Unsymmetrical Fault

Learning OutlineLearning Outline
Symmetrical Components TheorySymmetrical Components Theory
Sequence Impedance: Load, Line, Generator,
T fTransformer
Fault Analysis: Line‐Ground, Line‐Line, Line‐
Line‐Ground.
Examples and Class Exercisesp
1

Fault Analysisy ____________________
• Fault types:
– balanced faults (<5%)
• three‐phase to ground
• Three‐phase
– unbalanced faults
• single‐line to ground (60%‐75%)
d bl li t d (15% 25%)• double‐line to ground (15%‐25%)
• line‐to‐line faults (5%‐15%)line to line faults (5% 15%)
2

Example impact of faultExample impact of fault
The second largest blackout in the history of TEPCO
(The Tokyo Electric Power Company Inc ) hit central(The Tokyo Electric Power Company, Inc.) hit central
Tokyo area at about 7:38 a.m. on August 14, 2006. It
was caused by a floating crane on a barge going
upstream on a river on the eastern edge of the city.
The workers on the boat did not realize that the 33
meter crane was raised too high, so it hit TEPCO's
275 kV double circuit transmission lines that run
across the riveracross the river.
As a result of the accident the transmission
lines were short‐circuited and the wires
damaged. The relay protection operated and
tripped both lines
3

Symmetrical Componentsy p __________
• Three phase voltage or current is in a balance condition if it has
the following characteristic:the following characteristic:
– Magnitude of phase a,b, and c is all the same
– The system has sequence of a,b,cThe system has sequence of a,b,c
– The angle between phase is displace by 120 degree
• If one of the above is character is not satisfied, unbalanced
occur. Example:
4

• For unbalanced system, power system analysis cannot be
analyzed using per phase as in Load Flow analysis oranalyzed using per phase as in Load Flow analysis or
Symmetrical fault ‐>Symmetrical components need to be used.
• Symmetrical component allow unbalanced phase quantities
such as current and voltages to be replaced by three separate
balanced symmetrical componentsbalanced symmetrical components.
5

6

By convention, the direction of rotation of the phasors is taken
to be counterclock wiseto be counterclock‐wise.
Positive sequence:
111
0 aaa III =°∠=
1211
240 aab IaII =°∠=
111
120 III °∠
(10.1)
111
120 aac aIII =°∠=
Where we defined an operator a that causes a counterclockwise rotation of 120
degree such that:degree, such that:
866.05.0120sin120cos1201 jja +−=°+°=°∠=
866.05.02401)1201()1201(2
ja −−=°∠=°∠×°∠= 01 2
=++ aa
(10.2)
0136013
ja +=°∠=
(10.3)
7

Negative sequence:
°∠= 022
aa II
222
120 aab aIII =°∠=
2222
240 III °∠
(10.4)
2222
240 aac IaII =°∠=
000
cba III ==
Zero sequence:
(10.5)
8

Consider the three‐phase unbalanced current of cba III ,,
210
210
bbbb
aaaa
IIII
IIII
++=
++=
(10.6)
210
cccc IIII ++=
Based on (10.1), (10.4) and (10.5), (10.6) can be rewrite all in terms of phase a
components
2120
210
aaaa IIII ++=
⎥
⎤
⎢
⎡
⎥
⎤
⎢
⎡
⎥
⎤
⎢
⎡
1
0
2
111 aa II
2210
2120
aaac
aaab
IaaIII
aIIaII
++=
++= (10.7)
⎥
⎥
⎥
⎦
⎢
⎢
⎢
⎣⎥
⎥
⎥
⎦⎢
⎢
⎢
⎣
=
⎥
⎥
⎥
⎦⎢
⎢
⎢
⎣
2
1
2
2
1
1
a
a
c
b
I
I
aa
aa
I
I (10.8)
9

Equation 10.8 can be written as:
012
a
abc
AII =
Where A is known as symmetrical components transformation matrix,
(10.9)
which transforms phasor currents into components currents
and
abc
I 012
aI
⎥
⎤
⎢
⎡
2
111
(10 10)
⎥
⎥
⎥
⎦⎢
⎢
⎢
⎣
=
2
2
1
1
aa
aaA (10.10)
Solving (10.9) for the symmetrical components of currents:
abc
a IAI −
=012
(10.11)
The inverse of A is given by:
⎥
⎤
⎢
⎡ 111
1 (10 12)
⎥
⎥
⎥
⎦⎢
⎢
⎢
⎣
=−
aa
aa
2
2
1
1
3
1
A (10.12)
10

