--- title: "5th year peak load from consumer survey" subject: "Transmission and Distribution System Design" chapter: 6 key: "semester_8/transmission_and_distribution_system_design:fifth-year-peak-load-survey" source: https://www.bctengineer.com/q/semester_8/transmission_and_distribution_system_design/fifth-year-peak-load-survey site: BCT Engineering --- # 5th year peak load from consumer survey Transmission and Distribution System Design · Chapter 6: Electrical Load Characteristics and Load Forecast - Asked in: 2078 Kartik, 10 marks; 2075 Bhadra, 10 marks; 2070 Bhadra, 10 marks (asked 3 times) - 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-survey - 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 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 consumption growth factor of 5% in each year for each class and electrification coverage factor by 5th year is 70% for all, determine 5th year peak load of the load center in kW. ## Short answer Electrified consumers are 70% of potential; monthly consumption grows 5% yearly for 5 years (×1.276). Daily energy = monthly / effective days; peak per consumer = daily/(24×LF). Class peak × contribution factor yields contributions. Summing gives a 5th-year peak of 17.1 kW. ## Answer The 5th-year peak load of the load center is the sum of each consumer class's own peak, multiplied by its contribution (coincidence) factor to the system peak. #### Method For each class: 1. Consumers electrified by the 5th year: $N_5 = \text{coverage}\times N_{potential} = 0.7N$. 2. Monthly consumption per consumer in the 5th year, with 5% growth per year compounded over 5 years: $$ E_5 = E_0(1+0.05)^5 = 1.27628\,E_0 $$ 3. Daily energy per consumer: $$ E_d = \frac{E_5}{\text{effective days per month}} $$ 4. Peak demand per consumer, using the load factor: $$ P_{peak} = \frac{\text{average demand}}{LF} = \frac{E_d}{24\times LF} $$ 5. Class peak $= N_5\times P_{peak}$. Its contribution to the load-center peak = class peak × contribution factor. #### Domestic class (sample) $$ \begin{aligned} N_5 &= 0.7\times100 = 70\\ E_5 &= 30\times1.27628 = 38.288\ \text{kWh/month}\\ E_d &= 38.288/30 = 1.2763\ \text{kWh/day}\\ P_{peak} &= \frac{1.2763}{24\times0.25} = 0.2127\ \text{kW}\\ \text{Class peak} &= 70\times0.2127 = 14.890\ \text{kW}\\ \text{Contribution} &= 1.0\times14.890 = 14.890\ \text{kW} \end{aligned} $$ #### All classes | Class | Consumers by 5th yr | Monthly kWh in 5th yr | Daily kWh | Peak per consumer (kW) | Class peak (kW) | CF | Contribution (kW) | |---|---|---|---|---|---|---|---| | Domestic | 70 | 38.288 | 1.2763 | 0.2127 | 14.890 | 1.0 | 14.890 | | Commercial | 14 | 38.288 | 1.5315 | 0.2127 | 2.978 | 0.5 | 1.489 | | Non-commercial | 35 | 25.526 | 1.2763 | 0.2127 | 7.445 | 0.1 | 0.744 | #### 5th-year peak load $$ P_{peak,5} = 14.890 + 1.489 + 0.744 = 17.12\ \text{kW} $$ **Answer: the 5th-year peak load of the load center is about 17.1 kW.** If the survey year is counted as year 1, so that growth applies for only 4 years, the result is $17.12/1.05 = 16.31$ kW. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the 5th-year monthly consumption per consumer found with 5% yearly growth? A. E0 x (1 + 0.05 x 5) B. E0 x (1.05)^5 = 1.276 E0 C. E0 / (1.05)^5 D. E0 x 0.05 x 5 Answer: B. Growth is compounded: 1.05^5 = 1.27628, so 30 kWh becomes 38.288 kWh. (Key point: Growth factor) 2. How is the peak demand per consumer obtained from daily energy? A. Daily energy x 24 x LF B. Daily energy / 24 C. Daily energy / (24 x LF) D. Daily energy x LF Answer: C. Average demand = Ed/24 and peak = average/LF, so P = Ed/(24 x LF). (Key point: Peak from load factor) 3. How are the class peaks combined into the load-centre peak? A. Plain sum of class peaks B. Sum of class peak x contribution factor C. Average of class peaks D. Largest class peak only Answer: B. Classes peak at different times, so each class peak is weighted by its contribution factor before summing. (Key point: Contribution factor use) 4. What is the 5th-year peak load of the load centre in this answer (growth for 5 years)? A. About 17.1 kW B. About 14.9 kW C. About 21.4 kW D. About 31.3 kW Answer: A. 14.890 + 1.489 + 0.744 = 17.12 kW. (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, 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 with population growth](https://www.bctengineer.com/question.md?key=semester_8%2Ftransmission_and_distribution_system_design%3Afifth-year-peak-load-population-growth) — 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× ## Related in Transmission and Distribution System Design - [Peak demand of load center from load pattern](https://www.bctengineer.com/question.md?key=semester_8%2Ftransmission_and_distribution_system_design%3Apeak-demand-load-center-load-pattern) — Transmission and Distribution System Design, chapter 6, asked 2× - [Peak load, load factor and energy from load pattern](https://www.bctengineer.com/question.md?key=semester_8%2Ftransmission_and_distribution_system_design%3Aconsumer-class-load-pattern-transformer) — Transmission and Distribution System Design, chapter 6, asked 2×