Chapter 2
Power System Fundamentals: Balanced
Three-Phase Circuits
This chapter reviews the fundamentals of balanced three-phase alternating current
(ac) circuits. First, we define positive and negative balanced three-phase sequences.
Second, we analyze balanced three-phase voltages and currents. Third, the different
types of power are defined and measurements techniques for power are briefly
reviewed. Fourth, we provide an overview of the analysis of balanced three-phase
circuits using the per-unit system. This chapter provides an appropriate background
of three-phase power for the unfamiliar reader, establishing the link between
the physical reality and analytical techniques. It can be skipped by readers with
knowledge of three-phase circuit analysis.
2.1 Introduction
Power systems are generally based on three-phase alternating current (ac) circuits.
This chapter describes the fundamentals of this type of circuits and is organized as
follows. Section 2.2 defines balanced three-phase sequences. Section 2.3 describes
balanced three-phase voltage and currents, as well as the two different symmetrical
connections of system components and the equivalence among them. Section 2.4
defines instantaneous, active, reactive, and apparent powers and explains how to
measure them. Section 2.5 clarifies why three-phase power is generally preferred
over single phase-phase power. Section 2.6 defines the per-unit system, which is
used in the remaining chapters of this book. Section 2.7 summarizes the chapter
and suggests some references for further study. Finally, Sect. 2.8 proposes some
exercises for further comprehending the concepts addressed in this chapter.
© Springer International Publishing AG 2018
A.J. Conejo, L. Baringo, Power System Operations, Power Electronics and Power
Systems, https://doi.org/10.1007/978-3-319-69407-8_2
17
18
2 Power System Fundamentals: Balanced Three-Phase Circuits
2.2 Balanced Three-Phase Sequences
There are two ways of representing an ac source:
1. Using a sinusoidal representation:
a.t/ D
p
2Asin .!t C
/;
(2.1)
where:
• A is the root mean square (RMS) value of the source,
• ! is its angular frequency (also known as angular speed) measured in radians
per second, and
•
is its initial phase angle.
The RMS value of the source is computed as:
s
AD
1
T
Z
T
a2 .t/dt;
(2.2)
0
where T is the period (measured in seconds).
The angular frequency ! is defined as the rate of change of the phase of the
sinusoidal source and is computed as:
!D
2
D 2f ;
T
(2.3)
where f is the ordinary frequency (measured in Hertz).
2. Using a phasorial representation:
AN D A† :
(2.4)
Figure 2.1 illustrates the relationship between a sinusoidal ac source (left plot)
and a rotating vector or phasor (right plot). Observe that the projection of the rotating
vector on the imaginary axisp(right-hand-side of the figure) renders the sinusoidal
form of the source: a.t/ D 2Asin .!t C /, shown on the left-hand side of the
figure.
If three ac sinusoidal sources (or phasors) have equal magnitude and equal angle
ı
separation . 2
3 rad or120 /, then they constitute a balanced three-phase sequence.
For example, the following three ac sources constitute a balanced three-phase
sequence:
2.2 Balanced Three-Phase Sequences
19
a(t)
Im
√
2A
(w t + y )
y
wt
Re
y
Fig. 2.1 Relationship between a sinusoidal AC source (left) and a rotating vector (right)
8
p
ˆ
C /;
ˆ
<aA .t/ D p2Asin .!t
a .t/ D 2Asin !t C 2
;
3
ˆ B
:̂a .t/ D p2Asin !t C C 2 :
C
3
(2.5)
Since aA .t/, aB .t/, and aC .t/ constitute a balanced three-phase sequence, then we
have:
aA .t/ C aB .t/ C aC .t/ D 0:
(2.6)
The reader is encouraged to verify that (2.6) is correct.
We may also represent the balanced sinusoidal sources in (2.5) using phasors,
i.e.:
8
ˆ
N
ˆ
<AA D A† ;
(2.7)
;
AN D A† 2
3
ˆ B
:̂AN D A† C 2 :
C
3
Figure 2.2 shows a balanced three-phase sequence using phasors, where the
initial phase is 0. Note that the phasor denoted by AN A is leading 2=3 rad the
phasor denoted by AN B and lagging 2=3 rad the phasor denoted by AN C . In this case,
the balanced three-phase sequence is denominated positive sequence.
If phases B and C are swapped, we obtain the so-called negative sequence that is
shown in Fig. 2.3 and represented by the following phasors:
20
2 Power System Fundamentals: Balanced Three-Phase Circuits
Fig. 2.2 Balanced
three-phase positive sequence
A¯C
2p
3
2p
3
rad
A¯A
rad
2p
3
rad
2p
3
rad
A¯B
Fig. 2.3 Balanced
three-phase negative
sequence
A¯B
2p
3
rad
AA
2p
3
rad
AC
8
ˆ
N
ˆ
<AA D A†0;
;
AN D A† C 2
3
ˆ B
:̂AN D A† 2 :
C
3
(2.8)
In power systems, the reference phasor is generally indicated using the letter R,
rad) using the letter S, and the phasor leading 120ı
the phasor lagging 120ı (or 2
3
2
(or 3 rad) using the letter T. That is:
2.3 Balanced Three-Phase Voltages and Currents
21
8
ı
ˆ
N
ˆ
<AR D A†0 ;
AN D A† 120ı ;
ˆ S
:̂AN D A† C 120ı :
T
Therefore, in power systems, a balanced three-phase positive sequence is generally represented as RST, while a negative one as RTS.
2.3 Balanced Three-Phase Voltages and Currents
In this section, we analyze voltages and currents in balanced three-phase circuits
with different connections.
2.3.1 Balanced Three-Phase Voltages
In a balanced three-phase circuit, we identify two balanced voltage sequences,
namely phase voltages and line voltages, as described below and illustrated in
Fig. 2.4.
Phase voltages are defined as the voltages between each phase and a reference
point known as “common star point,” usually denoted as N (upper plot of Fig. 2.4).
That is:
Fig. 2.4 Phase (upper plot)
and line (lower plot) voltages
in a balanced three-phase
network
R
S
T
+
+
ŪR
−
R
S
T
ŪS
−
+
ŪT
−
N
+
−
ŪRS
−
ŪT R
+
ŪST
+
−
22
2 Power System Fundamentals: Balanced Three-Phase Circuits
8
ı
ˆ
N
N
ˆ
<UR D URN D UF †0 ;
N SN D UF † 120ı ;
N DU
U
ˆ S
:̂U
N TN D UF † C 120ı ;
NT D U
(2.9)
where UF is the magnitude of the phase voltage.
Since phase voltages constitute a balanced sequence, we have:
NR CU
NS CU
N T D 0:
U
(2.10)
On the other hand, each line voltage is defined as the difference of two phase
voltages (lower plot of Fig. 2.4). That is:
8
p
ı
ˆ
N
N
N
ˆ
<URS D UR US D p3UF †30 ;
NS U
N T D 3UF † 90ı ;
N DU
U
ˆ ST
p
:̂U
NT U
N R D 3UF †150ı :
N TR D U
(2.11)
Note that:
N ST C U
N TR D 0;
N RS C U
U
(2.12)
i.e., the line voltages also constitute a balanced sequence.