From (10.10) and (10.12), we conclude that
(10.13)*
3
1
AA =−
Substituting for A‐1 in (10.11), we have:Substituting for A in (10.11), we have:
⎥
⎥
⎥
⎤
⎢
⎢
⎢
⎡
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
=
⎥
⎥
⎥
⎤
⎢
⎢
⎢
⎡
b
a
a
a
I
I
I
aa
I
I
I
2
2
2
1
0
1
1
111
3
1 (10.14)
⎥
⎦
⎢
⎣⎥⎦⎢⎣⎥
⎦
⎢
⎣ ca IaaI 22
1
or in component form, the symmetrical components are:
)(
3
1
)(
3
1
21
0
cbaa
cbaa
IaaIII
IIII
++=
++=
(10.15)
)(
3
1 22
cbaa aIIaII ++=
11

Similar expressions exists for voltage:
(10.16)
2210
2120
210
aaac
aaab
aaaa
VaaVVV
aVVaVV
VVVV
++=
++=
++=
012
a
abc
AVV = (10.17)
aaac
1
The symmetrical components in terms of unbalanced voltages are:
)(
3
1
)(
3
1
21
0
cbaa
cbaa
VaaVVV
VVVV
++=
++=
(10.18)
abc
a VAV −
=012
(10.19)
)(
3
1 22
cbaa aVVaVV ++=
12

The apparent power may also be expressed in terms of the symmetrical
componentscomponents.
*
)3(
abcabcT
S IV=φ (10.20)
*
*012012
)3( )()( a
T
a
T
S AIAV=φ
Substituting (10.9) and (10.17) in (10.20), we obtain:
(10.21)*
012*012
a
T
a
T
IAAV=
3, *
== AAAA TT
Since complex power becomes
(10.21)
*
012012T
***
221100
012012
)3(
333
)(3
aaaaaa
T
S
IVIVIV
IV
++=
=φ
Total power for unbalance 3 phase system can be obtained from the sum of
(10.22)
Total power for unbalance 3‐phase system can be obtained from the sum of
symmetrical components powers.
13

Example 1p _______________________
One conductor of a three‐phase line is open. The current flowing to delta‐
connected load through line a is 10 A With the current in line a asconnected load through line a is 10 A. With the current in line a as
reference and assuming that line c is open, find the symmetrical
components of the line currents.
a
AIa °∠= 010
)(
3
1
)(
3
1
21
0
cbaa
cbaa
IaaIII
IIII
++=
++=
b AIb °∠= 18010
AI 0
)(
3
1
)(
3
22
cbaa
cbaa
aIIaII ++=
c AIc 0=
14

Solution________________________
The line current are:
°∠= 010aI °∠= 18010bI 0=cI
1
From (10.15): )(
3
1
)(
3
1
21
0
cbaa
cbaa
IaaIII
IIII
++=
++=
0)018010010(
3
1)0(
=+°∠+°∠=aI
Ia +°+°∠+°∠= )0)120180(10010(
3
1)1(
0 Sequence
+ Sequence
)(
3
1
3
22
cbaa aIIaII ++=
Aj °−∠=−= 3078.589.25
Aj
Ia
°∠=+=
+°+°∠+°∠=
307858925
)0)240180(10010(
3
1)2(
‐ Sequence
Aj ∠=+= 3078.589.25
From (10.4) 0)0(
=bI 0)0(
=cI
AIb °−∠= 15078.5)1( AIc °∠= 9078.5)1(
AIb °∠= 15078.5)2( AIc °−∠= 9078.5)2(
15

Example 2p _______________________
16

Exercise 1_______________________
:thatShow
°∠=
+
+
)(1
1201
)a(1
a)(1
(a)
2
2
°−∠=
+
−
1803
a)(1
a)(1
(b) 2
2
17

Exercise 2_______________________
Obtain the symmetrical components for the set of unbalanced voltages
°−∠=°∠=°−∠= 30100,90200,120300 cba VVV
)(
3
1
)(
3
1
21
0
cbaa
cbaa
VaaVVV
VVVV
++=
++=
°−∠
=
1202650.42
012V
The symmetrical components of a set of unbalanced three‐phase currents are
)(
3
1
3
22
cbaa aVVaVV ++=°−∠
°−∠
8961.849473.86
1351852.193
The symmetrical components of a set of unbalanced three phase currents are
Obtain the original unbalanced phasors.
°∠=°∠=°−∠= 304,905,303 210
aaa III
210
aaaa IIII ++=
2210
2120
aaac
aaab
aaaa
IaaIII
aIIaII
++=
++=
°−∠
°∠
=
304
2163.421854.8
Iabc
°−∠ 2163.1021854.8
18

