---
title: "Insulator discs per string, 180 MW over 200 km"
subject: "Transmission and Distribution System Design"
chapter: 3
key: "semester_8/transmission_and_distribution_system_design:insulator-discs-180mw-200km"
source: https://www.bctengineer.com/q/semester_8/transmission_and_distribution_system_design/insulator-discs-180mw-200km
site: BCT Engineering
---

# Insulator discs per string, 180 MW over 200 km

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

- Asked in: 2075 Bhadra, 16 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/insulator-discs-180mw-200km
- Whole chapter: https://www.bctengineer.com/md/semester_8/transmission_and_distribution_system_design/chapter-3

## Question

In design of a transmission line for delivering power of 180 MW for a distance of 200 km. Considering that the system is to be operated at power factor of 0.85, find the required numbers of insulator disc per string to be selected.

[Given: Most economical voltage empirical formula: V = [Lt/1.6 + (P × 1000)/(cosφ × Nc × 150)]^0.5 kV, with Lt in km and P in MW. Standard voltages: 66 kV, 132 kV, 220 kV, 400 kV.] [In this paper's appendix the formula is printed without the 5.5 multiplier: V = [L/1.6 + (P × 1000)/(cosφ × Nc × 150)]^0.5.]

[Given Table A-1: Transmission line capability curve with assumption of single circuit transmission line surge impedance of 400 Ω (m.f. = power transmission capability/SIL):]

| Length (km) | Multiplying factor |
|---|---|
| 80 | 2.75 |
| 160 | 2.25 |
| 240 | 1.75 |
| 320 | 1.35 |
| 480 | 1.0 |
| 640 | 0.75 |

[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 (printed as 100) | 710 | 1665 |
| 20 | 1045 | 740 | 1745 |

[Other given data: Minimum air clearance: 6.5 inch per 10 kV (rms) and factor of safety = 8 inch. Maximum insulator string swing: 45°. Minimum ground clearance Hg = (V − 33)/33 + 17 feet, V is L-L voltage in kV.]

## Short answer

Using Still's formula (5.5 factor): 215.6 kV → select 220 kV single circuit. Required withstands: wet395 kV, impulse900 kV. From tables,10 discs of254×154 mm satisfy all criteria; add one spare →11 discs per suspension string.

## Answer

The number of discs depends on the line voltage. So the voltage level and number of circuits are fixed first, and then the string is sized against the power-frequency (wet and dry), impulse and switching-surge withstand levels from Tables A-2 and A-3.

#### Step 1: Most economical voltage

The appendix prints Still's formula without its multiplier. That form gives about 39 kV, which is clearly not a transmission voltage. So the standard form with the factor 5.5 is used:

$$
V = 5.5\sqrt{\frac{L}{1.6} + \frac{P\times 1000}{\cos\phi \times N_c \times 150}}\ \text{kV}
$$

With $L = 200$ km, $P = 180$ MW and $\cos\phi = 0.85$:

| Circuits $N_c$ | Value under root | $V$ (kV) | Nearest standard |
|---|---|---|---|
| 1 | $125 + 1411.76 = 1536.76$ | 215.6 | 220 kV |
| 2 | $125 + 705.88 = 830.88$ | 158.5 | 132 kV or 220 kV |

#### Step 2: Technical check with SIL (Table A-1)

$$
SIL = \frac{V^2}{Z_c} = \frac{V^2}{400}
$$

For 200 km, interpolate between 160 km (2.25) and 240 km (1.75):

$$
m.f. = 2.25 - \frac{200-160}{240-160}(2.25-1.75) = 2.0
$$

| Option | SIL (MW) | Capability = m.f. × SIL (MW) | ≥ 180 MW? |
|---|---|---|---|
| 132 kV, 1 circuit | 43.56 | 87.1 | No |
| 132 kV, 2 circuits | 2 × 43.56 | 174.2 | No |
| 220 kV, 1 circuit | 121.0 | 242.0 | Yes |
| 220 kV, 2 circuits | 2 × 121.0 | 484.0 | Yes (over-designed) |

**Selected: 220 kV, single circuit.** It is the economic voltage for $N_c = 1$ and it carries 180 MW with a margin.

#### Step 3: Withstand levels required (Table A-2)

For 220 kV nominal, the table row is the 255 kV maximum system voltage:

| Test | Required withstand (kV) |
|---|---|
| 1 min dry, power frequency | 435 |
| 1 min wet, power frequency | 395 |
| Impulse (BIL) | 900 |

The string's flashover voltage must be at least the required withstand.

