Learning Outline
™Symmetrical Components Theory
™Sequence Impedance: Load, Line, Generator,
T
Transformer
f
™Fault Analysis: Line‐Ground, Line‐Line, Line‐
Line‐Ground.
™Examples
p and Class Exercises
1
Fault Analysis
y ____________________
• Fault types:
– balanced faults (<5%)
• three‐phase to ground
• Three‐phase
– unbalanced faults
• single‐line to ground (60%‐75%)
• double‐line
d bl li to
t ground
d (15%‐25%)
(15% 25%)
• line
line‐to‐line
to line faults (5%
(5%‐15%)
15%)
2
Example impact of fault
The second largest blackout in the history of TEPCO
(The Tokyo Electric Power Company,
Company Inc.)
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 river.
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
y
Components
p
__________
• 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 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
Symmetrical
y
Components
p
__________
• 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.
components
5
Symmetrical
y
Components
p
__________
6
Symmetrical
y
Components
p
__________
By convention, the direction of rotation of the phasors is taken
to be counterclock‐wise.
counterclock wise
Positive sequence:
I a1 = I a1∠0° = I a1
I b1 = I a1∠240° = a 2 I a1
(10.1)
I c1 = I a1∠120° = aII a1
Where we defined an operator a that causes a counterclockwise rotation of 120
degree such that:
degree,
a = 1∠120 ° = cos 120 ° + j sin 120 ° = −0.5 + j 0.866
(10.2)
1 + a + a2 = 0
a 2 = (1∠120°) × (1∠120°) = 1∠240° = −0.5 − j 0.866
a 3 = 1∠360° = 1 + j 0
(10.3)
7
Symmetrical
y
Components
p
__________
Negative sequence:
I a2 = I a2∠0°
I b2 = I a2∠120° = aI a2
(10.4)
I c2 = I a2∠ 240 ° = a 2 I a2
Zero sequence:
I a0 = I b0 = I c0
(10.5)
8
Symmetrical
y
Components
p
__________
Consider the three‐phase unbalanced current of I a , I b , I c
I a = I a0 + I a1 + I a2
Ib = I + I + I
0
b
1
b
2
b
(10.6)
I c = I c0 + I c1 + I c2
Based on (10.1), (10.4) and (10.5), (10.6) can be rewrite all in terms of phase a
components
I a = I a0 + I a1 + I a2
I b = I a0 + a 2 I a1 + aI a2
I c = I a0 + aI a1 + a 2 I a2
(10.7)
⎡ I a ⎤ ⎡1 1
⎢ I ⎥ = ⎢1 a 2
⎢ b⎥ ⎢
⎣⎢ I c ⎥⎦ ⎢⎣1 a
1 ⎤ ⎡ I a0 ⎤
⎢ ⎥
a ⎥⎥ ⎢ I a1 ⎥
a 2 ⎥⎦ ⎢⎣ I a2 ⎥⎦
(10.8)
9
Symmetrical
y
Components
p
__________
Equation 10.8 can be written as:
I abc = AI 012
a
(10.9)
Where A is known as symmetrical components transformation matrix,
which transforms phasor currents I abc into components currents I 012
a
and
⎡1 1 1 ⎤
(10 10)
(10.10)
A = ⎢⎢1 a 2 a ⎥⎥
2
⎣⎢1 a a ⎥⎦
Solving (10.9) for the symmetrical components of currents:
I 012
= A − I abc
a
The inverse of A is given by:
(10.11)
⎡1 1
1
A − = ⎢⎢1 a
3
⎢⎣1 a 2
1⎤
a 2 ⎥⎥
a ⎥⎦
(10 12)
(10.12)
10
Symmetrical
y
Components
p
__________
From (10.10) and (10.12), we conclude that
1
A − = A*
3
(10.13)
Substituting for A‐1 in (10.11), we have:
⎡ I a0 ⎤
⎡1 1
⎢ 1⎥ 1⎢
⎢ I a ⎥ = 3 ⎢1 a
2
⎢ I a2 ⎥
⎢
1
a
⎣
⎣ ⎦
1 ⎤ ⎡I a ⎤
⎢ ⎥
a 2 ⎥⎥ ⎢ I b ⎥
a ⎥⎦ ⎢⎣ I c ⎥⎦
(10.14)
or in component form, the symmetrical components are:
1
I a0 = ( I a + I b + I c )
3
1
I a1 = ( I a + aI b + a 2 I c )
3
1
I a2 = ( I a + a 2 I b + aI c )
3
(10.15)
11
Symmetrical
y
Components
p
__________
Similar expressions exists for voltage:
Va = Va0 + Va1 + Va2
Vb = Va0 + a 2Va1 + aVa2
(10.16)
V abc = AVa012
(10.17)
Vc = Va0 + aVa1 + a 2Va2
The symmetrical components in terms of unbalanced voltages are:
1
Va0 = (Va + Vb + Vc )
3
1
Va1 = (Va + aVb + a 2Vc )
3
1
Va2 = (Va + a 2Vb + aVc )
3
(10.18)
Va012 = A − V abc
(10.19)
12
Symmetrical
y
Components
p
__________
The apparent power may also be expressed in terms of the symmetrical
components.