Exercise 3_______________________
The line‐to‐line voltages in an unbalanced three‐phase supply are
°∠=°−∠=°∠= 120500,1500254.866,01000 cabcab VVV
Determine the symmetrical components for line and phase voltages, then find the
phase voltages Van, Vbn, and Vcn.
°∠
°∠
=
8934107626763
300.0
VL 012
°∠
=
00.0
Va 012
°−∠
=
1066.199586.440
Vabc
°∠
°−∠
306751.288
8934.107626.763
°∠
°−∠
606667.166
8934.409586.440
°∠
°−∠
603333.333
1021.1669252.600
19

Sequence Impedanceq p ______________
• The impedance of an equipment or component to theThe impedance of an equipment or component to the
current of different sequences.
iti i d (Z1) I d th t• positive‐sequence impedance (Z1): Impedance that
causes a positive‐sequence current to flow
• negative‐sequence impedance (Z2): Impedance that
causes a negative‐sequence current to flow
• zero‐sequence impedance (Z0): Impedance that causes
a zero‐sequence current to flow
20

Sequence Impedance of Y‐Connected Loadq p
Line to ground voltages are:
(10.23)
nncsbmamc
nncmbsamb
nncmbmasa
IZIZIZIZV
IZIZIZIZV
IZIZIZIZV
++++=
+++=
+++=
(10.24)cban IIII ++=
Kirchhoff’ current law:
(10 25)
Substituting In into (10.23):
⎥
⎤
⎢
⎡
⎥
⎤
⎢
⎡ +++
⎥
⎤
⎢
⎡ anmnmnsa IZZZZZZV )(
(10.25)
(10.26)
⎥
⎥
⎥
⎦
⎢
⎢
⎢
⎣⎥
⎥
⎥
⎦⎢
⎢
⎢
⎣ +++
+++=
⎥
⎥
⎥
⎦⎢
⎢
⎢
⎣ c
b
nsnmnm
nmnsnm
c
b
I
I
ZZZZZZ
ZZZZZZ
V
V
)(
)(
abcabcabc
IZV = (10.26)IZV =
21

(10 27)
⎥
⎥
⎤
⎢
⎢
⎡
+++
+++
)(
)( nmnmns
abc
ZZZZZZ
ZZZZZZ
Z (10.27)
⎥
⎥
⎥
⎦⎢
⎢
⎢
⎣ +++
+++=
)(
)(
nsnmnm
nmnsnm
ZZZZZZ
ZZZZZZZ
Writing Vabc and Iabc in terms of their symmetrical components:
(10.28)
012012
a
abc
a AIZAV =
Multiplying (10.28) by A‐1 :
(10.29)
012012
012012
)(
a
a
abc
a
IZ
IAZAV
=
= −
AZAZ abc−
=012where (10.30)
(10.31)
⎥
⎥
⎤
⎢
⎢
⎡
⎥
⎥
⎤
⎢
⎢
⎡
+++
+++
⎥
⎥
⎤
⎢
⎢
⎡
22012
1
111
)(
)(
1
111
1
aaZZZZZZ
ZZZZZZ
aaZ
nmnmns
Substituting for Zabc, A and A‐1 from (10.27), (10.10) and (10.12):
⎥
⎥
⎦⎢
⎢
⎣⎥
⎥
⎦⎢
⎢
⎣ +++
+++
⎥
⎥
⎦⎢
⎢
⎣
=
22
1
1
)(
)(
1
1
3
aa
aa
ZZZZZZ
ZZZZZZ
aa
aaZ
nsnmnm
nmnsnm
22

⎤⎡
Performing the multiplication in (10.31):
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
−
−
++
=
)(00
0)(0
00)23(
012
ms
ms
mns
ZZ
ZZ
ZZZ
Z (10.32)
⎥
⎤
⎢
⎡ +
0)(0
00)3(
012
ns
Z
ZZ
Z
When there is no mutual coupling, Zm = 0, and the impedance matrix becomes
(10.33)
⎥
⎥
⎥
⎦⎢
⎢
⎢
⎣
=
)(00
0)(0012
s
s
Z
ZZ (10.33)
23

Sequence Impedance of Transmission Linesq p
For sequence impedance transmission line, Z1 = Z2, whereas Z0 is different
and larger approximately 3 times than positive and negative sequence.
S I d f S h M hiSequence Impedance of Synchronous Machine
The positive-sequence generator impedance is the value found when positive-
sequence current flows from the action of an imposed positive-sequence set of
"2
q p p q
voltages.
The negative-sequence reactance is close to the positive-sequence
substransient reactance i e : "2
dXX ≈substransient reactance, i.e :
Zero-sequence reactance is approximated to the leakage reactance, i.e :
lXX ≈0
24

