Symmetrical Components
à Fortescue’s Theorem
o 3 unbalanced phasors of a 3-phase system can be
resolved into 3 balanced systems of phasors.
The balanced sets of components are:
à Positive-sequence components
+
3 balanced phasors
ð equal in magnitude
ð displaced from each other by 120°
ð same phase sequence as the original phasors
(for example a-b-c)
à Negative-sequence components
+
3 balanced phasors
ð equal in magnitude
ð displaced from each other by 120°
ð opposite phase sequence to the original
phasors (for example a-c-b)
à Zero-sequence components
+
ELEC 371
Short Circuit Studies
67
3 equal phasors
ð equal in magnitude
ð zero phase displacement from each other
Industrial Power Systems
© Salvador Acevedo, 2000
Symmetrical Components
à Original voltages:
à
Va
Vb
Vc
à
à
à
à
Positive-sequence components:
Va1
Vb1 Vc1
or
Va+
Vb+ Vc+
à
à
à
à
Negative-sequence components:
Va2
Vb2 Vc2
or
Va- Vb- Vc-
à Zero-sequence components:
à Va0
Vb0 Vc0
à ORIGINAL PHASORS ARE THE SUM OF
THEIR COMPONENTS:
à Va = Va0 + Va1 + Va2
à Vb = Vb0 + Vb1 + Vb2
à Vc = Vc0 + Vc1 + Vc2
ELEC 371
Short Circuit Studies
68
Industrial Power Systems
© Salvador Acevedo, 2000
Example 1
NEGATIVE-SEQUENCE
COMPONENTS
Vb2
POSITIVE-SEQUENCE
COMPONENTS
Vc1
Va1
Vc2
Va2
ZERO-SEQUENCE
COMPONENTS
Vb1
Va0
Vc0
Vb0
Vc0
Vc
Vc2
Vc1
Va1
Va2
Va
Va0
Vb
Vb0
Vb1
Vb2
ELEC 371
Short Circuit Studies
69
Industrial Power Systems
© Salvador Acevedo, 2000
Example 2
Vb2
Vb1
Va2
Vc2
Va1
Vc1
Va0
Vb0
Vc0
Vb
Va
Vc
ELEC 371
Short Circuit Studies
70
Industrial Power Systems
© Salvador Acevedo, 2000
Example 3
Vb2
Va2
Vb1
Vc2
Vc1
Va1
Va0=Vb0=Vc0=0
Vb
Va=Vc
ELEC 371
Short Circuit Studies
71
Industrial Power Systems
© Salvador Acevedo, 2000
Operators
à Lets define a phasor:
a = 1 Ð 120° = -0.5 + j 0.8666
120°
à The following relations are true:
a2 = (1Ð120°)(1Ð120°) = 1 Ð240°= 1 Ð -120°
a3 = 1 Ð360°= 1 Ð 0°
à
1 + a + a2 = 0
a = 1 120
a3=1
a 2 = 1 -120
ELEC 371
Short Circuit Studies
72
Industrial Power Systems
© Salvador Acevedo, 2000
Symmetrical Components Relations
Vb2
Vc1
Va1
Va0
Vb0
Vc0
Vc2
Vb1
Va2
The positive-sequence components can be written as:
Vb1 = (1 ∠ −120°) Va1 = a2 Va1
Vc1 = (1 ∠ 120°) Va1 = a Va1
For the negative-sequence, we have:
Vb2 = (1 ∠ 120°) Va2 = a Va2
Vc2 = (1 ∠ −120°) Va2 = a2 Va2
And for the zero-sequence:
Va0 = Vb0 = Vc0
The totals are:
Va= Va0 + Va1 + Va2
Vb=Vb0 + Vb1 + Vb2
Vc=Vc0 + Vc1 + Vc2
ELEC 371
Short Circuit Studies
73
= Va0 + Va1 + Va2
= Va0 + a2 Va1 + a Va2
= Va0 + a Va1 + a2 Va2
Industrial Power Systems
© Salvador Acevedo, 2000
Symmetrical Components Relations
In m atrix form :
 V a 


