Symmetrical Components and Sequence Networks

1. 1
Prof Chandrabhan sharma

2. SYMMETRICAL COMPONENTS
End of this topic, you should be able to:
- Explain the importance of symmetrical
component.
- Construct positive sequence networks for a
power system.
- Construct negative sequence networks for a
power system.
- Construct zero sequence networks for a
power system.
- Use the networks to calculate fault current and voltages
under SLG, DL, DLG and OC conditions
2

3. SYMMETRICAL COMPONENTS
Symmetrical Components:
- Theory originally expounded by Stokvis L.G.
(1915) but was given practical application to
Electrical Engineering by C.L. Fortescue in 1918.
- A mathematical model for representing an
unbalanced system of ‘n’ phasors as ‘n’ systems of
balanced phasors known as symmetrical
components.
N.B. : this is only a model for analysis since symmetrical
components do not exist in real life
3

4. Used for analyzing unbalanced fault conditions in a power system,
since fault on a balanced 3 system is usually NOT symmetrical.
They are usually:
- SLG (Single Line – Ground)
- DL (Double Line or Line - Line)
- DLG (Double Line – Ground)
- Open conductor
4

5. If either SLG, DLG, LL, OC occur then
- they will cause currents and voltages to be of
unequal magnitude, and
- the usual 3 phasor diagram is no longer displaced
by 120apart from each other.
It is imperative that values of short circuit currents be obtained
for these types of faults since the information is necessary for:
(a) Ordering protective equipment
(b) the determination of protective relay settings
(b) transient stability studies for interconnected systems
5

6. THEORY
The theory postulates that a 3 unbalanced system of voltage and
current phasors can be resolved into three balanced system of phasors.
These are :
A positive phase sequence system
- comprising three phasors of equal magnitude
- displaced by 120 from each other
- and with the same phase sequence as the original phasors.
6

7. A negative phase sequence system
- comprising three phasors of equal magnitude
- displaced by 120 from each other
- but with a phase sequence opposite to the original phasors.
A zero phase sequence system
- comprising three phasors equal in magnitude and phase,
- and revolving in the positive phase rotation.
7

8. Equivalent unbalanced system
B1
  
= + +
B
Y
R R1
Y1
R2
Y2 B2
Y0
B0
R0
120
120
120
120
120 120



 

Faulted System +ve sequence -ve sequence zero sequence
From the theory:
VR = VR1 + VR2 + VR0………………(1)
VY = VY1 + VY2 + VY0………………(2)
VB = VB1 + VB2 + VB0……………….(3)
8

9. For a 3 system, the theory states that:
R1 Y1 B1 V  V  V
R2 Y2 B2 R0 Y0 B0 V V V and V V V
Also,
   
The ‘a’ operator (or ‘h’)
Since we will be dealing with three phasors displaced by 120,
it would be helpful/convenient to define a new operator ‘a’ such
that:
a  1120 = ej2/3 = - 0.5 + j0.866
9

10. And:
a2 = 1240 = ej4/3 = - 0.5 – j0.866, and
a3 = 1360 = 1 + j0
Note, unlike the ‘j’ operator, -a ≠ 1-120
Recall -j = 1-90  = (j2)(j) = j3 = 1-90 
but -a = (j2)(a) = 1(120 + 180) = 1300  = 160 
Note also a2 + a + 1 = 0 = a3 + a2 + a
10

11. From the positive sequence diagram we have:
VY1 = a2 VR1……………………(4)
VB1 = a VR1 …………………...(5)
From the negative sequence,
VY2 = a VR2…………………….(6)
VB2 = a2 VR2 ……………………(7)
And from the zero sequence,
VR0 = VY0 = VB0 ………………..(8)
11

12. Substitute eqs. 4-8 into eqs. 1-3 we obtain:
VR = VR0 +VR1 + VR2
VY = VR0 + a2 VR1 + aVR2
VB = VR0 + aVR1 + a2 VR2
Written in matrix :


