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Per-Unit Fault Level Calculator

Full Per-Unit Calculation — Auditable Working

Build a simple radial network — utility source plus one or more transformers or impedance elements — and calculate three-phase fault current using the per-unit system on a common MVA and kV base. Unlike a black-box fault calculator, every conversion step is shown: each element's impedance converted to per-unit, the running cumulative total, and the final fault current and MVA derivation.

CALCULATOR

Per-Unit Fault Level

NETWORK ELEMENTS (SERIES)
Add elements in the order they appear from the source to the fault — a cable can be the first element if it sits before any transformer.
FAULT CURRENT
347.8
kA
FAULT LEVEL
250
MVA
BASE CURRENT
13912.1 A
Z TOTAL (PU)
0.04
CALCULATION STEPS
SourceZ=0.04 pu · Σ=0.04 pu

Results for guidance only. Verify against current standards and manufacturer data.
Not a substitute for engineering judgement or a licensed professional.

Methodology

The per-unit system expresses all impedances as a fraction of a chosen base impedance, allowing elements at different voltage levels to be combined directly.

Base Current
Ibase=MVAbase×1063kVbase×103I_{base} = \frac{\mathrm{MVA}_{base} \times 10^6}{\sqrt{3}\,\mathrm{kV}_{base} \times 10^3}
Base Impedance
Zbase=kVbase2MVAbaseZ_{base} = \frac{\mathrm{kV}_{base}^2}{\mathrm{MVA}_{base}}
Transformer Impedance on System Base
Zpu=%Z100×MVAbaseMVAtransformerZ_{pu} = \frac{\%Z}{100} \times \frac{\mathrm{MVA}_{base}}{\mathrm{MVA}_{transformer}}
Source Impedance on System Base
Zsource(pu)=MVAbaseMVAfaultsourceZ_{source(pu)} = \frac{\mathrm{MVA}_{base}}{\mathrm{MVA}_{fault\,source}}

All series impedances are summed on the common base, then fault current is derived:

Fault Current
Ifault=IbaseZtotal(pu)I_{fault} = \frac{I_{base}}{Z_{total(pu)}}

For a cable or line element, the ohmic impedance is converted to per-unit on the base impedance at that element's own voltage level:

Cable Impedance
Ractual=RΩ/km×LXactual=XΩ/km×LR_{actual} = R_{\Omega/km} \times L \qquad X_{actual} = X_{\Omega/km} \times L
Parallel Runs
Zeq=(R+jX)LNfor N parallel runsZ_{eq} = \frac{(R + jX)\,L}{N} \quad \text{for } N \text{ parallel runs}
Cable Per-Unit
Zcable(pu)=R+jXZbaseZbase=Vbase2SbaseZ_{cable(pu)} = \frac{R + jX}{Z_{base}} \qquad Z_{base} = \frac{V_{base}^{2}}{S_{base}}

The voltage base steps at each transformer, so a cable downstream of a 33/11 kV transformer is converted on 11 kV, not on the source-side 33 kV. Using the wrong base here would misstate the cable's per-unit impedance by the square of the turns ratio.

Cables offer two impedance input modes. Detailed mode uses genuine complex addition — resistance and reactance are combined with the rest of the network as perpendicular vector components, correct when real R and X are separately known (e.g. from a cable datasheet or AS/NZS 3008). Direct Ω is for a single flat "line impedance" figure with no resistance/reactance breakdown, common in simplified per-unit exercises. In that case the value is placed on the reactance axis alongside the source and transformer impedances, which are themselves dominated by reactance — this matches the standard simplified convention of summing all given impedance magnitudes directly, rather than silently treating an unknown-phase figure as pure resistance, which would understate the network total once combined with reactive elements. The two modes can give a genuinely different answer for the same numeric "0.25 Ω" — this is a real modelling choice, not a rounding difference, and the one that matches a specific worked reference has been verified to reproduce it to 4 significant figures.

Worked Example

Base 10 MVA, 0.415 kV. Source fault level 250 MVA → Z_source = 10/250 = 0.04 pu. Transformer 0.5 MVA, 5%Z → Z_tx = 0.05 × (10/0.5) = 1.0 pu. Total Z = 1.04 pu. Ibase=10×1063×415=13,912 AI_{base} = \frac{10\times10^{6}}{\sqrt{3}\times415} = 13{,}912\ \mathrm{A}. Fault current = 13,912/1.04 ≈ 13,377A ≈ 13.4kA.

Assumptions & Limitations

  • Magnitude-only calculation — does not track impedance angle (X/R ratio), so asymmetrical peak fault current is not derived
  • Assumes simple series (radial) network — parallel transformer or multi-source contributions require a more advanced fault study
  • Cable and line elements are modelled as series R+jXR + jX impedance on the voltage base at their own position in the network. For a multi-bus study with several fault locations, use the Distribution Fault Study Builder
  • Source and transformer impedances are taken as purely reactive, the standard assumption for a utility fault level quoted in MVA; cable resistance is carried separately so the total is a complex sum rather than an arithmetic one

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Built with reference to AS/NZS 3000, AS/NZS 3008, AS/NZS 1359
Results for guidance only — not a substitute for engineering judgement
Does not reproduce copyrighted Standards Australia material. Always consult current standards for compliance.