symmetrical components and fault...

11

Symmetrica l Components and Fault Currents

Terrence Smith

22

Review of Phasors

33

Origin of Phasors>Rotating rotors = alternating currents &

voltagesA

VA

A

B

B C

CVB

VC

N

S

> Phasors are well established means of representing ac circuits

Charles Proteus Steinmetz (1865-1923)Complex Quantities and their use in Electrical Engineering; Charles Proteus Steinmetz; Proceedings of the International Electrical Congress, Chicago, IL; AIEE Proceedings, 1893; pp.33-74.

= θt

Mt

t

4

Phasors

t

A

θ

A

Sine wave Phasor representation of sine wave

5

Addition of Phasors

θ1

A1

θ2A2

A3θ3

6

Multiplication of phasors

A1 x A2

θ1+θ2

θ1

A1

θ2A2

Multiplying two phasors results in another phasor. The magnitude of the new phasor is the product of the magnitude of the two original phasors. The angle of the new phasor is the sum of the angles of the two original phasors.

77

Review Of Symmetrica l Components

8

Symmetrical and Non -Symmetrical Systems:

Symmetrica l System:• Counter-clockwise rota tion

• All current vectors have equal amplitude

• All voltage phase vectors have equal amplitude

• All current and voltage vectors have 120 degrees phase shifts and a sum of 0v.

3 units

3 units

3 units

120°120°

120 °

Non-Symmetrica l System:• Fault or Unbalanced condition

• If one or more of the symmetrica l system conditions is not met

3 units

9 units

4 units 120°100°

140°

9

Symmetrical Components:

• A-B-CCounter-clockwisephase rotation

• All phasors with equal magnitude

• All phasors displaced 120 degrees apart

• No Rotation Sequence


• All phasors are in phase

• A-C-B counter -clockwisephase rotation


• All phasors displaced 120 degrees apart

Positive Sequence (Always Present)

C

A

B

Zero Sequence Negative Sequence

C

A

B

ABC

120°

120°

120°120°

120°

120°

10

Symmetrical Components:

I0 = ⅓ (Ia + Ib + Ic)

I1 = ⅓ (Ia + αIb + α2Ic)

I2 = ⅓ (Ia + α2Ib + αIc)

V0 = ⅓ (Va + Vb + Vc)

V1 = ⅓ (Va + αVb + α2Vc)

V2 = ⅓ (Va + α2Vb + αVc)

Unbalanced Line-to-Neutra l Phasors:

ABC

Zero Sequence Component:

PositiveSequence Component:

NegativeSequence Component:

α2 = 240°

CA

B α = 120°B

A

C

α2 = 240°α = 120°

Ia = I1 + I2 + I0

Ib = α2I1 + αI2 + I0

Ic = αI1 + α2I2 + I0

Va = V1 + V2 + V0

Vb = α2V1 + αV2 + V0

Vc = αV1 + α2V2 + αV0

α =Phasor @ +120°

α2 =Phasor @ 240°

11

Calculating Symmetrical Components:

Three-Phase Balanced / Symmetrica l System

α*VcVa

α2* Vb

3V2 =0

VcVa

Vb

3V0 =0

α2*VcVa α* Vb

3V1

Open-Phase Unbalanced / Non-Symmetrica l System

Vc

Va

Vb

Ic

IaIb α2*Ic

Iaα* Ib

3I1

α*Ic

Iaα2* Ib

3I2 Ic

Ia

Ib3I0

Positive Negative Zero

Positive Negative Zero

Vc

Va

Vb

Ic

IaIb

1212

Symmetrica l ComponentsExample: Perfectly Balanced & ABC Rota tion

Result: 100% I1 (Positive Sequence Component)

1313

I2

Symmetrica l ComponentsExample: B-Phase Rolled & ABC Rota tion

Result: 33% I1, 66% I0 and 66% I2

1414

Symmetrica l ComponentsExample: B-Phase & C-Phase Rolled & ABC Rota tion

Result: 100% I2 (Negative Sequence Component)

