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Unsymmetrical Short Circuit
Submitted By:Santosh Kumar GuptaAssistant ProfessorEE DepartmentSIT Sitamarhi
FaultAnalysis
2
• Fault types:
– balanced faults (<5%)
• three‐phase toground
• Three‐phase
– unbalanced faults
• single‐line to ground(60%‐75%)
• double‐line to ground(15%‐25%)
• line‐to‐line faults (5%‐15%)
Example impact offault
3
The second largest blackout in the history of TEPCO
(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 theriver.
As a result of the accident the transmission
lines were short‐circuited and the wires
damaged. The relay protection operatedand
tripped both lines
SymmetricalComponents
4
• Three phase voltage or current is in a balance condition if it has
the following characteristic:
– Magnitude of phase a,b, and c is all the same
– The system has sequence ofa,b,c
– The angle between phase is displace by 120 degree
• If one of the above is character is not satisfied, unbalanced
occur. Example:
SymmetricalComponents
5
• For unbalanced system, power system analysis cannot be
analyzed 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 components.
SymmetricalComponents
6
SymmetricalComponents
7
By convention, the direction of rotation of the phasors is taken
to becounterclock‐wise.
Positive sequence:
I1 = I 10 = I 1
a a a
I1 = I1240 = a2 I1
b a a
(10.1)
I1 = I 1120 = aI 1
c a a
Where we defined an operator a that causes a counterclockwise rotation of 120
degree, such that:
a = 1120 = cos120 + j sin 120 = −0.5 + j0.866
a2 = (1120) (1120) = 1240 = −0.5 − j0.866 1 + a + a2 =0
(10.2)
a3 = 1360 = 1 + j0(10.3)
SymmetricalComponents
8
Negative sequence:
I 2 = I 20a a
I 2 = I 2120 = aI 2
b a a
(10.4)
I 2 = I 2240 = a2 I 2
c a a
a b cI 0 = I 0 = I 0
Zero sequence:
(10.5)
SymmetricalComponents
9
Consider the three‐phase unbalancedcurrent of Ia , Ib , Ic
I = I 0 + I 1 + I 2
a a a a
I = I 0 + I 1 + I2
b b b b
(10.6)
I = I 0 + I1 + I 2
c c c c
Based on (10.1), (10.4) and (10.5), (10.6) can be rewrite all in terms of phase a
components
I = I 0 + I 1 + I 2
a a a a
a
a I 0
I = I 0 + aI 1 + a2 I 2
c a a a
I = I 0 + a2 I 1 + aI 2
b a a a(10.7)
22
1 a
a b Ia a
I 1 1 1
a 2 a I 1
Ic
I = 1 (10.8)
SymmetricalComponents
10
Equation 10.8 can be writtenas:
aIabc = AI012
WhereA is known as symmetrical components transformation matrix,
(10.9)
anda
which transforms phasor currents Iabc into components currents I012
1
a2 a
1 1 1
a2 a A = 1 (10.10)
Solving (10.9) for the symmetrical components of currents:
aI012 = A−Iabc (10.11)
The inverse of A is givenby: 1 1 1 1
a
a
a2
a2
1
13
A− =(10.12)
SymmetricalComponents
11
From (10.10) and (10.12), we concludethat
(10.13)A− =1
A*
3
Substituting for A‐1 in (10.11), wehave:
b
a
a
a
Ia2I 1
I 0
= 13
(10.14)
c a
1 1 1 I1 a
a 2 I 2 1 a I
or in component form, the symmetrical components are:
3
1
3
1
1
0
a b ca
a b ca
I = (I + aI + a2 I )
I = (I + I + I )
(10.15)
3
12
a b caI = (I + a2 I + aI )
SymmetricalComponents
12
Similar expressions exists forvoltage:
(10.16)
V = V 0 + aV 1 + a2V 2
c a a a
20 2 1
V = V 0 +V 1 +V 2
a a a a
ab a aV = V + a V + aV a
V abc = AV 012 (10.17)
The symmetrical components in terms of unbalanced voltages are:
+ aV + a2V )b c
3
1
3
1
1
0
aa
aa
V = (V
+ Vb + Vc )V = (V
(10.18) aV012 =A− Vabc (10.19)
+ a2V + aV )b c3
12
aaV = (V
SymmetricalComponents
13
The apparent power may also be expressed in terms of the symmetrical
components.
abcT abc*
S(3 ) = V I (10.20)
S = (AV012 )T (AI012)*
(3 ) a a
Substituting (10.9) and (10.17) in (10.20), weobtain:
(10.21)012
a
T * 012
a
T *
A A I= V
= A,ATSince AT A* = 3 complex powerbecomes
*** 2 20 0 1 1
a aa aa a= 3V I + 3V I + 3V I
012T 012*
S(3 ) = 3(V I )
(10.22)
Total power for unbalance 3‐phase system can be obtained from the sum of
symmetrical componentspowers.
