Introduction To Symmetrical Components - MIT OpenCourseWare

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13 Οκτ 2013 (πριν από 3 χρόνια και 9 μήνες)

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The transformation is de ned as:
2
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1 a a
2

1
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V
a
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6
V
76 7
4
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= 1 a a (4)
3
6
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5
where the complex number a is:
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2
a
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This transformation may be used for both voltage and current, and works for variables in
ordinary form as well as variables that have been normalized and are in per-unit form. The inverse
of this transformation is:
2 3 2 32 3
V
a
1 1 1 V
1
6 7 6
2
76 7
4
V
b
5
=
4
a a 1
54
V
2
5
(8)
V a a
2
1 V
0
c
The three component variables V
1
, V
2
, V
0
are called, respectively, positive sequence, negative
sequence and zero sequence. They are called symmetrical components because, taken separately,
they transform into symmetrical sets of voltages. The properties of these components can be
demonstrated by tranforming each one back into phase variables.
Consider rst the positive sequence component taken by itself:
V
1
= V (9)
V
2
= 0 (10)
V
0
= 0 (11)
yields:
V
a
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a
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2
V = a
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V or v
b
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) (13)
b
3
2
V
c
= aV or v
c
= V cos(!t +
) (14)
3
This is the familiar balanced set of voltages: Phase b lags phase a by 120

, phase c lags phase
b and phase a lags phase c.
The same transformation carried out on a negative sequence voltage:
V
1
= 0 (15)
V
2
= V (16)
V
0
= 0 (17)
2
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s
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s
s
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n
:
V
p
h1

V
p
h2
=
j
!
L
ph
I
ph
(
2
7
)
w
h
e
r
e
2
3
V
a
6
7
V
ph
=
4
V
b
5
(
2
8
)
V
c
2
3
I
a
6
7
I
ph
=
4
I
b
5
(
2
9
)
I
c
2
3
L
M
M
6
7
L
ph
=
4
M
L
M
5
(
3
0
)
M
M
L
N
o
t
e
t
h
a
t
t
h
e
s
y
m
m
e
t
r
i
c
a
l
c
o
m
p
o
n
e
n
t
t
r
a
n
s
f
o
r
m
a
t
i
o
n
(
4
)
m
a
y
b
e
w
r
i
t
t
e
n
i
n
c
o
m
p
a
c
t
f
o
r
m
:
V
s
=
T
V
p
(
3
1
)
4
where
2
1 a a
2
1

3
T =
6
1 a
2
a
7
(32)
3
4
1 1 1
5
and V
s
is the vector of sequence voltages:
2
V
1
3
V
s
=
6
V
2
(33)
4
V
7
0
5
Rewriting (27) using the inverse of (31):
T
1
V
s1
T
1
V
1
s2
= j!L T I (34)
ph
s
Then transforming to get sequence voltages:
V
s1

1
V

s2
= j!TL T I
s
(35)
ph
The sequence inductance matrix is de ned by carrying out the operation indicated:
L = TL T
1
(36)
s ph
which is:
2
L M 0 0
3
L =
s
6
0 L M 0 (37)

4
0 0 L+2M
7
5
Thus the coupled set of expressions which described the transmission line in phase variables becomes
an uncoupled set of expressions in the symmetrical components:
V
11
V
12
= j!(L M)I
1
(38)
V
21
V
22
= j!(L M)I
2
(39)
V
01
V
02
= j!(L+2M)I
0
(40)
The positive, negative and zero sequence impedances of the balanced transmission line are then:
Z
1
= Z
2
= j!(L M) (41)
Z
0
= j!(L+2M) (42)
So,in analysis of networks with transmission lines,it is now possible to replace the lines with three
independent, single- phase networks.
Consider next a balanced three-phase load with its neutral connected to ground through an
impedance as shown in Figure 3.
The symmetrical component voltage-current relationship for this network is found simply, by
assuming positive, negative and zero sequence currents and nding the corresponding voltages. If
this is done, it is found that the symmetrical components are independent, and that the voltage-
current relationships are:
V
1
= ZI
1
(43)
V
2
= ZI
2
(44)
V
0
= (Z +3Z
g
)I
0
(45)
5
4
a
b
c
Z
Z
Z
Z
g
F
i
g
u
r
e
3
:
B
a
l
a
n
c
e
d
L
o
a
d
W
i
t
h
N
e
u
t
r
a
l
I
m
p
e
d
a
n
c
e
U
n
b
a
l
a
n
c
ed
S
o
u
r
c
es
C
o
n
s
i
d
e
r
t
h
e
n
e
t
w
o
r
k
s
h
o
w
n
i
n
F
i
g
u
r
e
4
.
A
b
a
l
a
n
c
e
d
t
h
r
e
e
-
p
h
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s
e
r
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s
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d
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b
a
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a
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c
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i
n
e
(
w
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t
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m
u
t
u
a
l
c
o
u
p
l
i
n
g
b
e
t
w
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e
n
p
h
a
s
e
s
)
.
A
s
s
u
m
e
t
h
a
t
o
n
l
y
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h
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s
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o
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t
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t
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o
u
r
c
e
i
s
w
o
r
k
i
n
g
,
s
o
t
h
a
t
:
V
a
=
V
(
4
6
)
V
b
=
0
(
4
7
)
V
c
=
0
(
4
8
)
T
h
e
o
b
j
e
c
t
i
v
e
o
f
t
h
i
s
e
x
a
m
p
l
e
i
s
t
o

