---
title: "Initial line design for 200 MW over 120 km"
subject: "Transmission and Distribution System Design"
chapter: 2
key: "semester_8/transmission_and_distribution_system_design:line-design-200mw-120km"
source: https://www.bctengineer.com/q/semester_8/transmission_and_distribution_system_design/line-design-200mw-120km
site: BCT Engineering
---

# Initial line design for 200 MW over 120 km

Transmission and Distribution System Design · Chapter 2: Transmission Voltage Level and Number of Circuits

- Asked in: 2067 Mangsir (old course), 6+6+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/line-design-200mw-120km
- Whole chapter: https://www.bctengineer.com/md/semester_8/transmission_and_distribution_system_design/chapter-2

## Question

For a transmission line to deliver a power of 200 MW over a distance of 120 km carryout the following line design initial steps: a. Find the most economical voltage, check the design capability curve and hence suggest the standard voltage and no. of circuit. b. Determine the no of insulator discs. c. 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.

[Given: Most economical voltage empirical formula: V = 5.5 × [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.]

[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 | 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°. Toughest condition: 0°C, 100 kg/m² wind, no ice. Easiest condition: 60°C, no wind. Stringing condition: 27°C, no wind. Minimum ground clearance: (V − 33)/33 + 17 ft, where V is L-L maximum system voltage.]

## Short answer

Economical voltage ~217 kV, select standard 220 kV single circuit. Insulator string requires 15 discs, length 2.31 m. Cross-arm length is 4.27 m, vertical conductor separation 4.94 m, horizontal separation 8.53 m, and earth wire height above top conductor 7.39 m.

## Answer

Given: $P = 200$ MW, $L = 120$ km. Assumed $\cos\phi = 0.9$, $Z_c = 400\ \Omega$, factors FWR = 1.15, NACF = 1.1, FOS = 1.1 (usual design values), disc spacing 154 mm. Maximum system voltage of the 220 kV class = 255 kV (Table A-2).

#### a. Economical voltage, capability, standard voltage and circuits
$$
V = 5.5\sqrt{\frac{L}{1.6} + \frac{1000P}{\cos\phi\,N_c \times 150}}
$$
$N_c = 1$: $V = 5.5\sqrt{75 + 1481.48} = 5.5 \times 39.452 = 216.99$ kV
$N_c = 2$: $V = 5.5\sqrt{75 + 740.74} = 5.5 \times 28.561 = 157.09$ kV

m.f. at 120 km $= 2.75 - \frac{40}{80}(0.5) = 2.50$.

| Option | SIL (MW) | Capability (MW) | Remark |
|---|---|---|---|
| 132 kV D/C | 43.56 | 217.80 | Only 8.9% margin; 108.9 MW on N-1 |
| 220 kV S/C | 121 | 302.50 | Adequate, 51% margin |
| 220 kV D/C | 121 | 605.00 | Over-design |

$V_{econ}$ for S/C (217 kV) almost equals 220 kV. **Selected: 220 kV, single circuit** (vertical/triangular formation, one earth wire).

#### b. Number of insulator discs
$k = 1.15 \times 1.1 \times 1.1 = 1.3915$

| Over-voltage | Required FOV | Discs from Table A-3 |
|---|---|---|
| Lightning: $900 \times k$ | 1252.35 kV (impulse) | 14 (1265 kV) |
| Power freq. wet: $395 \times k$ | 549.64 kV | 14 (555 kV) |
| Power freq. dry: $435 \times k$ | 605.30 kV | 11 (635 kV) |
| Switching, SSR = 2.8 | 573.62 kV rms equiv. | 15 (590 kV wet) |

Switching: $V_{ss} = 2.8 \times \sqrt{2} \times 255/\sqrt{3} = 582.98$ kV peak; $\times k = 811.21$ kV peak $= 573.62$ kV rms. This exceeds the 14-disc wet FOV (555 kV), so 15 discs are needed.

