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Fuse selection for 250 kVA transformer

Short answer

Full-load currents: HV =13.12A, LV =347.8A. LV short-circuit current =7.32kA (276A on HV side). Select HV fuse rated 25A HRC (1.5–2x full-load) and LV fuse rated 400A HRC, ensuring HV fuse withstands 157.5A inrush for 0.1s and discrimination: LV total I²t < HV pre-arcing I²t.

A transformer having capacity of 250 KVA, 11/0.415 KV, percentage impedance of 4.75% is required to be protected with fuses. Suggest guide line for selection of HV and LV side fuses. State any assumptions you make.

Given: 250 kVA, 11/0.415 kV, 3-phase, impedance 4.75 %.

Assumptions: delta–star (Dyn11) transformer; infinite source on 11 kV side (gives the highest fault current); HRC (gG/back-up) fuses used on both sides; magnetising inrush ≈ 12 × full-load for 0.1 s.

Step 1: Full-load currents

 I_FL(HV) = 250 000 / (√3 × 11 000) = 13.12 A
 I_FL(LV) = 250 000 / (√3 × 415)    = 347.80 A

Step 2: Fault current for a fault at LV terminals

 I_sc(LV) = I_FL(LV) / Z_pu
          = 347.80 / 0.0475 = 7322 A ≈ 7.32 kA
 Seen on HV side = 13.12 / 0.0475 = 276.2 A

Step 3: Inrush current

 I_inrush ≈ 12 × 13.12 = 157.5 A for about 0.1 s (HV side)

Guidelines for HV fuse

  1. Rating must exceed full-load current with margin for permissible overload: about 1.5–2 × I_FL → 19.7–26.2 A. Choose standard 25 A HV HRC fuse.
  2. It must not melt on inrush: the 0.1 s point of its T–I curve must be above 157.5 A (for a 25 A fuse this is normally satisfied; otherwise use 31.5 A).
  3. It must clear a fault at the LV terminals (276 A on HV side, i.e. ≈ 11 × its rating) quickly, within the transformer short-circuit withstand time (2 s as per IEC 60076-5).
  4. Breaking capacity must exceed the 11 kV system fault level.
  5. It must discriminate with the LV fuse: for an LV feeder fault, the LV fuse must clear before the HV fuse melts (HV fuse curve, referred to LV via ratio 11/0.415 = 26.5, must lie above the LV fuse curve).

Guidelines for LV fuse

  1. Rating ≥ full-load current: 347.8 A → choose standard 400 A gG fuse (≈ 1.15 × I_FL), allowing normal overload.
  2. Breaking capacity ≥ 7.32 kA (a standard 80 kA LV HRC fuse is ample).
  3. It must clear an LV fault of 7.32 kA in a short time and protect the LV cables and busbars (its I²t below cable withstand).
  4. It must be selective with downstream feeder fuses (rating ratio ≈ 1.6:1 or more) and with the HV fuse.

Answer: HV fuse ≈ 25 A HRC (I_FL = 13.12 A); LV fuse ≈ 400 A HRC (I_FL = 347.8 A); LV terminal fault current = 7.32 kA (276 A on HV side). Final ratings are confirmed from manufacturer T–I curves.

✓ Answer checked by Claude Opus 5.5 (high effort)

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