Example1
14
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 as
reference and assuming that line c is open, find the symmetrical
components of the linecurrents.
aIa =100A
1
3
1
1
0
a b ca
aa
I = (I + aI + a2 I )
I = (I + Ib + Ic )
b Ib = 10180A 3
1
3
2
a b caI = (I + a2 I + aI )
c Ic = 0A
Solution
15
Ic = 0
1
The line currentare:
Ia = 100 Ib =10180
From (10.15):3
1
3
1
1
a
0
a
a b c
a b c
= (I + aI + a2 I )
= (I + I + I )
I
I
(100 + 10180 + 0) = 0a
I (0) =
a (100 + 10(180 +120) + 0)
3
1
3I (1) =
0 Sequence
+ Sequence
3
1a b c
2
aI = (I + a2 I + aI )
=5 − j2.89 = 5.78 − 30A
a ) + 0)0 + 10(180+ 240I = (103
1(2)
‐Sequence
=5 + j2.89 =5.7830A
From (10.4) bI (0) = 0 c
I (0) = 0
bI (1) = 5.78 −150 A c
I (1) = 5.7890A
bI = 5.78150A(2) cI (2) = 5.78 − 90A
Example2
16
Exercise1
17
Show that :
(a)(1+ a)
=1120(1+ a2 )
= 3−180(1+ a)2
(1− a)2
(b)
Exercise2
18
Obtain the symmetrical components for the set of unbalanced voltages
Va = 300 −120,Vb = 20090,Vc = 100 − 30
1
3
1
1
0
a b ca
a b ca
V = (V + aV + a2V )
V = (V +V + V )V 012 =
42.2650 − 120
193 .1852 − 135
86.9473 − 84.8961
The symmetrical components of a set of unbalanced three‐phase currents are
3
1
3
2
a b caV = (V + a2V + aV )
I 0 = 3 − 30, I 1 = 590, I 2 = 430a a a
Obtain the original unbalancedphasors. I = I 0 + I 1 + I 2
a a a a
I = I 0 + aI 1 + a2 I 2
c a a a
I = I 0 + a2 I 1 + aI 2
b a a a
Iabc =
8.1854 42.2163
4 − 30
8.1854 − 102 .2163
Exercise3
19
The line‐to‐line voltages in an unbalanced three‐phase supply areVab = 1000 0,Vbc = 866 .0254 −150,Vca = 500120
Determine the symmetrical components for line and phase voltages, then find the
phase voltages Van, Vbn, andVcn.
VL 012 =
0.030 0.00
Va 012 = Vabc =
440 .9586 − 19.1066
600 .9252 − 166 .1021
333 .3333 60288 .675130
763 .7626 − 10.8934 440 .9586 − 40.8934
166.6667 60
SequenceImpedance
20
• The impedance of an equipment or component to the current of differentsequences.