Sequence Impedances of Transformerq p
• Series Leakage Impedance.
– the magnetization current and core losses represented by the shunt branch t e ag et at o cu e t a d co e osses ep ese ted by t e s u t b a c
are neglected (they represent only 1% of the total load current)
– the transformer is modeled with the equivalent series leakage impedance
• Since transformer is a static device, the leakage impedance will not change
if the phase sequence is changedif the phase sequence is changed.
• Therefore, the positive and negative sequence impedance are the same;
Z0 = Z1 = Z2 = Zl
• Wiring connection always cause a phase shift. In Y‐Delta or Delta‐Y
transformer:
– Positive Sequence rotates by a +30 degrees from HV to LV side
– Negative Sequence rotates by a ‐30 degrees from HV to LV side
– Zero Sequence does not rotateZero Sequence does not rotate
• The equivalent circuit for zero‐sequence impedance depends on the
winding connections and also upon whether or not the neutrals are
grounded.
25

Connection diagram Zero‐sequence circuit
Figure (a)
Figure (b)g ( )
Figure (c)Figure (c)
Figure (d)Figure (d)
Figure (e)
26
Figure (e)

Description of Zero sequence Equivalent Circuit
(a) Y‐Y connections with both neutrals grounded – We know that the zero sequence current
equals the sum of phase currents. Since both neutrals are grounded, there is a path for the zero
sequence current to flow in the primary and secondary, and the transformer exhibits the
equivalent leakage impedance per phase as shown in Fig. (a).
(b) Y‐Y connections with primary the neutral grounded – The primary neutral is grounded, but( ) p y g p y g ,
since the secondary neutral is isolated, the secondary phase current must sum up to zero. This
means that the zero‐sequence current in the secondary is zero. Consequently, the zero
sequence current in the primary is zero, reflecting infinite impedance or an open circuit as
shown in Fig. (b).g ( )
27

c) Y‐Δ with grounded neutral – in this configuration, the primary currents can
flow because the zero‐sequence circulating current in the Δ‐connected
secondary and a ground return path for the Y‐connected primary. Note that noy g p p y
zero‐sequence current can leave the Δ terminals, thus there is an isolation
between the primary and secondary sides as shown in figure (c)
d) Y Δ ti ith i l t d t l i thi fi ti b thd) Y‐Δ connection with isolated neutral – in this configuration, because the
neutral is isolated, zero sequence current cannot flow and the equivalent
circuit reflects an infinite impedance or an open as shown in figure (d)
e) Δ‐Δ connection – in this configuration, zero‐sequence currents circulate in
the Δ‐connected windings, but no currents can leave the Δ terminals, and the
equivalent circuit is as shown in figure (e)
Notice that the neutral impedance plays an important part in the equivalent
circuit. When the neutral is grounded through an impedance Zn, because
In=3Io, in the equivalent circuit, the neutral impedance appears as 3Zn in the
28
path of Io.

Sequence Impedances of a Loaded Generatorq p
A synchronous machine generates balanced three‐phase internal voltages and is
represented as a positive‐sequence set of phasors
a
abc
E
a
a
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
= 2
1
E
(10.44)
a ⎥⎦⎢⎣
29

The machine is supplying a three‐phase balanced load. Applying kirchhoff’s voltage
law to each phase we obtain:
(10.45)nnbsbb
nnasaa
IZIZEV
IZIZEV
−−=
−−=
nncscc IZIZEV −−=
Substituting for In = Ia + Ib + Ic into (10.45):
⎤⎡⎤⎡⎤⎡⎤⎡ IZZZZEV )(
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
+
+
+
−
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
=
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
c
b
a
nsnn
nnsn
nnns
c
b
a
c
b
a
I
I
I
ZZZZ
ZZZZ
ZZZZ
E
E
E
V
V
V
)(
)(
)(
(10.46)
In compact form:
abcabcabcabc
IZEV −= (10.47)
30

Transforming the terminal voltages and currents phasors into their symmetrical
components:
(10.48)012012012
a
abc
aa AIZAEAV −=
Multiplying (10.48) by A‐1:
(10.49)012012012
012012012
)(
aa
a
abc
aa
IZE
IAZAEV
−=
−= −
h ⎤⎡⎤⎡ +⎤⎡ 111)(111 ZZZZ
(10.50)
Where:
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
+
+
+
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
=
2
2
2
2012
1
1
111
)(
)(
)(
1
1
111
3
1
aa
aa
ZZZZ
ZZZZ
ZZZZ
aa
aa
nsnn
nnsn
nnns
Z
Performing the above multiplication:
⎥
⎥
⎤
⎢
⎢
⎡
=⎥
⎥
⎤
⎢
⎢
⎡ +
= 1
0
012
00
00
00
00)3(
Z
Z
Z
ZZ
s
ns
Z
(10.51)
⎥
⎥
⎦
⎢
⎢
⎣⎥
⎥
⎦⎢
⎢
⎣
2
0000 ZZs
s
31

