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Improvements in Bidirectional Power Flow Balancing and Electric Power Quality of a Microgrid with Unbalanced Distributed Generators and Loads by Using Shunt Compensators.pdf


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Energies 2018, 11, x FOR PEER REVIEW

3 of 15

P DG 2  jQ DG 2
Energies 2018, 11, 3305
Energies 2018, 11, x FOR PEER REVIEW

Bus 0

Bus 1

DG

bc

Bus 2

P DG 4  jQ DG 4

Bus 3

Bus 4

DG

bc

Bus 5

P DG 5  jQ DG 5
DG

bc

3 of 14
3 of 15

Bus 6

abc
abc
abc
Three-Phase
abc
abc
abc
DG
5
 jQ DG 5
P DG 2  jQ DG 2
P DG 4  jQ DG 4 P
Power
Source
Cable
bc
ab DG
bc
ab DG
bc DG
Bus 1Line Bus 2
Bus 0
Bus 5
Bus 6ab
Bus 3ab
Bus 4
Three-Phase
Power
Source

abc

abc

abc

abc

P L 2  jQ L 2

Cable

P L 3  jQ L 3

ab

abc

abc

P L 4  jQ L 4

P L 6  jQ L6

ab

ab

ab
Figure 1. Radial-type, three-phase
microgrid with unbalanced distributed generators (DGs)
and
Line
loads.
L2
L2
L6
L6
L3
L3
L4
L4

P

 jQ

P

 jQ

P

P

 jQ

 jQ

The
symmetrical
components
method with
can unbalanced
simplify the
unbalanced
microgrid
Figure
1. Radial-type,
three-phase microgrid
distributed
generators
(DGs) andsystem
loads. for
Figure 1. Radial-type, three-phase microgrid with unbalanced distributed generators (DGs) and
conducting
a
steady-state
analysis.
The
required
compensation
principle
of
the
proposed
shunt
loads.
The
symmetrical
components
method
can
simplify
the
unbalanced
microgrid
system
for
compensator can also be derived. Figure 2 presents the equivalent circuit model between two
conducting
a
steady-state
analysis.
The
required
compensation
principle
of
the
proposed
shunt
neighboring
buses in Figure
1. Equation
(1) can
obtained the
by applying
Kirchhoff’s
Voltage
Law.
The symmetrical
components
method
canbesimplify
unbalanced
microgrid
system
for
compensator can also be derived. Figure 2 presents the equivalent circuit model between
two
l
l
Zaa , shunt
Z
Equation
(2) apresents
the impedance
matrix
of the three-phase
distribution
conducting
steady-state
analysis. The
required
compensation
principle lines,
of thewhere
proposed
bb ,
neighboring
buses in Figure 1. Equation
(1) can bel obtained by applying Kirchhoff’s Voltage Law.
l
compensator
can also be derived.
2 presents
equivalent
circuit model
between
two
Zccl are
Zbcl , and
Zca three-phase
l
l
and
and Zab Figure
, matrix
arethe
mutual
impedance.
general,
theZmutual
Equation
(2) self-impedance
presents the impedance
of the
distributionInlines,
where
aa , Zbb ,
neighboring
buses
in
Figure
1.
Equation
(1)
can
be
obtained
by
applying
Kirchhoff’s
Voltage
Law.
l arecan
l
l distribution
l
impedance
be neglected and
in a Z
power
system [25]. By combining Equations (1) and
and Zcc
self-impedance
l
ab , Zbc , and Zca are mutual impedance. In general, the lmutual
Zaaand
Equation
(2)
presents
the
impedance
matrix
of
the
three-phase
lines,
where (1)
, Z(2),
bb ,
(2),
Equationcan
(3)be
is neglected
obtained. in
Byausing
symmetrical
components
the
sequence
networks
impedance
powerthe
distribution
system
[25]. distribution
By method,
combining
Equations
l
are
by
using
Equation
(4). In
(4),ZTcomponents
is the symmetrical
components
transformation
Zccl (3)
Zabl Equation
Zbcl , and
andderived
are
and
,symmetrical
impedance.
In general,
the mutual
Equation
isself-impedance
obtained.
By using
the
method,
the sequence
networks
are
ca are mutual
−1 is the inverse symmetrical components transformation matrix, as presented in
matrix
and
T
derived
by using
Equation
(4). in
In aEquation
(4), T is the symmetrical
transformation
matrix
impedance
can be
neglected
power distribution
system [25].components
By combining
Equations (1)
and
−1 is
Equation
(5).
and
T
the
inverse
symmetrical
components
transformation
matrix,
as
presented
in
Equation
(5).
(2), Equation (3) is obtained. By using the symmetrical components method, the sequence networks
are derived by using Equation (4). In Equation (4), T is the symmetrical components transformation
2 Bus 2
l
1 Bus 1
V acomponents
matrix and T−1 is the
symmetrical
transformation matrix, as presented in
I a2
V a inverseZ aa
Equation (5).
l
2
L2
l
l
Z bb

