El-Hawary M.E. Electrical Energy Systems
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Figure 7.14 Zero Sequence Equivalent Circuit for a Three-Phase Transformer Bank Connected in
Wye-Wye with Neutrals Grounded.
Wye-wye Bank, Both Neutrals Grounded
With both wyes grounded, zero sequence current can flow. The
presence of the current in one winding means that secondary current exists in the
other. Figure 7.14 illustrates the situation.
Sequence Impedances for Synchronous Machines
For a synchronous machine, sequence impedances are essentially
reactive. The positive, negative, and zero sequence impedances have in general
different values.
Positive Sequence Impedance
Depending on the time interval of interest, one of three reactances may
be used:
1. For the subtransient interval, we use the subtransient reactance:
d
XjZ
′′
=
+
2. For the transient interval, we use the corresponding reactance:
d
XjZ
′
=
+
3. In the steady state, we have
d
jXZ
=
+
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Negative Sequence Impedance
The MMF produced by negative sequence armature current rotates in a
direction opposite to the rotor and hence opposite to the dc field winding.
Therefore the reactance of the machine will be different from that for the
positively rotating sequence.
Zero Sequence Impedance
The zero sequence impedance of the synchronous machine is quite
variable and depends on the nature of the stator windings. In general, these will
be much smaller than the corresponding positive and negative sequence
reactance.
Sequence Impedances for a Transmission Link
Consider a three-phase transmission link of impedance Z
L
per phase.
The return (or neutral) impedance is Z
N
. If the system voltages are unbalanced,
we have a neutral current I
N
. Thus,
CBAN
IIII
++=
The voltage drops
∆
V
A
,
∆
V
B
, and
∆
V
C
across the link are as shown below:
NNLCC
NNLBB
NNLAA
ZIZIV
ZIZIV
ZIZIV
+=∆
+=∆
+=∆
In terms of sequence voltages and currents, we have
()
NL
L
L
ZZIV
ZIV
ZIV
3
00
+=∆
=∆
=∆
−−
++
Therefore the sequence impedances are given by:
L
L
NL
ZZ
ZZ
ZZZ
=
=
+=
+
−
3
0
The impedance of the neutral path entered into the zero sequence impedance in
addition to the link’s impedance Z
L
. However, for the positive and negative
sequence impedances, only the link’s impedance appears.
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Figure 7.15 System for Example 7.3.
Example 7.3
Draw the zero sequence network for the system shown in Figure 7.15.
Solution
The zero sequence network is shown in Figure 7.16.
Figure 7.16 Zero Sequence Network for Example 7.3.
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Example 7.4
Obtain the sequence networks for the system shown in Figure 7.17. Assume the
following data in p.u. on the same base.
Generator G
1
: X
+
= 0.2 p.u.
X
-
= 0.12 p.u.
X
0
= 0.06 p.u.
Generator G
2
: X
+
= 0.33 p.u.
X
-
= 0.22 p.u.
X
0
= 0.066 p.u.
Transformer T
1
: X
+
= X
-
= X
0
= 0.2 p.u.
Transformer T
2
: X
+
=X
-
= X
0
= 0.225 p.u.
Transformer T
3
: X
+
= X
-
= X
0
= 0.27 p.u.
Transformer T
4
: X
+
= X
-
= X
0
= 0.16 p.u.
Line L
1
: X
+
= X
-
= 0.14 p.u.
X
0
= 0.3 p.u.
Line L
2
: X
+
= X
-
= 0.20 p.u.
X
0
= 0.4 p.u.
Line L
3
: X
+
= X
-
= 0.15 p.u.
X
0
= 0.2 p.u.
Load: X
+
= X
-
= 0.9 p.u.
X
0
= 1.2 p.u.
Assume an unbalanced fault occurs at F. Find the equivalent sequence networks
for this condition.
Figure 7.17 Network for Example 7.4.
Solution
The positive sequence network is as shown in Figure 7.18(A). One step in the
reduction can be made, the result of which is shown in Figure 7.18(B).Toavoid
tedious work we utilize Thévenin’s theorem to obtain the positive sequence
network in reduced form. We assign currents I
1
, I
2
, and I
3
as shown in Figure
7.18(B) and proceed to solve for the open-circuit voltage between F
+
and N
+
.
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Figure 7.18 Positive Sequence Network for Example 7.4.
Consider loop A. We can write
()()
[]
21311
9.036.02.001 IIIIIj
++−+=∠
For loop B, we have
()()
[]
31323
36.042.0565.00 IIIIIj
−−++=
For loop C, we have
()()
[]
21322
9.042.033.001 IIIIIj
++++=∠
The above three equations are rearranged to give
()
()
321
321
321
42.065.19.001
345.142.036.00
36.09.046.101
IIIj
III
IIIj
++=∠
−−=
−+=∠
Solving we obtain
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© 2000 CRC Press LLC
0247.0
3357.0
4839.0
3
2
1
jI
jI
jI
−=
−=
−=
Figure 7.19 Steps in Positive Sequence Impedance Reduction.
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Figure 7.19 (Cont.)
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© 2000 CRC Press LLC
Figure 7.20 Positive Sequence Network Equivalent for Example 7.4.
Figure 7.21 Steps in Reduction of the Negative Sequence Network for Example 7.4.
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Figure 7.21 (Cont.)
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© 2000 CRC Press LLC
Figure 7.22 Steps in Reducing the Zero Sequence Network for Example 7.4.
As a result, we get