Since the generated emf is balanced, there is only positive‐sequence voltage, i.e:
(10 52)⎥
⎤
⎢
⎡ 0
(10.52)
⎥
⎥
⎥
⎦⎢
⎢
⎢
⎣
=
0
012
aa EE
( )⎥
⎤
⎢
⎡
⎥
⎤
⎢
⎡
⎥
⎤
⎢
⎡
⎥
⎤
⎢
⎡ 000
000 aa IZV
Substituting for and in (10.49):
012
aE 012
Z
111
000
0 aa IZV −=
(10.54)
⎥
⎥
⎥
⎦
⎢
⎢
⎢
⎣
⎥
⎥
⎥
⎦
⎢
⎢
⎢
⎣
−
⎥
⎥
⎥
⎦⎢
⎢
⎢
⎣
=
⎥
⎥
⎥
⎦
⎢
⎢
⎢
⎣
2
1
2
1
2
1
00
00
0 a
aa
a
a
I
I
Z
ZE
V
V (10.53) or
222
111
0 aa
aaa
IZV
IZEV
−=
−=
32

The three equations in (10.54) can be represented by the three equivalent
sequence networks:
• Important observations:
– The three sequences are independent.
Th i i k i h h li di d i– The positive‐sequence network is the same as the one‐line diagram used in
studying balance three‐phase currents and voltages.
– Only the positive‐sequence network has a source and no voltage source for
other sequences.
h l f h h f f d– The neutral of the system is the reference for positive‐ and negative‐sequence
networks, but ground is the reference for zero‐sequence networks. Thus, zero
sequence current can only flow if the circuit from the system neutrals to
ground is complete.
The grounding impedance is reflected in the zero sequence network as 3Zn– The grounding impedance is reflected in the zero sequence network as 3Zn
– The three‐sequence systems can be solved separately on a per phase basis.
The phase currents and voltages can then be determined by superposing their
symmetrical components of current and voltage respectively. 33

Single Line‐To‐Ground Faultg
Three‐phase generator with neutral grounded through impedance Zn and SLGF
occurs at phase a through impedance Zf .
Assuming the generator is initially on no‐load, the boundary conditions at the
fault point are:
afa IZV = (10.55)
0== cb II (10.56)
34

Substituting for Ib = Ic = 0, the symmetrical components of currents from (10.14)
are:are:
⎥
⎥
⎥
⎤
⎢
⎢
⎢
⎡
⎥
⎥
⎥
⎤
⎢
⎢
⎢
⎡
=
⎥
⎥
⎥
⎤
⎢
⎢
⎢
⎡
01
111
3
1
2
2
2
1
0
a
a
a I
aaI
I
(10.57)
⎥⎦⎢⎣⎥⎦⎢⎣⎥
⎦
⎢
⎣ 01
3 22
a aaI
From the above equation, we find that:
1 (10 58) 111
000
0 aa IZV −=
aaaa IIII
3
1210
===
Phase a voltage in terms of symmetrical components is :
(10.58)
222
111
0 aa
aaa
IZV
IZEV
−=
−=
210
aaaa VVVV ++=
:notingand(10.54)fromand,ngSubstituti 210210
aaaaaa IIIVVV ==
(10.59)
0210
)( aaa IZZZEV ++−= (10. 60)
35

:getwe,3notingand(10.55),fromforngSubstituti.3Where 0
a
0
aans IIVZZZ =+=
(10 61)02100
)(3 aaaf IZZZEIZ ++−=
or
(10.61)
f
a
a
ZZZZ
E
I
3210
0
+++
=
h f l
(10.62)
The fault current is
f
a
aa
ZZZZ
E
II
3
3
3 210
0
+++
== (10.63)
f
In order to obtain symmetrical voltage at the point of fault Equation, (10.63) is
substituted into Eq. (10.54)
36

Eq. (10.58) and (10.62) can be represented by connecting the sequence
networks in series as shown in the following figure.
aaaa IIII
3
1210
===
f
a
a
ZZZZ
E
I
3210
0
+++
=(10.58) (10.62)
37

Line‐To‐Line Fault
Three‐phase generator with a fault through an impedance Zf between phase b
and c.
I 0
Zs
Ia=0
N
ZsZs
Ea
Eb
Ec
Z
Ib
Va
ZfIc
Vb
Vc
bfb IZVV =− I = 0
Assuming the generator is initially on no‐load, the boundary conditions at the
fault point are:
(10.64) (10.66)bfcb IZVV Ia 0
0=+ cb II
(10.64)
(10.65)
(10.66)
38

