ELE B7 Power System Engineering
Unbalanced Fault Analysis
Analysis of Unbalanced Systems
z
z
z
ELE B7
Except for the balanced three-phase fault, faults
result in an unbalanced system.
The most common types of faults are single lineground (SLG) and line-line (LL). Other types are
double line-ground (DLG), open conductor, and
balanced three phase.
The easiest method to analyze unbalanced system
operation due to faults is through the use of
symmetrical components
Slide # 1
Symmetrical Components
z
z
The key idea of symmetrical component analysis is
to decompose the unbalanced system into three
sequence of balanced networks. The networks are
then coupled only at the point of the unbalance (i.e.,
the fault)
The three sequence networks are known as the
–
–
–
ELE B7
positive sequence (this is the one we’ve been using)
negative sequence
zero sequence
Slide # 2
Symmetrical Components
Balance
Systems
Unbalance Currents
Unsymmetrical
Fault
IA
IC
IB
zero sequence
Zero
Sequence
Symmetrical
components
Unbalance
System
Sequence Currents
Positive
Sequence
positive sequence
Three
balanced
Systems
Negative
Sequence
negative sequence
ELE B7
Slide # 3
Symmetrical Components
Assuming three unbalance voltage phasors, VA, VB and VC having a positive
sequence (abc). Using symmetrical components it is possible to represent each
phasor voltage as:
VA = VA0 + VA+ + VA−
+
B
−
B
VB = V + V + V
0
B
VC = VC0 + VC+ + VC−
Zero Sequence Component
Positive Sequence Component
Negative Sequence Component
Where the symmetrical components are:
ELE B7
Slide # 4
Symmetrical Components
+
+
+
The Positive Sequence Components (
VA , VB , VC )
Three phasors
Equal in magnitude
Displaced by 120o in phase
Having the same sequence as the original phasors (abc)
The Negative Sequence Components (
VA− , VB− , VC− )
Three phasors
Equal in magnitude
Displaced by 120o in phase
Having the opposite sequence as the original phasors (acb)
0
0
0
The zero Sequence Components ( VA , VB , VC
Three phasors
Equal in magnitude
Having the same phase shift ( in phase)
ELE B7
)
VA+
VC+
120 o
120 o 120 o
VB+
VA−
−
B
V
120 o
120 o 120 o
VC−
VB0
VA0
VC0
Slide # 5
Example
VA = VA0 + VA+ + VA−
VB = VB0 + VB+ + VB−
V A0
VB 0
VC 0
Zero
Sequence
VA
VC = VC0 + VC+ + VC−
V A0
V A+
Positive
Sequence
VC = 0
V A−
VC +
120 o
V A+
120 o
VB +
VA
VC = 0
V A−
VB
Unbalance
Voltage
ELE B7
Negative
Sequence
120 o
120 o VC −
VB −
VB
Synthesis Unsymmetrical phasors
using symmetrical components
Slide # 6
Sequence Set Representation
z
Any arbitrary set of three phasors, say Ia, Ib, Ic can
be represented as a sum of the three sequence sets
I a = I a0 + I a+ + I a−
I b = I b0 + I b+ + I b−
I c = I c0 + I c+ + I c−
where
I a0 , I b0 , I c0 is the zero sequence set
I a+ , I b+ , I c+ is the positive sequence set
I a− , I b− , I c− is the negative sequence set
ELE B7
Slide # 7
Conversion Sequence to Phase
