88
CHAPTER 5
THREE-PHASE POWER FLOW ANALYSIS IN SEQUENCE
COMPONENT FRAME
5.1
INTRODUCTION
Power flow is an important tool in power system studies. Usually, a
power system is assumed to be balanced and the power flow analysis is
carried out for the single-phase system. In single-phase power flow analysis,
the power flow problem is solved for any one of the three-phases by assuming
that all the three-phase voltages are equal in magnitude and displaced by 120
degrees. For an ‘n’ bus system, the single-phase power flow analysis solves a
set of 2n power equations.
Practical power system networks are highly unbalanced and it is
required to obtain load flow solution for such unbalanced networks. The
unbalancing in power system is unavoidable due to the un-transposed
transmission lines and imbalance in loading. Hence, a power flow analysis
dealing with three-phase configuration of power systems is necessary for
proper planning and stability studies. In three-phase power flow problems,
unknown three-phase voltage magnitudes and angles of each bus have to be
solved for the given values of three-phase active and reactive power injections
or certain specification of voltages. Therefore, as in the single-phase case, a
solution to a set of non linear equations is needed.
89
Three-phase power flow problem is either formulated in phase
frame or sequence component frame. A variety of three-phase power-flow
algorithms in phase frame (Birt et al 1976, Wasley and Shlash 1974 and Chen
et al 1990) and in sequence component frame (Lo and Chang 1993) have been
studied for solving unbalanced power systems. Jose Mauro Marinho and
Glauco Nery Taranto (2008) solved the three-phase power flow problem using
single-phase and three-phase models. The advantage of the application of
sequence components is that the size of the problem is effectively reduced in
comparison to the phase component approach (Akher et al 2005). In all these
works, the load flow problem is solved using the Newton Raphson method.
The fast decoupled method has also been used to solve the same (Arrillaga
and Harker 1978). There are two main disadvantages in the existing threephase power flow approaches. Firstly, it might be computationally
unacceptable to solve a set of 6n, or two sets of 3n, non linear simultaneous
equations for three-phase power flows in relatively large systems. Secondly,
since the solution is obtained by taking the inverse of Jacobian matrix, the
conventional methods fail to solve the load flow problem in some cases.
This chapter presents the solution algorithm for the three-phase load
flow problem in symmetrical component frame using EAs. A sequence
coupled transformer model and a sequence line model are used. The objective
function is formulated using the sequence models of three-phase power
system and is solved using the EAs. The proposed methods eliminate the
formation of Jacobian matrix and its inversion and thus reduce the complexity
of the problem.
90
5.2
THREE-PHASE POWER FLOW PROBLEM
The
symmetrical
component
transformation
is
a
general
mathematical technique developed by Fortescue. According to this technique,
any set of three-phase voltages or currents can be transformed into three
symmetrical systems of three vectors each. Three-phase currents and voltages
are related by
[Iabc ] [Y abc ][V abc ]
[Iabc ] [Ia
[Y abc ]
Ib
Ic ]T
and
Y aa
Y ab
Y ac
Y ba
Y ca
Y bb
Y cb
Y bc
Y cc
[V abc ] [V a
Vb
V c ]T
The phase component (indicated by the superscripts abc) is
transformed into sequence component (indicated by the superscripts 012 for
the zero, positive and negative sequences respectively) as follows
[V 012 ] T 1[V abc ]
[I012 ] T 1[Iabc ]
[I012 ] T 1[Y abc ]T[V 012 ]
where T is the symmetrical component transformation matrix and is given by
1
T
1
1
1 a2
a
1
a2
a
where a 1 120 0
91
In case of balanced power networks, voltages, currents and
admittances in all three-phases are equal, that is, Va = Vb = Vc and Ia = Ib = Ic;
Yaa = Ybb = Ycc; and off diagonal elements are zero. Thus, analysis of balanced
networks requires the study of positive sequence alone. Hence, each element
in power system is represented only by its positive sequence admittance. As a
result, for single-phase balanced system, a set of equations defining the power
system network is obtained in matrix form as I=YV; I and V being current and
voltage vectors and Y is the bus admittance matrix.
For a power system consists of 2 buses, the nodal equation is
written as:
I1
Y
11
Y
12
V1
I2
Y
21
Y
22
V
2
For an ‘n’ bus system the current equation is written as
n
Ii =
V j Y ij;
i =1:n
j 1
where Ii is the current at bus i; Vi is the voltage at bus i;
The diagonal element (Ykk) of bus admittance matrix is the sum of
the self admittances terminating on node k.
The off diagonal element (Yik) is the negative of mutual admittance
between the nodes i and k.
