Dashboard PE Power Exam Prep Transmission and Distribution Fault analysis Three-phase fault analysis — symmetrical and unbalanced

Three-phase fault analysis — symmetrical and unbalanced

Thevenin reduction, 3-φ symmetrical fault, sequence networks for SLG / LL / DLG, fault impedance, breaker sizing, motor contribution.

Sophomore ~12 min

Step 1 — Thevenin reduction: system → V_th + Z_th at the fault bus

0.55×
fault 3-φ I_fault — pu S_SC — MVA

Reference notes

Use Next → on the narrator above to step through fault analysis: Thevenin reduction, symmetrical 3-φ faults, component sequence impedances, single line-to-ground faults (the most common type), other unbalanced faults, and practical fault-impedance + breaker-rating + motor-contribution considerations.

The Thevenin reduction

Every fault calculation starts by reducing the network behind the fault to a Thevenin equivalent at the fault bus: a pre-fault voltage source Vth in series with a Thevenin impedance Zth. Vth is typically taken as 1.0 pu (the pre-fault nominal voltage). Zth is computed by combining generator subtransient reactances, transformer leakage impedances, and line impedances in the appropriate parallel-series fashion, all expressed in per-unit on a common base.

Symmetrical three-phase fault

All three phases short to each other through a low impedance. By symmetry only the positive-sequence network is active:

Ifault,pu = Vth / (Z+ + Zf)
SSC = Vth² · Sbase / Z+,pu ≈ Sbase / Z+,pu (Vth ≈ 1 pu)

Worked example on a 100 MVA, 13.8 kV system with Z_+ = 0.10 pu, bolted fault:

Component sequence impedances

Each piece of equipment has three impedances:

ComponentZ+ZZ0
Static (transformer, line)X_leakage= Z_+depends on grounding
Synchronous turbo generatorX_d″ ≈ 0.15–0.25 pu≈ X_d″ to 0.1·X_d″very low (0.02–0.05 pu)
Synchronous hydro generatorX_d″ ≈ 0.20–0.40 pu~0.2·X_d″low
Induction motorX_lr (locked-rotor)≈ X_lrno contribution

Zero-sequence behaviour is dominated by the grounding scheme:

Single line-to-ground (SLG) fault

The most common fault type (70–80%). Boundary conditions on a phase-a SLG fault: I_b = I_c = 0 and V_a = Z_f · I_a. Substituting into the Fortescue transform:

I+ = I = I0 = Vth / (Z+ + Z + Z0 + 3·Zf)
Ifault,a = I+ + I + I0 = 3·I+

The three sequence networks connect in series. The 3·Zf appears because all three sequence currents flow through the same physical fault impedance.

Other unbalanced faults

Practical considerations

Fault impedance Zf

Bolted metallic fault: Zf ≈ 0. Arcing fault: typically a few ohms to tens of ohms (the arc-voltage drop divided by arc current). Tree contact, contact resistance, soil resistance: highly variable. High-impedance ground faults often sit below overcurrent settings and are detected only by sensitive ground-fault relays.

Breaker interrupting rating

The breaker must clear the maximum first-cycle fault current at its location. Standard ratings: 22, 42, 65, 100, 200 kA RMS symmetrical. The asymmetrical (peak with DC offset) rating is roughly 2.6× the symmetrical RMS rating immediately after fault initiation.

Motor contribution

Every running motor over ~50 hp briefly contributes to fault current for the first ~6 cycles via stored kinetic energy and air-gap flux. The contribution magnitude is approximately 4–6× the motor's rated current (governed by X_d″ or locked-rotor reactance), decaying as the motor decelerates. IEEE Std 141 (Red Book) and IEC 60909 give standardised procedures for including motor contribution in industrial fault studies.

Take-away. Reduce the system to V_th + Z_th at the fault bus. For 3-φ: I = V_th / (Z_+ + Z_f). For SLG: I_a = 3·V_th / (Z_+ + Z_− + Z_0 + 3·Z_f) with the three sequence networks in series. For LL and DLG: other sequence-network interconnections. Always check both 3-φ and SLG — on well-grounded systems SLG can EXCEED the 3-φ fault. Include motor contribution for breaker sizing.

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