Analysis I, chapter 6
- [Capacitor for zero regulation, 70 km 220 kV line](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3A70km-220kv-zero-regulation-capacitor) — Power System Analysis I, chapter 6
- [450 km 400 kV line, 420 MW, 0.95 pf](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3A450km-400kv-420mw-095) — Power System Analysis I, chapter 5
- [450 km 400 kV line, 420 MW, 0.9 pf](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3A450km-400kv-420mw-090) — Power System Analysis I, chapter 5