Static analysis of power systems

Static Analysis
of
Power Systems
Lennart S¨oder and Mehrdad Ghandhari
Electric Power Systems
Royal Institute of Technology
August 2010

Contents
1 Introduction 1
2 Power system design 3
2.1 The development of the Swedish power system . . . . . . . . . . . . . . . . . 3
2.2 The structure of the electric power system . . . . . . . . . . . . . . . . . . . 5
3 Alternating current circuits 9
3.1 Single-phase circuit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
3.1.1 Complex power . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
3.2 Balanced three-phase circuit . . . . . . . . . . . . . . . . . . . . . . . . . . . 14
3.2.1 Complex power . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16
4 Models of power system components 21
4.1 Electrical characteristic of an overhead line . . . . . . . . . . . . . . . . . . . 21
4.1.1 Resistance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22
4.1.2 Shunt conductance . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22
4.1.3 Inductance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22
4.1.4 Shunt capacitance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25
4.2 Model of a line . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26
4.2.1 Short lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26
4.2.2 Medium long lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27
4.3 Single-phase transformer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27
4.4 Three-phase transformer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30
iii

iv
5 Important theorems in power system analysis 31
5.1 Bus analysis, admittance matrices . . . . . . . . . . . . . . . . . . . . . . . 31
5.2 Millman’s theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34
5.3 Superposition theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36
5.4 Reciprocity theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37
5.5 Thévenin-Helmholtz’s theorem . . . . . . . . . . . . . . . . . . . . . . . . . 38
6 Analysis of balanced three-phase systems 41
6.1 Single-line and impedance diagrams . . . . . . . . . . . . . . . . . . . . . . . 44
6.2 The per-unit (pu) system . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46
6.2.1 Per-unit representation of transformers . . . . . . . . . . . . . . . . . 47
6.2.2 Per-unit representation of transmission lines . . . . . . . . . . . . . . 49
6.2.3 System analysis in the per-unit system . . . . . . . . . . . . . . . . . 50
7 Power transmission to impedance loads 53
7.1 Twoport theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53
7.1.1 Symmetrical twoports . . . . . . . . . . . . . . . . . . . . . . . . . . 54
7.1.2 Application of twoport theory to transmission line and transformer
and impedance load . . . . . . . . . . . . . . . . . . . . . . . . . . . 55
7.1.3 Connection to network . . . . . . . . . . . . . . . . . . . . . . . . . . 57
7.2 A general method for analysis of linear balanced three-phase systems . . . . 62
7.3 Extended method to be used for power loads . . . . . . . . . . . . . . . . . . 69
8 Power flow calculations 73
8.1 Power flow in a line . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73
8.1.1 Line losses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76
8.1.2 Shunt capacitors and shunt reactors . . . . . . . . . . . . . . . . . . . 77
8.1.3 Series capacitors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78
8.2 Non-linear power flow equations . . . . . . . . . . . . . . . . . . . . . . . . 78
8.3 Power flow calculations of a simple two-bus system . . . . . . . . . . . . . . 82
8.3.1 Slack bus + PU-bus . . . . . . . . . . . . . . . . . . . . . . . . . . . 83

v
8.3.2 Slack bus + PQ-bus . . . . . . . . . . . . . . . . . . . . . . . . . . . 84
8.4 Newton-Raphson method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87
8.4.1 Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87
8.4.2 Application to power systems . . . . . . . . . . . . . . . . . . . . . . 91
8.4.3 Newton-Raphson method for solving power flow equations . . . . . . 94
9 Analysis of three-phase systems using linear transformations 103
9.1 Linear transformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104
9.1.1 Power invarians . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104
9.1.2 The coefficient matrix in the original space . . . . . . . . . . . . . . . 105
9.1.3 The coefficient matrix in the image space . . . . . . . . . . . . . . . . 106
9.2 Examples of linear transformations that are used in analysis of three-phase
systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107
9.2.1 Symmetrical components . . . . . . . . . . . . . . . . . . . . . . . . . 107
9.2.2 Clarke’s components . . . . . . . . . . . . . . . . . . . . . . . . . . . 109
9.2.3 Park’s transformation . . . . . . . . . . . . . . . . . . . . . . . . . . . 110
9.2.4 Phasor components . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111
10 Symmetrical components 115
10.1 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115
10.2 Power calculations under unbalanced conditions . . . . . . . . . . . . . . . . 118
11 Sequence circuits of transmission lines 121
11.1 Series impedance of single-phase overhead line . . . . . . . . . . . . . . . . . 121
11.2 Series impedance of a three-phase overhead line . . . . . . . . . . . . . . . . 122
11.2.1 Symmetrical components of the series impedance of a three-phase line 124
11.2.2 Equivalent diagram of the series impedance of a line . . . . . . . . . . 125
11.3 Shunt capacitance of a three-phase line . . . . . . . . . . . . . . . . . . . . . 128
12 Sequence circuits of transformers 131
13 Analysis of unbalanced three-phase systems 135

vi
13.1 Symmetrical components of impedance loads . . . . . . . . . . . . . . . . . 135
13.2 Connection to a system under unbalanced conditions . . . . . . . . . . . . . 136
13.3 Single line-to-ground fault . . . . . . . . . . . . . . . . . . . . . . . . . . . . 137
13.4 Analysis of a linear three-phase system with one unbalanced load . . . . . . 139
13.5 A general method for analysis of linear three-phase systems with one unbal-
anced load . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 144
14 Power system harmonics 157
A MATLAB-codes for Examples 7.2, 7.3, 13.2 and 13.3 163
B Matlab-codes for Example 8.10 169
C MATLAB-codes for Example 8.12 171
D MATLAB-codes for Example 8.13 173
E MATLAB-codes for Example 8.14 175
F MATLAB-codes for Example 8.15 181

Chapter 1
Introduction
In this compendium, models and mathematical methods for static analysis of power systems
are discussed.
In chapter 2, the design of the power system is described and in chapter 3, the fundamental
theory of alternating current is presented. Models of overhead power lines and transformers
are given in chapter 4, and in chapter 5 some important theorems in three-phase analysis
are discussed. In chapter 6 and 7, power system calculations in symmetrical conditions are
performed by using the theorems presented earlier.
Chapter 5–7 are based on the assumption that the power system loads can be modeled as
impedance loads. This leads to a linear formulation of the problem, and by that, relatively
easy forms of solution methods can be used. In some situations, it is more accurate to model
the system loads as power loads. How the system analysis should be carried out in such
conditions is elaborated on in chapter 8.
In chapter 9, an overview of linear transformations in order to simplify the power system
analysis, is given. In chapter 10–13, the basic concepts of analysis in un-symmetrical con-
ditions are given. The use of symmetrical components is presented in detail in chapter 10.
The modeling of lines, cables and transformers must be more detailed in un-symmetrical
conditions, this is the topic of chapter 11–12. In chapter 13, an un-symmetrical, three-phase
power system with impedance loads is analyzed.
In chapter 3–13, it is assumed that the power system frequency is constant and the system
components are linear, i.e. sinusoidal voltages gives sinusoidal currents. With non-linear
components in the system, as high power electronic devices, non-sinusoidal currents and
voltages will appear. The consequences such non-sinusoidal properties can have on the
power system will be discussed brieﬂy in chapter 14.
1

Chapter 2
Power system design
2.1 The development of the Swedish power system
The Swedish power system started to develop around a number of hydro power stations,
Porjus in Norrland, Älvkarleby in eastern Svealand, Motala in the middle of Svealand and
Trollhättan in Götaland, at the time of the first world war. Later on, coal fired power
plants located at larger cities as Stockholm, Göteborg, Malmö and Väster˚as came into oper-
ation. At the time for the second world war, a comprehensive proposal was made concerning
exploitation of the rivers in the northern part of Sweden. To transmit this power to the
middle and south parts of Sweden, where the heavy metal industry were located, a 220 kV
transmission system was planned.
Today, the transmission system is well developed with a nominal voltage of 220 or 400 kV.
In rough outline, the transmission system consists of lines, transformers and sub-stations.
A power plant can have an installed capacity of more than 1000 MW, e.g. the nuclear power
plants Forsmark 3 and Oskarshamn 3, whereas an ordinary private consumer can have an
electric power need of some kW. This implies that electric power can be generated at some
few locations but the consumption, which shows large variations at single consumers, can be
spread all over the country. In Figure 2.1, the electricity supply in Sweden between 1947 and
1945 1950 1955 1960 1965 1970 1975 1980 1985 1990 1995 2000
0
50
100
150
TWh/year
Hydro power
Conv. therm. pow.
Nuclear power
Figure 2.1. Electricity supply in Sweden 1947–2000
2000 is given. The hydro power was in the beginning of this period the dominating source of
electricity until the middle of the 1960s when some conventional thermal power plants (oil
fired power plants, industrial back pressure, etc.) were taken into service. In the beginning
of the 1970s, the first nuclear power plants were taken into operation and this power source
3

4
has ever after being the one showing the largest increase in generated electric energy. In
recent years, the trend showing a continuous high increase in electric power consumption
has been broken. On 30th November 1999, the 600 MW nuclear power plant Barsebäck 1
was closed due to a political decision. In Table 2.1, the electricity supply in Sweden during
2000 is given.
Source of power Energy generation Installed capacity 00-12-31
TWh = 109
kWh MW
Hydro 77.8 16 229
Nuclear 54.8 9 439
Industrial back pressure 4.3 932
Combined heat and power 4.2 2 264
Oil fired condensing power 0.2 448
Gas turbine 0.003 1 341
Wind power 0.4 241
Total 141.9 30 894
Table 2.1. Electricity supply in Sweden 2000
The total consumption of electricity is usually grouped into different categories. In Figure
2.2, the consumption during the latest 45 years is given for different groups.
1945 1950 1955 1960 1965 1970 1975 1980 1985 1990 1995
0
50
100
150
TWh/year
Communication
Industry
Miscellaneous
Space heating
Losses
Figure 2.2. Consumption of electricity in Sweden 1947–1997
As shown in the figure, the major increase in energy need has earlier been dominated by
the industry. When the nuclear power was introduced in the early 1970s, the electric space
heating increased significantly. Before 1965, the electric space heating was included in the
group miscellaneous. Communication, i.e. trains, trams and subway, has increased its
consumption from 1.4 TWh/year in 1950 to 2.2 TWh/year in 1995.

5
In proportion to the total electricity consumption, the communication group has decreased
from 7.4 % to 1.6 % during the same period. The losses on the transmission and distri-
bution systems have during the period 1950–1995 decreased from more than 10 % of total
consumption to approximately 7 %.
2.2 The structure of the electric power system
A power system consists of generation sources which via power lines and transformers trans-
mits the electric power to the end consumers.
The power system between the generation sources and end consumers is divided into different
parts according to Figure 2.3.
Transmission network
400 – 200 kV
(Svenska Kraftnät)
Sub-transmission network
130 – 40 kV
Distribution network
primary part
40 – 10 kV
Distribution network
secondary part
low voltage 230/400 V
Figure 2.3. The structure of the electric power system
The transmission network, connects the main power sources and transmits a large amount
of electric energy. The Swedish transmission system consists of approximately 15250 km
power lines, and is connected to other countries on 23 different locations. In Figure 2.4, a
general map of the transmission system in Sweden and neighboring countries is given. The
primary task for the transmission system is to transmit energy from generation areas to load
areas. To achieve a high degree of efficiency and reliability, different aspects must be taken
into account. The transmission system should for instance make it possible to optimize the
generation within the country and also support trading with electricity with neighboring
countries. It is also necessary to withstand different disturbances such as disconnection of
transmission lines, lightning storms, outage of power plants as well as unexpected growth in

6
power demand without reducing the quality of the electricity services. As shown in Figure
2.4, the transmission system is meshed, i.e. there are a number of closed loops in the
transmission system.
Figure 2.4. Transmission system in north-western Europe
A new state utility, Svenska Kraftnät, was launched on January 1, 1992, to manage the
national transmission system and foreign links in operation at date. Svenska Kraftnät owns
all 400 kV lines, all transformers between 400 and 220 kV and the major part of the 220 kV
lines in Sweden. Note that the Baltic Cable between Sweden and Germany was taken into
operation after the day Svenska Kraftnät was launched and is therefore not owned by them.
Sub-transmission network, in Sweden also called regional network, has in each load region the
same or partly the same purpose as the transmission network. The amount of energy trans-
mitted and the transmission distance are smaller compared with the transmission network
which gives that technical-economical constraints implies lower system voltages. Regional
networks are usually connected to the transmission network at two locations.

7
Distribution network, transmits and distributes the electric power that is taken from the sub-
stations in the sub-transmission network and delivers it to the end users. The distribution
network is in normal operation a radial network, i.e. there is only one path from the sub-
transmission sub-station to the end user.
The electric power need of diﬀerent end users varies a lot as well as the voltage level where
the end user is connected. Generally, the higher power need the end user has, the higher
voltage level is the user connected to.
The nominal voltage levels (Root Mean Square (RMS) value for tree-phase line-to-line (LL)
voltages) used in distribution of high voltage electric power is normally lower compared with
the voltage levels used in transmission. In Figure 2.5, the voltage levels used in Sweden are
given. In special industry networks, except for levels given in Figure 2.5, also the voltage 660
V as well as the non-standard voltage 500 V are used. Distribution of low voltage electric
power to end users is usually performed in three-phase lines with a zero conductor, which
gives the voltage levels 400/230 V (line-to-line (LL)/line-to-neutral (LN) voltage).
transmission
network
sub-transmission
network
distribution network
high voltage
distribution network
low voltage
Nominal
voltage
kV
1000
800
400
220
800
400
200
Notation
132
66
45
130
70
50
33
22
11
6.6
3.3
30
20
10
6
3
400/230 V
ultra high
voltage (UHV)
extra high
voltage (EHV)
high voltage
industry network
only
low voltage
Figure 2.5. Standard voltage level for transmission and distribution. In Sweden,
400 kV is the maximum voltage

Chapter 3
Alternating current circuits
In this chapter, instantaneous and also complex power in an alternating current (AC) circuit
is discussed. Also, the fundamental properties of AC voltage, current and power in a balanced
(or symmetrical)three-phase circuit are presented.
3.1 Single-phase circuit
Assume that an AC voltage source with a sinusoidal voltage supplies an impedance load as
shown in Figure 3.1.
Z R jX= +( )u t
( )i t
+
-
Figure 3.1. A sinusoidal voltage source supplies an impedance load.
Let the instantaneous voltage and current be given by
u(t) = UM cos(ωt + θ)
i(t) = IM cos(ωt + γ)
(3.1)
where
UM = the peak value of the voltage
IM =
UM
|Z|
=
UM
Z
= the peak value of the current
ω = 2πf , and f is the frequency of the voltage source
θ = the voltage phase angle
γ = the current phase angle
φ = θ − γ = arctan
X
R
= phase angle between voltage and current
The single-phase instantaneous power consumed by the impedance Z is given by
p(t) = u(t) · i(t) = UM IM cos(ωt + θ) cos(ωt + γ) =
=
1
2
UM IM [cos(θ − γ) + cos(2ωt + θ + γ)] =
=
UM
√
2
IM
√
2
[(1 + cos(2ωt + 2θ)) cos φ + sin(2ωt + 2θ) sin φ] =
= P(1 + cos(2ωt + 2θ)) + Q sin(2ωt + 2θ)
(3.2)
9

10
where
P =
UM
√
2
IM
√
2
cos φ = U I cos φ = active power
Q =
UM
√
2
IM
√
2
sin φ = U I sin φ = reactive power
U and I are the RMS-value of the voltage and current, respectively. The RMS-values are
deﬁned as
U =
1
T
T
0
u(t)2dt (3.3)
I =
1
T
T
0
i(t)2dt (3.4)
With sinusoidal voltage and current, according to equation (3.1), the corresponding RMS-
values are given by
U =
1
T
T
0
U2
M cos2(ωt + θ) = UM
1
T
T
0
1
2
+
cos(2ωt + 2θ)
2
=
UM
√
2
(3.5)
I =
1
T
T
0
I2
M cos2(ωt + γ) =
IM
√
2
(3.6)
As shown in equation (3.2), the instantaneous power has been decomposed into two com-
ponents. The ﬁrst component has a mean value P, and pulsates with the double frequency.
The second component also pulsates with double frequency with a amplitude Q, but it has
a zero mean value. In Figure 3.2, the instantaneous voltage, current and power are shown.
time (t)
i(t)
u(t)
p(t)
UIcosφ
p(t)
time (t)
I
II
UIsinφ
UIcosφ
φ
Figure 3.2. Voltage, current and power versus time.

