Symmetrical components

EEE2416: POWER SYSTEMS II ➢ Symmetrical components are operators

which enable conversion of an
unsymmetrical condition to an equivalent
symmetrical condition.
Symmetrical components ➢ They are applicable to a system with any
number of phases n.
By ➢ An unbalanced 3- system can be resolved
into 3 balanced system of phasors:
K. Kaberere

1
2

If the unbalanced system voltage phasors are Va, Vb ,

1. Positive-sequence components – 3 phasors and Vc ;
equal in magnitude, 120° out of phase with ✓ the positive sequence components are Va1, Vb1 , and Vc1
each other, and having the same phase ✓ negative sequence components Va2, Vb2 , and Vc2
sequence as the original phasor.
✓ zero sequence components Va0, Vb0, and Vc0.
2. Negative-sequence components – 3 phasors
equal in magnitude, 120° out of phase with
each other, and phase sequence opposite to Vc1 Va1
that of the original phasor.
3. Zero-sequence components – 3 phasors
equal in magnitude and all in phase.
Vb1
Positive sequence Negative sequence Zero sequence

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Va = Va1 + Va 2 + Va 0  Substitute (2) in (1)


Vb = Vb1 + Vb 2 + Vb 0  ......(1) Va = Va1 + Va 2 + Va 0
Vc = Vc1 + Vc 2 + Vc 0  Vb = a 2Va1 + aVa 2 + Va 0
Vc = aVa1 + a 2Va 2 + Va 0
The operator a = 1120
a2 = 1240
a3 = 1360
Va  1 1 1  Va 0 
Vb  = 1 a 2 a  Va1  ………………….. (3)
From the sequence phasors Vc  1 a a 2  Va 2 
Vb1 = a2Va1 Vb2 = aVa2
A
Vc1 = aVa1 Vc2 = a2Va2 ……………(2)
1
1 1 1
A−1 = 1 a a 2 
Vb0 = Va0
3 2 
Vc0 = Va0 1 a a 
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1
Va 0  1 1 1 1  Va  Similarly for the currents
Va1  = 1 a a 2  Vb  ………………….. (4)  I a  1 1 1   I a 0   I a 0  1 1 1 1   I a 
Va 2  3 1 a 2 a  Vc   I b  = 1 a 2 a   I a1  and  I a1  = 1 a a 2   I b 
 I c  1 a a 2   I a 2   I a 2  3 1 a 2 a   I c 
3( a
Va 0 = 1 V + Vb + Vc )
In a 3- system Ia + Ib + Ic = In
Va1 =1
3 a (
V + aVb + a 2Vc ) But 1/3(Ia + Ib + Ic) = Ia0
(
Va 2 = 1 Va + a 2Vb + aVc
3 ) In = 3Ia0

❑ The sum of the line-to-line voltage phasors in a 3- ❑ In the absence of a neutral path, In = 0 and no zero
system is always zero i.e. VAB + VBC + VCA = 0. sequence current flows in the line current e.g. line current
❑ Hence, zero sequence components are never present in supplied to a –connected load or 3-wire Y-connected
the line voltage no matter the amount of imbalance. with an ungrounded neutral.
❑ However, voltages to neutral may contain zero sequence ❑ In a balanced Y-connected system, In = 0. Hence, no zero
components. sequence current component.
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Power in terms of symmetrical components Va  1 1 1  Va 0 

Vb  = 1 a 2 a  Va1   V ph  =  AVs  ……. (7b)
Vc  1 a a  Va 2 
The power consumed a 3- system is given by 2

S3 = P + jQ = Va I a* + Vb I b* + Vc I c* ………………….. (5)

Substitute (7a) and (7b) in (6)
Where the voltages and currents are phase values.
S3 =  AVs   AI s *
T
* T *
 Ia  Va   I a 
S3 = Va Vb Vc   I b  = Vb   I b 
But  AVs  = Vs   AT
T T
 I c  Vc   I c 

 S3 = Vs   AT  A*  I s *

T * T
= V ph   I ph  …………..………………...….. (6)

 AT =  A and a* = a 2 and a 2* = a

 I a  1 1 1   I a 0 
But  I b  = 1 a 2 a   I a1    I ph  =  A I s  ……. (7a)
 I c  1 a a 2   I a 2 
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Example 1
1 0 0
 A A = 3 0 1 0
T *
The following currents were recorded under fault conditions
 0 0 1  in a three-phase system:
*
 Ia 0   I A   15045 
S3 = 3 Va 0 Va1 Va 2   I a1   I B  =  250150  A
 I a 2   I C  100 300 

Calculate the values of the positive, negative, and zero phase

S3 = Va I a* + Vb I b* + Vc I c* sequence components for each line.

