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EE 0308 POWER SYSTEM ANALYSIS
CHAPTER 4
1
CHAPTER 4
SEQUENCE NETWORKS AND UNSYMMETRICAL FAULTS ANALYSIS
SEQUENCE NETWORKS AND UNSYMMETRICAL FAULTS ANALYSIS
1 SYMMETRICAL COMPONENTS AND SEQUENCE NETWORKS
2 UNSYMMETRICAL FAULTS AT THE GENERATOR TERMINALS
2
3 UNSYMMETRICAL FAULTS ON POWER SYSTEMS
4 CONSTRUCTION OF BUS IMPEDANCE MATRICES OF SEQUENCE
NETWORK
5 UNSYMMETRICAL FAULTS ANALYSIS
1 SYMMETRICAL COMPONENTS AND SEQUENCE NETWORKS When a symmetrical three phase fault occurs in a three phase system, the power system remains in the balanced condition. Hence single phase representation can be used to solve symmetrical three phase fault analysis. But various types of unsymmetrical faults can occur on power systems. In such cases, unbalanced currents flow in the system and this in turn makes the bus voltages unbalanced. Now the power system is in unbalanced
3
the bus voltages unbalanced. Now the power system is in unbalanced condition and single phase representation can not be used. Three phase unbalanced currents and voltages can be conveniently handled by Symmetrical Components. Therefore unsymmetrical faults are analyzed using symmetrical components. Some of the important aspects of symmetrical components are presented in brief.
aI (1)I
Sequence voltages and currents According to symmetrical components method, a three phase unbalanced system of voltages or currents may be represented by three separate system of balanced voltages or currents known as zero sequence, positive sequence and negative sequence as shown in Fig. 1 (1)I
(2)I
4
bI
cI
(1)bI
(1)aI
(2)cI
(2)bI
(0)cI
(0)bI
(0)aI (1)
cI = + +
Fig. 1
(2)aI
Defining operator ‘ a ‘ as a = 01201 ∠ (1) it is to be noted that
13601a;2401a 0302 =∠=∠= (2)
5
Also a = - 0.5 + j 0.866 ; 0.866j0.5a2 −−= (3) Hence 0aa1 2 =++ (4)
bI
cI
(1)bI
(1)aI
(2)cI
(2)bI
(0)cI
(0)bI
(0)aI
aI
(1)
cI = + + Further referring Fig. 1
(1)a
2(1)b IaI =
(1)a
(1)c IaI =
(2)a
(2)b IaI =
(2)2(2) =
(5)
(2)aI
6
(2)a
2(2)c IaI =
Therefore
(2)a
(1)a
(0)aa IIII ++=
(2)a
(1)a
2(0)a
(2)b
(1)b
(0)bb IaIaIIIII ++=++=
(2)a
2(1)a
(0)a
(2)c
(1)c
(0)cc IaIaIIIII ++=++=
Thus
(2)a
(1)a
(0)aa IIII ++=
(2)a
(1)a
2(0)ab IaIaII ++= (6)
(2)a
2(1)a
(0)ac IaIaII ++=
i.e.
c
b
a
III
=
2
2
aa1aa1111
(2)a
(1)a
(0)a
III
i.e. 2,1,0c,b,a IAI = (7)
The inverse form of the above is
(2)a
(1)a
(0)a
III
= 31
aa1aa1111
2
2
c
b
a
III
i.e. c,b,a1
2,1,0 IAI −= (8)
7
aI aa1 cI
Similarly, corresponding to voltage phasors 2,1,0c,b,a VAV = (9) and c,b,a
12,1,0 VAV −= (10)
Matrix A is known as symmetrical component transformation matrix. Similar expressions can be written for line to line voltages and phase currents also.
Sequence impedances and sequence networks The impedance of any three phase element is of the form
cb,a,z =
cccbca
bcbbba
acabaa
zzzzzzzzz
(11)
Then element voltage and element current are related as cb,a,cb,a,cb,a, izv = i.e. 0,1,2cb,a,0,1,2 iAzvA =
1 iAzAv −=
8
0,1,2cb,a,1
0,1,2 iAzAv −= 0,1,20,1,20.1,2 izv =
where z 0,1,2 = AzA cb,a,1−
Thus for any three phase element having the impedance cb,a,z the corresponding sequence impedance 0,1,2z can be obtained from
0,1,2z = AzA cb,a,1− (12)
It is to be noted that
for static loads and transformers cb,a,z =
smm
msm
mms
zzzzzzzzz
m2m1s zzz
9
for transmission lines cb,a,z =
sm3m2
m3sm1
m2m1s
zzzzzzzzz
(symmetric)
and for synchronous machines cb,a,z =
sm2m1
m1sm2
m2m1s
zzzzzzzzz
(cyclic symmetric)
For power system components, sequence impedance 0,1,2z will be decoupled as
0,1,2z =
(2)
(1)
(0)
z000z000z
(13)
For static loads and transformers (2)(1)(0) zzz == . For transmission lines (2)(1) zz = and (0)z > (1)z . For rotating machines (1)(0) z,z and (2)z will have different values.
10
The single phase equivalent circuit composed of sequence voltages, sequence currents and impedance to current of any one sequence is called the sequence network for that particular sequence. The sequence network includes any generated emf of like sequence. Consider a star connected generator with its neutral grounded through an impedance nZ as shown in Fig. 2. It is assumed that the generator is designed to generate balanced voltage.
nI
+
+
+
b
nZ
c
a
ncE
nbE
I
bI
aI
naE
11
cI
bE
aE
cE
Fig. 2 Let anE be its generated voltage in phase a . Then
c
b
a
EEE
=
aa1
2 anE This gives
Z
)1(aI a
)1(aI
(2)a
(1)a
(0)a
EEE
= 31
aa1aa1111
2
2
aa1
2 anE =
0E0
an (14)
This shows that there is no zero sequence and negative sequence generated voltages. The sequence networks of the generator are shown in Fig. 3.
12
1Z 1Z
1Z
1Z
Reference bus ( Neutral )
+
+
+
b
nZ
c
ncE
nbE
) 1 (cI
0In =
naE )1(
bI
__
+
)1(aV
Positive sequence network
naE
Note that In = 0
2Z
2Z
2Z
)2(aI
2Z
Reference bus ( Neutral ) b
nZ
c
a
0In =
)2(aI
)2(bI
)2(cI
)2(aV
Negative sequence network Note that In = 0
13
n
nZ3
0gZ nZ
)0(an I3I =
0gZ
0gZ
0gZ
)0(aI
0Z
Reference bus ( Ground ) b
c
a )0(aI
)0(bI
)0(cI
Fig. 3
)0(aV
Zero sequence network Note that In = 3 Ia(0)
1Z and 2Z are the positive sequence and negative sequence impedance of the generator. 0gZ is the zero sequence impedance of the generator. Total zero sequence
impedance 0Z = 0gZ + nZ3 . Sequence components of the terminal voltage are
)1()1(
)0(a0
)0(a
IZEV
IZV
−=
−=
(15)
14
)2(a2
)2(a
)1(a1na
)1(a
IZV
IZEV
−=
−= (15)
As far as zero sequence currents are concerned, the three phase system behaves as a single phase system. This is because of the fact that at any point the zero sequence currents are same in magnitude and phase. Therefore, zero sequence currents will flow only if a return path exists and hence extra care is needed while drawing the zero sequence network.
nZ
The connection diagram and the zero sequence equivalent circuit for star connected load is shown in Fig. 4. Z Z Z
Reference
Z
nZ3
15
Z
Fig. 4 The connection diagram and the zero sequence circuit for delta connected load is shown in Fig. 5. Z Z Fig. 5
Reference
Reference
Z
Special attention is required while obtaining the zero sequence network of three phase transformers. The zero sequence network will be different for various combination of connecting the windings and also by the manner in which the neutral is connected. The zero sequence networks are drawn remembering that no current flows in the primary of a transformer unless current flows in the secondary ( neglecting the small magnetizing current ).