components
S ( 3φ ) = V
abc T
I
abc *
(10.20)
Substituting (10.9) and (10.17) in (10.20), we obtain:
*
S ( 3φ ) = ( AVa012 )T ( AI 012
a )
=V
012T
a
Since
T
(10.21)
* 012*
a
A AI
A T = A, A T A * = 3 complex power becomes
S ( 3φ ) = 3( V 012 I 012 )
T
*
= 3Va0 I 0a + 3Va1I1a + 3Va2 I 2a
*
*
*
(10.22)
Total power for unbalance 3
3‐phase
phase system can be obtained from the sum of
symmetrical components powers.
13
Example
p 1_______________________
One conductor of a three‐phase line is open. The current flowing to delta‐
connected load through line a is 10 A.
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
b
c
1
I a0 = ( I a + I b + I c )
3
1
I a1 = ( I a + aI b + a 2 I c )
3
1
I a2 = ( I a + a 2 I b + aI c )
3
I a = 10∠0° A
I b = 10∠180° A
Ic = 0 A
14
Solution________________________
The line current are:
I a = 10∠0°
I b = 10∠180°
Ic = 0
From (10.15):
0 Sequence
1
I a( 0 ) = (10∠0° + 10∠180° + 0) = 0
3
+ Sequence
1
I a(1) = (10∠0° + 10∠(180° + 120°) + 0)
3
= 5 − j 2.89 = 5.78∠ − 30° A
‐ Sequence
1
I a( 2 ) = (10∠0° + 10∠(180° + 240°) + 0)
3
= 5 + j 2.89 = 5.78∠30° A
From (10.4)
(0)
b
I
=0
I b(1) = 5.78∠ − 150 ° A
I
( 2)
b
= 5.78∠150° A
1
I a0 = ( I a + I b + I c )
3
1
I a1 = ( I a + aI b + a 2 I c )
3
1
I a2 = ( I a + a 2 I b + aI c )
3
I c( 0 ) = 0
I c(1) = 5.78∠90 ° A
I c( 2 ) = 5.78∠ − 90° A
15
Example
p 2_______________________
16
Exercise 1_______________________
Show that :
(1 + a)
(a)
= 1∠120°
2
(1 + a )
(1 − a)) 2
(b)
= 3∠ − 180°
2
(1 + a)
17
Exercise 2_______________________
Obtain the symmetrical components for the set of unbalanced voltages
V a = 300 ∠ − 120 °, Vb = 200 ∠ 90 °, Vc = 100 ∠ − 30 °
V 012 =
42 .2650 ∠ − 120 °
193 .1852 ∠ − 135 °
86 .9473 ∠ − 84 .8961 °
1
Va0 = (Va + Vb + Vc )
3
1
Va1 = (Va + aVb + a 2Vc )
3
1
Va2 = (Va + a 2Vb + aVc )
3
The symmetrical components of a set of unbalanced three
three‐phase
phase currents are
I a0 = 3∠ − 30 °, I a1 = 5∠90 °, I a2 = 4∠30 °
Obtain the original unbalanced phasors.
I a = I a0 + I a1 + I a2
Iabc =
8 .1854 ∠ 42 .2163 °
I b = I a0 + a 2 I a1 + aI a2
4 ∠ − 30 °
8 .1854 ∠ − 102 .2163 °
I c = I a0 + aI a1 + a 2 I a2
18
Exercise 3_______________________
The line‐to‐line voltages in an unbalanced three‐phase supply are
Vab = 1000 ∠ 0 °, Vbc = 866 .0254 ∠ − 150 °, Vca = 500 ∠120 °
Determine the symmetrical components for line and phase voltages, then find the
phase voltages Van, Vbn, and Vcn.
