AEE Electrical Engineering Core · Chapter 9

Faults, Symmetrical Components and Stability

What to remember

Symmetrical components split any unbalanced three-phase set into positive, negative and zero sequence sets. Fault current is found by joining the three sequence networks in the way that suits the fault type.
Fault formulas: three-phase If = E/Z1; line-to-ground If = 3E/(Z0+Z1+Z2); line-to-line If = √3 E/(Z1+Z2). The three-phase fault is the most severe but the least frequent; the line-to-ground fault is the most frequent.
Stability is the ability to remain in synchronism after a disturbance. The swing equation and the equal-area criterion are the main tools.

Types of faults

Type	Nature	Frequency of occurrence
Single line-to-ground (LG)	Unsymmetrical, shunt	Most common
Line-to-line (LL)	Unsymmetrical, shunt	Less common
Double line-to-ground (LLG)	Unsymmetrical, shunt	Less common
Three-phase (LLL or LLLG)	Symmetrical, balanced	Least common, most severe
Open conductor	Unsymmetrical, series	Caused by broken conductor or blown fuse

A symmetrical fault keeps the three phases balanced, so a single-phase (per-phase) analysis is enough. Unsymmetrical faults need symmetrical components.

Fault level (short-circuit MVA) = Sbase / X(pu) = √3 × V(kV) × If(kA). If a system has a Thevenin reactance of 0.2 pu on 100 MVA, fault MVA = 100/0.2 = 500 MVA and fault current is 5 pu.

Machine reactances

Subtransient reactance Xd'' acts in the first few cycles (damper winding effect).
Transient reactance Xd' acts after that for a short time.
Synchronous reactance Xd acts in steady state.
Order: Xd'' < Xd' < Xd. Circuit breaker rating uses the subtransient value; relay settings often use transient value.

Symmetrical components

Any set of three unbalanced phasors Va, Vb, Vc can be written as the sum of:

Positive sequence: three equal phasors, 120° apart, same phase order as the system (a-b-c).
Negative sequence: three equal phasors, 120° apart, opposite phase order (a-c-b).
Zero sequence: three equal phasors in phase with each other.

Operator a = 1∠120° = −0.5 + j0.866. Also a² = 1∠240° and a³ = 1. The identity 1 + a + a² = 0 is important.

Formulas:

V0 = (Va + Vb + Vc)/3
V1 = (Va + aVb + a²Vc)/3
V2 = (Va + a²Vb + aVc)/3
Va = V0 + V1 + V2; Vb = V0 + a²V1 + aV2; Vc = V0 + aV1 + a²V2

Key results

A balanced set has only positive sequence.
Zero sequence current is one-third of the sum of phase currents: I0 = (Ia + Ib + Ic)/3. Neutral current In = 3I0.
Zero sequence current needs a return path: a grounded neutral. A delta winding or an ungrounded star blocks it from flowing in the line.
Ia = 30 A, Ib = Ic = 0 gives I0 = I1 = I2 = 10 A.
Line-to-line voltages have no zero-sequence component.

Sequence impedances

Element	Z1 and Z2	Z0
Static items: transformer, transmission line	Z1 = Z2	Transformer: equal to leakage impedance (depends on connection); line: about 2 to 3.5 times Z1
Rotating machine	Z1 ≠ Z2	Different again, usually smaller
Generator neutral impedance Zn	Does not appear	Appears as 3Zn

Sequence emf exists only in the positive sequence network. Negative and zero networks have no source.

Transformer zero-sequence connection

Grounded star on one side and delta on the other: zero-sequence current can flow on the star side, and circulates inside the delta. It cannot pass to the delta line.
Ungrounded star: zero-sequence current cannot flow on that side. The branch is open.
Star-delta transformers also shift the positive sequence by 30° and the negative sequence by 30° in the opposite direction.

Fault calculation using sequence networks

Let E be the pre-fault voltage and Zf the fault impedance.

Fault	Network connection	Fault current
Three-phase	Only positive sequence	If = E / (Z1 + Zf)
Single line-to-ground	All three in series, with 3Zf	Ia1 = E / (Z0 + Z1 + Z2 + 3Zf); If = 3 Ia1
Line-to-line	Positive and negative in parallel (zero network not used)	If = √3 E / (Z1 + Z2) (when Zf = 0)
Double line-to-ground	All three in parallel	Ia1 = E / (Z1 + Z2 ∥ Z0)

Worked example 1. Z1 = Z2 = Z0 = j0.2 pu, E = 1 pu.

