--- title: "Most economical conductor, five options (Rs 7.2)" subject: "Transmission and Distribution System Design" chapter: 4 key: "semester_8/transmission_and_distribution_system_design:economic-conductor-five-options-rs" source: https://www.bctengineer.com/q/semester_8/transmission_and_distribution_system_design/economic-conductor-five-options-rs site: BCT Engineering --- # Most economical conductor, five options (Rs 7.2) Transmission and Distribution System Design · Chapter 4: Conductor and Support Selection - Asked in: 2071 Bhadra, 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/economic-conductor-five-options-rs - 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 conductors cost/km (1000 Rs.) | Annual energy loss/km (1000 Rs.) | |---|---|---|---| | Goat | 890 | 1254 | 32 | | Sheep | 936 | 1450 | 27 | | Zebra | 974 | 1585 | 24 | | Deer | 964 | 1663 | 24 | | Elk | 968 | 1847 | 21 | Using the following data: Energy rate = Rs. 7.2 per unit, Interest rate = 10%, Project life = 20 years. ## Short answer Zebra is the most economical conductor. Its total annual cost (annual capital charge plus energy loss cost) is the lowest at Rs 473.38 thousand per km. ## 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. **Reading of the data:** the energy rate (Rs 7.2/unit) is given, so the loss column is taken as **energy lost per km per year in 1000 kWh**, as in the 2071 Magh version of this question. A cost in 1000 Rs would make the rate unnecessary. So the annual loss cost is the loss in 1000 kWh × 7.2, 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 × 7.2 = 230.40 | 482.23 | | Sheep | 2386 | 280.26 | 27 × 7.2 = 194.40 | 474.66 | | Zebra | 2559 | 300.58 | 24 × 7.2 = 172.80 | 473.38 | | Deer | 2627 | 308.57 | 24 × 7.2 = 172.80 | 481.37 | | Elk | 2815 | 330.65 | 21 × 7.2 = 151.20 | 481.85 | Sample (Goat): $2144\times0.11746 = 251.83$ and $32\times7.2 = 230.40$, total 482.23 thousand Rs/km per year. Zebra has the lowest total. Its higher capital cost compared with Goat and Sheep is more than recovered by its lower losses. Elk saves more energy, but its extra conductor cost is not repaid. If the loss column were instead read literally as cost in 1000 Rs, the totals would be Goat 283.83, Sheep 307.26, Zebra 324.58, Deer 332.57 and Elk 351.65 thousand Rs, and Goat would be chosen. With the energy rate given, the kWh reading above is the intended one. **Answer: Zebra is the most economical conductor, with a minimum total annual cost of Rs 473.38 thousand per km.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the most economical conductor chosen? A. Lowest conductor cost B. Lowest energy loss C. Lowest total annual cost = annual capital charge + annual energy loss cost D. Lowest tower cost Answer: C. Kelvin's law: balance capital charge against loss cost. (Key point: Selection criterion) 2. What does CRF = i(1+i)^n/((1+i)^n-1) give in this problem? A. The factor for the yearly energy loss B. The present value of tower cost C. The interest on loss cost D. The factor 0.11746 that converts capital cost into annual charge Answer: D. CRF at 10% and 20 years is 0.11746. (Key point: Capital recovery factor) 3. Why is the loss column read as 1000 kWh? A. Because an energy rate is given, which would be unnecessary if losses were already costs B. Because kWh is always used for losses C. Because the table heading says kWh D. Because Rs 7.2 is the interest rate Answer: A. Loss x 7.2 gives cost in 1000 Rs. (Key point: Reading of loss column) 4. Which conductor is chosen and why not Elk? A. Elk, since it has lowest loss B. Zebra at Rs 473.38 thousand/km; Elk's extra capital is not repaid by its loss saving C. Goat, since it has lowest capital cost D. Deer, since its loss is same as Zebra and cost lower Answer: B. Zebra total 473.38; Elk 481.85. (Key point: Result) ## Close variants set in other exams - [Most economical conductor, five options (Rs 9)](https://www.bctengineer.com/question.md?key=semester_8%2Ftransmission_and_distribution_system_design%3Aeconomic-conductor-five-options-kwh) — 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×