VL 012 =
Va 012 =
0 .0 ∠ 30 °
0 .0 ∠ 0 °
763 .7626 ∠ − 10 .8934 ° 440 .9586 ∠ − 40 .8934 °
288 .6751 ∠ 30 °
166 .6667 ∠ 60 °
Vabc =
440 .9586 ∠ − 19 .1066 °
600 .9252 ∠ − 166 .1021 °
333 .3333 ∠ 60 °
19
Sequence
q
Impedance
p
______________
• The impedance of an equipment or component to the
current of different sequences.
• positive‐sequence
iti
i
impedance
d
(Z1):
) Impedance
I
d
th t
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
q
Impedance
p
of Y‐Connected Load
Line to ground voltages are:
Va = Z s I a + Z m I b + Z m I c + Z n I n
Vb = Z m I a + Z s I b + Z m I c + Z n I n
(10.23)
Vc = Z m I a + Z m I b + + Z s I c + Z n I n
Kirchhoff’ current law:
In = Ia + Ib + Ic
(10.24)
Substituting In into (10.23):
Zm + Zn ⎤ ⎡I a ⎤
⎡Va ⎤ ⎡( Z s + Z n ) Z m + Z n
⎢V ⎥ = ⎢ Z + Z
⎥ ⎢I ⎥
Z
+
Z
Z
+
Z
(
)
b
m
n
s
n
m
n
⎥⎢ b⎥
⎢ ⎥ ⎢
Z m + Z n ( Z s + Z n )⎥⎦ ⎣⎢ I c ⎥⎦
⎣⎢Vc ⎦⎥ ⎣⎢ Z m + Z n
(10 25)
(10.25)
V abc = Z abc I abc
(10.26)
21
Sequence
q
Impedance
p
of Y‐Connected Load
Zm + Zn ⎤
⎡( Z s + Z n ) Z m + Z n
Z abc = ⎢⎢ Z m + Z n ( Z s + Z n ) Z m + Z n ⎥⎥
⎢⎣ Z m + Z n
Z m + Z n ( Z s + Z n )⎥⎦
(10 27)
(10.27)
Writing Vabc and Iabc in terms of their symmetrical components:
AVa012 = Z abc AI 012
a
(10.28)
Multiplying (10.28) by A‐1 :
Va012 = ( A − Z abc A )I 012
a
=Z I
012 012
a
where
Z 012 = A − Z abc A
(10.29)
(10.30)
Substituting for Zabc, A and A‐1 from (10.27), (10.10) and (10.12):
⎡1 1
1
Z 012 = ⎢⎢1 a
3
⎢⎣1 a 2
1 ⎤ ⎡( Z s + Z n ) Z m + Z n
Z m + Z n ⎤ ⎡1 1
a 2 ⎥⎥ ⎢⎢ Z m + Z n ( Z s + Z n ) Z m + Z n ⎥⎥ ⎢⎢1 a 2
a ⎥⎦ ⎣⎢ Z m + Z n
Z m + Z n ( Z s + Z n )⎦⎥ ⎢⎣1 a
1⎤
a ⎥⎥
a 2 ⎥⎦
(10.31)
22
Sequence
q
Impedance
p
of Y‐Connected Load
Performing the multiplication in (10.31):
Z
012
0
0
⎡( Z s + 3Z n + 2 Z m )
⎤
⎢
⎥
=⎢
0
(Z s − Z m )
0
⎥
⎢⎣
0
0
( Z s − Z m )⎥⎦
(10.32)
When there is no mutual coupling, Zm = 0, and the impedance matrix becomes
0
0 ⎤
⎡ ( Z s + 3Z n )
Z 012 = ⎢⎢
0
(Z s )
0 ⎥⎥
0
0
( Z s )⎥⎦
⎣⎢
(10.33)
23
Sequence
q
Impedance
p
of Transmission Lines
For sequence impedance transmission line, Z1 = Z2, whereas Z0 is different
and larger approximately 3 times than positive and negative sequence.
S
Sequence
IImpedance
d
off SSynchronous
h
M
Machine
hi
The positive-sequence generator impedance is the value found when positivesequence
q
current flows from the action of an imposed
p
positive-sequence
p
q
set of
voltages.
