--- title: "Most economical conductor from cost data" subject: "Transmission and Distribution System Design" chapter: 4 key: "semester_8/transmission_and_distribution_system_design:economic-conductor-goat-sheep-zebra-deer" source: https://www.bctengineer.com/q/semester_8/transmission_and_distribution_system_design/economic-conductor-goat-sheep-zebra-deer site: BCT Engineering --- # Most economical conductor from cost data Transmission and Distribution System Design · Chapter 4: Conductor and Support Selection - Asked in: 2079 Chaitra, 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-goat-sheep-zebra-deer - Whole chapter: https://www.bctengineer.com/md/semester_8/transmission_and_distribution_system_design/chapter-4 ## Question For the transmission line design following computations has been made. Select the most economical conductor. | Conductor | A: Tower cost/km (1000 Rs.) | B: Total conductor cost/km (1000 Rs.) | C: Annual energy loss/km (1000 kWh) | |---|---|---|---| | Goat | 890 | 1254 | 32 | | Sheep | 946 | 1450 | 30 | | Zebra | 984 | 1585 | 28 | | Deer | 910 | 1663 | 20 | Using the following data: Energy rate = Rs 9 per unit, Interest rate = 14%, Project life = 25 years. [Column C is printed as "Annual energy loss/km (1000 kW/hr.)".] ## Short answer With CRF=0.1455, annual cost = (A+B)×CRF +9×C. Deer has the lowest total annual cost of Rs554.37 thousand/km, making it the most economical conductor because its higher capital is offset by much lower energy losses. ## Answer The most economical conductor is the one with the lowest **total annual cost per km**. This is the annual charge on the capital (towers plus conductor), spread over the project life at the given interest rate, plus the annual cost of energy lost. #### Capital recovery factor With $i = 14\%$ and $n = 25$ years: $$ \begin{aligned} CRF &= \frac{i(1+i)^n}{(1+i)^n - 1} \\ &= \frac{0.14(1.14)^{25}}{(1.14)^{25} - 1} = 0.14550 \end{aligned} $$ #### Annual cost of energy loss Column C is in thousands of kWh per km per year, and the energy rate is Rs 9 per kWh. So the loss cost in thousand Rs is $9C$. For Goat: $32 \times 9 = 288$ thousand Rs. #### Comparison (per km) | Conductor | Capital A+B (1000 Rs) | Annual capital charge (1000 Rs) | Annual loss cost (1000 Rs) | Total annual cost (1000 Rs) | |---|---|---|---|---| | Goat | 2144 | 311.95 | 288 | 599.95 | | Sheep | 2396 | 348.61 | 270 | 618.61 | | Zebra | 2569 | 373.79 | 252 | 625.79 | | Deer | 2573 | 374.37 | 180 | 554.37 | Sample calculation for Deer: $(910 + 1663) \times 0.14550 = 374.37$, and $374.37 + 20 \times 9 = 554.37$ thousand Rs. **Answer: Deer is the most economical conductor**, with a total annual cost of about Rs 554.37 thousand per km. Its higher capital cost is more than recovered by its much lower energy loss. If the energy loss is instead worked as a present worth (annual loss cost ÷ CRF, added to the capital), the ranking is the same, because both methods differ only by the factor CRF. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The most economical conductor is the one with: A. Lowest capital cost B. Lowest energy loss C. Lowest total annual cost (capital charge plus loss cost) D. Highest tower cost Answer: C. Annual capital charge plus annual loss cost is minimised. (Key point: Criterion) 2. The capital recovery factor for 14% and 25 years is about: A. 0.0400 B. 0.1455 C. 0.2500 D. 0.5600 Answer: B. CRF = 0.14(1.14)^25/((1.14)^25 - 1) = 0.1455. (Key point: CRF) 3. Annual loss cost for Deer (20 thousand kWh/km at Rs 9/kWh) is: A. 180 thousand Rs B. 20 thousand Rs C. 29 thousand Rs D. 360 thousand Rs Answer: A. 20 x 9 = 180 thousand Rs. (Key point: Loss cost) 4. Which conductor is selected? A. Goat B. Sheep C. Zebra D. Deer Answer: D. Deer has the lowest total annual cost, 554.37 thousand Rs/km. 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