...........................(9)
V
R0
V
R1
V
1 1 1




1 a a

1 a a
V
R
V
Y
V
R2
2
2
B
























12

13. 1 1 1



1 a a

1 a a
T
Let,
2
2







'T' is known as the symmetrical component transformation matrix
1 1 1


-1 2
1 a a

1 a a
1
3
T
2








13
From matrix inversion

14. Applying T-1 to eq. (9):


.............................1(0)
V
R
V
Y
V
1 1 1


1 a a

1 a a
1
3
V
R0
V
R1
V
B
2
2
R2


























1
 
 
V a V aV ............................1(3)
3
  
R0 R Y B
1
  
1
V
V aV a V ............................1(2)
3
V
V V V .................................(11)
3
V
OR
Y B
2
R2 R
B
2
R1 R Y
  
14

15. Similarly:


.............................1(4)
I
I
I
1 1 1


1 a a

1 a a
1
3
I
I
I
R
Y
B
2
2
R0
R1
R2


























1
 
 
I a I aI ............................1(7)
3
  
R0 R Y B
1
  
1
I
I aI a I ............................1(6)
3
I
I I I .................................(15)
3
I
OR
Y B
2
R2 R
B
2
R1 R Y
  
15

16. Analyzing equations (11) and (15) …..
(11) → for balanced 3 system VR +VY + VB = 0
Balanced  voltage magnitude and angle normal
(current unbalanced)
(15) → IR + IY + IB = IN
 IN = 3IR0
 if the system is ungrounded  IN = 0 = IR0
→ no zero sequence currents
Hence  connected loads have no zero sequence currents.
16

17. Example:
(Higher Electrical Engineering – Sheppard, Morton and Spence)
The currents flowing in an unbalanced 4-wire system are
IR = 1000 A; IY = 200-90 A ; IB = 100120 A
Find the +ve, -ve and zero sequence components of these currents.
17

18. Solution:





R0
R1
1
I
I
 
  
I I I
R0 R Y B









1 1 1


1 a a

R
Y
I
I
 
1
3
1
        
100 0 200 90 100 120
3
From which
   
I 41.3 66.2 A
I
R0














B
2
2
R2
I
1 a a
3
I
Re call :
18

19.  
1
  
I aI a I
3
 
            
100 0 (1.0 120 ).(200 90 ) (1.0 240 ).(100 120 )
3
I 128.79 15 A
1
Also,
I
R1
B
2
R1 R Y
   
2
    
R2 R Y B
 
             
   
R2
And,
1
I I a I aI
3
1
100 0 (1.0 240 ).(200 90 ) (1.0 120 ).(100 120 )
3
I 41.3 174 A
19

20. SEQUENCE NETWORKS
Since
- component currents of one phase sequence
cause
- voltage drops of like sequence only
and
- independent of the other sequences (in a
balanced system)
- currents of any one sequence may be
considered to flow in an independent
network composed of impedances to
the current of that sequence.
20

21. SEQUENCE NETWORKS
The single phase equivalent circuit composed of the
impedances to current of any one sequence only is called the
sequence network for that particular sequence.
In static symmetrical systems, conversion of a +ve sequence to
a –ve sequence network is accomplished by changing only the
impedances representing rotating machinery and by omitting
any generated e.m.fs
21

22. Synchronous Machines:
- +ve sequence ≠ -ve sequence
- Xd”, Xd’ and Xd are +ve sequence
- the generated e.m.fs are +ve sequence only
- the ref. bus for +ve and –ve sequence n/w is the
generator neutral
- the ref. Bus for the zero sequence n/w is the
ground of the generator or earth
- there are no generated e.m.fs in the –ve and zero
sequence n/w
22