1

Bus 1

l
2
Zl
Z ccl aa Z bc II a2
c

1

Z cal

V1
V ca
Vb
V

Z ab I 2
b

Z ca

1

Vb

1
c

Z bbl
Z

l
cc

l
Z ab
I b2

Z bcl

I

2
c

Ia

Vb

2 Bus
DG 2 2
2
VVc a I b

I aDG 2  0
I aL 2

2

Vb

I cDG22 DG 2
V I b I DG 2
DGc
bc

I cL 2  0
I aDG 2  0

I bL 2

Load

I cL 2 two
 0 buses.
Figure 2.
2. Equivalent
Equivalentcircuit
circuit model between
Figure
I cDG 2 model between two buses.
I bL 2
DG 2
I
DG
Load
bc

I abL 2

I abL 2

11
22
22
l
VVa,b,c
−ZZal ,a,b,c
I ,c
 V a,b,c
a ,b , c=
a ,b ,c 
b , c I a , ba,b,c

Figure 2. Equivalentcircuit model between
 two buses.
l
Zlaa
  Z aal
l
Za,b,cl 2=  Z1lba
ZVaa,b,b,c,c VZabal,b,c
 Zlca

ll
l l
ac

ab ZZ
ZZab
ac
l
l 
Zl bb
Z
l2bc

l Z
bbl
bc 
ZZZ
a ,b , c IZal, b, c
l cb
l cc
Z ca Z cb Z cc 
 2   1   l
 2 
l
l
Z aaZ l Z ab
lZ ac Z l
Va
Va
Ia
Zab
aa
ac


l
l
l
 2  2  Z 1l 1  
 2 
l Z bb
lZ bc
l 2
Z







=
a
,
b
,
c
ba
Z
Z
Z
 V b  V a  VVb a   l ba l bb l  bcI a  I b 
  1   Z laa l Zlab l Zlac l 
2 
2
V c V 2  VVc 1   ZZcal ZcaZZcbl ZcbZZccl ZccI 2  I c
b
b
b
ba
bb
bc 

   
  
 2 
 2 
l 
l
l
2
V c22 V1V 1c1  Z ca ZZcbl ZZccl   I Z
l
Va
Ia
c2
a 
aa
ac




ab





l
l
l  I a
−V1 
1 Va −
a
 −1  2 
2 
−1 
1aa Zlab Z ac

Z
l
l






T  Vb  = T 
V b 1 − T  Zba Zbb Z2bc  TT  I b 
2
l
2 V b   1V
1   Zl
l  IZb l  2 
2
2
Zlbbl l Z


Zbccb
l  cc I a
Vc
Ic
V a   VcV a b   balZ l Zca
  2    1   Z aa Z lZ ab Z lZ ac  2   
2 V
cc   

V1 c
 ca  cb
c 1  2 
 b  T 1  Z bal 1 Z bbl 1 Z bcl1ITT
 I b
T 11V b 1  cT11 V
 
 
 
1
l
l
l
1 −1
  2 2



 Z ca 1 Z cb a Z cca2 , a=2 e j2π/3
V, c1T

T =  1 V ca2
=
a


2


V 
V 
IIc a
l
l 2
l
3Z aa
1  aa a2  a 
1 Z ab
a Zaac 
 
2
1
2


T 1 V b   T 1 V b   T 1  Z bal Z bbl Z bcl  TT 1  I b 
 
 


 Z cal Z cbl Z ccl 
V c2 
V 1c 
 I c2 


 
 
 

(1)
(1)
(2)
(2)
(1)

(2)
(3)
(3)

(4)
(3)
(4)

(5)
(4)