Faults, Relays, and Circuit Breakers - Rowan Universityusers.rowan.edu/~krchnavek/Rowan_University/Power_files/Faults.pdfFaults, Relays, and Circuit Breakers Robert R. Krchnavek Rowan

Faults, Relays, and Circuit Breakers

Robert R. KrchnavekRowan University

Glassboro, New Jersey

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 Components

a-b-c sequence a-c-b sequence identical phase

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.

Symmetrical Three-Phase Faults

Single-Line to Ground Fault

Ib = Ic = 0

Va = ZfIa

and

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.

Need a numerical example.

Seems like this is not necessary when a difficult node voltage analysis would be enough. Need to show

this.