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Symmetrical components (Fortescue)
Enter the three phase currents or voltages with magnitude and angle and get the zero, positive and negative sequences, or go the other way. You see the phasor diagrams, the fault pattern and which protection functions respond to each sequence.
Fortescue (1918)symmetrical componentsIEC 60909-0§ 5.3.2symmetrical components, a = 1∠120°IEC 61000-2-2voltage unbalance V2/V1
Results
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Assumptions
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Charts
- a
- b
- c
- A
- B
- C
Calculation and details
Values in polar and rectangular form
| Phasor | Polar | Rectangular |
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What the protection sees
What it is for
Numerical relays do not look only at the phase currents: they compute the symmetrical components and decide with them. Use this tool to check an oscillography or event report by hand, understand why a sequence element operated (or did not), prepare the values for a secondary injection test, or show a crew what the relay sees for each fault type.
- Positive sequence (1): present in normal load and in every fault. Distance protection 21 and most metering use it.
- Negative sequence (2): appears with any unbalance: asymmetrical faults, an open phase, unbalanced loads. It is used by 46 (unbalance and broken conductor, with I2/I1), directional 67Q and voltage element 47.
- Zero sequence (0): exists only with a path to ground or a neutral. It is measured by 50N/51N as residual 3I0, by 67N (polarized with 3V0 or with neutral current) and by 59N as 3V0.
How it is calculated
Fortescue (1918)IEC 60909-0§ 5.3.2symmetrical components
I0 = (Ia + Ib + Ic) / 3I1 = (Ia + a·Ib + a²·Ic) / 3I2 = (Ia + a²·Ib + a·Ic) / 3With the operator a = 1∠120° = −½ + j·√3/2 and ABC phase rotation (Anderson, Analysis of Faulted Power Systems; Blackburn, Symmetrical Components for Power Systems Engineering). For ACB rotation, b and c are swapped.
Inverse transformation
Ia = I0 + I1 + I2 · Ib = I0 + a²·I1 + a·I2 · Ic = I0 + a·I1 + a²·I2The same equations apply to voltages. The residual current measured by a relay with the three CTs in wye is Ia + Ib + Ic = 3I0.
Fault patterns (no pre-fault load, referred to phase A)
- Single-phase A-ground: I0 = I1 = I2. For B-ground, I1 = a·I0 and I2 = a²·I0; for C-ground, the other way round. The tool compares phasors, not only magnitudes.
- Phase-to-phase B-C without ground: I0 = 0 and I2 = −I1. For C-A, I2 = −a·I1; for A-B, I2 = −a²·I1. An open phase with a load and no neutral gives the same pattern as a phase-to-phase fault on the other two phases.
- Phase-to-phase-to-ground: the healthy phase carries no fault current (for example, Ia = 0 for B-C-ground) and there are I0 and I2.
IEC 61000-2-2compatibility level of unbalance
For voltages, the tool compares V2/V1 with the 2 % compatibility level of public low-voltage networks. Some networks with many single-phase loads reach 3 %, so between 2 and 3 % the indicator shows a warning.
Limits of the calculation
- This is the mathematical transformation, exact for the phasors you enter, but not a short-circuit study: it does not calculate how much current a real fault will draw, which depends on the sequence impedances of the network.
- The fault patterns are the ideal ones, without pre-fault load current. With load, the current in the healthy phase is not zero and the relations hold only in part.
- The phasors must be fundamental-frequency values, taken at the same instant and with the same angle reference. For an oscillography with DC offset or harmonics, use the filtered values from the relay software.
- Some manufacturers refer the sequences to another phase or measure 3I0 with a neutral CT: check the relay manual before comparing values.
Frequently asked questions
How are the symmetrical components of three currents calculated?
With the Fortescue equations: I0 = (Ia + Ib + Ic) / 3, I1 = (Ia + a·Ib + a²·Ic) / 3 and I2 = (Ia + a²·Ib + a·Ic) / 3, where a = 1∠120°. The sums are phasor sums: magnitude and angle of each phase matter, and so does the phase rotation of the system (ABC; for ACB, b and c are swapped).
What is the difference between I0 and 3I0?
I0 is the zero-sequence component of one phase. The sum of the three phase currents, which is what flows in the neutral or what a residually connected relay measures, equals 3I0. That is why 50N/51N and 67N are set in 3I0 on most relays; check in the manual which one yours uses.
Why does an open phase produce so much negative-sequence current?
With one phase open, the other two carry currents that no longer form a balanced set. In a load with no path to ground they become equal and opposite, which gives I2 equal to I1 in magnitude and no I0. That is why the I2/I1 ratio is the basis of broken-conductor detection (46BC) and of the 46 unbalance protection on motors and generators.
We analyze your relay trip
Send us the oscillography and the relay event report. Our team analyzes the sequences, the directionality and the settings, and tells you whether the protection operated correctly. If settings need correcting, we load them remotely.