Phasor diagrams for both phase voltages and line voltages are shown in Fig. 2.5.
Fig. 2.5 Balanced phase and
line voltages
ŪT R
ŪRS
ŪT
30◦
90◦
30◦
ŪR
90◦
ŪS
90◦
30◦
ŪST
2.3 Balanced Three-Phase Voltages and Currents
23
2.3.2 Balanced Three-Phase Currents
In a balanced three-phase network, the line currents constitute a balanced sequence,
i.e.:
8
ı
ˆ
N
ˆ
<IR D IL †.' 0/ ;
(2.13)
IN D IL †.' 120/ı ;
ˆS
:̂IN D I †.' C 120/ı ;
T
L
where:
• IL is the magnitude of the line current and
• ' is the angle of a phase voltage with respect to the corresponding line current.
Line currents are shown in Fig. 2.6.
Note that:
INR C INS C INT D 0:
(2.14)
Illustrative Example 2.1 Currents in a balanced three-phase delta-connected load
We consider the balanced three-phase delta-connected load (impedance ZN per
phase) depicted in Fig. 2.7. We show below that if this load is supplied by a balanced
three-phase line-current sequence (INR , INS , INT ), then the delta currents, i.e., the currents
“inside” the delta (INRS , INST , INTR ), constitute a balanced sequence as well.
From Fig. 2.7, we obtain:
Fig. 2.6 Balanced line
currents
-
Fig. 2.7 Balanced delta
currents
R
I¯R
I¯S
I¯T
I¯R
r
?I¯RS
Z̄
I¯S
S
-
T
-
I¯T
rI¯ST
Z̄
Z̄
¯
6
r IT R
24
2 Power System Fundamentals: Balanced Three-Phase Circuits
8
ˆ
N
N
N
ˆ
<IR C ITR D IRS ;
IN C INRS D INST ;
ˆS
:̂IN C IN D IN :
T
ST
TR
(2.15)
Since:
INR C INS C INT D 0
and:
INS C INT C INR D 0;
subtracting the above two expressions renders:
INRS C INST C INTR D 0:
Considering (2.15) and (2.16), we obtain:
8
D INS ;
< INRS C INST
INST C INTR D INT ;
: N
IRS C INST C INTR D 0;
which in matrix form is:
32 3 2 3
1 1 0
INS
INRS
4 0 1 1 5 4 INST 5 D 4 INT 5 :
1 1 1
0
INTR
2
Solving for the delta currents yields:
31 2 3
3 2
1 1 0
INS
INRS
4 INST 5 D 4 0 1 1 5 4 INT 5
1 1 1
0
INTR
2
or:
3
32 3
2
2 1 1
INRS
INS
4 INST 5 D 1 4 1 1 1 5 4 INT 5 :
3
0
INTR
1 2 1
2
(2.16)
2.3 Balanced Three-Phase Voltages and Currents
25
Thus, we can express delta currents as a function of line currents as:
8
1 N
ˆ
ˆ
INRS D
2IS INT ;
ˆ
ˆ
ˆ
3
ˆ
<
1 N
IS INT ;
IST D
ˆ
3
ˆ
ˆ
ˆ
ˆ
1 N
:̂ITR D
IS C 2INT :
3
Since INR C INS C INT D 0, we get:
8
ˆ
ˆ
ˆINRS D
ˆ
ˆ
ˆ
<
INST D
ˆ
ˆ
ˆ
ˆ
ˆ
:̂INTR D
1 N
IR INS ;
3
1 N
IS INT ;
3
1 N
IT INR ;
3
or:
8
ˆ
ˆ
INRS D
ˆ
ˆ
ˆ
ˆ
<
INST D
ˆ
ˆ
ˆ
ˆ
ˆ
:̂INTR D
1p N
3IR † C 30ı ;
3
1p N
3IS † C 30ı ;
3
1p N
3IT † C 30ı ;
3
or finally:
8
ˆ
ˆ
INRS D
ˆ
ˆ
ˆ
ˆ
<
INST D
ˆ
ˆ
ˆ
ˆ
ˆN
:̂ITR D
1
p IL † .' C 30/ı ;
3
1
p IL † .' 90/ı ;
3
1
p IL † .' C 150/ı :
3
We conclude that if the line currents used to supplied a balanced delta-connected
load constitute a balanced three-phase sequence, then the delta currents INRS , INST ,
and INTR have equal magnitude and equal angle separation, i.e., the delta currents
constitute a balanced three-phase sequence as well.
Figure 2.8 visualizes the relationship between line currents and delta currents in
a balanced three-phase delta-connected load.
26
2 Power System Fundamentals: Balanced Three-Phase Circuits
Fig. 2.8 Balanced line and
delta currents in a balanced
delta-connected load
I¯T
I¯T R
I¯RS
30◦
30◦
I¯R
30◦
I¯S
I¯ST
R
I¯R
I¯T R
I¯RS
I¯S
S
I¯ST
I¯T
T
Fig. 2.9 Balanced three-phase delta-connected generator
Illustrative Example 2.2 Currents in a balanced three-phase delta-connected generator
We obtain below the relationships between the line and delta currents in a
balanced delta-connected generator as the one in Fig. 2.9. Note that a simplified
circuit diagram is used to show exclusively the reference directions for the currents.
If instead of a delta-connected load we consider a delta-connected generator, the
current balance equations become (Fig. 2.9):
8
ˆ
N
N
N
ˆ
<IR C ITR D IRS ;
IN C INRS D INST ;
ˆ S
:̂IN C IN D IN :
T
ST
TR
Since INRS C INST C INTR D 0, we get (see (2.16)):
8
ˆ
ˆ
INRS D
ˆ
ˆ
ˆ
ˆ
<
INST D
ˆ
ˆ
ˆ
ˆ
ˆ
:̂INTR D
1 N
IS INR ;
3
1 N
IT INS ;
3
1 N
IR INT ;
3
2.3 Balanced Three-Phase Voltages and Currents
27
or:
8
ˆ
ˆ
ˆINRS D
ˆ
ˆ
ˆ
<
INST D
ˆ
ˆ
ˆ
ˆ
ˆ
:̂INTR D
1p N 3 IR † C 30ı ;
3
1p N 3 IS † C 30ı ;
3
1p N 3 IT † C 30ı ;
3
or similarly:
8
ˆ
ˆ
INRS D
ˆ
ˆ
ˆ
ˆ
<
INST D
ˆ
ˆ
ˆ
ˆ
ˆ
:̂INTR D
1p N
3IR † 150ı ;
3
1p N
3IS † 150ı ;
3
1p N
3IT † 150ı ;
3
8
ˆ
ˆ
INRS D
ˆ
ˆ
ˆ
ˆ
<
INST D
ˆ
ˆ
ˆ
ˆ
ˆ
:̂INTR D
1p N
3IS † 30ı ;
3
1p N
3IT † 30ı ;
3
1p N
3IR † 30ı :
3
(2.17)
and finally:
Considering:
8
ı
ˆ
N
ˆ
<IR D IG †0 ;
IN D IG † 120ı ;
ˆS
:̂IN D I † C 120ı ;
T
(2.18)
G
where IG is the magnitude of the line current, we have:
8
ˆ
ˆ
INRS D
ˆ
ˆ
ˆ
ˆ
<
INST D
ˆ
ˆ
ˆ
ˆ
ˆN
:̂ITR D
1
p IG † 150ı ;
3
1
p IG † C 90ı ;
3
1
p IG † 30ı :
3
(2.19)
From Eqs. (2.18) and (2.19) above, we conclude that the line and delta currents
in a balanced three-phase delta-connected generator are balanced sequences.
