--- title: "Air clearances for 132 kV double circuit line" subject: "Transmission and Distribution System Design" chapter: 4 key: "semester_8/transmission_and_distribution_system_design:air-clearance-132kv-double-circuit" source: https://www.bctengineer.com/q/semester_8/transmission_and_distribution_system_design/air-clearance-132kv-double-circuit site: BCT Engineering --- # Air clearances for 132 kV double circuit line Transmission and Distribution System Design · Chapter 4: Conductor and Support Selection - Asked in: 2074 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-132kv-double-circuit - Whole chapter: https://www.bctengineer.com/md/semester_8/transmission_and_distribution_system_design/chapter-4 ## Question For a double circuit, 132 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. [Given in appendix: Minimum air clearance: I. 1 cm for 1 kV (maximum per phase peak) and factor of safety is 30 cm; II. 6.5 inch per 10 kV (maximum per phase rms) and factor of safety is 8 inch.] ## Short answer Air clearance is 1.6 m (based on 145 kV peak/rms criteria). Insulator string length is 1.4 m (7 discs). Cross-arm length is 2.6 m (clearance plus swing). Vertical spacing is 3.0 m, horizontal spacing is 6.4 m. Earth wire height above top conductor is 5.6 m for a 30° shielding angle. ## Answer The dimensions follow from the minimum phase-to-earth air clearance, the insulator string length and the insulator swing. **Assumptions:** maximum system voltage 145 kV; swing 45°; disc 254 × 154 mm; tower width at cross-arm level 1.2 m. #### Minimum air clearance $$ \begin{aligned} \text{I:}\quad & V_{ph,peak} = \frac{145\sqrt{2}}{\sqrt{3}} = 118.4\ \text{kV} \Rightarrow a = 118.4 + 30 = 148.4\ \text{cm} = 1.48\ \text{m} \\ \text{II:}\quad & V_{ph,rms} = \frac{145}{\sqrt{3}} = 83.72\ \text{kV} \Rightarrow a = 6.5 \times 8.372 + 8 = 62.4\ \text{in} = 1.585\ \text{m} \end{aligned} $$ Take the larger value: **a ≈ 1.6 m**. #### ii) Insulator string length For 145 kV, Table A-3 gives: - wet withstand 230 kV needs 6 discs (250 kV), - impulse 550 kV needs 6 discs (610 kV), - dry withstand 265 kV needs 4 discs. Adding one spare disc gives **7 discs**: $$ l = 7 \times 0.154 + 0.30\ (\text{hardware}) = 1.38 \approx 1.4\ \text{m} $$ Common practice uses 9 discs (about 1.6 m) where pollution is a concern. #### i) Cross-arm length $$ L_{ca} = a + l\sin45° = 1.6 + 1.4 \times 0.707 = 2.59 \approx 2.6\ \text{m} $$ This keeps clearance $a$ from the tower body when the string swings by 45°. #### iii) Separation of conductors - **Vertical** (between cross-arms on the same side): $$ V = l + a = 1.4 + 1.6 = 3.0\ \text{m} $$ The hanging conductor keeps clearance $a$ from the cross-arm below. - **Horizontal** (between the two circuits across the tower): $$ H = 2L_{ca} + w = 2 \times 2.6 + 1.2 = 6.4\ \text{m} $$ This is well above the phase-to-phase clearance of about $1.15a = 1.8$ m. ```text ew (earth wire) | o---------+---------o top \ | | 3.0 m o---------+---------o middle / | o---------+---------o bottom |<-2.6->|1.2|<-2.6->| ``` #### iv) Height of earth wire above the top conductor One earth wire at the peak is enough for a 132 kV double-circuit vertical tower. For a 30° shielding angle, the horizontal distance to the outer conductor is $2.6 + 0.6 = 3.2$ m: $$ y = \frac{3.2}{\tan 30°} = 5.54 \approx 5.6\ \text{m} $$ | Item | Value | |---|---| | Air clearance | 1.6 m | | Cross-arm length | 2.6 m | | String length | 1.4 m (7 discs) | | Vertical / horizontal spacing | 3.0 m / 6.4 m | | Earth wire above top conductor | 5.6 m | ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the final phase-to-earth air clearance of about 1.6 m chosen? A. Average of the two methods B. The larger of the two methods (1.48 m and 1.585 m) C. The smaller of the two methods D. Fixed by insulator string length Answer: B. Taking the larger value is the safer design, giving a of about 1.6 m. (Key point: Minimum air clearance) 2. What does the cross-arm length account for? A. L = a + l sin45°, keeping clearance a from the tower when the string swings 45° B. L = a only C. L = l + a D. L = 2a + w Answer: A. The swung string moves the conductor inward by l sin(swing), so the arm must be longer by that amount. (Key point: Cross-arm length) 3. How is vertical separation between conductors found? A. V = l + a, string length plus air clearance to the cross-arm below B. V = 2a C. V = 1.15a D. V = 2 L_ca + w Answer: A. The hanging conductor must keep clearance a from the cross-arm below, so V = 1.4 + 1.6 = 3.0 m. (Key point: Vertical separation) 4. How is the height of the earth wire above the top conductor obtained? A. Horizontal distance to the outer conductor divided by tan of shielding angle B. Horizontal distance times tan of shielding angle C. Equal to cross-arm length D. Equal to string length Answer: A. y = 3.2/tan30° = 5.54 m, so the shielding angle protects the outer conductor. (Key point: Earth wire height) ## Related in Transmission and Distribution System Design - [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, 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 - [Tower dimensions for 220 kV double circuit line](https://www.bctengineer.com/question.md?key=semester_8%2Ftransmission_and_distribution_system_design%3Atower-dimensions-220kv-double-circuit) — Transmission and Distribution System Design, chapter 4, asked 2× - [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 - [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× - [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 ## Related in other subjects - [True/false: cross-arm length and string efficiency](https://www.bctengineer.com/question.md?key=semester_8%2Fhigh_voltage_engineering%3Atf-cross-arm-string-efficiency) — High Voltage Engineering, chapter 5 - [132 kV horizontal line: susceptance and charging](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Ahorizontal-132kv-4-5m-susceptance) — Power System Analysis I, chapter 4 - [Charging current of 60 Hz horizontal 132 kV line](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Ahorizontal-60hz-132kv-charging) — Power System Analysis I, chapter 4 - [Inductive reactance of vertical 132 kV line](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Avertical-132kv-12mm-5m) — Power System Analysis I, chapter 4