16
( neglecting the small magnetizing current ). Five different cases are considered and the corresponding zero sequence network are shown in Fig. 6. The arrows in the connection diagram show the possible path for the flow of zero sequence current. Absence of arrow indicates that the zero sequence current can not flow there. Impedance 0Z accounts for the leakage impedance Z and the neutral impedances NZ3 and
nZ3 where applicable.
NZ nZ
Connection diagrams Zero sequence equivalent circuit P Q P Q
Reference
0Z
17
NZ
P Q P Q
Reference
0Z
P Q P Q
Reference
0Z
18
P Q P Q NZ
Reference
0Z
P Q P Q Fig. 6 Example 1 For the power system shown in Fig. 7, with the data given, draw the zero sequence, positive sequence and negative sequence networks.
Reference
0Z
19
sequence, positive sequence and negative sequence networks. 1T 2T Fig. 7
G
1M
2M
Per unit reactances are: Generator 0.32X;0.05X ng0 == ; 0.25X;0.2X 21 == Transformer 1T 0.08XXX 210 === Transformer 2T 0.09XXX 210 ===
20
Transmission line 0.18XX0.52;X 210 === Motor 1 0.27XX0.22;X0.06;X 21nom ==== Motor 2 0.55XX0.12;X 21om ===
1T 2T
G
1M
2M
Positive sequence network
21
+ + + gE
2mE 1mE
j0.18 j0.09 j0.08
j0.55 j0.27 j0.2
Reference
1T 2T
G
1M
2M
Negative sequence network
22
j0.09 j0.18 j0.08
j0.55 j0.27
j0.25
Reference
1T 2T
G
1M
2M
Zero sequence network
23
j0.52 j0.09 j0.08
j0.12 j0.06
j0.66
j0.05
j0.96
Reference
2 UNSYMMETRICAL FAULTS AT GENERATOR TERMINALS Single line to ground fault ( LG fault ), Line to line fault ( LL fault ) and Double line to ground ( LLG fault ) are unsymmetrical faults that may occur at any point in a power system. To understand the unsymmetrical fault analysis, let us first consider these faults at the terminals of unloaded generator. This treatment can be extended to unsymmetrical fault analysis when the fault occurs at any point in a power system.
24
Consider a three phase unloaded generator generating balanced three phase voltage. The sequence components of the terminal voltages are
1(1)ana
(1)a ZIEV −= (16)
2(2)a
(2)a ZIV −= (17)
0(0)a
(0)a ZIV −= (18)
1
(1)ana
(1)a ZIEV −= (16)
2(2)a
(2)a ZIV −= (17)
0(0)a
(0)a ZIV −= (18)
The above three equations apply regardless of the type of fault occurring at the terminals of the generator. i) For each type of fault there will be three relations in terms of phase components of currents and voltages.
25
components of currents and voltages. ii) Using these, three relations in terms of sequence components of currents and voltages can be obtained. iii) These three relations and the eqns. (16), (17) and (18) are used to solve for the sequence currents (2)
a(1)a
(0)a I,I,I and sequence voltages (2)
a(1)a
(0)a V,V,V . Sequence
components relationship will enable to interconnect the sequence networks to represent the particular fault.
aI a
Single line to ground fault ( LG fault ) The circuit diagram is shown in Fig. 9. nZ + fZ
26
+ nbE
b c
cI
bI
nZ _ Fig. 9
+ ncE naE
The fault conditions are
0Ib = (19) 0Ic = (20)
afa IZV = (21)
/3I)III(1/3I acba(0)a =++=
/3I)IaIa(I1/3I ac2
ba(1)a =++=
/3I)IaIaI(1/3I acb2
a(2)a =++=
Thus (2)
a(1)a
(0)a III == (22)
27
Further from eqn. (21)
(1)af
(2)a
(1)a
(0)af
(2)a
(1)a
(0)a IZ3)III(ZVVV =++=++ (23)
Using eqns. (16) to (18) in the above
(1)af2
(2)a1
(1)ana0
(0)a IZ3ZIZIEZI =−−+− i.e.
(1)af2
(1)a1
(1)ana0
(1)a IZ3ZIZIEZI =−−+− i.e.
f021
na(1)a Z3ZZZ
EI
+++= (24)
1Z
(1)aI
Z
naE
f021
na(1)a Z3ZZZ
EI
+++= (24)
Then the sequence networks are to be connected as shown in Fig. 10. + _
+ (1)aV
_
28
(2)aI
0Z
2Z
(0)aI
Fig. 10
+ (0)aV
_
+ (2)aV
_ fZ3
aI
nZ
a
nbE
I
Line to line fault The circuit diagram is shown in Fig. 11 _ E
naE
+
+
29
c b
bI
cI
Fig. 11 The fault conditions are
0Ia = (25) 0II cb =+ (26)
cbfb VIZV =− (27)
+ ncE
fZ
0Ia = (25) 0II cb =+ (26)
cbfb VIZV =− (27) Then 0)III(1/3I cba
(0)a =++= (28)
)aa(/3I)IaIa(I1/3I 2bc
2ba
(1)a −=++=
)aa(/3I)IaIaI(1/3I 2bcb
2a
(2)a −=++=
Since (0)aI = 0 , (0)
aV = - Z0 (0)aI = 0 (29)
Further (1)a
(2)a II −= (30)
From eqn. (27)
(2)2(1)(0)(2)(1)2(2)(1)2(0) VaVaV)IaIa(ZVaVaV ++=+−++
30
(2)a
2(1)a
(0)a
(2)a
(1)a
2f
(2)a
(1)a
2(0)a VaVaV)IaIa(ZVaVaV ++=+−++
(2)a
2(1)a
2f
(1)a
2 Va)a(I)aa(ZV)aa( −=−−− Thus (2)
a(1)af
(1)a VIZV =− (31)
From the above eqn. (1)
a2(2)a2
(1)af
(1)a1na IZIZIZIZE =−=−−
i.e. (1)af21na I)ZZZ(E ++=
Therefore f21
na(1)a ZZZ
EI
++= (32)
1Z 2Z
Therefore f21
na(1)a ZZZ
EI
++= (32)
(2)aI = - (1)
aI
and (0)aI = 0; (0)
aV = 0 Sequence networks are to be connected as shown in Fig. 12.
fZ
Z
31
(1)aI (2)
aI
1Z 2Z
naE
+ _ Fig. 12
Va(0) = 0
(2)aV (1)
aV
0Z
(0)aI
aI
nZ
a
nbE
Double line to ground fault The circuit diagram is shown in Fig. 13. _ +
+
naE
32
c b cI
bI Fig. 13 The fault conditions are
0Ia = ; )II(ZV cbfb += and )II(ZV cbfc += (33)
+
+ ncE
fZ
The fault conditions are
0Ia = ; )II(ZV cbfb += and )II(ZV cbfc += (33) Because of )III(1/3I cba
(0)a ++= , (0)
acb I3II =+ Therefore
(0)afb IZ3V = (34)
33
afb IZ3V = (34) (0)afc IZ3V = (35)
]V)aa(V[1/3)VaVaV(1/3V b
2ac
2ba
(1)a ++=++=
]V)aa(V[1/3)VaVaV(1/3V b2
acb2
a(2)a ++=++=
Therefore (2)
a(1)a VV = (36)
Further )VVV(1/3V cba(0)a ++= i.e.