11
Example 3.1 A resistor of 1210 Ω is fed by an AC voltage source with frequency 50 Hz and
voltage 220 V (RMS). Find the mean value power (i.e. the active power) consumed by the
resistor.
Solution
The consumed mean value power over one period can be calculated as
P =
1
T
T
0
p(t) dt =
1
T
T
0
R · i2
(t)dt =
1
T
T
0
R
u2
(t)
R2
dt =
1
R
1
T
T
0
u2
(t)dt
which can be rewritten according to equation (3.3) as
P =
1
R
U2
=
2202
1210
= 40 W
3.1.1 Complex power
The complex method is a powerful tool for calculation of electrical power, and can oﬀer
solutions in an elegant manner.
The single-phase phasor voltage and current are expressed by
U = Uej arg(U)
I = Iej arg(I)
(3.7)
where
U = phasor voltage
U = UM /
√
2 = the RMS magnitude of voltage
I = phasor current
I = IM /
√
2 = the RMS magnitude of current
The complex power (S) is expressed by
S = Sej arg(S)
= P + jQ = U I
∗
= UIej(arg(U)−arg(I))
(3.8)
With the voltage and current phase angles given in equation (3.1), we have
S = P + jQ = U I
∗
= UIej(θ−γ)
= UIejφ
= UI(cos φ + j sin φ) (3.9)
which implies that
P = S cos φ = UI cos φ
Q = S sin φ = UI sin φ
(3.10)
where, P is the active power and Q is the reactive power.

12
Example 3.2 Calculate the complex power consumed by an inductor with the inductance
of 3.85 H which is fed by an AC voltage source with the phasor U = U θ = 220 0 V. The
circuit frequency is 50 Hz.
Solution
The impedance is given by
Z = jωL = j · 2 · π · 50 · 3.85 = j1210 Ω
Next, the phasor current through the impedance can be calculated as
I =
U
Z
=
220
j1210
= −j0.1818 A = 0.1818 e−j π
2 A
Thus, the complex power is given by
S = U I
∗
= UIej(θ−γ)
= UIej(φ)
= 220 (0.1818) ej(0+ π
2
)
= 220(j0.1818) = j40 VA
i.e. P = 0 W, Q = 40 VAr.
Example 3.3 Two series connected impedances are fed by an AC voltage source with the
phasor U1 = 1 0 V as shown in Figure 3.3.
1 2
1 1U V=
1 0.1 0.2Z j= + Ω
2 2 2U U θ= ∠
2 0.7 0.2Z j= + Ω
I
Figure 3.3. Network used in Example 3.3.
a) Calculate the power consumed by Z2 as well as the power factor (cosφ) at bus 1 and 2
where φk is the phase angle between the voltage and the current at bus k.
b) Calculate the magnitude U2 when Z2 is capacitive : Z2 = 0.7 − j0.5 Ω
Solution
a)
U1 = U1 θ1 = 1 0 V and I =
U1
Z1 + Z2
= I γ = 1.118 − 26.57◦
A
Thus, φ1 = θ1 − γ = 26.57◦
, and cos φ1 = 0.8944 lagging, since the current lags the voltage.
Furthermore,
U2 = Z2 · I = U2 θ2 = 0.814 − 10.62◦

13
Thus, φ2 = θ2 − γ = −10.62◦
+ 26.57◦
= 15.95◦
, and cos φ2 = 0.9615, lagging. The equation
above can be written on polar form as
U2 = Z2 · I
θ2 = arg(Z2) + γ
i.e. φ2 = arg(Z2) = arctan X2
R2
= 15.95◦
U1
−R1 · I
U2 = U1 − R1 · I − jX1 · I
I
γ
φ2
θ2
Figure 3.4. Solution to Example 3.3 a).
The power consumption in Z2 can be calculated as
S2 = P2 + jQ2 = Z2 · I2
= (0.7 + j0.2)1.1182
= 0.875 + j0.25 VA
or
S2 = P2 + jQ2 = U2I
∗
= U2 I φ2 = 0.814 · 1.118 15.95◦
= 0.875 + j0.25 VA
1 2
1 1U V=
1 0.1 0.2Z j= + Ω
2 2 2U U θ= ∠
2 0.7 0.5Z j= − Ω
I
Figure 3.5. Solution to Example 3.3 b).
b)
U2 =
Z2
Z1 + Z2
U1 =
|0.7 − j0.5|
|0.8 − j0.3|
=
√
0.49 + 0.25
√
0.64 + 0.09
=
√
0.74
√
0.73
= 1.007 V
Conclusions from this example are that
• a capacitance increases the voltage - so called phase compensation,

14
• active power can be transmitted towards higher voltage,
• the power factor cos φ may be diﬀerent in diﬀerent ends of a line,
• the line impedances are load impedances.
3.2 Balanced three-phase circuit
In a balanced (or symmetrical) three-phase circuit, a three-phase voltage source consists of
three AC voltage sources (one for each phase) which are equal in amplitude (or magnitude)
and displaced in phase by 120◦
. Furthermore, each phase is equally loaded.
Let the instantaneous phase (also termed as line-to-neutral (LN)) voltages be given by
ua(t) = UM cos(ωt + θ)
ub(t) = UM cos(ωt + θ −
2π
3
) (3.11)
uc(t) = UM cos(ωt + θ +
2π
3
)
Variations of the three voltages versus time are shown in Figure 3.6.
0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08
−1
0
1
ua(t)
0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08
−1
0
1
ub(t)
0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08
−1
0
1
uc(t)
0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08
−1
0
1
uab(t)
Figure 3.6. ua(t), ub(t), uc(t) and uab(t) versus time with f = 50 Hz, UM = 1 and θ = 0.
For analysis of a balanced three-phase system, it is very common to use the voltage between
two phases. This voltage is termed as line-to-line (LL) voltage. The line-to-line voltage uab

15
is given by
uab(t) = ua(t) − ub(t) = UM cos(ωt + θ) − UM cos(ωt + θ −
2π
3
) = (3.12)
=
√
3 UM cos(ωt + θ +
π
6
)
As shown in equation (3.12), in a balanced three-phase circuit the line-to-line voltage leads
the line-to-neutral voltage by 30◦
, and is
√
3 times larger in amplitude (or magnitude, see
equation (3.5)). For instance, at a three-phase power outlet the magnitude of a phase is 230
V, but the magnitude of a line-to-line voltage is
√
3 · 230 = 400 V, i.e. ULL =
√
3 ULN . The
line-to-line voltage uab is shown at the bottom of Figure 3.6.
Next, assume that the voltages given in equation (3.11) supply a balanced (or symmetrical)
three-phase load whose phase currents are
ia(t) = IM cos(ωt + γ)
ib(t) = IM cos(ωt + γ −
2π
3
) (3.13)
ic(t) = IM cos(ωt + γ +
2π
3
)
Then, the total instantaneous power is given by
p3(t) = pa(t) + pb(t) + pc(t) = ua(t)ia(t) + ub(t)ib(t) + uc(t)ic(t) =
=
UM
√
2
IM
√
2
[(1 + cos 2(ωt + θ)) cos φ + sin 2(ωt + θ) sin φ] +
+
UM
√
2
IM
√
2
[(1 + cos 2(ωt + θ −
2π
3
)) cos φ + sin 2(ωt + θ −
2π
3
) sin φ] +
+
UM
√
2
IM
√
2
[(1 + cos 2(ωt + θ +
2π
3
)) cos φ + sin 2(ωt + θ +
2π
3
) sin φ] = (3.14)
= 3
UM
√
2
IM
√
2
cos φ + cos 2(ωt + θ) + cos 2[ωt + θ −
2π
3
] + cos 2[ωt + θ +
2π
3
]
=0
+
+ sin 2(ωt + θ) + sin 2[ωt + θ −
2π
3
] + sin 2[ωt + θ +
2π
3
]
=0
=
= 3
UM
√
2
IM
√
2
cos φ = 3 ULN I cos φ
Note that the total instantaneous power is equal to three times the active power of a single
phase, and it is constant. This is one of the main reasons why three-phase systems have
been used.

16
3.2.1 Complex power
The corresponding phasor voltages are deﬁned as:
Ua = ULN θ
Ub = ULN (θ − 120◦
) (3.15)
Uc = ULN (θ + 120◦
)
Figure 3.7 shows the phasor diagram of the three balanced line-to-neutral voltages, and also
the phasor diagram of the line-to-line voltages.
+120o
-120o
Ua
Ub
Uc
Uab
Uca
Ubc
Figure 3.7. Phasor diagram of the line-to-neutral and line-to-line voltages.
The phasor of the line-to-line voltages can be determined as follows
Uab =Ua − Ub =
√
3 ULN (θ + 30◦
) =
√
3 Ua ej30◦
Ubc =Ub − Uc =
√
3 ULN (θ − 90◦
) =
√
3 Ub ej30◦
Uca =Uc − Ua =
√
3 ULN (θ + 150◦
) =
√
3 Uc ej30◦
(3.16)
Obviously, the line-to-line voltages are also balanced. Equation (3.16) also shows that the
line-to-line phasor voltage leads the line-to-neutral phasor voltage by 30◦
, and it is
√
3 times
the line-to-neutral phasor voltage.
Next, let the balanced phasor currents be deﬁned as
Ia = I γ
Ib = I (γ − 120◦
) (3.17)
Ic = I (γ + 120◦
)

17
Then, the total three-phase power (S3Φ) is given by :
S3Φ = Sa + Sb + Sc = UaI
∗
a + UbI
∗
b + UcI
∗
c =
= 3 ULN I cos φ + j3 ULN I sin φ =
= 3 ULN I ejφ
(3.18)
Obviously, for a balanced three-phase system Sa = Sb = Sc and S3Φ = 3 S1Φ, where S1Φ is
the complex power of a single phase.
Example 3.4 The student Elektra lives in a house situated 2 km from a transformer having
a completely symmetrical three-phase voltage (Ua = 220V 0◦
, Ub = 220V − 120◦
, Uc =
220V 120◦
). The house is connected to this transformer via a three-phase cable (EKKJ,
3×16 mm2
+ 16 mm2
). A cold day, Elektra switches on two electrical radiators to each
phase, each radiator is rated 1000 W (at 220 V with cosφ = 0.995 lagging (inductive)).
Assume that the cable can be modeled as four impedances connected in parallel (zL = 1.15 +
j0.08 Ω/phase,km, zL0 = 1.15 + j0.015 Ω/km) and that the radiators also can be considered
as impedances. Calculate the total thermal power given by the radiators.
LZ
aI aU′
bU′
cU′
0U′
aU
bU
cU
0U
bI
cI
0I
LZ
0LZ
aZ
bZ
cZ
LZ
Figure 3.8. Single line diagram for example
Solution
Ua = 220 0◦
V, Ub = 220 − 120◦
V, Uc = 220 120◦
V
ZL = 2(1.15 + j0.08) = 2.3 + j0.16 Ω
ZL0 = 2(1.15 + j0.015) = 2.3 + j0.03 Ω
Pa = Pb = Pc = 2000 W (at 220 V, cos φ = 0.995)
sin φ = 1 − cos2 φ = 0.0999
Qa = Qb = Qc = S sin φ = P
cos φ
sin φ = 200.8 VAr
Za = Zb = Zc = U
I
= U·U
∗
I·U
∗ = U2
/S
∗
= U2
/(Pa − jQa) = 23.96 + j2.40 Ω
Ia = Ua−U0
ZL+Za
Ib = Ub−U0
ZL+Zb
Ic = Uc−U0
ZL+Zc
Ia + Ib + Ic = U0−U0
ZL0
= U0−0
ZL0
⇒ U0
1
ZL0
+ 1
ZL+Za
+ 1
ZL+Zb
+ 1
ZL+Zc
= Ua
ZL+Za
+ Ub
ZL+Zb
+ Uc
ZL+Zc

18
⇒ U0 = 0.0
⇒ Ia = 8.34 − 5.58◦
A, Ib = 8.34 − 125.58◦
A, Ic = 8.34 114.42◦
A
The voltage at the radiators can be calculated as :
Ua = U0 + IaZa = 200.78 0.15◦
V
Ub = 200.78 − 119.85◦
V
Uc = 200.78 120.15◦
V
Finally, the power to the radiators can be calculated as
Sza = ZaI2
a = 1666 + j167 VA
Szb = ZbI2
b = 1666 + j167 VA
Szc = ZaI2
c = 1666 + j167 VA
Thus, the total consumed power is
Sza + Szb + Szc = 4998 + j502 VA, i.e. the thermal power = 4998 W
Note that since we are dealing with a balanced three-phase system, Sza = Szb = Szc.
The total transmission losses are
ZL(I2
a + I2
b + I2
c ) + ZL0|Ia + Ib + Ic|2
) = ZL(I2
a + I2
b + I2
c ) = 480 + j33 VA
i.e. the active losses are 480 W, which means that the eﬃciency is 91.2 %.
In a balanced three-phase system, Ia + Ib + Ic = 0. Thus, no current ﬂows in the neutral
conductor (i.e. I0 = 0), and the voltage at the neutral point is zero, i.e. U0 = 0. Therefore,
for analyzing a balanced three-phase system, it is more common to analyze only a single
phase (or more precisely only the positive-sequence network of the system, see Chapter 13).
Then, the total three-phase power can be determined as three times the power of the single
phase.
Example 3.5 Use the data in Example 3.4, but in this example the student Elektra connects
one 1000 W radiator (at 220 V with cosφ = 0.995 lagging) to phase a, three radiators to
phase b and two to phase c. Calculate the total thermal power given by the radiators, as well
as the system losses.
Solution
Ua = 220 0◦
V, Ub = 220 − 120◦
V, Uc = 220 120◦
V
ZL = 2(1.15 + j0.08) = 2.3 + j0.16 Ω
ZL0 = 2(1.15 + j0.015) = 2.3 + j0.03 Ω
Pa = 1000 W (at 220 V, cos φ = 0.995)
sin φ = 1 − cos2 φ = 0.0999
Qa = S sin φ = P
cos φ
sin φ = 100.4 VAr
Za = U2
/S
∗
a = U2
/(Pa − jQa) = 47.9 + j4.81 Ω

19
Zb = Za/3 = 15.97 + j1.60 Ω
Zc = Za/2 = 23.96 + j2.40 Ω
U0
1
ZL0
+ 1
ZL+Za
+ 1
ZL+Zb
+ 1
ZL+Zc
= Ua
ZL+Za
+ Ub
ZL+Zb
+ Uc
ZL+Zc
⇒ U0 = 12.08 − 155.14◦
V
⇒ Ia = 4.58 − 4.39◦
A, Ib = 11.45 − 123.62◦
A, Ic = 8.31 111.28◦
A
The voltages at the radiators can be calculated as :
Ua = U0 + IaZa = 209.45 0.02◦
V
Ub = U0 + IbZb = 193.60 − 120.05◦
V
Uc = U0 + IcZc = 200.91 129.45◦
V
Note that these voltages are not local phase voltages since they are calculated as Ua − U0
etc. The power to the radiators can be calculated as :
Sza = ZaI2
a = 1004 + j101 VA
Szb = ZbI2
b = 2095 + j210 VA
Szc = ZaI2
c = 1655 + j166 VA
The total amount of power consumed is
Sza + Szb + Szc = 4754 + j477 VA, i.e. the thermal power is 4754 W
The total transmission losses are
ZL(I2
a + I2
b + I2
c ) + ZL0|Ia + Ib + Ic|2
) = 572.1 + j36 VA, i.e. 572.1 W
which gives an eﬃciency of 89.3 %.
As shown in this example, an unsymmetrical impedance load will result in unsymmetrical
phase currents, i.e. we are dealing with an unbalanced three-phase system. As a consequence,
a voltage can be detected at the neutral point (i.e. U0 = 0) which gives rise to a current
in the neutral conductor, i.e. I0 = 0. The total thermal power obtained was reduced by
approximately 5 % and the line losses increased partly due to the losses in the neutral
conductor. The eﬃciency of the transmission decreased. It can also be noted that the power
per radiator decreased with the number of radiators connected to the same phase. This owing
to the fact that the voltage at the neutral point will be closest to the voltage in the phase
with the lowest impedance, i.e. the phase with the largest number of radiators connected.