28
2 Power System Fundamentals: Balanced Three-Phase Circuits
Fig. 2.10 Voltage sources:
wye (upper plot) and delta
(lower plot) connected
ĒR
Z̄y
+
I¯S
ĒT
− ∼ +
I¯T
−
N
Z̄y
∼
+
∼
I¯T R
E¯A
R
S
T
I¯R
R
−
Z̄ d
−
I¯R
∼
ĒS
Z̄y
−
∼ E¯ C
+
+
−
Z̄d
∼
+
I¯RS
Z̄ d
E¯B
I¯ST
I¯S
I¯T
S
T
2.3.3 Equivalence Wye-Delta
To preserve balance, three balanced voltage sources can be either wye (y) or delta
(d) connected, as shown in the upper and lower plots of Fig. 2.10, respectively. In
the wye connection, point N, known as the common start point, is considered as the
reference for phase voltages.
We consider the balanced voltage source sequence:
8
ı
ˆ
N
ˆ
<ER D EF †0 ;
EN D EF † 120ı ;
ˆ S
:̂EN D E † C 120ı ;
T
(2.20)
F
where EF is the magnitude of each voltage source, while the balanced voltage-source
current sequence is:
8
ı
ˆ
N
ˆ
<IR D IL † .' 0/ ;
IN D IL † .' 120/ı ;
ˆS
:̂IN D I † .' C 120/ı ;
T
L
(2.21)
2.3 Balanced Three-Phase Voltages and Currents
29
where IL is the magnitude of the line current. We derive below equivalence
conditions for the wye and delta connections considering the circuits in Fig. 2.10.
On the one hand, these circuits should be equivalent under no load conditions,
i.e., if currents are null:
INR D INS D INT D 0:
(2.22)
Then:
8
p
p
ı
ı
ˆ
N
N
N
N
ˆ
<ER ES D EA D p3ER †30 D p 3EF †30 ;
EN EN T D EN B D 3EN S †30ı D 3EF † 90ı ;
ˆ S
:̂EN EN D EN D p3EN †30ı D p3E †150ı :
T
R
C
T
F
(2.23)
On the other hand, these circuits should be equivalent under load conditions as
well. Thus:
8
ˆ
N
N
N
ˆ
<ITR IRS D IR ;
(2.24)
IN INST D INS ;
ˆ RS
:̂IN IN D IN :
ST
TR
T
Considering (2.17) and (2.18) renders:
8
ˆ
INRS D
ˆ
ˆ
ˆ
ˆ
ˆ
<
INST D
ˆ
ˆ
ˆ
ˆ
ˆ
:̂INTR D
1
p INS † 30ı ;
3
1 N
p IT † 30ı ;
3
1 N
p IR † 30ı ;
3
(2.25)
1
p IG † 150ı ;
3
1
p IG † C 90ı ;
3
1
p IG † 30ı :
3
(2.26)
or:
8
ˆ
ˆ
INRS D
ˆ
ˆ
ˆ
ˆ
<
INST D
ˆ
ˆ
ˆ
ˆ
ˆN
:̂ITR D
Under load conditions, line voltages in both connections should be equal. Thus,
from Fig. 2.10, we have:
EN A C INRS ZN d D EN S C ZN y INS ZN y INR C EN R ;
(2.27)
30
2 Power System Fundamentals: Balanced Three-Phase Circuits
or:
EN A C INRS ZN d D EN R EN S C INS INR ZN y :
(2.28)
Considering (2.23), we obtain:
p
3INR † 150ı ZN y ;
(2.29)
p
1
p INR †150ı ZN d D 3INR † 150ı ZN y ;
3
(2.30)
p
1
p ZN d D 3ZN y ;
3
(2.31)
ZN d D 3ZN y :
(2.32)
EN A C INRS ZN d D EN A C
or:
as well as:
and finally:
We conclude that the equivalence conditions for the wye and delta connections
in the circuits in Fig. 2.10 are as follows:
8
p
ı
ˆ
N
N
ˆ
<EA D p3ER †30 ;
ı
EN D 3EN †30 ;
ˆ B p S
:̂EN D 3EN †30ı ;
C
T
(2.33)
and:
ZN d D 3ZN y ;
(2.34)
1
p EN A † 30ı ;
3
1 N
p EB † 30ı ;
3
1 N
p EC † 30ı ;
3
(2.35)
or alternatively:
8
ˆ
ˆEN R D
ˆ
ˆ
ˆ
ˆ
<
EN S D
ˆ
ˆ
ˆ
ˆ
ˆ
:̂EN T D
2.3 Balanced Three-Phase Voltages and Currents
Fig. 2.11 Illustrative
Example 2.3: balanced
three-phase circuit
31
Z̄iy
I¯1
ĒR
¯
− ∼ + ŪR I R
I¯RS
Ē S
Z̄iy
Ē T
¯
− ∼ + Ū T I T
I¯T R
I¯2
Z̄cd
¯
− ∼ + Ū S I S
Z̄iy
Z̄cd
I¯ST
Z̄cd
I¯3
and:
1
ZN y D ZN d :
3
(2.36)
Illustrative Example 2.3 Balanced three-phase circuit
We consider the circuit depicted in Fig. 2.11 in which voltage sources constitute a
known balanced three-phase positive sequence and impedances ZN iy and ZN cd are also
known. We compute below:
1. Currents INR , INS , and INT .
2. Currents INRS , INST , and INTR .
N R, U
N S , and U
N T.
