---
title: "Loss and voltage drop, 16 km uniformly varying load"
subject: "Transmission and Distribution System Design"
chapter: 7
key: "semester_8/transmission_and_distribution_system_design:uniformly-varying-load-16km-feeder"
source: https://www.bctengineer.com/q/semester_8/transmission_and_distribution_system_design/uniformly-varying-load-16km-feeder
site: BCT Engineering
---

# Loss and voltage drop, 16 km uniformly varying load

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

- Asked in: 2074 Magh, 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/uniformly-varying-load-16km-feeder
- Whole chapter: https://www.bctengineer.com/md/semester_8/transmission_and_distribution_system_design/chapter-7

## Question

Compute the power loss and maximum voltage drop for a 16 km long 11 kV, 3 phase primary distribution line having sending end current of 100 A with uniformly varying load. (Use R = 0.6 Ω/km and X = 0.3 Ω/km)

## Short answer

For uniformly increasing load, loss factor is 8/15, drop factor 2/3. With I_s=100 A, l=16 km, R=9.6 Ω, |z|=0.6708 Ω/km: three-phase power loss = 3×(8/15)×100²×9.6 = 153.6 kW; voltage drop per phase = (2/3)×100×0.6708×16 = 715.5 V, line drop 1239 V (11.3% of 11 kV).

## Answer

Assumption: "uniformly varying load" is taken as a **uniformly increasing load**, i.e. the load density rises linearly from zero at the sending end to a maximum at the far end (Gonen's case). No power factor is given, so the **maximum** possible drop is found using $|z|$ (the drop $I(R\cos\phi+X\sin\phi)$ is largest when it equals $I|Z|$).

#### Current distribution
Let load density $= kx$ (A/km). The current at distance $x$ from the source is the load beyond $x$:
$$
\begin{aligned}
I(x) &= \int_x^{l} k\,u\,du = \frac{k}{2}(l^2-x^2),\quad I_s = I(0) = \frac{kl^2}{2}\\
I(x) &= I_s\left(1-\frac{x^2}{l^2}\right)
\end{aligned}
$$

#### Voltage drop (per phase)
$$
V_D = \int_0^l z\,I_s\left(1-\frac{x^2}{l^2}\right)dx = z I_s\left(l-\frac{l}{3}\right) = \frac{2}{3}I_s z l
$$

#### Power loss (three-phase)
$$
P_L = 3\int_0^l r I_s^2\left(1-\frac{x^2}{l^2}\right)^2dx = 3 r I_s^2 l\left(1-\frac{2}{3}+\frac{1}{5}\right) = 3\cdot\frac{8}{15}I_s^2 r l
$$

#### Substitution
$I_s = 100$ A, $l = 16$ km, $R = rl = 0.6\times16 = 9.6\ \Omega$, $|z| = \sqrt{0.6^2+0.3^2} = 0.6708\ \Omega$/km.

$$
\begin{aligned}
P_L &= 3\times\frac{8}{15}\times100^2\times9.6 = 153{,}600\ \text{W} = 153.6\ \text{kW}\\
V_{D,ph} &= \frac{2}{3}\times100\times0.6708\times16 = 715.5\ \text{V}\\
V_{D,L} &= \sqrt3\times715.5 = 1239.4\ \text{V}\\
\%V_D &= \frac{1239.4}{11000}\times100 = 11.27\ \%
\end{aligned}
$$

#### Comparison with other load patterns
| Load type | Drop factor | Loss factor |
|---|---|---|
| Lumped at end | 1 | 1 |
| Uniformly distributed | 1/2 | 1/3 |
| Uniformly increasing | 2/3 | 8/15 |

**Answer: Power loss = 153.6 kW; maximum voltage drop ≈ 715.5 V per phase = 1239 V line (≈ 11.3 % of 11 kV).**

## Check yourself

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

1. Maximum voltage drop uses impedance because:
   A. Drop is largest when it equals I|Z| (no pf given)
   B. pf is 1
   C. Reactance is ignored
   D. Load is concentrated
   Answer: A. I(R cos + X sin) is maximum = I|Z|. (Key point: Assumption)

2. Power loss = 3 x (8/15) x Is^2 x R with R = 9.6 ohm is:
   A. 51.2 kW
   B. 153.6 kW
   C. 288 kW
   D. 96 kW
   Answer: B. 3 x 0.5333 x 10^4 x 9.6 = 153.6 kW. (Key point: Loss)

3. The per-phase drop is:
   A. (2/3) Is |z| l = 715.5 V
   B. Is |z| l = 1073 V
   C. (1/2) Is z l = 536 V
   D. 1239 V
   Answer: A. (2/3)(100)(0.6708)(16) = 715.5 V. (Key point: Drop)

4. Line drop and % of 11 kV:
   A. 715.5 V, 6.5%
   B. 1239 V, 11.27%
   C. 1239 V, 6.5%
   D. 2000 V, 18%
   Answer: B. sqrt3 x 715.5 = 1239 V = 11.27%. (Key point: Percent)

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