The negative-sequence reactance is close
substransient reactance,
reactance i.e
ie:
X2 ≈ X"
to
the
positive-sequence
d
Zero-sequence reactance is approximated to the leakage reactance, i.e :
X 0 ≈ Xl
24
Sequence
q
Impedances
p
of Transformer
• Series Leakage Impedance.
– tthee magnetization
ag et at o cu
current
e t aand
d co
coree losses
osses represented
ep ese ted by tthee sshunt
u tb
branch
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 changed.
changed
• Therefore, the positive and negative sequence impedance are the same;
Z 0 = 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 rotate
• The equivalent circuit for zero‐sequence impedance depends on the
winding connections and also upon whether or not the neutrals are
grounded.
25
Sequence
q
Impedances
p
of Transformer
Connection diagram
Zero‐sequence circuit
Figure (a)
Figure
g
(b)
( )
Figure (c)
Figure (d)
Figure (e)
26
Sequence
q
Impedances
p
of Transformer
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).
( ) Y‐Y connections with p
(b)
primaryy the neutral ggrounded – The p
primaryy neutral is ggrounded,, 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.
g ((b).
)
27
Sequence
q
Impedances
p
of Transformer
c) Y‐Δ with grounded neutral – in this configuration, the primary currents can
flow because the zero‐sequence circulating current in the Δ‐connected
secondaryy and a gground return p
path for the Y‐connected p
primary.
y 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
Y Δ connection
ti with
ith isolated
i l t d neutral
t l – in
i this
thi configuration,
fi
ti
b
because
th
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.
28
Sequence
q
Impedances
p
of a Loaded Generator
A synchronous machine generates balanced three‐phase internal voltages and is
represented as a positive‐sequence set of phasors
⎡1⎤
E abc = ⎢⎢a 2 ⎥⎥ Ea
⎢⎣ a ⎥⎦
(10.44)
29
Sequence
q
Impedances
p
of a Loaded Generator
The machine is supplying a three‐phase balanced load. Applying kirchhoff’s voltage
law to each phase we obtain:
Va = Ea − Z s I a − Z n I n
(10.45)
Vb = Eb − Z s I b − Z n I n
Vc = Ec − Z s I c − Z n I n
Substituting for In = Ia + Ib + Ic into (10.45):
Zn
Z n ⎤ ⎡I a ⎤
⎡Va ⎤ ⎡ E a ⎤ ⎡( Z s + Z n )
⎢V ⎥ = ⎢ E ⎥ − ⎢ Z
⎥ ⎢I ⎥
(
Z
+
Z
)
Z
b
b
n
s
n
n
⎢ ⎥ ⎢ ⎥ ⎢
⎥⎢ b⎥
⎢⎣Vc ⎥⎦ ⎢⎣ Ec ⎥⎦ ⎢⎣ Z n
Zn
( Z s + Z n ) ⎥⎦ ⎢⎣ I c ⎥⎦
In compact form:
V abc = E abc − Z abc I abc
(10.46)
(10.47)
30
Sequence
q
Impedances
p
of a Loaded Generator
Transforming the terminal voltages and currents phasors into their symmetrical
components:
abc
AVa012 = AE 012
AI 012
a −Z
a
(10.48)
Multiplying (10.48) by A‐1:
− abc
Va012 = E 012
A)I 012
a − (A Z
a
(10.49)
012 012
= E 012
Ia
a −Z
Where:
h
⎡1 1
1
Z 012 = ⎢⎢1 a
3
⎢⎣1 a 2
1 ⎤ ⎡( Z s + Z n )
Zn
Z n ⎤ ⎡1 1
a 2 ⎥⎥ ⎢⎢ Z n
(Z s + Z n )
Z n ⎥⎥ ⎢⎢1 a 2
a ⎥⎦ ⎢⎣ Z n
( Z s + Z n )⎥⎦ ⎢⎣1 a
Zn
1⎤
a ⎥⎥
a 2 ⎥⎦
(10.50)
Performing the above multiplication:
⎡( Z s + 3Z n ) 0
Z 012 = ⎢⎢
0
Zs
⎢⎣
0
0
0 ⎤ ⎡Z 0
⎢
0 ⎥⎥ = ⎢ 0
Z s ⎦⎥ ⎣⎢ 0
0
Z1
0
0⎤
⎥
0⎥
Z 2 ⎥⎦
(10.51)
31
Sequence
q
Impedances
p
of a Loaded Generator
Since the generated emf is balanced, there is only positive‐sequence voltage, i.e:
⎡0⎤
= ⎢⎢ Ea ⎥⎥
E 012
a
⎢⎣ 0 ⎥⎦
(10 52)
(10.52)
012
Substituting for E a and Z 012 in (10.49):
⎡V ⎤ ⎡ 0 ⎤ ⎡ Z
⎢ ⎥ ⎢ ⎥ ⎢
⎢V ⎥ = ⎢ E a ⎥ − ⎢ 0
⎢V ⎥ ⎢⎣ 0 ⎦⎥ ⎢ 0
⎣ ⎦
⎣
0
a
1
a
2
a
0
0
Z1
0
0 ⎤ ⎡I ⎤
⎥⎢ ⎥
0 ⎥ ⎢I ⎥
Z 2 ⎥⎦ ⎣⎢ I ⎥⎦
0
a
1
a
2
a
Va0 = 0 − Z 0 I a0
(10.53)
or
Va1 = Ea − Z 1 I a1
(
(10.54)
)
Va2 = 0 − Z 2 I a2
32
Sequence
q
Impedances
p
of a Loaded Generator
The three equations in (10.54) can be represented by the three equivalent
sequence networks:
• Important observations:
– The three sequences are independent.