23. Positive Sequence Networks:
- For a given power system the positive sequence network shows
all the paths for the flow of positive sequence currents in the
system.
- The one-line diagram of the system is converted to an
impedance diagram that shows the equivalent circuit of each
component under balanced operating conditions
- Each generator in the system is represented by a source voltage
in series with the appropriate reactance and resistance.
- To simplify the calculations, all resistance and the
magnetizing current for each transformer are neglected.
23

24. Positive Sequencenetwork of Synchronous generator:
24

25. Negative Sequence Networks:
- Three-phase generators and motors have only positive
sequence generated voltages. Thus, the negative sequence
network model will not contain voltage sources associated
with rotating machinery.
- The negative sequence impedance will in general be
different from the positive sequence values.
- For static devices such as transmission lines and
transformers, the negative sequence impedances have the
same values as the corresponding positive sequence
impedances.
25

26. Negative Sequence network of Synchronous generator:
26

27. Zero Sequence Networks:
- The zero sequence network of a system depends on the
nature of the connections of the three-phase windings for
each of the system’s components.
27

28. Zero Sequence Networks:
28

29. Zero Sequence Networks:
29

30. Zero Sequence Networks:
30

31. Lines and cables:
- +ve and –ve sequence impedances are the
normal balanced values.
- zero sequence impedances depends on the nature
of the return path through earth.
The following factors can be used in the absence of
manufacturer data:
Cable Zero Seq. Impedance
3 – core Z0 =3Z1 to 5Z1
1 – core Z0 =1.5Z1
31

32. Single Circuit Double Circuit
No ground wire  Z0 =3.5Z1 Z0 = 5.5Z1
Steel ground wire  Z0 =3.5Z1 Z0 =5Z1
Non-metallic ground wire
 Z0 =3.5Z1 Z0 =3Z1
Lines:
32

33. Transformers:
- +ve and –ve sequence impedances are the same i.e.
(Z1 = Z2)
- zero sequence cct depends on whether a return path exists
through which a completed circuit can be provided
Recall the equivalent cct for a transformer is:
33

34. The equivalent circuit becomes:
For +ve and –ve sequence Z1 = Z2
For zero sequence the equivalent circuit changes depending on
the configuration of the transformer.
34

35. Zero Sequence Network (Transformer)
- Close ‘a’ contact whenever zero sequence currents
can flow, i.e. earth connection exists.
- Close ‘b’ contact whenever zero sequence currents
can circulate, i.e.  connection.
- circuit is left open when neither condition exists,
for example unearthed star connection.
35

36. Transformer zero sequence models
36

37. Transformer zero sequence models
37

38. Static Loads
38

39. Static Loads
39

40. The A-phase of the unloaded generator shown below is subjected
to a SLG fault. The pre-faulted L-N voltage of the generator is EA
Volts. The generator has positive, negative and zero sequence
impedance of Z1, Z2 and Zo respectively. Find the fault current
and phase voltages under the fault condition.
40
Line to Ground Fault (SLG)

41. Conditions under SLG fault:
VA = 0 ; and IB = IC = 0 (i.e. neglecting load currents)










I
I 
0






1 1 1


1 a a















B
I 0
1 a a
1
3
I
I
I
C
A
2
2
A0
A1
A2
1
A0 A1 A2 A I
3
I  I  I 
Since IA1 = IA2 = IA0 then the seq. n/ws must be connected in series
41

42. From the interconnected networks
42
A
E
1
A0 A1 A2 A Z Z Z
1 2 0
I
3
I I I
 
   
Also
VA = VA0 +VA1 + VA2
VB= VA0 + a2 VA1 + aVA2
VC= VA0 + aVA1 + a2 VA2
Where:
VA1= EA - IA1 Z1
VA2= 0 - IA2 Z2
VA0= 0 - IA0 Z0

43. A 3-phase star connected solidly earthed, unloaded generator has it’s A-phase
subjected to a solidly grounded earth fault. Given that the pre-faulted
L-N voltage of the generator is 100 V and Z1 = j1.0Ω, Z2 = j0.5Ω
and Zo = j0.1Ω, calculate the fault currents developed and the L-L
voltages attained due to a fault.
43
Example

