Skip to main content

Selecting distribution transformer size using Ku and Ks

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.

Explain with examples, select the size of distribution transformer for different application considering factor of maximum utilization and factor of simultaneity.

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

BoardLoadkVAKuUtilized kVA
SDB-14 motors × 18 kVA720.7554.0
SDB-2Lighting121.012.0
SDB-2Socket outlets150.23.0
SDB-33 AC units × 5 kVA151.015.0
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.

✓ Answer checked by Claude Opus 5.5 (high effort)

All Industrial Power Distribution and Illumination chapter 4 questions →