MV-506 · Phase 2 Design & analysis

Short Circuit & Load Flow Studio

Build a supply chain from the 11 kV source to a final distribution board. The studio calculates IEC 60909 fault levels at every bus, animates the fault current path, checks your switchgear ratings and plots the voltage profile under load.

In short

IEC 60909 uses an equivalent voltage source c·U_n/√3 at the fault, with c = 1.1 for maximum MV faults and 1.05 or 1.10 for LV.
Initial fault current I_k" = c·U_n / (√3·|Z_k|), and peak current i_p = κ·√2·I_k", with κ = 1.02 + 0.98e^−3R/X.
At the LV board the transformer impedance dominates. A 1500 kVA, 6.25% unit gives about 33 to 35 kA at 400 V.
Busduct and cable impedance bring the fault level down quickly, which is why final DBs can use lower-rated breakers.
Running motors add to the fault current for the first few cycles, so include them when checking peak and making ratings.

By Eng. Mannan Maqsood, Electrical EngineerPart of the MV Hub

Fault Level Studio 2D Animated

Set each element in the chain. Click a bus in the diagram (or use the buttons) to put a fault there and watch the fault current flow from the source.

MV (kV)

MV fault level (kA)

Source X/R

Transformer kVA

Impedance uk (%)

Load loss Pk (kW)

LV (V)

Busduct length (m)

Busduct rating (A)

Cable to SMDB (m)

SMDB cable (mm² Cu, per phase)

Cable to final DB (m)

Final DB cable (mm² Cu)

Running motors (% of TX)

LV c factor

Results

Switchgear rating check

LV main board Icw (kA, 1 s)

SMDB breaker Icu (kA)

Final DB MCCB Icu (kA)

Fault level at each bus

Load Flow & Voltage Profile 2D

Add load at each board and see how the voltage falls along the chain. The transformer tap can lift the whole profile.

Load at LV main (kW)

Load at SMDB (kW)

Load at final DB (kW)

Power factor

Transformer tap

MV voltage (pu)

Green band: ±6% around nominal, a common design target. Statutory supply limits are often wider (for example +10%/-6% or ±10%). Many designs also limit the drop from the origin to the final circuit to about 4 to 5%.

What is the IEC 60909 method?

IEC 60909 replaces all the sources with one equivalent voltage source at the fault, c·U_n/√3, and adds up the positive-sequence impedances of everything between the source and the fault. The voltage factor c covers the fact that the real network voltage can be higher than nominal. For maximum currents, c_max is 1.10 for MV and 1.05 (or 1.10) for LV systems. For minimum currents (used for protection sensitivity), c_min is 0.95 at LV and 1.00 at MV.

ExpertThe full method also applies an impedance correction factor K_T = 0.95·c_max/(1 + 0.6·x_T) to network transformers, and uses cable resistances at 20 °C for maximum currents and at the end-of-fault temperature for minimum currents. This studio applies K_T in Expert mode.

Why is peak current important?

The first peak of a fault current can be up to 2.8 times the RMS value because of the DC offset. Busbars, breakers' making capacity and cable supports have to survive these electromagnetic forces. The ratio κ depends on R/X: close to a large transformer, X/R is high and κ approaches 1.8 to 1.9.

How do earth fault levels differ?

In a TN system with a Dyn11 transformer, a phase-to-earth fault at the LV terminals can be slightly higher than the three-phase fault, because the delta winding gives a low zero-sequence impedance. Further out, the protective conductor adds impedance and the earth fault current falls faster. The minimum earth fault current at the end of a circuit decides whether the breaker trips fast enough for automatic disconnection.

Regional notes

GCC

Utilities publish the maximum MV fault level for design (for example 20 to 31.5 kA at 11 kV) and often specify LV main boards at 50 to 65 kA for large transformers. Check the utility's design values for LV fault level at the incoming terminals.

Australia and New Zealand

AS/NZS 3000 requires the fault level at the main switchboard to be determined. Network operators publish prospective fault currents at the point of supply. AS/NZS 3008.1 helps with cable impedances.

Asia

IEC 60909 is the common basis. Utilities publish the prospective fault level at the point of supply for consumer substations.

Test yourself

0 / 8

8 questions. Pick an answer to see the explanation. Practise about 50 more MV-506 questions in the MV Question Bank.

Frequently asked questions

How do I calculate the fault level at a transformer LV terminal?

A quick estimate is Ik = In / uk, where In is the LV full-load current and uk is the per-unit impedance. A 1500 kVA, 6% transformer at 400 V has In = 2165 A, so Ik is about 36 kA with an infinite source. The MV source impedance and the c factor change this slightly, which the studio includes.

What is the c factor in IEC 60909?

The voltage factor c scales the nominal voltage to cover the highest likely operating voltage. For maximum fault currents it is 1.10 at MV and 1.05 or 1.10 at LV, depending on the voltage tolerance.

What is the difference between Icu, Ics and Icw?

Icu is a breaker's ultimate breaking capacity, Ics its service breaking capacity (it can still be used afterwards), and Icw the short-time withstand current a board or breaker can carry for a set time, usually 1 s, so downstream devices can clear first.

Do motors contribute to fault current?

Yes. Running induction motors feed fault current for the first few cycles, roughly their locked-rotor current. IEC 60909 includes them in Ik" and ip, but they decay quickly.

Related MEPVirtual labs

MV-500 Medium Voltage Hub MV-504 Distribution Transformer Lab MV-507 MV Protection & Relay Coordination E-127 LV Switchgear Design Lab

Standards referenced

IEC 60909-0 Short-circuit currents
IEC 60947-2 LV circuit-breakers
IEC 61439-1 LV switchgear assemblies
AS/NZS 3000 and AS/NZS 3008.1 (AU/NZ)