44. 100
j1.6
From analysis of the ckt we obtain
E
A
A0 A1 A2 A 
Z Z Z
1
I
3
I I I
 
1 2 0
   
    
I I I j62.5 A
A0 A1 A2
     
I I 3( j62.5) A j187.5 A
F A















we also know that

1 1 1




1 a a











V
A0
V
A1
A2
2
2
V
A
V
B
C
V
1 a a
V
44

45.  VA = VA0 +VA1 + VA2………………(1)
VB= VA0 + a2 VA1 + aVA2…………..(2)
VC= VA0 + aVA1 + a2 VA2…………...(3)
From the ckt: VA1= EA - IA1 Z1= 100 – [(-j62.5)(j1.0)] = 37.5 V/gn
Similarly,
VA2= 0 - IA2 Z2 = - (-j62.5) (j0.5) = -31.25 V/gn
And,
VA0= 0 – IA0 Z0 = - (-j62.5) (j0.1) = -6.25 V/gn
45

46. Substitute V1, V2 and V0 into equations (1), (2) and (3)
VA = VA0 +VA1 + VA2 = 0
And
VB= VA0 + a2 VA1 + aVA2 = - 9.37 – j59.5 = 60.23-98.9 V
VC= VA0 + aVA1 + a2 VA2 = - 9.37 + j59.5 = 60.23 98.9 V
Also,
VAB= VA - VB = 9.37 + j59.5 = 60.23 81.1 V
VBC= VB - VC = - j119 = 119 -90 V
VCA= VC - VA = - 9.37 + j59.5 = 60.23 98.9 V
46

47. DOUBLE LINE FAULT (DL)
Condition when fault occur
IA = 0 ; IB + IC = 0 ; and phase voltage VB = VC
47

48. 1
1 1 1


1 a a
1




 
 
 B B 
A0
A1
I
I





   
A0 B B
2
A2
B
2
1
 
A1 B
I 
0
B
A
I
C B
2
2
A2
1
0 a I - aI
3
I



0 aI - a I and
3
I
0 I - I 0;
3
I
I I
1 a a
3
I
 







 










1
a - a 2
 I
3
A0 A1 A2    
I 0 a nd I - I B
48
j
But a - a j1.732 j 3 I - I A1 A2 B
I
3
2     
Since IA0 = 0 then the zero sequence ckt is open

49. Equivalent sequence ckt becomes
From which we obtain
E
A
A1 
Z Z
I
1 2

IA1 = - j66.67
IA2 = j66.67
- 114.94 A

A1 B
- j 66.67
   
j0.58
I - I
j
I
3
But I
B C
49

50. Also,
VA1 = EA - IA1Z1 = 100 - (-j66.67)(j1.0)
= 100 - 66.67
= 33.33 V/gn
Therefore, VA1 = VA2 = 33.33 V/gn
Now















1 1 1




1 a a












V
A0
V
A1
A2
2
2
V
A
V
B
C
V
1 a a
V
50

51.  VA =VA1 + VA2 = 66.67 V/gn
VB =VA1(a2 + a) = VC
But a2 + a = -1  VB = -33.33 V/gn = VC
51

52. DOUBLE LINE TO GROUND FAULT (DLG)
Condition when fault occur
IA = 0 ; VB = VC = 0 ; IB + IC = IF
52

53. From inspection IA = 0 and VB = VC










V
V 
0






1 1 1


1 a a














B
V 0
1 a a
1
3
V
0
V
1
V
C
A
2
2
2
1
0 1 2 A V
3
V  V  V 
Hence, the sequence networks are in parallel
53