 V b  =
 V c 
1

1
 1
1
a
a
1
a
a 2
2




V

V
 V
a 0
a 1
a 2




T h is defin es the tra n s f o r m atio n m atrix:
A
=
1

1
 1
1 

a 
a 2 
1
a 2
a
Its in v e r s e is:
A
-1
=
1
1 
1
3 
 1
1
a
a 2
1
a 2
a




T h e r e f o r e :
 V

 V
 V
a 0
a 1
a 2


 =

1
1 
1
3 
 1
1
a
a 2
1 

a 2 
a 
V a 


V b 
 V c 
or:
V
a 0
=
V
a 1
=
V
a 2
=
ELEC 371
Short Circuit Studies
74
1
(V a + V b + V c )
3
1
(
V a + a V b + a 2 V c
3
1
(
V a + a 2V b + a V c
3
)
)
Industrial Power Systems
© Salvador Acevedo, 2000
Example
Ia=10 A.
Ib=-Ia
Ic=0
Ia = 10∠0° A. Ib = 10∠180° A.
Ic = 0 A.
1
Ia0 = [10∠0°+10∠180°+0] = 0
3
1
Ia1 = [10∠0°+ (10∠180° )(1∠120° ) + 0] = 5.78∠ − 30°
3
1
Ia2 = [10∠ 0°+(10∠180° )(1∠ − 120° ) + 0] = 5.78∠30°
3
Ib0 = Ia0 = 0
Ib1 = 5.78∠ − 30°−120° = 5.78∠ − 150°
Ib2 = 5.78∠ + 30°+120° = 5.78∠ + 150°
Ic0 = Ia0 = 0
Ic1 = 5.78∠ − 30°+120° = 5.78∠ + 90°
Ic2 = 5.78∠ + 30°−120° = 5.78∠ − 90°
ELEC 371
Short Circuit Studies
75
Industrial Power Systems
© Salvador Acevedo, 2000
Unloaded Generator
a
+
Ea
-
jXn
n
+
Ec
- Eb
+
b
c
Generator is grounded through a grounding reactor
Internal voltages are balanced.
Find the symmetrical components for the internal voltages.
Ea0 
1 1

 1
Ea1

 = 3 1 a
Ea2 
1 a 2
1 1
1  Ea 
  1 
a 2  Eb  = 1 a
3
1 a 2
a  Ec 
1   Ea 


a 2  a 2 Ea 
a  aEa 
Ea0  0 

  
Ea1

 =  Ea 
Ea2  0 
Only positive-sequence voltage exists!!!
ELEC 371
Short Circuit Studies
76
Industrial Power Systems
© Salvador Acevedo, 2000
Generator Equivalent
Positive-sequence
jX1
+
Ea1
-
Negative-sequence
jX2
jX0
Zero-sequence
Xn=Impedance from
neutral to ground
j3Xn
The current in the neutral of the generator is:
In = Ia + Ib + Ic = ( Ia1 + Ib1 + Ic1) + (Ia2 + Ib2 + Ic2 ) + Ia0 + Ib0 + Ic0
The positive and negative sequence components add to zero:
Ia1 + Ib1 + Ic1 = 0
Ia2 + Ib2 + Ic2 = 0
This means that the neutral does not carry positive or negative sequence
components.
However, the zero-sequence components are in phase, and their sum is:
In = Ia0 + Ib0 + Ic0 = 3 Ia0
Therefore the zero sequence equivalent has a grounding impedance of
value: Zg = 3 j Xn
ELEC 371
Short Circuit Studies
77
Industrial Power Systems
© Salvador Acevedo, 2000
Single-line to ground fault
Unloaded generator (with balanced internal voltages)
If
a
b
G
c
jXn
Generator is Y-connected
grounded using a grounding reactor
Ia = If
Ib = 0
Ic = 0
 Ia 0 
1
1



 Ia 1  = 3  1
 Ia 2 
 1
Va = 0
Vb=?
Vc=?
1
a
a2
1
1   Ia 
1
 

a 2   Ib  =  1
3
 1
a   Ic 
1
a
a2
1   If

a 2  0
a   0




 Ia 0   I f / 3 

 

Ia
1
=
I
f
/
3

 

 Ia 2   I f / 3 
Ia 0 = Ib 0 = Ic 0 =
ELEC 371
Short Circuit Studies
78
If
3
Industrial Power Systems
© Salvador Acevedo, 2000
Single-line to ground fault
We have Ia1=Ia2=Ia0 and Va=0.
This situation can be represented in the following way:
Ia1=Ia2=Ia0
jX1
+
Ea1
-
+
+
Va1
Ia2
jX2
Va = Va0 + Va1 + Va2 = 0
+
Va2
Ia0
jX0
j3Xn
+
Va0
-
-
F r o m t h e c ircuit:
Ia0 = Ia1 = Ia2 =
Ea1
jX1 + jX2 + jX0 + j3Xn
V a 1 = Ea1 - (jX1)(Ia1)
V a 2 = 0 - (jX2)(Ia2)
Va0 = 0 - j(X0 + 3Xn)(Ia0)
ELEC 371
Short Circuit Studies
79
With this term
we include the
value of the
grounding
reactor
Industrial Power Systems
© Salvador Acevedo, 2000
Example: Line-to-ground fault in an unloaded generator
Assume:
X 1 = X 2 = 0.12 p. u.
X 0 = 0.06 p. u. , X n = 0
Ea1 = 1 p. u.
1∠ 0 °
= − j 3.33 p. u.
j ( 0.12 + 0 .12 + 0 .06 )
Ia = If = 3Ia1 = -j10 p. u.
Ib = 0
Ic = 0
Ia0 = Ia1 = Ia2 =
V a 0 = -jX0(Ia0) = -j0.06(-j3.33) = -0.2
V a 1 = Ea1 - jX1(Ia1) = 1 - j0.12(-j3.3 3 ) = 1 - j0.4 = 0.6
V a 2 = -jX2(Ia2) = -j0.12(-j3 .33) = -0.4
 V a  1

