lec6 power system
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POWER SYSTEMS ILecture 6
06-88-590-68
Electrical and Computer Engineering
University of Windsor
Dr. Ali Tahmasebi
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Analysis of Unsymmetric Systems
l Except for the balanced three-phase fault, faults
result in an unbalanced system.
l The most common types of faults are single line-
ground (SLG) and line-line (LL). Other types aredouble line-ground (DLG), open conductor, and
balanced three phase.
l System is only unbalanced at point of fault!
l The easiest method to analyze unbalanced system
operation due to faults is through the use of
symmetrical components
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Symmetric Components
l The key idea of symmetrical component analysis is
to decompose the system into three sequence
networks. The networks are then coupled only at
the point of the unbalance (i.e., the fault)
l The three sequence networks are known as the
– positive sequence (this is the one we’ve been using)
– negative sequence
– zero sequence
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Positive Sequence Sets
l The positive sequence sets have three phase
currents/voltages with equal magnitude, with phase
b lagging phase a by 120°, and phase c lagging
phase b by 120°.l We’ve been studying positive sequence sets (also
known as ‘abc’ sequence)
Positive sequencesets have zero
neutral current
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Negative Sequence Sets
l The negative sequence sets have three phase
currents/voltages with equal magnitude, with phase
b leading phase a by 120°, and phase c leading
phase b by 120°.l Negative sequence (also known as ‘acb’ sequence)
sets are similar to positive sequence, except the
phase order is reversed
Negative sequence
sets have zero
neutral current
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Sequence Set Representation
Any arbitrary set of three phasors (vectors), say Ia, I b, Iccan be represented as a sum of the three sequence sets
0
0
0
0 0 0
where
, , is the zero sequence set
, , is the positive sequence set
, , is the negative sequence set
a a a a
b b b b
c c c c
a b c
a b c
a b c
I I I I
I I I I
I I I I
I I I
I I I
I I I
+ -
+ -
+ -
+ + +
- - -
= + +
= + +
= + +
I a
I b
I c
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Conversion from Sequence to Phase
0a
2 3 3
0 0 0a b c
2
Only three of the sequence values are unique,
I , , ; the others are determined as follows:
1 120 0 1
I I I (since by definition they are all equal)
a a
b a c a b a c
I I
I I I I I I I
a a a a a
a a a
+ -
+ + + + - - +
Ð ° + + = =
= =
= = = =
2
0
0 + 2 2a a
2 2
1 1 1 11 1I 1 I 1
1 1
a
aa
b a a
c a
I
I I I I I
I I
a
a a a a
a a a a
-
- +
-
é ùé ù é ùé ùé ù é ù ê úê ú ê úê úê ú ê ú= + + = ê úê ú ê úê úê ú ê ú
ê úê ú ê úê úê ú ê úë û ë û ë ûë û ë ûê úû
=
-
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Conversion Sequence to Phase
2
2
0 0
Define the symmetrical components transformation
matrix
1 1 1
1
1
Then
aa
b a s
c a
I I I
I I I
I I I
a a
a a
+ +
- -
é ù
ê ú= ê úê úë û
é ù é ùé ù
ê ú ê úê ú= = = =ê ú ê úê úê ú ê úê úë û ê ú ê úû ë û
A
I A A A I
Is :
sequence
currents
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Conversion Phase to Sequence
1
1 2
2
By taking the inverse we can convert from the
phase values to the sequence values
1 1 11
with 13
1
Sequence sets can be used with voltages as well
as with currents
s
a a
a a
-
-
=
é ùê ú
= ê úê úë û
I A I
A
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Conversion Phase to Sequence
It can be easily derived that: =1
3 + +
In any Y-connected 3-phase system: I n = I a + I b + I c , so:
I n = 3 I 0
If this system is also balanced, then I n = I 0 = 0
(Also, any system with no neutral path will have I n = I 0 = 0)
NOTE: Some books use the notation 0, 1 and 2 instead of ±
signs. So for example instead of I 0, I 1 and I 2, we will have I 0, I 1and I 2, respectively.