**Number of discs = 15** per suspension string (switching surge governs).

#### c. Air clearances
**Minimum air clearance** (6.5 in per 10 kV of maximum phase rms voltage + 8 in):
$$
\begin{aligned} V_{ph} &= \frac{255}{\sqrt{3}} = 147.22\ \text{kV} \\ a &= 6.5 \times \frac{147.22}{10} + 8 = 103.70\ \text{in} = 2.63\ \text{m} \end{aligned}
$$

**ii. Insulator string length** (15 discs × 154 mm, fittings neglected):
$$
l = 15 \times 0.154 = 2.31\ \text{m}
$$

**i. Cross-arm length.** With a 45° string swing, the conductor moves $l\sin45°$ toward the tower and must still be $a$ away from the tower body:
$$
CA = l\sin45° + a = 2.31 \times 0.7071 + 2.634 = 1.633 + 2.634 = 4.27\ \text{m}
$$

**iii. Separation of conductors**
- Vertical separation between cross-arms: the conductor hangs $l$ below its cross-arm and must be $a$ above the cross-arm below:
$$
d_v = l + a = 2.31 + 2.634 = 4.94\ \text{m}
$$
- Horizontal separation between conductors on opposite sides of the tower (tower body width neglected):
$$
d_h = 2 \times CA = 2 \times 4.267 = 8.53\ \text{m}
$$

**iv. Height of earth wire above the top conductor.** With a single earth wire on the tower axis and a shielding angle of 30°:
$$
h = \frac{CA}{\tan30°} = 4.267 \times 1.732 = 7.39\ \text{m}
$$

```text
          o  earth wire
          |\
          | \ 30 deg
     7.39 m  \
          |   \
   -------+----o  top phase
   4.27 m |    | string 2.31 m
          |
   4.94 m |
   o------+------o  lower phases
     8.53 m apart
          |
       tower
```

(Also: minimum ground clearance $= (255-33)/33 + 17 = 23.73$ ft $= 7.23$ m.)

(If the IEC value $V_{max} = 245$ kV were used instead, the switching requirement would be 551.12 kV rms, 14 discs would just suffice, and the clearances would be slightly smaller: $a = 2.54$ m, $l = 2.16$ m, $CA = 4.06$ m.)

**Summary:** 220 kV S/C; 15 discs; string length 2.31 m; cross-arm 4.27 m; vertical spacing 4.94 m; horizontal spacing 8.53 m; earth wire 7.39 m above the top conductor.

## Check yourself

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

1. Why is 220 kV single circuit selected for 200 MW over 120 km?
   A. V_econ (217 kV) is about 220 kV and capability is 302.5 MW
   B. 132 kV D/C has 100 percent margin
   C. 400 kV is the only adequate level
   D. 220 kV S/C carries only 121 MW
   Answer: A. V_econ for S/C is 217 kV, and 2.5 x 121 = 302.5 MW exceeds 200 MW with a 51% margin. (Key point: Selection)

2. Which over-voltage conditions decide 14 discs, and how is the factor k found?
   A. Only dry power frequency; k = 1
   B. Lightning, wet power-frequency and switching all give 14; k = FWR x NACF x FOS = 1.3915
   C. Only switching; k = 2.8
   D. Only dry; k = 1.15
   Answer: B. Lightning (1252 kV), wet (550 kV) and switching (551 kV rms) all give 14 discs. Dry gives only 11. (Key point: Disc number)

3. How is the minimum air clearance a computed?
   A. 6.5 in per 10 kV of phase rms voltage plus 8 in (2.54 m for 245 kV)
   B. 6.5 in per 10 kV of line voltage only
   C. String length divided by 2
   D. Ground clearance minus string length
   Answer: A. a = 6.5 x 141.45/10 + 8 = 99.94 in = 2.54 m. (Key point: Minimum air clearance)

4. How is the cross-arm length obtained?
   A. l cos45 + a
   B. l sin45 + a (string swing plus air clearance)
   C. 2 x a
   D. l + a
   Answer: B. At 45 degree swing the conductor moves l sin45 toward the tower and must still keep clearance a: 1.52 + 2.54 = 4.06 m. (Key point: Cross-arm length)

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