Power Systems Lab · 18 modules · all computed in your browser
Symmetrical components & faults
Fortescue decomposer, sequence networks, SLG/LL/LLG/3φ.
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Power Systems Lab · 18 modules · all computed in your browser
Fortescue decomposer, sequence networks, SLG/LL/LLG/3φ.
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Fortescue decomposer · drag the phases
V0
0.000 ∠ 0.0°
V1
1.000 ∠ -0.0°
V2
0.000 ∠ 0.0°
VUF = |V2|/|V1|
0.0 %
Fault analysis · sequence networks
All three networks in SERIES (plus 3Zf): I1 = I2 = I0 = E/(Z1+Z2+Z0+3Zf).
Phase currents
Phase voltages at the fault
I0 pu
2.000 ∠ -90.0°
I1 pu
2.000 ∠ -90.0°
I2 pu
2.000 ∠ -90.0°
In = 3I0
6.000 pu
| Phase | pu | kA | V pu |
|---|---|---|---|
| A | 6.00 ∠ -90.0° | 1.51 | 0.000 |
| B | 0.00 ∠ 0.0° | 0.00 | 0.917 |
| C | 0.00 ∠ 0.0° | 0.00 | 0.917 |
I_base = MVA·1000/(√3·kV) = 251 A.
IEC 60909 short-circuit calculation · voltage factor c and peak current
c max / c min
1.10 / 1.00
R/X → κ
0.077 → 1.798
I"k3 max
5.60 kA
I"k3 min
5.09 kA
Peak ip = κ√2·I″k
14.24 kA
I"k1 max (Z0 ≈ 1.5R + jX0)
5.42 kA
Use c_max for equipment duty and breaker ratings, and c_min for protection sensitivity at minimum generation. I″k = c·Un/(√3·|Zk|). The equivalent voltage source at the fault replaces every network EMF, which is why it differs from the load-flow-based (superposition) result in the SLD simulator.
Learn it
An unbalanced three-phase set can always be split into three balanced sets: positive (normal rotation, what motors like), negative (reverse rotation, it heats rotors and signals unbalance), and zero (all three in phase, it can only flow if there is a ground or neutral path).
Each sequence sees its own impedance network. A fault is just a particular way of connecting those three networks at the fault point. That is why a ground fault needs Z0 and a line-to-line fault doesn’t.
Quiz · 0/3
1. Which fault type does not involve the zero-sequence network?
2. For a bolted SLG fault, Ia equals…
3. A delta winding on the fault side makes Z0 seen from that side…