(0)af
(0)af
(2)a
(1)a
(0)a
(0)a IZ3IZ3VVVV3 ++++= i.e. (0)
af(1)a
(0)a IZ6V2V2 +=
i.e. (0)af
(0)a
(1)a IZ3VV −= (0)
af(0)a0 IZ3IZ −−= (0)
af0 I)Z3Z( +−= i.e. (1)
aV (0)af0 I)Z3Z( +−= (37)
From eqn. (33) 0III (2)a
(1)a
(0)a =++ i.e.
0ZV
I3ZZ
Vi.e.0
ZV
I3ZZ
V
2
(1)a(1)
af0
(1)a
2
(2)a(1)
af0
(1)a =−+
+−=−+
+−
34
Therefore )3ZZ(Z
3ZZZV)
Z1
3ZZ1(VI
f02
f02(1)a
2f0
(1)a
(1)a +
++=+
+=
i.e. (1)a
f02
f02(1)a I
Z3ZZ)Z3Z(Z
V++
+= (38)
i.e. 1(1)ana ZIE − (1)
af02
f02 IZ3ZZ
)Z3Z(Z++
+=
Thus
f02
f021
na(1)a
Z3ZZ)Z3Z(Z
Z
EI
+++
+= (39)
Thus
f02
f021
na(1)a
Z3ZZ)Z3Z(Z
Z
EI
+++
+= (39)
From eqn. (38) (2)a2
(2)a
(1)a IZVV −== (1)
af02
f02 IZ3ZZ
)Z3Z(Z++
+=
Therefore (1)a
(2)a II −=
f02
f0
Z3ZZZ3Z
+++
(40)
Again substituting eqn. (37) in eqn. (38) (0)af0 I)Z3Z( +− (1)
af02 I
Z3ZZ)Z3Z(Z
+++
= Thus 2(1)a
(0)a Z3ZZ
ZII
++−= (41)
35
0Z
(0)aI (2)
aI (1)aI
1Z 2Z
naE
f02 Z3ZZ ++ f02 Z3ZZ ++For this fault, the sequence networks are to be connected as shown in Fig. 14. _ Fig. 14
+
fZ3
(0)aV (1)
aV (2)aV
1Z
3 SUMMARY OF UNSYMMETRICAL FAULTS AT THE GENERATOR TERMINALS
For any unsymmetrical fault (1)
aV = aE - 1Z (1)aI aI = (0)
aI + (1)aI + (2)
aI (2)
aV = - 2Z (2)aI bI = (0)
aI + 2a (1)aI + a (2)
aI (0)
aV = - 0Z (0)aI cI = (0)
aI + a (1)aI + 2a (2)
aI Single line to ground fault I
36
(2)aI
(1)aI
0Z
2Z
naE
(0)aI
Fault conditions are:
bI = 0
cI = 0
aV = fZ aI
cI
bI
aI
fZ
+ (2)aV
_ fZ3
+ (0)aV
_
+ (1)aV
_
f021
na(1)a Z3ZZZ
EI
+++= ;
(1)a
(0)a
(1)a
(2)a II;II ==
Corresponding phase components are cba IandI,I
(1)==
37
Fault current (1)aaf I3II ==
(1)aV = aE - 1Z (1)
aI (2)
aV = - 2Z (2)aI
(0)aV = - 0Z (0)
aI
Corresponding phase components are cb,a VandVV
Line to line fault Fault conditions are:
aI = 0 cI = - bI bV - fZ bI = cV
cI
bI
aI
fZ
38
0Z
naE (0)aI (2)
aI (1)aI
1Z 2Z
fZ
(1)aV
(2)aV
0I;II;ZZZ
EI (o)
a(1)a
(2)a
f21
na(1)a =−=
++=
Corresponding phase components are cba IandI,I Fault current
(1)I3j
(1)Ia)2a
(2)Ia
(1)I2a
(0)III −=−=++== (
39
(1)aI3j
(1)aIa)2a
(2)aIa
(1)aI2a
(0)aIbIfI −=−=++== (
(1)aV = aE - 1Z (1)
aI (2)aV = - 2Z (2)
aI (0)aV = - 0Z (0)
aI Corresponding phase components are cb,a VandVV
Double line to ground fault Fault conditions are:
fZ)II(V
Z)II(V
0I
cbc
fcbb
a
+=
+=
=
cI
bI
aI
fZ
40
naE
0Z
(0)aI
(2)aI (1)
aI
1Z 2Z
fZ)II(V cbc += -
+
fZ3
(0)aV (1)
aV (2)aV
f02
f021
na(1)a
Z3ZZ)Z3Z(ZZ
EI
++++
= f02
f0(1)a
(2)a Z3ZZ
Z3ZII++
+−=
f02
2(1)a
(0)a Z3ZZ
ZII++
−=
Corresponding phase components are cba IandI,I
41
Fault current cbf III += Since Ia
(0) = 1/3 (Ia + Ib + Ic) and Ia = 0 If = 3 Ia
(0)
(1)aV = aE - 1Z (1)
aI (2)aV = - 2Z (2)
aI (0)aV = - 0Z (0)
aI Corresponding phase components are cb,a VandVV
Example 2 The reactances of an alternator rated 10 MVA, 6.9 kV are
1X = 2X = 15 % and g0X = 5 %. The neutral of the alternator is grounded through a reactance of 0.38 ΩΩΩΩ. Single line to ground fault occurs at the terminals of the alternator. Determine the line currents, fault current and the terminal voltages. Solution
1X = 2X = 0.15 p.u. 10
42
nX = 0.38 x 26.910
= 0.0798 p.u.
0X = g0X +3 nX = 0.05 + 0.2394 = 0.2894 p.u. (1)aI = (2)
aI = (0)aI = 1.0 / j ( 0.2894 + 0.15 + 0.15 ) = - j 1.6966 p.u.
Corresponding phase components are
aI = -j 5.0898 p.u. bI = cI = 0
Base current = 6.9x31000x10
= 836.7 A
Line currents are aI = - j 4258.8 A ; bI = cI = 0
Fault current, fI = aI = - j 4258.8 A
(1)aV = 1.0 – ( j 0.15 ) (- j 1.6966 ) = 1.0 – 0.2545 = 0.7455 p.u. (2)aV = - ( j 0.15 ) (- j 1.6966 ) = - 0.2545 p.u.
43
a = - ( j 0.15 ) (- j 1.6966 ) = - 0.2545 p.u. (0)aV = - ( j 0.2894 ) (- j 1.6966 ) = - 0.491 p.u.
Corresponding phase components are
aV = 0 ; bV = 1.1386 ∠ 0130.38− p.u. ; cV = 1.1386 0130.38∠ p.u. Multiplying by
36.9
aV = 0; bV = 4.5359 0130.38−∠ kV ; cV = 4.5359 0130.38∠ kV
Example 3 The reactances of an alternator rated 10 MVA, 6.9 kV are
1X =15 %; 2X = 20 % and g0X = 5 %. The neutral of the alternator is grounded through a reactance of 0.38 ΩΩΩΩ. Line to line fault, with fault impedance j 0.15 p.u. occurs at the terminals of the alternator. Determine the line currents, fault current and the terminal voltages. Solution
44
Solution
1X = 0.15 p.u. ; 2X = 0.2 p.u. ; FX = 0.15 p.u. 0X = ? (1)aI = 1.0 / j ( 0.15 + 0.2 + 0.15 ) = - j 2 p.u. (2)aI = - (1)
aI = j 2 p.u. and (0)aI = 0
Corresponding phase components are
aI = 0 ; bI = - 3.4641 p.u. ; cI = 3.4641 p.u.
