---
title: "Loss and drop, 15 km 11 kV feeder"
subject: "Transmission and Distribution System Design"
chapter: 7
key: "semester_8/transmission_and_distribution_system_design:udl-feeder-15km-11kv"
source: https://www.bctengineer.com/q/semester_8/transmission_and_distribution_system_design/udl-feeder-15km-11kv
site: BCT Engineering
---

# Loss and drop, 15 km 11 kV feeder

Transmission and Distribution System Design · Chapter 7: Distribution System Design

- Asked in: 2070 Magh, 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/udl-feeder-15km-11kv
- Whole chapter: https://www.bctengineer.com/md/semester_8/transmission_and_distribution_system_design/chapter-7

## Question

Compute the power loss & voltage drop for a 15 km long 11 kV three phase distribution feeder having sending end current of 100 A with uniformly distributed load. Use z = (0.6 + j3) Ω/km. Derive any expression used in above calculation.

## Short answer

For a uniformly distributed load, power loss = I_s² r l and voltage drop = ½ I_s z l. With given values, loss = 90 kW, drop = 2294.6 V per phase (3974 V line).

## 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.

#### Substitution
$I_s = 100$ A, $l = 15$ km, $z = 0.6 + j3\ \Omega$/km, $|z| = \sqrt{0.6^2+3^2} = 3.059\ \Omega$/km. No load p.f. is given, so the drop is taken as $\frac12 I_s|z|l$ (the maximum value).

$$
\begin{aligned}
P_L &= 100^2\times0.6\times15 = 90{,}000\ \text{W} = 90\ \text{kW}\\
V_{D,ph} &= \frac{1}{2}\times100\times3.059\times15 = 2294.6\ \text{V}\\
V_{D,L} &= \sqrt3\times2294.6 = 3974.3\ \text{V}\ (36.1\ \%\ \text{of } 11\ \text{kV})
\end{aligned}
$$

Note: a reactance of 3 Ω/km is unusually high for an 11 kV line (typical is about 0.3–0.4 Ω/km). If $z = 0.6 + j0.3$ was intended, $|z| = 0.6708$, $V_{D,ph} = 503.1$ V, $V_{D,L} = 871.4$ V (7.92 %); the loss is unchanged at 90 kW since it depends only on $r$.

**Answer: Power loss = 90 kW; voltage drop ≈ 2294.6 V per phase (3974 V line) with the given z.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. In the derivation, the current at distance x along a UDL feeder is:
   A. I_s
   B. I_s (1 - x/l)
   C. I_s x/l
   D. I_s (1 - x/l)^2
   Answer: B. Load is tapped uniformly so current falls linearly to zero at the far end. (Key point: UDL current profile)

2. Why does the UDL loss expression contain 1/3?
   A. The integral of (1 - x/l)^2 over l equals l/3
   B. There are three phases
   C. Loss occurs only in one third of the line
   D. The reactance is divided by 3
   Answer: A. Integrating I^2 gives l/3; with 3 phases the result is I_s^2 r l. (Key point: Loss factor)

3. A UDL feeder behaves like the whole load lumped at:
   A. The far end for both drop and loss
   B. The middle for drop and one-third length for loss
   C. One-third for drop and middle for loss
   D. The sending end for both
   Answer: B. Drop uses half the length, loss uses one third of the length. (Key point: Equivalent lumping)

4. With I_s = 100 A, r = 0.6 Ω/km and 15 km, the power loss is:
   A. 270 kW
   B. 30 kW
   C. 135 kW
   D. 90 kW
   Answer: D. P_L = I_s^2 r l = 10^4 x 0.6 x 15 = 90 kW; reactance does not affect it. (Key point: Loss value)

## Close variants set in other exams

- [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×

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