Only three of the sequence values are unique,
I0a , I a+ , I a− ; the others are determined as follows:
α = 1∠120°
0
0
0
Ia = I b = Ic
I b+ = α 2 I a+
α +α2 + α3 = 0
α3 = 1
(since by definition they are all equal)
I c+ = α I a+
I b− = α I a−
I c+ = α 2 I a−
0⎤
⎡
1
1
1
1
I
⎡ ⎤
⎤ a
⎡1⎤ ⎡
⎡ Ia ⎤
⎡1⎤
⎢ I ⎥ = I0 ⎢1⎥ + I + ⎢α 2 ⎥ + I − ⎢ α ⎥ = ⎢1 α 2 α ⎥ ⎢ I + ⎥
⎢
⎥
b
a
a
a
a
⎢
⎥
⎢
⎥
⎢
⎥
⎢ ⎥
⎢⎥
⎢α ⎥
⎢α 2 ⎥ ⎢1 α α 2 ⎥ ⎢ I − ⎥
⎢⎣ I c ⎥⎦
⎢⎣1⎥⎦
⎣ ⎦ ⎣
⎣ ⎦
⎦ ⎢⎣ a ⎥⎦
ELE B7
Slide # 8
Conversion Sequence to Phase
Define the symmetrical components transformation
matrix
1⎤
⎡1 1
⎢
⎥
2
α⎥
A = ⎢1 α
⎢1 α α 2 ⎥
⎣
⎦
0⎤
0⎤
⎡
⎡
I
I
a
⎡ Ia ⎤
⎢ +⎥
⎢ +⎥
⎢
⎥
Then I = I b = A ⎢ I a ⎥ = A ⎢ I ⎥ = A I s
⎢ ⎥
⎢ −⎥
⎢ −⎥
⎢⎣ I c ⎥⎦
⎢⎣ I a ⎥⎦
⎢⎣ I ⎥⎦
ELE B7
Slide # 9
Conversion Phase to Sequence
By taking the inverse we can convert from the
phase values to the sequence values
I s = A −1I
1⎤
⎡1 1
−1 1 ⎢
2⎥
with A = ⎢1 α α ⎥
3
⎢1 α 2 α ⎥
⎣
⎦
Sequence sets can be used with voltages as well
as with currents
ELE B7
Slide # 10
Example
If the values of the fault currents in a three phase system are:
I A = 150 ∠45 I B = 250 ∠150
I C = 100 ∠300
Find the symmetrical components?
Solution:
VO
ELE B7
Slide # 11
Example
If the values of the sequence voltages in a three phase system are:
Vo = 100
V+ = 200 ∠60
V− = 100 ∠120
Find the three phase voltages
Solution:
V A = 200 ∠60 + 100 ∠120 + 100
V A = 300 ∠60
V B = 1∠ 240( 200 ∠60 ) + 1∠120( 100 ∠120 ) + 100
V B = 300 ∠ − 60
VC = 1∠120( 200 ∠60 ) + 1∠ 240( 100 ∠120 ) + 100
VC = 0
ELE B7
Slide # 12
Use of Symmetrical Components
Consider the following wye-connected load:
I n = I a + Ib + I c
Vag = I a Z y + I n Z n
Vag = ( ZY + Z n ) I a + Z n I b + Z n I c
Vbg = Z n I a + ( ZY + Z n ) I b + Z n I c
Vcg = Z n I a + Z n I b + ( ZY + Z n ) I c
ELE B7
⎡Vag ⎤
⎡Z y + Zn
⎢ ⎥
⎢
⎢Vbg ⎥ = ⎢ Z n
⎢V ⎥
⎢ Z
n
⎣ cg ⎦
⎣
Zn
Z y + Zn
Zn
⎤ ⎡ Ia ⎤
⎥⎢ ⎥
Z n ⎥ Ib
⎢ ⎥
Z y + Z n ⎥⎦ ⎢⎣ I c ⎥⎦
Zn
Slide # 13
Use of Symmetrical Components
⎡Vag ⎤
⎡Z y + Zn
Zn
⎢ ⎥
⎢
Z y + Zn
⎢Vbg ⎥ = ⎢ Z n
⎢V ⎥
⎢ Z
Zn
n
⎣ cg ⎦
⎣
= Z I V = A Vs
V
A Vs = Z A I s
→
⎡ Z y + 3Z n
⎢
−1
A ZA = ⎢
0
⎢
0
⎣
ELE B7
⎤ ⎡ Ia ⎤
⎥⎢ ⎥
Z n ⎥ Ib
⎢ ⎥
Z y + Z n ⎥⎦ ⎢⎣ I c ⎥⎦
I = A Is
Zn
Vs = A −1 Z A I s
0
Zy
0
0⎤
⎥
0⎥
Z y ⎥⎦
Slide # 14
Networks are Now Decoupled
⎡V 0 ⎤
⎡ Z y + 3Z n 0
⎢ +⎥
⎢
0
Zy
⎢V ⎥ = ⎢
⎢ −⎥
⎢
0
0
⎢⎣V ⎥⎦
⎣
Systems are decoupled
V 0 = ( Z y + 3Z n ) I 0
−
V = Zy I
ELE B7
0⎤
⎡
I
0⎤
⎥⎢ +⎥
0 ⎥ ⎢I ⎥
⎢ −⎥
⎥
Z y ⎦ ⎢I ⎥
⎣ ⎦
V+
= Zy I+
−
Slide # 15
Grounding
z
z
z
ELE B7
When studying unbalanced system operation how a
system is grounded can have a major impact on the
fault flows
Ground current only impacts zero sequence system
In previous example if load was ungrounded the
zero sequence network is (with Zn equal infinity):
Slide # 16
Sequence diagrams for lines
z
z
ELE B7
Similar to what we did for loads, we can develop
sequence models for other power system devices,
such as lines, transformers and generators
