--- title: "Conductor selection, 100 MW 80 km, no corona" subject: "Transmission and Distribution System Design" chapter: 4 key: "semester_8/transmission_and_distribution_system_design:conductor-selection-100mw-80km-corona" source: https://www.bctengineer.com/q/semester_8/transmission_and_distribution_system_design/conductor-selection-100mw-80km-corona site: BCT Engineering --- # Conductor selection, 100 MW 80 km, no corona Transmission and Distribution System Design · Chapter 4: Conductor and Support Selection - Asked in: 2073 Magh, 10 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/conductor-selection-100mw-80km-corona - Whole chapter: https://www.bctengineer.com/md/semester_8/transmission_and_distribution_system_design/chapter-4 ## Question For a 3-phase 50 Hz transmission line to deliver 100 MW of power over 80 km, 132 kV single circuit line with following air clearances of phase conductors has been decided to design. Suggest the best suitable conductor which meets following technical criterion: (i) Satisfy the thermal limit (ii) Efficiency of line not to be less than 94% (iii) No corona in fair weather condition. [Figure: conductor arrangement — three phase conductors, top and bottom on the right and the middle one on the left; vertical spacing 2.5 m from top to middle level and 2.5 m from middle to bottom level; horizontal distance between the left (middle) conductor and the right conductors is 9 m.] [The conductor table referred to as attached is not reproduced with this paper.] ## Short answer ACSR "Goat" (30/7/3.71 mm, 400 mm²) meets all criteria. It carries 486 A within its 543 A rating, gives line efficiency of 95.2% (>94%), and has a critical disruptive voltage of 148 kV per phase against 83.7 kV operating, so no fair-weather corona. ## Answer The conductor is chosen as the smallest standard ACSR that passes all three checks. **Assumptions:** power factor 0.9 lagging; the standard ACSR table supplied with IOE papers (DC resistance at 20°C, current rating at 80°C conductor temperature); fair weather with $\delta = 1$ and irregularity factor $m = 0.85$; maximum system voltage for 132 kV is 145 kV. #### Spacing (GMD) from the figure The middle conductor is 9 m horizontally and 2.5 m vertically from the top and bottom conductors. The top and bottom conductors are 5 m apart vertically. $$ \begin{aligned} D_{12} &= D_{23} = \sqrt{9^2+2.5^2} = 9.341\ \text{m}, \quad D_{13} = 5\ \text{m}\\ GMD &= \sqrt[3]{9.341\times 9.341\times 5} = 7.584\ \text{m} \end{aligned} $$ #### (i) Thermal limit $$ I = \frac{P}{\sqrt3\, V\cos\phi} = \frac{100\times10^6}{\sqrt3\times132\times10^3\times0.9} = 485.99\ \text{A} $$ - Bear is rated 480 A, which is less than 486 A, so Bear is **not** acceptable. - Goat is rated 543 A, which is more than 486 A, so Goat passes. Goat is the smallest conductor that does. #### (ii) Efficiency of at least 94% $$ \begin{aligned} P_{loss,max} &= P\left(\frac{1}{0.94}-1\right) = 6.383\ \text{MW}\\ R_{max} &= \frac{6.383\times10^6}{3\times485.99^2} = 9.009\ \Omega \;\Rightarrow\; 0.1126\ \Omega/\text{km} \end{aligned} $$ | Conductor | R (Ω/km) | R for 80 km (Ω) | Loss (MW) | η (%) | |---|---|---|---|---| | Bear | 0.1093 | 8.744 | 6.196 | 94.17 | | Goat | 0.0891 | 7.128 | 5.051 | 95.19 | | Sheep | 0.0771 | 6.168 | 4.370 | 95.81 | Here $\eta = P/(P+P_{loss})$ and $P_{loss} = 3I^2R$. Goat meets the target with a margin. Even if its resistance is raised by 10% for temperature and skin effect, η is still about 94.7%. #### (iii) No corona in fair weather The disruptive critical voltage (phase, rms) for Goat, with $r = 25.97/2 = 1.2985$ cm, is: $$ \begin{aligned} V_c &= 21.1\, m\,\delta\, r \ln\frac{GMD}{r}\\ &= 21.1\times0.85\times1\times1.2985\times\ln\frac{758.4}{1.2985}\\ &= 148.3\ \text{kV (phase)} \end{aligned} $$ The highest operating phase voltage is $145/\sqrt3 = 83.7$ kV. Since 148.3 kV is greater than 83.7 kV, there is no corona in fair weather. This also holds for Bear, whose $V_c = 136.1$ kV. #### Selection | Check | Bear | Goat | |---|---|---| | Thermal (486 A) | Fails (480 A) | Passes (543 A) | | Efficiency ≥ 94% | 94.17% | 95.19% | | No corona | Passes | Passes | **Answer: ACSR "Goat" (30/7/3.71 mm, 400 mm² total area) is the most suitable conductor. It carries 486 A within its 543 A rating, gives an efficiency of 95.2%, and has a corona voltage of 148 kV per phase against 83.7 kV operating.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the GMD computed for this unsymmetrical arrangement? A. Arithmetic mean of spacings B. Cube root of D12 x D23 x D13 = 7.584 m C. The largest spacing D. The vertical spacing 2.5 m Answer: B. GMD is the cube root of the product of the three phase spacings (9.341, 9.341 and 5 m). (Key point: GMD from figure) 2. Why does Bear satisfy efficiency but not the thermal criterion? A. Efficiency 94.17% is above 94%, but its 480 A rating is below the 486 A load B. Its efficiency is 90% C. Its resistance is too high D. It corona-fails Answer: A. Bear passes efficiency and corona but fails the thermal limit, so Goat is the smallest valid conductor. (Key point: Efficiency check) 3. What condition ensures no corona in fair weather? A. Vc (phase) exceeds the highest operating phase voltage 145/sqrt(3) B. Vc is below operating voltage C. GMD is below 1 m D. δ = 0 Answer: A. Goat's Vc is 148.3 kV against 83.7 kV operating, so there is no corona. (Key point: Fair-weather corona criterion) 4. How is the maximum allowable loss for 94% efficiency found? A. P x (1/0.94 - 1) = 6.383 MW B. P x 0.06 = 6 MW exactly C. P x 0.94 D. P/0.06 Answer: A. η = P/(P + loss) so loss = P(1/η - 1). 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