15

Fault Analysis - Example:

300 A900 A

Ia αIb

300 A

α2Ic

I1= 1/3(Ia+ αIb+ α2Ic)Positive Sequence Component , I1:For Fault Condition:

α2 = 240°α = 120°

Power System Faults

16

Fault Analysis - Example

I1= 1/3(Ia+ αIb+ α2Ic)

= 1/3(900 A + 300 A + 300 A)

= 1/3(1500 A)

I1 = 500 Amps

Positive Sequence Component , I1:For Fault Condition:

αIb=300 A

α2Ic= 300 A

Ia= 900 A

I1 = 500 AI1= 1/3(1500 A)

1500 A

Power System Faults

17

Fault Analysis – Example:

Negative Sequence Component, I2:For Fault Condition:

I2= 1/3(Ia+α2Ib+αIc)

900 A

300 A

α2Ib

300 A

αIcIa

α2 = 240°

α = 120°

Power System Faults

18

Fault Analysis - Example :

Negative Sequence Component, I2:For Fault Condition:

I2 = 200 A

α2Ib = 300 A

αIc = 300 A

I2= 1/3(600 A)Ia = 900 A

I2 = 1/3(Ia+α2Ib+αIc)

= 1/3(600 A)

I2 = 200 Amps

600 A

Power System Faults

19

Fault Analysis - Example:

Zero Sequence Component, I0:For Fault Condition:

300 A

900 A

300 A

Ib Ic

I0= 1/3(Ia+Ib+Ic)

Power System Faults

20

Fault Analysis – Example:

Zero Sequence Component, I0:For Fault Condition:

Ib= 300 A

Ic= 300 A

Ia = 900 A

I0= 1/3(Ia+Ib+Ic)

= 1/3(600 A)

I0= 200 Amps

I0 = 200 AI0= 1/3(600 A)

600 A

Power System Faults

2121

Fault Currents

2222

Power System

2323

Sequence Networks

• Where is sequence voltage highest?

• What genera tes negative and zero sequence currents?

2424

How do we connect the sequence networks for a particular fault?

• Assuming a Phase to Ground fault on a solidly grounded system, what do we know about the fault current?

• Faulted phase is very large

• Un-faulted phases are small, a lmost zero.

2525


• Ia = x

• Ic = Ib = 0

• Ia+Ib+Ic = x = I0

• Ia+a*Ib+a 2*IC = x = I1

• Ia + a 2Ib+aIc = x = I2

• I1=I2=I0

2626

How do we connect so tha t I1=I2=I0?

• The sequence networks have to be in series for a phase to ground fault on a solidly grounded system.

2727

Phase to phase fault

• Assuming a Phase to Phase fault , what do we know about the fault current?

• Faulted phases are very large

• Un-faulted phase is small, a lmost zero.

2828


• Ic = 0

• Ia = -Ib = x

• Ia+Ib+Ic = I0 = 0

• Ia+a*Ib+a 2*IC = sqrt(3)x = I1

• Ia + a 2Ib+aIc = sqrt(3)x = I2

• I1=I2 and I0 is not involved

2929

How do we connect so tha t I1=I2 and I0=0

• The positive and negative sequence networks have to be in series for a phase to phase fault .

3030

Three Phase Fault

• Assuming a three phase, what do we know about the fault current?

• All three phase are la rge and equal

3131


3232

How do we connect so tha t I2=I0=0

• The negative and zero sequence networks are not involved.

3333

Line-to-line-to-ground fault

3434

Common Fault Types:

3535

Transformer Interconnections:

3636

Transformer Interconnections:

3737

Networks with Transformers

3838

What a re my sequence impedances

• In transmission and distribution lines: Z1=Z2. Z0 is a lways different from Z1 because it is a loop impedance (conductor + earth and or ground.) For transmission lines, X1 is typica lly 2 to 6 times X0.

