IEC 60502-2 / IEC 609498 min read

Worked Example: 11 kV Transformer Feeder Where the Short-Circuit Rating, Not the Load, Sets the Size

Circuit C01 of the worked example project: a 2 MVA transformer needs 105 A, but 25 kA for 0.47 s needs a 120 mm² conductor. Includes the MV screen check.

Scenario

Load11 kV feeder to a 2 000 kVA transformer
CableSingle-core Cu/XLPE 6/10 kV, trefoil, buried direct, 180 m, two circuits 0.2 m apart
Ground40 °C, soil 2.0 K·m/W, depth 0.8 m
Fault level25 kA three-phase; earth fault 1 kA (low-resistance earthed)
Clearing timeRelay 0.40 s + breaker 0.07 s = 0.47 s
Screen25 mm² copper wire screen

Step-by-step calculation

Step 1: Design current

Ib = S / (√3 · U)
2 000 / (√3 × 11)
Ib = 105.0 A

Step 2: Correction factors (IEC 60502-2 Annex B)

FactorBasisValue
Ground temperature40 °C, Table B.11 (√((90−40)/(90−20)))0.845
Grouping2 trefoil circuits, 0.2 m, Table B.190.83
Soil resistivity2.0 K·m/W, Table B.14 (lowest value over sizes)0.88
Depth0.8 m (reference)1.00

Step 3: Current capacity

Table B.2, Cu single-core trefoil buried: 70 mm² 239 A, 95 mm² 285 A, 120 mm² 323 A.

Step 4: Short-circuit withstand (adiabatic)

S ≥ Ik · √t / k, k = 226 · √ln((250 + 234.5)/(90 + 234.5)) = 143.1
25 000 × √0.47 / 143.1
S ≥ 119.8 mm² → 120 mm²

95 mm² fails this check only; it is fine for current and voltage drop.

Step 5: Voltage drop

Step 6: Screen short-circuit check

S_screen ≥ Ik1 · √t / k, k = 128.5 (Cu, 80 → 200 °C)
1 000 × √0.47 / 128.5
5.3 mm² required ≤ 25 mm² fitted

Result summary

CheckRequirementActualStatus
Current capacityIz ≥ 105 A199.4 A (120 mm²)✓ PASS
Conductor short circuitS ≥ 119.8 mm²120 mm²✓ PASS
Screen short circuit≥ 5.3 mm²25 mm²✓ PASS
Voltage drop≤ 2 %0.07 %✓ PASS
3 × 1C 120 mm² Cu/XLPE 6/10 kV, 25 mm² copper screen. The suite names short circuit as the governing criterion: 95 mm² carries the load easily but cannot withstand 25 kA for 0.47 s.

Key insight: On MV feeders close to the source the fault level and the backup clearing time decide the conductor size. Use the actual primary clearing time from the coordination study — the backup time here (0.8 s relay + 0.07 s breaker) would require 163 mm², i.e. 185 mm².

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Frequently asked questions

Why is the soil factor 0.88 and not interpolated by size?

IEC 60502-2 gives soil factors per conductor size. The suite uses the lowest value over all sizes at each resistivity, which is safe side (the spread is under 0.02 at 2.0 K·m/W).

Which clearing time should be used?

The suite's Design Basis lets you choose primary or backup clearing; the example uses primary (relay + breaker). Many specifications require the backup time for feeders.

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