Chapter 4
Models of power system components
Electric energy is transmitted from power plants to consumers via overhead lines, cables and
transformers. In the following, these components will be discussed and mathematical models
to be used in the analysis of symmetrical three-phase systems will be derived. In chapter 11
and 12, analysis of power systems under unsymmetrical conditions will be discussed.
4.1 Electrical characteristic of an overhead line
Overhead transmission lines need large surface area and are mostly suitable to be used in
rural areas and in areas with low population density. In areas with high population density
and urban areas cables are a better alternative. For a certain amount of power transmitted,
a cable transmission system is about ten times as expensive as an overhead transmission
system.
Power lines have a resistance (r) owing to the resistivity of the conductor and a shunt con-
ductance (g) because of leakage currents in the insulation. The lines also have an inductance
(l) owing to the magnetic flux surrounding the line as well as a shunt capacitance (c) because
of the electric field between the lines and between the lines and ground. These quantities
are given per unit length and are continuously distributed along the whole length of the line.
Resistance and inductance are in series while the conductance and capacitance are shunt
quantities.
l l l lr r r r
c c c cg g g g
Figure 4.1. A line with distributed quantities
Assuming symmetrical three-phase, a line can be modeled as shown in Figure 4.1. The
quantities r, g, l, and c determine the characteristics of a line. Power lines can be modeled
by simple equivalent circuits which, together with models of other system components, can
be formed to a model of a complete system or parts of it. This is important since such
models are used in power system analysis where active and reactive power flows in the
network, voltage levels, losses, power system stability and other properties at disturbances
as e.g. short circuits, are of interest.
For a more detailed derivation of the expressions of inductance and capacitance given below,
more fundamental literature in electro-magnetic theory has to be studied.
21

22
4.1.1 Resistance
The resistance of a conductor with the cross-section area A mm2
and the resistivity ρ
Ωmm2
/km is
r =
ρ
A
Ω/km (4.1)
The conductor is made of copper with the resistivity at 20◦
C of 17.2 Ωmm2
/km, or aluminum
with the resistivity at 20◦
C of 27.0 Ωmm2
/km. The choice between copper or aluminum is
related to the price difference between the materials.
The effective alternating current resistance at normal system frequency (50–60 Hz) for lines
with a small cross-section area is close to the value for the direct current resistance. For
larger cross-section areas, the current density will not be equal over the whole cross-section.
The current density will be higher at the peripheral parts of the conductor. That phenomena
is called current displacement or skin effect and depends on the internal magnetic flux of
the conductor. The current paths that are located in the center of the conductor will be
surrounded by the whole internal magnetic flux and will consequently have an internal self
inductance. Current paths that are more peripheral will be surrounded by a smaller magnetic
flux and thereby have a smaller internal inductance.
The resistance of a line is given by the manufacturer where the influence of the skin effect is
taken. Normal values of the resistance of lines are in the range 10–0.01 Ω/km.
The resistance plays, compared with the reactance, often a minor role when comparing the
transmission capability and voltage drop between different lines. For low voltage lines and
when calculating the losses, the resistance is of significant importance.
4.1.2 Shunt conductance
The shunt conductance of an overhead line represents the losses owing to leakage currents
at the insulators. There are no reliable data over the shunt conductances of lines and these
are very much dependent on humidity, salt content and pollution in the surrounding air. For
cables, the shunt conductance represents the dielectric losses in the insulation material and
data can be obtained from the manufacturer.
The dielectric losses are e.g. for a 12 kV cross-linked polyethylene (XLPE) cable with a
cross-section area of 240 mm2
/phase 7 W/km,phase and for a 170 kV XLPE cable with the
same area 305 W/km,phase.
The shunt conductance will be neglected in all calculations throughout this compendium.
4.1.3 Inductance
The inductance is in most cases the most important parameter of a line. It has a large
influence on the line transmission capability, voltage drop and indirectly the line losses. The
inductance of a line can be calculated by the following formula :

23
l = 2 · 10−4
ln
a
d/2
+
1
4n
H/km,phase (4.2)
where
a = 3
√
a12a13a23 m, = geometrical mean distance according to Figure 4.2.
d = diameter of the conductor, m
n = number of conductors per phase
Ground level
H1
H2 H3
a12
a23
a13
A1
A2
A3
Figure 4.2. The geometrical quantities of a line in calculations of inductance and capacitance
The calculation of the inductance according to equation (4.2), is made under some assump-
tions, viz. the conductor material must be non-magnetic as copper and aluminum together
with the assumption that the line is transposed. The majority of the long transmission lines
are transposed, see Figure 4.3.
Transposing cycle
Locations of transposing
Figure 4.3. Transposing of three-phase overhead line
This implies that each one of the conductors, under a transposing cycle, has had all three
possible locations in the transmission line. Each location is held under equal distance which
implies that all conductors in average have the same distance to ground and to the other

24
conductors. This gives that the mutual inductance between the three phases are equalized
so that the inductance per phase is equal among the three phases.
In many cases, the line is constructed as a multiple conductor, i.e. more than one conductor
is used for each phase, see Figure 4.4. Multiple conductors implies both lower reactance of
D
2
d
Figure 4.4. Cross-section of a multiple conductor with three conductors per phase
the line and reduced corona effect (glow discharge). The radius d/2 in equation (4.2) must
in these cases be replaced with the equivalent radius
(d/2)eq = n
n(D/2)n−1 · (d/2) (4.3)
where
n = number of conductors per phase
D/2 = radius in the circle formed by the conductors
By using the inductance, the reactance of a line can be calculated as
x = ω l = 2πf l Ω/km,phase (4.4)
and is only dependent on the geometrical design of the line if the frequency is kept constant.
The relationship between the geometrical mean distance a and the conductor diameter d
in equation (4.2) varies within quite small limits for different lines. This due to the large
distance between the phases and the larger conductor diameter for lines designed for higher
system voltages. The term 1
4n
has, compared with ln( a
d/2
), usually a minor influence on the
line inductance.
At normal system frequency, the reactance of an overhead line can vary between 0.3 and 0.5
Ω/km,phase with a typical value of 0.4 Ω/km,phase. For cables, the reactance vary between
0.08 and 0.17 Ω/km,phase where the higher value is valid for cables with a small cross-section
area. The reactance for cables is considerably lower than the reactance of overhead lines. The
difference is caused by the difference in distance between the conductors. The conductors
are more close to one another in cables which gives a lower reactance. See equation (4.2)
which gives the inductance of overhead lines.
Example 4.1 Determine the reactance of a 130 kV overhead line where the conductors are
located in a plane and the distance between two closely located conductors is 4 m. The
conductor diameter is 20 mm. Repeat the calculations for a line with two conductors per
phase, located 30 cm from one another.

25
Solution
a12 = a23 = 4, a13 = 8
d/2 = 0.01 m
a = 3
√
4 · 4 · 8 = 5.04
x = 2π · 50 · 2 · 10−4
ln 5.04
0.01
+ 1
4
= 0.0628 (ln(504) + 0.25) = 0.41 Ω/km,phase
Multiple conductor (duplex)
(d/2)eq = 2
2(0.3/2)0.01 = 0.055
x = 0.0628 ln 5.04
0.055
+ 1
8
= 0.29 Ω/km,phase
The reactance is in this case reduced by 28 %.
4.1.4 Shunt capacitance
For a three-phase transposed overhead line, the capacitance to ground per phase can be
calculated as
c =
10−6
18 ln 2H
A
· a
(d/2)eq
F/km,phase (4.5)
where
H = 3
√
H1H2H3 = geometrical mean height for the conductors according to Figure 4.2.
A = 3
√
A1A2A3 = geometrical mean distance between the conductors and their image con-
ductors according to Figure 4.2.
As indicated in equation (4.5), the ground has some influence on the capacitance of the line.
The capacitance is determined by the electrical field which is dependent on the characteristics
of the ground. The ground will form an equipotential surface which has an influence on the
electric field.
The degree of influence the ground has on the capacitance is determined by the factor 2H/A
in equation (4.5). This factor has usually a value near 1.
Assume that a line mounted on relatively high poles (⇒ A ≈ 2H) is considered and that the
term 1
4n
can be neglected in equation (4.2). By multiplying the expressions for inductance
and capacitance, the following is obtained
l · c = 2 · 10−4
ln
a
(d/2)eq
·
10−6
18 ln a
(d/2)eq
=
1
(3 · 105)2
km
s
−2
=
1
v2
(4.6)
where v = speed of light in vacuum in km/s. Equation (4.6) can be interpreted as the
inductance and capacitance are the inverse of one another for a line. Equation (4.6) is a
good approximation for an overhead line.
The shunt susceptance of a line is
bc = 2πf · c S/km,phase (4.7)

26
A typical value of the shunt susceptance of a line is 3 · 10−6
S/km,phase. Cables have
considerable higher values between 3 · 10−5
– 3 · 10−4
S/km,phase.
Example 4.2 Assume that a line has a shunt susceptance of 3 · 10−6
S/km,phase. Use
equation (4.6) to estimate the reactance of the line.
Solution
x = ωl ≈
ω
cv2
=
ω2
bv2
=
(100π)2
3 · 10−6(3 · 105)2
= 0.366 Ω/km
which is near the standard value of 0.4 Ω/km for the reactance of an overhead line.
4.2 Model of a line
Both overhead lines and cables have their electrical quantities r, x, g and b distributed along
the whole length. Figure 4.1 shows an approximation of the distribution of the quantities.
Generally, the accuracy of the calculation result will increase with the number of distributed
quantities.
At a first glance, it seems possible to form a line model where the total resistance/inductance
is calculated as the product between the resistance/inductance per length unit and the length
of the line. This approximation is though only valid for short lines and lines of medium
length. For long lines, the distribution of the quantities r, l, c and g must be taken into
account. Such analysis can be carried out with help of differential calculus.
There are no absolute limits between short, medium and long lines. Usually, lines shorter
than 100 km are considered as short, between 100 km and 300 km as medium long and lines
longer than 300 km are classified as long. For cables, having considerable higher values of
the shunt capacitance, the distance 100 km should be considered as medium long. In the
following, models for short and medium long lines are given.
4.2.1 Short lines
In short line models, the shunt parameters are neglected, i.e. conductance and susceptance.
This because the current flowing through these components is less than one percent of the
rated current of the line. The short line model is given in Figure 4.5. This single-phase
model of a three-phase system is valid under the assumption that the system is operating
under symmetrical conditions.
The impedance of the line between bus k and bus j can be calculated as
Zkj = Rkj + jXkj = (rkj + jxkj) L Ω/phase (4.8)
where L is the length of the line in km.

27
kU kI kj kj kjZ R jX= +
jU
Figure 4.5. Short line model of a line
4.2.2 Medium long lines
For lines having a length between 100 and 300 km, the shunt capacitance cannot be neglected.
The model shown in Figure 4.5 has to be extended with the shunt susceptance, which
results in a model called the π-equivalent shown in Figure 4.6. The impedance is calculated
jU
2
sh kjY −
2
sh kjY −
shI
I
or
jU
sh kjy −
shI
I
sh kjy −
Figure 4.6. Medium long model of a line
according to equation (4.8) and the admittance to ground per phase is obtained by
Y sh−kj
2
= j
bc L
2
= ysh−kj = jbsh−kj S (4.9)
i.e. the total shunt capacitance of the line is divided into two equal parts, one at each end
of the line. The π-equivalent is a very common and useful model in power system analysis.
4.3 Single-phase transformer
The principle diagram of a two winding transformer is shown in Figure 4.7. The fundamental
principles of a transformer are given in the figure. In a real transformer, the demand of
a strong magnetic coupling between the primary and secondary sides must be taken into
account in the design.
Assume that the magnetic flux can be divided into three components. There is a core flux Φm
passing through both the primary and the secondary windings. There are also leakage fluxes,
Φl1 passing only the primary winding and Φl2 which passes only the secondary winding. The
resistance of the primary winding is r1 and for the secondary winding r2. According to the
law of induction, the following relationships can be given for the voltages at the transformer

28
Iron core
m
Φ
1l
Φ 2l
Φ
1
u 2
u1
i 2
i′
Primary
Winding
N1 turns
Secondary
Winding
N2 turns
Figure 4.7. Principle design of a two winding transformer
terminals :
u1 = r1i1 + N1
d(Φl1 + Φm)
dt
(4.10)
u2 = r2i2 + N2
d(Φl2 + Φm)
dt
Assuming linear conditions, the following is valid
N1Φl1 = Ll1i1 (4.11)
N2Φl2 = Ll2i2
where
Ll1 = inductance of the primary winding
Ll2 = inductance of the secondary winding
Equation (4.10) can be rewritten as
u1 = r1i1 + Ll1
di1
dt
+ N1
dΦm
dt
(4.12)
u2 = r2i2 + Ll2
di2
dt
+ N2
dΦm
dt
With the reluctance R of the iron core and the deﬁnitions of the directions of the currents
according to Figure 4.7, the magnetomotive forces N1i1 and N2i2 can be added as
N1i1 + N2i2 = RΦm (4.13)

29
Assume that i2 = 0, i.e. the secondary side of the transformer is not connected. The current
now flowing in the primary winding is called the magnetizing current and the magnitude can
be calculated using equation (4.13) as
im =
RΦm
N1
(4.14)
If equation (4.14) is inserted into equation (4.13), the result is
i1 = im −
N2
N1
i2 = im +
N2
N1
i2 (4.15)
where
i2 = −i2 (4.16)
Assuming linear conditions, the induced voltage drop N1
dΦm
dt
in equation (4.12) can be
expressed by using an inductor as
N1
dΦm
dt
= Lm
dim
dt
(4.17)
i.e. Lm = N2
1 /R. By using equations (4.12), (4.15) and (4.17), the equivalent diagram of a
single-phase transformer can be drawn, see Figure 4.8.
2
2
1
N
i
N
mi
1i 2i2r1r
1u 2u2e1e
1lL 2lL
1N 2N
mL
ideal
Figure 4.8. Equivalent diagram of a single-phase transformer
In Figure 4.8, one part of the ideal transformer is shown, which is a lossless transformer
without leakage fluxes and magnetizing currents.
The equivalent diagram in Figure 4.8 has the advantage that the different parts represents
different parts of the real transformer. For example, the inductance Lm represents the
assumed linear relationship between the core flux Φm and the magnetomotive force of the
iron core. Also the resistive copper losses in the transformer are represented by r1 and r2.
In power system analysis, where the transformer is modeled, a simplified model is often used
where the magnetizing current is neglected.

30
4.4 Three-phase transformer
There are three fundamental ways of connecting single-phase transformers into one three-
phase transformer. The three combinations are Y-Y-connected, ∆-∆-connected and Y-∆-
connected (or ∆-Y-connected). In Figure 4.9, the different combinations are shown.
a
b
c
n N
A
B
C
a
b
c
A
B
C
a
b
c
n
A
B
C
Y-Y-connected Y- -connected- -connected
Figure 4.9. Standard connections for three-phase transformers
When the neutral (i.e. n or N) is grounded, the Y-connected part will be designated by
Y0. The different consequences that these different connections imply, will be discussed in
Chapter 12.