Three-phase: If = 1/0.2 = 5 pu.
LG: If = 3/(0.6) = 5 pu.
LL: If = 1.732/0.4 = 4.33 pu.
LLG: Ia1 = 1/(0.2 + 0.1) = 3.33 pu; Ia0 = −3.33 × 0.2/0.4 = −1.67 pu; ground current = 3Ia0 = 5 pu (magnitude).

Remarks

When Z0 is small compared with Z1, the LG fault current can exceed the three-phase fault current. This can happen near a solidly grounded generator or transformer.
Grounding the neutral through a resistor or reactor (Zn) reduces LG fault current.
In a line-to-line fault, no zero-sequence current flows. In a three-phase fault, only positive sequence current flows.

Power system stability

Definition. Stability is the ability of the system to return to normal operation (synchronism) after a disturbance.

Type	Disturbance	Study
Steady-state stability	Small, slow load change	Maximum power transfer; limit at δ = 90°
Transient stability	Large, sudden: fault, switching, loss of generation	First swing, a few seconds
Dynamic stability	Small disturbance with automatic controls (AVR, governor)	Longer period

Power angle equation. For a generator connected to an infinite bus through reactance X:

P = (E V / X) sin δ

The maximum power Pmax = EV/X occurs at δ = 90° (steady-state stability limit). With E = 1.2, V = 1, X = 0.6 pu, Pmax = 2 pu; at δ = 30°, P = 1 pu.

Inertia constant H = kinetic energy stored at rated speed (MJ) / machine rating (MVA). Unit: MJ/MVA (or second). A 200 MVA machine with H = 5 stores 1000 MJ.

Swing equation. (H / (π f)) d²δ/dt² = Pm − Pe (in per-unit, δ in electrical radian). It is similar to Newton's law for rotation: accelerating power Pa = Pm − Pe. If Pm = 1 pu and Pe = 0.4 pu, accelerating power is 0.6 pu and the rotor speeds up. In M = SH/(180f) form, M is in MJ-s per electrical degree.

Equal-area criterion. For a single machine on an infinite bus, stability is kept if the accelerating area (A1) can be matched by an equal decelerating area (A2) before the angle reaches a limit. It avoids solving the swing equation. It applies only to one machine against an infinite bus (or two machines reduced to this).

Critical clearing angle: the largest angle at which the fault can be cleared with the system still stable.
Critical clearing time: time corresponding to the critical clearing angle.

Ways to improve stability

Fast fault clearing and high-speed circuit breakers.
Single-pole auto-reclosing.
High-speed excitation systems and AVR.
Reducing line reactance: bundled conductors, double circuits, series capacitors.
Fast valving of turbines, braking resistors.
Lower transformer reactance.
Higher system voltage and more parallel paths.

Related terms

Synchronising power coefficient: dP/dδ = (EV/X) cos δ. Positive for stable operation.
Loss of synchronism (pole slipping): the machine falls out of step. Out-of-step relays are used.
Infinite bus: a bus of constant voltage and frequency; its inertia is very large.

Exam traps

Three-phase faults are the most severe, but LG faults are the most common.
Z0 of a transmission line is larger than Z1; Z1 and Z2 of a line are equal.
For rotating machines Z1 and Z2 differ; for static equipment they are equal.
Zero-sequence network: Zn appears as 3Zn.
A delta winding traps zero-sequence current; it does not carry it to the line.
LG fault: networks in series. LL fault: positive and negative in parallel. LLG: all three in parallel.
Subtransient, transient, synchronous: Xd'' is the smallest.
Steady-state limit is 90°, but equal-area transient swings may pass beyond this and still be stable.
Fault level is in MVA; fault current is in kA.
Inertia constant H has units of MJ/MVA, not MW.
Equal-area criterion is for a single machine and infinite bus.
Stability studies do not apply to a pure fault level calculation.

One-liners

1. Operator a = 1∠120°; 1 + a + a² = 0.
2. A balanced system has only positive sequence components.
3. Zero-sequence current needs a ground return path.
4. In-phase current in all three lines (equal zero-sequence) flows back through the neutral.
5. Neutral current = 3 × zero-sequence current.
6. Fault MVA = base MVA / pu reactance.
7. Xd'' < Xd' < Xd.
8. Three-phase fault uses only the positive sequence network.
9. LG fault current = 3E/(Z0 + Z1 + Z2).
10. Pmax = EV/X, at δ = 90°.
11. H = stored kinetic energy in MJ divided by MVA rating.
12. Equal-area criterion: accelerating area equals decelerating area at the stability limit.