For an n bus single-phase system, n nodal equations are written.
92
Similarly, for a 2 bus three-phase system, the above matrix equation
can be written as
I 10
I 02
Y1010
Y 2010
Y1020
Y 2020
Y1011
Y 2011
Y1021
Y 2021
Y1012
Y 2012
Y1022
Y 2022
V10
V 20
I 11
Y 1110
Y1120
Y1111
Y1121
Y1112
Y1122
V11
I 12
Y 2110
Y 2120
Y 2111
Y 2121
Y 2112
Y 2122
V 21
I 12
Y1210
Y1220
Y1211
Y1221
Y1212
Y1222
V12
I 22
Y 2210
Y 2220
Y 2211
Y 2221
Y 2212
Y 2222
V 22
Superscript indicates the sequence component and subscript
indicates the node number. Thus, for a three-phase system with ‘n’ buses, 3n
such nodal equations are written and solved.
5.3
MODELING
OF
THREE-PHASE
POWER
SYSTEM
COMPONENTS
A power system network comprises of four major components generator, transformer, transmission line and load. This section describes the
derivation of equivalent circuits of each of these power components and the
formation of system admittance matrix relating the current and voltage at
every node (busbar) of the power system. Modeling is detailed as follows:
5.3.1
Generator Model
Figure 5.1 shows the generator model for three-phase power flow
problems in symmetrical component coordinate. The model is represented by
an injected power with a specified bus voltage. Internal voltages of all
generators are well balanced. The positive sequence reactance and EMF
behind it are not introduced (Akher et al 2005). V1, V2 and V0 are positive,
negative and zero sequence voltages respectively. Z2 and Z0 are negative and
zero sequence reactances of generator. For unbalance case, current will flow
in the negative sequence and zero sequence components. Hence, the generator
terminal bus voltages become unbalanced and unequal.
93
V1
V2
Z2
V0
Z0
Figure 5.1 Generator model in sequence component frame of reference
5.3.2
Transformer Model
The transformer model used in this work is grounded star and delta
connection as shown in Figure 5.2. The models of different connections of
transformers are given in (Chen and Dillon 1974). In Figure 5.2, y = 1/zt;
where zt is the transformer impedance;
t
is the phase shift determined by
transformer connection; and ys is a very small admittance used to avoid the
singularity of bus admittance matrix.
V1
e-
V2
t
e
V1Y
y
t
ys
y
V2Y
y
Figrue 5.2 Transformer Model for positive, negative and zero sequences.
5.3.3
Transmission Line Model
The series sequence admittance matrix can be formed by
transforming the overall series admittance matrix in phase components to its
counterparts in sequence components frame as follows:
YZ012
T 1YZabc T
94
Similarly the sequence shunt admittance matrix is formed by
YS012
T 1YSabc T
If the transmission line is unbalanced or un-transposed, the
sequence admittance matrix will be full and unsymmetrical. Figure 5.3 depicts
the model of a transmission line.
Ii012
Ij012
Bus i
Yz
Vi012
Bus j
012
Ys012
Ys012
Vj012
2
2
Figure 5.3 Sequence coupled line model
5.3.4 Load Model
Loads are specified by constant real and reactive powers Pa, Qa, Pb,
Qb, Pc and Qc.
5.3.5
Formation
of
System
Admittance
Matrix
in
sequence
component frame
The system admittance matrix written in the sequence component
frame can be obtained by combining all the sub matrices formed for the above
individual power system components. The self admittance of any busbar is the
sum of the individual self admittance matrices at that bus bar. The mutual
95
admittance between any two branches is the sum of the individual mutual
admittance matrices from all the subsystems containing those two nodes. The
mutual sequence admittance matrix between sequence networks will be a null
matrix, if the network is balanced.
[Y 012 ]
Y 00
Y 01
Y 02
Y10
Y 20
Y11
Y 21
Y12
Y 22
Ykm (k, m = 0,1,2) is the sequence sub matrix with the size of nxn.
The size of Y01 2 is 3nx3n.
5.4
CLASSIFICATION OF BUSES FOR THREE-PHASE LOAD
FLOW PROBLEM
The complete definition of single-phase power flow requires the
knowledge of four variables at each ith node (bus) as given below.
real or active power of ith node
reactive power of ith node
voltage magnitude at ith node
voltage phase angle at ith node.
Only two variables are known a priori to solve the problem, and the
aim of the load flow is to solve the remaining two variables at a bus. For
three-phase power flow analysis, buses are classified into four types based on
the known variables.