Chapter 5
Important theorems in power system analysis
In many cases, the use of theorems can simplify the analysis of electrical circuits and systems.
In the following sections, some important theorems will be discussed and proofs will be given.
5.1 Bus analysis, admittance matrices
Consider an electric network which consists of four buses as shown in Figure 5.1. Each bus
is connected to the other buses via an admittance ykj where the subscript indicates which
buses the admittance is connected to. Assume that there are no mutual inductances between
o o
o
o
y12
y23
y13
y14
y24
y34
1
3
4
2 I2I1
I4
I3
Figure 5.1. Four bus network
the admittances and that the buses voltages are U1, U2, U3 and U4. The currents I1, I2, I3
and I4 are assumed to be injected into the buses from external current sources. Application
of Kirchhoﬀ’s current law at bus 1 gives
I1 = y12(U1 − U2) + y13(U1 − U3) + y14(U1 − U4) (5.1)
or
I1 = (y12 + y13 + y14)U1 − y12U2 − y13U3 − y14U4 = (5.2)
= Y 11U1 + Y 12U2 + Y 13U3 + Y 14U4
where
Y 11 = y12 + y13 + y14 and Y 12 = −y12, Y 13 = −y13, Y 14 = −y14 (5.3)
Corresponding equations can be formed for the other buses. These equations can be put
31

32
together to a matrix equation as :
I =




I1
I2
I3
I4



 =




Y 11Y 12Y 13Y 14
Y 21Y 22Y 23Y 24
Y 31Y 32Y 33Y 34
Y 41Y 42Y 43Y 44








U1
U2
U3
U4



 = YU (5.4)
This matrix is termed as the bus admittance matrix or Y-bus matrix.whicha has the following
properties :
• It can be uniquely determined from a given admittance network.
• The diagonal element Y kk = the sum of all admittances connected to bus k.
• Non-diagonal element Y kj = −ykj = − 1
Zkj
where ykj is the admittance between bus k
and bus j.
• This gives that the matrix is symmetric, i.e. Y kj = Y jk (one exception is when the
network includes phase shifting transformers).
• It is singular since I1 + I2 + I3 + I4 = 0
If the potential in one bus is assumed to be zero, the corresponding row and column in the
admittance matrix can be removed which results in a non-singular matrix. Bus analysis using
the Y-bus matrix is the method most often used when studying larger, meshed networks in
a systematic manner.
Example 5.1 Re-do Example 3.5 by using the Y-bus matrix of the network in order to
calculate the power given by the radiators.
2I
1U
2U
3U
3I
0I 4I
4U
1I
LZ
0U
LZ
0LZ
aZ
bZ
cZ
LZ
Figure 5.2. Network diagram used in example
Solution
According to the task and to the calculations performed in Example 3.5, the following is valid;
ZL = 2.3 + j0.16 Ω, ZL0 = 2.3 + j0.03 Ω, Za = 47.9 + j4.81 Ω, Zb = 15.97 + j1.60 Ω, Zc =

33
23.96 + j2.40 Ω. Start with forming the Y-bus matrix. I0 and U0 are neglected since the
system otherwise will be singular.
I =




I1
I2
I3
I4



 =





1
ZL+Za
0 0 − 1
ZL+Za
0 1
ZL+Zb
0 − 1
ZL+Zb
0 0 1
ZL+Zc
− 1
ZL+Zc
− 1
ZL+Za
− 1
ZL+Zb
− 1
ZL+Zc
Y 44









U1
U2
U3
U4



 = YU (5.5)
where
Y 44 =
1
ZL + Za
+
1
ZL + Zb
+
1
ZL + Zc
+
1
ZL0
(5.6)
In the matrix equation above, U1, U2, U3 and I4 (I4=0) as well as all impedances, i.e. the
Y-bus matrix, are known. If the given Y-bus matrix is inverted, the corresponding Z-bus
matrix is obtained :
U =




U1
U2
U3
U4



 = ZI = Y−1
I =




Z11Z12Z13Z14
Z21Z22Z23Z24
Z31Z32Z33Z34
Z41Z42Z43Z44








I1
I2
I3
I4



 (5.7)
Since the elements in the Y-bus matrix are known, all the elements in the Z-bus matrix can
be calculated. Since I4=0 the voltages U1, U2 and U3 can be expressed as a function of the
currents I1, I2 and I3 by using only a part of the Z-bus matrix :


U1
U2
U3

 =


Z11Z12Z13
Z21Z22Z23
Z31Z32Z33




I1
I2
I3

 (5.8)
Since the voltages U1, U2 and U3 are known, the currents I1, I2 and I3 can be calculated as :


I1
I2
I3

 =


Z11Z12Z13
Z21Z22Z23
Z31Z32Z33


−1 

U1
U2
U3

 = (5.9)
= 10−3


19.0 − j1.83 −1.95 − j0.324 −1.36 + j0.227
−1.95 + j0.324 48.9 − j4.36 −3.73 − j0.614
−1.36 + j0.227 −3.73 + j0.614 35.1 − j3.25




220 0◦
220 − 120◦
220 120◦

 =
=


4.58 − 4.39◦
11.5 − 123.6◦
8.31 111.3◦

 A
By using these currents, the power given by the radiators can be calculated as :
Sza = ZaI2
1 = 1004 + j101 VA
Szb = ZbI2
2 = 2095 + j210 VA = 4754 + j477 VA (5.10)
Szc = ZcI2
3 = 1655 + j166 VA
i.e. the thermal power obtained is 4754 W.

34
5.2 Millman’s theorem
Millman’s theorem (the parallel generator-theorem) gives that if a number of admittances
Y 1, Y 2, Y 3 . . . Y n are connected to a common bus k, and the voltages to a reference bus
U10, U20, U30 . . . Un0 are known, the voltage between bus k and the reference bus, Uk0 can
be calculated as
Uk0 =
n
i=1
Y iUi0
n
i=1
Y i
(5.11)
Assume a Y-connection of admittances as shown in Figure 5.3. The Y-bus matrix for this
20U
10U 0nU
0kU
1I
nI
nY1Y
2Y2I
1
0
n
k
2
Figure 5.3. Y-connected admittances
network can be formed as







I1
I2
...
In
Ik







=







Y 1 0 . . . 0 −Y 1
0 Y 2 . . . 0 −Y 2
...
...
...
...
...
0 0 . . . Y n −Y n
−Y 1 −Y 2 . . . −Y n (Y 1 + Y 2 + . . . Y n)














U10
U20
...
Un0
Uk0







(5.12)
This equation can be written as





I1
I2
...
Ik





=





U10Y 1 − Uk0Y 1
U20Y 2 − Uk0Y 2
...
−U10Y 1 − U20Y 2 − . . . + n
i=1 Y iUk0





(5.13)
Since no current is injected at bus k (Ik = 0), the last equation can be written as
Ik = 0 = −U10Y 1 − U20Y 2 − . . . +
n
i=1
Y iUk0 (5.14)

35
This equation can be written as
Uk0 =
U10Y 1 + U20Y 2 + . . . + Un0Y n
n
i=1
Y i
(5.15)
and by that, the proof of the Millman’s theorem is completed.
Example 5.2 Find the solution to Example 3.5 by using Millman’s theorem, which will be
the most eﬃcient method to solve the problem so far.
2I
1U
2U
3U
3I
0I 4I
4U
1I
LZ
0U
LZ
0LZ
aZ
bZ
cZ
LZ
Figure 5.4. Diagram of the network used in the example.
Solution
According to the task and to the calculations performed in Example 3.5, the following is valid;
ZL = 2.3 + j0.16 Ω, ZL0 = 2.3 + j0.03 Ω, Za = 47.9 + j4.81 Ω, Zb = 15.97 + j1.60 Ω, Zc =
23.96 + j2.40 Ω.
By using Millman’s theorem, the voltage at bus 4 can be calculated as
U40 =
n
i=1
Y iUi0
n
i=1
Y i
=
U0
1
ZL0
+ U1
1
Za+ZL
+ U2
1
Zb+ZL
+ U3
1
Zc+ZL
1
ZL0
+ 1
Za+ZL
+ 1
Zb+ZL
+ 1
Zc+ZL
=
= 12.08 − 155.1◦
V
(5.16)
The currents through the impedances can be calculated as
I1 =
U1 − U4
Za + ZL
= 4.58 − 4.39◦
A
I2 =
U2 − U4
Zb + ZL
= 11.5 − 123.6◦
A (5.17)
I3 =
U3 − U4
Zc + ZL
= 8.31 111.3◦
A

36
By using these currents, the power from the radiators can be calculated in the same way as
earlier :
Sza = ZaI2
1 = 1004 + j101 VA
Szb = ZbI2
2 = 2095 + j210 VA = 4754 + j477 VA (5.18)
Szc = ZcI2
3 = 1655 + j166 VA
i.e. the thermal power is 4754 W.
5.3 Superposition theorem
According to section 5.1, each admittance network can be described by a Y-bus matrix, i.e.
I = YU (5.19)
where
I = vector with currents injected into the buses
U = vector with the bus voltages
The superposition theorem can be applied to variables with a linear dependence, as shown in
equation (5.19). This implies that the solution is obtained piecewise, e.g. for one generator
at the time. The total solution is obtained by adding all the part solutions found :
I =





I1
I2
...
In





= Y





U1
U2
...
Un





= Y





U1
0
...
0





+ Y





0
U2
...
0





+ . . . + Y





0
0
...
Un





(5.20)
It can be noted that the superposition theorem cannot be applied to calculations of the
power ﬂow since they cannot be considered as linear properties since they are the product
between voltage and current.
Example 5.3 Use the conditions given in Example 5.1 and assume that a fault at the feeding
transformer gives a short circuit of phase 2. Phase 1 and 3 are operating as usual. Calculate
the thermal power obtained in the house of Elektra.
Solution
According to equation (5.9) in Example 5.1, the phase currents can be expressed as a function
of the feeding voltages as


I1
I2
I3

 =


Z11Z12Z13
Z21Z22Z23
Z31Z32Z33


−1 

U1
U2
U3

 (5.21)

37
2I
~
~ 2U−
1U
2U
3U
3I
0I 4I
4U
1I
LZ
0U
LZ
0LZ
aZ
bZ
cZ
LZ
Figure 5.5. Diagram of the network used in the example.
A short circuit in phase 2 is equivalent with connecting an extra voltage source in reverse
direction in series with the already existing voltage source. The phase currents in the changed
system can be calculated as :


I1
I2
I3

 =


Z11Z12Z13
Z21Z22Z23
Z31Z32Z33


−1 

U1
U2
U3

 +


Z11Z12Z13
Z21Z22Z23
Z31Z32Z33


−1 

0
−U2
0

 =
=


4.58 − 4.39◦
11.5 − 123.6◦
8.31 111.3◦

 +


Z11Z12Z13
Z21Z22Z23
Z31Z32Z33


−1 

0
−220 − 120◦
0

 =
=


4.34 − 9.09◦
0.719 − 100.9◦
7.94 − 116.5◦

 A (5.22)
Sza = ZaI2
1 = 904 + j91 VA
Szb = ZbI2
2 = 8.27 + j0.830 VA = 2421 + j243 VA (5.23)
Szc = ZcI2
3 = 1509 + j151 VA
i.e. the thermal power is 2421 W
As shown in this example, the superposition theorem can, for instance, be used when studying
changes in the system. But it should once again be pointed out that this is valid under the
assumption that the loads (the radiators in this example) can be modeled as impedances.
5.4 Reciprocity theorem
Assume that a voltage source is connected to a terminal a in a linear reciprocal network and
is giving rise to a current at terminal b. According to the reciprocity theorem, the voltage
source will cause the same current at a if it is connected to b. The Y-bus matrix (and by
that also the Z-bus matrix) are symmetrical matrices for a reciprocal electric network.

38
Assume that an electric network with n buses can be described by a symmetric Y-bus matrix,
i.e.





I1
I2
...
In





= I = YU =





Y 11 Y 12 . . . Y 1n
Y 21 Y 22 . . . Y 2n
...
...
...
...
Y n1 Y n2 . . . Y nn










U1
U2
...
Un





(5.24)
Assume that all voltages are = 0 except Ua. The current at b can now be calculated as
Ib = Y baUa (5.25)
Assume instead that all voltages are = 0 except Ub. This means that the current at a is
Ia = Y abUb (5.26)
If Ua = Ub, the currents Ia and Ib will be equal since the Y-bus matrix is symmetric, i.e.
Y ab = Y ba. By that, the proof of the reciprocity theorem is completed.
5.5 Thévenin-Helmholtz’s theorem
This theorem is often called the Thévenin’s theorem (after Léon Charles Thévenin, telegraph
engineer and teacher, who published the theorem in 1883). But 30 years earlier, Hermann
von Helmholtz published the same theorem in 1853, including a simple proof. The theorem
can be described as
• Thévenin-Helmholtz’s theorem states that from any output terminal in a linear electric
network, no matter how complex, the entire linear electric network seen from the
output terminal can be modelled as an ideal voltage source UTh (i.e. the voltage will
be constant (or unchanged) regardless of how the voltage source is loaded) in series
with an impedance ZTh. According to this theorem, when the output terminal is not
loaded, its voltage is UTh, and the impedance ZTh is the impedance seen from the
output terminal when all voltage sources in the network are short circuited and all
current sources are disconnected.

39
Proof :
Assume that the voltage at an output
terminal is UTh. Loading the output
terminal with an impedance Zk, a cur-
rent I will ﬂow through the impedance.
This connection is similar to have a net-
work with a voltage source UTh con-
necting to the output terminal in se-
ries with the impedance Zk, together
with having a network with the voltage
source −UTh connecting to the output
terminal and the other voltage sources
in the network shortened. By using
the superposition theorem, the current
I can be calculated as the sum of I1
and I2. The current I1 = 0 since the
voltage is equal on both sides of the
impedance Zk. The current I2 can be
calculated as
I2 = −(−UTh)/(Zk + ZTh)
since the network impedance seen from
the output terminal is ZTh. The con-
clusion is that
~
I
ThU
kZ
Linear
electric
network
=
Linear
electric
network
1I
+
~ ThU−
Voltage
sources
shortened
2I
kZ
kZ
Output
terminal
I = I1 + I2 =
UTh
Zk + ZTh
(5.27)
which is the same as stated by Th´evenin-Helmholtz’s theorem, viz.
~ThU
Linear
electric
network
=
ThZ
ThU
Output
terminal

Chapter 6
Analysis of balanced three-phase systems
Consider the simple balanced three-phase system shown in Figure 6.1, where a symmetric
three-phase Y0-connected generator supplies a symmetric Y0-connected impedance load.
The neutral of the generator (i.e. point N) is grounded via the impedance ZNG. However,
the neutral of the load (i.e. point n) is directly grounded. Since we are dealing with a
balanced (or symmetrical) system, IN = Ia + Ib + Ic = In = 0, i.e. UN = Un = 0 and ZNG
has no impact on the system. Note that also in case of connecting point n directly to point
N via the impedance ZNG, the neutrals n and N have the same potential, i.e. UN = Un,
since in a balanced system Ia + Ib + Ic = 0.
~
NGZ
nN
aI
~
~
NI bI
cI
nI
GZ
GZ
GZ
LDZ
LDZ
LDZ
aU
bU
cU
Figure 6.1. A simple three-phase system.
Therefore, the analysis of a balanced three-phase system can be carried out by studying only
one single phase where the components can be connected together by a common neutral
conductor as shown in Figure 6.1 a).
~
aIGZ
LDZaU ~
IGZ
LDZU
a) b)
Figure 6.2. Single-phase equivalent of a symmetric three-phase system.
Based on Figure 6.1 a), the total three-phase supplied power is given by
Ia = I ejγ
=
Ua
ZG + ZL
=
ULN ejθ
ZG + ZL
S3Φ = 3 Ua Ia = 3 ULN I ej(θ−γ)
= 3 ULN I ejφ
=
√
3
√
3 ULN I ejφ
=
√
3 ULL I ejφ
(6.1)
For analysis of balanced three-phase systems, it is common to use the line-to line voltage
magnitudes, i.e. the voltage Ua in Figure 6.1 a) is replaced by U = U ejθ
(as shown in
41