Practice questions

A system has a Thevenin reactance of 0.2 pu on a 100 MVA base. The three-phase fault level is:
1. 20 MVA
2. 200 MVA
3. 500 MVA
4. 50 MVA
Answer
C. 500 MVA
Fault MVA = base MVA / X = 100/0.2 = 500 MVA.
Phase currents are Ia = 30 A, Ib = 0, Ic = 0. The zero-sequence current is:
1. 0 A
2. 90 A
3. 30 A
4. 10 A
Answer
D. 10 A
I0 = (30 + 0 + 0)/3 = 10 A.
A balanced set Va = 10∠0°, Vb = 10∠−120°, Vc = 10∠120° has zero-sequence voltage V0 equal to:
1. 3.33 V
2. 0
3. 10 V
4. 30 V
Answer
B. 0
V0 = (Va + Vb + Vc)/3 = 0 for a balanced set; only positive sequence exists.
Z0 = Z1 = Z2 = j0.2 pu and E = 1 pu. The single line-to-ground fault current is:
1. 5 pu
2. 3.33 pu
3. 4.33 pu
4. 1.67 pu
Answer
A. 5 pu
If = 3E/(Z0 + Z1 + Z2) = 3/0.6 = 5 pu.
With Z1 = Z2 = j0.2 pu and E = 1 pu, the line-to-line fault current is about:
1. 1.73 pu
2. 5 pu
3. 4.33 pu
4. 2.5 pu
Answer
C. 4.33 pu
If = √3 E/(Z1 + Z2) = 1.732/0.4 = 4.33 pu.
For Z0 = Z1 = Z2 = j0.2 pu and E = 1 pu, the earth current in a double line-to-ground fault is (magnitude):
1. 3.33 pu
2. 10 pu
3. 1.67 pu
4. 5 pu
Answer
D. 5 pu
Ia1 = 1/(0.2 + 0.1) = 3.33; Ia0 = 3.33 × 0.2/0.4 = 1.67; earth current = 3Ia0 = 5 pu.
A generator neutral is grounded through Zn = 0.05 pu. In the zero-sequence network it appears as:
1. 0.05 pu
2. 0.0167 pu
3. 0.45 pu
4. 0.15 pu
Answer
D. 0.15 pu
The neutral impedance carries 3I0, so Z0 includes 3Zn = 0.15 pu.
A 200 MVA machine has an inertia constant H = 5 MJ/MVA. The kinetic energy stored at rated speed is:
1. 1000 MJ
2. 100 MJ
3. 5000 MJ
4. 40 MJ
Answer
A. 1000 MJ
KE = H × S = 5 × 200 = 1000 MJ.
A generator has E = 1.2 pu, V = 1 pu and X = 0.6 pu. The maximum steady-state power transfer is:
1. 0.5 pu
2. 2 pu
3. 0.72 pu
4. 1.2 pu
Answer
B. 2 pu
Pmax = EV/X = 1.2 × 1/0.6 = 2 pu.
For the same machine (Pmax = 2 pu), the power transferred at δ = 30° is:
1. 2 pu
2. 1.73 pu
3. 1 pu
4. 0.5 pu
Answer
C. 1 pu
P = Pmax sin 30° = 2 × 0.5 = 1 pu.
A generator has Pm = 1 pu and Pe = 0.4 pu after a fault. The accelerating power is:
1. 0.6 pu
2. −0.6 pu
3. 0.4 pu
4. 1.4 pu
Answer
A. 0.6 pu
Pa = Pm − Pe = 1 − 0.4 = 0.6 pu; the rotor speeds up.
Which relation is correct for the operator a?
1. a³ = −1
2. 1 + a + a² = 0
3. 1 + a + a² = 1
4. a = 1∠90°
Answer
B. 1 + a + a² = 0
a = 1∠120°, a³ = 1, and 1 + a + a² = 0.
The operator a² equals:
1. 1∠−60°
2. 1∠180°
3. 1∠240°
4. 1∠120°
Answer
C. 1∠240°
a² = (1∠120°)² = 1∠240°.
Equal in-phase currents of 5 A flow in each of the three lines (zero sequence). The current in the neutral is:
1. 8.66 A
2. 0 A
3. 5 A
4. 15 A
Answer
D. 15 A
In = 3I0 = 15 A.
A fault level of 500 MVA at 11 kV corresponds to a fault current of about:
1. 45.5 kA
2. 15.1 kA
3. 26.2 kA
4. 5.25 kA
Answer
C. 26.2 kA
I = 500/(√3 × 11) = 26.2 kA.
Which fault occurs most frequently on overhead lines?
1. Three-phase
2. Single line-to-ground
3. Line-to-line
4. Double line-to-ground
Answer
B. Single line-to-ground
Single line-to-ground faults form the largest share of all faults.
Which fault is generally the most severe but the least frequent?
1. Single line-to-ground
2. Open conductor
3. Line-to-line
4. Three-phase
Answer
D. Three-phase
A three-phase fault usually gives the highest current in a system with a small zero-sequence reactance.
The correct order of generator reactances is:
1. Xd'' < Xd' < Xd
2. Xd'' = Xd' = Xd
3. Xd < Xd' < Xd''
4. Xd' < Xd'' < Xd
Answer
A. Xd'' < Xd' < Xd
Reactance rises from subtransient to transient to synchronous.
Zero-sequence current can flow in the lines only if there is:
1. An isolated neutral
