---
title: "5th year peak load with population growth"
subject: "Transmission and Distribution System Design"
chapter: 6
key: "semester_8/transmission_and_distribution_system_design:fifth-year-peak-load-population-growth"
source: https://www.bctengineer.com/q/semester_8/transmission_and_distribution_system_design/fifth-year-peak-load-population-growth
site: BCT Engineering
---

# 5th year peak load with population growth

Transmission and Distribution System Design · Chapter 6: Electrical Load Characteristics and Load Forecast

- Asked in: 2077 Chaitra, 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/fifth-year-peak-load-population-growth
- Whole chapter: https://www.bctengineer.com/md/semester_8/transmission_and_distribution_system_design/chapter-6

## Question

From the initial survey of particular load center, the following consumer data has been obtained.

| Consumer class | Potential consumer | Avg. monthly consumption | No. of effective days/month | 5th year contribution factor to peak | 5th year load factor |
|---|---|---|---|---|---|
| Domestic | 100 | 30 kWh | 30 | 1.0 | 0.25 |
| Commercial | 20 | 30 kWh | 25 | 0.5 | 0.30 |
| Non-commercial | 50 | 20 kWh | 20 | 0.1 | 0.25 |

Assuming population growth factor of 1.8%, consumption growth factors of 5% in each year for each class and electrification coverage factor by 5th year is 70% for all consumers classes, determine 5th year peak load of the load center in kW. Also, project implementation period be 2 years.

## Short answer

With 7 years of 1.8% population and 5% consumption growth, 70% coverage, individual peak 0.2345 kW, and contribution factors 1.0, 0.5, 0.1, the 5th-year peak load is 21.4 kW.

## Answer

The load centre peak is found by projecting the number of consumers and their consumption to the 5th year, converting energy to peak using the load factor, and combining the class peaks with their contribution factors.

**Assumptions:** The 5th year is counted after project completion, so with a 2-year implementation period the growth period from the survey is $n = 5 + 2 = 7$ years for both population and consumption.

#### Formulae

$$
\begin{aligned}
N_5 &= N_0 (1 + p)^n \times \text{coverage} \\
E_5 &= E_0 (1 + g)^n \\
P_{ind} &= \frac{E_5 / \text{effective days}}{24 \times LF} \\
P_{LC} &= \sum \text{CF} \times N_5 \times P_{ind}
\end{aligned}
$$

#### Growth multipliers

$$
\begin{aligned}
(1.018)^7 &= 1.13301 \\
(1.05)^7 &= 1.40710
\end{aligned}
$$

#### Number of consumers in 5th year

| Class | $N_0$ | $N_0 \times 1.13301 \times 0.7$ |
|---|---|---|
| Domestic | 100 | 79.31 |
| Commercial | 20 | 15.86 |
| Non-commercial | 50 | 39.66 |

#### Consumption and individual peak

| Class | $E_5$ (kWh/month) | Daily (kWh) | LF | $P_{ind}$ (kW) |
|---|---|---|---|---|
| Domestic | 30 × 1.4071 = 42.213 | 42.213/30 = 1.4071 | 0.25 | 0.23452 |
| Commercial | 42.213 | 42.213/25 = 1.6885 | 0.30 | 0.23452 |
| Non-commercial | 20 × 1.4071 = 28.142 | 28.142/20 = 1.4071 | 0.25 | 0.23452 |

Example (domestic): $P_{ind} = \dfrac{1.4071}{24 \times 0.25} = 0.23452$ kW.

#### Class peak and contribution to load centre peak

| Class | Class peak (kW) | CF | Contribution (kW) |
|---|---|---|---|
| Domestic | 79.31 × 0.23452 = 18.600 | 1.0 | 18.600 |
| Commercial | 15.86 × 0.23452 = 3.720 | 0.5 | 1.860 |
| Non-commercial | 39.66 × 0.23452 = 9.300 | 0.1 | 0.930 |
| **Total** | | | **21.39** |

**Answer: 5th year peak load of the load centre ≈ 21.4 kW.**

(If the implementation period were ignored, i.e. $n = 5$, the result would be lower; the 2-year period is included because consumers keep growing while the project is being built.)

## Check yourself

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

1. With a 2-year implementation period, how many years of growth are applied to reach the 5th year?
   A. 5
   B. 2
   C. 7
   D. 3
   Answer: C. The 5th year counts after completion, so n = 5 + 2 = 7. (Key point: Growth period)

2. How is the number of domestic consumers in the 5th year computed?
   A. 100 x 1.018^7 x 0.7 = 79.3
   B. 100 x 1.05^7 x 0.7
   C. 100 x 0.7 = 70
   D. 100 x 1.018 x 5 x 0.7
   Answer: A. Population growth 1.8% over 7 years gives 1.133, then 70% coverage. (Key point: Consumer count)

3. Which multiplier applies to monthly consumption per consumer?
   A. 1.018^7 = 1.133
   B. 1.05^7 = 1.407
   C. 1.05^5 = 1.276
   D. 1.068^7
   Answer: B. Consumption grows at 5% per year over n = 7 years. (Key point: Consumption growth)

4. The 5th-year load centre peak found is about:
   A. 17.1 kW
   B. 18.6 kW
   C. 30.0 kW
   D. 21.4 kW
   Answer: D. 18.600 + 1.860 + 0.930 = 21.39 kW using contribution factors 1.0, 0.5 and 0.1. (Key point: Result)

## Close variants set in other exams

- [5th year peak load (commercial 30 days)](https://www.bctengineer.com/question.md?key=semester_8%2Ftransmission_and_distribution_system_design%3Afifth-year-peak-load-30-days) — Transmission and Distribution System Design, chapter 6, asked 7×
- [5th year peak load from consumer survey](https://www.bctengineer.com/question.md?key=semester_8%2Ftransmission_and_distribution_system_design%3Afifth-year-peak-load-survey) — Transmission and Distribution System Design, chapter 6, asked 7×
- [5th year peak load, 4% growth, 60% coverage](https://www.bctengineer.com/question.md?key=semester_8%2Ftransmission_and_distribution_system_design%3Afifth-year-peak-load-4-percent) — Transmission and Distribution System Design, chapter 6, asked 7×
- [5th year peak load, incremental growth rate](https://www.bctengineer.com/question.md?key=semester_8%2Ftransmission_and_distribution_system_design%3Afifth-year-peak-load-incremental-growth) — Transmission and Distribution System Design, chapter 6, asked 7×

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