--- title: "5th year peak load, incremental growth rate" subject: "Transmission and Distribution System Design" chapter: 6 key: "semester_8/transmission_and_distribution_system_design:fifth-year-peak-load-incremental-growth" source: https://www.bctengineer.com/q/semester_8/transmission_and_distribution_system_design/fifth-year-peak-load-incremental-growth site: BCT Engineering --- # 5th year peak load, incremental growth rate Transmission and Distribution System Design · Chapter 6: Electrical Load Characteristics and Load Forecast - Asked in: 2072 Asoj, 9 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-incremental-growth - Whole chapter: https://www.bctengineer.com/md/semester_8/transmission_and_distribution_system_design/chapter-6 ## Question From the initial survey of a 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 | 30 | 0.5 | 0.30 | | Non-commercial | 50 | 20 kWh | 20 | 0.1 | 0.25 | Determine the 5th year peak load of load center in kW. Data for load forecast (for all consumer classes): Load growth factor: 5% for first year and an increment of 1% from 2nd year. Population growth factor: 2% each year. Percentage coverage factor: 70% by 5th year. ## Short answer Apply consumption growth multiplier (1.05×1.06×1.07×1.08×1.09=1.4019) and population multiplier (1.02^5=1.1041) with 70% coverage to get 5th-year consumers and monthly energy. Individual peak = (daily kWh)/(24×LF). Sum CF×N×P_ind yields peak demand ≈20.5 kW. ## Answer The 5th year peak is found by projecting consumer numbers and consumption, converting daily energy to individual peak with the load factor, and combining class peaks using contribution factors. #### Growth multipliers Consumption growth: 5% in year 1, increasing by 1% each year (6%, 7%, 8%, 9%): $$ \begin{aligned} G_c &= 1.05 \times 1.06 \times 1.07 \times 1.08 \times 1.09 = 1.40194 \\ G_p &= (1.02)^5 = 1.10408 \end{aligned} $$ #### Formulae $$ \begin{aligned} N_5 &= N_0 \times G_p \times \text{coverage} \\ E_5 &= E_0 \times G_c \\ P_{ind} &= \frac{E_5 / \text{effective days}}{24 \times LF} \\ P_{LC} &= \sum CF \times N_5 \times P_{ind} \end{aligned} $$ #### Number of consumers (5th year) | Class | $N_0$ | $N_5 = N_0 \times 1.10408 \times 0.7$ | |---|---|---| | Domestic | 100 | 77.29 | | Commercial | 20 | 15.46 | | Non-commercial | 50 | 38.64 | #### Consumption and individual peak | Class | $E_5$ (kWh/month) | Daily kWh | LF | $P_{ind}$ (kW) | |---|---|---|---|---| | Domestic | 30 × 1.40194 = 42.058 | 42.058/30 = 1.4019 | 0.25 | 0.23366 | | Commercial | 42.058 | 42.058/30 = 1.4019 | 0.30 | 0.19471 | | Non-commercial | 20 × 1.40194 = 28.039 | 28.039/20 = 1.4019 | 0.25 | 0.23366 | Example (commercial): $P_{ind} = \dfrac{1.4019}{24 \times 0.30} = 0.19471$ kW. #### Contribution to load centre peak | Class | Class peak (kW) | CF | Contribution (kW) | |---|---|---|---| | Domestic | 77.29 × 0.23366 = 18.058 | 1.0 | 18.058 | | Commercial | 15.46 × 0.19471 = 3.010 | 0.5 | 1.505 | | Non-commercial | 38.64 × 0.23366 = 9.029 | 0.1 | 0.903 | | **Total** | | | **20.47** | **Answer: 5th year peak load of the load centre ≈ 20.5 kW** (20.47 kW). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Growth is 5% in year 1 and increases by 1% each year. How is the 5th year consumption multiplier found? A. 1.05^5 B. 1.05 x 1.06 x 1.07 x 1.08 x 1.09 C. 1 + 5 x 0.09 D. 1.09^5 Answer: B. Each year has a different rate (5,6,7,8,9%), so the factors are multiplied: 1.40194. (Key point: Consumption growth multiplier) 2. How are 5th year consumer numbers obtained? A. N0 x consumption growth x coverage B. N0 x population growth (1.02)^5 x coverage factor C. N0 x coverage only D. N0 x 1.05^5 Answer: B. N5 = N0 x 1.10408 x 0.7; population growth, not consumption growth, increases consumer numbers. (Key point: Number of consumers) 3. How is the individual consumer peak calculated? A. E5 / (24 x LF) using daily energy E5/effective days B. Monthly energy x LF C. Daily energy x 24 x LF D. Monthly energy / 30 Answer: A. P_ind = (E5/effective days) / (24 x LF), since average = LF x peak. (Key point: Individual peak from load factor) 4. What do the contribution factors do? A. Convert kW to kVA B. Grow the consumers C. Weight each class peak by how much it contributes at the system peak, before summing D. Give the load factor Answer: C. P_LC = sum(CF x class peak) accounts for classes not peaking together; result 20.47 kW. (Key point: Combining classes) ## Close variants set in other exams - [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 (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× ## 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×