--- title: "5th year peak load, 4% growth, 60% coverage" subject: "Transmission and Distribution System Design" chapter: 6 key: "semester_8/transmission_and_distribution_system_design:fifth-year-peak-load-4-percent" source: https://www.bctengineer.com/q/semester_8/transmission_and_distribution_system_design/fifth-year-peak-load-4-percent site: BCT Engineering --- # 5th year peak load, 4% growth, 60% coverage Transmission and Distribution System Design · Chapter 6: Electrical Load Characteristics and Load Forecast - Asked in: 2067 Mangsir (old course), 12 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-4-percent - 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 no. | Avg. monthly consumption | No. of effective days/month | 5th year contribution factor to peak | 5th year load factor | |---|---|---|---|---|---| | Domestic | 200 | 30 kWhr | 30 | 1.0 | 0.25 | | Commercial | 50 | 30 kWhr | 30 | 0.5 | 0.30 | | Non-commercial | 40 | 20 kWhr | 25 | 0.1 | 0.25 | Assuming a uniform consumption growth factors of 4% in each year and electrification coverage factor by 5th year is 60% for all consumer classes determine the 5th year peak load of the load center in kW. ## Short answer Project consumption with (1.04)⁵ = 1.21665, apply 60% coverage to consumer numbers, compute individual peak as daily energy ÷ (24 × LF), and sum contributions. The 5th year peak load of the load centre is approximately 27.26 kW. ## Answer The method: project each class's consumption to the 5th year, apply the coverage factor to the consumer numbers, convert daily energy to individual peak demand using the load factor, then combine the class peaks using their contribution factors. **Assumption:** no population growth is given, so potential consumer numbers stay constant; consumption grows for $n = 5$ years. #### Formulae $$ \begin{aligned} N_5 &= N_0 \times \text{coverage} \\ E_5 &= E_0 (1+g)^5 \\ P_{ind} &= \frac{E_5/\text{effective days}}{24 \times LF_5} \\ P_{LC} &= \sum CF_i \times N_{5,i} \times P_{ind,i} \end{aligned} $$ #### Step 1: Growth multiplier $$ (1.04)^5 = 1.21665 $$ #### Step 2: Connected consumers in 5th year (60% coverage) | Class | $N_0$ | $N_5 = 0.6 N_0$ | |---|---|---| | Domestic | 200 | 120 | | Commercial | 50 | 30 | | Non-commercial | 40 | 24 | #### Step 3: Consumption and individual peak | Class | $E_5$ (kWh/month) | Daily kWh | $LF_5$ | $P_{ind}$ (kW) | |---|---|---|---|---| | Domestic | 30 × 1.21665 = 36.500 | 36.500/30 = 1.2167 | 0.25 | 0.20278 | | Commercial | 36.500 | 36.500/30 = 1.2167 | 0.30 | 0.16898 | | Non-commercial | 20 × 1.21665 = 24.333 | 24.333/25 = 0.9733 | 0.25 | 0.16222 | Sample (non-commercial): $$ P_{ind} = \frac{0.9733}{24 \times 0.25} = 0.16222 \text{ kW} $$ #### Step 4: Class peaks and contribution | Class | Class peak (kW) | CF | Contribution (kW) | |---|---|---|---| | Domestic | 120 × 0.20278 = 24.333 | 1.0 | 24.333 | | Commercial | 30 × 0.16898 = 5.069 | 0.5 | 2.535 | | Non-commercial | 24 × 0.16222 = 3.893 | 0.1 | 0.389 | | **Total** | 33.296 | | **27.257** | #### Step 5: Load centre peak $$ P_{LC} = 24.333 + 2.535 + 0.389 = 27.257 \text{ kW} $$ The non-coincident sum (33.30 kW) is higher because commercial and non-commercial classes do not peak with the domestic evening peak. **Answer: 5th year peak load of the load centre ≈ 27.26 kW.** At about 0.85 p.f. this is roughly 32 kVA; allowing for losses and later growth, a 50 kVA standard transformer would be selected. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The 5th year consumption multiplier for 4% growth is: A. 1.20 B. 1.04^5 = 1.21665 C. 1.04 x 5 D. 1.04^4 Answer: B. Compounded over 5 years: 1.04^5 = 1.21665. (Key point: Growth multiplier) 2. With 60% coverage, domestic, commercial and non-commercial consumers are: A. 120, 30, 24 B. 200, 50, 40 C. 80, 20, 16 D. 60, 30, 24 Answer: A. N5 = 0.6 x N0 = 0.6 x 200, 50, 40. (Key point: Connected consumers) 3. Why is the non-commercial individual peak lower (0.162 kW) than domestic (0.203 kW) although LF is the same? A. Its monthly energy is 20 kWh and it is spread over 25 effective days, giving lower daily energy (0.973 kWh) B. Its load factor is lower C. Its coverage is lower D. Its contribution factor is lower Answer: A. Daily energy = 24.333/25 = 0.9733 kWh versus 1.2167 kWh for domestic, so P_ind = 0.9733/6 = 0.1622 kW. (Key point: Effective days effect) 4. The 5th year load centre peak is: A. 33.30 kW B. 24.33 kW C. 27.26 kW D. 50 kW Answer: C. 24.333 + 0.5(5.069) + 0.1(3.893) = 27.257 kW; 33.30 kW is the non-coincident sum. (Key point: Final 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 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×