Substituting for Ia = 0, and Ic = ‐Ib, the symmetrical components of the currents
from (10.14) are:from (10.14) are:
⎥
⎥
⎤
⎢
⎢
⎡
⎥
⎥
⎤
⎢
⎢
⎡
=⎥
⎥
⎤
⎢
⎢
⎡
ba
a
IaaI
I 0
1
111
1 21
0
(10.67)
⎥
⎥
⎦⎢
⎢
⎣−⎥
⎥
⎦⎢
⎢
⎣⎥
⎥
⎦
⎢
⎢
⎣ b
b
a
a
IaaI 1
3 22
From the above equation we find that:From the above equation, we find that:
00
=aI
1
(10.68)
(10 69)
ba IaaI )(
3
1 21
−=
ba IaaI )(
3
1 22
−=
(10.69)
(10.70)
ba
3
39

Also, from (10.69) and (10.70), we note that:
21
(10 71)21
aa II −=
From (10.16), we have:
VVVVVVVV )()( 22102120
(10.71)
2120
210
aaab
aaaa
aVVaVV
VVVV
++=
++=
(10.16)
bf
aa
aaaaaacb
IZ
VVaa
VaaVVaVVaVVV
=
−−=
++−++=−
))((
)()(
212
22102120
(10.72)
2210
aaac VaaVVV ++=
000
0 aa IZV −=
:getwe,notingand(10.54)fromandforngSubstituti 1221
a aaa IIVV −=
bfaa IZIZZEaa =+−− ])()[( 1212
(10.73)
(10.54)
222
111
0
0
aa
aaa
aa
IZV
IZEV
V
−=
−=
:getwe(10.69),fromforngSubstituti bI
))((
3
)( 22
1
121
aaaa
I
ZIZZE a
faa
−−
=+− (10.74) ba IaaI )(
3
1 21
−= (10.69)
))((
40

:inresultsforsolving,3Since 1
aIa))(aa(a 22
=−−
)( 21
1
f
a
a
ZZZ
E
I
++
= (10.75)
The phase currents are
⎥
⎤
⎢
⎡
⎥
⎤
⎢
⎡
⎥
⎤
⎢
⎡ 0111aI
(10 76)
⎥
⎥
⎥
⎦⎢
⎢
⎢
⎣−⎥
⎥
⎥
⎦⎢
⎢
⎢
⎣
=
⎥
⎥
⎥
⎦⎢
⎢
⎢
⎣
1
1
2
2
1
1
a
a
c
b
I
I
aa
aa
I
I
(10.76)
12
)( acb IaaII −=−= or 1
3 ab IjI −=(10.77) (10.78)
41

Eq. (10.71) and (10.75) can be represented by connecting the positive and negative –
sequence networks as shown in the following figure.
21
aa II −= 1 aE
I =
42
aa
)( 21
f
a
ZZZ
I
++
=

Double Line‐To‐Ground Fault
Figure 10.14 shows a three‐phase generator with a fault on phases b and c
through an impedance Zf to ground. Assuming the generator is initially on no‐
load the boundary conditions at the fault point areload, the boundary conditions at the fault point are
0
)(
210
=++=
+==
aaaa
cbfcb
IIII
IIZVV (10.79)
(10.80)
From (10.16), the phase voltages Vb and Vc are
Figure 10 14Figure 10.14
Double line‐to‐ground fault
43

Double Line‐To‐Ground Fault
(10.81)
2210
2120
aaab
VaaVVV
aVVaVV
++=
++=
(10 82)
21 ( )
aaac VaaVVV ++ (10.82)
thatnoteweabovefrom,VSinceVb c=
21
aa VV =
Substituting for the symmetrical components of current in (10.79), we get
(10.83)
)2(
)(
210
22102120
)(
aaaf
aaaaaafb
IIIZ
IaaIIaIIaIZV
−−=
+++++=
0
3 af IZ= (10.84)
44

:havewe(10.81),into(10.83)fromforand(10.84)fromforngSubstituti 2
b aVV
10
1200
)(3
aa
aaaf
VV
VaaVIZ
−=
++=
( ),( )( )g b a
(10.85)
11
0 aa IZE
I
−
:getwe,forsolvingand
(10.85)into(10.54)fromvoltageofcomponentslsymmetricafor thengSubstituti
0
aI
(10 86)
)3( 0
0
f
aa
a
ZZ
I
+
−=
Also, substituting for the symmetrical components of voltage in (10.83), we obtain
11
IZE − (10 87)
(10.86)
2
2
Z
IZE
I aa
a
−
−=
E
:getwe,forsolvingand(10.80)intoIandforngSubstituti 120
a aa II
(10.87)
f
f
a
a
ZZZ
ZZZ
Z
E
I
3
)3(
02
02
1
1
++
+
+
= (10.88)
45

