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 with equal magnitude
• All phasors are in phase
• A-C-B counter -clockwisephase rotation
• All phasors with equal magnitude
• 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
How do we connect the sequence networks for a particular fault?
• 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
How do we connect the sequence networks for a particular fault?
• 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
How do we connect the sequence networks for a particular fault?
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
• Ibase=50MVA/(sqrt(3)*13KV)=2221A
• 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
First , build the sequence network
• Zt1=5% on 50MVA base• I=V/Z=1pu/0.05pu
• I=20pu
• Ibase=50MVA/(sqrt(3)*13KV)=2221A
• 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
First , build the sequence network
• 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
• Ibase=50MVA/(sqrt(3)*13KV)=2221A
• I=Ipu*Ibase=7.51KA
• Ia=I1+I2+I3=22KA
4646
What does this look like in the
rea l world
4747
4848
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
5757
58
Positive and Negative Sequence Current are equal, No Zero Sequence current
Problem #2
5959
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.