--- title: "Actual load estimation in industrial plants" subject: "Industrial Power Distribution and Illumination" chapter: 1 key: "semester_6/industrial_power_distribution_and_illumination:industrial-load-estimation-method" source: https://www.bctengineer.com/q/semester_6/industrial_power_distribution_and_illumination/industrial-load-estimation-method site: BCT Engineering --- # Actual load estimation in industrial plants Industrial Power Distribution and Illumination · Chapter 1: Introduction - Asked in: 2079 Chaitra, 4 marks; 2068 Bhadra (old course), 8 marks (asked 2 times) - Video: [Understanding Electrical Load Calculations & Load balancing](https://www.youtube.com/watch?v=xVm03u9xSRo) — Engineering Master - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/industrial_power_distribution_and_illumination/industrial-load-estimation-method - Whole chapter: https://www.bctengineer.com/md/semester_6/industrial_power_distribution_and_illumination/chapter-1 ## Question How actual load estimation is done in the industrial plants? Explain with the help of proper examples. ## Short answer Estimation uses utilization factor ku and simultaneity factor ks. For each load, apparent power = Pn/(η×cosφ). Apply ku (motors 0.75), then ks per group (4 motors: 0.8). Sum at each level, apply main board ks (0.9), add 25% margin. Example: 4×7.5 kW motors, 4 kVA lighting, 10×1 kVA sockets yields 27.06 kVA demand, so a 50 kVA transformer is chosen. ## Answer **Actual load estimation** means finding the maximum power (kVA) that the plant will really draw, not the sum of nameplate ratings. Machines do not run at full load all the time and do not all run together, so the installed power is reduced by suitable factors. The common method (IEC / Schneider Electrical Installation Guide) works level by level. #### Steps 1. **List the installed power Pₙ (kW)** of each load from nameplates. 2. **Find the installed apparent power** of each load: ```text Pa = Pn / (η × cos φ) (kVA) ``` For lamps, add ballast losses. 3. **Apply the factor of maximum utilization ku** (≤ 1): the ratio of the actual maximum power a device takes to its rating. Typical: motors 0.75, lighting and heating 1. 4. **Apply the factor of simultaneity ks** (≤ 1) for each group, because not all loads in a group run at the same time. Typical for distribution boards (IEC 61439): 2–3 circuits 0.9, 4–5 circuits 0.8, 6–9 circuits 0.7, 10 or more 0.6. 5. **Add up level by level:** load → sub-board → main board → transformer. At each level, multiply the sum by that level's ks. 6. **Add a margin for future growth** (about 20–25%) and select the next standard transformer size. #### Example A small workshop has: - 4 lathe motors, 7.5 kW each, η = 0.88, cos φ = 0.85 - Lighting load 4 kVA - 10 socket outlets of 1 kVA each ```text Motor: Pa = 7.5/(0.88×0.85) = 10.03 kVA with ku = 0.75 = 7.52 kVA 4 motors, ks = 0.8: 4 × 7.52 × 0.8 = 24.06 kVA Light: 4 × ku 1 × ks 1 = 4.00 kVA Socket: 10 × 1 × ks 0.2 = 2.00 kVA Sum = 30.06 kVA Main board ks = 0.9: 30.06 × 0.9 = 27.06 kVA Future margin 25%: 27.06 × 1.25 = 33.82 kVA ``` The installed load is 4 × 10.03 + 4 + 10 = 54.1 kVA, but the estimated demand is only about 27 kVA. **A 50 kVA transformer** (next standard size above 33.8 kVA) is enough, instead of a much larger one sized on connected load. #### Why it matters - Avoids oversized transformers, cables and switchgear (high cost, poor efficiency at light load). - Avoids undersizing, which causes overheating and tripping. - Gives the maximum demand used for tariff and contract demand with the utility. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What does actual load estimation find? A. The sum of all nameplate ratings B. The average energy per year only C. The cost of the cables D. The maximum power the plant will really draw, not the sum of nameplate ratings Answer: D. Machines do not run at full load or together, so factors reduce the installed power to a realistic demand. (Key point: Purpose of load estimation) 2. How is the installed apparent power Pa of a motor found? A. Pa = Pn x efficiency x cos phi B. Pa = Pn / cos phi only, ignoring efficiency C. Pa = Pn / (efficiency x cos phi) D. Pa = Pn x ku Answer: C. Input kVA is rated kW divided by efficiency and power factor. (Key point: Installed apparent power) 3. How are ku and ks applied? A. ks to each load, then ku to groups B. ku to each individual load first, then ks to groups at each distribution level C. Both only at the transformer level D. Neither is used for motors Answer: B. ku accounts for a device not running at full rating, ks for loads of a group not running together. (Key point: Order of ku and ks) 4. In the workshop example, why is a 50 kVA transformer chosen? A. Estimated demand with 25% margin is about 33.8 kVA, so the next standard size above is used B. The connected load is 54.1 kVA, so 50 kVA is just below it C. The motors alone need exactly 50 kVA D. Standard sizes below 100 kVA are not made Answer: A. 27.06 kVA x 1.25 = 33.82 kVA, so the next standard size, 50 kVA, is enough. 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