Base current = 836.7 A
Line currents are aI = 0 ; bI = - 2898.4 A ; cI = 2898.4 A
Fault current fI = bI = - 2898.4 A (1)aV = 1.0 – ( j 0.15 ) (- j 2 ) = 0.7 p.u. (2)aV = - ( j 0.3 ) ( j 2 ) = 0.4 p.u. (0)aV = 0
Corresponding phase components are
45
Corresponding phase components are
aV = 1.1 ; bV = 0.6083 ∠ 0154.72− p.u. ; cV = 0.6083 0154.72∠ p.u.
Multiplying by 36.9
, aV = 4.3821 kV
bV = 2.4233 0154.72−∠ kV cV = 2.4233 0154.72∠ kV
Example 4 An unloaded, solidly grounded 10 MVA, 11 kV generator has positive, negative and zero sequence impedances as j 1.2 , j 0.9 and j 0.04 respectively. A double line to ground fault occurs at the terminals of the generator. Calculate the currents in the faulted phases and voltage of the healthy phase. Solution
Base impedance = 10112
= 12.1 ;
Z = j 0.09917 p.u. ; Z = j 0.07438 p.u. ; Z = j 0.00331 p.u.
46
1Z = j 0.09917 p.u. ; 2Z = j 0.07438 p.u. ; 0Z = j 0.00331 p.u.
1Z + 02
02
ZZZZ
+ = j 0.10234 p.u.
(1)aI = 1.0/ j 0.10234 = -j 9.7714 p.u.
(2)aI = j 9.7714 0.07769
0.00331 = j 0.4163 p.u.
(0)aI = j 9.7714 0.07769
0.07438 = j 9.3551 p.u.
(1)aI = -j 9.7714 p.u.; (2)
aI = j 0.4163 p.u.; (0)aI = j 9.3551 p.u.
Corresponding phase components are
aI = 0 ; bI = 16.5758 0122.16∠ p.u. ; cI = 16.5758 057.84∠ p.u.
Base current = 11x3
1000x10 = 542.86 A
47
Current in faulted phases are bI = 8998.3 0122.16∠ A
cI = 8998.3 057.84∠ A
(1)aV = (2)
aV = (0)aV = - ( j 0.07438 ) ( j 0.4163 ) = 0.03096 p.u.
Voltage of the healthy phase aV = 0.09288 x 3
11 = 0.5899 kV
4 UNSYMMETRICAL FAULTS ON POWER SYSTEMS In a general power system fault can occur at any bus p. In such case, the fault analysis discussed in previous section can be extended following one-to-one correspondence shown below.
Fault at the terminals of the
generator
Fault occurs at bus p in the power system
Positive sequence pre-fault voltage is naE Positive sequence pre-fault voltage is fV
Thevenin’s equivalent impedance between the fault point and the
48
Positive sequence impedance is 1Z between the fault point and the reference bus in the positive sequence network is 1Z
Negative sequence impedance is 2Z
Thevenin’s equivalent impedance between the fault point and the reference bus in the negative sequence network is 2Z
Zero sequence impedance is 0Z
Thevenin’s equivalent impedance between the fault point and the reference bus in the zero sequence network is 0Z
Note that the Thevenin’s equivalent circuit of different sequence networks will be similar to the sequence networks of the generator. Thevenin’s equivalent circuit of the sequence networks are interconnected, much similar to the case of fault occurring at the generator terminals, to represent different types of faults.
49
the generator terminals, to represent different types of faults. This method is not suitable for large scale power systems as it involves network reduction in positive, negative and zero sequence networks.
5 UNSYMMETRICAL FAULT ANALYSIS USING busZ matrix When an unsymmetrical fault occurs in a power system, three phase network has to be considered. Any three phase element can be represented as shown in Fig. 15.
cb,a,
qpz
q p
bV apV
bV aqV
50
Fig. 15 It can be described as cb,a,
qpcb,a,
qpcb,a,
qp izv = i.e. (42)
cqp
bqp
aqp
vvv
=
ccqp
bcqp
acqp
cbqp
bbqp
abqp
caqp
baqp
aaqp
zzzzzzzzz
cqp
bqp
aqp
iii
(43)
cqV c
pV
bpV b
qV qV
Voltages at bus p and q can be denoted as
=cp
bp
ap
cb,a,p
VVV
V ;
=cq
bq
aq
cb,a,q
VVV
V (44)
Considering the impedance of each three phase element as cb,a,qpz , using building
algorithm, the bus impedance matrix of transmission-generator can be obtained as 1 2 N
51
1 2 N
=cb,a,busZ
cb,a,NN
cb,a,N2
cb,a,N1
cb,a,2N
cb,a,22
cb,a,21
cb,a,1N
cb,a,12
cb,a,11
ZZZ
ZZZZZZ
where =cb,a,jiZ i
ccji
bcji
acji
cbji
bbji
abji
caji
baji
aaji
ZZZZZZZZZ
The bus impedance matrix cb,a,busZ will be normally full with non-zero entries.
Since impedance of any element in sequence frame, 0,1,2qpz is decoupled,
computationally it is advantage to use the matrix 0,1,2busZ instead of cb,a,
busZ .
N
1 2
j
2 1
2 1
2 1
For two bus system 1 2 1 2 1 2
=0busZ
(0)22
(0)21
(0)12
(0)11
ZZZZ
; =1busZ
(1)22
(1)21
(1)12
(1)11
ZZZZ
; =2busZ
(2)22
(2)21
(2)12
(2)11
ZZZZ
Then
=0,1,2Z (2)(2)
(1)12
(1)11
(0)12
(0)11
ZZ2ZZ1
ZZ0210210
(45)
1
1
2
52
=0,1,2busZ
(2)22
(2)21
(1)22
(1)21
(0)22
(0)21
(2)12
(2)11
ZZ2ZZ1
ZZ0ZZ2 (45)
Normally 2
bus1bus
0bus ZandZ,Z are constructed and stored independently. It is
evident that as compared to cb,a,busZ , construction of 0,1,2
busZ requires less computer time and less core storage. For a 100 bus system, cb,a,
busZ will be a 300 x 300 full matrix; whereas for 0,1,2
busZ , we need 3 numbers of 100 x 100 matrices. Thus only 1/3 rd of the core storage is required for 0,1,2
busZ as compared to cb,a,busZ . Hence for
unsymmetrical fault analysis, use of 0,1,2busZ is more advantages than cb,a,
busZ .
2
Further, when unsymmetrical faults occur, the currents and voltages are unbalanced and using symmetrical components transformation, we can handle them conveniently. Therefore, symmetrical components are invariably used in the study of unsymmetrical fault analysis. In order to obtain 0,1,2Z , first the three sequence networks
53
In order to obtain 0,1,2busZ , first the three sequence networks
are be drawn as discussed earlier. Considering the zero sequence, positive sequence and negative sequence networks separately, using bus impedance building algorithm, are to be constructed independently . Of course special attention is necessary while drawing the zero sequence network.
Example 5 Consider the system described in Example 1. Obtain the matrices
2bus
1bus
0bus ZandZ,Z .
Solution Required bus impedance matrices can be constructed using bus impedance building algorithm. First consider the zero sequence network shown in Fig. 8(a).