42
Figure 6.1 b)) where, U = ULL, however the phase angle of this voltage is the phase angle
of the phase voltage. Furthermore, the other components in Figure 6.1 b) are per phase
components. Based on Figure 6.1 b), we have then
U =
√
3 I (ZG + ZL)
S3Φ =
√
3 U I =
√
3 U I ejφ
(6.2)
Single-phase equivalent of three-phase transformers
Figure 6.3 shows the single-phase equivalent of a Y-Y-connected three-phase transformer. In
the ﬁgure, Uan and UAN are the line-to-neutral phasor voltages of the primary and secondary
sides, respectively. However, Uab and UAB are the line-to-line phasor voltages of the primary
and secondary sides, respectively. As shown in Figure 6.3 b), the ratio of line-to-neutral
voltages is the same as the ratio of line-to-line voltages.
aIa
b
c
nabU
anU 1N
N
ANU
A
C
B
ABU
AI
1 2:N N
anU ANU
aI AI
a) b)
1 1
2 2
;an ab
AN AB
U UN N
U N U N
= =
2N
Figure 6.3. Single-phase equivalent of a three-phase Y-Y-connected transformer.
Figure 6.4 shows the single-phase equivalent of a ∆-∆-connected three-phase transformer.
For a ∆-∆-connected transformer the ratio of line-to-neutral voltages is also the same as
the ratio of line-to-line voltages. Furthermore, for Y-Y-connected and ∆-∆-connected trans-
formers Uan is in phase with UAN (or Uab is in phase with UAB).
aI
a
b
c
n
abU
anU
1N
N
ANU
A
C
B
ABU
AI
a) b)
2N
1 2
:
3 3
N N
anU ANU
aI AI
1 1
22
/ 3
;
/ 3
an ab
AB AB
U UN N
U U NN
= =
1 / 3N 2 / 3N
Figure 6.4. Single-phase equivalent of a three-phase ∆-∆-connected transformer.
It should be noted that ∆ windings have no neutral, and for analysis of ∆-connected trans-
formers it is more convenient to replace the ∆-connection with an equivalent Y-connection

43
as shown with the dashed lines in the ﬁgure. Since for balanced operation, the neutrals of the
equivalent Y-connections have the same potential the single-phase equivalent of both sides
can be connected together by a neutral conductor. This is also valid for Y-∆-connected (or
∆-Y-connected) three-phase transformer.
Figure 6.5 shows the single-phase equivalent of a Y-∆-connected three-phase transformer.
aIa
b
c
nabU
anU 1N
N
ANU
A
C
B
ABU
AI
a) b)
2N
2 / 3N
2
1 :
3
N
N
anU ANU
aI AI
1 1
2 2
; 3an ab
AB AB
U UN N
U N U N
= =
Figure 6.5. Single-phase equivalent of a three-phase Y-∆-connected transformer.
It can be shown that Uan = N1
N2
UAB =
√
3 N1
N2
UAN ej 30◦
, i.e. Uan leads UAN by 30◦
(see also
equation (3.16)).
In this compendium, this phase shift is not of concern. Furthermore, in this compendium
the ratio of rated line-to-line voltages (rather than the turns ration) will be used. Therefore,
regardless of the transformer connection, the voltage and current can be transferred from the
voltage level on one side to the voltage level on the other side by using the ratio of rated line-
to-line voltages as multiplying factor. Also, the transformer losses and magnetizing currents
(i.e. im in Figure 4.8) are neglected.
Figure 6.6 shows the single-line diagram of a lossless three-phase transformer which will be
used in this compendium. In the ﬁgure, U1n is the rated line-to-line voltage (given in kV) of
1U
1
2
; ;n
nt t
n
U
S x
U
2U
Figure 6.6. Single-line diagram of a three-phase transformer.
the primary side and U2n is the rated line-to-line voltage (given in kV) of the secondary side.
U1n/U2n is the ratio of rated line-to-line voltages. Snt is the transformer three-phase rating
given in MVA, and xt is the transformer leakage reactance, normally given as a percent based
on the transformer rated (or nominal) values. Finally, U1 and U2 are the line-to-line phasor
voltages of the transformer terminals.

44
6.1 Single-line and impedance diagrams
A single-line diagram of a balanced three-phase power system shows the main components
as well as the connections between them. A component is only given in the diagram if it is
of interest for the analysis. Figure 6.7 shows the single-line diagram of a simple balanced
three-phase power system. The system consists of four buses (or nodes) numbering from one
to four, two generators G1 and G2, two transformers T1 and T2, two loads LD1 and LD2,
and a transmission line between bus2 and bus3.
~
~
1 2 3 4
Line
LD1 LD2
G1
G2T1 T2
Figure 6.7. Single-line diagram of a small power system
Here-onward, if not otherwise explicitly stated, the following is valid in this compendium:
• all system quantities (power, voltage, current, impedances and admittances) are given
in the complex form,
• power is given as three-phase power in MVA, MW and/or MVAr,
• for the phasor voltage U = U θ, the magnitude U is a line-to-line voltage given in kV,
however the phase angle θ is the phase angle of a line-to-neutral voltage,
• currents (given in kA), impedances (given in Ω) and admittances (given in S) are per
phase quantities.
Consider again the system shown in Figure 6.7. A typical system data can be given as
follows:
• Generator G1 : Sng=30 MVA, Ung=10 kV, xg=10%
• Generator G2 : Sng=15 MVA, Ung=6 kV, xg=8%
• Transformer T1 : Snt=15 MVA,
U1n
U2n
=
10 kV
30 kV
, xt=10%
• Transformer T2 : Snt=15 MVA,
U1n
U2n
=
30 kV
6 kV
, xt=10%
• Line : r = 0.17 Ω/km, x = 0.3 Ω/km, bc = 3.2 × 10−6
S/km and L = 10 km
• Load LD1 : impedance load, PLD = 15 MW, Un = 30 kV, cos φ = 0.9 inductive
• Load LD2 : impedance load, PLD = 40 MW, Un = 6 kV, cos φ = 0.8 inductive

45
Comments:
Sng is the generator three-phase rating, Ung is the generator rated (or nominal) line-to-line
voltage and xg is the generator reactance given as a percent based on the generator rated
values. The actual value of the generator reactance can be determined by
Xg =
xg
100
U2
ng
Sng
Ω and Zg = j Xg
In a similar way the actual value of the transformer leakage reactance can be determined,
however, depending on which side of the transformer it will be calculated. Having the
reactance on the primary side, then it is determined by
Xtp =
xt
100
U2
1n
Snt
Ω and Ztp = j Xtp
Having the reactance on the secondary side, then it is determined by
Xts =
xt
100
U2
2n
Snt
Ω and Zts = j Xts
For the line, using the model shown in Figure 4.6, we have
Z12 = L (r + jx) Ω and Y sh−12 = jbc L S
For the load, P is the consumed three-phase active power with the power factor cos φ at the
nominal (or rated) voltage Un. Thus, the impedance load can be determined by
ZLD =
U2
n
S
∗
LD
=
U2
n
SLD(cos φ − j sin φ)
=
U2
n
SLD
(cos φ + j sin φ) where SLD =
PLD
cos φ
Figure 6.8 shows the single-phase impedance diagram corresponding to the single-line dia-
gram shown in Figure 6.7.
tpZ
~
gZ
1LDZ
G1 T1 LD1
LineZ
2
sh LineY −
2
sh LineY −
~
gZ
2LDZ
Line T2 LD2
tsZ
G2
Figure 6.8. Impedance network of a small power system.
The simple system shown in Figure 6.8 has three different voltage levels (6, 10 and 30 kV).
The analysis of the system can be carried out by transferring all impedances to a single
voltage level. This method gives often quite extensive calculations, especially dealing with
large systems with several different voltage levels. To overcome this difficulty, the so called
per-unit system was developed, and it will be presented in the next section.

46
6.2 The per-unit (pu) system
A common method to express voltages, currents, powers and impedances in an electric
network is in per-unit (or percent) of a certain base or reference value. The per-unit value
of a certain quantity is defined as
Per-unit value =
true value
base value of the quantity
(6.3)
The per-unit method is very suitable for power systems with several voltage levels and
transformers. In a three-phase system, the per-unit value can be calculated using the corre-
sponding base quantity. By using the base voltage
Ub = base voltage, kV (line-to-line voltage) (6.4)
and a base power,
Sb = three-phase base power, MVA (6.5)
the base current
Ib =
Sb
√
3 Ub
= base current/phase, kA (6.6)
as well as a base impedance
Zb =
U2
b
Sb
=
Ub
√
3 Ib
= base impedance, Ω (6.7)
can be calculated. In expressions given above, the units kV and MVA have been assumed,
which imply units in kA and Ω. Of course, different combinations of units can be used, e.g.
V, VA, A, Ω or kV, kVA, A, kΩ.
There are several reasons for using a per-unit system:
• The percentage voltage drop is directly given in the per-unit voltage.
• It is possible to analyze power systems having different voltage levels in a more efficient
way.
• When having different voltage levels, the relative importance of different impedances
is directly given by the per-unit value.
• When having large systems, numerical values of the same magnitude are obtained
which increase the numerical accuracy of the analysis.
• Use of the constant
√
3 is reduced in three-phase calculations.

47
6.2.1 Per-unit representation of transformers
Figure 6.9 shows the single-phase impedance diagram of a symmetrical three-phase trans-
former. In Figure 6.9 a), the transformer leakage impedance is given on the primary side,
and in Figure 6.9 b), the transformer leakage impedance is given on the secondary side. Fur-
thermore, α is the ratio of rated line-to-line voltages. Thus, based on transformer properties
we have
U1n
U2n
=
1
α
and
I1
I2
= α (6.8)
Let the base power be Sb. Note that Sb is a global base value, i.e. it is the same in all
diﬀerent voltages levels. Let also U1b and U2b be the base voltages on the primary side and
secondary side, respectively. The base voltages have been chosen such that they have the
same ratio as the ratio of the transformer, i.e.
U1b
U2b
=
1
α
(6.9)
Furthermore, since Sb =
√
3 U1b I1b =
√
3 U2b I2b, by virtue of equation (6.9) we ﬁnd that
I1b
I2b
= α (6.10)
where, I1b and I2b are the base currents on the primary side and secondary side, respectively.
The base impedances on both sides are given by
Z1b =
U2
1b
Sb
=
U1b
√
3 I1b
and Z2b =
U2
2b
Sb
=
U2b
√
3 I2b
(6.11)
tpZ
1U 2U
1I 2I1:α
2U
α
tsZ
1U 2U
1I 2I1:α
1Uα
a) b)
Figure 6.9. single-phase impedance diagram of a symmetrical three-phase transformer.
Now consider the circuit shown in Figure 6.9 a). The voltage equation is given by
U1 =
√
3 I1 Ztp +
U2
α
(6.12)

48
In per-unit (pu), we have
U1
U1b
=
√
3 I1 Ztp
√
3 I1b Z1b
+
U2
α U1b
=
I1
I1b
Ztp
Z1b
+
U2
U2b
⇒ U1pu = I1pu Ztppu + U2pu (6.13)
Next, consider the circuit shown in Figure 6.9 b). The voltage equation is given by
α U1 =
√
3 I2 Zts + U2 (6.14)
In per-unit (pu), we have
α U1
U2b
=
α U1
α U1b
=
√
3 I2 Zts
√
3 I2b Z2b
+
U2
U2b
=
I2
I2b
Zts
Z2b
+
U2
U2b
⇒ U1pu = I2pu Ztspu + U2pu (6.15)
By virtue of equations (6.13) and (6.15), we ﬁnd that
I1pu Ztppu = I2pu Ztspu
Furthermore, based on equations (6.8) and (6.10) it can be shown that I1pu = I2pu (show
that). Thus,
Ztppu = Ztspu (6.16)
Equation (6.16) implies that the per-unit impedance diagram of a transformer is the same
regardless of whether the actual impedance is determined on the primary side or on the
secondary side. Based on this property, the single-phase impedance diagram of a three-phase
transformer in per-unit can be drawn as shown in Figure 6.10, where Ztpu = Ztppu = Ztspu.
tpuZ
1puU
puI
2 puU or tpuZpuI1puU 2 puU
Figure 6.10. Per-unit impedance diagram of a transformer.
Example 6.1 Assume that a 15 MVA transformer has a voltage ratio of 6 kV/30 kV and
a leakage reactance of 8 %. Calculate the pu-impedance when the base power of the system
is 20 MVA and the base voltage on the 30 kV-side is 33 kV.
Solution
Based on given data, Snt = 15 MVA, U1n/U2n = 6/30, xt = 8% and U2b = 33 kV. We ﬁrst
calculate the transformer impedance in ohm on the 30 kV-side and after that, the per-unit
value.
Z30kv =
Z%
100
Ztb30 =
Z%
100
U2
2n
Snt
=
j8 · 302
100 · 15
= j4.8 Ω
Ztpu =
Z%
100
Ztb30
Z2b
=
Z30kV
Z2b
=
Z30kV · Sb
U2
2b
=
j4.8 · 20
332
= j0.088 pu

49
The given leakage reactance in percent can be considered as the per unit value of reactance
based on the transformer ratings, i.e. Ztb. To convert this per unit value to the system per
unit value, we may apply the following equation
Ztpu−new = Ztpu−given
U2
b−given
Sb−given
Sb−new
U2
b−new
In our case, Ztpu−given = j 8/100, Ub−given = U2n = 30, Sb−given = Snt = 15, Sb−new = 20,
and Ub−new = 33.
Thus,
Ztpu−new =
j8
100
302
15
20
332
= j0.088 pu (6.17)
The pu-value of the reactance can be also determined based on the base values on the primary
side. From equation (6.9), we have
U1b
U2b
=
1
α
=
6
30
⇒ U1b =
6
30
U2b =
6
30
33
Z1b =
U2
1b
Sb
=
1
α
2
U2
2b
Sb
=
1
α
2
Z2b
Thus,
Z6kv =
Z%
100
Ztb6 =
Z%
100
U2
1n
Snt
=
j8
100
62
15
Ztpu =
Z6kV
Z1b
=
j8
100
62
15
30
6
2
20
332
=
j8
100
302
15
20
332
= j0.088 pu
6.2.2 Per-unit representation of transmission lines
Figure 6.11 shows the π-equivalent model of a line, where ysh−kj = Y sh−kj/2.
jU
2
sh kjY −
2
sh kjY −
shI
I
or
jU
sh kjy −
shI
I
sh kjy −
Figure 6.11. π-equivalent model of a line.
The voltage at bus k in kV is given by
Uk =
√
3 Zkj I + Uj, where I = Ik − Ish = Ik − ysh−kj
Uk
√
3

50
Let Sb, Ub, Ib and Zb be the base values for the line. Note that the base admittance is given
by Yb = 1/Zb. Then, the above equations i per unit are given by
Uk
Ub
=
√
3 Zkj I
√
3 Zb Ib
+
Uj
Ub
⇒ Ukpu = Zkjpu Ipu + Ujpu
where
Ipu =
Ik
Ib
− ysh−kj
Uk
√
3
√
3 Zb
Ub
= Ikpu − ysh−kj Zb
Uk
Ub
= Ikpu − ysh−kjpu Ukpu
Figure 6.12 shows the per-unit impedance diagram of a transmission line.
kpuU
kjpuZ jpuU
sh kjpuy − sh kjpuy −
Figure 6.12. Per-unit impedance diagram of a transmission line.
6.2.3 System analysis in the per-unit system
To analysis a three-phase power system, it is more convenient and eﬀective to convert the
physical quantities into the per-unit system as follows:
1. Choose a suitable base power for the system. It should be in the same range as the
rated power of the installed system equipments.
2. Choose a base voltage at one section (or voltage level) of the system. The system is
divided into diﬀerent sections (or voltage levels) by the transformers.
3. Calculate the base voltages in all sections of the system by using the transformer ratios.
4. Calculate all per-unit values of all system components that are connected.
5. Draw the per-unit impedance diagram of the system.
6. Perform the system analysis (in the per-unit system).
7. Convert the per-unit results back to the physical values.
Example 6.2 Consider the power system shown in Figure 6.13, where a load is fed by a
generator via a transmission line and two transformers. Based on the given system data
below, calculate the load voltage as well as the active power of the load.