• positive‐sequence impedance (Z1): Impedance that causes a positive‐sequence current toflow
• negative‐sequence impedance (Z2): Impedance that causes a negative‐sequence current toflow
• zero‐sequence impedance (Z0): Impedance that causes a zero‐sequence current toflow
Sequence Impedance of Y‐ConnectedLoad
21
(10.23)
Line to ground voltagesare:
Va = Zs Ia + Zm Ib + Zm Ic + Zn In
(10.24)I n = I a + I b + I c
Vb = Zm Ia + Zs Ib + Zm Ic + Zn In
Vc = Zm Ia + Zm Ib + +Zs Ic + Zn In
Kirchhoff’ current law:
Substituting In into (10.23):
Zm + Zn IaVa (Zs + Zn ) Zm +Zn
(10.25)
(10.26)
m n b mb
I
Zm + Zn Zm +ZnVc
V = Z
(Zs + Zn )Ic
+ Zn (Zs +Zn ) Z + Z
Vabc = Zabc Iabc
Sequence Impedance of Y‐ConnectedLoad
22
Zm + Zn Zm + Zn
(Zs + Zn )
(10.27) m n m n
Zm +Zn
+ Z (Zs +Zn ) Z + Z
Zm + Zn (Zs + Zn )
Zabc = Z
Writing Vabc and Iabc in terms of their symmetrical components:
(10.28)AV012 = Zabc AI012
a a
Multiplying (10.28) by A‐1:
(10.29)a
a a
= Z012I012
V012 = (A− Zabc A)I012
Z012 = A− Zabc Awhere (10.30)
(10.31)
Substituting for Zabc,A andA‐1 from (10.27), (10.10) and (10.12):
1
13
a a2 Z + Z (Z + Z )1
+ Z (Z + Z ) Z + Z
a Z + Za2
a2 Z 1
1 1 1 (Zs + Zn ) Zm + Zn Zm + Zn 1 1 1 1
a a2 a Z 012 =
n sm n m n
n ms nm n
Sequence Impedance of Y‐ConnectedLoad
23
Performing the multiplication in(10.31):
0
0s m
0 (Zs − Zm )
0 (Z − Z )
0
(Zs + 3Zn +2Zm ) 0
Z012 = (10.32)
When there is no mutual coupling, Zm = 0, and the impedance matrix becomes
(Zs +3Zn ) 0 0 (10.33)
0
s 0 (Zs )
0 (Z )Z012 = 0
Sequence Impedance of TransmissionLines
24
For sequence impedance transmission line, Z1 = Z2, whereas Z0 is different
and larger approximately 3 times than positive and negative sequence.
Sequence Impedance of SynchronousMachineThe positive-sequence generator impedance is the value found when positive-
sequence current flows from the action of an imposed positive-sequence set of
voltages.
The negative-sequence reactance is close to the positive-sequence
dX 2
X "substransient reactance, i.e :
Zero-sequence reactance is approximated to the leakage reactance, i.e :
X lX 0
Sequence Impedances ofTransformer
25
• Series Leakage Impedance.– the magnetization current and core losses represented by the shunt branch
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 notchange if the phase sequence ischanged.
• Therefore, the positive andnegative 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 notrotate
• The equivalent circuit for zero‐sequence impedance depends on thewinding connections and also upon whether or not the neutrals aregrounded.
Sequence Impedances ofTransformer
26
Connection diagram Zero‐sequence circuit
Figure (a)
Figure (b)
Figure (c)
Figure (d)
Figure (e)
Sequence Impedances ofTransformer
27
Description of Zero sequence EquivalentCircuit
(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
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).
Sequence Impedances ofTransformer
28
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 no
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 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
path of Io.
Sequence Impedances of a LoadedGenerator
29
A synchronous machine generates balanced three‐phase internal voltages and is
represented as a positive‐sequence set ofphasors
a a
1
Eabc = a2E(10.44)
Sequence Impedances of a LoadedGenerator
30
The machine is supplying a three‐phase balanced load.Applying kirchhoff’s voltage
law to each phase weobtain:
(10.45)
Va = Ea − Zs Ia − Zn In
Vb = Eb − Zs Ib − Zn In
Vc = Ec − Zs Ic −Zn In
Substituting for In = Ia + Ib + Ic into(10.45):
b I Ia
(Zs + Zn )Ic
nb b Vc
Ec Zn Zn
V = E −Va Ea
Zn (Zs + Zn ) Z
(Zs + Zn ) Zn Zn
(10.46)
In compactform: V abc = Eabc −Zabc Iabc (10.47)
Sequence Impedances of a LoadedGenerator
31
Transforming the terminal voltages and currents phasors into their symmetrical
components:
(10.48)AV012 = AE012 − Zabc AI012
a a a
Multiplying (10.48) byA‐1:
(10.49)a a
a a a
= E012 −Z012I012
V012 = E012 − (A−Zabc A)I012
(10.50)Where:
a
1 1
a2
a a2 1
1
(Z + Z )1
13
Z
a Z
a2
Z
(Z + Z ) Z
s n
ZnZn
nn
ns nn
1 1 1 (Zs + Zn )1
a
a2
Z012 = 1
Performing the abovemultiplication:
00 = 0
0 Z0 0 0
Z1 0 Z
(Z + 3Z ) 0
s
s n
Z012 = (10.51)
0 00 Z2 0 Z
s
Sequence Impedances of a LoadedGenerator
32
Since the generated emf is balanced, there is only positive‐sequence voltage, i.e:
0
(10.52)0
012 aaE = E
0 aaZ0 0 0 I 0 V 0
Substituting for and Z012 in (10.49):012
aE
V 0 = 0 − Z 0I 0
a a
(10.54)
0
1
0
a V 2
1 1
a
0 Ia Z 2 I 2
Va = Ea − 0 Z
0
(10.53) orV 2 = 0 − Z 2I 2
a a
V 1 = E − Z 1I 1
a a a
Sequence Impedances of a LoadedGenerator
The three equations in (10.54) can be represented by the three equivalent
sequence networks:
• Important observations:– The three sequences are independent.
