--- title: "T/F: feeder average consumption near peak" subject: "Transmission and Distribution System Design" chapter: 6 key: "semester_8/transmission_and_distribution_system_design:tf-average-consumption-near-peak" source: https://www.bctengineer.com/q/semester_8/transmission_and_distribution_system_design/tf-average-consumption-near-peak site: BCT Engineering --- # T/F: feeder average consumption near peak Transmission and Distribution System Design · Chapter 6: Electrical Load Characteristics and Load Forecast - Asked in: 2080 Chaitra, 4 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/tf-average-consumption-near-peak - Whole chapter: https://www.bctengineer.com/md/semester_8/transmission_and_distribution_system_design/chapter-6 ## Question State whether the following statement is TRUE or FALSE and justify your answer with brief explanation: The average consumption in the feeder must be almost near to the peak demand. ## Short answer FALSE. Feeder load factors are typically 0.3–0.7, so average demand is well below peak because loads vary markedly through the day. Equipment is sized for peak, and energy loss uses loss factor, not assuming peak load all day. ## Answer **FALSE.** The ratio of average demand to peak demand of a feeder is its **load factor**: $$ LF = \frac{\text{Average demand}}{\text{Peak demand}} = \frac{\text{Energy in period } T}{P_{peak} \times T} $$ If average consumption were almost equal to the peak, the load factor would be close to 1, which happens only for a flat, round-the-clock load (e.g. a continuous-process industry). **Justification:** - A distribution feeder mainly supplies domestic and commercial consumers whose use varies strongly through the day: low at night, moderate in the day, and a sharp evening peak (lighting, cooking, TV). - Typical feeder load factors are about 0.3–0.5 for residential/rural feeders and 0.5–0.7 for mixed urban feeders. So the average demand is usually only 30–60% of the peak. - Example: a feeder with a 1 MW peak and 9,600 kWh/day has an average of 400 kW, i.e. LF = 0.4. - This is why the feeder and transformer must be sized for the peak, not the average, and why energy loss is computed using the loss of load factor ($LLF = 0.3LF + 0.7LF^2$) rather than assuming peak loss all day. A high load factor is desirable (better use of equipment, lower loss per kWh), and utilities try to raise it by demand-side management, but in practice the average is well below the peak. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. 'Average consumption in a feeder must be almost near the peak demand' is: A. True, since feeders run flat B. False, since this would mean a load factor near 1, which is rare for domestic and commercial feeders C. True for residential feeders D. False because average always exceeds peak Answer: B. Average/peak is the load factor; feeders have strongly varying load, so LF is well below 1. (Key point: Verdict) 2. Typical feeder load factors are about: A. 0.9 to 1.0 B. 0.3 to 0.7 C. 1.5 to 2 D. 0.01 to 0.05 Answer: B. About 0.3-0.5 for residential or rural and 0.5-0.7 for mixed urban feeders. (Key point: Typical values) 3. A feeder has 1 MW peak and 9,600 kWh/day. Its load factor is: A. 0.4 B. 0.96 C. 0.04 D. 0.24 Answer: A. Average = 9600/24 = 400 kW, so LF = 400/1000 = 0.4. 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