--- title: "Selecting distribution transformer size using Ku and Ks" subject: "Industrial Power Distribution and Illumination" chapter: 4 key: "semester_6/industrial_power_distribution_and_illumination:transformer-size-ku-ks" source: https://www.bctengineer.com/q/semester_6/industrial_power_distribution_and_illumination/transformer-size-ku-ks site: BCT Engineering --- # Selecting distribution transformer size using Ku and Ks Industrial Power Distribution and Illumination · Chapter 4: Distribution Substation of Industrial Plant - Asked in: 2077 Chaitra, 4 marks; 2075 Baisakh, 8 marks; 2071 Magh, 8 marks (asked 3 times) - 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/transformer-size-ku-ks - Whole chapter: https://www.bctengineer.com/md/semester_6/industrial_power_distribution_and_illumination/chapter-4 ## Question Explain with examples, select the size of distribution transformer for different application considering factor of maximum utilization and factor of simultaneity. ## Short answer Multiply each load's rated kVA by its maximum utilization factor Ku, sum per board, multiply by simultaneity factor Ks, then apply the main board Ks. Add 20% future margin and select next higher standard rating. Example: 114 kVA connected load yields 75.82 kVA after Ku/Ks and margin, so a 100 kVA transformer is chosen. ## Answer The transformer for a building or industry is sized on the **maximum demand**, not on the sum of connected loads, because loads rarely run at full rating and never all at the same time. Two factors (IEC/Schneider Electric method) are used: - **Factor of maximum utilization, Ku** (≤ 1): the ratio of the power actually drawn by a load to its rated power. Typical: motors 0.75, lighting 1, heating 1, socket outlets 0.1–0.2 (depends on use), EV/AC as per duty. - **Factor of simultaneity (diversity), Ks** (≤ 1): accounts for the fact that all loads of a group do not run together. Typical Ks for a distribution board (IEC 61439): 2–3 circuits 0.9, 4–5 circuits 0.8, 6–9 circuits 0.7, 10 or more circuits 0.6. For residential buildings Ks falls with number of consumers (e.g. 1.0 for 2–4 flats, 0.78 for 5–9 flats, 0.4 for 50 or more flats). #### Procedure 1. List every load with its rated power in kVA (kVA = kW / (η·cosφ)). 2. Multiply each load by Ku to get its utilized power. 3. Add the loads of each SDB and multiply by the Ks of that board. 4. Add the SDB demands at the MDB and multiply by the Ks of the MDB. 5. Add a margin for future expansion (about 20–25 %). 6. Choose the next higher standard transformer rating (25, 50, 100, 160, 200, 250, 315, 400, 500, 630, 800, 1000 kVA ...). #### Example: a small factory | Board | Load | kVA | Ku | Utilized kVA | |---|---|---|---|---| | SDB-1 | 4 motors × 18 kVA | 72 | 0.75 | 54.0 | | SDB-2 | Lighting | 12 | 1.0 | 12.0 | | SDB-2 | Socket outlets | 15 | 0.2 | 3.0 | | SDB-3 | 3 AC units × 5 kVA | 15 | 1.0 | 15.0 | ```text SDB-1: 54.0 × Ks(4 circuits = 0.8) = 43.2 kVA SDB-2: (12 + 3) × Ks(2 circuits = 0.9) = 13.5 kVA SDB-3: 15 × Ks(3 circuits = 0.9) = 13.5 kVA Sum at MDB = 43.2 + 13.5 + 13.5 = 70.2 kVA MDB demand = 70.2 × Ks(3 boards = 0.9) = 63.18 kVA With 20 % future margin = 63.18 × 1.2 = 75.82 kVA ``` **Selected transformer: 100 kVA, 11/0.4 kV** (next standard size above 75.82 kVA). Compare: the total connected load is 114 kVA, but the transformer is only 100 kVA, and it will run at about 63 % load at peak, near its best efficiency region. #### Other applications - **Apartment building**: flats have high connected load but low Ks (e.g. 50 flats × 8 kVA × Ks 0.4 = 160 kVA plus common services such as lifts and pumps). - **Office**: lighting and AC Ku ≈ 1, computers and sockets lower; Ks of floors about 0.7–0.8. - **Industry**: motor-dominated; Ku ≈ 0.75, Ks depends on shifts and process; large motors' starting must be checked. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. On what is a distribution transformer sized according to the answer? A. The maximum demand, found using Ku and Ks B. The total connected load in kVA C. The rating of the largest motor only D. The average monthly energy consumption Answer: A. Loads rarely run at full rating and never all together, so sizing uses maximum demand rather than connected load. (Key point: Basis of sizing) 2. What does the factor of maximum utilization Ku represent? A. Ratio of loads running at the same time to total loads B. Ratio of average load to peak load C. Ratio of power actually drawn by a load to its rated power D. Ratio of transformer rating to connected load Answer: C. Ku is applied to each load individually (e.g. 0.75 for motors); Ks handles the not-all-at-once effect. (Key point: Meaning of Ku) 3. Why is the factor of simultaneity Ks applied to a group of loads on a board? A. Because not all loads of the group run together B. Because motors draw less than rated power C. Because the transformer is oversized D. Because cable resistance reduces the load Answer: A. Ks (diversity) accounts for loads of a group not operating at the same time; it falls as the number of circuits rises. (Key point: Meaning of Ks) 4. In the factory example, the MDB demand is 63.18 kVA. How is the transformer rating selected? A. Pick 63.18 kVA directly as a 63 kVA unit B. Use the 114 kVA connected load and pick 160 kVA C. Add 20% margin to get 75.82 kVA and pick the next standard size, 100 kVA D. Add 20% margin and pick the nearest lower size, 50 kVA Answer: C. The 20% expansion margin gives 75.82 kVA, and the next higher standard rating is 100 kVA. 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