--- title: "Peak demand of load center from load pattern" subject: "Transmission and Distribution System Design" chapter: 6 key: "semester_8/transmission_and_distribution_system_design:peak-demand-load-center-load-pattern" source: https://www.bctengineer.com/q/semester_8/transmission_and_distribution_system_design/peak-demand-load-center-load-pattern site: BCT Engineering --- # Peak demand of load center from load pattern Transmission and Distribution System Design · Chapter 6: Electrical Load Characteristics and Load Forecast - Asked in: 2073 Magh, 10 marks - Video: [Solved Problem 1 - Load Curve & Load Duration Curve Made Easy⚡(Average Load + Load Factor Explained)](https://www.youtube.com/watch?v=z005OTiWeqk) — Power Systemz - 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/peak-demand-load-center-load-pattern - Whole chapter: https://www.bctengineer.com/md/semester_8/transmission_and_distribution_system_design/chapter-6 ## Question The consumer data for particular distribution transformer for a specified year is as shown as follows. Table-1: | Consumer class | Class A | Class B | Class C | |---|---|---|---| | Consumer number | 80 | 30 | 40 | | Monthly energy consumption | 30 kWh | 22 kWh | 25 kWh | | No. of effective days per month | 30 | 24 | 26 | | Coincidence factor | 1 | 0.9 | 0.95 | | Power factor | 0.9 | 0.8 | 0.85 | | Load pattern | Table 2 | Table 2 | Table 2 | Table-2 (load pattern): | Time (hrs.) | 0:00-6:00 | 6:00-10:00 | 10:00-14:00 | 14:00-18:00 | 18:00-21:00 | 21:00-24:00 | |---|---|---|---|---|---|---| | Class A | 0.1 | 0.4 | 0.2 | 0.2 | 1 | 0.4 | | Class B | 0.1 | 0.1 | 0.5 | 1 | 0.5 | 0.1 | | Class C | 0.1 | 0.1 | 1 | 1 | 0.1 | 0.1 | Determine the peak demand of the load center. ## Short answer Compute each class’s equivalent full-load hours from the load pattern; class peak = (daily kWh / equivalent hours) × consumers × coincidence factor. Sum class contributions per time block. Peak occurs 18:00–21:00 at 11.79 kW (13.29 kVA at ~0.89 pf). ## Answer The load pattern gives each class's load in per unit of its own peak. The class peak is found from its daily energy and the equivalent full-load hours of the pattern; the load centre peak is then the maximum of the summed class load curves. #### Step 1: Equivalent full-load hours per day $$ h_{eq} = \sum(\text{p.u. load} \times \text{duration}) $$ - Class A: $0.1(6)+0.4(4)+0.2(4)+0.2(4)+1(3)+0.4(3) = 8.0$ h - Class B: $0.1(6)+0.1(4)+0.5(4)+1(4)+0.5(3)+0.1(3) = 8.8$ h - Class C: $0.1(6)+0.1(4)+1(4)+1(4)+0.1(3)+0.1(3) = 9.6$ h #### Step 2: Class peak demand $$ P_{class} = N \times \frac{\text{Monthly kWh} / \text{Effective days}}{h_{eq}} \times \text{Coincidence factor} $$ | Class | Daily kWh | $P_{ind}$ (kW) | $P_{class}$ (kW) | |---|---|---|---| | A | 30/30 = 1.000 | 1/8 = 0.1250 | 80 × 0.1250 × 1 = 10.000 | | B | 22/24 = 0.9167 | 0.9167/8.8 = 0.1042 | 30 × 0.1042 × 0.9 = 2.813 | | C | 25/26 = 0.9615 | 0.9615/9.6 = 0.1002 | 40 × 0.1002 × 0.95 = 3.806 | #### Step 3: Combined load curve (kW) | Time | A | B | C | Total kW | |---|---|---|---|---| | 0–6 | 1.000 | 0.281 | 0.381 | 1.66 | | 6–10 | 4.000 | 0.281 | 0.381 | 4.66 | | 10–14 | 2.000 | 1.406 | 3.806 | 7.21 | | 14–18 | 2.000 | 2.813 | 3.806 | 8.62 | | 18–21 | 10.000 | 1.406 | 0.381 | **11.79** | | 21–24 | 4.000 | 0.281 | 0.381 | 4.66 | The maximum occurs during **18:00–21:00**. #### Step 4: Peak in kVA Reactive power at peak, $Q = P\tan\phi$: $$ \begin{aligned} Q &= 10(0.4843) + 1.406(0.75) + 0.381(0.6197) \\ &= 4.843 + 1.055 + 0.236 = 6.134 \text{ kVAR} \\ S &= \sqrt{11.787^2 + 6.134^2} = 13.29 \text{ kVA} \end{aligned} $$ The contribution factors at peak are A = 1.0, B = 0.5, C = 0.1. **Answer: Peak demand of the load centre ≈ 11.79 kW (≈ 13.29 kVA at about 0.89 p.f.), occurring between 18:00 and 21:00.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is the equivalent full-load hours h_eq of a class used for? A. To convert daily energy to the class's individual peak demand B. To find the power factor C. To find the number of consumers D. To find the coincidence factor Answer: A. h_eq = sum(p.u. load x duration); individual peak = daily kWh / h_eq. (Key point: Equivalent full-load hours) 2. Which expression gives the class peak demand? A. N x monthly kWh x h_eq B. N x (monthly kWh / effective days) / h_eq x coincidence factor C. N x h_eq / daily kWh D. Monthly kWh x power factor / 30 Answer: B. Daily energy divided by h_eq gives the individual peak; multiply by number of consumers and coincidence factor. (Key point: Class peak formula) 3. How is the peak demand of the load centre obtained? A. Sum of the three class peaks B. Peak of the largest class only C. Maximum of the summed class load curves over the day D. Average of the class peaks Answer: C. The class curves peak at different times, so the maximum of their sum (11.79 kW at 18-21 h) is the load centre peak. (Key point: Finding the load centre peak) 4. In which period and at what value does the peak occur? A. 14:00-18:00, 8.62 kW B. 18:00-21:00, 11.79 kW C. 10:00-14:00, 7.21 kW D. 21:00-24:00, 4.66 kW Answer: B. Class A is at its full 10 kW in 18-21, with B at 1.406 and C at 0.381 kW, total 11.79 kW. 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