– The positive‐sequence network is the same as the one‐line diagram used in studying balance three‐phase currents andvoltages.
– Only the positive‐sequence network has a source and no voltage source for other sequences.
– 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 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 andvoltage respectively. 33
Single Line‐To‐GroundFault
34
Three‐phase generator with neutral grounded through impedance Zn andSLGF
occurs at phase a through impedanceZf.
Assuming the generator is initially on no‐load, the boundary conditions at the
fault point are:
Va = Z f I a(10.55)
Ib = Ic = 0 (10.56)
Single Line‐To‐GroundFault
35
Substituting for Ib = Ic =0, the symmetrical components of currents from (10.14)
are:
0= 13
a
a
a
1 I
a2 I 1 (10.57)
a 1 a 0
I0 1 1 1 a
a2 I 2
From the above equation, we findthat: V 0 = 0 − Z 0I 0
a a
aaaa3
1210I = I = I = I
Phase a voltage in terms of symmetrical components is :
(10.58) 1
V 2 = 0 − Z 2I 2
a a
a a aV = E − Z 1I 1
V = V 0 + V 1 + V2
a a a a
Substituting V 0 ,V 1and V 2 from (10.54) and noting I 0 = I 1 = I 2 :a a a a a a
(10.59)
V = E − (Z 0 + Z 1 + Z 2 )I 0
a a a(10.60)
Single Line‐To‐GroundFault
36
Where Z 0 = Z + 3Z .Substituting for V from (10.55), and noting I = 3I 0 , we get :s n a a a
3Z I 0 = E − (Z 0 + Z 1 + Z 2 )I 0
f a a a
or
(10.61)
f
Ea
aZ 0 + Z 1 + Z 2 + 3Z
I 0=
(10.62)
The fault current is
f
3Ea
Z 0 + Z1 + Z 2 + 3ZI = 3I 0 =
a a
(10.63)
In order to obtain symmetrical voltage at the point of fault Equation, (10.63) is
substituted into Eq.(10.54)
Single Line‐To‐GroundFault
37
Eq. (10.58) and (10.62) can be represented by connecting the sequence
networks in series as shown in the following figure.
aaaa3
1210I = I = I = I
f
Ea
aZ 0 + Z 1 + Z 2 + 3Z
I 0=(10.58) (10.62)
Line‐To‐LineFault
38
Three‐phase generator with a fault through an impedance Zf between phaseband c.
Ia=0
Zs
N
ZsZs
EaEb
Ec
Ib
Va
ZfIcVb
Vc
Assuming the generator is initially on no‐load, the boundary conditions at the
fault point are:
(10.64) (10.66)Vb − Vc = Z f Ib Ia = 0
Ib + I c =0 (10.65)
Line‐To‐LineFault
39
Substituting for Ia = 0, and Ic = ‐Ib, the symmetrical components of the currents
from (10.14) are:
a
I= 1
1 0
a 2 I 1 (10.67)
b
b
a
a
a −I13
I0 1 1 1 a
a2 I 2
From the above equation, we findthat:
= 0aI 0 (10.68)
ba3
11 2I = (a − a )I
ba
12 2I = (a − a)I
(10.69)
(10.70)
3
Line‐To‐LineFault
40
Also, from (10.69) and (10.70), we notethat:
I 1 = −I 2
a a
From (10.16), wehave:
(10.71)
V = V 0 + a2V 1 + aV 2
b a a a
V = V 0 +V 1 +V 2
a a a a
(10.16)
= Z f Ib
V −V = (V 0 + a2V 1 + aV 2 ) − (V 0 + aV 1 + a2V 2 )b c a a a a a a
= (a2 − a)(V 1 −V 2 )a a (10.72)
V = V 0 + aV 1 + a2V 2
c a a a
Substituti ng for V 1 and V 2 from (10.54) and noting I 2 = −I1 , we get :a a a a
(10.73)
(10.54)
V 0 = 0 − Z 0I 0
a a
V 2 = 0 − Z 2I 2
a a
V 1 = E − Z 1I 1
a a a
(a 2 − a)[E − (Z 1 + Z 2 )I 1 ] = Z Ia a f b
Substituti ng for I b from (10.69), we get :
3I 1a
(a − a 2)(a 2 − a)f
E − (Z 1 + Z 2 )I1 = Za a
(10.74) ba3
11 2I = (a − a )I (10.69)
Line‐To‐LineFault
41
aSince (a − a2 )(a2 − a) = 3,solving for I 1 results in :
f
Ea
a(Z 1 + Z 2 + Z )
I 1 =(10.75)
The phase currentsare
1 0 I a
a
1 a
c
b I
a 2 −I 1 I 1
1 1 I = 1 a 2 a
a
(10.76)
The fault current is
I = −I = (a 2 − a)I 1
b c aor
abI = − j 3I 1(10.77) (10.78)
Line‐To‐LineFault
42
Eq. (10.71) and (10.75) can be represented by connecting the positive and negative –
sequence networks as shown in the following figure.
