Transformer short-circuit current calculator

Result

Initial symmetrical short-circuit current I″kRMS value at the low-voltage terminals: —kA
Rated current: —A
Ratio I″k / Iᵣ: —

A guide, not a design to standard. We give no warranty for the correctness of the results. The applicable standards and case-by-case verification govern.

Calculated on

Short-circuit current at the transformer

Inputs

Rated apparent power: kVA
Nominal voltage, low-voltage side: V
Short-circuit voltage uₖ: %

Result

Initial symmetrical short-circuit current I″k: kA
Rated current: A
Ratio I″k / Iᵣ

Formula

Iᵣ = Sᵣ / (√3 · Uᵣ)
I″k = Iᵣ / (uₖ / 100)

Assumptions and standards

The calculation assumes an infinitely strong upstream source, neglecting its impedance, the real network limits the current further, so the result is on the safe side.
Only the transformer is considered, not cables, busbars or motor contribution.
The result is the RMS value I″k, which governs the rated breaking capacity of protective devices. The peak short-circuit current iₚ, the crest of the first half-cycle, governing the dynamic strength of busbars, follows from iₚ = κ · √2 · I″k with κ ≈ 1.8 and is therefore about 2.5 times as high.
A binding design requires calculation to IEC 60909.

A guide, not a design to standard. We give no warranty for the correctness of the results. The applicable standards and case-by-case verification govern.

FISEKON GmbH · Fischer Elektrokonstruktion · https://fisekon.com

The grey values are an example. Simply enter your own.

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Calculating is the easy part.

A formula gives you a number. Designing a plant also demands installation method, grouping, discrimination, the standards in force and a look at the installed base. That is what we take on.

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Method and questions

Short-circuit current

Approximate short-circuit current at the low-voltage terminals.

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Worked example

The default case the calculator starts with, worked through once. Change the values above and it recalculates immediately.

Inputs

Rated apparent power: 630 kVA
Nominal voltage, low-voltage side: 400 V
Short-circuit voltage uₖ: 6 %

Result

Initial symmetrical short-circuit current I″k: 15.2 kA
Rated current: 909 A
Ratio I″k / Iᵣ: 16.7

Formula and standards

Formula

Iᵣ = Sᵣ / (√3 · Uᵣ)
I″k = Iᵣ / (uₖ / 100)

Assumptions and standards

The calculation assumes an infinitely strong upstream source, neglecting its impedance, the real network limits the current further, so the result is on the safe side.
Only the transformer is considered, not cables, busbars or motor contribution.
The result is the RMS value I″k, which governs the rated breaking capacity of protective devices. The peak short-circuit current iₚ, the crest of the first half-cycle, governing the dynamic strength of busbars, follows from iₚ = κ · √2 · I″k with κ ≈ 1.8 and is therefore about 2.5 times as high.
A binding design requires calculation to IEC 60909.

Why is the value higher than in the network study?

Because the upstream network impedance is neglected here. In practice it reduces the short-circuit current, often by 5 to 15 %. For selecting breaking capacity, erring high is the right direction.

What does uk on the nameplate mean?

The voltage that has to be applied on the high-voltage side so that rated current flows on the low-voltage side under short circuit, given as a percentage of nominal voltage. A small uk means a high short-circuit current.