96
5.4.1
Load or PQ Buses
In these buses, the active and the reactive power injections are
specified (load buses and some generator buses with fixed active and reactive
power outputs). This type is the same as the conventional PQ type. The total
number of buses of this type in a network is denoted by ‘npq’.
5.4.2
Generator Terminal Buses
Buses similar to conventional PV buses but with bus reactive power
injection specified are referred as Generator Terminal buses. (The latter makes
up the deficiency of specified conditions at the corresponding internal buses).
Only the terminal buses of generators belong to this type. Incidentally, the
specified voltages are the positive sequence voltages and are specified directly
or obtained from simple calculations based on other kinds of voltage
specifications. The total number of buses of this type is denoted by ‘ng’.
5.4.3
Generator Internal Buses
These are buses with only bus active power injections specified.
Only the internal buses of generators, except the slack bus, belong to this type.
In these buses, positive sequence voltage magnitude alone is specified and the
negative and zero sequence voltages are zero, since the internal voltages are
balanced. The total number of buses of this type is denoted by ‘ng’.
5.4.4
Slack Bus
One of the generator buses is referred as slack bus (swing bus). At
the slack bus bar, the positive sequence voltage magnitude and the angle are
specified and negative and zero sequence voltages are zero. This bus is
numbered last.
Total number of buses in the system : n = npq+ng+1
97
5.5
FORMULATION OF OBJECTIVE FUNCTION
The objective of the three-phase load flow is to determine the zero,
positive and negative sequence voltages at all PQ buses and zero and negative
sequence voltages at generator buses.
Let Vik and Iik (k = 0, 1, 2) be the bus voltage and bus current
injections of sequence k at bus i.
i = 1,2,…n;
I
YV
The above equation is decomposed into 3 equations
I0
Y 00 V 0 Y 01V1 Y 02V 2
I1
Y10 V 0 Y11V1 Y12 V 2
I2
Y 20 V 0
Y 21V1 Y 22 V 2
where Ik and Vk (k = 0,1,2) are current and voltage vectors consist of n
elements each.
Ik
Vk
I1k
I k2
V1k
I 3k
V2k
....
V3k
I kn
....
t
V nk
t
Yks is sequence component bus admittance submatrix; (k, s = 0,1,2)
Total three-phase power
S ia b c
Siabc
V ia I ia *
Piabc
jQiabc
V ib I ib *
V ic I ic *
(5.1)
98
Taking conjugate on both sides
S ia b c *
P ia b c
jQ ia b c
Piabc
Pia
Pib
Pic
Q iabc
Q ia
Q ib
Qic
Let
SRPi
where
Pim
Pia
PGim
Pib
V ia * I ia
Pic / 3 and SRQi
PDim and Q im
V ib * I ib
Qia
V ic * I ic
Qib Qic / 3
m
m
Q Gi
Q Di
m
PGim and QGi
are the specified real and reactive power generation
respectively at bus i on phase m (m = a, b, c)
PDim and Q mDi are the specified real and reactive power demand
respectively at bus i on phase m (m = a, b, c)
Power is invariant
Per phase power = Power in sequence frame
1 a* a
Vi Ii Vib*Iib Vic*Iic
3
Vi0*Ii0 Vi1*I1i Vi2*Ii2
Calculated Real Power
CRP i
realpart Vi0* I 0i
Vi1* I1i
Vi2* I i2
99
n
n
Vi0* (
Yij00 Vj0
j 1
Yij02 Vj2 )
j 1
n
1*
= RP Vi (
n
Yij01V1j
j 1
n
n
Yij10 Vj0
Yij11Vj1
j 1
Yij12 Vj2 )
j 1
n
j 1
n
Vi2* (
Yij20 Vj0
n
Yij21Vj1
j 1
(5.2)
Yij22 Vj2 )
j 1
j 1
Calculated Reactive Power
imaginary part Vi0* I i0
CRQ i
n
Vi0* (
n
Yij00 Vj0
Vi1* I1i
n
Yij01Vj1
Yij02Vj2 )
j 1
j 1
j 1
n
n
n
= IP Vi1* (
Yij10 Vj0
j 1
Yij11Vj1
j 1
n
Vi2* (
j 1
Yij12 Vj2 )
n
Yij21Vj1
j 1
(5.3)
j 1
n
Yij20 Vj0
Vi2* I i2
Yij22 Vj2 )
j 1
Power mismatch equations are
Fpi
CRPi SRPi
(5.4)
Fqi
CRQi SRQi
(5.5)
Y is derived from modeling of the three-phase system. The
objective is to find the bus sequence voltages (V i0, Vi1, Vi2, i is the bus
number) such that the calculated real and reactive powers are equal to the
specified real and reactive powers (CRPi = SRPi and CRQi = SRQi ) . Therefore
the sequence voltages which make Fpi and Fqi are zero (<0.001) are the results
of the load flow problem.
