---
title: "Monthly energy loss in four secondary lines"
subject: "Transmission and Distribution System Design"
chapter: 7
key: "semester_8/transmission_and_distribution_system_design:monthly-loss-four-secondary-lines"
source: https://www.bctengineer.com/q/semester_8/transmission_and_distribution_system_design/monthly-loss-four-secondary-lines
site: BCT Engineering
---

# Monthly energy loss in four secondary lines

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

- Asked in: 2073 Bhadra, 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/monthly-loss-four-secondary-lines
- Whole chapter: https://www.bctengineer.com/md/semester_8/transmission_and_distribution_system_design/chapter-7

## Question

A 11 kV/0.4 kV, 100 kVA distribution transformer has 4 outgoing secondary distribution lines of 1.5 km each. The conductor used has resistance of 1 Ω/km. Determine the monthly energy loss in the lines if the peak demand of the load centre has 80 kVA at the load factor of 0.4. Given LLF = 0.3LF + 0.7LF².

## Short answer

Each line carries 20 kVA, current 28.87 A, resistance 1.5 Ω, peak loss per line I²R=1250 W; total peak loss 5 kW. LLF = 0.3×0.4+0.7×0.4² = 0.232. Monthly energy loss = 5×0.232×720 = 835.2 kWh.

## Answer

Assumptions: the 80 kVA peak is shared **equally** by the four lines, load on each line is **uniformly distributed**, lines are 3-phase at 400 V, and a month = 30 days = 720 h.

#### Step 1: Current per line
Peak load per line $= 80/4 = 20$ kVA.
$$
I = \frac{20\,000}{\sqrt3\times400} = 28.87\ \text{A}
$$

#### Step 2: Peak power loss per line (UDL)
For a uniformly distributed load, 3-phase loss $= 3\cdot\frac13 I^2 R = I^2 R$, where $R = 1\times1.5 = 1.5\ \Omega$ per phase.
$$
P_{L,line} = 28.87^2\times1.5 = 1250\ \text{W}
$$
Total peak loss in 4 lines:
$$
P_{L,peak} = 4\times1250 = 5000\ \text{W} = 5\ \text{kW}
$$

#### Step 3: Loss load factor
$$
\begin{aligned}
LLF &= 0.3LF + 0.7LF^2\\
&= 0.3(0.4) + 0.7(0.4)^2 = 0.12 + 0.112 = 0.232
\end{aligned}
$$

#### Step 4: Monthly energy loss
$$
\begin{aligned}
E_{loss} &= P_{L,peak}\times LLF\times720\\
&= 5\times0.232\times720 = 835.2\ \text{kWh}
\end{aligned}
$$

#### Summary
| Quantity | Value |
|---|---|
| Current per line | 28.87 A |
| Peak loss per line | 1.25 kW |
| Total peak loss | 5 kW |
| LLF | 0.232 |
| Monthly loss | 835.2 kWh |

Note: if the whole line load were assumed lumped at the far end, the loss would be three times larger (2505.6 kWh); the uniform-load assumption is the usual one for LT distribution lines serving houses along the route.

**Answer: Monthly energy loss in the lines ≈ 835 kWh.**

## Check yourself

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

1. Per-line load 20 kVA at 400 V gives:
   A. 28.87 A
   B. 50 A
   C. 115 A
   D. 14.4 A
   Answer: A. 20000/(sqrt3 x 400) = 28.87 A. (Key point: Current)

2. Peak loss per line for UDL:
   A. 3 I^2 R
   B. I^2 R = 1250 W
   C. I^2 R/3
   D. 2500 W
   Answer: B. 3 x (1/3) I^2 R = 28.87^2 x 1.5 = 1250 W. (Key point: UDL loss)

3. LLF at LF = 0.4:
   A. 0.4
   B. 0.232
   C. 0.16
   D. 0.3
   Answer: B. 0.12 + 0.112 = 0.232. (Key point: LLF)

4. Monthly loss is:
   A. 835.2 kWh
   B. 2505.6 kWh
   C. 5000 kWh
   D. 3600 kWh
   Answer: A. 5 kW x 0.232 x 720 = 835.2 kWh. (Key point: Monthly)

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