For transmission lines, assume we have the
following, with mutual impedances
Slide # 17
Sequence diagrams for lines, cont’d
Assume the phase relationships are
⎡ ΔVa ⎤
⎡ Zs Zm Zm ⎤ ⎡ Ia ⎤
⎢ ΔV ⎥ = ⎢ Z
⎥ ⎢I ⎥
Z
Z
s
m⎥⎢ b⎥
⎢ b⎥
⎢ m
⎢⎣ ΔVc ⎥⎦
⎢⎣ Z m Z m Z s ⎥⎦ ⎢⎣ I c ⎥⎦
where
Z s = self impedance of the phase
Zm
= mutual impedance between the phases
Writing in matrix form we have
ΔV = Z I
ELE B7
Slide # 18
Sequence diagrams for lines, cont’d
Similar to what we did for the loads, we can convert
these relationships to a sequence representation
ΔV
= Z I ΔV = A ΔVs
I = A Is
A ΔVs
= Z A Is
→
ΔVs
= A −1 Z A I s
0
0 ⎤
⎡ Z s + 2Z m
A −1 Z A = ⎢
Zs − Zm
0
0 ⎥
⎢
⎥
Z s − Z m ⎥⎦
0
0
⎢⎣
Thus the system is again decoupled. A rule of thumb
is that Z+ = Z − and Z0 is approximate 3 times Z+ .
ELE B7
Slide # 19
Sequence diagrams for generators
z
Key point: generators only produce positive
sequence voltages; therefore only the positive
sequence has a voltage source
During a fault Z+ ≈ Z− ≈ Xd”. The zero
sequence impedance is usually substantially
smaller. The value of Zn depends on whether
the generator is grounded
ELE B7
Slide # 20
Sequence diagrams for Transformers
z
z
ELE B7
The positive and negative sequence diagrams for
transformers are similar to those for transmission
lines.
The zero sequence network depends upon both how
the transformer is grounded and its type of
connection. The easiest to understand is a double
grounded wye-wye
Slide # 21
Transformer Sequence Diagrams
ELE B7
Slide # 22
Unbalanced Fault Analysis
z
z
The first step in the analysis of unbalanced faults is
to assemble the three sequence networks.
Consider the following example
VT =1.05
T1
G1
T2
Transmission Line
fault
Δ
Δ
ELE B7
VT =1.05
MVA
Voltage
X+
G1
100
11 kV
0.15
0.17
0.05
G2
100
11 kV
0.20
0.21
0.1
T1
100
11/220kV
0. 1
0.1
0.1
T2
100
11/220kV
0.1
0.1
0.1
Line
100
220kV
0.105
0.105
0.315
X-
Xo
G2
J0.0
5
Slide # 23
Sequence Diagrams for Example
Positive Sequence Network
J0.15
J0.1
J0.105
J0.1
1.05∠0o
J0.2
1.05∠0o
Negative Sequence Network
J0.17
ELE B7
J0.1
J0.105
J0.1
J0.21
Slide # 24
Sequence Diagrams for Example
Zero Sequence Network
J0.05
J0.1
J0.315
J0.1
J0.1
J0.15
ELE B7
Slide # 25
Create Thevenin Equivalents
z
Second is to calculate the Thevenin equivalents as
seen from the fault location. In this example the
fault is at the terminal of the right machine so the
Thevenin equivalents are:
j 0 . 1389
1 .05 ∠ 0 0
E
I a+
I a−
++
Va
I a0
+
j 0 .1456
-
V a−
j 0 . 25
V a0
-
Zth+ = j 0.2 in parallel with j0.455
Zth− = j 0.21 in parallel with j0.475
ELE B7
Slide # 26
Single Line-to-Ground (SLG) Faults
z
z
ELE B7
Unbalanced faults unbalance the network, but only
at the fault location. This causes a coupling of the
sequence networks. How the sequence networks
are coupled depends upon the fault type. We’ll
derive these relationships for several common
faults.
