Fault level
Three-phase symmetrical fault current I″k (IEC 60909)
- where
- magnitude of total short-circuit impedance (Ω)
- total positive-sequence resistance and reactance to the fault (Ω)
- initial symmetrical short-circuit current (kA)
- IEC 60909 voltage factor
- nominal line-to-line voltage (kV)
- initial symmetrical short-circuit power (MVA)
- peak short-circuit current (kA)
Explain this equation
Gives the initial symmetrical three-phase fault current and fault level at a point, from the total impedance back to the source. Use it to check switchgear breaking and making duty and to set the starting point for protection studies.
Zk is the phasor sum of all series resistance Rk and reactance Xk between the IEC 60909 equivalent source and the fault. c·Un/√3 is the equivalent phase voltage driving the fault: the √3 turns line voltage into phase voltage, and c (1.10 for maximum currents above 1 kV) covers pre-fault voltage, taps and loads that are not modelled.
The current is simply phase voltage over impedance. S″k = √3·Un·I″k is the conventional fault MVA. The peak ip multiplies the r.m.s. by √2 and by κ, which grows from 1.02 to 2 as the circuit becomes more inductive, because a high X/R means a slowly decaying d.c. offset.
Sense check: 11 kV through 0.5 Ω gives about 14 kA, or roughly 265 MVA. Halving Zk doubles I″k. An X/R of 10 gives κ ≈ 1.75, so ip is about 2.5 times I″k.
Watch out: this is a radial calculation. For meshed networks IEC 60909 needs method B or C for κ, and motor or generator infeeds must be added separately.
Initial symmetrical three-phase short-circuit current from the total impedance between the equivalent voltage source and the fault, using the IEC 60909 voltage factor c.
Enter your values to see the result.
Good to know
Add all series impedances (source, transformers with KT applied, cables, lines) referred to the fault voltage first. Above 1 kV use cmax = 1.10 for maximum currents and cmin = 1.00 for minimum currents. S″k is the conventional fault level √3·Un·I″k. The peak shown uses κ from the R/X of the total impedance, which is correct for a radial (single-path) feed.
Common mistake
Using the measured or rated voltage instead of Un with c, or forgetting that c already allows for pre-fault voltage, so adding a separate +10% doubles the margin.
Limits
Far-from-generator, balanced, bolted fault in a radial network. Does not add motor or generator contributions, and for meshed networks κ needs IEC 60909 method B or C. Impedances must all be referred to the voltage Un.
Sources
- IEC 60909-0:2016 Short-circuit currents in three-phase a.c. systems – Part 0: Calculation of currents
- Schneider Electric Cahier Technique 158: Calculation of short-circuit currents
- ENA Engineering Recommendation G74 Issue 2: Procedure to meet the requirements of IEC 60909 for the calculation of short-circuit currents
- Grainger & Stevenson, Power System Analysis, McGraw-Hill 1994 (per-unit, symmetrical faults, symmetrical components)
For competent high-voltage engineers. Results are calculations from the published methods shown and depend entirely on the data you enter. They are not a substitute for a system study, manufacturer data, the applicable standards and Engineering Recommendations, or your organisation’s safety rules. Never use a result to justify working on or near live HV equipment.