--- title: "Peak loss and voltage drop for LT feeder options" subject: "Transmission and Distribution System Design" chapter: 7 key: "semester_8/transmission_and_distribution_system_design:lt-feeder-configuration-options" source: https://www.bctengineer.com/q/semester_8/transmission_and_distribution_system_design/lt-feeder-configuration-options site: BCT Engineering --- # Peak loss and voltage drop for LT feeder options Transmission and Distribution System Design · Chapter 7: Distribution System Design - Asked in: 2075 Bhadra, 10 marks - Video: [How to Calculate Line Drop and Line Loss of a Circuit](https://www.youtube.com/watch?v=2lWm-fgZ3GI) — Engineer4Free - 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/lt-feeder-configuration-options - Whole chapter: https://www.bctengineer.com/md/semester_8/transmission_and_distribution_system_design/chapter-7 ## Question A particular load center has a peak demand of 50 kW at power factor of 0.8 lagging. Compute % peak power loss and voltage drop in 3 phase LT feeder configuration with rated transformer of 11/0.4 kV. | Option | Feeder configuration | |---|---| | I | 4 feeders each of length 3 km | | II | 2 feeders each of 4 km | | III | 3 feeders each of 2 km | Impedance per phase = (1.1 + j0.25) Ω/km. Assume uniformly distributed load along feeders. ## Short answer Option III (3 feeders × 2 km) gives the lowest peak power loss (11.94%) and voltage drop (13.41%). All options exceed typical LT limits; practical solutions include larger conductor, additional transformer, or extending a higher voltage closer. ## Answer **Assumptions:** load is shared equally among the feeders of each option and uniformly distributed along each feeder; LT line voltage = 400 V (transformer secondary); $\cos\phi = 0.8$, $\sin\phi = 0.6$; $z = 1.1 + j0.25$ Ω/km per phase. #### Formulae (uniformly distributed load) $$ \begin{aligned} I &= \frac{P_f}{\sqrt{3} \times 400 \times 0.8} \\ P_{loss} &= n \times I^2 R \quad (R = 1.1\,l) \\ \Delta V_L &= \frac{\sqrt{3}\, I (R\cos\phi + X\sin\phi)}{2} \end{aligned} $$ where $n$ = number of feeders, $P_f = 50/n$ kW per feeder, $l$ = feeder length. #### Option I: 4 feeders × 3 km $$ \begin{aligned} P_f &= 12.5 \text{ kW}, \quad I = \frac{12{,}500}{\sqrt{3}(400)(0.8)} = 22.553 \text{ A} \\ R &= 3.3\ \Omega, \quad X = 0.75\ \Omega, \quad R\cos\phi + X\sin\phi = 3.09\ \Omega \\ P_{loss} &= 4 \times 22.553^2 \times 3.3 = 6713.9 \text{ W} \Rightarrow 13.43\% \\ \Delta V &= \frac{\sqrt{3}(22.553)(3.09)}{2} = 60.35 \text{ V} \Rightarrow 15.09\% \end{aligned} $$ #### Option II: 2 feeders × 4 km $$ \begin{aligned} P_f &= 25 \text{ kW}, \quad I = 45.105 \text{ A} \\ R &= 4.4\ \Omega, \quad X = 1.0\ \Omega, \quad R\cos\phi + X\sin\phi = 4.12\ \Omega \\ P_{loss} &= 2 \times 45.105^2 \times 4.4 = 17{,}903.6 \text{ W} \Rightarrow 35.81\% \\ \Delta V &= \frac{\sqrt{3}(45.105)(4.12)}{2} = 160.94 \text{ V} \Rightarrow 40.23\% \end{aligned} $$ #### Option III: 3 feeders × 2 km $$ \begin{aligned} P_f &= 16.667 \text{ kW}, \quad I = 30.070 \text{ A} \\ R &= 2.2\ \Omega, \quad X = 0.5\ \Omega, \quad R\cos\phi + X\sin\phi = 2.06\ \Omega \\ P_{loss} &= 3 \times 30.070^2 \times 2.2 = 5967.9 \text{ W} \Rightarrow 11.94\% \\ \Delta V &= \frac{\sqrt{3}(30.070)(2.06)}{2} = 53.65 \text{ V} \Rightarrow 13.41\% \end{aligned} $$ #### Summary | Option | Feeders | I per feeder (A) | Peak loss (W) | % loss | ΔV (V) | % ΔV | |---|---|---|---|---|---|---| | I | 4 × 3 km | 22.55 | 6713.9 | 13.43 | 60.35 | 15.09 | | II | 2 × 4 km | 45.11 | 17,903.6 | 35.81 | 160.94 | 40.23 | | III | 3 × 2 km | 30.07 | 5967.9 | 11.94 | 53.65 | 13.41 | **Answer:** Option III (3 feeders × 2 km) gives the lowest peak loss (11.94%) and voltage drop (13.41%); option II is the worst. All options exceed the usual LT voltage drop limit (about 5–10%), so in practice the area should be split with an additional transformer, a larger conductor, or a higher voltage (11 kV) brought closer to the loads. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which option has the lowest loss and voltage drop? A. I (4 x 3 km) B. II (2 x 4 km) C. III (3 x 2 km) D. All equal Answer: C. Option III gives 11.94% loss and 13.41% drop. (Key point: Best option) 2. Which is worst and why? A. II, long feeders carry more current each B. I, too many feeders C. III, shortest D. None Answer: A. 2 x 4 km has the longest feeders with highest current: 35.8% loss. (Key point: Worst option) 3. Current per feeder for option I is: A. 50/(sqrt3 x 400 x 0.8) B. 12500/(sqrt3 x 400 x 0.8) = 22.55 A C. 50000/400 D. 12500/400 Answer: B. Load shared equally: 12.5 kW per feeder. (Key point: Current formula) 4. What does the answer recommend since all exceed limits? A. Ignore them B. Add a transformer, larger conductor or higher voltage C. Lower voltage D. Longer feeders Answer: B. All options exceed 5-10% drop, so split the area, upsize or raise the voltage. (Key point: Remedy) ## Close variants set in other exams - [LT feeder options: loss and voltage drop, 50 kW](https://www.bctengineer.com/question.md?key=semester_8%2Ftransmission_and_distribution_system_design%3Alt-feeder-two-options-50kw) — Transmission and Distribution System Design, chapter 7, asked 3× ## Related in Transmission and Distribution System Design - [LT loss and voltage drop, 60 kW four feeders](https://www.bctengineer.com/question.md?key=semester_8%2Ftransmission_and_distribution_system_design%3Alt-feeders-60kw-1-5km-380v) — Transmission and Distribution System Design, chapter 7, asked 3× - [Distribution loss and voltage drop, four LT feeders](https://www.bctengineer.com/question.md?key=semester_8%2Ftransmission_and_distribution_system_design%3Aload-center-40kw-four-lt-feeders) — Transmission and Distribution System Design, chapter 7 - [LT loss and voltage drop, four 2 km feeders](https://www.bctengineer.com/question.md?key=semester_8%2Ftransmission_and_distribution_system_design%3Alt-feeders-40kw-2km-380v) — Transmission and Distribution System Design, chapter 7, asked 2× - [Loss and drop, 15 km 11 kV feeder](https://www.bctengineer.com/question.md?key=semester_8%2Ftransmission_and_distribution_system_design%3Audl-feeder-15km-11kv) — Transmission and Distribution System Design, chapter 7, asked 2× - [Power loss, 10 km 11 kV UDL feeder](https://www.bctengineer.com/question.md?key=semester_8%2Ftransmission_and_distribution_system_design%3Audl-feeder-10km-11kv) — Transmission and Distribution System Design, chapter 7, asked 2× - [Peak demand of load center from load pattern](https://www.bctengineer.com/question.md?key=semester_8%2Ftransmission_and_distribution_system_design%3Apeak-demand-load-center-load-pattern) — Transmission and Distribution System Design, chapter 6, asked 2×