Transformer size for given industry (figure)

The figure is not available, so the method is shown on an assumed industry with two sub-distribution boards; the same steps apply to any given layout.

Assumed loads

• SDB-A: 5 motors, each 11 kW, η = 0.88, pf = 0.85
• SDB-B: lighting 8 kVA, two heaters of 10 kVA each

Factors used (IEC / Schneider Electric guide)

• Ku (factor of maximum utilization): motors 0.75; lighting and heating 1.0.
• Ks (factor of simultaneity) for a board with n circuits: 2–3 → 0.9, 4–5 → 0.8, 6–9 → 0.7, 10 and above → 0.6.

Step 1: rated kVA of each motor

kVA = kW / (η × pf) = 11 / (0.88 × 0.85) = 14.71 kVA

Step 2: utilized power with Ku

Step 3: apply Ks at each board

SDB-A = 55.15 × Ks(5 circuits = 0.8)  = 44.12 kVA
SDB-B = (8 + 20) × Ks(3 circuits = 0.9) = 25.20 kVA
MDB   = (44.12 + 25.20) × Ks(2 = 0.9)  = 69.32 × 0.9
      = 62.39 kVA

Step 4: future expansion and selection

Required = 62.39 × 1.2 (20 % margin) = 74.87 kVA

Answer: select a 100 kVA, 11/0.4 kV transformer (next standard rating above 74.87 kVA; standard sizes 50, 100, 160, 200, 250 kVA ...).

Remarks

• Connected load is 101.5 kVA but maximum demand is only about 62 kVA; sizing on connected load would over-size the transformer and lower its efficiency.
• If the figure gives Ku and Ks values directly, use those instead of the typical values.
• Check that the largest motor's starting kVA does not cause an excessive dip (usually < 10–15 %) on the chosen transformer.