3. Voltages U
Since voltage sources constitute a balanced three-phase positive sequence, we
have:
8
ˆ
N
ˆ
<ER ;
EN D ˛N 2 EN R ;
ˆ S
:̂EN D ˛N EN ;
T
R
where ˛N D 1†120ı . Additionally note that ˛N 2 D ˛N D 1†120ı
The circuit in Fig. 2.11 is solved below using the mesh-current method [5]:
2
2ZN iy C ZN cd
4 ZN cd
ZN iy
32 3 2
3
ZN cd
ZN iy
EN R EN S
IN1
5
3ZN cd
ZN cd 5 4 IN2 5 D 4
0
ZN cd 2ZN iy C ZN cd
IN3
EN S EN T
32
2 Power System Fundamentals: Balanced Three-Phase Circuits
or:
2
2ZN iy C ZN cd
4 ZN cd
ZN iy
3
32 3 2 ZN cd
ZN iy
EN R 1 ˛N 2
IN1
5:
3ZN cd
ZN cd 5 4 IN2 5 D 4 0
2
N
N
N
N
N
Zcd 2Ziy C Zcd
ER ˛N ˛N
I3
Solving for the currents:
3 2
2ZN iy C ZN cd
IN1
4 IN2 5 D 4 ZN cd
ZN iy
IN3
2
3
31 2 ZN cd
ZN iy
EN R 1 ˛N 2
5;
3ZN cd
ZN cd 5 4 0
2
N
N
N
N
Zcd 2Ziy C Zcd
ER ˛ ˛
and:
2
3
3
2
3
2
1
1
NI1
1 ˛N 2
6
7
N iy
1
Z
6
7 EN 4 0 5 ;
4 IN2 5 D
N iy C ZN cd 4 1 1 C ZN cd 1 5 R
3
Z
IN3
˛N 2 ˛N
1
1
2
2
or:
8
3EN R
ˆ
ˆ
ˆIN1 D
;
ˆ
ˆ
N
3Ziy C ZN cd
ˆ
ˆ
ˆ
ˆ
<
.1 ˛/
N EN R
;
IN2 D
ˆ
3ZN iy C ZN cd
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆIN D 3˛N EN R :
:̂ 3
3ZN iy C ZN cd
Thus, currents INR , INS , and INT are:
8
EN R
ˆ
NR D IN1 D
ˆ
I
;
ˆ
ˆ
ˆ
ZN iy C 13 ZN cd
ˆ
ˆ
ˆ
ˆ
<
.˛N 1/ EN R
˛N 2 EN R
D
;
INS D IN3 IN1 D
ˆ
ZN iy C 13 ZN cd
ZN iy C 13 ZN cd
ˆ
ˆ
ˆ
ˆ
ˆ
˛N EN R
ˆ
ˆINT D IN3 D
;
:̂
ZN iy C 13 ZN cd
(2.37)
2.3 Balanced Three-Phase Voltages and Currents
33
or:
8
EN R
ˆ
ˆINR D
;
ˆ
ˆ
ˆ
ZN iy C 13 ZN cd
ˆ
ˆ
ˆ
ˆ
<
EN S
;
INS D
N iy C 1 ZN cd
ˆ
Z
ˆ
ˆ
3
ˆ
ˆ
ˆ
EN T
ˆ
ˆINT D
:
:̂
ZN iy C 13 ZN cd
On the other hand, currents INRS , INST , and INTR are computed as:
8
ˆ
N
N
N
ˆ
ˆIRS D I1 I2
ˆ
ˆ
ˆ
ˆ
ˆ
<
INST D IN3 IN2
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆN
:̂ITR D IN2 D
D
N
ER .3 1 C ˛/
;
N
N
3Ziy C Zcd
D
N
EN R .3˛N 1 C ˛/
;
N
N
3Ziy C Zcd
EN R .˛N 1/
;
3ZN iy C ZN cd
or:
8
ˆ
ˆ
ˆINRS D
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
<
INST D
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆN
:̂ITR D
EN R 1 ˛N 2
;
3ZN iy C ZN cd
1
˛N 2 EN R 1 ˛
EN R ˛N 2 ˛N
EN S 1 ˛N 2
N
D
D
;
3ZN iy C ZN cd
3ZN iy C ZN cd
3ZN iy C ZN cd
1
˛N EN R 1 ˛
NER .˛N 1/
EN T 1 ˛N 2
N
D
D
:
3ZN iy C ZN cd
3ZN iy C ZN cd
3ZN iy C ZN cd
Note that the relationship between currents INRS , INST , INTR and INR , INS , INT is as follows:
INST
INTR
1 ˛N 2
INRS
D
D
D
D
3
INR
INS
INT
p
1
3†30ı
D p †30ı :
3
3
N S , and U
N T are computed as:
N R, U
Finally, voltages U
8
ˆ
N
N N
N
ˆ
<UR D ER IR Ziy ;
N D EN S INS ZN iy ;
U
ˆ S
:̂U
N D EN IN ZN ;
T
T
T iy
(2.38)
34
2 Power System Fundamentals: Balanced Three-Phase Circuits
or:
8
N N
ˆ
N R D EN R ER Ziy ;
ˆ
U
ˆ
ˆ
ˆ
ZN iy C 13 ZN cd
ˆ
ˆ
ˆ
ˆ
ˆ
<
N N
N S D EN S ES Ziy ;
U
ˆ
ZN iy C 13 ZN cd
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
EN T ZN iy
ˆ
ˆU
N
N
;
:̂ T D ET N
Ziy C 13 ZN cd
and finally:
0
1
8
ˆ
N
ˆ
Ziy
ˆ
N R D EN R @1 A;
ˆ
U
ˆ
ˆ
1 ZN
ˆ
N
C
Z
ˆ
iy
cd
ˆ
3
ˆ
ˆ
ˆ
0
1
ˆ
ˆ
<
N iy
Z
N S D EN S @1 A;
U
ˆ
ˆ
ZN iy C 13 ZN cd
ˆ
ˆ
ˆ
ˆ
0
1
ˆ
ˆ
ˆ
ˆ
N iy
Z
ˆ
ˆ
N D EN T @1 A:
ˆU
:̂ T
ZN iy C 13 ZN cd
N R; U
N S; U
N T ) constitute balanced threeNote that (INR ; INS ; INT ), (INRS ; INST ; INTR ), and (U
phase positive sequences.
Currents and voltages have been obtained by analyzing the three-phase circuit
depicted in Fig. 2.11. However, note that the resulting equations for line currents (2.37) and phase voltages (2.38) are decoupled per phase. Thus, instead of
using the three-phase circuit in Fig. 2.11, it is possible to use the three equivalent
single-phase circuits, one per phase, depicted in Fig. 2.12.
Illustrative Example 2.4 Equivalent single-phase circuits
We consider the balanced three-phase circuit depicted in Fig. 2.13. Impedances
ZN id and ZN cy , as well as voltage sources are known. Moreover, voltage sources
constitute a balanced three-phase positive sequence and, thus:
8
ˆ
N
ˆ
<EA ;
EN D ˛N 2 EN A ;
ˆ B
:̂EN D ˛N EN :
C
A
(2.39)
2.3 Balanced Three-Phase Voltages and Currents
35
Fig. 2.12 Illustrative
Example 2.3: equivalent
single-phase circuits
I¯R
Z̄ iy
1
Z̄
3 cd
+
∼ E¯R
−
I¯S
Z̄iy
1
Z̄
3 cd
+
∼ E¯S
−
I¯T
Z̄iy
1
Z̄
3 cd
+
∼ E¯ T
−
+
∼ ĒA
−
−
∼
I¯T R
+ ∼ −
ŪR I¯R
Z̄ cy
+
I¯1
Z̄ id
ĒB
I¯RS
Z̄ id
Ē C
I¯2
Z̄ id
I¯ST ¯
I3
ŪS
I¯S
Z̄ cy
ŪT
I¯T
Z̄ cy
Fig. 2.13 Illustrative Example 2.4: balanced three-phase circuit
We compute below:
1. Currents INR , INS , and INT .
2. Currents INRS , INST , and INTR .
N R, U
N S , and U
N T.
