Fault level
Network feeder (source) impedance from fault level
- where
- network feeder impedance (Ω)
- IEC 60909 voltage factor
- nominal voltage at the connection point (kV)
- fault level at the connection point (MVA)
- source reactance and resistance (Ω)
- source X/R ratio
Explain this equation
Converts the fault level declared by the network operator at your point of connection into an equivalent source impedance, ready to add to transformers and cables in a fault build-up.
S″kQ is the initial symmetrical fault level at the connection, Un the nominal voltage there, and c the same voltage factor you will use in the final fault current. ZQ is the Thevenin impedance of everything upstream; X/R tells you how that impedance divides into reactance XQ and resistance RQ.
From S″k = √3·Un·I″k and I″k = c·Un/(√3·ZQ), rearranging gives ZQ = c·Un²/S″kQ. Including c here means that when c reappears in the fault current, the connection point gives back exactly the declared fault level. The split uses XQ = ZQ·(X/R)/√(1 + (X/R)²), which is just the impedance triangle.
Sense check: 500 MVA at 33 kV gives about 2.4 Ω. With X/R = 10, XQ is 99.5% of ZQ, which is the IEC 60909 default when RQ is unknown (RQ = 0.1·XQ).
Watch out: use the operator's maximum fault level for ratings and the minimum for protection sensitivity; they can differ widely with network running arrangements.
Turns a declared fault level (from the DNO or a system study) into the equivalent network feeder impedance for building up a fault calculation.
Enter your values to see the result.
Good to know
The DNO normally declares S″kQ (or I″kQ) and X/R at the point of connection; ask for both maximum and minimum values. If only S″kQ is known, IEC 60909 allows RQ = 0.1·XQ with XQ = 0.995·ZQ (X/R = 10). UK 11 kV and 33 kV X/R values are often 5–15, higher near grid supply points.
Common mistake
Dropping c: IEC 60909 defines ZQ = c·UnQ²/S″kQ, and the same c then reappears in the fault current formula. Omitting it on one side only shifts results by 10%.
Limits
Represents the upstream network as a single Thevenin impedance at the given X/R. Not valid if large local generation or motors changes the upstream contribution; model those separately.
Sources
- IEC 60909-0:2016 Short-circuit currents in three-phase a.c. systems – Part 0: Calculation of currents
- ENA Engineering Recommendation G74 Issue 2: Procedure to meet the requirements of IEC 60909 for the calculation of short-circuit currents
- Schneider Electric Cahier Technique 158: 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.