100
The load flow problem of three-phase unbalanced networks is
formulated as an optimization problem as enumerated below.
Minimize
npq 2ng
Fpi2
Fqi2
f (V 0 , V1 , V 2 )
(5.6)
i 1
Subjected to reactive power limit and voltage limit
Real and reactive power mismatch are calculated for load and
generator terminal buses.
Real Power mismatch alone is calculated for generator internal
buses.
Voltage mismatch is calculated for generator terminal buses.
5.6
PROPOSED ALGORITHMS FOR THREE-PHASE POWER
FLOW PROBLEM
5.6.1
MHPSO Algorithm
1.
Form System admittance matrix in sequence component
frame.
2.
Initialization: Initialize a population of particles for positive,
negative and zero sequence voltages (real part and imaginary
part) with random positions and velocities. (For better results,
the initial random values chosen for real part of V0 and V2 are
between 0 and 0.1. For real part of V1, the random values are
varying between 0.9 and 1.0 and for the imaginary parts the
values are varying between 0.0 and 0.05 );
X = [V0 V1 V2]
101
V0 = Vid0, V1 = Vid1; V2 = Vid2; i = 1,2,3, …n; d=1,2,3, …Np;
V0 =
For example
[V110 V120 V130 …V1n0]t;
[V210 V220 V230 …V2n0]t;
.
.
.
[Vnp10 Vnp20 Vnp30 …Vnpn0]t ;
where n = number of buses; Np = population size;
Similarly for V1 and V2 the random values are chosen.
3.
Set iteration count G = 1
4.
Fitness Evaluation: For each particle set, calculate the fitness
value
n
n
Vid0* (
CRP id
Yij01V jd1
Yij02 V jd2 )
j 1
j 1
j 1
n
n
n
RP Vid1* (
Yij10 V jd0
j 1
Yij11V 1jd
j 1
n
n
Yij20 V jd0
j 1
Yij21V jd1
j 1
n
Vid0* (
Yij12 V jd2 )
j 1
n
Vid2* (
CRQ id
n
Yij00 V jd0
j 1
n
Yij00 V jd0
Yij22 V jd2 )
n
Yij01V jd1
Yij02 V jd2 )
j 1
j 1
j 1
n
n
n
IP Vid1* (
Yij10 V jd0
Yij11V jd1
j 1
j 1
n
n
Vid2* (
Yij20 V jd0
j 1
Yij12 V jd2 )
j 1
n
Yij21 V jd1
j 1
Yij22 V jd2 )
j 1
FPid = CRPid – (Pia + Pib + Pic )/3;
FQid = CRQid – (Qia + Qib + Qic )/3;
Pim and Qim (m=a,b,c) are specified in the load data
102
Fitness Fd = (FPid) 2 + (FQid )2
5.
Selection of pbest and gbest: From the population set, select
pbest and gbest by comparing the fitness values of present
particles and over all particles. The voltage vector set produces
the minimum fitness value is considered as the best particle.
pbest is the position of the best particle at that particular
iteration and gbest is the position of the best particle up to that
iterations.
6.
Breeding: From the randomly chosen parent particles,
produce child1 and child2 using Equations (2.3) and (2.4)
7.
Updating of voltage vector: To increase the convergence
speed, replace the particle which produces very high fitness
value by the average of the best two particles. Calculate the
velocity using Equations (2.1), (2.5) and (2.6) and update the
position of the particle (voltage) according to Equation (2.2).
8.
Check Convergence: If the fitness value (power mismatch) is
lesser than the limit (0.001), then stop the iteration. Calculate
phase voltages from sequence voltages, other wise increase the
iteration count G and go to step 4
5.6.2
MDEPSO Algorithm
MDEPSO algorithm explained in section 3.3.2 is applied to solve
the three-phase load flow problem. The voltage vector consists of three
sequence voltage vectors namely positive, negative and zero sequences.
103
5.7
CASE STUDY
The above said algorithms are tested on a sample 5 bus three-phase
system (Lo and Zhang 1993). In this system, both the load and network are
unbalanced. The system network is shown in Figure A 5.1. The data for this
sample system are given in Appendix 4. The population size chosen for the
simulation of this system is 50. The number of epochs in the EAs is 500.
