--- title: "5th year peak load (commercial 30 days)" subject: "Transmission and Distribution System Design" chapter: 6 key: "semester_8/transmission_and_distribution_system_design:fifth-year-peak-load-30-days" source: https://www.bctengineer.com/q/semester_8/transmission_and_distribution_system_design/fifth-year-peak-load-30-days site: BCT Engineering --- # 5th year peak load (commercial 30 days) Transmission and Distribution System Design · Chapter 6: Electrical Load Characteristics and Load Forecast - Asked in: 2068 Bhadra (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-30-days - 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 | 100 | 30 kWhr | 30 | 1.0 | 0.25 | | Commercial | 20 | 30 kWhr | 30 | 0.5 | 0.30 | | Non-commercial | 50 | 20 kWhr | 20 | 0.1 | 0.25 | Assuming consumption growth factors of 5% in each year and electrification coverage factor by 5th year is 70% for all consumer classes determine the 5th year peak load of the load center in kW. ## Short answer Grow each class's monthly consumption by (1.05)⁵ = 1.27628, apply 70% coverage to consumer numbers, divide daily energy by (24 × LF) for individual peak, then sum class peaks weighted by contribution factors. The 5th year peak load is approximately 16.88 kW. ## Answer The 5th year peak of the load centre is found by growing each class's consumption to the 5th year, applying the coverage factor to the consumer number, converting daily energy to individual peak using the 5th year load factor, and adding class peaks weighted by their contribution factors. **Assumption:** no population growth is given, so the number of potential consumers stays the same; growth is applied for $n = 5$ years. #### Formulae $$ \begin{aligned} N_5 &= N_0 \times \text{coverage factor} \\ E_5 &= E_0 (1 + g)^5 \\ E_{day} &= \frac{E_5}{\text{effective days per month}} \\ P_{ind} &= \frac{E_{day}}{24 \times LF_5} \\ P_{class} &= N_5 \times P_{ind} \\ P_{LC} &= \sum CF_i \times P_{class,i} \end{aligned} $$ #### Step 1: Consumption growth multiplier $$ (1 + 0.05)^5 = 1.27628 $$ #### Step 2: Effective number of consumers | Class | $N_0$ | $N_5 = 0.7 N_0$ | |---|---|---| | Domestic | 100 | 70 | | Commercial | 20 | 14 | | Non-commercial | 50 | 35 | #### Step 3: Consumption and individual peak demand | Class | $E_5$ (kWh/month) | $E_{day}$ (kWh) | $LF_5$ | $P_{ind}$ (kW) | |---|---|---|---|---| | Domestic | 30 × 1.27628 = 38.288 | 38.288/30 = 1.2763 | 0.25 | 0.21271 | | Commercial | 38.288 | 38.288/30 = 1.2763 | 0.30 | 0.17726 | | Non-commercial | 20 × 1.27628 = 25.526 | 25.526/20 = 1.2763 | 0.25 | 0.21271 | Sample calculation (domestic): $$ P_{ind} = \frac{1.2763}{24 \times 0.25} = \frac{1.2763}{6} = 0.21271 \text{ kW} $$ #### Step 4: Class peak and contribution to load centre peak | Class | Class peak (kW) | CF | Contribution (kW) | |---|---|---|---| | Domestic | 70 × 0.21271 = 14.890 | 1.0 | 14.890 | | Commercial | 14 × 0.17726 = 2.482 | 0.5 | 1.241 | | Non-commercial | 35 × 0.21271 = 7.445 | 0.1 | 0.745 | | **Total** | 24.817 | | **16.875** | #### Step 5: Load centre peak $$ P_{LC} = 14.890 + 1.241 + 0.745 = 16.875 \text{ kW} $$ Note that the plain sum of class peaks (24.82 kW) would overestimate the demand by about 47%, because commercial and non-commercial loads peak at other hours. **Answer: 5th year peak load of the load centre ≈ 16.88 kW.** For transformer selection, this would be converted to kVA (e.g. at 0.85 p.f., $16.88/0.85 \approx 19.9$ kVA), plus a loss allowance, so a standard 25 kVA transformer would be suitable. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. With 5% growth each year, consumption in the 5th year is E0 times: A. 1.25 B. 1.05^5 = 1.27628 C. 1.05 x 5 D. 1.05^4 Answer: B. Compound growth over n = 5 years gives (1.05)^5 = 1.27628. (Key point: Growth multiplier) 2. With no population growth given, the 5th year number of consumers is: A. N0 x 1.05^5 B. N0 x 0.7 (coverage) C. N0 D. N0 x 0.3 Answer: B. Only the 70% coverage factor is applied, so Domestic = 70, Commercial = 14, Non-commercial = 35. (Key point: Consumer number) 3. The domestic individual peak is: A. 1.2763/(24 x 0.25) = 0.2127 kW B. 38.288/(24 x 0.25) = 6.4 kW C. 1.2763 x 0.25 = 0.319 kW D. 1.2763/0.25 = 5.1 kW Answer: A. Daily energy 38.288/30 = 1.2763 kWh divided by 24 x LF = 6 gives 0.2127 kW. (Key point: Individual peak) 4. Why is the load centre peak 16.88 kW rather than the 24.82 kW sum of class peaks? A. Contribution factors weight each class, as classes do not peak together B. Losses were subtracted C. Coverage was applied twice D. Power factor was applied Answer: A. P = 14.890 + 0.5(2.482) + 0.1(7.445) = 16.875 kW. 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