• In transformers: Z0 = Z1=Z2 or Z0= infinity depending on the connection.

3939

Ca lcula te Phase to Ground Fault current on secondary of transformer, assuming an infinite bus

4040

First , build the sequence network

• Zt1=5% on 50MVA base

• Zt2=5% on 50MVA base

• Zt3=5% on 50MVA base• I=V/Z=1pu/(0.05pu+0.05pu+0.05pu)

• I=6.667pu

• Ibase=50MVA/(sqrt(3)*13KV)=2221A

• I1=I2=I0=Ipu*Ibase=14.8KA

• Ia=I1+I2+I0=44KA

4141

What would be the fault current with two transformers in para llel?

• Zt1=2.5% on 50MVA base

• Zt2=2.5% on 50MVA base

• Zt3=2.5% on 50MVA base• I=V/Z=1pu/(0.025pu+0.025pu+0.025pu)

• I=13.3pu


• I1=I2+I3=Ipu*Ibase=30KA

• Ia=90KA

4242

Ca lcula te Three Phase Fault current on secondary of transformer, assuming an infinite bus

4343


• Zt1=5% on 50MVA base• I=V/Z=1pu/0.05pu

• I=20pu


• I1=Ipu*Ibase=44KA

• Ia=I1+I2+I0=44KA Phase to ground current is la rger than phase to phase current a t the transformer secondary terminals.

4444

Ca lcula te Phase to Ground Fault current on secondary of transformer, given a utility ava ilable fault current of 215000MVA and Z0=3.5*Z1 for system

4545


• Zt1=Zt2=Zt0=5% on 50MVA base• Zs1=1pu on 215000MVA base

• Zs1=1*(230/13)̂ 2*50/215000=.0728pu@13&50

• I=V/Z=1pu/(0.05pu+0.05pu+0.05pu+0.0728+0.0728+)

• I=3.38pu


• I=Ipu*Ibase=7.51KA

• Ia=I1+I2+I3=22KA

4646

What does this look like in the

rea l world

4949

F1 Symmetrical components

M1 Symmetrical components

Pre-fault all positive seq.

Pre-Fault Values

5050

Fault Values

F1 Pos, Neg, and no Zero

M1 Pos, Neg, and Zero

Why are the sequence values of M1 and F1 different??????

5151

The Fault Network as Seen From F1

5252

The Fault Network as Seen From M1

5353

What Effect Does this Fault Have on Voltage?

5454

What Effect Does this Fault Have on Voltage?

5555

Identifica tion of Fault Types

Quiz

56

Positive and Negative Sequence Current are equal, Zero Sequence is a different values, Positive, Negative and Zero sequence Voltage Equal.

Problem #1

58

Positive and Negative Sequence Current are equal, No Zero Sequence current

Problem #2

6060

The V0 and V2 conundrum

6161

POTT Scheme

POTT – Permissive Over-reaching Transfer Trip

End Zone

Communication Channel120

6262

Local Relay Remote Relay

Remote Relay FWD IGND

Ground Dir OC Fwd

OR

Local Relay – Z2

ZONE 2 PKP

Local Relay FWD IGND

Ground Dir OC Fwd

OR

TRIP

Remote Relay – Z2

POTT TX

ZONE 2 PKP

POTT RX

Communication Channel

POTT Scheme

121

63

Polarizing Quantities

ZL

R

X

Zs

Steady State

Dynamic Characteristic

64

Arc Impedance

ZL

No Load Fault ResistanceLoad into Relay Bus

Load From Relay Bus

R

X

65

Solve the Arc Impedance Problem with Directional Overcurrent

66

Sequence V Generated by a fault

6767

Faults Genera te V0 and V1

6868

SolutionIn a weak in feed, not enough current for polarizationFor remote faults not enough V0 for polarization.Solution: use dual polarization.

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