Short-circuit, bus by bus: an 11/0.4 kV substation in calculator #002

Fault current is not one number. It is four — I″k3, I″k2, I″k1, ip — at every busbar, in a maximum and a minimum regime, and each of them answers a different design question. This guide takes calculator #002 through a substation everyone has built: 11 kV utility supply, a 1000 kVA transformer, an LV cable to the main board. All figures below come from that run.

What the tool models

A circuit in #002 is a chain of elements between busbars, fed by one or more sources:

• sources — an infinite bus with a stated Sk″, a transformer, a generator, a motor contribution, or an impedance you type in directly. Several sources can feed the same circuit, and the contributions table shows what each one brings;
• elements — cable, overhead line or transformer, each of which creates a new busbar downstream of itself;
• regimes — maximum and minimum are computed independently, not scaled from one another.

Every busbar gets its own row: positive-sequence impedance Z1, zero-sequence Z0, the three fault currents, the κ factor and the peak ip. That structure matters because equipment is selected against different numbers at different points, and a single "fault level at the board" hides the ones you need.

Where the numbers come from

The calculation follows IEC 60909-0: an equivalent voltage source c·Un/√3 at the fault point, all network elements reduced to sequence impedances, and the fault currents from the symmetrical-component networks:

I″k3 = c·Un / (√3 · |Z1|)                    three-phase
I″k2 = c·Un / |Z1 + Z2|                      line-to-line
I″k1 = √3·c·Un / |Z1 + Z2 + Z0|              line-to-earth
ip   = κ · √2 · I″k3

Two details decide whether the result is usable:

The voltage factor c. IEC 60909-0 uses cmax for the maximum regime and cmin for the minimum. The calculator applies the pair for the voltage level automatically — that is why the maximum and minimum currents are not simply proportional to each other.

κ is not a constant. It comes from the R/X ratio at the fault point, so it changes bus by bus. In the run below κ falls from 1.746 at the 11 kV bus to 1.407 at the main board, because the LV cable adds resistance faster than reactance. Take κ = 1.8 everywhere and you overstate the peak on the LV side by a third.

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Worked example: TS-1, 11/0.4 kV

Utility short-circuit level 250 MVA at 11 kV (200 MVA in the minimum regime). A 1000 kVA 11/0.4 kV transformer, uk = 6.0 %, load losses 10.5 kW, Dyn11, solidly earthed neutral. From the LV terminals, 50 m of 2 × 240 mm² cable to the main switchboard.

Step 1 — the supply

An infinite bus with a stated Sk″ is how a utility normally gives you the data. The tool converts it to an impedance at the circuit voltage; the alternative source types exist for when you have the actual machine data instead.

Step 2 — transformer and cable

Each element names the busbar it creates — LV-MAIN after the transformer, MSB after the cable — which is what makes the results table readable later.

Step 3 — fault levels at every bus

Reading it

The earth fault at LV-MAIN is larger than the three-phase fault. 24.08 kA against 23.54 kA. That is not an error, and it is the single most useful thing this table tells you. The transformer is Dyn11 with a solidly earthed star point, so the zero-sequence path sees only the transformer winding — Z0 = 9.60 mΩ against Z1 = 10.30 mΩ. With Z0 < Z1 the line-to-earth current exceeds the three-phase one. Size the switchgear on I″k3 alone and you have understated the duty on a single-pole fault.

The 50 m cable takes 16 % off the fault level. 23.54 kA at the transformer terminals, 19.71 kA at the board 50 m away. Cable impedance is not a rounding error at LV — it is the reason a breaker at the board can be a frame size smaller than one at the transformer, and the reason protection at the far end may not see what you assumed.

Check the transformer figure by hand:

I_r  = 1000 kVA / (√3 × 400 V)      = 1 443 A
I″k  ≈ I_r / uk = 1 443 / 0.06      ≈ 24.1 kA   (transformer alone)
with the 11 kV network in series     = 23.54 kA

The utility contribution costs about 2 %: at 250 MVA the 11 kV network is stiff compared with a 1000 kVA transformer. On a weaker supply — a long rural feeder, a generator island — that difference grows, which is exactly when the minimum regime starts to matter.

Minimum regime is the protection case. I″k3 min = 17.47 kA at the board against 19.71 kA maximum. Breaking capacity is chosen on the maximum; sensitivity and disconnection time are checked on the minimum, and I″k1 min = 15.15 kA is what an earth-fault element has to detect. The calculator computes both regimes independently, with their own c factor and their own source levels, so you can read them straight off the same table.

κ and ip go together. 1.604 at LV-MAIN gives ip = 53.40 kA; at the board κ drops to 1.407 and ip to 39.22 kA. ip is what the busbar bracing and the breaker's making capacity are chosen against — and it is the number most often carried over unchanged from the wrong bus.

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Hand check against a published IEC example

A fault-level number is only as good as the impedance chain behind it, and there is a public way to test that chain: IEC TR 60909-4:2000 publishes fully worked examples with every intermediate impedance printed. Clause 3 of that report is a 400 V system fed from a 20 kV network — so we can rebuild it here, element by element, and compare.

The published data

From clause 3 of IEC TR 60909-4:

• Network feeder Q — U_nQ = 20 kV, I″_kQ = 10 kA, R_Q = 0,1 X_Q, c_Q = c_max = 1,1;
• Transformer T1 — 630 kVA, Dyn5, 20 kV / 410 V, u_kr = 4 %, P_krT = 6,5 kW;
• Line L1 — two parallel four-core cables, 10 m, 4 × 240 mm² Cu, Z′ = (0,077 + j 0,079) Ω/km.