3. Phase voltages U
N
36
2 Power System Fundamentals: Balanced Three-Phase Circuits
We solve the circuit in Fig. 2.13 using the mesh-current method [5]:
2
32 3 2
3
3ZN id
0
ZN id
ZN id
IN1
4 ZN id 2ZN cy C ZN id
ZN cy 5 4 IN2 5 D 4 ˛N 2 EN A 5 :
ZN id
ZN cy
2ZN cy C ZN id
˛N EN A
IN3
Solving for the currents:
2
3 2
31 2
3
3ZN id
0
IN1
ZN id
ZN id
4 IN2 5 D 4 ZN id 2ZN cy C ZN id
ZN cy 5 4 ˛N 2 EN A 5 ;
ZN id
ZN cy
2ZN cy C ZN id
˛N EN A
IN3
and:
2
3
2
3
3
ZN cy
0
IN1
1
C
1
1
6
7
NZid
1
6
7 4 ˛N 2 EN A 5 ;
4 IN2 5 D
1
2 15
N id C 3ZN cy 4
Z
˛N EN A
IN3
1
1 2
2
and finally:
8
1
ˆ
ˆ
EN ;
ˆIN1 D N
ˆ
N cy A
C
3
Z
Z
ˆ
id
ˆ
ˆ
<
1
IN2 D
EN A 1 ˛N 2 ;
N
N
ˆ
Zid C 3Zcy
ˆ
ˆ
ˆ
ˆ
1
ˆN
N :
EN .1 ˛/
:̂I3 D
NZid C 3ZN cy A
Then, we can compute currents INR , INS , and INT as follows:
8
EN A 1 ˛N 2
ˆ
ˆ
ˆ
INR D IN2 D
;
ˆ
ˆ
ZN id C 3ZN cy
ˆ
ˆ
ˆ
ˆ
<
EN A ˛N 2 ˛N
;
INS D IN3 IN2 D
ˆ
ZN id C 3ZN cy
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
EN A .˛N 1/
ˆN
;
:̂IT D IN3 D N
Zid C 3ZN cy
or:
2.3 Balanced Three-Phase Voltages and Currents
8
ˆ
ˆINR D
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
<
INS D
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆN
:̂IT D
37
EN A 1 ˛N 2
;
ZN id C 3ZN cy
EN B 1 ˛N 2
;
ZN id C 3ZN cy
EN C 1 ˛N 2
:
ZN id C 3ZN cy
Next, mesh currents INRS , INST , and INTR are computed as:
8
˛N 2 EN A
ˆ
ˆ
;
ˆINRS D IN1 IN2 D
ˆ
ˆ
ZN id C 3ZN cy
ˆ
ˆ
ˆ
<
˛N EN A
INST D IN1 IN3 D
;
N
ˆ
Zid C 3ZN cy
ˆ
ˆ
ˆ
ˆ
ˆ
EN A
ˆN
;
:̂ITR D IN1 D
ZN id C 3ZN cy
or:
8
EN B
ˆ
ˆ
INRS D
;
ˆ
ˆ
ˆ
ZN id C 3ZN cy
ˆ
ˆ
ˆ
<
EN C
INST D
;
ˆ
ZN id C 3ZN cy
ˆ
ˆ
ˆ
ˆ
ˆ
EN A
ˆN
:
:̂ITR D
NZid C 3ZN cy
Note that line currents (INR , INS , INT ) and mesh currents (INRS , INST , INTR ) constitute
balanced three-phase positive sequences. The relationship between line and mesh
currents is as follows:
INRS
INST
INTR
˛N 2
1
D
D
D
D p † 150ı :
NIS
NIT
IR
1 ˛N 2
3
Note that phases R, S, and T are decoupled. Thus, instead of analyzing the threephase circuit in Fig. 2.13, it is possible to consider the three equivalent single-phase
circuits depicted in Fig. 2.14. Using these equivalent single-phase circuits, we obtain
that phase voltages are equal to:
8
ˆ
N
N N
ˆ
<UR D IR Zcy ;
N D INS ZN cy ;
U
ˆ S
:̂U
N D IN ZN ;
T
T cy
38
2 Power System Fundamentals: Balanced Three-Phase Circuits
Fig. 2.14 Illustrative
Example 2.4: equivalent
single-phase circuits
1
Z̄
3 id
+
∼
−
1 ¯
E
3 A
1
Z̄
3 id
+
∼
−
1 ¯
E
3 B
1
Z̄
3 id
+
∼
−
1 ¯
E
3 C
I¯R
1 −a¯
2
Z̄ cy
I¯S
1 − a¯
2
Z̄cy
I¯T
1 − a¯
2
Z̄cy
or:
8
ZN cy
ˆ
ˆ
N R D EN A 1 ˛N 2
U
;
ˆ
ˆ
ˆ
ZN id C 3ZN cy
ˆ
ˆ
ˆ
<
ZN cy
N S D ˛N 2 EN A 1 ˛N 2
U
;
ˆ
ZN id C 3ZN cy
ˆ
ˆ
ˆ
ˆ
ˆ
ZN cy
ˆN
:
:̂UT D ˛N EN A 1 ˛N 2
NZid C 3ZN cy
N R, U
N S, U
N T constitute
As in the case of line and mesh currents, the phase voltages U
also a balanced three-phase positive sequence.
2.3.4 Common Star Connection
We consider the balanced three-phase network depicted in Fig. 2.15. In this circuit,
we have:
2.3 Balanced Three-Phase Voltages and Currents
39
− ∼ +
Fig. 2.15 Common star
connection: analysis of circuit
with connection NN 0
ĒR
Z̄
I¯R
Z̄
I¯S
Z̄
I¯T
− ∼ +
ĒS
− ∼ +
ĒT
N
I¯N
8
ˆ
N
NN
ˆ
<ER Z IR D 0;
NES ZN INS D 0;
ˆ
:̂EN ZN IN D 0;
T
T
N
(2.40)
and:
8
ˆ
ˆ
ˆ
INR D
ˆ
ˆ
ˆ
ˆ
ˆ
<
INS D
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆN
:̂IT D
EN R
;
ZN
EN S
;
ZN
EN T
:
ZN
(2.41)
Adding these currents, we obtain:
1
INN D INR C INS C INT D .EN R C EN S C EN T /:
ZN
(2.42)
Since voltage sources constitute a balanced sequence, we obtain:
INN D 0
(2.43)
N NN 0 D 0:
U
(2.44)
and, thus:
That is, connection NN 0 is immaterial.