With a SLG fault only one phase has non-zero fault
current -- we’ll assume it is phase A.
Slide # 27
SLG Faults, cont’d
Ignoring prefault currents, the
SLG fault can be described by
the following voltage and
current relationships:
Ib = 0
Ic = 0
&
Ia
Va
Zf
Ic
Ib
Vb
Vc
Va = I a Z f
The terminal unbalance currents at the fault point can be
transferred into their sequence components as follows:
⎡1
⎡ I a0 ⎤
⎢
⎢ ⎥
⎢ ⎥
1⎢
⎢ I + ⎥ = ⎢1
3⎢
⎢ a⎥
⎢
⎢ ⎥
⎢⎣1
⎢ −⎥
I
⎢ a ⎦⎥
ELE B7 ⎣
1
α
α2
1 ⎤ ⎡I a ⎤
⎥ ⎢ ⎥
⎥ ⎢ ⎥
2
α ⎥ ⎢0 ⎥
⎥ ⎢ ⎥
⎥ ⎢ ⎥
α ⎥⎦ ⎢0 ⎥
⎣ ⎦
→
I a0 = I a+ =
I a− =
Ia
3
Slide # 28
SLG Faults, cont’d
During fault,
Ia =
Va
Zf
and
I ao =
Va
3Z f
The terminal voltage at phase “a” can be transferred into
its sequence components as:
Va = Va0 + Va+ + Va−
0
+
−
V
V
V
V
+
+
a
a
I a0 = a = a
3Z f
3Z f
ELE B7
Slide # 29
SLG Faults, cont’d
The only way that these two constraint can be satisfied
is by coupling the sequence networks in series
I ao
Zero
Sequence
Circuit
Ia
Va
Zf Vb
Vc
Zero
Sequence
Circuit
I a+
Ic
Ib
Vao
I ao
Positive
Sequence
Circuit
Va +
+
a
I = I = I
−
a
Negative
Sequence
Circuit
Va −
Vao
Zo
Ia+
Positive
Sequence
Circuit
I a−
0
a
I ao
Va +
Ia+
3Z f
1 ∠0
Z+
E
Va −
Va +
3Z f
Ia−
Ia−
Negative
Sequence
Circuit
Vao
Z−
Va −
Va
Va0 + Va+ + Va−
=
I =
3Z f
3Z f
0
a
ELE B7
Slide # 30
Example:
z
Consider the following system
VT =1.05
T1
T2
Transmission Line
G1
Δ
Δ
z
ELE B7
fault
G2
J0.05
Its Thevenin equivalents as seen from the fault
location are:
j 0 . 1389
1 .05 ∠ 0 0
VT =1.05
E
I a+
I a−
++
Vf
-
I a0
+
j 0 .1456
V f−
j 0 . 25
V f0
-
Slide # 31
Example, cont’d
I
j 0 . 1389
E
With the sequence networks in
series, we can solve for the fault
currents
1 . 05 ∠ 0
+
a
V
0
I
−
a
+
+
a
+
V
j 0 . 1456
−
a
-
I
j 0 . 25
0
1
.
05
∠
0
I a+ = I a− = I a0 =
= − j1.964
j (0.1389 + 0.1456 + 0.25)
0
a
V
0
a
I = AI s → I a = − j 5.8 (of course, I b = I c = 0)
NOTE 1: These are the currents at the SLG fault point.
The currents in the system during the SLG fault should be
computed by analyzing the sequence circuits.
ELE B7
Slide # 32
Example, cont’d
From the sequence currents we can find the sequence
voltages as follows:
Va+ = 1.05∠00 − I a+ Z + , Va− = − I a− Z − , Va0 = − I ao Z o
V = AVs → Va = 0, Vb = 1.166 − j 0.178 , Vc = 1.166 + j 0.178
NOTE 2: These are the voltages at the SLG fault point.