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Symmetrical Component Example 1
1 2s
2
s
10 0
Let 10 Then
10
1 1 1 10 01
1 10 10 03
10 01
10 0 0
If 10 0
10 10 0
a
b
c
I
I
I
a a
a a
-
Ð °é ù é ùê ú ê ú= = Ð - 120°ê ú ê ú
Ð 120°ê ú ê úë û ë û
é ù Ð ° 0é ù é ùê úê ú ê ú= = Ð - 120° = Ð °ê úê ú ê úê ú Ð 120°ê ú ê úû ë ûë û
Ð °é ù é ùê ú ê ú= Ð +120° ® =ê ú ê ú
Ð - 120° Ð °ê ú ê úû ë û
I
I A I
I I
(‘abc’ or positive sequence), then:
(‘acb’ or negative sequence)
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Symmetrical Component Example 2
1 2s
2
0
Let
Then
1 1 1 0 01
13 6.121
a
b
c
V
V
V
a a
a a
-
5Ð 9 °é ù é ùê ú ê ú= = 8Ð 150°ê ú ê ú
8Ð - 30°ê úê ú ë ûë û
é ù 5Ð 9 ° 1.67Ð 9 °é ù é ùê úê ú ê ú= = 8Ð 150° = 3.29Ð - 135°ê ú
ê ú ê úê ú 8Ð - 30° Ð 68°ê ú ê úû ë ûû
V
V A V
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Symmetrical Component Example 3
0
2
2
10 0
Let 10
Then
1 1 1 10 0
1 10
1
s
s
I
I
I
a a
a a
+
-
é ù Ð °é ùê ú ê ú= = - Ð 0°ê ú ê úê ú 5Ð 0°ê ú
ë ûê úë û
é ù Ð ° 5.0Ð 0°é ù é ùê úê ú ê ú
= = - Ð 0° = 18.0Ð 46.1°ê úê ú ê úê ú 5Ð 0° 18.0Ð - 46.1°ê ú ê úû ë ûë û
I
I AI
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Symmetrical Component Example 4
Phase b open:
=
=
10Ð 0°
0
10Ð 120°
(this is obviously an
unbalanced system)
=
=
3.33Ð 60°
6.66Ð 0°3.33Ð − 60°
I n = I a + I b + I c = 10Ð 0° + 0 + 10Ð 120° = 10Ð 60° = 3 I 0
(as we expected!)
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Use of Symmetrical Components
Consider this balanced, wye-connected impedance load:
( )
( )
( )
n a b c
ag a y n n
ag Y n a n b n c
bg n a Y n b n c
cg n a n b Y n c
I I I I
V I Z I Z
V Z Z I Z I Z I
V Z I Z Z I Z I
V Z I Z I Z Z I
= + +
= +
= + + += + + +
= + + +
ag y n n n a
bg n y n n b
ccg n n y n
V Z Z Z Z I
V Z Z Z Z I
I V Z Z Z Z
é ù é ù+ é ùê ú ê úê ú= +ê ú ê úê úê ú ê úê ú+ ûû ë û
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Use of Symmetrical Components
1
13 0 0
0 0
0 0
ag y n n n a
bg n y n n b
ccg n n y n
s s
s s s s
y n
y
y
V Z Z Z Z I
V Z Z Z Z I
I V Z Z Z Z
Z Z
Z
Z
-
-
é ù é ù+ é ùê ú ê úê ú= +ê ú ê úê úê ú ê úê ú+ ûë û ë û
= = =
= ® =
é ù+ê ú= ê ú
ê úû
V Z I V A V I A I
A V Z A I V A Z A I
A Z A
( Z is called the phase impedance matrix)
= Zs = sequence
impedance
matrix
® Vs = Zs Is
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Networks are Now Decoupled
0 0
0 0
3 0 0
0 00 0
Systems are decoupled
( 3 )
y n
y
y
y n y
y
V I Z Z
V Z I Z V I
V Z Z I V Z I
V Z I
+ +
- -
+ +
- -
é ù é ùé ù+ê ú ê úê ú
=ê ú ê úê úê ú ê úê úê ú ê úë ûë û ë û
= + ==
(because Zs is a diagonal matrix):
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Sequence Networks
® I0
+
-
V0Zy
3Zn
Z0 = Zy + 3ZnZero-sequence network:
® I+
+
-V+ Zy Z
+ = ZyPositive-sequence network:
®I
-
+
-V- Zy Negative-sequence network: Z- = Zy
Z0 = zero-sequence
impedance
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Grounding
l When studying unbalanced system operation how a
system is grounded can have a major impact on the
fault flows
l Ground current only impacts zero sequence system
l In previous example if load was ungrounded the
zero sequence network is (with Zn equal infinity):
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Grounding, cont’d
l Voltages are always defined as a voltage
difference. The ground is used to establish the
zero voltage reference point
– ground need not be the actual ground (e.g., an airplane)
l During balanced system operation we can ignore
the ground since there is no neutral current
l There are two primary reasons for grounding
electrical systems1. safety
2. protect equipment
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How good a conductor is dirt?
l There is nothing magical about an earth ground. All
the electrical laws, such as Ohm’s law, still apply.
l Therefore to determine the resistance of the ground
we can treat it like any other resistive material:
8
8
onductor lengthResistance
cross sectional area
2.65 10 -m for aluminum1.68 10 -m for copper
where is the resistivity
c R
r
r
r
r
-
-
´=
= ´ W
= ´ W
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How good a conductor is dirt?