Next, we analyze the same network, but in this case without connection NN 0 ,
as depicted in Fig. 2.16. Solving this circuit by the mesh-current method [5], we
obtain:
40
2 Power System Fundamentals: Balanced Three-Phase Circuits
− ∼ +
Fig. 2.16 Common star
connection: analysis of circuit
without connection NN 0
Ē R
Z̄
− ∼ +
N
Ē S
Z̄
− ∼ +
Ē T
ZN
2
1
1
2
IN1
IN2
1 ˛N 2
N
D ER
:
˛N 2 ˛N
Z̄
I¯1
I¯2
I¯R
I¯S
N
I¯T
(2.45)
Thus:
IN1
IN2
EN R 2
D
ZN 1
1
2
1 1 ˛N 2
;
˛N 2 ˛N
(2.46)
and:
IN1
IN2
D
EN R 2 1
1 ˛N 2
;
˛N 2 ˛N
3ZN 1 2
(2.47)
and finally:
8
ˆ
EN
ˆ
<IN1 D R ;
ZN
EN
EN
ˆN
:̂I2 D ˛N R D T :
Z
ZN
(2.48)
Then, line currents can be computed as:
8
1
ˆ
INR D IN1 D EN R ;
ˆ
ˆ
N
ˆ
Z
<
NIS D IN2 IN1 D 1 EN R .˛N 1/ ;
ˆ
ZN
ˆ
ˆ
:̂IN D IN D 1 EN ;
T
2
T
ZN
(2.49)
2.3 Balanced Three-Phase Voltages and Currents
41
or:
8
ˆ
INR D
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
<
INS D
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
:̂INT D
1N
ER ;
ZN
1N
ES ;
ZN
(2.50)
1N
ET ;
ZN
which is the same result previously obtained in Eq. (2.41) for the circuit with NN 0
connection.
Note also that:
N N 0 N D EN R INR ZN D EN R 1 EN R ZN D 0;
U
ZN
(2.51)
as expected.
We note once more that phase equations are decoupled: the equation of a given
phase depends only on variables and constants of that phase. That is:
8
ˆ
NN
N
ˆ
<ER D Z IR ;
EN D ZN INS ;
ˆ S
:̂EN D ZN IN ;
T
(2.52)
T
or, in matrix form:
3 2
32 3
ZN 0 0
EN R
INR
4 EN S 5 D 4 0 ZN 0 5 4 INS 5 :
EN T
INT
0 0 ZN
2
(2.53)
In other words, the impedance matrix is diagonal, which verifies phase decoupling.
Thus, we can consider three independent single-phase networks as depicted in
Fig. 2.17.
Typically, the R equivalent single-phase circuit is used to represent the balanced
three-phase circuit. Note that such single-phase circuit includes all required information to characterize the balanced three-phase circuit. The other two equivalent
single-phase circuits replicate the R one; circuit S lagging 120ı , and circuit T leading
120ı .
42
2 Power System Fundamentals: Balanced Three-Phase Circuits
Fig. 2.17 Equivalent
single-phase circuits
I¯R
Z̄
+
∼ Ē R
−
I¯S
Z̄
+
∼ Ē S
−
I¯T
Z̄
+
∼ Ē T
−
2.4 Instantaneous, Active, Reactive, and Apparent Power
One of the most important magnitudes in three-phase circuits is the power, which is
analyzed in this section.
2.4.1 Definitions
The instantaneous power at any point of a three-phase circuit is defined as:
4
pR .t/ C pS .t/ C pT .t/
p.t/ D
D uR .t/iR .t/ C uS .t/iS .t/ C uT .t/iT .t/:
Thus, considering a balanced three-phase circuit, p.t/ can be computed as:
p.t/ D
p
2UF sin.!t/ 2IL sin.!t '/
p
2 p
2
'
C 2UF sin !t 2IL sin !t 3
3
p
(2.54)
2.4 Instantaneous, Active, Reactive, and Apparent Power
p
2
C 2UF sin !t C
3
p
2
' ;
2IL sin !t C
3
43
(2.55)
where:
• UF is the RMS value of the phase voltage and
• IL is the RMS value of the line current.
Rearranging terms:
p.t/ D UF IL cos' cos .2Œ!t 0 '/
2
C UF IL cos' cos 2 !t '
3
2
C UF IL cos' cos 2 !t C
'
3
(2.56)
and:
UL
p.t/ D 3UF IL cos' D 3 p IL cos':
3
(2.57)
That is:
p.t/ D
p
3UL IL cos';
(2.58)
where UL is the RMS value of the line voltage. Thus, the instantaneous power at
any point of a balanced three-phase circuit is time invariant.
The three-phase active power, denoted by P, is equal to the instantaneous power
and, thus:
p
4
p.t/ D 3UL IL cos':
PD
(2.59)
The fact that in three-phase power systems the three-phase active power is timeinvariant makes these systems preferable over single-phase systems, in which the
active power has a nonzero average value, but is alternating. Alternating active
power results in vibrations and long-term mechanical issues, while time-invariant
active power does not. This is indeed a key reason for using three-phase systems
instead of single-phase ones.
The per-phase complex power is computed as:
8
UL
ˆ
N R INR D p
ˆ
SN R D U
IL †';
ˆ
ˆ
ˆ
3
ˆ
ˆ
ˆ
<
UL
2
2
UL
NSS D U
N S INS D p
IL † C
C ' D p IL †';
ˆ
3
3
3
3
ˆ
ˆ
ˆ
ˆ
ˆ
UL
2 2
UL
ˆN
N T INT D p
C ' D p IL †':
IL †
:̂ST D U
3
3
3
3
(2.60)
44
2 Power System Fundamentals: Balanced Three-Phase Circuits
N is computed as:
Then, the three-phase complex power, S,
SN D SN R C SN S C SN T D
D
p
p
3UL IL †'
p
3UL IL cos' C j 3UL IL sin '
D P C jQ;
(2.61)
where:
4
QD
p
3UL IL sin'
(2.62)
is the three-phase reactive power.
The magnitude of the three-phase complex power (S) is the so-called three-phase
apparent power.
2.4.2 How to Measure Power?
Note that the active power can be computed using either of the two expressions
below:
P D UR IR cos'R C US IS cos'S C UT IT cos'T ;
(2.63)
or:
PD
p
3UL IL cos':
(2.64)
Equation (2.64) requires the system to be balanced, while Eq. (2.63) does not.
Active power is generally measured using a watt-meter that multiplies three
terms: the RMS value of current, the RMS value of the voltage, and the cosine
of the angle between these two signals (see (2.64)).
On the other hand, reactive power is measured as active power, but using varmeters that, instead of multiplying by the cosine, multiply by the sine.
Finally, apparent power is measured using a volt-meter and an amp-meter.
If we need to measure energy, then we should use an energy meter, which is a
watt-meter that integrates over time.
2.5 Why Three-Phase Power?
We illustrate below the economic advantage of three-phase power versus singlephase power through an example.
2.5 Why Three-Phase Power?
45
We consider the transfer of apparent power S [MVA] over a distance d [km]
with a phase-to-neutral voltage U [kV]. Additionally, we consider that the available
conductor admits a maximum current density ı [A=cm2 ].
Using single-phase ac power, we have:
S1 D S;
U1 D U;
(2.65)
S
:
U
I1 Then, the conductor section should be:
A1 D
S
I1
D
ı
ıU
(2.66)
and the required material is:
Sd
I1
M1 D 2A1 d D 2 d D 2 ;
ı
ıU
(2.67)
while losses are:
PL1 2I12 S
d
S2 ıUd
D 2 ıd;
D 2 2
A1
U
S
U
(2.68)
where is the resistivity of the material used.