– The
Th positive‐sequence
ii
networkk is
i the
h same as the
h one‐line
li diagram
di
used
d iin
studying balance three‐phase currents and voltages.
– Only the positive‐sequence network has a source and no voltage source for
other sequences.
– The
h neutrall off the
h system is the
h reference
f
for
f positive‐ and
d 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
33
symmetrical components of current and voltage respectively.
Single
g Line‐To‐Ground Fault
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:
(10.55)
Va = Z f I a
Ib = Ic = 0
(10.56)
34
Single
g Line‐To‐Ground Fault
Substituting for Ib = Ic = 0, the symmetrical components of currents from (10.14)
are:
⎡ I a0 ⎤
⎡1 1
⎢ 1⎥ 1⎢
⎢ I a ⎥ = 3 ⎢1 a
⎢ I a2 ⎥
⎢⎣1 a 2
⎣ ⎦
1 ⎤ ⎡I a ⎤
a 2 ⎥⎥ ⎢⎢ 0 ⎥⎥
a ⎥⎦ ⎢⎣ 0 ⎥⎦
(10.57)
From the above equation, we find that:
I a0 = I a1 = I a2 =
1
Ia
3
Va0 = 0 − Z 0 I a0
(10 58) V 1 = E − Z 1 I 1
(10.58)
a
a
a
Phase a voltage in terms of symmetrical components is :
V a = V a0 + V a1 + V a2
Va2 = 0 − Z 2 I a2
(10.59)
Substituti ng Va0 , Va1and Va2 from (10.54) and noting I a0 = I a1 = I a2 :
V a = E a − ( Z 0 + Z 1 + Z 2 ) I a0
(10. 60)
35
Single
g Line‐To‐Ground Fault
Where Z 0 = Z s + 3Z n . Substituti ng for Va from (10.55), and noting I a = 3I a0 , we get :
3Z f I a0 = E a − ( Z 0 + Z 1 + Z 2 ) I a0
(10 61)
(10.61)
or
Ea
I = 0
Z + Z 1 + Z 2 + 3Z f
0
a
(10.62)
The
h ffault
l current is
I a = 3I a0 =
3E a
Z 0 + Z 1 + Z 2 + 3Z f
(10.63)
In order to obtain symmetrical voltage at the point of fault Equation, (10.63) is
substituted into Eq. (10.54)
36
Single
g Line‐To‐Ground Fault
Eq. (10.58) and (10.62) can be represented by connecting the sequence
networks in series as shown in the following figure.
I a0 = I a1 = I a2 =
1
Ia
3
(10.58)
I a0 =
Ea
Z 0 + Z 1 + Z 2 + 3Z f
(10.62)
37
Line‐To‐Line Fault
Three‐phase generator with a fault through an impedance Zf between phase b
and c.