n
d
c
u
r
r
e
n
t
s
i
n
t
h
e
t
h
r
e
e
p
h
a
s
e
s
.
L
V
a
V
b
V
c
R
R
R
M
M
M
L
L
+
+
-
-
+
-
F
i
g
u
r
e
4
:
B
a
l
a
n
c
e
d
L
o
a
d
,
B
a
l
a
n
c
e
d
L
i
n
e
,
U
n
b
a
l
a
n
c
e
d
S
o
u
r
c
e
T
o
s
t
a
r
t
,
n
o
t
e
t
h
a
t
t
h
e
u
n
b
a
l
a
n
c
e
d
v
o
l
t
a
g
e
s
o
u
r
c
e
h
a
s
t
h
e
f
o
l
l
o
w
i
n
g
s
e
t
o
f
s
y
m
m
e
t
r
i
c
a
l
c
o
m
p
o

n
e
n
t
s
:
V
V
1
=
3
(
4
9
)
V
V
2
=
3
(
5
0
)
V
V
0
=
3
(
5
1
)
6
Figure 5: Sequence Networks
Currents are:
V
I
1
=
3(j!(L M) +R)
V
I
2
=
3(j!(L M) +R)
I
0
= 0
Phase currents may now be re-assembled:
I
a
= I
1
+ I
2
+ I
0
2
I
b
= aI
1
+ aI
2
+ I
0
2
I
c
= aI
1
+ aI
2
+ I
0
or:
2V
I
a
=
3(j!(L M) +R)

(a
2
+ a)V
I
b
=
3(j!(L M) +R)
7




Next, the network facing the source consists of the line, with impedances:
Z
1
= j!(L

M) (52)
Z
2
= j!(L

M) (53)
Z
0
= j!(L+2M) (54)
and the three- phase resistor has impedances:
Z
1
= R (55)
Z
2
= R (56)
Z
0
= (57)
1
Note that the impedance to zero sequence is in nite because the neutral is not connected back
to the neutral of the voltage source. Thus the sum of line currents must always be zero and this
in turn precludes any zero sequence current. The problem is thus described by the networks which
appear in Figure 5.
j!(L

M) j!(L

M) j!(L+2M)
R

^ ^ ^
_ _
\\\\
\\\\
\\\\
<
>
<
>
<
<
>
<
>
<

+

V
3
+
V
+
V
R
R
3
3

Positive Negative Zero
5
6

V
=
3
(
j
!
(
L

M
)
+
R
)
(
a
+
a
2
)
V
I
=
c
3
(
j
!
(
L

M
)
+
R
)

V
=
3
(
j
!
(
L

M
)
+
R
)
(
N
o
t
e
t
h
a
t
w
e
h
a
v
e
u
s
e
d
a
2
+
a
=

1
)
.
R
o
t
a
t
i
ng
M
a
c
h
i
n
es
S
o
m
e
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t
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e
m
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s
a
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m
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r
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a
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d
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w
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d
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w
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s
.
T
h
i
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t
h
e
c
a
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,
f
o
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x
a
m
p
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f
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a
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d
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v
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a
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a
v
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m
i
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p
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.
F
o
r
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s
w
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b
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p
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b
y
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p
h
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r
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,
b
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q
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b
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w
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S
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c
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,
a
s
s
h
o
w
n
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n
F
i
g
u
r
e
6




+

E
1
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7
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1
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2
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0
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0
(
5
8
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7
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D
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F
a
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c
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d
,
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t
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f
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t
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:
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b
=
0
V
=
0
c
I
=
0
a
T
h
e
n
,
u
s
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n
g
t
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t
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1
V
1
=
V
2
=
V
0
=
a
V
3
C
o
m
b
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n
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n
g
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f
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m
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t
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a
=
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1
+
I
2
+
I
0
=
0
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h
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d
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s
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a
t
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h
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n
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n
F
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g
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r
e
1
2
.
1
0
Sequence
Positive
I I
Sequence
Negative
I
1 2
0
+
+
Sequence
Zero
+
V
1
V
2
V
0
- - -
 
