Fortescue’s symmetrical components (1918)

An unbalanced three-phase system, such as one with a fault to ground, couples all three phases and resisted hand calculation. Charles LeGeyt Fortescue of Westinghouse showed in 1918 that any three phasors are the sum of three balanced sets: a zero-sequence set (three equal phasors), a positive-sequence set (a-b-c rotation) and a negative-sequence set (a-c-b) (symmetrical_components()). In a symmetrical network the three sequences do not interact, so each is solved as a single-phase circuit and the results recombined (phase_components()). This example decomposes an unbalanced set, computes a single-line-to-ground fault the way protection engineers still do and compares the four fault types (fault_currents()).

import matplotlib.pyplot as plt
import numpy as np

from engineeringkit.electrical_engineering import fault_currents, phase_components, phasor, symmetrical_components
from engineeringkit.electrical_engineering.visualizers import plot_phasors

Decomposing an unbalanced set

va, vb, vc = phasor(1.0, 0.0), phasor(0.8, np.radians(-125)), phasor(1.1, np.radians(115))
v0, v1, v2 = symmetrical_components(va, vb, vc)
print(f"|V0| = {abs(v0):.3f}, |V1| = {abs(v1):.3f}, |V2| = {abs(v2):.3f} pu; unbalance V2/V1 = {abs(v2 / v1):.1%}")
zero, positive, negative = phase_components(v0, 0, 0), phase_components(0, v1, 0), phase_components(0, 0, v2)
assert np.allclose(np.add(np.add(zero, positive), negative), (va, vb, vc))

|V0| = 0.117, |V1| = 0.966, |V2| = 0.060 pu; unbalance V2/V1 = 6.2%

A single-line-to-ground fault

Phase a shorts to ground at a bus whose sequence impedances are \(Z_1 = Z_2 = j0.2\) and \(Z_0 = j0.1\) pu, prefault voltage 1 pu. The boundary conditions \(I_b = I_c = 0\), \(V_a = 0\) connect the three sequence networks in series: \(I_0 = I_1 = I_2 = E/(Z_0 + Z_1 + Z_2)\).

E, Z1, Z2, Z0 = 1.0, 0.2j, 0.2j, 0.1j
I_seq = E / (Z0 + Z1 + Z2)
Ia, Ib, Ic = phase_components(I_seq, I_seq, I_seq)
V_seq = (-Z0 * I_seq, E - Z1 * I_seq, -Z2 * I_seq)
Va, Vb, Vc = phase_components(*V_seq)
print(f"fault current |Ia| = {abs(Ia):.2f} pu (a three-phase fault would give {abs(E / Z1):.2f} pu)")
print(f"healthy-phase voltages |Vb| = {abs(Vb):.3f}, |Vc| = {abs(Vc):.3f} pu")
assert np.isclose(Ia, 3 * E / (Z0 + Z1 + Z2))
assert abs(Ib) < 1e-12 and abs(Ic) < 1e-12 and abs(Va) < 1e-12
assert np.allclose(fault_currents(E, Z1, Z2, Z0, "slg"), (Ia, Ib, Ic))

fault current |Ia| = 6.00 pu (a three-phase fault would give 5.00 pu)
healthy-phase voltages |Vb| = 0.917, |Vc| = 0.917 pu

The four fault types

The other faults connect the sequence networks differently. Here the single-line-to-ground fault is the most severe, because the zero-sequence impedance is below the positive-sequence one.

for kind in ("3ph", "slg", "ll", "llg"):
    currents = fault_currents(E, Z1, Z2, Z0, kind)
    print(f"{kind:>4}: |Ia|, |Ib|, |Ic| = " + ", ".join(f"{abs(i):.2f}" for i in currents) + " pu")
assert np.isclose(abs(fault_currents(E, Z1, Z2, Z0, "ll")[1]), np.sqrt(3) / 2 * abs(E / Z1))

fig, axes = plt.subplots(1, 4, figsize=(14, 3.6))
labels = ["a", "b", "c"]
plot_phasors([va, vb, vc], labels=labels, ax=axes[0])
plot_phasors(positive, labels=labels, ax=axes[1])
plot_phasors(negative, labels=labels, ax=axes[2])
plot_phasors(zero, labels=labels, ax=axes[3])
for ax, title in zip(axes, ["unbalanced set", "positive sequence", "negative sequence", "zero sequence"]):
    ax.set_title(title)
    ax.set(xlim=(-1.2, 1.2), ylim=(-1.2, 1.2))
fig.tight_layout()

plt.show()

3ph: |Ia|, |Ib|, |Ic| = 5.00, 5.00, 5.00 pu
slg: |Ia|, |Ib|, |Ic| = 6.00, 0.00, 0.00 pu
 ll: |Ia|, |Ib|, |Ic| = 0.00, 4.33, 4.33 pu
llg: |Ia|, |Ib|, |Ic| = 0.00, 5.73, 5.73 pu