Ia=0
0
Zs
Zs
Eb
N
Ea
Va
Zs
Ec
Ib
Zf
Ic
Vb
Vc
Assuming the generator is initially on no‐load, the boundary conditions at the
fault point are:
Vb − Vc = Z f I b
(10.64)
Ib + Ic = 0
(10.65)
Ia = 0
(10.66)
38
Line‐To‐Line Fault
Substituting for Ia = 0, and Ic = ‐Ib, the symmetrical components of the currents
from (10.14) are:
⎡ I a0 ⎤
⎡1 1
⎢ 1⎥ 1⎢
⎢ I a ⎥ = 3 ⎢1 a
2
⎢ 2⎥
⎣⎢1 a
⎣I a ⎦
1 ⎤⎡ 0 ⎤
a 2 ⎥⎥ ⎢⎢ I b ⎥⎥
a ⎥⎦ ⎢⎣− I b ⎥⎦
(10.67)
From the above equation
equation, we find that:
I a0 = 0
(10.68)
1
I a1 = (a − a 2 ) I b
3
(10 69)
(10.69)
1
I a2 = (a 2 − a) I b
3
(10.70)
39
Line‐To‐Line Fault
Also, from (10.69) and (10.70), we note that:
I a1 = − I a2
(10 71)
(10.71)
Va = Va0 + Va1 + Va2
From (10.16), we have:
Vb = Va0 + a 2Va1 + aVa2 (10.16)
Vb − Vc = (Va0 + a 2Va1 + aV
Va2 ) − (Va0 + aV
Va1 + a 2Va2 )
= ( a 2 − a )(Va1 − Va2 )
= Z f Ib
Vc = Va0 + aVa1 + a 2Va2
(10.72)
Substituti ng for Va1 and Va2 from (10.54) and noting I a2 = − I a1 , we get :
(a 2 − a)[ E a − ( Z 1 + Z 2 ) I a1 ] = Z f I b
Va0 = 0 − Z 0 I a0
Va1 = Ea − Z 1 I a1
(10.73)
Va2 = 0 − Z 2 I a2
(10.74)
1
I a1 = (a − a 2 ) I b
3
(10.54)
Substituti ng for I b from (10.69), we get :
3I a1
1
2
1
Ea − (Z + Z ) I a = Z f
( a − a 2 )( a 2 − a)
(10.69)
40
Line‐To‐Line Fault
Since (a − a 2 )(a 2 − a) = 3, solving for I a1 results in :
I a1 =
Ea
(Z 1 + Z 2 + Z f )
(10.75)
The phase currents are
⎡ I a ⎤ ⎡1 1
⎢ I ⎥ = ⎢1 a 2
⎢ b⎥ ⎢
⎢⎣ I c ⎦⎥ ⎢⎣1 a
1 ⎤⎡ 0 ⎤
a ⎥⎥ ⎢⎢ I a1 ⎥⎥
a 2 ⎥⎦ ⎢⎣− I a1 ⎥⎦
(10 76)
(10.76)
The fault current is
I b = − I c = ( a 2 − a ) I a1
(10.77)
or
I b = − j 3I a1
(10.78)
41
Line‐To‐Line Fault
Eq. (10.71) and (10.75) can be represented by connecting the positive and negative –
sequence networks as shown in the following figure.
I a1 = − I a2
I a1 =
Ea
(Z 1 + Z 2 + Z f )
42
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 are
load,
Vb = Vc = Z f ( I b + I c )
I a = I a0 + I a1 + I a2 = 0
(10.79)
(10.80)
From (10.16), the phase voltages Vb and Vc are
Figure 10.14
10 14
Double line‐to‐ground fault
43
Double Line‐To‐Ground Fault
Vb = Va0 + a 2Va1 + aVa2
(10.81)
Vc = Va0 + aVa1 + a 2Va2
(10 82)
(10.82)
SinceVb = Vc , from above we note that
Va1 = Va2
(
(10.83)
)
Substituting for the symmetrical components of current in (10.79), we get
V( b ) = Z f ( I a0 + a 2 I a1 + aI a2 + I a0 + aI a1 + a 2 I a2 )
= Z f ( 2 I a0 − I a1 − I a2 )
= 3 Z f I a0
(10.84)
44
Substituti ngg for Vb from ((10.84)) and for Va2 from ((10.83)) into ((10.81),
), we have :
3Z f I a0 = Va0 + (a 2 + a )Va1
= Va0 − Va1
(10.85)
Substituti ng for the symmetrica l components of voltage from (10.54) into (10.85)
and solving for I a0 , we get :
E a − Z 1 I a1
I =− 0
( Z + 3Z f )
0
a
(10 86)
(10.86)
Also, substituting for the symmetrical components of voltage in (10.83), we obtain
E a − Z 1 I a1
I =−
Z2
2
a
(10 87)
(10.87)
Substituti ng for I a0 and I 2a into (10.80) and solving for I a1 , we get :
I a1 =
Ea
Z ( Z 0 + 3Z f )
2
Z +
1
(10.88)
Z 2 + Z 0 + 3Z f
45
Equation (10.86) - (10.88) can be represente d by connecting the positive - sequence
impedance in series with the paralel combinatio n of the negative - sequence
and zero - sequence networks as shown in the equivalent circuit of figure 10.15.