54
j0.52 j0.09 j0.08
j0.12 j0.06
j0.66
j0.05
j0.96
Reference
1 2 3 4
0
1 2
2 1
Element 0-1 is added: =0busZ j 1 [ ]1.01
1 2
Element 0 – 2 is added: =0busZ j
0.08001.01
1 2 3
Element 0 – 3 is added: =0busZ j
0.090000.080001.01
55
3
0.0900 Element 2 – 3 is added: With th
bus 1 2 3
=0busZ j
−−0.690.090.0800.090.0900
0.0800.0800001.01
1
2 3
3
1 2
1 2 3
Eliminating the th bus, =0
busZ j
0.080.0100.010.070
001.01
Element 0 – 4 is added: The final 0
busZ is obtained as 1 2 3 4
56
1 2 3 4
=0busZ j
0.7200000.080.01000.010.0700001.01
Consider the positive sequence network shown in Fig. 8(b)
1
2 3 4
j0.18 j0.09 j0.08
j0.55 j0.27 j0.2
+ + +
1 2 3 4
gE Element 0 – 1 is added: =1
busZ j 1 [ ]0.2
2mE 1mE
Reference
57
Element 0 – 1 is added: =busZ j 1 [ ]0.2 1 2
Element 1 – 2 is added: =1busZ j
0.280.20.20.2
1 2 3
Element 2 – 3 is added: =1busZ j
0.460.280.20.280.280.20.20.20.2
2
1
1
3
2
1 2 3 4
Element 3 – 4 is added: =1busZ j
0.550.460.280.20.460.460.280.20.280.280.280.20.20.20.20.2
Element 0 – 4 is added: It has an impedance of j0.18. With the th
bus 1 2 3 4
−−0.280.280.280.280.20.20.20.20.20.2
1
2 3 4
1
2
58
=1busZ j
−−−−−−−
0.730.550.460.280.20.550.550.460.280.20.460.460.460.280.20.280.280.280.280.2
Eliminating the th
bus, final 1busZ is obtained as
1 2 3 4
1busZ = j
0.13560.11340.06900.04930.11340.17010.10360.07400.06900.10360.17260.12330.04930.07400.12330.1452
2 3 4
1
2 3 4
j0.55 j0.27
j0.25
Similarly, considering the negative sequence network shown in Fig. 8(c) j0.08 j0.18 j0.09
1 2 3 4
59
Its bus impedance 2
busZ can be obtained as 1 2 3 4
2busZ = j
0.13840.11770.07610.05770.11770.17650.11430.08660.07610.11430.19040.14420.05770.08660.14420.1699
1
2 3 4
Reference 0
6 UNSYMMETRICAL FAULT ANALYSIS USING 0,1,2busZ MATRIX
For unsymmetrical fault analysis using 0,1,2
busZ the first step is to construct 1busZ ,
2busZ and 0
busZ by considering the positive sequence, negative sequence and zero sequence network of the power system. They are 1 2 p N
(1)(1)(1)(1) ZZZZ
60
N
=1busZ
(1)NN
(1)Np
(1)N2
(1)N1
(1)pN
(1)pp
(1)p2
(1)p1
(1)2N
(1)2p
(1)22
(1)21
(1)1N
(1)1p
(1)12
(1)11
ZZZZ
ZZZZ
ZZZZZZZZ
(46)
1 2
p
1 2 p N
=2busZ
(2)NN
(2)Np
(2)N2
(2)N1
(2)pN
(2)pp
(2)p2
(2)p1
(2)2N
(2)2p
(2)22
(2)21
(2)1N
(2)1p
(2)12
(2)11
ZZZZ
ZZZZ
ZZZZZZZZ
and (47)
1
2
p
N
61
1 2 p N
=0busZ
(0)NN
(0)Np
(0)N2
(0)N1
(0)pN
(0)pp
(0)p2
(0)p1
(0)2N
(0)2p
(0)22
(0)21
(0)1N
(0)1p
(0)12
(0)11
ZZZZ
ZZZZ
ZZZZZZZZ
(48)
1
2
p
N
Suitable assumptions are made so that prior to the occurrence of the fault, there will not be any current flow in the positive, negative and zero sequence networks and the voltages at all the buses in the positive sequence network are equal to
fV . The currents flowing out of the original balanced system from phases candba,at the fault point are designated as cfbfaf IandI,I . We can visualize these currents by referring to Fig. 16 which shows the three lines candba, of the three phase system where the fault occurs.
62
Fig. 16
P
bfI
afI
cfI
c
a
b
The currents flowing out in hypothetical stub are cfbfaf IandI,I . The
corresponding sequence currents are (0)afI , (1)
afI and (2)afI . These sequence currents
(1)afI , (2)
afI and (0)afI are flowing out as shown in Fig. 17. The line to ground voltages
at any bus j of the system during the fault are ajV , bjV and cjV . Corresponding
sequence components of voltages are (0)ajV , (1)
ajV and (2)ajV .
1 - V f + Positive
63
2
(1)afI
- V f +
- V f +
Positive sequence network having bus impedance matrix (1)
busZ
N
p
2 2
1
Negative sequence network having bus impedance matrix N
p
1
Zero sequence network having bus impedance matrix N
p
64
Fig. 17
(2)afI
matrix (2)
busZ
N (0)afI
matrix (0)
busZ
N
Consider the Positive Sequence Network: In the faulted system, there are two types of sources. 1 Current injection at the faulted bus. 2 Pre-fault voltage sources. The bus voltages in the faulted system namely
(1)
(1)1a
VV1
2
65
=(1)busV
(1)
aN
(1)ap
(1)2a
V
V
V
(49)
can be obtained using Superposition Theorem.
N
2
p
It is to be noted that
=(1)busI
−
0
0I0
00
(1)af
and pre-fault voltage =
1
111
11
fV (50) p
1 2
N
66
Using these we get
(1)aN
(1)ap
(1)2a
(1)1a
V
V
VV
=
(1)NN
(1)Np
(1)N2
(1)N1
(1)pN
(1)pp
(1)p2
(1)p1
(1)2N
(1)2p
(1)22
(1)21
(1)1N
(1)1p
(1)12
(1)11
ZZZZ
ZZZZ
ZZZZZZZZ
−
0
I
00
(1)fa
+
1
1
11
fV =
−
−
−−
(1)af
(1)pNf
(1)af
(1)ppf
(1)af
(1)2pf
(1)af
(1)1pf
IZV
IZV
IZVIZV
(51)
Consider the Negative Sequence and Zero Sequence Networks: In a much similar manner, the negative sequence and the zero sequence bus voltages in the faulted system, namely (2)
busV and (0)busV , can be obtained considering
the negative sequence and the zero sequence networks. Knowing the pre-fault voltages are zero in the negative and zero sequence networks we get
67
(2)aN
(2)ap
(2)2a
(2)1a
V
V
VV
=
−
−
−−
(2)af
(2)pN
(2)af
(2)pp
(2)af
(2)2p
(2)af
(2)1p
IZ
IZ
IZIZ
and
(0)aN
(0)ap
(0)2a
(0)1a
V
V
VV
=
−
−
−−
(0)af
(0)pN
(0)af
(0)pp
(0)af
(0)2p
(0)af
(0)1p
IZ
IZ
IZIZ
(52)
When the fault occurs at bus p , it is to be noted that only the thp column of 1busZ , 2
busZ and 0busZ are involved in the calculations. If the symmetrical
components of the fault currents , namely (1)afI , (2)
afI and (0)afI , are known, than the
sequence voltages at any bus j can be computed from
(0)af
(0)pj
(0)aj IZV −= (53)
(1)af
(1)pjf
(1)aj IZVV −= (54)
(2)(2)(2) −=
68
(2)af
(2)pj
(2)aj IZV −= (55)
It is important to remember that the (1)
afI , (2)afI and (0)
afI are the symmetrical component currents in the stubs hypothetically attached to the system at the fault point. These currents take on values determined by the particular type of fault being studied, and once they are calculated, they can be regarded as negative injection into the corresponding sequence networks.
General procedure for unsymmetrical fault analysis
69
when fault occurs at a point in a
power system
PRELIMINARY CALCULATIONS 1. Draw the positive sequence, negative sequence and zero sequence
networks.