I
1
+
Positive
Sequence
V
1
-
I
2
+
Negative
Sequence
V
2
-
I
0
+
Zero
Sequence
V
0
-
Figure 11: Sequence Connection For A Single-Line-To-Ground Fault
Figure 12: Sequence Connection For A Double-Line-To-Ground Fault
7.3
Line-Line Fault
If phases b and c are shorted together but not grounded,
V
b
=
V
c
I
b
=
I
c
I
a
=
0
Expressing these in terms of the symmetrical components:
V
1
= V
2
1
2
=
a + a V
b
3
I
0
= I
a
+ I
b
+ I
c
= 0
I
a
= I
1
+ I
2
11
=
0
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p
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k
s
,
a
s
s
h
o
w
n
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n
F
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g
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r
e
1
3
.
V
2
Sequence
Negative
-
+
I
2
Sequence
Positive
I
1
V
1
-
+
F
i
g
u
r
e
1
3
:
S
e
q
u
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n
c
e
C
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n
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=
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(
j
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1
5
j
j
j
:
1
2
)
+
j
:
2

j
3
:
0
1
3
-
i
i
1B
5
?



j:1



j:12
 

+




1
i
2B
?

j:15

 
 
j:12
 
i
0B
?
j:2




Figure 16: Completed Network For Single Line-Ground Fault
Then the sequence currents at the breaker are:
i
1B
= i
2B
j:12
= i

j:12 +j:15
=

j1:33
i
0B
= i
=

j3:0
The phase currents are re-constructed using:
i
a
= i
1B
+ i
2B
+ i
0B
i
b
= a
2
i
1B
+ ai
2B
+ i
0B
i
c
= ai
1B
+ a
2
i
2B
+ i
0B
These are:
i =

j5:66 per-unit
a
i
b
=

j1:67 per-unit
i =

j1:67 per-unit
c
7.4.3
Double Line-Ground Fault
For the double line-ground fault, the networks are in parallel, as shown in Figure 17.
14
i
1B
6
j:15 j:12 i
2B
6
j:15 j:12 i
0B
6
j:2


+

1




















6
i

To start, nd the source current i:
1
i =
j(:15jj:12) +j(:15jj:12jj:2)
= j8:57
Then the sequence currents at the breaker are:
j:12
i
1B
= i 
j:12 +j:15
= j3:81
j:12 j
i
2B
= i
jj:2

j:12jjj:2 +j:15
= j2:86
j:12
i
0B
= i
jjj:15

j:2 +j:12jjj:15
= j2:14
Reconstructed phase currents are:
i
a
= j1:19
1
p
3
i
b
= i
0B
(i
1B
+ i
2
2
B
) j(i
1B
i
2B
)
2
= j2:67 5:87
1
p
3
i
c
= i
0B
(i
1B
+ i
2B
) + j(i
1
2 2
B
i
2B
)
= j2:67 +5:87
ji
a
j = 1:19 per-unit
ji
b
j = 6:43 per-unit
ji
c
j = 6:43 per-unit
15
Figure 17:Completed Network For Double Line-Ground Fault
7.4.4 Line-Line Fault
The situation is even easier here, as shown in Figure 18
i
1B
6

j:15

 
 
j:12 i
 
2B
6



j:15



j:12
 

+


1
i
6

Figure 18: Completed Network For Line-Line Fault
The source current i is:
1
i =
2 j(:15jj:12)
= j7:50
and then:
i
1B
= i
2B
j:12
= i
j:12 +j:15
= j3:33
Phase currents are:
i
a
= 0
1
p
3
i
b
= (i
1B
+ i
2B
) j (i
1B
i
2

2B
)
2
ji
b
j = 5:77 per-unit
ji
c
j = 5:77 per-unit
7.4.5 Conversion To Amperes
Base current is:
P
B
I
B
=
p
= 418:4A
3V
Bl l
Then current amplitudes are, in Amperes, RMS:
16
P
h
a
s
e
A
P
h
a
s
e
B
P
h
a
s
e
C
T
h
r
e
e
-
P
h
a
s
e
F
a
u
l
t
2
7
9
1
2
7
9
1
2
7
9
1
S
i
n
g
l
e
L
i
n
e
-
G
r
o
u
n
d
,

a
2
3
6
8
6
9
9
6
9
9
D
o
u
b
l
e
L
i
n
e
-
G
r
o
u
n
d
,

b
,

c
4
9
8
2
6
9
0
2
6
9
0
L
i
n
e
-
L
i
n
e
,

b
,

c
0
2
4
1
4
2
4
1
4
1
7
MIT OpenCourseWare
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6.061 / 6.690 Introduction to Electric Power Systems
Spring 2011
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