8
16
2.65 10 -m for aluminum
5 10 -m for quartz (insulator!)
160 -m for top soil
900 -m for sand/gravel
20 -m for salt marsh
What is resistance of a mile long, one inch diameter,
circular wir
r
r
r
r
r
-= ´ W
= ´ W
= W
= W
= W
8
2
e made out of aluminum ?
2.65 10 1609 R= 0.083
mile0.0128
-´ ´ W=´
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How good a conductor is dirt?
6
2
What is resistance of a mile long, one inch diameter,
circular wire made out of topsoil?
160 1609 R= 500 10
mile0.0128
In order to achieve 0.08 with our dirt wiremile
we would need a cross sectio
p
´ W= ´
´W
6 2
nal area of
160 1609 3.2 10 (i.e., a radius of about 1000 m)0.08
But what the ground lacks in , it makes up for in A!
m
r
´ = ´
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Calculation of grounding resistance
• Because of its large cross sectional area the earth is
actually a pretty good conductor.
• Devices are physically grounded by having aconductor in physical contact with the ground;
having a fairly large area of contact is important.
• Most of the resistance associated with establishing
an earth ground comes within a short distance of
the grounding point.
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Sequence Networks of Balanced- load
AD-connected
balanced load and its
equivalent Y-connected
load ( =∆
).
® I0 = 0
+
-
V0∆
3
Z0 = ¥
® I++
-V+
Z+ = ∆
® I-+
-V-
Z- = ∆
Sequence networks:
∆
3∆
3
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Sequence Networks of Generators
Key point: generators only
produce positive sequence
voltages; therefore only the
positive sequence has avoltage source.
Sequence networks:
¬ I0
+
-
V0
¬I
+
+
-
V+
¬ I-
+
-V-
Zg0
3Zn Zg+
Eg+
Zg-
Zero-sequence network Positive-sequence network Negative-sequence network
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• During a fault Zg+ » Zg
-» Xd
².
• The zero sequence impedance is usually
substantially smaller.
• The value of Zn depends on whether the generator
is grounded.
Sequence Networks of Generators, cont’d
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Sequence Networks of Motors
• The sequence networks of synchronous motors are the
same as the synchronous generators (with a source Em+ in
the positive-sequence network), except the currents go
into the networks.
• The sequence networks of the induction motors are the
same as the synchronous motors, except there is no
voltage source (Em+ = 0).
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Sequence Networks of Transformers
Ideal Y-Y transformer:
Sequence networks:® I
H
+ or IH
-
+
-
EH+ or EH
-
+
-
Positive and Negative-sequence network (p.u.)Zero-sequence network (p.u.)
® IX+ or IX
-
EX+ or EX
-
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• When IA0 = IB
0 = IC0 = I0 is applied to an ideal
transformer, I N = 3I0 flows through Z N and also
through Zn.
• Note that if either one of the neutrals is ungrounded,
then no zero sequence can flow in either the high- or
low-voltage windings. For example, if the high-
voltage winding has an open neutral, then I N = 3I0 = 0,
which in turns forces I0
= 0 on the low-voltage side.
Sequence Networks of Transformers, cont’d
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Sequence
Networks of
Practical
Transformers
(p.u.)
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Sequence Networks of Transformers, cont’d
Notes on sequence networks of practical transformers:
• In (a) double-grounded Y-Y: positive- and negative
sequence networks are the same as p.u. transformer model,
and 3Z N and 3Zn are added only in zero-sequence network.
• In (b) grounded-Y/D: positive- and negative sequence
networks are the same as transformer model. Zero-
sequence currents can enter Y side if it is grounded and
flow through the winding and also the Dwinding (since itwill be inducted) but cannot leave the Dside.
• In (c) D- D: zero-sequence currents cannot enter or leave D
windings, but can circulate within them.
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Power in Sequence Networks
Total complex power delivered to a 3-phase load, in
terms of phase voltages and currents is:
= ∗ +
∗ + ∗
= .∗ = ∗ = ( ∗)∗
∗ =
3 0 0
0 3 0
0 0 3
→ = 3
∗
→ = 3 ∗ + ∗ + ∗ = 3
Ss = total complex power delivered to sequence networks.
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