Short-Circuit Formulas: Fault Level by MVA
Short-circuit formulas using the MVA method: transformer fault level, adding the source, three-phase fault current, peak current and the impedance method.
The formulas
| To find | Formula |
|---|---|
| Fault level of a transformer | MVAsc = S (MVA) / (Z% / 100) |
| Source and transformer in series | 1 / MVAtotal = 1 / MVAsource + 1 / MVAtransformer |
| Three-phase fault current | Isc = MVAsc × 10⁶ / (√3 × V) |
| Peak fault current | ip = κ × √2 × Isc |
| Impedance method | Isc = V / (√3 × Z), with Z = √(R² + X²) |
S is the transformer rating in MVA, Z% its impedance, V the line-to-line voltage in volts and κ a factor that depends on the R/X ratio of the circuit, between 1 and 2. A full study to IEC 60909 also includes a voltage factor and motor contributions.
Worked example: 1 MVA, 5% transformer on a 250 MVA source
The secondary is 415 V and κ is taken as 1.8.
- Transformer: MVAsc = 1 / 0.05 = 20 MVA
- 1 / MVAtotal = 1 / 20 + 1 / 250, so MVAtotal = 18.52 MVA
- Isc = 18.52 × 10⁶ / (√3 × 415) = 25.76 kA
- ip = 1.8 × √2 × 25.76 = 65.58 kA
Worked example: the impedance method
- A fault point with a total impedance of 0.02 Ω on 415 V gives Isc = 415 / (√3 × 0.02) = 11.98 kA
Fault current at 415 V (5% transformers on a 250 MVA source)
| Transformer | Fault level of the transformer | Fault current |
|---|---|---|
| 500 kVA | 10 MVA | 13.38 kA |
| 1000 kVA | 20 MVA | 25.76 kA |
| 1600 kVA | 32 MVA | 39.47 kA |
| 2500 kVA | 50 MVA | 57.97 kA |
Frequently asked questions
Why add the source and transformer as reciprocals?
The impedances are in series, and the fault level is inversely proportional to the impedance, so the reciprocals of the fault levels add.
Why does the breaker need the peak current as well?
The first half-cycle of a fault carries a peak larger than the rms value because of a DC component. Breakers are rated for the peak making current as well as the breaking current.
Is the fault level of the supply a fixed number?
No. It changes with the network configuration. Ask the utility for the maximum and minimum fault levels at the point of supply.
Do motors add to the fault current?
Yes. Running motors feed current into a fault for a few cycles. Large motors should be included in a proper study.
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