51
~
1 2 3 4
Line
LD
G T1 T2
Figure 6.13. Single-line diagram of the system for Example 6.2.
System data:
Generator G : Ug=13.8 kV,
Transformer T1 : Snt=10 MVA,
U1n
U2n
=
13.8 kV
69 kV
, Xtp=1.524 Ω (on 13.8 kV-side),
Transformer T2 : Snt=5 MVA,
U1n
U2n
=
66 kV
13.2 kV
, xt=8%,
Line : x = 0.8 Ω/km and L = 10 km
Load LD : impedance load, PLD = 4 MW, Un = 13.2 kV, cos φ = 0.8 inductive.
Solution
1. Let the base power be Sb=10 MVA.
2. Let the base voltage at the generator be U1b=13.8 kV.
3. The transformer ratio gives the base voltage U2b=69 kV for the line and
U3b = 69 · 13.2/66 = 13.8 kV for the load.
In Figure 6.14, the diﬀerent sections of the system are given.
~
1 2 3 4
Line
LD
G
1
10 MVA
13.8 kV
b
b
S
U
=
= 2
10 MVA
69 kV
b
b
S
U
=
= 3
10 MVA
13.8 kV
b
b
S
U
=
=
Figure 6.14. Diﬀerent sections of the system given in Example 6.2.
4. Calculate the per-unit values of the system components.
G: Ugpu = U1pu =
Ug
U1b
=
13.8
13.8
= 1.0 pu
T1: Zt1pu =
Ztp
Z1b
= j1.524
10
13.82
= j0.080 pu
T2: Zt2pu = j
8
100
13.22
5
1
Z3b
= j
8
100
13.22
5
10
13.82
= 0.146 pu
Line: Z23pu =
L z23
Z2b
= 10 · j0.8
10
692
= j0.017 pu

52
LD: ZLD =
U2
n
S
∗
LD
=
U2
n
SLD(cos φ − j sin φ)
=
U2
n
SLD
(cos φ + j sin φ) =
=
13.22
4/0.8
(0.8 + j0.6) = 27.88 + j20.91 Ω
ZLDpu =
ZLD
Z3b
= (27.88 + j20.91) ·
10
13.82
= 1.464 + j1.098 pu
5. By using these values, an impedance diagram can be drawn as shown in Figure 6.15.
1 1 0puU = ∠
1t puZ 23 puZ
2 puU 3 puU
2t puZ
4 puU
LDpuZ
~
puI
Figure 6.15. Impedance network in per-unit
6. The current through the network can be calculated as
Ipu =
1 + j0
j0.08 + j0.017 + j0.146 + 1.464 + j1.098
= 0.3714 − j0.3402 pu (6.18)
The load voltage is
U4pu = ULDpu = Ipu ZLDpu = 0.9173 − j0.0903 pu (6.19)
The load power is
SLDpu = ULDpuI
∗
pu = 0.3714 + 0.2785 pu (6.20)
7. The load voltage and active load power in physical units can be obtained by multiplying
the per-unit values with corresponding base quantities.
ULD = ULDpu U3b =
√
0.91732 + 0.09032 13.8 = 12.72 kV (6.21)
PLD = Real(SLDpu) Sb = 0.371 · 10 = 3.71 MW (6.22)
Note that the PLD given in the system data (i.e. PLD=4 MW) is the consumed active power
at the rated (or nominal) voltage Un=13.2 kV. However, the actual voltage at bus 4 is 12.72
kV. Therefore, the actual consumed power is 3.71 MW.

Chapter 7
Power transmission to impedance loads
Transmission lines and cables are normally operating in balanced (or symmetrical) condi-
tions, and as shown in Figure 6.12 a three-phase transmission line (or cable) can be rep-
resented with a single-phase line equivalent (or more precisely, with a positive-sequence
network, see chapter 11). This equivalent can be described by a twoport.
7.1 Twoport theory
Assume that a linear, reciprocal twoport is of interest, where the voltage and current in one
end are Uk and Ik whereas the voltage and current in the other end are Uj and Ij. The
conditions valid for this twoport can be described by constants ABCD as
Uk
Ik
=
A B
C D
Uj
Ij
(7.1)
Assume that the twoport is shortened in the receiving end, (i.e. Uj = 0) according to Figure
7.1, and that the voltage U is applied to the sending end.
U
A B
C D
1kI
1jI
Figure 7.1. Twoport, shortened in the receiving end
For the system shown in Figure 7.1, we have
U = A · 0 + B · Ij1 = B · Ij1 (7.2)
Ik1 = C · 0 + D · Ij1 = D · Ij1 (7.3)
If it is assumed that the twoport is shortened in the sending end instead, (Uk = 0) as shown
in Figure 7.2, and the voltage U is applied to the receiving end. Then according to Figure
U
A B
C D
2kI
2jI
Figure 7.2. Twoport, shortened in the sending end
53

54
7.2, we have
0 = A · U − B · Ij2 (7.4)
−Ik2 = C · U − D · Ij2 (7.5)
The reciprocity theorem gives that
Ik2 = Ij1 = I (7.6)
From the equations given above, the following expressions can be derived :
eq. (7.4) ⇒ Ij2 =
A
B
U (7.7)
eq. (7.7)+(7.6)+(7.2) ⇒ Ij2 = A · I (7.8)
eq. (7.2)+(7.5)+(7.8) ⇒ −I = C · B · I − D · A · I (7.9)
eq. (7.9), I = 0 ⇒ A · D − B · C = 1 (7.10)
i.e. the determinant of a reciprocal twoport is equal to 1. This implies that if several
reciprocal twoports are connected after one another, the determinant of the total twoport
obtained is also equal to 1. With three reciprocal twoports F1, F2 and F3 connected after
one another, the following is always valid :
det(F1F2F3) = det(F1) det(F2) det(F3) = 1 · 1 · 1 = 1 (7.11)
7.1.1 Symmetrical twoports
Assume that a symmetrical linear reciprocal twoport is of interest. If the deﬁnitions of
directions given in Figure 7.3 is used, a current injected in the sending end Ik at the voltage
1U
A B
C D
kI 1I
kU
Figure 7.3. Symmetrical twoport, connection 1
Uk gives rise to a current I1 at the voltage U1 in the receiving end. This can be written in
an equation as
Uk
Ik
=
A B
C D
U1
I1
(7.12)
Suppose that the circuit is fed in the opposite direction, i.e. U1 and I1 are obtained in the
sending end according to Figure 7.4. This connection can mathematically be formulated as :
U1
−I1
=
A B
C D
Uj
−Ij
(7.13)

55
1U
A B
C D
jI1I
jU
Figure 7.4. Symmetrical twoport, connection 2
By changing the position of the minus sign inside the matrix, equation (7.13) can be rewritten
as
U1
I1
=
A −B
−C D
Uj
Ij
(7.14)
The matrix in equation (7.14) can be inverted which gives that
Uj
Ij
=
1
A · D − B · C
=1
D B
C A
U1
I1
(7.15)
Since the twoport is symmetrical, the following is valid
Uj
Ij
≡
Uk
Ik
(7.16)
The equations (7.12), (7.15) and (7.16) give together that
A B
C D
=
D B
C A
(7.17)
This concludes that for symmetrical twoports A = D.
7.1.2 Application of twoport theory to transmission line and trans-
former and impedance load
Note that all variables in this subsection are expressed in (pu).
Figure 7.5 shows the π-equivalent model of a line.
kU
kjZ jU
kI jI
sh kjy − sh kjy −

56
From the ﬁgure, we have
Uk = Uj + Ij + Uj · ysh−kj Zkj (7.18)
Ik = Uk · ysh−kj + Ij + Uj · ysh−kj
These equations can be rewritten as
Uk = 1 + Zkj · ysh−kj Uj + Zkj · Ij (7.19)
Ik = ysh−kj 1 + 1 + Zkj · ysh−kj Uj + Zkj · ysh−kj + 1 Ij
and by using the matrix notation, this can be written as a twoport equation
Uk
Ik
=








A
1 + ysh−kj · Zkj
B
Zkj
ysh−kj(2 + ysh−kj · Zkj)
C
1 + ysh−kj · Zkj
D








Uj
Ij
(7.20)
As shown in equation (7.20), a line is symmetrical which gives that A = D. A line is also
reciprocal which gives that A · D − B · C = 1.
Using the short line model, then Y sh−kj = 0. Therefore, the twoport equation for a short
line model is given by
Uk
Ik
=
1 Zkj
0 1
Uj
Ij
(7.21)
The per-unit impedance diagram of a transformer is similar to the per-unit impedance di-
agram of a short line. Therefore, the twoport equation for a transformer is similar to the
twoport equation of a short line model, i.e.
Uk
Ik
=
1 Zt
0 1
Uj
Ij
(7.22)
Figure 7.6 shows the per-unit impedance diagram of an impedance load.
kU
LDZ
jU
kI jI
LDI
Figure 7.6. Impedance diagram of an impedance load.

57
From the figure, the following can be easily obtained.
Uk = Uj
Ik = Ij + ILD = Ij +
Uj
ZLD
(7.23)
Therefore, the twoport equation for an impedance load is given by
Uk
Ik
=
1 0
1
ZLD
1
Uj
Ij
(7.24)
7.1.3 Connection to network
As discussed in section 5.5, based on Thévenin-Helmholtz’s theorem from any output ter-
minal in a linear electric network the entire linear electric network seen from the output
terminal can be modelled as an ideal voltage source UTh in series with an impedance ZTh.
Considering any bus in a linear electric network as an output terminal, seen from any bus
k the network can be replaced with a Thévenin equivalent as shown in Figure 7.7, where
Uk = UTh. Assume that a solid three-phase short circuit (i.e. Zk = 0) is applied to bus k.
ThkZ
ThU
~
k
Figure 7.7. Thévenin equivalent of the network as seen from bus k.
This model implies that the short circuit current is
Isck =
UTh kV
√
3 ZThk Ω
kA or Isck =
UTh p.u
ZThk p.u
p.u (7.25)
The question is now how well this model can be adapted to real conditions. For instance,
consider the simple system shown in Figure 7.8, where LD is an impedance load and short
line model is used for the lines.
~
A B C D
Line1
LD
G T1 T2
Line2
E
Figure 7.8. A simple system.
Assume that the initial voltage at bus D is known, i.e. UDi = UDi θDi (p.u). If the pu-values
of all components are known, then as seen from bus D the following Thévenin equivalent can
be obtained,

58
ThDZ
ThU
~
D
Figure 7.9. Thévenin equivalent seen from bus D.
where,
UTh = UDi and ZThD =
Zt1 ZLD
Zt1 + ZLD
+ ZBC + Zt2
If the pu-values of all components are not known, by applying a solid three-phase short circuit
to bus D, the short circuit current can be measured and converted to per unit (i.e. Isck (p.u)
will be known). Then, the Thévenin impedance as seen from bus D can be calculated as
ZThD =
UTh
Isck
p.u
Having connected an impedance load ZLDD (p.u) to bus D, the voltage at bus D will be
UD =
ZLDD
ZThD + ZLDD
UTh ⇒ UD =
ZLDD
ZThD + ZLDD
UTh (7.26)
i.e., the voltage magnitude at bus D will drop with
ZLDD
ZThD + ZLDD
.100 %
Now assume that the transformer T2 has a regulator to automatically regulate the voltage
magnitude at bus D to its initial value, i.e. UD0 (p.u). This kind of transformer is known as
On Load Tap Changer (OLTC). When the load is connected to bus D, the OLTC regulates
the voltage at bus D to UDi, i.e. UD = UDi not the voltage given in equation (7.26). The
Thévenin equivalent is not valid in this case. Thévenin-Helmholtz’s theorem is applied to
linear circuits with passive components (static linear circuits), and an OLTC is not a passive
component.
Next, seen from bus E the following Thévenin equivalent can be obtained,
ThEZ
ThU
~
E
Figure 7.10. Thévenin equivalent as seen from bus E.
where, ZThE = ZThD + ZDE and UTh = UEi.

59
Having connected an impedance load ZLDE (p.u) to bus E, the voltage at bus E will be
UE =
ZLDE
ZThE + ZLDE
UTh ⇒ UE =
ZLDE
ZThE + ZLDE
UTh (7.27)
i.e., the voltage magnitude at bus E will drop with
ZLDE
ZThE + ZLDE
.100 %
If transformer T2 is an OLTC, the voltage at bus D will be recovered to its initial value (i.e.
UD = UDi), but not the voltage at bus E. Therefore, the voltage at bus E when the load i
connected will be
UE =
ZLDE
ZDE + ZLDE
UDi =
ZLDE
(ZThE − ZThD) + ZLDE
UDi (7.28)
The conclusion is that the equivalent impedance from a bus located out in a distribution
system (with a fairly weak voltage) to the closest bus with regulated voltage can be calculated
as the difference between the Thévenin impedance from the bus with weak voltage and the
Thévenin impedance from the bus with voltage regulation. To calculate the voltage drop
at the connection of the load, the calculated equivalent impedance and the voltage at the
regulated bus will be used in the Thévenin equivalent model.
In some cases, the term short circuit capacity Ssck at a bus k is used. It is defined as
Ssck = UThI
∗
sck = UTh Isck φsck p.u (7.29)
which gives the power that is obtained in the Thévenin impedance. Since this impedance
often is mostly reactive, φsck ≈ 90◦
. The short circuit capacity is of interest when the
loadability of a certain bus is concerned. The short circuit capacity indicates how much the
bus voltage will change for different loading at that bus. The voltage increase at generator
buses can be also calculated.
Example 7.1 At a bus with a pure inductive short circuit capacity of 500 MVA (i.e. cos φsck =
0) an impedance load of 4 MW, cos φLD = 0.8 at nominal voltage, is connected. Calculate
the change in the bus voltage when the load is connected.
Solution
Assume a voltage of 1 pu and a base power Sb = 500 MVA, i.e. Sscpu = 1 90◦
. The network
can then be modeled as shown in Figure 7.11.
The Thévenin impedance can be calculated according to equation (7.25) and (7.29) :
ZThpu =
UThpu
Iscpu
=
U2
Thpu
S
∗
scpu
=
1
1 − 90◦
= j1 (7.30)
The load impedance can be calculated as
ZLDpu =
U2
npu
S
∗
LDpu
=
U2
npu
PLD
Sb·cos φLD
(cos φLD + j sin φLD) =
12
4
500·0.8
(0.8 + j0.6) (7.31)

60
ThpuZ
ThpuU
~
LDpuZ
LDpuU
Figure 7.11. Single-phase model of system given in example.
Thus, the voltage ULD at the load is
ULDpu =
ZLDpu
ZThpu + ZLDpu
UThpu =
80 + j60
j1 + 80 + j60
1 0 = 0.9940 − 0.4556◦
(7.32)
i.e. the voltage drop is about 0.6 %.
Conclusion : A load with an apparent power of 1 % of the short circuit capacity at the bus
connected, will cause a voltage drop at that bus of ≈ 1 %.
Example 7.2 As shown in Figure 7.12, a small industry (LD) is fed by a power system via
a transformer (5 MVA, 70/10, x = 4 %) which is located at a distance of 5 km. The electric
power demand of the industry is 400 kW at cosφ=0.8, lagging, at a voltage of 10 kV. The
industry can be modeled as an impedance load. The 10 kV line has an series impedance of
0.9+j0.3 Ω/km and a shunt admittance of j3 × 10−6
S/km. Assume that the line is modeled
by the π-equivalent. When the transformer is disconnected from bus 3, the voltage at this
bus is 70 kV, and a three-phase short circuit applied to this bus results in a pure inductive
short circuit current of 0.3 kA.
Calculate the voltage at the industry as well as the power fed by the transformer into the
line.
123
Line
LD
T
Power
system
70/10 5 km
Figure 7.12. Single-line diagram of the system in Example 7.2.
Solution
Choose the base values (MVA, kV, ⇒ kA, Ω) :
Sb = 0.5 MVA, Ub10 = 10 kV ⇒ Ib10 = Sb/
√
3Ub10 = 0.0289 kA, Zb10 = U2
b10/Sb = 200 Ω
Ub70 = 70 kV ⇒ Ib70 = Sb/
√
3Ub70 = 0.0041 kA