The report publishes the resulting impedances in its Table 3, referred to the 410 V side, in mΩ.

The same chain in the calculator

Element by element

Written out, the two steps that people most often skip:

Z_Q  = c U_nQ / (sqrt(3) I"_kQ) = 1,1 x 20 000 V / (1,732 x 10 000 A) = 1,270 ohm
Z_Qt = Z_Q (U_rTLV / U_rTHV)^2 = 1,270 x (410/20 000)^2 = 0,534 mohm
       X_Qt = 0,995 Z_Qt = 0,531 mohm ; R_Qt = 0,1 X_Qt = 0,053 mohm

Z_T  = u_kr/100 x U_rTLV^2 / S_rT = 0,04 x 410^2 / 630 000 = 10,673 mohm
R_T  = P_krT x U_rTLV^2 / S_rT^2 = 6 500 x 410^2 / 630 000^2 = 2,753 mohm
X_T  = sqrt(Z_T^2 - R_T^2) = 10,312 mohm
x_T  = X_T / (U_rT^2/S_rT) = 10,312 / 266,8 = 0,03865
K_T  = 0,95 c_max / (1 + 0,6 x_T) = 0,95 x 1,05 / 1,02319 = 0,975

The fault current

At busbar A — the transformer LV terminals:

Z_k  = Z_Qt + Z_T1K = (0,053 + 2,684) + j(0,531 + 10,053) = 2,737 + j 10,584 mohm
|Z_k| = 10,932 mohm
I"k3 = c U_n / (sqrt(3) |Z_k|) = 1,05 x 400 / (1,732 x 0,010932) = 22,18 kA

and after the 10 m cable, at F1:

Z_k  = 3,122 + j 10,979 mohm ;  |Z_k| = 11,414 mohm
I"k3 = 1,05 x 400 / (1,732 x 0,011414) = 21,24 kA
R/X  = 0,284  ->  kappa = 1,02 + 0,98 e^(-3 R/X) = 1,438   (Formula 55)
ip   = kappa sqrt(2) I"k3 = 1,438 x 1,414 x 21,24 = 43,19 kA

The calculator returns 22,18 kA and 21,24 kA, κ = 1,471 and 1,438, i_p = 46,15 kA and 43,19 kA — the same numbers, because it is the same chain.

For reference, the report's own answer for its full network — both transformers T1 and T2 feeding the busbar in parallel through the shared feeder — is Z_k = (1,881 + j 6,746) mΩ and I″_k3 = 34,62 kA, with κ = 1,445. That is the case a radial chain cannot express: two branches share one upstream feeder, so it belongs in Electrical Networks, where the topology is drawn rather than stacked.

Three things this check pins down

• The impedance correction factor K_T is not optional. IEC 60909-0, 6.3.3 requires transformer impedances to be multiplied by K_T = 0,95 c_max /(1 + 0,6 x_T) whenever maximum short-circuit currents are calculated — and by nothing at all for minimum currents. It is a modest 2,5 % here, but it moves the number in the unsafe direction if omitted: the fault current comes out lower than it really is, and switchgear gets selected against it.
• The rated voltage and the nominal voltage are different numbers. The transformer is 20 kV / 410 V, and 410 V is what refers the impedances. The fault current is computed on the nominal voltage of the network behind it, U_n = 400 V, with c = 1,05 for LV (Formula 29 and Table 1). Mixing them up inflates every downstream fault level by 2,5 % — which is why the calculator asks for both.
• κ is not a constant. It follows R/X at the fault location, and R/X changes at every busbar: 0,10 at the 20 kV feeder, 0,26 at the transformer terminals, 0,28 after 10 m of cable — giving κ = 1,75 / 1,47 / 1,44. A peak current taken with a "typical" κ is a guess; taken with the local R/X it is a calculation. Note also 4.3.1.2 b) of IEC 60909-0: in meshed networks, method (b) carries an additional factor of 1,15 unless R/X is checked branch by branch.

Which number goes where

FREE and PRO: where the line runs

Same structure as #001, and the same single dividing line — project size.

FREE — open access, no account:

• the complete IEC 60909 calculation: multiple parallel sources at BB0, transformer / cable / OHL elements, both MAX and MIN regimes, all three fault types (I″k3, I″k2, I″k1), κ and i_p per busbar, the sequence-impedance build-up;
• the per-busbar results table and the warnings;
• the full .docx report with the step-by-step derivation;
• up to 3 circuits in one project.

PRO:

• more than three circuits in a single project file (unlimited_projects).

Nothing in the physics is behind the paywall here: the engine, the regimes and the report are the same on both tiers.

Where this sits next to the other tools

• The thermal withstand check on the cable takes I″k from here and the size from #004 Cable Ampacity, which also gives the conductor temperature the cable starts the fault from.
• Voltage drop picks the size from the other side — #001 — and the larger of the two requirements wins.
• Earthing design needs the earth-fault current from this table: #003 Substation Grounding uses it as the grid current for touch and step voltages.
• For a whole network with several sources and rings, Electrical Networks runs the same IEC 60909 engine over a schematic instead of a chain.

Open calculator #002 →