The voltages at other locations in the system (during the
SLG fault) should be computed by analyzing the
sequence circuits.
ELE B7
Slide # 33
Line-to-Line (LL) Faults
z
The second most common fault
is line-to-line, which occurs
when two of the conductors
come in contact with each
other. With out loss of
generality we'll assume phases
b and c.
Ia
Va
Ib
Ic
Vb Z f Vc
Ib = − Ic
&
Current relationships: I a = 0
Voltage relationships: Vb = Vc + I b Z f
ELE B7
Slide # 34
LL Faults, cont'd
Using the current relationships, we get
⎡ I a0 ⎤
⎢ ⎥
⎢ ⎥
⎢I + ⎥
⎢ a⎥
⎢ ⎥
⎢ −⎥
⎢⎣ I a ⎥⎦
⎡1
⎢
1⎢
⎢1
=
3⎢
⎢
⎢1
⎣
1
α
α2
1 ⎤
⎥
⎥
2⎥
α
⎥
⎥
α ⎥⎦
⎡ 0 ⎤
⎢
⎥
⎢
⎥
⎢ Ib ⎥
⎢
⎥
⎢
⎥
⎢− I ⎥
⎣ b⎦
NOTE
Therefore,
I =0
0
a
1
I = (α − α 2 ) I b
3
1 2
−
I a = (α − α ) I b
3
+
a
Hence
ELE B7
I a− = − I a+
α = 1∠120
α = −0.5 + j 0.866
α 2 = 1∠240
α 2 = −0.5 − j 0.866
α 2 −α = − j 3
α −α 2 = j 3
Slide # 35
LL Faults, con'td
Therefore, it is obvious that, during a LL
Faults there is no zero sequence components
in the sequence circuit that represents this
fault.
During LL fault,
we have:
I =0
Vb = Vc + I b Z f
Using the symmetrical components,
then:
Vb = Va0 + α 2Va+ + α Va−
Vao = 0
Zo
I a0 = 0
Va0 = − I a0 Z 0 = 0
0
a
Vc = Va0 + α Va+ + α 2Va−
I b Z f = Z f ( I a0 + α 2 I a+ + α I a− )
ELE B7
Ia
Ib
+
Zf
+
Va
+
Vb
Vc
_
_
_
Slide # 36
LL Faults, con'td
Therefore,
Va0 + α 2Va+ + α Va− = Va0 + αVa+ + α 2Va− + Z f ( I a0 + α 2 I a+ + α I a− )
I a+ = − I a−
Substitute for I a0 = 0
Va0 = − I a0 Z 0 = 0
(α 2 − α )Va+ = (α 2 − α )Va− + (α 2 − α ) I a+ Z f
Then,
Va+ = Va− + I a+ Z f
To satisfy I a− = − I a+ , Va+ = Va− + I a+ Z f and I a0 = 0 ,
the positive and negative sequence networks must be
connected in parallel
I ao
Zero
Sequence
Circuit
I =0
0
a
V ao
Ia+
I a+ = − I a−
Ia
Va
ELE B7
Ib
Positive
Sequence
Circuit
Ia−
Ic
Vb Z f Vc
Va+
+
a
−
a
+
a
V =V + I Zf
Negative
Sequence
Circuit
Va−
I ao
Zero
Sequence
Circuit
Ia+
1∠0
Vao
Ia+
Positive
Sequence
Circuit
Va +
Ia−
Va +
Ia−
Negative
Sequence
Circuit
Z+
E
Zf
Z−
Zf
Va −
Va −
Slide # 37
LL Faults-Example
In the previous example, assume a phase-b-to-phase-c
fault occurs at the busbar of generator 2 (G2)
j 0 .1389
1.05∠00
I a+
j 0 . 1389
Va+
E
I
j 0.1456
Zf
−
a
Va−
Va0
Note: Zf = 0
ELE B7
Va+
E
Zf
I a−
Va−
j 0 . 1456
Solving the network for the currents, we get
I a0
j0.25
1 .05 ∠ 0 0
I a+
1.05∠00
= 3.691∠ − 900
I =