sequence-negativetheofncombinatioparalelwith theseriesinimpedance
sequence-positivetheconnectingbydrepresentebecan(10.88)-(10.86)Equation
(10 8)fromfoundthenarecurrentphaseThefoundareIandIand
(10.87),and(10.86)indsubstituteis(10.86)fromfoundIofvalueThe
10.15.figureofcircuitequivalentin theshownasnetworkssequence-zeroand
20
1
a
fromobtainediscurrentfaulttheFinally,
(10.8).fromfoundthenarecurrentphaseThefound.areIandIand aa
0
3bf IIII =+= (10.89)3 acbf IIII + ( )
Figure 10.15 Sequence network connection for double line‐to‐ground fault
46

The one-line diagram of a simple power system is show in Figure 10 16 The neutral of each
EXAMPLE
The one-line diagram of a simple power system is show in Figure 10.16. The neutral of each
generator is grounded through a current-limiting reactor of 0.25/3 per unit on a 100-MVA
base. The system data expressed in per unit on a common 100-mva base tabulated below. The
generators are running on no-load at their rated voltage and rated frequency with their emfs in
phasephase.
Determine the fault current for the following faults
a. A balaced three-phase fault at bus 3 through a fault impedance = j 0.1 per unit
b. A single line-to-ground fault at bus 3 through a fault impedance = j 0.1 per unit
c A line-to-line fault at bus 3 through a fault impedance = j 0 1 per unit
fZ
fZ
fZc. A line-to-line fault at bus 3 through a fault impedance j 0.1 per unit
d. A double line-to-ground fault at bus 3 through a fault impedance = j 0.1 per unit
fZ
fZ
Item Base Rated X1 X2 X0
MVA Voltage
G1 100 20-kV 0.15 0.15 0.05
G2 100 20 kV 0.15 0.15 0.05
T1 100 20/220 kV 0.10 0.10 0.10T1 100 20/220 kV 0.10 0.10 0.10
T2 100 20/220 kV 0.10 0.10 0.10
L12 100 220 kV 0.125 0.125 0.30
L13 100 220 kV 0.15 0,15 0.35
L23 100 220 kV 0 25 0 25 0 7125L23 100 220 kV 0.25 0.25 0.7125
47

Figure 10 16Figure 10.16
Fault
Item Base
MVA
Rated
Voltage
X1 X2 X0
G1 100 20-kV 0.15 0.15 0.05
G2 100 20 kV 0 15 0 15 0 05G2 100 20 kV 0.15 0.15 0.05
T1 100 20/220 kV 0.10 0.10 0.10
T2 100 20/220 kV 0.10 0.10 0.10
L12 100 220 kV 0.125 0.125 0.30
48
L13 100 220 kV 0.15 0,15 0.35
L23 100 220 kV 0.25 0.25 0.7125

To find Thevenin impedance viewed from the faulted bus (bus 3), we convert the delta
formed by buses 123 to an equivalent Y as shown belowformed by buses 123 to an equivalent Y as shown below
03571430
)15.0)(125.0(
j
jj
Z )250)(1250( jj
Fig. 10.17Positive-sequence impedance
0357143.0
525.0
1 j
j
Z S ==
0595238.0
525.0
)25.0)(125.0(
2 j
j
jj
Z S ==
0714286.0
5250
)25.0)(15.0(
3 j
j
jj
Z S ==
525.0j
49

0714286.0
5952381.0
)3095238.0)(2857143.0(1
33 j
j
jj
Z +=
22.0j
j
=
50

To find thevenin impedance viewed from the faulted bus (bus 3), we convert the
delta formed by buses 123 to an equivalent Y as shown in figure 10.19(b)
)350)(300( jj
0770642.0
3625.1
)35.0)(30.0(
1 j
j
jj
Z S ==
15688070
)7125.0)(30.0(
j
jj
Z ==
j0.077064
1568807.0
3625.1
2 j
j
Z S ==
1830257.0
3625.1
)7125.0)(35.0(
3 j
j
jj
Z S ==
51
Fig. 10.19Zero-sequence impedance
j

Combining the parallel branches, the zero‐sequence thevenin impedance isg p q p
1830275.0
73394490
)2568807.0)(4770642.0(0
33 j
j
jj
Z +=
35.0
7339449.0
33
j
j
j
=
j0.077064
So, the zero‐sequence impedance diagram is show in fig. 10.20
52
Fig. 10.20
Zero‐sequence network