The value of I1a found from (10.86) is substitute d in (10.86) and (10.87),
and I a0 and I a2 are found.
found The phase current are then found from (10.8).
(10 8)
Finally, the fault current is obtained from
I f = I b + I c = 3I a0
((10.89))
Figure 10.15 Sequence network connection for double line‐to‐ground fault
46
EXAMPLE
The one-line diagram of a simple power system is show in Figure 10.16.
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.
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
b. A single line-to-ground fault at bus 3 through a fault impedance Z f = j 0.1 per unit
c A line-to-line fault at bus 3 through a fault impedance Z f = j 0.1
c.
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
G1
G2
T1
T2
L12
L13
L23
Base
MVA
100
100
100
100
100
100
100
Rated
X1
Voltage
20-kV
0.15
20 kV
0.15
20/220 kV 0.10
20/220 kV 0.10
220 kV
0.125
220 kV
0.15
220 kV
0 25
0.25
X2
X0
0.15
0.15
0.10
0.10
0.125
0,15
0 25
0.25
0.05
0.05
0.10
0.10
0.30
0.35
0 7125
0.7125
47
Figure 10.16
10 16
Fault
Item
G1
G2
T1
T2
L12
L13
L23
Base
MVA
100
100
100
100
100
100
100
Rated
X1
Voltage
20-kV
0.15
20 kV
0 15
0.15
20/220 kV 0.10
20/220 kV 0.10
220 kV
0.125
220 kV
0.15
220 kV
0.25
X2
X0
0.15
0 15
0.15
0.10
0.10
0.125
0,15
0.25
0.05
0 05
0.05
0.10
0.10
0.30
0.35
0.7125
48
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
Z 1S =
( j 0.125)( j 0.15)
= j 0.0357143
j 0.525
Z 3S =
Z 2S =
( j 0.125)( j 0.25)
= j 0.0595238
j 0.525
( j 0.15)( j 0.25)
= j 0.0714286
j 0.525
49
( j 0 .2857143 )( j 0.3095238 )
+ j 0 .0714286
j 0.5952381
= j 0.22
1
Z 33
=
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)
( j 0.30)( j 0.35)
Z 1S =
= j 0.0770642
j1.3625
Z 2S =
( j 0.30)( j 0.7125)
= j 0.1568807
j1.3625
Z 3S =
( j 0.35)( j 0.7125)
= j 0.1830257
j1.3625
Fig. 10.19Zero-sequence impedance
j0.077064
51
Combiningg the p
parallel branches, the zero‐sequence
q
thevenin impedance
p
is
( j 0.4770642 )( j 0.2568807 )
+ j 0.1830275
j 0.7339449
= j 0.35
Z 330 =
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 bus 3
Assuming the no‐load generated emfs are equal to 1.0 per unit, the fault
current is
I 3 (F) =
V3a( 0)
Z133 + Z f
=
1.0
= -j3.125
j3 125 pu = 820.1
820 1∠ - 90° A
j0.22 + j 0.1
(b) Single line‐to ground fault at bus 3
F
From
(10
(10.62),
62) the
th sequence componentt off the
th fault
f lt currentt are
I 30 = I13 = I 32 =
V3a( 0)
2
0
Z133 + Z 33
+ Z 33
+ 3Z f