j (0.1389 + 0.1456)
+
a
⎡I a ⎤
⎡1
⎢ ⎥
⎢
⎢ ⎥
1⎢
⎢I ⎥
⎢ b ⎥ = 3 ⎢1
⎢
⎢ ⎥
⎢
⎢ ⎥
⎢⎣1
⎢⎣ I c ⎥⎦
1
α
α2
1⎤ ⎡ 0
⎤ ⎡ 0 ⎤
⎥ ⎢
⎥ ⎢
⎥
⎥ ⎢
⎥ ⎢
⎥
0⎥
2
⎢
α ⎥ 3.691∠ − 90 = ⎢ − 6.39 ⎥
⎥ ⎢
⎥
⎥ ⎢⎢
⎥ ⎢
⎥
⎥ ⎢
0
⎥
⎢
6.39⎥⎦
3.691∠90
⎣
⎣
⎦
⎥
α ⎦
Slide # 38
LL Faults-Example, cont'd
Solving the network for the voltages we get
+
V fa
V fa−
= 1.05∠0° − j 0.1389 × 3.691∠ − 90° = 0.537∠0°
= − j 0.1452 × 3.691∠90° = 0.537∠0°
⎡Vaf ⎤
1 ⎤ ⎡ 0 ⎤ ⎡ 1.074 ⎤
⎡1 1
⎢ f⎥
⎢
⎥⎢
2
α ⎥ 0.537 ⎥ = ⎢ −0.537 ⎥
⎢Vb ⎥ = ⎢1 α
⎥
⎢
⎥ ⎢
⎢ f⎥
⎢1 α α 2 ⎥ ⎢⎣0.537 ⎥⎦ ⎢⎣ −0.537 ⎥⎦
V
⎣
⎦
⎢⎣ c ⎥⎦
ELE B7
Slide # 39
Double Line-to-Ground Faults
z
With a double line-to-ground (DLG) fault two line
conductors come in contact both with each other
and ground. We'll assume these are phases b and c.
The voltage and the current relationships are:
Vb = Vc
Vb = Vc = ( I b + I c ) Z f
Ia = 0
+
a
Ia
−
a
Ia = I + I + I = 0
0
a
Va
Ib
Vb
Ic
Vc
Zf
Note, because of the path to ground the zero sequence
current is no longer zero.
ELE B7
Slide # 40
DLG Faults, cont'd
Using the symmetrical components, the terminal
voltages are:
0
+
−
Vb = Vb + Vb + Vb
Vb = Va0 + α 2Va+ + α Va−
Vc = Va0 + α Va+ + α 2Va−
Vb = Vc
Va0 + α 2Va+ + α Va− = Va0 + α Va+ + α 2Va−
(α 2 − α )Va+ = (α 2 − α )Va−
Va+ = Va−
ELE B7
Slide # 41
DLG Faults, cont'd
Using the symmetrical components, the terminal
currents are:
I b = I a0 + α 2 I a+ + α I a−
I c = I a0 + α I a+ + α 2 I a−
The voltage between fault terminal and ground is:
Vb = Vc = ( I b + I c ) Z f
Express the above equation in terms of its symmetrical
components:
Va0 + α 2Va+ + α Va− = ( I a0 + α 2 I a+ + α I a− + I a0 + α I a+ + α 2 I a− ) Z f
Using Va+ = Va− , 1 + α + α 2 = 0 &
Then
Va0 − Va+ = 3I a0 Z f
ELE B7
I a = I a0 + I a+ + I a− = 0
Slide # 42
DLG Faults, cont'd
To satisfy I a = I a0 + I a+ + I a− = 0 , Va+ = Va− &
the three symmetrical circuits, during a double line to
ground fault, are connected as follows:
I ao
I ao
I ao
Zero
Sequence
Circuit
Vao
Zero
Sequence
Circuit
Ia+
I a+
Ia
Va
Ib
Ic
Positive
Sequence
Circuit
Va +
Ia−
Vb
Vc
Zf
I a = I a0 + I a+ + I a− = 0
Va0 − Va+ = 3I a0 Z f
Negative
Sequence
Circuit
Va −
Vao
Positive
Sequence
Circuit
Va +
Zo
3Z f
Va −
3Z f
Ia+
1 ∠0
Z+
E
Va+ = Va−
Va +
Ia−
Ia−
Negative
Sequence
Circuit
Vao
Z−
Va −
Va0 − Va+ = 3I a0 Z f
ELE B7
Slide # 43
DLG Faults-Example
In previous example, assume DLG fault occurred at
G2 bus.
