--- title: "Power loss, 10 km 11 kV UDL feeder" subject: "Transmission and Distribution System Design" chapter: 7 key: "semester_8/transmission_and_distribution_system_design:udl-feeder-10km-11kv" source: https://www.bctengineer.com/q/semester_8/transmission_and_distribution_system_design/udl-feeder-10km-11kv site: BCT Engineering --- # Power loss, 10 km 11 kV UDL feeder Transmission and Distribution System Design · Chapter 7: Distribution System Design - Asked in: 2068 Bhadra (old course), 4+4 marks - Video: [Lecture video 18EE43 Module 5 Uniformely distributed loads Malini k v](https://www.youtube.com/watch?v=S1Xfhg3sef0) — Sairam EE Video - 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/udl-feeder-10km-11kv - Whole chapter: https://www.bctengineer.com/md/semester_8/transmission_and_distribution_system_design/chapter-7 ## Question Compute the power loss for a 10 km long 11 kV, 3-phase distribution feeder having sending end current of 100 A with uniformly distributed load (Use R = 0.4 Ω/km, X = 0.3 Ω/km). Derive any expression used. ## Short answer For a uniformly distributed feeder, three-phase power loss equals I_s^2 r l. With sending-end current 100 A, r = 0.4 Ω/km, l = 10 km, loss = 40 kW. ## Answer #### Derivation for uniformly distributed load Let a feeder of length $l$ carry sending-end current $I_s$, with load tapped uniformly so the current falls linearly to zero at the far end: $$ I(x) = I_s\left(1-\frac{x}{l}\right) $$ Per-phase voltage drop, with $z = r + jx$ per km: $$ V_D = \int_0^l z I_s\left(1-\frac{x}{l}\right)dx = \frac{1}{2}I_s z l $$ Three-phase power loss: $$ P_L = 3\int_0^l r I_s^2\left(1-\frac{x}{l}\right)^2dx = 3\cdot\frac{1}{3}I_s^2 r l = I_s^2 r l $$ So a UDL feeder behaves like the whole load lumped at the **middle** for voltage drop, and at **one-third** of the length for loss. #### Why $1/3$ Each small element $dx$ at distance $x$ carries current $I(x)$. Integrating $r I(x)^2$ gives $$ \int_0^l\left(1-\frac{x}{l}\right)^2dx = \frac{l}{3} $$ so the loss is only one-third of what it would be with the whole load at the far end. #### Substitution $I_s = 100$ A, $r = 0.4\ \Omega$/km, $l = 10$ km, so $R = 4\ \Omega$ per phase. $$ \begin{aligned} P_L &= 3\times\frac{1}{3}I_s^2 R = I_s^2 R\\ &= 100^2\times0.4\times10 = 40{,}000\ \text{W} \end{aligned} $$ (For comparison, with the same current lumped at the end, loss $=3\times100^2\times4 = 120$ kW.) The reactance $X = 0.3\ \Omega$/km does not affect real power loss; it only enters the voltage drop, $V_D = \frac12 I_s(R\cos\phi + X\sin\phi)$. **Answer: Power loss = 40 kW.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For a 3-phase UDL feeder, the power loss is: A. 3 I_s^2 R B. I_s^2 R C. I_s^2 R / 3 D. I_s^2 R / 2 Answer: B. 3 x (1/3) I_s^2 R = I_s^2 R. (Key point: UDL loss expression) 2. Why is the UDL loss one third of the end-lumped loss per phase? A. The integral of (1 - x/l)^2 over the length equals l/3 B. Load is at one third of the feeder voltage C. Reactance reduces loss D. Only a third of the feeder is energised Answer: A. Current decreases linearly, so integrating I^2 gives l/3 of the lumped value. (Key point: Why one third) 3. What is the effect of X = 0.3 Ω/km on the power loss? A. Loss increases by X/R B. None; X only enters the voltage drop C. Loss is halved D. Loss is multiplied by sqrt3 Answer: B. Real power loss is I^2R; reactance affects voltage drop only. (Key point: Role of X) 4. With I_s = 100 A, r = 0.4 Ω/km, 10 km, the loss is: A. 120 kW B. 40 kW C. 16 kW D. 4 kW Answer: B. 100^2 x 0.4 x 10 = 40 kW (the end-lumped value would be 120 kW). 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