Methods / Short-Circuit Duty
How RokBench checks this

Short-Circuit Duty: ANSI vs IEC

Two standard families answer the same question — can this breaker handle the fault? — with different currents and different ratings.

1. What this check does

Builds the impedance of one radial path (utility source, transformer, cable) and any motor contribution at the fault bus, then calculates a bolted three-phase fault two ways. ANSI/IEEE C37: symmetrical current at 1.0 pu prefault voltage, its X/R from separate R and X networks, and a multiplying factor when the system X/R is above the breaker's test X/R; the adjusted duty is compared with the symmetrical interrupting rating, and the first-cycle peak with the close-and-latch rating. IEC 60909: initial symmetrical current Ik″ with the voltage factor c and the transformer correction KT, peak ip, breaking current Ib, DC component and thermal equivalent Ith; compared with Icu or Isc, Icm and Icw.

2. Inputs used

3. Formulas and assumptions

Impedances on the fault-bus voltage

Source, transformer
ZQ=V2SscZT=z%100 V2STR=Z1+(X/R)2Z_Q = \dfrac{V^2}{S_{sc}} \qquad Z_T = \dfrac{z\%}{100}\,\dfrac{V^2}{S_T} \qquad R = \dfrac{Z}{\sqrt{1 + (X/R)^2}}

Cable: R and X per conductor × length ÷ parallel sets. Motors: Z_M = Xd″ on the motor kVA.

ANSI / IEEE C37

Symmetrical
I=V3 ∣Z∣,XR=XX-only networkRR-only networkI = \dfrac{V}{\sqrt{3}\,|Z|}, \qquad \dfrac{X}{R} = \dfrac{X_{\text{X-only network}}}{R_{\text{R-only network}}}

1.0 pu prefault voltage; motor impedance × the first-cycle or interrupting multiplier.

First cycle
Iasym=I1+2e−2π/(X/R)Ipeak=2 I(1+e−π/(X/R))I_{\text{asym}} = I\sqrt{1 + 2e^{-2\pi/(X/R)}} \qquad I_{\text{peak}} = \sqrt{2}\,I\left(1 + e^{-\pi/(X/R)}\right)
LV factor
MFPCB=1+e−π/(X/R)1+e−π/(X/R)tMFMCCB, fused=1+2e−2π/(X/R)1+2e−2π/(X/R)t\mathrm{MF}_{\text{PCB}} = \dfrac{1 + e^{-\pi/(X/R)}}{1 + e^{-\pi/(X/R)_t}} \qquad \mathrm{MF}_{\text{MCCB, fused}} = \dfrac{\sqrt{1 + 2e^{-2\pi/(X/R)}}}{\sqrt{1 + 2e^{-2\pi/(X/R)_t}}}

(X/R)t = breaker test circuit X/R. MF is not taken below 1.

MV factor
MF=1+2e−4πt/(X/R)1+2e−4πt/(X/R)t\mathrm{MF} = \dfrac{\sqrt{1 + 2e^{-4\pi t/(X/R)}}}{\sqrt{1 + 2e^{-4\pi t/(X/R)_t}}}

Remote source, t = contact parting time in cycles, (X/R)t = 17 at 60 Hz.

Duty
Iduty=MF⋅I≤IinterruptingIpeak≤Iclose & latch≈2.6 IinterruptingI_{\text{duty}} = \mathrm{MF}\cdot I \le I_{\text{interrupting}} \qquad I_{\text{peak}} \le I_{\text{close \& latch}} \approx 2.6\,I_{\text{interrupting}}

IEC 60909

Correction
KT=0.95 cmax⁡1+0.6 xTZQ←c ZQK_T = 0.95\,\dfrac{c_{\max}}{1 + 0.6\,x_T} \qquad Z_{Q} \leftarrow c\,Z_Q

c = 1.05 (LV, ±6 %) or 1.10 (LV ±10 %, and MV); xT = transformer reactance in per unit.

Initial current
Ik′′=c Un3 ∣Zk∣I_k'' = \dfrac{c\,U_n}{\sqrt{3}\,|Z_k|}
Peak
ip=∑κ 2 Ik′′ ,κ=1.02+0.98 e−3R/Xi_p = \sum \kappa\,\sqrt{2}\,I_k''\ ,\qquad \kappa = 1.02 + 0.98\,e^{-3R/X}

Summed over the network branch and the motor branch.

