---
title: "Number of insulator discs for 220 kV line"
subject: "Transmission and Distribution System Design"
chapter: 3
key: "semester_8/transmission_and_distribution_system_design:insulator-discs-220kv-appendix"
source: https://www.bctengineer.com/q/semester_8/transmission_and_distribution_system_design/insulator-discs-220kv-appendix
site: BCT Engineering
---

# Number of insulator discs for 220 kV line

Transmission and Distribution System Design · Chapter 3: Overhead Line Insulator Design

- Asked in: 2080 Chaitra, 8 marks
- Video: [Number of disc insulator ll calculation ll](https://www.youtube.com/watch?v=FldL8adG_gI) — ELECTRICAL ENGINEER
- 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/insulator-discs-220kv-appendix
- Whole chapter: https://www.bctengineer.com/md/semester_8/transmission_and_distribution_system_design/chapter-3

## Question

Compute the number of insulator discs required for a single circuit, 220 kV, 50 Hz overhead transmission line considering the various external and internal over voltages. Use the related data given in the appendix.

[Given Table A-2: Withstand voltage capability for different system voltages:]

| Maximum system voltage (kV) | 1 minute dry withstand (kV) | 1 minute wet withstand (kV) | Impulse withstand (kV) |
|---|---|---|---|
| 123 | 215 | 185 | 450 |
| 145 | 265 | 230 | 550 |
| 255 | 435 | 395 | 900 |
| 420 | 760 | 680 | 1550 |

[Given Table A-3: Flashover voltages for 254 × 154 mm disc insulators:]

| No. of discs | 1 minute dry FOV (kV) | 1 minute wet FOV (kV) | Impulse FOV (kV) |
|---|---|---|---|
| 1 | 80 | 50 | 150 |
| 2 | 155 | 90 | 255 |
| 3 | 215 | 130 | 355 |
| 4 | 270 | 170 | 440 |
| 5 | 325 | 210 | 525 |
| 6 | 380 | 250 | 610 |
| 7 | 435 | 290 | 695 |
| 8 | 485 | 330 | 780 |
| 9 | 535 | 370 | 860 |
| 10 | 585 | 410 | 945 |
| 11 | 635 | 450 | 1025 |
| 12 | 685 | 485 | 1105 |
| 13 | 730 | 520 | 1185 |
| 14 | 775 | 555 | 1265 |
| 15 | 820 | 590 | 1345 |
| 16 | 865 | 620 | 1425 |
| 17 | 910 | 650 | 1505 |
| 18 | 955 | 680 | 1585 |
| 19 | 1000 | 710 | 1665 |
| 20 | 1045 | 740 | 1745 |

[Minimum air clearance: 6.5 inch per 10 kV (rms) and factor of safety = 8 inch.]

## Short answer

After applying correction factors (FWR=1.15, NACF=1.1, FOS=1.1) to the required withstands and using the given disc flashover data, lightning requires 14 discs, power-frequency wet requires 14, and switching surge requires 15 discs. The governing number is 15 discs per suspension string (string length about 2.31 m).

## Answer

Data: 220 kV, single circuit, 50 Hz. From Table A-2 (220 kV class row): maximum system voltage 255 kV, 1-min dry withstand 435 kV, 1-min wet withstand 395 kV, impulse withstand 900 kV. Usual design factors assumed: flashover withstand ratio (FWR) = 1.15, non-atmospheric condition factor (NACF) = 1.1, factor of safety (FOS) = 1.1, so

$$
k = 1.15 \times 1.1 \times 1.1 = 1.3915
$$

#### 1. External over-voltage: lightning
$$
V_{req} = 900 \times 1.3915 = 1252.35\ \text{kV}
$$
Table A-3 (impulse FOV): 13 discs = 1185 kV < 1252.35; **14 discs = 1265 kV**. → 14 discs