I a0
V a0
j 0 . 25
I a+
1 . 05 ∠ 0 0 E
j 0 . 1389
3Z f
I a−
I a+
j 0 . 1389
V a+
V
+
a
V
−
a
1 . 05 ∠ 0 0
V a−
V a0
j 0 . 25
E
I a−
j 0 . 1456
j 0 . 1456
I a0
3Z f
Assuming Zf=0, then
+
0
V
1
.
05
∠
0
a
=
I a+ = +
−
0
Z + Z //( Z + 3Z f ) j (0.1389 + j 0.092)
= 4.547∠ − 900
ELE B7
Slide # 44
DLG Faults, cont'd
I a−
I a+
j 0 . 1389
V a+
1 . 05 ∠ 0 0
I a0
j 0 . 1456
V a−
V a0
j 0 . 25
E
3Z f
V fa+ = 1.05 − 4.547∠ − 90° × j 0.1389 = 0.4184
I −fa = −0.4184 / j 0.1456 = j 2.874
I 0fa = − I +fa − I −fa = j 4.547 − j 2.874 = j1.673
Converting to phase: I bf = −1.04 + j 6.82
I cf = 1.04 + j 6.82
ELE B7
Slide # 45
Unbalanced Fault Summary
z
z
z
ELE B7
SLG: Sequence networks are connected in series,
parallel to three times the fault impedance
LL: Positive and negative sequence networks are
connected in parallel; zero sequence network is not
included since there is no path to ground
DLG: Positive, negative and zero sequence
networks are connected in parallel, with the zero
sequence network including three times the fault
impedance
Slide # 46
Generalized System Solution
z
z
1.
2.
3.
ELE B7
Assume we know the pre-fault voltages
The general procedure is then
Calculate Zbus for each sequence
For a fault at bus i, the Zii values are the Thevenin
equivalent impedances; the pre-fault voltage is the
positive sequence Thevenin voltage
Connect and solve the Thevenin equivalent
sequence networks to determine the fault current;
how the sequence networks are interconnected
depends upon the fault type
Slide # 47
Generalized System Solution, cont’d
4.
Sequence voltages throughout the system are given
by
⎡ 0 ⎤
⎢ M ⎥
This is solved
⎢
⎥
⎢ 0 ⎥
for each
⎢
⎥
sequence
V = V prefault + Z ⎢ − I f ⎥
network!
⎢ 0 ⎥
⎢
⎥
⎢ M ⎥
⎢⎣ 0 ⎥⎦
5. Phase values are determined from the sequence values
ELE B7
Slide # 48
Unbalanced System Example
Bus 1
Bus 2
G1
Δ
G2
fault
Bus 3
For the generators assume Z+ = Z− = j0.2; Z0 = j0.05
For the transformers assume Z+ = Z− =Z0 = j0.05
For the lines assume Z+ = Z− = j0.1; Z0 = j0.3
Assume unloaded pre-fault, with voltages =1.0 p.u.
ELE B7
Slide # 49
Positive/Negative Sequence Network
Bus 1
j0.2
j0.05
Bus 2
j0.3
j0.05
1.0∠0o
j0.2
1.0∠0o
j0.3
fault
j0.3
Bus 3
⎡−24 10 10 ⎤
⎡0.1397 0.1103 0.125⎤
+
+
Ybus
= j ⎢ 10 −24 10 ⎥ Zbus
= ⎢0.1103 0.1397 0.125⎥
⎢
⎥
⎢
⎥
⎢⎣ 10 10 −20⎥⎦
⎢⎣0.1250 0.1250 0.175⎥⎦
Negative sequence is identical to positive sequence
ELE B7
Slide # 50
Zero Sequence Network
Bus 1
j0.05
j0.05
Bus 2
j0.3
j0.3
fault
j0.05
j0.05
j0.3
Bus 3
⎡−16.66 3.33 3.33 ⎤
⎡0.0732 0.0148 0.0440 ⎤
0
0
= ⎢ 0.0148 0.0435 0.0.292 ⎥
Ybus
= j ⎢ 3.33 −26.66 3.33 ⎥ Zbus
⎢
⎥
⎢
⎥
⎢⎣0.0440 0.0292 0.1866 ⎥⎦
3.33 −6.66⎥⎦
⎢⎣ 3.33
ELE B7
Slide # 51
For a SLG Fault at Bus 3
The sequence networks are created using the pre-fault
voltage for the positive sequence thevenin voltage,
and the Zbus diagonals for the thevenin impedances
j 0 . 1750
1 .0 ∠ 0 0
E
Positive Seq.
j 0 .1750
Negative Seq.
j 0 . 1866
Zero Seq.