Breaking, DC
Ib=Ik′′iDC=2 Ik′′ e−2πf tmin⁡ R/XI_b = I_k'' \qquad i_{DC} = \sqrt{2}\,I_k''\,e^{-2\pi f\,t_{\min}\,R/X}

Far from generators; full motor contribution kept.

Thermal
Ith=Ik′′m+n,m=e4fTkln⁡(κ−1)−12fTkln⁡(κ−1),n=1I_{th} = I_k''\sqrt{m + n}, \quad m = \dfrac{e^{4fT_k\ln(\kappa - 1)} - 1}{2fT_k\ln(\kappa - 1)}, \quad n = 1
Duty
Ib≤Icu (Isc)ip≤IcmIth2Tk≤Icw2twI_b \le I_{cu}\ (I_{sc}) \qquad i_p \le I_{cm} \qquad I_{th}^2 T_k \le I_{cw}^2 t_w
Symbols
V, Un, fV,\ U_n,\ fFault-bus line-to-line voltage; frequencyV, Hz
Ssc, ST, z%S_{sc},\ S_T,\ z\%Utility short-circuit power; transformer rating and impedanceMVA, kVA, %
Z, R, XZ,\ R,\ XImpedance at the fault and its partsΩ
I, Iasym, IpeakI,\ I_{\text{asym}},\ I_{\text{peak}}ANSI symmetrical, first-cycle asymmetrical rms and peak currentkA
(X/R)t(X/R)_tBreaker test circuit X/R (from its test power factor)
MF, t\mathrm{MF},\ tANSI multiplying factor; MV contact parting time—, cycles
c, KT, xTc,\ K_T,\ x_TIEC voltage factor; transformer correction; transformer reactanceper unit
Ik′′, ip, IbI_k'',\ i_p,\ I_bIEC initial symmetrical, peak and breaking currentkA
κ\kappaPeak factor from R/X
iDC, tmin⁡i_{DC},\ t_{\min}DC component at the minimum break timekA, s
Ith, Tk, m, nI_{th},\ T_k,\ m,\ nThermal equivalent current over the fault duration; DC and AC heat factorskA, s, —
Icu, Isc, Icm, Icw, twI_{cu},\ I_{sc},\ I_{cm},\ I_{cw},\ t_wIEC breaking (LV, MV), making, short-time withstand ratings and its durationkA, s
Fault current waveform starting fully offset, with its decaying DC component, the first peak, the symmetrical envelope, and the contact parting time marked.
Figure 1. An asymmetrical fault current: the DC offset decays with the X/R time constant; the peak comes in the first half cycle.
Two columns: ANSI symmetrical current times multiplying factor against interrupting rating and first-cycle peak against close and latch; IEC Ik double prime with c and KT, ip, Ib and Ith against Icm, Icu or Isc and Icw.
Figure 2. Same fault, two standard families: what each calculates and which rating it is compared with.

4. What is NOT checked

5. Reference standards

Pointers only; consult the edition adopted by your authority having jurisdiction.

6. Validation cases

These worked examples run through the engine on every change; the expected values come from the cited source. All cases: Validation.

480 V bus behind 1500 kVA, 5.75 % Z, X/R 6 — ANSI and IEC shortcircuit/duty · Hand calculation

Pass

Source: Hand calculation from the ANSI (IEEE C37.13 practice) and IEC 60909-0 equations on the Methods page — Zt = 0.0575 · 480² / 1.5 MVA = 8.832 mΩ. ANSI: I = 480/(√3·Zt) = 31.38 kA; asym = I·√(1+2e^(−2π/6)) = 40.93 kA; peak = √2·I·(1+e^(−π/6)) = 70.66 kA; MCCB MF = 1.3046/1.2469 = 1.046. IEC: c = 1.05; xT = 0.0567; KT = 0.95·1.05/(1+0.6·0.0567) = 0.9647; Ik″ = 1.05·480/(√3·KT·Zt) = 34.15 kA; κ = 1.02+0.98e^(−0.5) = 1.614; ip = κ·√2·Ik″ = 77.98 kA.

QuantityExpectedRokBenchErrorResult
ansi.firstCycleKa31.3831.380%Pass
ansi.asymRmsKa40.9340.930%Pass
ansi.peakKa70.6670.660%Pass
ansi.multiplyingFactor1.0461.0460%Pass
iec.kt0.96470.96470%Pass
iec.ikssKa34.1534.150%Pass
iec.kappa1.6141.6140%Pass
iec.ipKa77.9877.980%Pass

7. Known issues and changes

See the Changelog. Try the tool: Short-Circuit Duty: ANSI vs IEC.

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