#### 2. Internal over-voltage: switching surge
Assume switching surge ratio SSR = 2.8 (usual for 220 kV).
$$
\begin{aligned} V_{ss} &= SSR \times \frac{\sqrt{2}\,V_{max}}{\sqrt{3}} = 2.8 \times \frac{\sqrt{2} \times 255}{\sqrt{3}} = 582.98\ \text{kV (peak)} \\ V_{req} &= 582.98 \times 1.3915 = 811.21\ \text{kV (peak)} \\ &= 811.21/\sqrt{2} = 573.62\ \text{kV (rms)} \end{aligned}
$$
Compared with 1-min wet FOV: 14 discs = 555 kV < 573.62, **15 discs = 590 kV**. → 15 discs

#### 3. Power-frequency over-voltage
Wet: $395 \times 1.3915 = 549.64$ kV → wet FOV 14 discs = 555 kV → **14 discs**
Dry: $435 \times 1.3915 = 605.30$ kV → dry FOV 11 discs = 635 kV → 11 discs

#### Result
| Condition | Required (kV) | Discs |
|---|---|---|
| Lightning impulse | 1252.35 | 14 |
| Switching surge | 573.62 (rms eq.) | 15 |
| Power freq., wet | 549.64 | 14 |
| Power freq., dry | 605.30 | 11 |

The largest requirement governs: **15 discs (254 × 154 mm) per suspension string** (switching surge governs). String length $\approx 15 \times 0.154 = 2.31$ m. (Tension strings commonly use one extra disc.)

(If the IEC value $V_{max} = 245$ kV were used instead, the switching requirement would be 551.12 kV rms and 14 discs would just suffice.)

Check of air clearance: $a = 6.5 \times \frac{255/\sqrt{3}}{10} + 8 = 103.70$ in $\approx 2.63$ m.

## Check yourself

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

1. How is the overall multiplying factor k obtained for the 220 kV disc calculation?
   A. k = 1.1 x 1.1 = 1.21 only (safety factor squared)
   B. k = 1.15 + 1.1 + 1.1 = 3.35
   C. k = 1.15 x 1.1 x 1.1 = 1.3915 (FWR x NACF x FOS)
   D. k = sqrt(2) x 1.15 x 1.1 = 1.789
   Answer: C. The required withstand is multiplied by the product of flashover withstand ratio, non-atmospheric condition factor and factor of safety. (Key point: Overall design factor)

2. The required impulse level is 900 x 1.3915 = 1252.35 kV. Why are 14 discs needed rather than 13?
   A. 13 discs give 1265 kV, which is too high
   B. The impulse FOV is compared with the dry withstand value
   C. Impulse needs one disc per 90 kV exactly
   D. 13 discs give 1185 kV, which is below 1252.35 kV, while 14 give 1265 kV
   Answer: D. The string impulse FOV must be at least the required value; 1185 kV falls short, 1265 kV is the first that satisfies it. (Key point: Lightning impulse discs)

3. In the switching surge check, how is the peak requirement compared with the table?
   A. Compare the peak directly with the impulse FOV
   B. Divide the peak by sqrt(2) to get an rms equivalent and compare with the 1-min wet FOV
   C. Multiply by sqrt(3) and compare with the wet FOV
   D. Compare the peak directly with the dry FOV
   Answer: B. The 779.4 kV peak is divided by sqrt(2) to get 551.12 kV rms and compared with the wet FOV column, giving 14 discs (555 kV). (Key point: Switching surge conversion)

4. The requirements are 14 (impulse), 14 (switching), 14 (wet) and 11 (dry) discs. What is the design number per suspension string?
   A. 53 discs, the sum of all
   B. The average, about 13 discs
   C. 14 discs, because the largest requirement governs
   D. 11 discs, the smallest value
   Answer: C. The string must withstand all stresses, so the maximum number from all criteria is chosen. (Key point: Final selection)

## Related in Transmission and Distribution System Design

- [Insulator discs per string, 180 MW over 200 km](https://www.bctengineer.com/question.md?key=semester_8%2Ftransmission_and_distribution_system_design%3Ainsulator-discs-180mw-200km) — Transmission and Distribution System Design, chapter 3
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