2. A delta-connected generator only
3. A balanced load
4. A path to ground (grounded neutral)
Answer
D. A path to ground (grounded neutral)
Zero-sequence currents are in phase and need a return path through ground or neutral.
A delta winding in a transformer:
1. Blocks positive-sequence current
2. Allows zero-sequence current to circulate inside it but not to flow in the line
3. Has infinite positive-sequence impedance
4. Passes zero-sequence current to the line
Answer
B. Allows zero-sequence current to circulate inside it but not to flow in the line
Zero-sequence current circulates within the delta and does not appear in the lines.
For a transmission line, the sequence impedances satisfy:
1. Z1 = Z2 and Z0 is larger
2. Z0 < Z1 = Z2
3. Z0 = Z2 and Z1 is larger
4. Z1 = Z0 and Z2 is larger
Answer
A. Z1 = Z2 and Z0 is larger
Static equipment has Z1 = Z2; for lines Z0 is about two to three and a half times Z1.
In sequence networks, a source emf is present in:
1. Only the positive-sequence network
2. Only the zero-sequence network
3. Only the negative-sequence network
4. All three networks
Answer
A. Only the positive-sequence network
Generators produce positive-sequence voltage only.
In a line-to-line fault (no ground):
1. Negative-sequence current is zero
2. Zero-sequence current is one-third of fault current
3. Zero-sequence current is zero
4. Positive sequence current is zero
Answer
C. Zero-sequence current is zero
With no ground path, I0 = 0 and I2 = −I1.
For a single line-to-ground fault, the sequence networks are connected:
1. In parallel
2. Only the positive network is used
3. In series
4. Positive and negative in parallel, zero open
Answer
C. In series
I0 = I1 = I2, so the networks are in series.
For a double line-to-ground fault, the sequence networks are connected:
1. All three in series
2. All three in parallel
3. Positive and negative in parallel only
4. Only zero in series
Answer
B. All three in parallel
V0 = V1 = V2 means a parallel connection.
The steady-state stability limit of a simple generator-infinite bus system corresponds to a power angle of:
1. 180°
2. 45°
3. 30°
4. 90°
Answer
D. 90°
P = Pmax sin δ is maximum at 90°.
The equal-area criterion applies directly to:
1. Any large multi-machine network
2. Only to induction motors
3. Only to dynamic stability
4. A single machine connected to an infinite bus
Answer
D. A single machine connected to an infinite bus
It works for a one-machine system, or two machines reduced to such a system.
The critical clearing angle is:
1. The angle of 90° always
2. The angle at which the fault occurs
3. The largest angle at which the fault can be cleared without loss of stability
4. The angle at which Pe is zero
Answer
C. The largest angle at which the fault can be cleared without loss of stability
Beyond this angle the decelerating area is too small and the machine loses synchronism.
The unit of the inertia constant H is:
1. MJ/MVA
2. kg
3. MW
4. MVA/s
Answer
A. MJ/MVA
H = kinetic energy at rated speed (MJ) divided by rating (MVA).
Which measure does NOT improve transient stability?
1. Auto-reclosing
2. Increasing the reactance of the transmission line
3. Fast fault clearing
4. High-speed excitation
Answer
B. Increasing the reactance of the transmission line
A higher reactance reduces Pmax and the decelerating area.
Transient stability concerns the response of the system to:
1. Slow drift in frequency
2. A small gradual load change
3. Voltage flicker
4. A large, sudden disturbance such as a fault
Answer
D. A large, sudden disturbance such as a fault
