Symmetrical components and unbalanced faults on the PE Power exam

The PE Power exam asks two things here. One is to resolve a set of unbalanced phasors into sequence components, or to rebuild a phase quantity from them. The other is to solve an unbalanced fault by connecting sequence networks at the fault point. In the NCEES handbook, the Symmetrical Components section of the Circuits chapter gives both transformations and the value of the operator a. The Fault Current Analysis section of the Transmission and Distribution chapter gives the network connection for each fault type, the fault-impedance rule, and the zero-sequence models for wye connections and transformer banks.

What you actually need to own

• The operator. a = 1∠120° = −0.5 + j0.866, and a² = 1∠240°. Two identities do most of the work: 1 + a + a² = 0, and a − a² = j√3.
• Resolving phase quantities. I_a0 = (I_a + I_b + I_c)/3. I_a1 = (I_a + a·I_b + a²·I_c)/3. I_a2 = (I_a + a²·I_b + a·I_c)/3. The 1/3 belongs to this direction only.
• Rebuilding phase quantities. I_a = I_a0 + I_a1 + I_a2. I_b = I_a0 + a²·I_a1 + a·I_a2. I_c = I_a0 + a·I_a1 + a²·I_a2. There is no 1/3 here. Voltages follow the same rows.
• What the three sets are. Each set has three phasors of equal magnitude. The positive set is spaced 120° apart in abc order, the negative set 120° apart in acb order, and the three zero-sequence phasors are in phase. Because they are in phase, they add directly: neutral current and residual current both equal 3·I_a0. A zero residual means no ground return.
• The connection for each fault. E is the prefault line-to-neutral voltage.
  • Three-phase: positive network alone. I = E/Z1.
  • Single line-to-ground: all three networks in series. I_a1 = I_a2 = I_a0 = E/(Z1 + Z2 + Z0), and the faulted-phase current is 3·I_a0.
  • Line-to-line: positive and negative networks in parallel at the fault point, zero-sequence network unused. I_a1 = −I_a2 = E/(Z1 + Z2), I_a0 = 0, and each faulted phase carries √3·|I_a1|.
  • Double line-to-ground: all three networks in parallel. The sequence voltages are equal and the sequence currents sum to zero. I_a1 = E/(Z1 + Z2·Z0/(Z2 + Z0)), the zero-sequence share is |I_a0| = |I_a1|·Z2/(Z2 + Z0), and the ground current is 3·I_a0.
• Zero-sequence paths. A solidly grounded wye connects to the zero-sequence reference. A wye grounded through Z_n connects through 3·Z_n. An ungrounded wye is open. A grounded-wye to delta bank is open on its delta side: zero-sequence current flows in the grounded-wye winding and its neutral, circulates inside the delta, and does not appear on the delta-side lines.
• Fault impedance. For a fault involving ground, 3·Z_F goes in the zero-sequence network. For a fault without ground, Z_F goes in series in the positive-sequence network.

Where people lose points

• Misplacing the 1/3. Dropping it when resolving triples every component. Dividing a known I_a0 by 3 to get neutral current is the same slip in reverse.
• Swapping a and a². In the resolve step this returns the positive-sequence component when the negative was asked. In the rebuild it returns phase c in place of phase b.
• Stopping at I_a1. The line current comes from the rebuild: a factor of 3 for a single line-to-ground fault, √3 for a line-to-line fault.
• Driving the networks with line-to-line voltage. This inflates the answer by √3.
• Entering a neutral impedance once. It carries three times the zero-sequence current, so it counts three times.
• Inverting the current divider in a double line-to-ground fault. The larger share goes through the smaller impedance.
• Building the zero-sequence network from the data list. Build it from the one-line. An ungrounded star point, a delta winding or an open tie removes a path. Do not assume Z0 equals Z1.
• Reading "bolted" as "largest". Bolted means zero fault impedance. Which fault type gives the most current depends on Z0 relative to Z1.

How to study it

Take one bus with E = 1.0, Z1 = Z2 = j0.10 and Z0 = j0.05 per unit, and solve all four faults. You should get 10.0 per unit for three-phase, 12.0 for single line-to-ground, 8.66 in each faulted phase for line-to-line, and 15.0 of ground current for double line-to-ground. Repeat with Z0 = j0.30 and watch the ranking change. Keep the bases straight with the per-unit system, and see fault current analysis for three-phase duty and source contributions.

Part of the Circuit Analysis area of the PE Power exam. → Practice it on CastorPrep