61
Calculate the per-unit values of the Thévenin equivalent of the system:
UThpu =
UTh
Ub70
=
70 0
70
= 1 0◦
= 1 and Iscpu =
Isc
Ib70
=
0.3 − 90◦
0.00412
= 72.8155 − 90◦
ZThpu =
UThpu
Iscpu
= j0.0137
Calculate the per-unit values of the transformer:
Ztpu =
Zt%
100
Ztb10
Zb10
=
Zt%
100
U2
2n
Snt
Sb
U2
b10
=
j4
100
102
5
0.5
102
=
j4
100
0.5
5
= j0.004
Calculate the per-unit values of the line:
Z21pu =
5 · (0.9 + j0.3)
Zb10
= 0.0225 + j0.0075
ysh−21pu =
Y sh−21pu
2
=
5 · (j3 × 10−6
)
2
Zb10 =
j0.003
2
AL = 1 + ysh−21pu · Z21pu = 1.0000 + j0.0000
BL = Z21pu = 0.0225 + j0.0075
CL = ysh−21pu(2 + ysh−21pu · Z21pu) = 0.0000 + j0.0030
DL = AL = 1.0000 + j0.0000
Calculate the per-unit values of the industry impedance:
ZLDpu =
U2
n
S
∗
LD
1
Zb10
=
102
0.4
0.8
(0.8 + j0.6)
1
200
= 0.8 + j0.6
Figure 7.13 shows the per-unit impedance diagram of the entire system, where the power
system has been modelled by its Thévenin equivalent. Bus 4 (the terminal bus of the ideal
voltage source) is termed as infinite bus.
ThpuZ
ThpuU
~
LDpuZ
21puZtpuZ
123
4
Thévenin equivalent
of the power system
21sh puy − 21sh puy −
Figure 7.13. Per-unit impedance diagram of the system in Example 7.2.
The twoport of the above system (from the infinite bus to bus 1) can be formulated as
UThpu
I4pu
=
1 ZThpu + Ztpu
0 1
U2pu
I2pu
=
1 ZThpu + Ztpu
0 1
AL BL
CL DL
U1pu
I1pu
=
=
A B
C D
U1pu
I1pu
=
0.9999 + j0.0000 0.0225 + j0.0252
0.0000 + j0.0030 1.0000 + j0.0000
U1pu
I1pu

62
Seen from bus 4, the impedance of the entire system (including the industry) can be calcu-
lated as
Ztotpu =
UThpu
I4pu
=
A U1pu + B I1pu
C U1pu + D I1pu
=
A
U1pu
I1pu
+ B
C
U1pu
I1pu
+ D
=
A ZLDpu + B
C ZLDpu + D
= 0.8254 + j0.6244
⇒ I4pu = UThpu/Ztotpu = 0.9662 − 37.1035◦
The power fed by the transformer into the line can be calculated as
U2pu
I2pu
=
1 ZThpu + Ztrapu
0 1
−1
UThpu
I4pu
=
0.9898 − 0.7917◦
0.9662 − 37.1035◦
⇒ S2 = U2pu I
∗
2pu Sb = 0.3853 + j0.2832 MVA
the voltage at the industry can be calculated as
U1pu
I1pu
=
A B
C D
−1
UThpu
I4pu
=
0.9680 − 0.3733◦
0.9680 − 37.2432◦
⇒ U1 (kV) = U1pu Ub10 = 9.6796 kV
7.2 A general method for analysis of linear balanced
three-phase systems
When analyzing large power systems, it is necessary to perform the analysis in a systematic
manner. Below, a small system is analyzed with a method which can be used for large
systems. In Figure 7.14, an impedance load ZLD1 is fed from an inﬁnite bus (i.e. bus 3
~
1LDZ
21ZtZ
123
3I
2I 1I
Figure 7.14. Per unit impedance diagram of a balanced power system.
which is the terminal bus of the ideal voltage source) via a transformer with impedance Zt
and a line with impedance Z21. The voltage at the inﬁnite bus is U3. All variables are
expressed in per unit. The Y-bus matrix for this system can be formulated as


I1
I2
I3

 = I = YU =



1
ZLD1
+ 1
Z21
− 1
Z21
0
− 1
Z21
1
Z21
+ 1
Zt
− 1
Zt
0 − 1
Zt
1
Zt





U1
U2
U3

 (7.33)
The Y-bus matrix can be inverted which results in the corresponding Z-bus matrix :
U = Y−1
I = ZI (7.34)

63
Since I1 = I2 = 0, the third row in equation (7.34) can be written as
U3 = Z(3, 3) · I3 ⇒ I3 =
U3
Z(3, 3)
(7.35)
where Z(3, 3) is an element in the Z-bus matrix. With that value of the current inserted into
equation (7.34), all system voltages are obtained.
U1 = Z(1, 3) · I3 (7.36)
U2 = Z(2, 3) · I3
Corresponding calculations can be performed for arbitrarily large systems containing impedance
loads and one voltage source.
a) ~
1LDZ
21ZtZ
123
2LDI′
2LDZ
b) ~
1LDZ
21ZtZ
123
2 0LDI =
2LDZ
~2U
c)
1LDZ
21ZtZ
12
2LDZ
~2U−
2 2LDI I∆
′= −
1 0I∆ =
U′
preU
U∆
Figure 7.15. Total voltage obtained by using superposition
Assume that an impedance ZLD2 is added to the system at bus 2, as shown in Figure 7.15
a). This will change the voltage magnitudes at all buses with exception of the bus connected
to the voltage source (bus 3 in this example). Then, the actual voltages can be expressed by
U = Upre + U∆ (7.37)
where U is a vector containing the actual voltages due to the change, Upre is a vector
containing the voltages of all buses (with exception of the bus connected to the voltage

64
source) prior to the change and U∆ is the applied change. This equation can be illustrated
graphically as shown in Figure 7.15, i.e. the total voltage can be calculated as a superposition
of two systems with equal impedances but with diﬀerent voltage sources.
As indicated by the system in Figure 7.15 c), the feeding voltage is −U2 while the voltage
source at bus 3 is shortened (when the voltage source is short circuited the bus connected
to the voltage source (i.e. bus 3) is removed). The Y-bus matrix for this system can be
obtained by removing the row and column corresponding to bus 3 in Y (see equation (7.33))
since bus 3 is grounded and removed. If bus 3 was kept in the mathematical formulation,
Y(3, 3) = ∞ since the impedance to ground is zero.
I∆1
I∆2
= I∆ = Y∆U∆ =
1
ZLD1
+ 1
Z21
− 1
Z21
− 1
Z21
1
Z21
+ 1
Zt
U∆1
U∆2
(7.38)
The expression given above, can be inverted which gives the corresponding Z-bus matrix :
U∆ = Y−1
∆ I∆ = Z∆I∆ ⇒
U∆1
U∆2
=
Z∆(1, 1) Z∆(1, 2)
Z∆(2, 1) Z∆(2, 2)
I∆1
I∆2
(7.39)
In this equation, I∆1 = 0 which is shown in Figure 7.15 c). This gives that the second row
can be written as
U∆2 = Z∆(2, 2)I∆2 (7.40)
Figure 7.14 gives the same currents as Figure 7.15 b), since the voltage over ZLD2 in Figure
7.15 b) is zero. This implies that the current through ZLD2 is zero. Therefore, I∆2 = −ILD2.
At bus 2 in Figure 7.15 a) the following is valid
U2 = ILD2 · ZLD2 = −I∆2 · ZLD2 (7.41)
By combining equations (7.37), (7.40) and (7.41), the following can be obtained
I∆2 =
− U2
ZLD2 + Z∆(2, 2)
(7.42)
By inserting that value in the equations given above, all voltages after the system change
can be calculated as :
U2 =
ZLD2
ZLD2 + Z∆(2, 2)
U2 (7.43)
U1 = U1 −
Z∆(1, 2)
ZLD2 + Z∆(2, 2)
U2 (7.44)
The procedure given above can be generalized to be used for an arbitrarily large system.
Assume that an impedance Zr is connected to a bus r and an arbitrary bus is termed i. The
current Ir (=−I∆r) through Zr can be calculated as well as the voltages after connection of
the impedance Zr at bus r. The equations are as follows.
Ir =
Ur
Zr + Z∆(r, r)
(7.45)
Ur =
Zr
Zr + Z∆(r, r)
Ur (7.46)
Ui = Ui −
Z∆(i, r)
Zr + Z∆(r, r)
Ur (7.47)

65
Note that i = r, and bus r and bus i do not represent the bus connected to the voltage
source.
The Thévenin equivalent at a bus in a symmetrical network can be calculated by using
equations (7.37) and (7.39). At bus r (r=2 in this case), the equation will be as
U (r) = Upre(r) + Z∆(r, r)I∆(r) (7.48)
where
Upre(r) = UThr Thévenin voltage at bus r prior to the change, see Figure 7.16.
Z∆(r, r) = ZThr Thévenin impedance as seen from bus r, see Figure 7.16.
I∆(r) = −Ir The actual injected current into bus r.
U (r) = Ur the actual voltage at bus r.
( , )Z r r∆
preU ( )r
~
r
U ( )r′
I ( )r∆
ThrZ
ThrU
~
r
rU′
rI′
or
Figure 7.16. Thévenin equivalent at bus r in a symmetrical three-phase network.
As given by equation (7.48) and Figure 7.16, U (r) = Upre(r) if I∆(r) = 0. This formu-
lation shows that the Thévenin voltage at bus r can be calculated as the voltage at bus r
when the bus is not loaded, i.e. I∆(r) = 0. The Thévenin impedance is found as the r-th
diagonal element of the impedance matrix Z∆ which is determined when the voltage source
is shortened.
Example 7.3 In Figure 7.17, an internal network of an industry is given. Power is delivered
by an infinite bus with a nominal voltage at bus 1. Power is transmitted via transformer T1,
Line2 and transformer T2 to the load LD2. There is also a high voltage load LD1 connected
1
2 3
4
Line1
LD1T1
T2
Line2
5
LD2
Figure 7.17. Single-line diagram of an internal industry network
to T1 via Line1. The system data is given as follows:

66
• Transformer T1 : 800 kVA, 70/10, x = 7 %
• Transformer T2 : 300 kVA, 10/0.4, x = 8 %
• Line1 : r = 0.17 Ω/km, ωL = 0.3Ω/km, ωC = 3.2 × 10−6
S/km, L = 2 km
• Line2 : r = 0.17 Ω/km, ωL = 0.3Ω/km, ωC = 3.2 × 10−6
S/km, L = 1 km
• Load LD1 : impedance load, 500 kW, cosφ = 0.80, inductive at 10 kV
• Load LD2 : impedance load, 200 kW, cosφ = 0.95, inductive at 0.4 kV
The π-equivalent model is used for the lines.
Calculate the eﬃciency of the internal network as well as the short circuit current that is
obtained at a solid three-phase short circuit at bus 4.
Solution
Chose base values (MVA, kV, ⇒ kA, Ω) : Sb = 500 kVA = 0.5 MVA, Ub70 = 70 kV
1
2 3
4 5
1t puZ
23 puZ
23sh puy − 1LD puZ
24 puZ
2LD puZ
2t puZ
24sh puy −
Figure 7.18. Network in Example 7.3.
Ub10 = 10 kV ⇒ Ib10 = Sb/
√
3Ub10 = 0.0289 kA, Zb10 = U2
b10/Sb = 200 Ω
Ub04 = 0.4 kV ⇒ Ib04 = Sb/
√
3Ub04 = 0.7217 kA, Zb04 = U2
b04/Sb = 0.32 Ω
Calculate the per-unit values of the inﬁnite bus :
U1 = 70/Ub70 = 70/70 = 1
Calculate the per-unit values of the transformer T1 :
Zt1pu = (Zt1%/100) · Zt1b10/Zb10 = (ZT1%/100) · Sb/Snt1 = (j7/100) · 0.5/0.8 = j0.0438
Calculate the per-unit values of the transformer T2 :
Zt2pu = (Zt2%/100) · Sb/Snt2 = (j8/100) · 0.5/0.3 = j0.1333
Calculate the per-unit values of Line1 :
Z23pu = 2 · [0.17 + j0.3]/Zb10 = 0.0017 + j0.003
ysh−23pu = Y sh−23pu/2 = 2 · [3.2 × 10−6
] · Zb10/2 = j0.0013/2

67
Calculate the per-unit values of Line2 :
Z24pu = 1 · [0.17 + j0.3]/Zb10 = 0.0009 + j0.0015
ysh−24pu = Y sh−24pu/2 = 1 · [3.2 × 10−6
] · Zb10/2 = j0.00064/2
Calculate the per-unit values of the impedance LD1 :
ZLD1pu = (U2
LD1/S
∗
LD1)/Zb10 = (102
/[0.5/0.8]) · (0.8 + j0.6)/200 = 0.64 + j0.48
Calculate the per-unit values of the impedance LD2 :
ZLD2pu = (U2
LD2/S
∗
LD2)/Zb04 = (0.42
/0.2/0.95)·(0.95+j
√
1 − 0.952)/0.32 = 2.2562+j0.7416
Calculate the Y-bus matrix of the network. The grounding point is not included in the Y-bus
matrix since the system then is overdetermined.
Y =








1
Zt1pu
− 1
Zt1pu
0 0 0
− 1
Zt1pu
Y 22 − 1
Z23pu
− 1
Z24pu
0
0 − 1
Z23pu
Y 33 0 0
0 − 1
Z24pu
0 Y 44 − 1
Zt2pu
0 0 0 − 1
Zt2pu
1
Zt2pu
+ 1
ZLD2pu








(7.49)
where
Y 22 =
1
Zt1pu
+
1
Z23pu
+ ysh−23pu +
1
Z24pu
+ ysh−24pu
Y 33 =
1
Z23pu
+ ysh−23pu +
1
ZLD1pu
Y 44 =
1
Z24pu
+ ysh−24pu +
1
Zt2pu
Next, we have
I = YU (7.50)
which can be rewritten as






U1
U2
U3
U4
U5






= U = Y−1
I = ZI = Z






I1
I2
I3
I4
I5






(7.51)
The Z-bus matrix can be calculated by inverting the Y-bus matrix :
Z =






0.510+j0.375 0.510+j0.331 0.508+j0.329 0.510+j0.331 0.516+j0.298
0.510+j0.331 0.510+j0.331 0.508+j0.329 0.510+j0.331 0.516+j0.298
0.508+j0.329 0.508+j0.329 0.509+j0.330 0.508+j0.329 0.515+j0.296
0.510+j0.331 0.510+j0.331 0.508+j0.329 0.510+j0.332 0.517+j0.299
0.516+j0.298 0.516+j0.298 0.515+j0.296 0.517+j0.299 0.529+j0.397






(7.52)
Since all injected currents with exception of I1 are zero, I1 can be calculated using the ﬁrst
row in equation (7.51) :
U1 = Z(1, 1)I1 ⇒ I1 = U1/Z(1, 1) = 1.0/(0.510 + j0.375) = 1.58 − 36.33◦
(7.53)

68
The voltages at the other buses can be easily be solved by using equation (7.51) :
U2 = Z(2, 1)I1 = (0.510 + j0.331) · (1.58 − 36.33◦
) = 0.9606 − 3.324◦
U3 = Z(3, 1)I1 = (0.508 + j0.329) · (1.58 − 36.33◦
) = 0.9569 − 3.423◦
(7.54)
U4 = Z(4, 1)I1 = (0.510 + j0.331) · (1.58 − 36.33◦
) = 0.9601 − 3.350◦
U5 = Z(5, 1)I1 = (0.516 + j0.298) · (1.58 − 36.33◦
) = 0.9423 − 6.351◦
The total amount of power delivered to the industry is
S1 = U1 · I
∗
1 · Sb = 0.6367 + j0.4682 MVA (7.55)
The power losses in Line1 and Line2 can be calculated as
IZ23 =(U2 − U3)/Z23pu = 1.1957 − 40.27◦
IZ24 =(U2 − U4)/Z24pu = 0.3966 − 24.50◦
PfLine1 =Real(Z23pu)I2
Z23 · Sb = 0.0012 MW
PfLine2 =Real(Z24pu)I2
Z24 · Sb = 0.0000669MW
(7.56)
The eﬃciency for the network is then
η =
Real(S1) − PfLine1 − PfLine2
Real(S1)
= 0.9980 ⇒ 99.80% (7.57)
A solid short circuit at bus 4 can be calculated by connecting an impedance with Z4 = 0 at
bus 4. According to section 7.2, the current through the impedance Z4 can be determined by
removing the row and the column of the Y-bus matrix that corresponds to the bus connected
to the voltage source (i.e. bus 1 in this example). Thus,
Y∆ = Y(2 : 5, 2 : 5) =






Y 22 − 1
Z23pu
− 1
Z24pu
0
− 1
Z23pu
Y 33 0 0
− 1
Z24pu
0 Y 44 − 1
Zt2pu
0 0 − 1
Zt2pu
1
Zt2pu
+ 1
ZLD2pu