2. Using bus impedance building algorithm, construct ZBus (1), ZBus
(2) and
ZBus(0).
DATA REQUIRED Type of fault, fault location (Bus p) and fault impedance (Zf) TO COMPUTE FAULT CURRENTS I f a , I f b and I f c
70
TO COMPUTE FAULT CURRENTS I f a , I f b and I f c 1. Extract the columns of ZBus
(1), ZBus(2) and ZBus
(0) corresponding to the
faulted bus.
2. Depending on the type of fault interconnect the sequence networks.
3. Calculate I f a(1), I f a
(2) and I f a(0)
4. Compute the corresponding phase components I f a. I f b and I f c using
cf
bf
af
III
=
2
2
aa1aa1111
(2)af
(1)af
(0)af
III
TO COMPUTE FAULTED BUS VOLTAGES V p a, V p b and V p c 1. Compute the sequence components V p a
(1), V p a(2) and V p a
(0) from V p a
(1) = Vf – Z p p(1) I f a
(1) V p a
(2) = – Z p p(2) I f a
(2) V p a
(0) = – Z p p(0) I f a
(0)
71
V p a = – Z p p I f a 2. Calculate the corresponding phase components V p a, V p b and V p c from
cp
bp
ap
VVV
=
2
2
aa1aa1111
(2)ap
(1)ap
(0)ap
VVV
TO COMPUTE BUS VOLTAGES AT BUS j i.e V j a, V j b and V j c 1. Compute the sequence components V j a
(1), V j a(2) and V j a
(0) from V j a
(1) = Vf – Z j p(1) I f a
(1) V j a
(2) = – Z j p(2) I f a
(2) V (0) = – Z (0) I (0)
72
V j a(0) = – Z j p
(0) I f a(0)
2. Calculate the corresponding phase components V j a, V j b and V j c from
cj
bj
aj
VVV
=
2
2
aa1aa1111
(2)aj
(1)aj
(0)aj
VVV
fV
(2)afI
1)(afI
Single line to ground fault
(1)
afI = (2)afI = (0)
afI
(1)ppZ +
_
Z3
(1)apV
73
0)(afI
(2)ppZ
fZ3
(0)ppZ
(2)apV
(0)apV
(1)ppZ (2)
ppZ
Line to line fault
fZ
74
(0)afI (2)
afI (1)afI
(2)apV (1)
apV _
+ fV
(0)ppZ
(0)afI (2)
afI (1)afI
(1)ppZ (2)
ppZ
Double line to ground fault +
(0)V (1)V (2)V
75
afaf
fV _
fZ3
(0)apV (1)
apV (2)apV
SINGLE LINE TO GROUND FAULT For a single line to ground fault through impedance fZ , the hypothetical stubs on the three lines will be as shown in Fig. 18 The fault conditions are
a
afI fZ
P
76
Fig. 18
0I bf = 0I cf = (56)
affap IZV =
c
b bfI
cfI
Using the above conditions ( similar to conditions for the LG fault at generator terminals through impedance ) and also knowing that
(1)af
(1)ppf
(1)ap IZVV −= (57)
(2)af
(2)pp
(2)ap IZV −= (58)
(0)af
(0)pp
(0)ap IZV −= (59)
from eqns. (51) and (52), similar to eqns. (22) and (23), we can get the relations
(1)I = (2)I = (0)I and (60)
77
(1)afI = (2)
afI = (0)afI and (60)
(1)aff
(0)ap
(2)ap
(1)ap IZ3VVV −++ = 0 (61)
Therefore
f(0)
pp(2)
pp(1)
pp
f(1)af Z3ZZZ
VI
+++= (62)
The above relationships are satisfied by connecting the sequence networks as shown in Fig. 19
fV
(2)afI
1)(afI
(1)
afI = (2)afI = (0)
afI
(2)ppZ
(1)ppZ +
_
fZ3
(1)apV
(2)apV
78
2)(afI
Fig. 19 The series connection of Thevenin equivalents of the sequence networks, as shown in the above Fig. 18, is a convenient means of remembering the equations for the solution of single line to ground fault.
(0)ppZ
(0)apV
Once the currents (1)afI , (2)
afI and (0)afI are known, the sequence components of
voltage at the faulted bus are calculated as
(1)af
(1)ppf
(1)ap IZVV −=
(2)af
(2)pp
(2)ap IZV −= (63)
(0)af
(0)pp
(0)ap IZV −=
Thereafter the sequence components of voltage at any bus j can be calculated as
79
(1)
af(1)pjf
(1)aj IZVV −=
(2)af
(2)pj
(2)aj IZV −= (64)
(0)af
(0)pj
(0)aj IZV −=
Phase components of voltage and current can be calculated from the relations
2,1,0c,b,a VAV =
2,1,0c,b,a IAI =
pjN.1,2,......j
≠=
Example 6 The positive sequence, negative sequence and zero sequence bus impedance matrices of a power system are shown below. 1 2 3 4
== (2)bus
(1)bus ZZ j
0.14370.12110.07890.05630.12110.16960.11040.07890.07890.11040.16960.12110.05630.07890.12110.1437
1 2
4
3
80
1 2 3 4
=(0)busZ j
0.15530.14070.04930.03470.14070.19990.07010.04930.04930.07010.19990.14070.03470.04930.14070.1553
A bolted single line to ground fault occurs on phase ‘a’ at bus 3. Determine the fault current and the voltage at buses 3 and 4.
1 2
3 4
fV
(2)afI
1)(afI
(1)
afI = (2)afI = (0)
afI
(2)3 3Z
(1)3 3Z +
_
fZ3
(1)a3V
(2)apV
81
2)(afI
(0)3 3Z
apV
(0)apV
Solution (1)
afI = (2)afI = (0)
afI = (0)33
(2)33
(1)33
f
ZZZV
++
Let 0f 01.0V ∠=
Then (1)afI = (2)
afI = (0)afI =
)0.19990.1696(0.1696j01.0 0
++∠ = 1.8549j−
The fault current (1)
af(2)
af(1)
af(0)
afaf I3IIII =++= = - j 5.5648; bfI = cfI = 0
82
The sequence components of voltage at bus 3 are calculated as
0.3708)1.8549j()0.1999j(IZV (0)af
(0)33
(0)a3 −=−−=−=
0.6854)1.8549j()0.1696j(1.0IZVV (1)
af(1)33f
(1)a3 =−−=−=
0.3146)j1.8549()j0.1696(IZV (2)
af(2)33
(2)a3 −=−−=−=
Phase components of line to ground voltage of bus 3 are computed as
c3
b3
a3
VVV
=
2
2
aa1aa1111
−
−
0.31460.68540.3708
=
The sequence components of voltage at bus 4 are calculated as
0.2610)1.8549j()0.1407j(IZV (0)af
(0)43
(0)a4 −=−−=−=
0.7754)1.8549j()0.1211j(1.0IZVV (1)(1)(1) =−−=−=
0 1.0292 0122.71−∠ 1.0187 0122.71∠
83
0.7754)1.8549j()0.1211j(1.0IZVV (1)af
(1)43f
(1)a4 =−−=−=
0.2246)j1.8549()j0.1211(IZV (2)af
(2)43
(2)a4 −=−−=−=
Phase components of line to ground voltage of bus 4 are computed as
c4
b4
a4
VVV
=
2
2
aa1aa1111
−
−
0.22460.77540.2610
= 0.2898 00∠ 1.0187 0121.8−∠ 1.0187 0121.8∠
LINE TO LINE FAULT To represent a line to line fault through impedance fZ the hypothetical stubs on the three lines at the fault are connected as shown in Fig. 20.
b
a
bfI
afI
Z
P
84
Fig. 20 Fault conditions are
0I af = 0II cfbf =+ (65)
cpbffbp VIZV =−
( c
fZ
cfI
Using the above conditions along with the relations
(1)af
(1)ppf
(1)ap IZVV −=
(2)af
(2)pp
(2)ap IZV −= (66)
(0)af
(0)pp
(0)ap IZV −=
we can get the following relations.