(
= 3 Va 0 I a*0 + Va1 I a*1 + Va 2 I a*2 ) Solution

 I a 0   52.2112.7 
 I a1  =  48.0 − 87.6  A
 I a 2   163.240.4 

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2
 Example 2 Example 3
Va 0   100 
Given that: Va1  =  20060  V Using the voltages and currents in examples 1 and 2, find
Va 2  100120  the power using (i) symmetrical components and (ii) phase
parameters.
Find the phase voltages Va, Vb , Vc and line voltages VAB,
VBC , VCA.
Solution

Va   30060 
Vb  =  300 − 60  V
Vc   0 

V AB   52090 
VBC  =  300 − 60  V
VCA   300 240 

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Sequence impedances and sequence networks

➢ A sequence network is the 1- equivalent
➢ The voltage drop in a network caused by a current circuit consisting of sequence impedance.
of a given sequence depends on the network ➢ The network also includes any generated
impedance to the particular sequence current. emfs of like sequence.
➢ The 3 sequence currents may encounter different ➢ In analyzing asymmetrical faults, the various
impedances for the same network. sequence networks are inter-connected to
represent the fault condition.
➢ The impedance “seen” by the positive, negative,
and zero sequence currents when each of these
current flows independently is positive-sequence,
negative-sequence, and zero-sequence impedance,
respectively.

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Sequence networks of rotating machines

➢ Only +ve sequence voltages are generated – voltage

source is included only in +ve sequence network
➢ The generator positive sequence impedance is
1. Synchronous reactance – steady state conditions
2. Transient reactance – stability studies
3. Sub-transient reactance – short circuit studies
Synchronous generator The PPS component of the phase voltages at the generator
terminals
V1 = Eg1 – I1Zg1
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3
 Voltage drop across generator neutral impedance Vn = ZnIn .
But In = 3I0  Vn =I0 3Zn. Therefore, Z0 = 3Zn + Zg0
V0 = – I0Z0 where Z0 = 3Zn + Zg0
The NPS components of the phase voltage at the generator Zg0 < Zg2 < Zg1
terminals ▪Sequence networks for synchronous generator and motor are
V2 = – I2Zg2 similar except for the direction of current flow.
▪Induction motors have same sequence networks as synchronous
motors except for the +ve sequence voltage source (IM do not have
a DC source of magnetic flux in their rotor circuit, E M1 = 0)
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Sequence networks of 3- lines

➢ The PPS and NPS impedances are the

normal phase impedance values
➢ The ZPS impedance of overhead lines
is 2 – 3.5 times the PPS impedance –
higher values apply to double circuits
and lines without a ground wire

Series impedances of a completely transposed 3- line

21 22
Synchronous motor Induction motor

Sequence networks of impedance loads Sequence networks for a balanced Y impedance load
Vag = IaZY + InZn
➢ Note:
But In = Ia + Ib + Ic
➢ Zn only appears in the ZPS
 Vag = Ia(ZY + Zn ) + IbZn + network  PPS and NPS
IcZn
currents do not flow through
Similarly the neutral impedance
Vbg = Ia Zn + Ib (ZY + Zn ) + IcZn ➢ If the neutral is solidly
Vcg = Ia Zn+ IbZn + Ic (ZY + Zn ) grounded – Zn = 0
[Vp] = [Zp] [Ip]
➢ If neutral point is not
connected to ground – Zn =
 [A][Vs] = [Zp][A][Is]
, open circuit – no zero
[Vs] = [A]-1[Zp][A][Is] = [Zs] [Is] sequence current exists
Balanced Y impedance load Va 0  ( ZY + 3 Z n ) 0 0   I a 0 
Va1  =  0 ZY 0   I a1 
Va 2   0 0 ZY   I a 2 

Zs 23 24

4
 Sequence networks for a balanced  impedance load Sequence networks for a balanced  impedance load

VAB
VAB

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Sequence networks for transformers

Comments

➢ The PPS and NPS values are the normal balanced values 1- currents can
S
circulate in the  but not
➢ ZPS impedance depends on winding connections outside it
➢ For ZPS currents to flow through the windings on one side
of the transformer and into the connected lines, a S No flow of ZPS currents
complete circuit must exist
➢ There must be a path for the corresponding current in the
coupled windings on the other side S No flow of ZPS currents

S ZPS currents free to

flow in both primary and
secondary
S No path for ZPS
currents in secondary
circuit
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Three-winding transformers

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5
 Example 1
Draw the sequence networks for the system whose one-line diagram is shown
below.

31 Positive sequence network Zero sequence network 32

Example 3 Va 0  1 1 1 1  Va  Va 0   15.9162.1 

A Y-connected voltage source with the following unbalanced line to Va1  = 1 a a 2  Vb  Va1  =  277.09 − 1.77  V
ground voltage is applied to a balanced  load with Z = 3040 . Va 2  3 1 a 2 a  Vc  Va 2   9.22 − 143.42 
The line impedance between the source and the load is 185  for
each phase. The phase voltages are
Vag   2770 
V  =  260 − 120  V
 bg   295115 
Vcg   
The source neutral is solidly grounded. Determine the source line
currents Ia, Ib, and Ic.  I a  1 1 1   I a 0 
 Ia 0   0 
Solution  I a1  =  25.82 − 45.55  A  I b  = 1 a 2 a   I a1 
 I a 2   0.859 − 187.2   I c  1 a a 2   I a 2 

 I a   25.15 − 46.76 
 I b  =  25.71 − 163.66  A
 I c   26.6173.77 

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HOMEWORK
Draw the sequence networks for the systems whose one-line diagrams are
shown in figures E2 (a) and (b).

Figure E2(a)

Figure E2(b)
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