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Transformer sizing and 11 kV feeder SLD

Short answer

Size the transformer by plant maximum demand: sum (connected kVA × demand factor) divided by diversity factor plus20-25% margin. Example: 120 kVA(0,7), 20 kVA(0,9), 40 kVA(0,6) sum 126 kVA; /1,3 =96,9 kVA; +20% =116,3 kVA →160 kVA. SLD: 11 kV feeder, LA, AB switch, DO fuse, 11/0.4 kV Dyn11, LT MCCB, MDB, MCCB to SDB, MCB/RCCB to230 V.

Explain with example, how does the proper size of transformer determined for a industry. Draw single diagram of electrical connection of 11KV feeder up to end user at 230 V.

Determining the proper transformer size

The transformer is sized on the maximum demand, not on connected load:

  1. List all connected loads in kVA (kVA = kW / (η·pf)).
  2. Apply demand factor (or Ku) to each load group to get its maximum demand.
  3. Divide the sum of group maximum demands by the diversity factor (or multiply by Ks).
  4. Add 20–25 % for future expansion.
  5. Choose the next higher standard rating (50, 100, 160, 200, 250, 315, 400 kVA ...).

Example: an industry has motors 120 kVA (demand factor 0.7), lighting 20 kVA (DF 0.9) and other loads 40 kVA (DF 0.6); diversity factor 1.3.

Max demands: 120×0.7 = 84, 20×0.9 = 18, 40×0.6 = 24
Sum = 126 kVA
Plant max demand = 126 / 1.3 = 96.92 kVA
With 20 % margin = 96.92 × 1.2 = 116.31 kVA

Selected: 160 kVA, 11/0.4 kV transformer.

Single line diagram: 11 kV feeder to 230 V end user

 11 kV feeder
     |
   [LA]   lightning arrester
     |
   [AB]   isolator
     |
   [DO]   HT fuse
     |
   (TR)   11/0.4 kV, Dyn11
     |
   [MCCB] LT main + meter
     |
 === MDB 400 V ===
     |
   [MCCB] -> SDB (400/230 V)
     |
   [MCB + RCCB]
     |
   230 V end user (phase-neutral)

✓ Answer checked by Claude Opus 5.5 (high effort)

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