=
1.0
= -j0.9174 pu
j0.22 + j0.22 + j 0.35 + 3(j0.1)
The fault current is :
⎡ I 3a ⎤ ⎡1 1
⎢ b⎥ ⎢
2
=
I
1
a
⎢ 3⎥ ⎢
⎢ I 3c ⎥ ⎢⎣1 a
⎣ ⎦
1 ⎤ ⎡ I 30 ⎤ ⎡3I 30 ⎤ ⎡− j 2.7523⎤
⎢ ⎥ ⎢ ⎥
⎥ pu
a ⎥⎥ ⎢ I 30 ⎥ = ⎢ 0 ⎥ = ⎢⎢
0
⎥
2
0
⎥⎦
a ⎥⎦ ⎢⎣ I 3 ⎥⎦ ⎢⎣ 0 ⎥⎦ ⎢⎣
0
53
(c) Line‐to Line fault at bus 3
The zero‐sequence component of current is zero, i.e.,
I 30 = 0
The positive‐and negative‐sequence components of the fault current are
I13 = −I 32 =
V3a( 0)
2
0
Z133 + Z 33
+ Z 33
+ Zf
=
1.0
= -j1.8519 pu
j0.22 + j0.22 + j0.1
The fault current is
⎡ I 3a ⎤ ⎡1 1
⎢ b⎥ ⎢
2
1
I
a
=
3
⎢ ⎥ ⎢
⎢ c⎥
⎣ I 3 ⎦ ⎢⎣1 a
1⎤⎡
0
⎤ ⎡ 0
⎤
a ⎥⎥ ⎢⎢− j1.8519⎥⎥ = ⎢⎢− 3.2075⎥⎥ pu
a 2 ⎥⎦ ⎢⎣− j1.8519⎥⎦ ⎢⎣− 3.2075⎥⎦
(d) Double Line‐to Line‐fault at bus 3
The positive‐sequence component of the fault current is
I =
1
3
V3a( 0)
2
0
Z 33
( Z 33
+ 3Z f )
Z + 2
0
Z 33 + Z 33
+ 3Z f )
1
33
=
1.0
= -j2.6017 pu
j0.22 (j0.35 + j0.3)
j0.22 +
j0.22 + j0.35 + j0.3
54
The negative‐sequence component of current is :
I 32 = −
V3a( 0) − Z133 I133
2
Z 33
=−
1.0 − ( j 0.22)(− j 2.6017)
= j1.9438 pu
j0.22
Th zero‐sequence componentt off currentt is:
The
i
I 30 = −
V3a( 0 ) − Z133 I133
0
Z 33
+ 3Z f
=−
1.0 − ( j 0.22)(− j 2.6017)
= j0.6579 pu
j0 35 + j0.3
j0.35
j0 3
And the phase currents are :
⎡ I 3a ⎤ ⎡1 1
⎢ b⎥ ⎢
2
=
I
1
a
3
⎢ ⎥ ⎢
⎢ I 3c ⎥ ⎢⎣1 a
⎣ ⎦
1 ⎤ ⎡ j 0.6579 ⎤ ⎡
0
⎤
a ⎥⎥ ⎢⎢− j 2.6017 ⎥⎥ = ⎢⎢4.058∠165.93°⎥⎥ pu
a 2 ⎥⎦ ⎢⎣ j1.9438 ⎥⎦ ⎢⎣ 4.058∠14.07° ⎥⎦
And the fault currents is:
I 3 ( F ) = I 3b + I 3c = 1.9732∠90°
55
UNBALANCED FAULT ANALYSIS USING BUS IMPEDANCE MATRIX
Single Line‐to‐Ground Fault Using Zbus
Vk (0)
I =I =I = 1
Z kk + Z kk2 + Z kk0 + 3Z f
0
k
1
k
2
k
(10.90)
Where Z1kk , Z 2kk and Z okk are the diagonal elements in the k axis of the correspond ing
bus impedance matrix and Vk (0)is the prefault voltage at bus k.
The fault phase current is :
(10.91)
I kabc = A I k
012
Line‐to‐Line Fault Using Zbus
I k0 = 0
I k1 = − I k2 =
(10.92)
Vk (0)
Z kk1 + Z kk2 + Z f
(10.93)
56
Double Line‐to‐Ground Fault Using Zbus
I k1 =
Z kk1 +
Vk (0)
Z kk2 ( Z kk0 + 3Z f )
(10.94)
Z kk2 + Z kk0 + 3Z f
Vk (0) − Z kk1 I k1
I =−
Z kk2
(10.95)
Vk (0) − Z kk1 I k1
I =−
Z kk0 + 3Z f
(10.96)
2
k
0
k
Where Z1kk , and Z 2kk , and Z okk are the diagonal elements in the k axis of the correspond ing
bus impedance
p
matrix. The pphase currents are obtained from ((10.91),
), and the result current is
I k ( F ) = I kb + I kC
(10.97)
57