#### Step 4: Number of discs for each criterion (Table A-3, 254 × 154 mm discs)

1. **Dry power frequency:** 435 kV needs FOV ≥ 435 kV. 7 discs give 435 kV, so **7 discs**.
2. **Wet power frequency:** 395 kV needs FOV ≥ 395 kV. 9 discs give only 370 kV and 10 discs give 410 kV, so **10 discs**.
3. **Impulse (lightning):** 900 kV needs FOV ≥ 900 kV. 9 discs give 860 kV and 10 discs give 945 kV, so **10 discs**.
4. **Switching surge:** take a switching surge ratio of 2.8 (usual for a 220 kV line without closing resistors).

$$
\begin{aligned} V_{ph,peak} &= \frac{\sqrt{2}\times 245}{\sqrt{3}} = 200.0\ \text{kV} \\ V_{ss} &= 2.8\times 200.0 = 560.1\ \text{kV (peak)} \\ \text{Equivalent rms} &= \frac{560.1}{\sqrt{2}} = 396.1\ \text{kV} \end{aligned}
$$

Comparing 396.1 kV with the wet FOV column gives 10 discs (410 kV).

| Criterion | Discs needed |
|---|---|
| Dry 1-min | 7 |
| Wet 1-min | 10 |
| Impulse | 10 |
| Switching surge | 10 |

#### Step 5: Final selection

- Governing requirement: **10 discs**, set by the wet, impulse and switching-surge criteria.
- Add **one disc** to allow for a cracked or punctured disc in service, which keeps the withstand level intact.

**Answer: 220 kV single-circuit line with 11 discs (254 × 154 mm) per suspension string, and 12 discs per tension string.** The extra disc in the tension string covers its heavier mechanical duty and the greater exposure of dead-end and angle positions. The suspension string length is about $11 \times 0.154 = 1.69$ m plus hardware.

Assumptions: standard atmospheric conditions (no altitude correction) and light pollution. For heavy pollution or high-altitude routes in Nepal, add discs to meet creepage and air-density corrections.

## Check yourself

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

1. Why is Still's formula used with the factor 5.5 here?
   A. It is needed only for double circuits
   B. Without it the formula gives about 39 kV, which is not a transmission voltage
   C. The 5.5 converts MW to kW
   D. It converts line to phase voltage
   Answer: B. The standard form V = 5.5 sqrt(...) gives about 215.6 kV for a single circuit, matching 220 kV. (Key point: Economic voltage)

2. How is the line capability checked against 180 MW at 200 km?
   A. Capability = V^2/400 only, with no multiplier
   B. Capability = m.f. x V^2/400, with m.f. = 2.0 interpolated between 160 and 240 km
   C. Capability = m.f. x V, with m.f. = 1.0
   D. Capability = P / cos(phi)
   Answer: B. For 220 kV: SIL = 121 MW, times 2.0 gives 242 MW, which is at least 180 MW; 132 kV single circuit gives only 87 MW. (Key point: SIL check)

3. Dry, wet, impulse and switching criteria need 7, 10, 10 and 10 discs. What is the final suspension string?
   A. 10 discs
   B. 7 discs
   C. 11 discs: governing 10 plus one spare for a defective disc
   D. 27 discs
   Answer: C. The governing requirement is 10 discs; one extra is added for a cracked or punctured disc. (Key point: Governing criterion)

4. Why does the tension string get one more disc (12) than the suspension string?
   A. To reduce string length
   B. Heavier mechanical duty and greater exposure at dead-end and angle positions
   C. Because tension strings are vertical
   D. Because it has a lower voltage
   Answer: B. Tension strings carry the full conductor tension and are more exposed. (Key point: Tension string)

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