Short-circuit & equipment verification

This page runs the IEC 60909 short-circuit study of the scheme and turns it into selection verdicts. Four fault types are available (three-phase, two-phase, two-phase-to-earth, single-phase-to-earth), on the MV and the LV side, with a per-node table. The engine is engines/shortcircuit-60909.js (equivalent voltage source method, I"k = c·Un/(√3·Zk), voltage factor c per the standard, peak factor κ and ip, motor contribution) extended by engines/sequence-networks.js for the independent positive / negative / zero-sequence networks, engines/sc-maxmin.js for the maximum versus minimum operating case, and engines/selection-check.js for the device verdicts (Icw versus ip or Ik, transformer dynamic idyn, cable thermal minimum section).

The short-circuit current is the number every switchgear and cable decision is built on: it sets the rated short-time withstand and breaking capacity that must be ordered, it sets the minimum cable section that survives the thermal duty until the protection clears, it caps the protection settings that can still be selective, and it is the current that the arc-flash energy and the earth-fault loop impedance are computed from. Reviewers and switchgear suppliers read exactly this page, so getting the fault point, the line impedance and the operating case right matters more than the last digit.

Input: external conditions Ssc and X/R, transformer Sn and uk, the length / section / material of every branch on the path from the source to the fault point, motor capacity (largest motor and total motor), the fault type and the clearing time t_clear → chain: Z_Q from Ssc with the X/R split into R and X, folded to the LV side → Z_T = (uk/100)·Un²/Sn, with a low-voltage direct feed assuming uk = 4% for the upstream distribution transformer → Z_line from edge lenM, section and material, referred to the LV side → Zk = Z_Q + Z_T + Z_line → I"k = c·Un/(√3·Zk) → the peak factor κ = 1.02 + 0.98·e^(−3R/X) of IEC 60909 and ip = κ·√2·I"k including motor feedback → for single-phase-to-earth the sequence path is assembled (zero-sequence depends on the transformer vector group and on neutral earthing, so a Dyn11 gives Z0/Z1 ≈ 3.0, Yyn0 ≈ 8.0, YNd11 or Dd0 ≈ 10.0; with no zero-sequence path the result is 0 plus an explicit "zero-sequence path does not exist" verdict) → Ik1 = √3·c·Un/(Z1+Z2+Z0+3Zn) → per-node table → device checks (Icw versus ip by default, or versus Ik if the vendor basis is chosen; transformer idyn = 2.5·Ik; cable thermal S_min = Ik·√t/K with K = 143 copper / 94 aluminium and t defaulting to 0.5 s) → output: per-node Ik3 and Ik1, ip, total and split impedances, withstand verdicts with suggested standard tiers (25 / 31.5 / 40 / 50 / 65 kA) when no rating exists. Linkage: the section and length entered here are the same values the load flow uses, so widening a cable both lowers the voltage drop and raises the fault current that the switchgear must withstand; the transformer uk moves short circuit and load flow together; Ssc scales the whole MV-side table and re-runs the device verdicts in the max / min comparison. Approximations, all disclosed by the engine: the impedance is combined by magnitude without phase-angle difference (preliminary engineering grade); when no X/R is entered the default is R/X = 0.10 on the MV side and 0.50 on the LV side, that is X/R = 10 and 2; the k0 = 3 fallback is only used if the legacy zero-sequence model is explicitly selected; and a missing length uses the engine default.

10 kV / 0.4 kV, grid Ssc 500 MVA with X/R 10, 1000 kVA transformer (SCB13), 45 m MV cable and 200 m of 240 mm² to the load: the LV busbar sees I″k ≈ 23.6 kA with a first peak ip ≈ 41.5 kA and a steady Ik ≈ 22.5 kA at t_clear 0.1 s (short-circuit capacity ≈ 15.6 MVA). Raising the same transformer to Uk 8% lowers it to I″k ≈ 18.5 kA / ip ≈ 32.5 kA — which is exactly why 6% versus 8% changes the switchgear rating you have to order.

IEC 60909-0 (equivalent voltage source, c factor, μ and κ factors), IEC 60947-2 / IEC 61439 (Icw and peak withstand levels 25 / 31.5 / 40 / 50 / 65 kA), GB/T 16895.5 and IEC 60364-4-43 (S ≥ I·√t/k, k = 143 for copper), IEEE 1584-2002 (arc-flash incident energy, shown on the protection page).

Why is the short-circuit current at the end of the feeder lower than at the busbar?

Because the engine adds the cable impedance of every branch to the source impedance. The longer or thinner the feeder, the higher the series impedance and the lower the fault current at its far end — that is why the switchgear rating has to be checked at its own location, not only at the main busbar.

Is this IEC 60909 or a simplified estimate?

It is the IEC 60909 equivalent voltage source method: network impedance from Ssc and X/R, transformer impedance from Uk, plus line impedance from the real cable length and cross-section; the DC component and the peak factor are computed from X/R. Time-domain transient simulation is out of scope.

Does changing Uk affect the equipment I have to order?

Yes. Impedance is inversely proportional to the fault current, so the 6% to 8% step in the example drops the LV fault current by about 22% — it can move you from a 40 kA to a 31.5 kA switchgear line, or the other way round. The Icw / peak / thermal verdicts in the panel update with every run.