85
(2)ap
(1)aff
(1)ap
(2)af
(1)af
(0)af
VIZV
II
0I
=−
−=
=
(68)
and hence f
(2)pp
(1)pp
f(1)af ZZZ
VI
++= (70)
(67)
(69)
(0)afI (2)
afI (1)afI
(1)ppZ (2)
ppZ
To satisfy the above relations the sequence networks are to be connected as shown in Fig. 21. (0)
ppZ
(2)apV (1)
apV _
+ fV
fZ
86
Fig. 21 Once (1)
afI , (2)afI and (0)
afI are calculated, (2)ap
(1)ap V,V and (0)
apV can be computed from
eqn. (63). Thereafter (2)aj
(1)aj V,V and (0)
ajV for pj;N1,2,....,j ≠= can be calculated using eqn. (63). The corresponding phase components are then calculated using the symmetrical component transformation matrix.
Example 7 Consider the power system described in example 6. A bolted line to line fault occurs at bus 3. Determine the currents in the fault, voltages at the fault bus and the voltages at bus 4. Solution For line to line fault 0I(0)
af =
2.9481j0.1696j0.1696j
1.0ZZ
VII (2)
33(1)
33
f(2)af
(1)af −=
+=
+=−=
87
0.1696j0.1696jZZ 3333 ++The phase components of the currents in the fault are
=++= (2)af
(1)af
(0)afaf IIII 0
=−=+= (1)af
(2)af
(1)af
2bf I3jIaIaI - 5.1061
=−= bfcf II 5.1061 Sequence components of voltage at bus 3 are
0V(0)a3 = ; 0.5)2.9481(j)0.1696j(IZVV (2)
af(2)
33(2)a3
(1)a3 =−=−==
Phase components of voltage at bus 3 are
=a3V +(0)a3V =+ (2)
a3(1)a3 VV 1.0
=−=+== (1)a3
(2)a3
(1)a3
2c3b3 VVaVaVV -0.5
Sequence components of voltage at bus 4 are
0IZV (0)af
(0)43
(0)a4 =−=
0.643)2.9481j()0.1211j(1IZVV (1)af
(1)43f
(1)4a =−−=−=
88
0.643)2.9481j()0.1211j(1IZVV af43f4a =−−=−=
0.357)2.9481j()0.1211j(IZV (2)af
(2)43
(2)a4 =−−=
Phase components of voltage at bus 4 are
=++= (2)a4
(1)a4
(0)a44a VVVV 1.0
=+= (2)a4
(1)a4
2b4 VaVaV - 0.5
=+= (2)a4
2(1)a4c4 VaVaV - 0.5
DOUBLE LINE TO GROUND FAULT For a double line to ground fault, the hypothetical stubs are connected as shown in Fig. 22. ( c
b
a
bfI
afI
P
89
fZ
Fig. 22 The relations at the fault bus are
)II(ZV
)II(ZV
0I
cfbffcp
cfbffbp
af
+=
+=
=
(71)
cfI
( c
Further the relations are also applicable.
(1)af
(1)ppf
(1)ap IZVV −=
(2)af
(2)pp
(2)ap IZV −= (72)
(0)af
(0)pp
(0)ap IZV −=
Using eqns. (71) and (62) the following relations can be obtained.
(2)ap
(1)ap VV =
90
apap
(0)aff
(0)ap
(1)ap IZ3VV −=
0III (2)af
(1)af
(0)af =++
On further simplification we get
f(0)
pp(2)
pp
f(0)
pp(2)
pp(1)pp
f(1)af
Z3ZZ)Z3Z(Z
Z
VI
+++
+= (73)
(0)ppZ
(0)afI (2)
afI (1)afI
(1)ppZ (2)
ppZ
To represent the above relations, the sequence networks must be interconnected as shown in Fig. 23.
fV _
+
fZ3
(0)apV (1)
apV (2)apV
91
Fig. 23 The sequence currents (2)
afI and (0)afI can be obtained from
f(0)
pp(2)
pp
f(0)
pp(1)af
(2)af Z3ZZ
Z3ZII
+++
−= (74)
f(0)
pp(2)
pp
(2)pp(1)
af(0)
af Z3ZZZ
II++
−= (75)
fZ3
Knowing (1)afI , (2)
afI and (0)afI sequence components of voltage at fault point are
calculated from
(1)af
(1)ppf
(1)ap IZVV −=
(2)af
(2)pp
(2)ap IZV −= (76)
(0)af
(0)pp
(0)ap IZV −=
Thereafter, sequence components of voltage at any other bus can be obtained from
92
(1)
af(1)pjf
(1)aj IZVV −=
(2)af
(2)pj
(2)aj IZV −= (77)
(0)af
(0)pj
(0)aj IZV −=
Knowing the sequence components, corresponding phase conponents are obtained as
21,0,cb,a, IAI = or 21,0,cb,a, VAV = (78)
pjN.1,2,......j
≠=
Example 8 The positive sequence, negative sequence and zero sequence bus impedance matrices of a power system are shown below. 1 2 3 4
== (2)bus
(1)bus ZZ j
0.12110.16960.11040.07890.07890.11040.16960.12110.05630.07890.12110.1437
1 2 3
93
0.14370.12110.07890.05630.12110.16960.11040.0789
1 2 3 4
=(0)busZ j
0.1900000.580.08000.080.0800000.19
4
3
1 2 3 4
A double line to ground fault with 0Zf = occurs at bus 4. Find the fault current and voltages at the fault bus. Solution Sequence components of fault current are
4.4342j
0.19j0.1437j)0.19j()0.1437j(0.1437j
1.0
ZZZZ
Z
VI
(0)(2)
(0)44
(2)44(1)
44
f(1)af −=
++
=
++
=
94
0.19j0.1437jZZ (0)44
(2)44
44 ++
j2.5247j0.19j0.1437
j0.19)j4.4342(ZZ
ZII (0)
44(2)44
(0)44(1)
af(2)
af =+
−−=+
−=
j1.9095j0.19j0.1437
j0.1437)j4.4342(ZZ
ZII (0)
44(2)44
(2)44(1)
af(0)
af =+
−−=+
−=
Phase components of current at the fault bus are
=++= (2)af
(1)af
(0)afaf IIII 0
00(2)af
(1)af
2(0)afbf 2102.52471504.4342j1.9095IaIaII ∠+∠+=++=
= - 6.0266 + j 2.8642 0302.5247304.4342j1.9095IaIaII 0(2)
af2(1)
af(0)
afcf −∠+∠+=++= = 6.0266 + j 2.8642 Fault current =+= cfbff III j 5.7285 Sequence components of voltage at the faulted bus are calculated as follows.
95
Sequence components of voltage at the faulted bus are calculated as follows. Noting that 0Zf =
0.3628)j4.4342()j0.1437(1.0IZVVVV (1)af
(1)44f
(0)a4
(2)a4
(1)a4 =−−=−===
Phase components of faulted bus voltage are:
=++= (2)a4
(1)a4
(0)a4a4 VVVV 1.0884
=b4V 0 =c4V 0
PROBLEMS – UNSYMMETRICAL FAULTS
1. In an unbalanced circuit the three line currents are measured as
0
c
0b
0a
160.752.6810I
203.844.3733I
59.857.0311I
∠=
∠=
∠=
Obtain the corresponding sequence components of currents and draw them to scale.
