IEC 60909-010 min read

Worked Example: IEC 60909 Short-Circuit Currents in an 11/0.4 kV Network with a Generator and a Motor

Initial, peak, breaking and thermal short-circuit currents at three buses fed by the utility, a standby generator and an induction motor — and why the peak current needs the 1.15 factor here.

Scenario

Utility supply11 kV, Sk″ = 250 MVA, R/X = 0.1
Generator on the 11 kV board2.5 MVA, 11 kV, xd″ = 14%, cos φ = 0.8
Transformer1.6 MVA, 11/0.4 kV Dyn, uk = 6%, Pk = 15 kW
Cable to the MCC50 m of 3-core 240 mm² Cu (0.0754 + j0.078 Ω/km)
Motor on the MCC160 kW, 0.4 kV, ILR/IrM = 7, η = 0.95, cos φ = 0.88
Voltage factorcmax = 1.10 at 11 kV, 1.05 at 0.4 kV (+6% tolerance)

Step-by-step calculation

Step 1: Build the network impedances on a common base

Each element is converted to per-unit on 100 MVA: the utility from its Sk″, the generator with its correction factor KG, the transformer with KT = 0.95·cmax/(1 + 0.6·xT), the cable from its length, and the motor from its locked-rotor current.

ZQ = c·Un²/Sk″ KT = 0.95·cmax/(1 + 0.6·xT) ZM = (1/(ILR/IrM))·UrM²/SrM

Step 2: Find the initial symmetrical short-circuit current Ik″ at each bus

Ik″ = c·Un / (√3·|Zk|)
BusIk″ (kA)Utility shareGenerator shareMotor share
11 kV switchboard14.2013.121.010.07
0.4 kV LV board39.5835.002.701.95
Motor control centre23.4919.921.542.03

Shares are magnitudes from the network solution, so they add up only approximately.

Step 3: Peak current ip with method B

Several sources feed each fault, so IEC 60909-0 method B applies: κ from the R/X ratio at the fault, multiplied by 1.15 and limited to 2.0 (HV) or 1.8 (LV).

κ = 1.02 + 0.98·e^(−3R/X) κ(b) = 1.15·κ ip = κ(b)·√2·Ik″
BusR/Xκ(b)ip (kA)
11 kV switchboard0.0992.00 (limit)40.17
0.4 kV LV board0.1651.80 (limit)100.77
Motor control centre0.4581.45848.44

Step 4: Breaking and thermal currents

At the 11 kV board the generator's contribution decays by tmin = 0.1 s, so the breaking current Ib is slightly lower than Ik″. Motor decay is not credited, which is conservative. The thermal equivalent current uses Tk = 1 s and n = 1.

Step 5: Check the switchgear ratings

BusRatingLargest fault (3-phase or earth)Result
11 kV switchboard25 kA14.50 kA (earth fault)Adequate
0.4 kV LV board50 kA40.34 kA (earth fault)Adequate
Motor control centre36 kA23.49 kAAdequate

Result summary

CheckRequirementActualStatus
11 kV board short-circuit rating≥ 14.50 kA25 kA✓ PASS
LV board short-circuit rating≥ 40.34 kA50 kA✓ PASS
LV board making capacity (peak)≥ 100.8 kACheck breaker Icm✓ PASS
MCC short-circuit rating≥ 23.49 kA36 kA✓ PASS
All three switchboards are adequate for the symmetrical fault current. The LV board sees almost 40 kA with a 101 kA peak, so its main breaker needs a making capacity (Icm) of at least about 101 kA — a 50 kA breaker typically has Icm = 105 kA at 400 V, which just covers it.

Key insight: The motor adds about 2 kA at the MCC, nearly 9% of the fault current there, even though it is a modest 160 kW machine. Leaving motors out of a maximum short-circuit study underestimates both the breaking and the peak duty.

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Frequently asked questions

Why is the 11 kV earth-fault current higher than the three-phase current?

The generator's neutral is solidly earthed and its zero-sequence reactance (7%) is half its subtransient reactance (14%), while the utility's zero-sequence impedance equals its positive-sequence impedance in this model. The zero-sequence impedance at the bus is therefore lower than the positive-sequence impedance, and with Z0 < Z1 the single-phase fault current exceeds the three-phase current. (The transformer's delta winding is on the 11 kV side, so it adds no zero-sequence path there.)

When can I use κ without the 1.15 factor?

When a single source feeds the fault through a radial (non-meshed) path, κ from that path's R/X is used directly. With several sources in parallel, method B's 1.15 factor covers the different R/X ratios of the branches.

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