Asymmetry and duration
Thermal equivalent short-time current Ith (IEC 60909 m and n)
- where
- factor for the heat effect of the d.c. component
- system frequency (Hz)
- short-circuit duration (s)
- peak factor
- factor for the heat effect of the a.c. component (1 far from generators)
- thermal equivalent short-time current (kA)
- initial symmetrical short-circuit current (kA)
- Joule integral (kA²s)
Explain this equation
Gives a single r.m.s. current that, held for the fault duration Tk, heats conductors as much as the real fault current. Compare it with the short-time withstand rating of switchgear, CTs and cables.
I″k is the initial symmetrical current. m adds the heat from the decaying d.c. component and depends on κ, the frequency f and the duration Tk. n accounts for any decay of the a.c. component; it is 1 here because the calculator assumes a fault far from generators.
Heating follows the integral of i² over time. Integrating the squared exponential d.c. term over Tk gives the expression for m; ln(κ − 1) stands in for the decay rate implied by the peak factor. The Joule integral I²t = Ith²·Tk is the energy figure used in cable adiabatic checks.
Sense check: at 50 Hz with Tk = 1 s, m is only a few per cent, so Ith is barely above I″k. For a 0.1 s fault with κ = 1.8, m is about 0.44 and Ith is about 20% above I″k.
Watch out: m is only valid for κ below 2, and near generators n is less than 1.
Single r.m.s. current that, held for Tk, produces the same heating as the actual fault current including its d.c. component. Compare with the short-time withstand rating of switchgear, CTs and cables.
Enter your values to see the result.
Good to know
m implements IEC 60909-0 exactly. n is set to 1, which is the far-from-generator value (the a.c. component does not decay). Equipment rated for Ithr over Tkr is acceptable if Ith ≤ Ithr·√(Tkr/Tk) for Tk ≥ Tkr. For Tk ≥ 0.5 s at typical κ, m is small and Ith is close to I″k.
Common mistake
Using a short-time rating of, say, 25 kA for 3 s against a 1 s fault without rescaling, or forgetting that m is significant for short durations with high X/R.
Limits
n = 1 only: near-generator faults (where the a.c. component decays) need IEC 60909 n from I″k/Ik curves, which are not implemented here. For successive faults (auto-reclose) sum the Joule integrals.
Sources
- IEC 60909-0:2016 Short-circuit currents in three-phase a.c. systems – Part 0: Calculation of currents
- pandapower documentation: thermal short-circuit current (IEC 60909 m and n factors)
- ENA Engineering Recommendation G74 Issue 2: Procedure to meet the requirements of IEC 60909 for the calculation of short-circuit currents
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.