Using three-phase ac power, we have:
S3 D S;
p
U3 D 3U;
(2.69)
S3
S
:
D
I3 p
3U
3U3
In this case, the conductor section should be:
A3 D
S
I3
D
ı
3ıU
(2.70)
and the required material is:
M3 D 3A3 d D 3
Sd
S
dD
;
3ıU
ıU
(2.71)
while losses in this case are:
PL3 3I32 S
d
S2 3ıUd
D ıd:
D 3 2
A3
9U
S
U
(2.72)
46
2 Power System Fundamentals: Balanced Three-Phase Circuits
Note that, on the one hand:
M1
D 2;
M3
(2.73)
i.e., the material required to transmit apparent power S [MVA] over a distance d
[km] with a phase-to-neutral voltage U [kV] considering a single-phase ac line is
about twice the material needed if a three-phase ac line is used.
On the other hand, we have:
PL1
D 2;
PL3
(2.74)
i.e., the losses of transmitting apparent power S [MVA] over a distance d [km] with
a phase-to-neutral voltage U [kV] considering a single-phase ac line are about twice
the losses if a three-phase ac line is used.
This simple back-of-the-envelope analysis illustrates the economic advantage of
building/using a three-phase transmission line over a single-phase one.
2.6 Per-Unit System
This section defines and describes the per-unit system, which is important in power
systems spanning different voltage levels.
2.6.1 Motivation
Power transformers interconnect power system areas with different voltage levels.
This is a problem at the time of analyzing these systems since all magnitudes
need to be transformed to a single voltage level. However, if a per-unit analysis
is performed, this problem disappears and a unique voltage level is obtained. This
greatly simplifies the subsequent analysis.
2.6.2 Per-Unit System Definition
Any electrical variable or parameter (voltage, current, power, impedance) can be
expressed as a function of its own units or with respect to a reference value known
as base value, i.e.:
2.6 Per-Unit System
47
Fig. 2.18 Illustrative
Example 2.5: single-phase
circuit using real magnitudes
I
+
∼ V = 220 V
−
mD
M
;
MB
Z = 55 W
(2.75)
where:
• m is the per-unit value,
• M is the value of the variable/parameter in its own units, and
• M B is the base value.
Then, instead of analyzing a circuit using actual values, it is possible to analyze it
using per-unit values. This generally simplifies the subsequent analysis.
Illustrative Example 2.5 Illustration of per-unit analysis
We consider the single-phase circuit depicted in Fig. 2.18. Taking into account
that V D 220 V and Z D 55 , we obtain the current I. To do so, we analyze the
circuit using the per-unit system considering a base-voltage value of 220 V and a
base-current value of 2 A.
First, we obtain the equivalent per-unit circuit by transforming the voltage and
impedance values to per-unit values.
On the one hand, we compute the per-unit voltage v as follows:
vD
V
220
D 1 puV:
D
VB
220
In order to obtain the per-unit impedance, first we need to compute the baseimpedance value, which is obtained as the base-voltage value divided by the basecurrent value, i.e.:
ZB D
VB
220
D 110 :
D
IB
2
Then, we obtain the per-unit impedance as:
zD
Z
55
D 0:5 pu:
D
ZB
110
Finally, we derive the equivalent single-phase circuit using per-unit values and
depict it in Fig. 2.19.
48
2 Power System Fundamentals: Balanced Three-Phase Circuits
Fig. 2.19 Illustrative
Example 2.5: single-phase
circuit using per-unit
magnitudes
i
+
∼ v = 1 puV z = 0.5 pu W
−
Next we can compute the per-unit current i as:
iD
1
v
D
D 2 puA:
z
0:5
Finally, we can obtain the actual current I as:
I D i I B D 2 2 D 4 A:
Using per-unit analysis in Illustrative Example 2.5 is not convenient since it is
V
220
possible to directly compute current I from the circuit in Fig. 2.18 as
D
D
Z
55
4 A. However, the use of a per-unit system to analyze power systems with multiple
voltage levels is most convenient for two reasons (provided that the per-unit system
is properly defined):
1. Power transformers disappear from the equivalent single-phase circuit. This is
further analyzed and shown in Sect. 3.3 of Chap. 3.
2. Voltage values are close to 1 puV, which allows detecting errors.
Besides these two important advantages, there is an additional advantage of using
a per-unit analysis for three-phase power systems:
3. The per-unit impedances of machines generally take values within tight bounds,
independently of their nominal values, which facilitates their characterization.
2.6.3 Definition of Base Values
An appropriate definition of the base values is as follows:
1. We select a common single-phase base power, typically 1=3 of the rated power
of the component of highest rated power, i.e.:
SB D
1
SNk ;
3
(2.76)
2.6 Per-Unit System
49
where:
• SB is the single-phase base power and
• SNk is the three-phase rated power of component k.
2. Recalling that power transformers separate the voltage zones of the network, we
select the base voltage in one zone as the rated phase voltage of one component
in that zone, i.e.:
1
UBi D p UNj ;
3
(2.77)
where:
• UBi is the base-voltage value at zone i and
• UNj is the rated three-phase voltage of component j in zone i.
3. The base-voltage values in other zones are determined strictly complying with
the transformation ratios of the power transformers, which makes transformers
disappear from equivalent single-phase circuits, i.e.:
UBi D UBj
Ui
;
Uj
(2.78)
where:
• UBi is the base-voltage value in zone i,
• UBj is the base-voltage value in zone j, and
• Ui =Uj is the three-phase transformation ratio of the power transformer
coupling zones i and j.
4. We define the base-current value and the base-impedance value per zone as
IBi D
SB
UBi
(2.79)
ZBi D
2
UBi
;
SB
(2.80)
and:
respectively, where:
• IBi is the base-current value at zone i and
• ZBi is the base-impedance value at zone i.
5. We specify phase shifts due to power transformers including delta and zigzag
connections. We explain transformer phase shifts in Chap. 3.
50
2 Power System Fundamentals: Balanced Three-Phase Circuits
Fig. 2.20 Illustrative
Example 2.6: power system
G
∼
T1
L
T2
M
∼
Illustrative Example 2.6 Per-unit analysis: base values
We consider the three-phase power system depicted in Fig. 2.20. This system
comprises a generator, two power transformers, a transmission line, and a motor.
The rated powers and voltages of these components are provided below:
•
•
•
•
•
Generator G: 60 MVA, 11 kV.
Transformer T1 : 60 MVA, 132/12 kV.
Line L: 70 MVA, 132 kV.
Transformer T2 : 60 MVA, 125/5 kV.
Motor M: 80 MVA, 5 kV.
Using these data, we determine below the number of voltage zones, and the base
values of each zone.
Each power transformer divides the system into two voltage zones. Thus, we
have three voltage zones, namely, generator zone (zone 1), line zone (zone 2), and
motor zone (zone 3).
Next, we follow the procedure described above to determine the base values in
each zone:
1. We select the single-phase base power as 1/3 of the rated power of the motor,
80
which has the highest rated power, i.e., SB D
D 26:667 MVA. Note that this
3
base-power value is the same for all voltage zones.
2. We fix the base-voltage value of one of the zones as the rated phase voltage of one
of the components of that zone. For example, we select the base-voltage value in
11
the generator zone as the rated phase voltage of the generator, i.e., UB1 D p D
3
6:351 kV.