I 1 = −I 2
a a1 Ea
I =1 2
f
a(Z + Z + Z )
Double Line‐To‐GroundFault
43
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 pointare
I = I 0 + I1 + I 2 = 0a a a a
From (10.16), the phase voltages Vb and Vcare
Vb = Vc = Z f (Ib + Ic ) (10.79)
(10.80)
Figure 10.14Double line‐to‐ground fault
Double Line‐To‐GroundFault
44
(10.81)V = V 0 + a2V 1 + aV 2
b a a a
V = V 0 + aV 1 + a2V 2
c a a a (10.82)
SinceVb = Vc , from above we note that
V 1 = V 2
a a
Substituting for the symmetrical components of current in (10.79), we get
V = Z (I 0 + a 2 I 1 + aI 2 + I 0 + aI 1 + a 2 I 2 )(b ) f a a a a a a
= Z (2 I 0 − I 1 − I 2 )f a a a
(10.83)
= 3 Z f aI 0 (10.84)
Substituti ng for V from (10.84) and for V 2 from (10.83) into (10.81), we have :b a
3Z I 0 = V 0 + (a 2 + a)V 1
f a a a
= V 0 −V 1 (10.85)a a
Substituti ng for the symmetrical components of voltage from (10.54) into (10.85)
E − Z 1I 1
aand solving for I 0 , we get :
a(Z 0 + 3Z )
I 0 = − aa
f
Also, substituting for the symmetrical components of voltage in (10.83), we obtain
E − Z 1I 1
(10.86)
Z 2aI 2 = − a a
Substituti ng for I 0 and I2 into (10.80) and solving for I 1 , we get :a a a
(10.87)
f
45
f
Ea
a
Z 2 + Z 0 +3Z
Z 2 (Z 0 + 3Z )Z 1 +
I 1 = (10.88)
Equation (10.86) - (10.88) can be represented by connecting the positive -sequence
impedance in series with the paralel combination of the negative - sequence
46
The value of I1 found from (10.86) is substitute d in (10.86) and (10.87),
and zero - sequence networks as shown in the equivalent circuit of figure 10.15.
and I0 and I2 are found. The phase current are then found from (10.8).a a
Finally, the fault current is obtained from
a
= 3I 0 (10.89)acI f = Ib + I
Figure 10.15 Sequence network connection for double line‐to‐groundfault
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 phase.