 
V
b

 = 1
 V c  1
1
a2
a
1   Va0 


a   Va1 
a 2   Va2 
V a = V a 0 + Va1 + Va2 = -0.2 + 0.6 - 0.4 = 0
V b = Va0 + a 2 V a 1 + a V a 2
V b = -0.2 + 0.6 a 2 - 0.4 a = -0.2 + 0.6 ∠ - 120 ° -0.4 ∠ 120 °
V b = 0 .9165 ∠ − 109 .1°
V c = V a 0 + a Va1 + a 2 Va2 = -0.2 + 0.6 ∠ 120 ° -0.4 ∠ - 120 °
V c = 0 .9165 ∠ + 109 .1°
ELEC 371
Short Circuit Studies
80
Industrial Power Systems
© Salvador Acevedo, 2000
Line-to-line fault
Unloaded generator
G
a
0
b
If
c
jXn
-If
Ia = 0
Ib = If
Ic = -If
 Ia 0 
1
1


 Ia 1  = 3  1
 Ia 2 
 1
1
a
a2
Va = ?
Vb=Vc
1
1   Ia 
1




a 2   Ib  = 1
3



1
a   Ic 
1
a
a2
1  0

a 2   If
a   − I f




 Ia 0 
 1−1 
I
f



2 
 Ia 1  = 3  a − a 
 Ia 2 
 a 2 − a 
Ia 0 = 0
 If 
Ia 1 = j 

 3
 If 
Ia 2 = − j 

 3
ELEC 371
Short Circuit Studies
81
Industrial Power Systems
© Salvador Acevedo, 2000
Line-to-line fault
We have Ia1=-Ia2, Ia0=0.
This situation can be represented in the following way:
Ia1
jX1
+
Ea1
-
+
Va1
Ia2=-Ia1
jX2
+
Va2
Ia0=0
jX0
j3Xn
+
Va0
-
F r o m t h e c ircuit:
Ea1
Ia1 = -Ia2 =
jX1 + jX2
Va1 = Ea1 - (jX1)(Ia1)
Va2 = Va1
Va0 = 0
ELEC 371
Short Circuit Studies
82
Industrial Power Systems
© Salvador Acevedo, 2000
Example: Line-to-line fault in an unloaded generator
Assume:
X 1 = X2 = 0.12 p.u.
Ia1 = -Ia2 =
X 0 = 0.06 p.u ., X n = 0
Ea1 = 1 p. u.
1∠ 0 °
= − j 4 .1 7 p . u .
j ( 0 .1 2 + 0 .1 2 )
Ia0 = 0
P h a s e c u r re n t s d u r i n g t h e f a u lt a r e :
Ia = 0
I b = If = -j 3 I a 1 = - 7 . 2 2 p . u .
Ic = -If = 7 . 2 2 p . u.
Va0 = -jX0(Ia0) = 0
V a 1 = E a 1 - j X 1 ( I a 1 ) = 1 - j 0 . 1 2 ( - j 4 .1 7 ) = 1 - 0 . 5 = 0 . 5
V a 2 = - j X 2 ( I a 2 ) = - j 0 . 1 2 ( j 4 .1 7 ) = 0 . 5
P h a s e v o l t a g e s d u r i n g t h e f a u lt a r e :
V a = Va0 + Va1 + Va2
V a = 0 + 0.5 + 0.5 = 1 . 0 p . u .
Did not change!!!
V b = Va0 + a 2Va1 + aVa2
V b = 0 + 0 . 5 a 2 + 0.5 a = 0. 5 ( a
2
+ a ) = 0 .5 ( − 1 )
V b = − 0 .5 p . u .
V c = Va0 + aVa1 + a 2 Va2
V c = 0 . 5 ( a + a 2 ) = − 0 .5
V c = − 0 .5 p . u .
ELEC 371
Short Circuit Studies
83
Industrial Power Systems
© Salvador Acevedo, 2000
Protection in Industrial Electric Networks
à Objective: Isolation of the problem with the
minimum service disruption
à Based on the protective device Time-Current
characteristics
time
current
à Information needed for a protection study:
o
o
o
o
o
o
o
o
ELEC 371
Short Circuit Studies
84
Protective device manufacturer and type
Protective device ratings
Trip settings and ratings
Short circuit current at each bus
Full load current of all loads
Voltage level at each bus
Power transformers data
Instrumentation transformers ratios
Industrial Power Systems
© Salvador Acevedo, 2000