The fault type then determines how the networks are
interconnected
ELE B7
Slide # 52
Bus 3 SLG Fault, cont’d
I +f
+
If
V+
V−
ELE B7
1.0∠0°
=
= − j1.863
j (0.1750 + 0.1750 + 0.1866)
=
−
If
=
0
If
= − j1.863
⎡1.0∠0°⎤
⎡ 0 ⎤ ⎡ 0.7671⎤
+ ⎢
= ⎢1.0∠0°⎥ + Zbus
0 ⎥ = ⎢ 0.7671⎥
⎢
⎥
⎢
⎥ ⎢
⎥
⎢⎣1.0∠0°⎥⎦
⎢⎣ j1.863⎥⎦ ⎢⎣ 0.6740 ⎥⎦
⎡ 0 ⎤ ⎡ −0.2329 ⎤
− ⎢
= Zbus
0 ⎥ = ⎢ −0.2329 ⎥
⎥
⎢
⎥ ⎢
⎢⎣ j1.863⎥⎦ ⎢⎣ −0.3260 ⎦⎥
Slide # 53
Bus 3 SLG Fault, cont’d
⎡ 0 ⎤ ⎡ − 0.0820 ⎤
0 ⎢
0 ⎥ = ⎢ − 0.0544 ⎥
V 0 = Z bus
⎢
⎥ ⎢
⎥
⎢⎣ j1.863 ⎥⎦ ⎢⎣ − 0.3479 ⎥⎦
We can then calculate the phase voltages at any bus
0
⎡ − 0.3479 ⎤
⎡
⎤
V3 = A × ⎢ 0.6740 ⎥ = ⎢ − 0.522 − j 0.866 ⎥
⎢
⎥
⎢
⎥
⎢⎣ − 0.3260 ⎥⎦
⎢⎣ − 0.522 + j 0.866 ⎥⎦
0.4522
⎡ − 0.0820 ⎤
⎡
⎤
V1 = A × ⎢ 0.7671 ⎥ = ⎢ − 0.3491 − j 0.866 ⎥
⎢
⎥
⎢
⎥
⎢⎣ − 0.2329 ⎥⎦
⎢⎣ − 0.3491 + j 0.866 ⎥⎦
ELE B7
Slide # 54
Faults on Lines
z
z
ELE B7
The previous analysis has assumed that the fault is
at a bus. Most faults occur on transmission lines,
not at the buses
For analysis these faults are treated by including a
dummy bus at the fault location. How the
impedance of the transmission line is then split
depends upon the fault location
Slide # 55
Line Fault Example
Assume a SLG fault occurs on the previous system
on the line from bus 1 to bus 3, one third of the way
from bus 1 to bus 3. To solve the system we add a
dummy bus, bus 4, at the fault location
Bus 1
Bus 2
j0.1
j0.25
1.0∠0o
j0.25
j0.0333
1.0∠0o
j0.1
j0.0677
Dummy
fault bus
Bus 3
Bus 4
ELE B7
Slide # 56
Line Fault Example, cont’d
0 30 ⎤
⎡−44 10
The Ybus
⎢ 10 −24 10
⎥
0
+
now has
⎥
Ybus
= j⎢
4 buses
⎢ 0 10 −25 15 ⎥
⎢ 30
⎥
0 15 −45⎦
⎣
Adding the dummy bus only changes the new
row/column entries associated with the dummy bus
ELE B7
⎡0.1397
⎢0.1103
+
= j⎢
Zbus
⎢0.1250
⎢0.1348
⎣
0.1103
0.1397
0.1250
0.1152
0.1250
0.1250
0.1750
0.1417
0.1348⎤
0.1152⎥
⎥
0.1417⎥
0.1593⎥⎦
Slide # 57