About 20 independent trials are considered and the best results are considered
as the optimal solution. Almost all the trials provide the results with the
expected accuracy (power mismatch<0.001). To validate the performances of
EAs, the same problem is solved using conventional Newton Raphson (NR)
algorithm. The results are same as obtained by EAs and they prove the ability
of the proposed algorithms to solve the unbalanced load flow problem. NR
algorithm requires the formation and inversion of Jacobian matrix. This
process is complicated, since the Jacobian elements are derived by taking the
derivatives of bus power equations.
For three-phase systems, 6n power equations are required. Jacobian
elements are formed using 2*6n equations. Since the proposed algorithm
eliminates the formation of jacobian matrix and its inversion, it highly reduces
the complexity of the problem. Table 5.1, 5.2 and 5.3 show the phase voltages
obtained using DE, MHPSO and MDEPSO algorithms respectively. Table 5.5
illustrates the statistical measures obtained using these algorithms. Statistical
measures indicate the superior performance of the evolutionary algorithms.
Figure 5.4 shows the convergence characteristics.
104
Figure 5.4 Convergence characteristics for three-phase 5 bus system
Table 5.1 Phase voltages of three-phase 5 bus system using DE
Phase A
Phase B
Phase C
Bus
No
Magnitude Angle Magnitude
Angle
Magnitude
Angle
(p.u)
(deg)
(p.u)
(deg)
(p.u)
(deg)
1
1.0477
30.7
1.0531
-89.58
1.0432
150.89
2
1.0494
28.41
1.0529
-91.58
1.0479
148.41
3
1.0301
26.19
1.0468
-93.64
1.0263
147.12
4
1.0496
6.1
1.0509
-113.96
1.0494
126.00
5
1.0445
-0.10
1.0453
-120.10
1.0450
119.87
105
Table 5.2 Phase voltages of three-phase 5 bus system using MHPSO
Phase A
Phase B
Bus
Magnitude Angle Magnitude Angle
No
(p.u)
(p.u)
(deg)
(deg)
Phase C
Magnitude
(p.u)
Angle
(deg)
1
1.0477
30.7
1.0531
-89.58
1.0432
150.89
2
1.0494
28.22
1.0529
-91.67
1.0479
148.41
3
1.0308
26.17
1.0466
-93.64
1.0263
146.27
4
1.0496
6.08
1.0509
-113.96
1.0494
126.00
5
1.0447
-0.10
1.0453
-120.10
1.0450
119.87
Table 5.3 Phase voltages of three-phase 5 bus system using MDEPSO
Phase A
Bus
No
Phase B
Phase C
Angle
(deg)
Magnitude
(p.u)
1.0531
-89.58
1.0432
150.89
28.22
1.0529
-91.67
1.0479
148.41
1.0308
26.17
1.0466
-93.64
1.0263
146.27
4
1.0496
6.08
1.0509
-113.96
1.0494
126.00
5
1.0447
-0.10
1.0453
-120.10
1.0450
119.87
Angle
(deg)
Magnitude
(p.u)
1.0477
30.7
2
1.0494
3
Magnitude
(p.u)
Angle
1
(deg)
Table 5.4 Phase voltages of three-phase 5 bus system using NR
Phase A
Bus
No
Phase B
Phase C
Magnitude
Angle
Magnitude
Angle
Magnitude
Angle
(p.u)
(deg)
(p.u)
(deg)
(p.u)
(deg)
1
1.0477
30.7
1.0531
-89.58
1.0432
150.89
2
1.0494
28.22
1.0529
-91.67
1.0479
148.41
3
1.0308
26.17
1.0466
-93.64
1.0263
146.27
4
1.0496
6.08
1.0509
-113.96
1.0494
126.00
5
1.0447
-0.10
1.0453
-120.10
1.0450
119.87
106
Table 5.5
Statistical performances of Evolutionary Algorithms for
three-phase 5 bus system
Best
Worst
Mean
Standard
Deviation
DE
0.0001
0.0005
0.000165
0.0001268
MHPSO
0.0001
4.000*10-4
1.200*10-4
6.9585*10-5
1.000*10-4
0.0005
1.5500*10-4
1.3563*10-4
Algorithm
MDEPSO
5.8
CONCLUSION
In this chapter, three-phase power flow problem in sequence
component frame of reference has been formulated as an optimization
problem and evolutionary algorithms have been applied to solve this
unbalanced load flow problem. The problem considers the unbalancing in
both system and load side. These algorithms are tested on a sample threephase unbalanced power system. The algorithms are proven to give accurate
results with reduced complexity. The proposed algorithm eliminates the
formation of Jacobian matrix and hence reduces the problem complexity to a
greater extent.