--- title: "Most economical conductor, five options (Rs 9)" subject: "Transmission and Distribution System Design" chapter: 4 key: "semester_8/transmission_and_distribution_system_design:economic-conductor-five-options-kwh" source: https://www.bctengineer.com/q/semester_8/transmission_and_distribution_system_design/economic-conductor-five-options-kwh site: BCT Engineering --- # Most economical conductor, five options (Rs 9) Transmission and Distribution System Design · Chapter 4: Conductor and Support Selection - Asked in: 2071 Magh, 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/economic-conductor-five-options-kwh - Whole chapter: https://www.bctengineer.com/md/semester_8/transmission_and_distribution_system_design/chapter-4 ## Question For transmission line design following computations has been made. Select the most economical conductor. | Conductor | Total tower cost/km (1000 Rs) | Total conductor cost/km (1000 Rs) | Annual energy loss/km (1000 kWh) | |---|---|---|---| | Goat | 890 | 1254 | 32 | | Sheep | 936 | 1450 | 27 | | Zebra | 974 | 1585 | 24 | | Deer | 964 | 1663 | 24 | | Elk | 968 | 1847 | 21 | Using the following data choose the best conductor: Energy rate = Rs. 9.0 per unit, Interest rate = 10%, Project life = 20 years. ## Short answer The most economical conductor is Zebra, with a minimum total annual cost of Rs 516.58 thousand per km, computed as annual capital charge plus annual loss cost (energy lost times rate). ## Answer The most economical conductor is the one with the **minimum total annual cost per km**: the annual charge on capital (towers plus conductor) plus the annual cost of energy lost (Kelvin's law idea). #### Capital recovery factor $$ CRF = \frac{i(1+i)^n}{(1+i)^n-1} = \frac{0.1(1.1)^{20}}{(1.1)^{20}-1} = \frac{0.1\times6.7275}{5.7275} = 0.11746 $$ The annual capital charge is (tower cost + conductor cost) × 0.11746. The annual loss cost is the energy lost per year × the energy rate. The annual loss cost is (energy lost in 1000 kWh) × Rs 9.0, in 1000 Rs. #### Annual cost comparison | Conductor | Capital (1000 Rs/km) | Annual capital charge (1000 Rs) | Annual energy loss cost (1000 Rs) | Total annual cost (1000 Rs/km) | |---|---|---|---|---| | Goat | 2144 | 251.83 | 32 × 9.0 = 288.00 | 539.83 | | Sheep | 2386 | 280.26 | 27 × 9.0 = 243.00 | 523.26 | | Zebra | 2559 | 300.58 | 24 × 9.0 = 216.00 | 516.58 | | Deer | 2627 | 308.57 | 24 × 9.0 = 216.00 | 524.57 | | Elk | 2815 | 330.65 | 21 × 9.0 = 189.00 | 519.65 | Sample (Zebra): $2559\times0.11746 = 300.58$ and $24\times9.0 = 216.00$, total 516.58 thousand Rs/km per year. Deer has the same loss as Zebra but costs more, so it is ruled out. Elk has the lowest loss, but its extra capital charge (about Rs 30 thousand per year) exceeds its loss saving (Rs 27 thousand per year). Zebra gives the best balance. **Answer: choose Zebra, with a minimum total annual cost of Rs 516.58 thousand per km.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is capital included in the annual cost? A. Conductor cost only x CRF B. (Tower + conductor cost) / 20 C. (Tower + conductor cost) x CRF 0.11746 D. (Tower + conductor cost) x 0.10 Answer: C. Capital is spread as an annuity over life at the interest rate, using CRF. (Key point: Annual charge) 2. How is the annual loss cost computed? A. Energy lost x CRF B. Energy lost / 9.0 C. Energy lost x 20 years D. Energy lost in 1000 kWh x Rs 9.0 Answer: D. e.g. Zebra 24 x 9.0 = 216 thousand Rs. (Key point: Loss cost) 3. Why is Deer rejected against Zebra? A. Same energy loss but higher capital cost B. Higher energy loss C. Lower tower cost but higher loss D. Deer exceeds the project life Answer: A. Both lose 24, but Deer costs 2627 vs 2559. (Key point: Why Deer is ruled out) 4. Which conductor is best and what is its total? A. Elk, Rs 519.65 thousand/km B. Zebra, Rs 516.58 thousand/km C. Goat, Rs 539.83 thousand/km D. Sheep, Rs 523.26 thousand/km Answer: B. Elk's extra capital charge (about 30) exceeds its loss saving (27). (Key point: Result) ## Close variants set in other exams - [Most economical conductor, five options (Rs 7.2)](https://www.bctengineer.com/question.md?key=semester_8%2Ftransmission_and_distribution_system_design%3Aeconomic-conductor-five-options-rs) — Transmission and Distribution System Design, chapter 4, asked 3× - [Most economical conductor from cost data](https://www.bctengineer.com/question.md?key=semester_8%2Ftransmission_and_distribution_system_design%3Aeconomic-conductor-goat-sheep-zebra-deer) — Transmission and Distribution System Design, chapter 4, asked 3× ## Related in Transmission and Distribution System Design - [Economical span, 200 km, A/B/C tower types](https://www.bctengineer.com/question.md?key=semester_8%2Ftransmission_and_distribution_system_design%3Aeconomic-span-200km-three-tower-types) — Transmission and Distribution System Design, chapter 4, asked 6× - [Most economical span, 180 km single circuit](https://www.bctengineer.com/question.md?key=semester_8%2Ftransmission_and_distribution_system_design%3Aeconomic-span-180km-single-circuit) — Transmission and Distribution System Design, chapter 4 - [Most economical span, double circuit line](https://www.bctengineer.com/question.md?key=semester_8%2Ftransmission_and_distribution_system_design%3Aeconomic-span-double-circuit) — Transmission and Distribution System Design, chapter 4, asked 6×