Faults
• Short circuits occur due to equipment insulation failures.
• Often caused by overvoltage conditions due to lightning, switching surges, insulation contamination (e.g. salt spray) or other mechanical failures.
• The “fault” current is determined by generator voltage and impedances between the source and the fault.
• Fault currents may be orders of magnitude larger than normal currents.
• Fault currents can cause extensive thermal damage if not stopped. The high magnetic forces may also cause damage.
• Above approximately 300 kV, faults are cleared within 3 cycles (50 ms for 60 Hz.)
• Below 300 kV, faults may take up to 5-20 cycles to clear.
Symmetrical vs Unsymmetrical
• Symmetrical faults may be the easiest to analyze, but not very realistic.
• More than 80% of all faults involve only a single phase and ground.
• In 1918, C. L. Fortescue demonstrated that unbalanced currents (e.g., in an unsymmetrical fault) can be represented by sums of balanced and symmetrical components. A type of superposition.
Symmetrical Components
• An unsymmetrical fault. ia, ib, ic are not equal.• The rest of the system is balanced.• Assume the components have reached a
“pseudo-steady state.” Probably not true.• This allows the use of phasors. Replace ia with
Ia, etc.
Symmetrical ComponentsIa = Ia1 + Ia2 + Ia0
Ib = Ib1 + Ib2 + Ib0
Ic = Ic1 + Ic2 + Ic0
where the left side are the phasors representing the fault currents and the right side consists of 3 symmetrical components:
subscript 1: positive sequencesubscript 2: negative sequencesubscript 0: zero sequence
Symmetrical ComponentsBecause the components follow the standard sequences (positive, negative) and are balanced, we can relate the different components to each other.
Define the following operators:
a = 1 120� = �0.5 + |0.866
a2 = 1 240� = �0.5� |0.866
Symmetrical ComponentsThen, we can create the following relationships:
a = 1 120� = �0.5 + |0.866
a2 = 1 240� = �0.5� |0.866
Ib1 = a2Ia1Ic1 = aIa1
Ib2 = aIa2
Ic2 = a2Ia2
Symmetrical ComponentsThen, the fault currents can all be related to the a-phase currents:
Ia = Ia1 + Ia2 + Ia0
Ib = a2Ia1 + aIa2 + Ia0Ic = aIa1 + a2Ia2 + Ia0
Putting this in matrix form:2
4IaIbIc
3
5 =
2
41 1 1a2 a 1a a2 1
3
5
2
4Ia1Ia2Ia0
3
5
Symmetrical ComponentsInverting the matrix and solving for the a-phase sequence currents:
2
4Ia1Ia2Ia0
3
5 =1
3
2
41 a a2
1 a2 a1 1 1
3
5
2
4IaIbIc
3
5
What does this mean?
Symmetrical Components
2
4Ia1Ia2Ia0
3
5 =1
3
2
41 a a2
1 a2 a1 1 1
3
5
2
4IaIbIc
3
5
Best shown with an example.
Types of Faults• Symmetrical three-phase, and three-phase
to ground fault.
• Single-line to ground fault.
• Double-line to ground fault.
• Double line fault (without ground.)
• Fault with fault impedances.
• Other short-circuit faults.
• Open circuit faults.
Single-Line to Ground Fault
Ib = Ic = 0
2
4Ia1Ia2Ia0
3
5 =1
3
2
41 a a2
1 a2 a1 1 1
3
5
2
4IaIbIc
3
5with
Ia1 = Ia2 = Ia0 Ia1 =Ia3
and
and
Single-Line to Ground Fault
Ib = Ic = 0 Ia1 = Ia2 = Ia0 Ia1 =Ia3
,
Va = Va1 +Va2 +Va0 = ZfIa = Zf3Ia1
, Va = ZfIa, and
Single-Line to Ground Fault
Va = Va1 +Va2 +Va0 = ZfIa = Zf3Ia1
Ia1 = Ia2 = Ia0 =Ea1
Z1 + Z2 + Z0 + 3ZF
Single-Line to Ground Fault
Va = Va1 +Va2 +Va0 = ZfIa = Zf3Ia1
Ia1 = Ia2 = Ia0 =Ea1
Z1 + Z2 + Z0 + 3ZF
Once the sequence currents are known, all currents and voltages in the faulted network can be calculated.
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