Determine the fault current for the following faults
a. A balaced three-phase fault at bus 3 through a fault impedance Z f = j 0.1 per unit
47
b. A single line-to-ground fault at bus 3 through a fault impedance = j 0.1 per unitZ f
c. A line-to-line fault at bus 3 through a fault impedance Z f = j 0.1 per unit
d. A double line-to-ground fault at bus 3 through a fault impedance Z f = j 0.1 per unit
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 0.10
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 0.7125
Figure 10.16
Fault
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 0.10
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 0.7125
Item
48
Base Rated X1 X2 X0
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 below
Fig. 10.17Positive-sequence impedance
j0.525Z1S
=( j0.125)( j0.15)
= j0.0357143
j0.525Z
2S=
( j0.125)( j0.25)= j0.0595238
Z 3S=
( j0.15)( j0.25) = j0.0714286
j0.525
49
33j0.5952381
Z 1 =( j0.2857143 )( j0.3095238 )
+ j0.0714286
50
= j0.22
Tofind thevenin impedance viewed from the faulted bus (bus 3), we convert the
delta formed by buses 123 to an equivalent Yas shown in figure 10.19(b)
j1.3625
( j0.30)( j0.35)Z1S = j0.0770642=
Z
j0.077064
=( j0.30)( j0.7125)
= j0.15688072S
j1.3625
51
Z3S
j1.3625
=( j0.35)( j0.7125)
= j0.1830257
Fig. 10.19Zero-sequence impedance
Combining the parallel branches, the zero‐sequence thevenin impedanceis
33Z 0 =
( j0.4770642)( j0.2568807) + j0.1830275
j0.7339449
= j0.35
j0.077064
So, the zero‐sequence impedance diagram is show in fig. 10.20
Fig. 10.20
Zero‐sequence network
52
(a) Balanced three‐phase fault at bus3
Assuming the no‐load generated emfs are equal to 1.0 per unit, the fault
current is
1.03(0)V a
Z1
33 f
3 = -j3.125 pu =820.1 - 90 A=I (F)=+ Z j0.22 + j0.1
(b) Single line‐to ground fault at bus3
From (10.62), the sequence component of the fault current are
V a1.0
j0.22 + j0.22 + j0.35 +3(j0.1)= = -j0.9174 pu
f33Z1
33 33+ Z2 + Z0 + 3ZI0 = I1 = I2 =
3 3 3
3(0)
The fault current is:
I a 1 1 1 I 0 3I 0
pu
53
aI b
I c
0
0
−j2.7523
0
= 0 == 1
3
I 0
3
3
3
a 2
1 a a 2 I 03
3
3
(c) Line‐to Line fault at bus3
54
The zero‐sequence component of current is zero,i.e.,
I0 = 03
The positive‐and negative‐sequence components of the fault currentare
V a1.0
fZ1
33
3(0)= = -j1.8519 pu
+ Z2 + Z0 + Z j0.22 + j0.22 + j0.133 33
I1 = −I 2 =3 3
The fault current is
I b
I c
a − j1.8519 = − 3.2075 pu
0 0
1
= 1
1
a 2 − j1.8519 −3.2075
Ia 1 1 a2
a3
3
3
(d) Double Line‐to Line‐fault at bus3
The positive‐sequence component of the fault currentis
V a
j0.22 + j0.35 + j0.3
1.0
j0.22 (j0.35 + j0.3)j0.22 +
= = -j2.6017 pu
f33Z2
33
33
3(0)
3
+ Z0 + 3Z )
Z2 (Z0 + 3Z )Z1 + 33 33 f
I1 =
The negative‐sequence component of current is:
= j1.9438 pu1.0 − ( j0.22)(− j2.6017)V a
j0.22Z2
33
33 333(0)
3
− Z1 I1
I2 = − = −
The zero‐sequence component of currentis:
= j0.6579 pu1.0 − ( j0.22)(− j2.6017)
Z0
V a I1
3(0) 33 33
3+ 3Z
− Z1
I0 = − = −j0.35 + j0.3
33 f
And the phase currents are:
I a 1
a
55
I b
I c
− j2.6017 = 4.058165.93 pu= 1
a 2 j1.9438 4.05814.07
1 1 j0.6579 0 a2
1 a3
3
3
And the fault currents is:
I (F ) = I b + I c =1.9732903 3 3
UNBALANCED FAULTANALYSIS USING BUS IMPEDANCE MATRIX
Single Line‐to‐Ground Fault UsingZbus
Vk (0)I 0 = I 1 = I 2 =k k k
(10.90)
kkZ1 + Z 2 +Z 0 + 3Z
kk kk f
Where Z1 , Z2 and Zo are the diagonal elements in the k axis of the correspondingkk kk kk
bus impedance matrix and Vk (0)is the prefault voltage at busk.
The fault phase current is :
I abc
k k= A I 012 (10.91)
Line‐to‐Line Fault Using Zbus
kI 0 = 0 (10.92)
f
56
kk kk
Vk (0)I k = −I 2
kZ + Z 2 +Z
=1
1(10.93)
Double Line‐to‐Ground Fault UsingZbus
57
kk fkk
kk kk f
kk
Vk (0)k
Z 2 + Z 0 + 3Z
Z 2 (Z 0 + 3Z )Z1
I 1 =
+
(10.94)
k Z 2
V (0) − Z 1 I 1
I 2 =− k kk k (10.95)kk
Ik0 V (0) − Z 1 I 1
= − k kk k(10.96)
fkkZ 0 + 3Z
Where Z1 , and Z 2 , and Zo are the diagonal elements in the k axis of the correspond ingkk kk kk
bus impedance matrix. The phase currents are obtained from (10.91), and the result current is
I (F ) = I b + I C k k k
(10.97)
Thank You