3. We compute the base-voltage values in other zones by using the transformation
132 11
ratios of transformers that separate each zone, i.e., UB2 D
p D 69:859 kV
12 3
5 132 11
and UB3 D
p D 2:794 kV.
125 12 3
4. We define the base-current value and the base-impedance value per zone using
the corresponding base-power and base-voltage values, as well as Eqs. (2.79)
and (2.80), respectively. For example, the base-current value in the generator
26:667 106
zone is IB1 D
D 4:199 kA, while the base-impedance value in the
3
6:351
10
2
69:859 103
line zone is ZB2 D
D 183:013 .
26:667 106
Table 2.1 summarizes the base values in each zone.
2.6 Per-Unit System
51
Table 2.1 Illustrative
Example 2.6: base values
Value
SB [MVA]
VB [kV]
IB [kA]
ZB []
Generator zone
26.667
6.351
4.199
1.512
Line zone
26.667
69.859
0.382
183.013
Motor zone
26.667
2.794
9.544
0.293
Note that the base values defined in this section are single-phase base values.
However, three-phase base values, equivalent to single-phase ones, are similarly
defined.
If we define the three-phase base-power value as:
4
3SB ;
SB3 D
(2.81)
and the line base-voltage value in each zone as:
4
UB3i D
p
3UBi ;
(2.82)
then single-phase and three-phase base-current and single-phase and three-phase
base-impedance values coincide. That is:
IB3i D p
SB3
3UB3i
3SB
SB
Dp p
D
D IBi
UBi
3 3UBi
(2.83)
and:
ZB3i D
2
UB3
SB3
p
D
3UBi
3SB
2
D
2
UBi
D ZBi ;
SB
(2.84)
respectively.
2.6.4 Per-Unit Analysis Procedure
The procedure to analyze power systems using a per-unit system comprises the five
steps below:
1. Define base values as explained in Sect. 2.6.3.
2. Transform the three-phase power system into an equivalent single-phase circuit
in which impedances are expressed in per unit.
3. Apply the operating conditions (in per unit).
4. Solve the circuit (in per unit).
5. Obtain actual values by multiplying per-unit values by the corresponding base
values.
52
2 Power System Fundamentals: Balanced Three-Phase Circuits
A number of examples to illustrate this procedure to analyze power systems are
provided in Sect. 3.6 of Chap. 3.
2.7 Summary and Further Reading
This chapter provides an overview of the fundamentals of power systems. First, we
define balanced sequences, balanced voltages and currents, and powers. Second,
we illustrate why three-phase power systems are used instead of single-phase ones.
Finally, we provide an introduction to the analysis of power systems considering a
per-unit system, which is used and further analyzed in the following chapters of this
book.
Basic references regarding power system analysis include Kothari and Nagrath
[4] and Duncan Glover et al. [2]. Advanced references include Gómez-Expósito
et al. [3] and Bergen and Vittal [1].
2.8 End-of-Chapter Exercises
2.1 Why three-phase power systems are used instead of single-phase ones?
2.2 List the advantages of analyzing power systems using a per-unit system.
2.3 Consider the three-phase circuit depicted in Fig. 2.21. Voltage sources constitute a balanced three-phase positive sequence:
8
ı
ˆ
N
ˆ
<ER D 100†0 V;
EN D 100† 120ı V;
ˆ S
:̂EN D 100†120ı V:
T
Impedances ZN i are equal to j5 , impedances ZN ` are equal to j15 , and impedances
ZN RS D ZN ST D ZN TR are equal to j30 .
Using these data:
1. Compute currents INR , INS , and INT , as well as currents INRS , INST , and INTR .
N S1 , and U
N T1 , as well as voltages U
N R2 , U
N S2 , and U
N T2 .
N R1 , U
2. Compute voltages U
3. Recompute the above voltages and currents if an impedance of j20 is connected between nodes N and N 0 .
4. Compute the above voltages and currents if impedances ZN RS , ZN ST , and ZN TR are
equal to 10 , j10 , and j10 , respectively. Is it possible in this case to use
the equivalent single-phase circuit?
2.4 Consider a balanced three-phase load that is supplied at a line voltage of 400 kV
and absorbs a line current of 500† 10ı A:
2.8 End-of-Chapter Exercises
53
Ē R
¯
− ∼ + ŪR1 IR
Z̄ i
Z̄ i
ŪR2
¯
− ∼ + ŪS1 IS
I¯T R
I¯RS
Z̄ N
ŪS 2
Ē T
¯
− ∼ + ŪT 1 I T
Z̄ i
Z̄T R
Z̄RS
Ē S
N
Z̄ Z̄ ST
I¯ST ŪT 2
Z̄ Fig. 2.21 Exercise 2.3: three-phase circuit
G1
1
T1
2
∼
T2
3
L1
G2
∼
L3
L2
5
7
T3
T4
6
8
C
4
∼
M
Fig. 2.22 Exercise 2.5: power system
1. Compute the instantaneous power consumed in phases R, S, and T.
2. Compute the instantaneous three-phase power consumed by the load.
3. Compute the reactive and apparent power consumed by the load.
2.5 Consider the three-phase power system depicted in Fig. 2.22. The rated powers
and voltages of the system components are provided below:
•
•
•
•
Generators G1 and G2 : 500 MVA, 20 kV.
Transformers T1 and T2 : 200 MVA, 500/18 kV.
Transformers T3 and T4 : 150 MVA, 500/20 kV.
Motor M: 111 MW, cos D 0:8 (inductive), 20 kV.
54
2 Power System Fundamentals: Balanced Three-Phase Circuits
Fig. 2.23 Exercise 2.6:
power system
G1
T1
∼
G2
∼
T2
L1
L2
Using these data, determine the number of voltage zones and the base values of
each zone.
2.6 Consider the three-phase power system depicted in Fig. 2.23. The rated powers
and voltages of the system components are provided below:
•
•
•
•
Generator G1 : 50 MVA, 12 kV.
Generator G2 : 100 MVA, 15 kV.
Transformer T1 : 50 MVA, 10/138 kV.
Transformer T2 : 100 MVA, 15/138 kV.
Using as base values the rated parameters of generator G2 , determine the number
of voltage zones and the base values of each zone.
References
1. Bergen, A.R., Vittal, V.: Power Systems Analysis, 2nd edn. Prentice Hall, Upper Saddle River,
NJ (1999)
2. Duncan Glover, J., Sarma, M.S., Overbye, T.: Power System Analysis and Design, 5th edn.
Cengage Learning, Stamford, CT (2008)
3. Gómez-Expósito, A., Conejo, A.J., Cañizares, C.: Electric Energy Systems: Analysis and
Operation. Taylor and Francis, Boca Raton, FL (2008)
4. Kothari, D.P., Nagrath, I.J.: Modern Power System Analysis, 4th edn. Tata McGraw Hill
Education Private Limited, New Delhi (2011)
5. Nilsson, J.W., Riedel, S.A.: Electric Circuits, 10th edn. Pearson, Upper Saddle River, NJ (2014)
http://www.springer.com/978-3-319-69406-1