(a) Balanced three‐phase fault at bus 3
Assuming the no‐load generated emfs are equal to 1.0 per unit, the fault
current is
A90820 1j3 125
0.1
(F)I
)0(3
°∠
V a
A90-820.1pu-j3.125
0.1j0.22ZZ
(F)I
f
1
33
)0(3
3 °∠==
+
=
+
=
j
(b) Single line‐to ground fault at bus 3
F (10 62) th t f th f lt tFrom (10.62), the sequence component of the fault current are
pu
j
V a
-j0.9174
3(j0.1)0.35j0.22j0.22
0.1
3ZZZZ
III
f
0
33
2
33
1
33
)0(32
3
1
3
0
3 =
+++
=
+++
===
The fault current is :
jIII a
⎤⎡−⎤⎡⎤⎡⎤⎡⎤⎡ 752323111 0
3
0
33
pu
jI
I
I
I
aa
aa
I
I
I
c
b
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
=
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
=
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
=
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
0
0
7523.2
0
0
3
1
1
111 3
0
3
0
3
3
2
2
3
3
3
53

(c) Line‐to Line fault at bus 3
The zero‐sequence component of current is zero, i.e.,
0I0
3 =
The positive‐and negative‐sequence components of the fault current are
V a
pu
V a
-j1.8519
j0.1j0.22j0.22
0.1
ZZZZ
II
f
0
33
2
33
1
33
)0(32
3
1
3 =
++
=
+++
=−=
pu
j
j
aa
aa
I
I
I
c
b
a
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
−
−=
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
−
−
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
=
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
2075.3
2075.3
0
8519.1
8519.1
0
1
1
111
2
2
3
3
3
j ⎦⎣⎦⎣⎦⎣⎦⎣ 3
(d) Double Line‐to Line‐fault at bus 3
The positive‐sequence component of the fault current is
pu
V a
-j2.6017
j0.3j0.35j0.22
j0.3)(j0.35j0.22
j0.22
0.1
)Z3ZZ
)Z3Z(Z
Z
I
f
0
33
2
33
f
0
33
2
331
33
)0(31
3 =
++
+
+
=
++
+
+
=
)f3333
54

The negative‐sequence component of current is :
pu
jjV a
j1.9438
j0.22
)6017.2)(22.0(0.1
Z
IZ
I 2
33
1
33
1
33)0(32
3 =
−−
−=
−
−=
Th t f t iThe zero‐sequence component of current is:
pu
jjV a
j0.6579
j0 3j0 35
)6017.2)(22.0(0.1
3ZZ
IZ
I 0
1
33
1
33)0(30
3 =
+
−−
−=
+
−
−=
j0.3j0.353ZZ f33 ++
And the phase currents are :
jI a
⎤⎡⎤⎡⎤⎡
⎥
⎤
⎢
⎡ 06579.01113
pu
j
j
j
aa
aa
I
I
c
b
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
°∠
°∠=
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
−
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
=
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
07.14058.4
93.165058.4
0
9438.1
6017.2
6579.0
1
1
2
2
3
3
3
And the fault currents is:
°∠=+= 909732.1)( 333
cb
IIFI
55

UNBALANCED FAULT ANALYSIS USING BUS IMPEDANCE MATRIX
Single Line‐to‐Ground Fault Using Zbus
k
kkk
V
III
)0(
021
210
=== (10.90)
fkkkkkk
kkk
ZZZZ 3021
+++
k.busatoltageprefault vtheis)0(Vandmatriximpedancebus
ingcorrespondtheofaxiskin theelementsdiagonaltheareZandZ,ZWhere 2
kk
1
kk
k
o
kk
:iscurrentphasefaultThe
=abc
kI A
012
kI (10.91)
Line‐to‐Line Fault Using Zbus
00
=kI
V )0(
(10.92)
fkkkk
k
kk
ZZZ
V
II
++
=−= 21
21 )0(
(10.93)
56

Double Line‐to‐Ground Fault Using Zbus
fkkkk
fkkkk
kk
k
k
ZZZ
ZZZ
Z
V
I
3
)3(
)0(
02
02
1
1
++
+
+
= (10.94)
fkkkk
2
11
2 )0(
kk
kkkk
k
Z
IZV
I
−
−= (10.95)
kk
kkkk
k
ZZ
IZV
I
3
)0(
0
11
0
+
−
−= (10.96)
fkk ZZ 3+
iscurrentresulttheand(10.91),fromobtainedarecurrentsphaseThematrix.impedancebus
ingcorrespondtheofaxiskin theelementsdiagonaltheareZand,Zand,ZWhere 2
kk
1
kk
o
kk
( ),pp
C
k
b
kk IIFI +=)( (10.97)
57

Unsymmetrical Fault

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