54. Since Z2 //Z0 then the circuit reduces to:
Z Z
2 0
P Z Z
2 0
where Z


A
E
1 Z Z
1 P
I

 
54
2
Z


where I 
0 I
Z Z
Z


I and 
2 I
1
0
Z Z
0 2





1
0 2






55. Example:
The diagram shows two generators connected to a transmission system
comprising two parallel 66kV transmission lines via transformers. The data of
this system is given below
GA and GB – 13.8kV, 100 MVA, Z1 = j0.15, Z2 = j0.1 and Z0 = j0.03 (all p.u)
TFA – 13.8/66kV, 100 MVA, Z1 = Z2 = Z0 = j0.15 p.u
TFB – 13.8/66kV, 100 MVA, Z1 = Z2 = Z0 = j0.1 p.u
Each Transmission Line - Z1 = j0.4, Z2 = j0.4 and Z0 = j0.7 (all p.u)
All star points are solidly earthed, except that of generator B which is
unearthed. All p.u. values are on the 100 MVA base. Calculate the fault current
for a DLG fault at the feeder terminals of transformer A.
55

56. Soln
+ ve
- ve
zero
56

57. 

+ ve seq n/w
reduces to
- ve seq n/w
reduces to
zero seq n/w
reduces to
57

58. For DLG fault, the three sequence networks are connected in
parallel (as seen earlier).

j3.8 p.u.
1
I pu
1  
j(0.18 0.083)
V
Z

 
j2.05 p.u.
j0.18


2 1   
j0.334
(j3.8)
Z
0
Z Z
I I
0 2







58

59. j1.75 p.u.
j0.154


0 1   
j0.334
(j3.8)
Z
2
Z Z
I I
0 2






 
I
I
A
B








I I I I j1.75 ( j3.8 ) j2.05 0 A 0 1 2         
I
0
I
I
1 1 1

1 a a

1 a a
I
Substitute the seq current values in the matrix
1
2
2
2
C





















2
I  I  a I 
aIB 0
1 2
          
j1.75 (1 240 )(3.8 90 ) (1 120 )(2.05 90 )
  
5.72 152.7 p.u.
59

60. 874.8A
100 x 10
pu   
3 x 66 x 10
MVA
3V
But I
3
6
base
base
I 874.8 (5.72 152.7 ) 5003.9 152.7 A B       
Similarly, I 5003.9 27.37 A
And I  I  I 
j5.26 p.u.
F B C
I j4601.5 A
or
F
C

  
60

61. One line Open Circuit
61
Under this condition consider the elemental values of V.
From inspection IA = 0 and VB = VC

62. Applying these conditions:











V
 
V 0
B
 





1 1 1


1 a a














V 0
1 a a
1
3
V
0
V
1
V
C
A
2
2
2
1
0 1 2 A V
3
V  V  V 
Hence, the sequence networks are in parallel as shown below:
62

63. From the example before: EA = 1000
and Z1 = j1.0 ; Z2 = j0.5 ; Z0 = j0.1
Since Z2 //Z0 then the circuit reduces to:
j0.083
(j0.5)(j0. 1)
P  
j0.6
Z Z
2 0
Z Z
where Z

2 0

j92.31
100
1  
1 j0.083
E
A
Z Z
I
1 P



 
    
V E - I Z 100 [( j92.31)(j1.0)]
  
2 0
1 A 1 1
7.69 V/gn V V
63

64. j15.38 A
7.69
2     
j0.5
V
Z
Also, I
2
2
j76.9 A
7.69
0     
j0.1
V
Z
And, I
0
0
But, 3V1 = VA  VA = 23.07 V/gn















1 1 1




1 a a













I

I

0
I
I
1
2
2
2
A
B
C
B C
I
1 a a
I
To determine I and I we use:
I I I I 0 A 0 1 2     
Also I I a (-j92.31) a(j15.38) - 93.3 j115.4 2
B 0      
And I I a(-j92.31) a (j15.38) 93.3 j115.4 2
C 0      
64