2. For the sequence components calculated in Problem 1, find the
96
2. For the sequence components calculated in Problem 1, find the
corresponding phase components of line currents and verify the results
graphically.
3. A three phase transmission line has the phase impedance of
=216662166621
jcb,a,z
Calculate its sequence impedances.
4. A 20 MVA, 13.8 kV alternator has the following reactances:
X1 = 0.25 p.u. X2 = 0.35 p.u. Xg0 = 0.04 p.u. Xn = 0.02 p.u.
A single line to ground fault occurs at its terminals. Draw the interconnections of the sequence networks and calculate
i) the current in each line
ii) the fault current
iii) the line to neutral voltages
iv) the line to line voltages
Denoting the neutral point as and the ground as , draw the phasor
97
Denoting the neutral point as n and the ground as o , draw the phasor
diagram of line to neutral voltages.
5. Repeat Problem 4 for line to line fault.
6. Repeat Problem 4 for double line to ground fault.
7. Repeat Problem 4 for symmetrical three phase fault.
8. Consider the alternator described in Problem 4. It is required to limit the
fault current to 2500 A for single line to ground fault. Find the additional
reactance necessary to be introduced in the neutral.
9. Two synchronous machines are connected through three-phase
transformers to the transmission line as shown.
2 1
1
T1
4 3
2
T2
98
The ratings and reactances of the machines and transformers are:
Machines 1 and 2: 100 MVA, 20 kV, X1 = X2 = 20 %, Xm0 = 4 %, Xn = 5 %
Transformers T1 and T2: 100 MVA, 20 / 345 Y kV, X = 8 %
On a chosen base of 100 MVA, 345 kV in the transmission line circuit, the
line reactances are X1 = X2 = 15 % and X0 = 50 %. Draw each of the three
sequence networks and find Zbus0, Zbus
1 and Zbus2.
10. The one-line diagram of a power system is shown below.
The following are the p.u. reactances of different elements on a common base.
Generator 1: Xg0 = 0.075; Xn = 0.075; X1 = X2 = 0.25
6 5
4
3 2 1
1 3
2
T1
T2
T3
99
Generator 1: Xg0 = 0.075; Xn = 0.075; X1 = X2 = 0.25
Generator 2: Xg0 = 0.15; Xn = 0.15; X1 = X2 = 0.2
Generator 3: Xg0 = 0.072; X1 = X2 = 0.15
Transformer 1: X0 = X1 = X2 = 0.12
Transformer 2: X0 = X1 = X2 = 0.24
Transformer 3: X0 = X1 = X2 = 0.1276
Transmission line 2 – 3 X0 = 0.5671; X1 = X2 = 0.18
Transmission line 3 – 5 X0 = 0.4764; X1 = X2 = 0.12
Draw the three sequence networks and determine Zbus0, Zbus
1 and Zbus2.
11. The single line diagram of a small power system is shown below.
Generator: 100 MVA, 20 kV, X1 = X2 = 20 %, Xg0 = 4 %, Xn = 5 %
Transformers T1 and T2: 100 MVA, 20 / 345 Y kV, X = 10 %
On a chosen base of 100 MVA, 345 kV in the transmission line circuit, the
line reactances are:
From T to P: X = X = 20 %; X = 50%
2 1
1
T1
4 3
T2
P S
Switch open
100
From T1 to P: X1 = X2 = 20 %; X0 = 50%
From T2 to P: X1 = X2 = 10 %; X0 = 30%
A bolted single lone to ground occurs at P. Determine
i) fault current IfA, IfB and IfC.
ii) currents flowing towards P from T1.
iii) currents flowing towards P from T2.
iv) current supplied by the generator.
Note that the positive sequence current in winding of transformer lags
that in Y winding by 300; the negative sequence current in winding leads
that in Y winding by 300.
12. In the power system described in Problem 10, a single line to ground fault
occurs at bus 2 with a fault impedance of j0.1. Determine the bus currents
at the faulted bus and the voltages at buses 1 and 2.
ANSWERS
1. 0c0b0a0 1202III ∠=== ; 0
a1 303.5I ∠= ; 0b1 2703.5I ∠= ; 0
c1 1503.5I ∠=
0a2 603I ∠= ; 0
b2 1803I ∠= ; 0c2 3003I ∠=
2. 0a 59.857.0311I ∠= ; 0
b 203.844.3733I ∠= ; 0c 160.752.681I ∠=
101
3. 15jZZ;33jZ 210 ===
4. Ia = -j 3586.1 A Ib = 0 Ic = 0 If = -j 3586.1 A
Va = 0 Vb = 8.0694 0102.22−∠ kV Vc = 8.0694 0102.22∠ kV
Vab = 8.0694 ∠ 077.78 kV; Vbc= 15.7724 090−∠ kV; Vca= 8.0694 0102.22∠ kV
5. Ia = 0 Ib = -2415.5 A Ic = 2415.5 A If = -2415.5 A
Va = 9.2948 kV Vb = -4.6474 kV Vc = -4.6474 kV
Vab = 13.9422 kV Vbc = 0 Vca = 13.9422 0180∠ kV
6. Ia = 0 Ib = 4020.95 0132.22∠ A Ic = 4020 047.78∠ A If = 5956.08 090∠ A
Va = 5.6720 kV Vb = 0 Vc = 0
Vab = 5.6720 kV Vbc = 0 Vca = 5.6720 0180∠ kV
7. Ia = 3347 090−∠ A Ib = 3347 0150∠ A Ic = 3347 030∠ A
If = 3347 090−∠ A
Va = Vb = Vc = 0
V = V = V = 0
102
4 3 2 1
Vab = Vbc = Vca = 0
8. 0.9655
9. Zero sequence network:
1
+
2
-
3 4
-
+
Positive sequence network:
103
1 2
3 4
Negative sequence network:
4
3
2
1
j
4
3
2
1
j
1 2 3 4
Zbus0 =
0.1900000.580.08000.080.0800000.19
Zbus1 = Zbus
2 =
0.14370.12110.07890.05630.12110.16960.11040.07890.07890.11040.16960.12110.05630.07890.12110.1437
104
4
3 2 1
5 6
10. Zero sequence network:
2
3
4
5
Zbus0 = j
0.65427100.1778710.031065000.6000
0.17787100.1778710.03106500.03106500.0310650.1044680
00000.3
Negative sequence network:
105
4
3 2 1
5 6
1 2 3 4 5 6
1
2
3
4
5
6
Zbus1 = Zbus
2 =
j
0.1149560.0851450.0259590.0571090.0384190.0259590.0851450.1575740.0480410.1056900.0711010.0480410.0259590.0480410.1403670.0688080.0462890.0312760.0571090.1056900.0688080.1513770.1018360.0688080.0384190.0711010.0462890.1018360.1895000.1281080.0259590.0480410.0312760.0688080.1281080.167640
11. IfA = - j 2.4195 p.u.; IfB = 0; IfC = 0
IA = - j 1.9356 p.u.; IB = j 0.4839 p.u.; IC = j 0.4839 p.u.
106
IA = - j 1.9356 p.u.; IB = j 0.4839 p.u.; IC = j 0.4839 p.u.
IA = - j 0.4839 p.u.; IB = - j 0.4839 p.u.; IC = - j 0.4839 p.u.
Ia = - j 1.3969 p.u.; IB = j 1.4969 p.u.; IC = 0
12. I2a = j 3.828162 p.u.; I2b = 0; I2c = 0
V2a = 0.382815 p.u.; V2b = 0.950352 ∠ 245.680 p.u.
V2c = 0.950352 ∠ 114.320 p.u.
V1a = 0.673054 p.u.; V1b = 0.929112 ∠ 248.760 p.u.
V1c = 0.929112 ∠ 111.2360 p.u.