It studies the first swings after a major disturbance.
The synchronising power coefficient of a machine is:
1. EV/X²
2. EV X cos δ
3. (EV/X) cos δ
4. (EV/X) sin δ
Answer
C. (EV/X) cos δ
It is dP/dδ; it must be positive for stable operation.
Statements on symmetrical components: 1. A balanced three-phase set has only positive-sequence components. 2. Zero-sequence components are 120° apart.
1. 2 only
2. 1 only
3. Both 1 and 2
4. Neither 1 nor 2
Answer
B. 1 only
Zero-sequence components are equal and in phase.
Statements on faults: 1. The single line-to-ground fault is the most frequent fault. 2. The three-phase fault is generally the most severe fault.
1. Both 1 and 2
2. 1 only
3. 2 only
4. Neither 1 nor 2
Answer
A. Both 1 and 2
Both are standard facts.
Statements on transformer connections: 1. A delta winding traps the zero-sequence current. 2. An ungrounded star allows zero-sequence current to flow freely.
1. 2 only
2. Both 1 and 2
3. Neither 1 nor 2
4. 1 only
Answer
D. 1 only
An ungrounded star has no ground path, so zero-sequence current cannot flow.
Statements on generator reactances: 1. Xd'' is less than Xd'. 2. Xd' is less than Xd.
1. 1 only
2. 2 only
3. Both 1 and 2
4. Neither 1 nor 2
Answer
C. Both 1 and 2
Both are true.
Statements on stability: 1. The steady-state limit occurs at δ = 90°. 2. Transient stability deals with small, slow load changes.
1. 2 only
2. 1 only
3. Both 1 and 2
4. Neither 1 nor 2
Answer
B. 1 only
Transient stability concerns large sudden disturbances.
Statements on the equal-area criterion: 1. It avoids direct solution of the swing equation. 2. It can be applied directly to any multi-machine system.
1. 1 only
2. 2 only
3. Both 1 and 2
4. Neither 1 nor 2
Answer
A. 1 only
It is limited to a one-machine infinite-bus system.
Statements: 1. Fault level is measured in kA. 2. The inertia constant H is expressed in MW.
1. 1 only
2. 2 only
3. Both 1 and 2
4. Neither 1 nor 2
Answer
D. Neither 1 nor 2
Fault level is in MVA and H is in MJ/MVA.
Which pair is correctly matched?
1. Line-to-line fault – zero-sequence network in series
2. Three-phase fault – negative-sequence network only
3. Single line-to-ground fault – sequence networks in series
4. Double line-to-ground fault – networks in series
Answer
C. Single line-to-ground fault – sequence networks in series
Line-to-line uses only networks 1 and 2 in parallel; three-phase uses only network 1; LLG uses parallel.
Statements on zero-sequence quantities: 1. Neutral current equals three times the zero-sequence current. 2. Line-to-line voltages contain zero-sequence components.
1. 1 only
2. 2 only
3. Both 1 and 2
4. Neither 1 nor 2
Answer
A. 1 only
Zero-sequence components cancel in line-to-line voltages.
Statements on a three-phase fault: 1. Only positive-sequence current flows. 2. Zero-sequence network carries the fault current.
1. 2 only
2. Both 1 and 2
3. Neither 1 nor 2
4. 1 only
Answer
D. 1 only
A symmetrical fault involves only the positive-sequence network.
Statements on the line-to-line fault: 1. Ia0 = 0. 2. Ia2 = −Ia1.
1. 1 only
2. 2 only
3. Both 1 and 2
4. Neither 1 nor 2
Answer
C. Both 1 and 2
With no ground path, I0 = 0 and the negative current equals the negative of positive current.
Statements on stability improvement: 1. Double-circuit lines reduce transfer reactance. 2. Faster circuit breakers shorten fault duration.
1. 1 only
2. Both 1 and 2
3. 2 only
4. Neither 1 nor 2
Answer
B. Both 1 and 2
Both help improve transient stability.

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