(7.58)
The inverse of this matrix is
Z∆ = Y−1
∆ =




0.0024+j0.0420 0.0025+j0.0418 0.0025+j0.0419 0.0046+j0.0410
0.0025+j0.0418 0.0043+j0.0446 0.0025+j0.0418 0.0046+j0.0408
0.0025+j0.0419 0.0025+j0.0418 0.0033+j0.0434 0.0055+j0.0424
0.0046+j0.0410 0.0046+j0.0408 0.0055+j0.0424 0.0144+j0.1719



 (7.59)
The short circuit current at bus 4 can then be calculated according to equation (7.45).
Isc4 =
U4
Z4 + Z∆(4, 4)
element (3,3) in Z∆
Ib10 =
0.9601 − 3.350◦
0 + (0.0033 + j0.0434)
0.0289 =
= 0.6366 − 88.97◦
kA (7.60)

69
7.3 Extended method to be used for power loads
The method described in section 7.2 is valid when all system loads are modeled as impedance
loads, i.e. the power consumed is proportional to the voltage squared. In steady-state
conditions, an often used load model is the constant power model. The method described
in section 7.2 can be used in an iterative way, described as follows:
1. Calculate the per-unit values of all components that are of interest. Loads that are
modeled with constant power (independent of the voltage) are replaced by impedances.
The impedance of a load at bus k can be calculated as ZLDk = U2
n/S
∗
LDk where Un = 1
pu is the rated (or nominal) voltage, and SLDk is the rated power of the load.
2. Calculate the Y-bus matrix and the corresponding Z-bus matrix of the network as well
as the load impedances. By using the method described in section 7.2 (equation (7.35)
and (7.36)), the voltage at all buses can be calculated.
3. Calculate the load demand at all loads. The power demand SLDk−b at load LDk is
obtained as SLDk−b = U2
k /Z
∗
LDk where Uk is the actual calculated voltage at bus k.
4. Calculate the difference between the actual calculated and specified load demand for
all power loads :
∆PLDk = |Re(SLDk−b) − Re(SLDk)| (7.61)
∆QLDk = |Im(SLDk−b) − Im(SLDk)| (7.62)
5. If ∆PLDk and/or ∆QLDk are too large for a certain bus :
(a) Calculate new load impedances according to ZLDk = U2
k /S
∗
LDk where Uk is the
actual calculated voltage at bus k obtained in step 3,
(b) Go back to step 2 and repeat the calculations.
If ∆PLDk and ∆QLDk are found to be acceptable for all power loads, the iteration
process is finished.
A simple example will be given to clarify this method.
Example 7.4 Assume a line operating with a voltage of U1 = 225 0◦
kV in the sending
end, i.e. bus 1, and with a load of PLD = 80 MW and QLD = 60 MVAr in the receiving end,
i.e. bus 2. The line has a length of 100 km with x = 0.4 Ω/km, r = 0.04 Ω/km and bc =
3 × 10−6
S/km. Calculate the receiving end voltage.
Solution
In Figure 7.19, the network modeled by impedance loads is given.
Assume Sb = 100 MVA and Ub = 225 kV which gives
Zb = U2
b /Sb = 506.25 Ω

70
1 2
12Z
12 12sh shy jb− −= LD LD LDS P jQ= +
12I
Figure 7.19. Impedance diagram for Example 7.4.
This gives the following per-unit values of the line
U1pu = 225/Ub = 1.0, PLDpu = PLD/Sb = 0.8, QLDpu = QLD/Sb = 0.6
Z12pu = L (r + j x)/Zb = 100 (0.04 + j 0.4)/Zb = 0.0079 + j 0.0790
Based on equation (4.9),
bsh−12pu = L bc Zb/2 = 100 (3 × 10−6
) Zb/2 = 0.0759
The iteration process can now be started :
1. U2pu = 1, ZLDpu = U2
2pu/(PLDpu − jQLDpu) = 0.8 + j0.6
2. I12pu = U1pu/(Z12pu + 1
jbsh−12pu
ZLDpu) = 0.7330 - j0.5415 ⇒ U2pu = |U1pu − I12puZ| =
0.9529
3. SLDpu = U2
2pu/Z
∗
LDpu = 0.7265 + j0.5448
4. ∆PLD = 0.0735, ∆QLD = 0.0552
1. ZLDpu = 0.7265 + j0.5448
2. U2pu = 0.9477
4. ∆PLD = 0.0087, ∆QLD = 0.0066
1. ZLDpu = 0.7185 + j0.5389
2. U2pu = 0.9471
4. ∆PLD = 0.0011, ∆QLD = 0.00079
1. ZLDpu = 0.7176 + 0.5382i
2. U2pu = 0.9470
4. ∆PLD = 0.0001, ∆QLD = 0.0001
This is found to be acceptable, which gives a voltage magnitude in the sending end of
U2 = 0.9470 · Ub = 213.08 kV. This simple example can be solved exactly by using a non-
linear expression which will be shown in Example 8.4.
Example 7.5 Consider the system in Example 7.2, but let the short line model be used for
the line (i.e. bc = 0), and the load be considered as a constant power load.
Calculate the voltage level at the industry.
Solution
1 : From Example 7.2, we have the following:
UThpu = 1, ZThpu = j0.0137, Ztpu = j0.004, Z21pu = 0.0225 + j0.0075

71
totpuZ
ThpuU
LD LD LDS P jQ= +
LDpuU
Figure 7.20. Network used in Example 7.5.
The total impedance between bus 1 and bus 4 is given by:
Ztotpu = ZThpu + Ztpu + Z21pu = 0.0225 + j0.0252
Calculate the per-unit values of the power demand of the industry as well as the correspond-
ing impedance at nominal voltage :
SLDpu = (PLD + j[PLD/ cos φ] · sin φ)/Sb = 0.8000 + j0.6000
ZLDpu = (U2
n/S
∗
LDpu)/U2
b10 = 0.8 + j0.6
2 : The Y-bus matrix of the network can be calculated as :
Y =
1
Ztotpu
− 1
Ztotpu
− 1
Ztotpu
1
Ztotpu
+ 1
ZLDpu
=
19.67 − j22.08 −19.67 + j22.08
−19.67 + j22.08 20.47 − j22.68
(7.63)
The Z-bus matrix is calculated as the inverse of the Y-bus matrix :
Z = Y−1
=
0.82 + j0.63 0.80 + j0.60
0.80 + j0.60 0.80 + j0.60
(7.64)
The voltage at the industry is now calculated according to equation (7.36) :
ULDpu = Z(2, 1) · UThpu/Z(1, 1) = 0.9679 − 0.3714◦
(7.65)
3 : The power delivered to the industry can be calculated as :
SLDpu−b = U2
LD/Z
∗
LDpu = 0.7495 + j0.5621 (7.66)
4 : The diﬀerence between calculated and speciﬁed power can be calculated as :
∆PLD = |Re(SLDpu−b) − Re(SLDpu)| = 0.0505 (7.67)
∆QLD = |Im(SLDpu−b) − Im(SLDpu)| = 0.0379 (7.68)
5 : These deviations are too large and the calculations are therefore repeated and a new
industry impedance is calculated by using the new voltage magnitude :
ZLDpu = (U2
LDpu/S
∗
LDpu) = 0.7495 + j0.5621 (7.69)
Repeat the calculations from step 2.
2, 3 : ⇒ SLDpu−b = 0.7965 + j0.5974
4 : ∆PLD = 0.0035, ∆QLD = 0.0026

72
Unacceptable ⇒
5 : ZLDpu = 0.7462 + j0.5597
Continue from step 2.
2, 3 : ⇒ SLDpu−b = 0.7998 + j0.5998
4 : ∆PLD = 0.00024, ∆QLD = 0.00018
Unacceptable ⇒
5 : ZLDpu = 0.7460 + j0.5595
Continue from step 2.
2, 3 : ⇒ SLDpu−b = 0.8000 + j0.6000
4 : ∆PLD = 0.000016, ∆QLD = 0.000012
Acceptable ⇒
ULD = ULDpu · Ub10 = 9.6565 − 0.3974◦

Chapter 8
Power flow calculations
In chapter 6–7, it was assumed that the network had only one voltage source (or generator
bus), and the loads were modelled as impedances. These assumptions resulted in using a
linear set of equations which could be easily solved.
In this chapter, the loads are modelled as constant power loads, and the system has more
than one generator bus (i.e. a multi-generator system).
First, the power flow in a transmission line will be derived, and then a more general power
flow calculations (commonly known as load flow) will be presented.
8.1 Power flow in a line
Consider the the π-equivalent model of a line shown in Figure 8.1, where all variables ex-
pressed in per-unit.
kjZ
sh kjjb −
IkU jU
sh kjjb −
shI
kjS
Let
Uk = Ukejθk
, Uj = Ujejθj
Z = R + jX , Z =
√
R2 + X2
θkj = θk − θj
(8.1)
The power Skj in the sending end k is given by
Skj = Uk I
∗
sh + I
∗
= Uk j bsh−kj Uk
∗
+
U
∗
k − U
∗
j
Z
∗
kj
=
= −j bsh−kj U2
k +
U2
k
R − jX
−
UkUj
R − jX
ej(θk−θj)
=
= −j bsh−kj U2
k +
U2
k
X2
(R + jX) −
UkUj
X2
(R + jX) (cos θkj + j sin θkj)
(8.2)
By dividing equation (8.2) into a real and an imaginary part, expressions for the active and
73

74
reactive power can be obtained, respectively, as
Pkj =
R
Z2
U2
k +
Uk Uj
Z2
(X sin θkj − R cos θkj)
=
R
Z2
U2
k +
Uk Uj
Z
sin θkj − arctan
R
X
(8.3)
Qkj = −bsh−kj U2
k +
X
Z2
U2
k −
UkUj
Z2
(R sin θkj + X cos θkj)
= −bsh−kj +
X
Z2
U2
k −
Uk Uj
Z
cos θkj − arctan
R
X
(8.4)
From equations (8.3) and (8.4), it can be concluded that if the phasor voltages (i.e. the
voltage magnitude and phase angle) at both ends of the line are known, the power flow can
be uniquely determined. This implies that if the phasor voltages of all buses in a system are
known, the power flows in the whole system are known, i.e the phasor voltages define the
system state.
Example 8.1 Assume a line where the voltage in the sending end is U1 = 225 0◦
kV and
in the receiving end U2 = 213.08 − 3.572◦
kV. The line has a length of 100 km and has
x = 0.4 Ω/km, r = 0.04 Ω/km and b = 3 × 10−6
S/km. Calculate the amount of power
transmitted from bus 1 to bus 2.
Solution
Assume Sb = 100 MVA and Ub = 225 kV, this gives that
Zb = U2
b /Sb = 506.25 Ω
The per-unit values for the line are
U1 = 225/Ub = 1.0 pu, U2 = 213.08/Ub = 0.9470 pu, θ12 = 0-(-3.572) = 3.572◦
R = 0.04 · 100/Zb = 0.0079 pu, X = 0.4 · 100/Zb = 0.0790 pu,
bsh−12 = 3 × 10−6
· 100 · Zb/2 = 0.0759 pu, Z =
√
R2 + X2 = 0.0794 pu
The power flow in per-unit can be calculated by using equation (8.3) and (8.4) :
P12 =
0.0079
0.07942
1.02
+
1.0 · 0.9470
0.0794
sin 3.572◦
− arctan
0.0079
0.0790
=
= 0.8081 pu
Q12 = −0.0759 +
0.0790
0.07942
∗ 1.02
−
1.0 · 0.9470
0.0794
cos 3.572◦
− arctan
0.0079
0.0790
=
= 0.5373 pu
expressed in nominal values
P12 = 0.8081 · Sb = 80.81 MW
Q12 = 0.5373 · Sb = 53.73 MVAr

75
For this simple system, the calculations can be performed without using the per-unit system.
By using equation (6.7), equation (8.3) can be rewritten as
Pkj(MW) = Pkj(pu)Sb =
U2
b
Zb
R
Z2
U2
k +
Uk Uj
Z
sin θkj − arctan
R
X
=
=
R(Ω)
Z2(Ω)
U2
k (kV) +
Uk(kV) · Uj(kV)
Z(Ω)
sin θkj − arctan
R
X
i.e. this equation is the same independent on if the values are given as nominal or per-unit
values. Note that arctan(R/X) = arctan(R(Ω)/X(Ω)).
For a high voltage overhead line (U > 70 kV), the line reactance is normally considerably
higher than the resistance of the line, i.e. R X in equation (8.3). An approximate form
of that equation is (i.e. R ≈ 0)
Pkj ≈
UkUj
X
sin θkj (8.5)
i.e. the sign of θkj determines the direction of the active power flow on the line. Normally,
the active power will flow towards the bus with the lowest voltage angle. This holds also for
lines having a pronounced resistivity.
Assume that the voltages Uk and Uj are in phase and that the reactance of the line is
dominating the line resistance (i.e. R ≈ 0). This implies that the active power flow is very
small. Equation (8.4) can be rewritten as
Qkj = −bsh−kj U2
k +
Uk(Uk − Uj)
X
(8.6)
Equation (8.6) indicates that this type of line gives a reactive power flow towards the bus
with the lowest voltage magnitude. The equation shows that if the difference in voltage
magnitude between the ends of the line is small, the line will generate reactive power. This
since the reactive power generated by the shunt admittances in that case dominates the
reactive power consumed by the series reactance. The “rule of thumb” that reactive power
flows towards the bus with lowest voltage is more vague than the rule that active power
flows towards the bus with lowest angle. The fact that overhead lines and especially cables,
generates reactive power when the active power flow is low, is important to be aware of.
Example 8.2 Using the approximate expressions (8.5) and (8.6), respectively, calculate the
active and reactive power flow in the line in Example 8.1.
Solution
P12 ≈
1.0 · 0.9470
0.0790
sin 3.572◦
= 0.7468 pu ⇒ 74.68 MW
Q12 ≈ −0.0759 · 1.02
+
1.0(1.0 − 0.9470)
0.0790
= 0.5948 pu ⇒ 59.48 MVAr
The answers are of right dimension and have correct direction of the power flow but the
active power flow is about 8 % too low and the reactive power flow is 11 % too large.

76
8.1.1 Line losses
The active power losses on a three-phase line are dependent on the line resistance and the
actual line current. By using physical units (i.e. not in per-unit), the losses can be calculated
as
Pf = 3RI2
(8.7)
The squared current dependence in equation (8.7) can be written as
I2
= Iej arg(I)
Ie−j arg(I)
= I I
∗
=
S
∗
√
3U
∗
S
√
3U
=
S2
3U2
=
P2
+ Q2
3U2
(8.8)
The active power losses for the line given in Figure 8.1 can be calculated as
Pf = R
P2
kj + (Qkj + bsh−kj U2
k )2
U2
k
(8.9)
where
bsh−kj U2
k is the reactive power generated by the shunt capacitance. at bus k
The expression given by (8.9) is valid both for nominal and for per-unit values. This equation
shows that a doubling of transmitted active power will increase the active power losses by a
factor of four. If the voltage is doubled, the active power losses will decrease with a factor
of four.
Assume that the active power injections at both ends of the line are known, i.e. both Pkj
and Pjk have been calculated using equation (8.3). The active power losses can then be
calculated as
Pf = Pkj + Pjk (8.10)
The reactive power losses can be obtained in the corresponding manner
Qf = 3XI2
= X
P2
kj + (Qkj + bsh−kj U2
k )2
U2
k
(8.11)
Equations (8.8) and (8.9) shows that the losses are proportional to S2
and that the losses
will increase if reactive power is transmitted over the line. A natural solution to that is to
generate the reactive power as close to the consumer as possible. Of course, active power
is also generated as close to the consumer as possible, but the generation costs are of great
importance.
Example 8.3 Use the same line as in Example 8.1 and calculate the active power losses.
Solution
The losses on the line can be calculated by using equation (8.9) and the conditions that
apply at the sending end
Pf (MW) = R
P2
12 + (Q12 + bsh−12U2
1 )2
U2
1
Sb =
= 0.0079
0.80812
+ (0.5373 + 0.0759 · 1.02
)2
1.02
100 = 0.81 MW

77
The losses can also be calculated by using the receiving end conditions
Pf (MW) = R
P2
21 + (Q21 + bsh−12U2
2 )2
U2
2
Sb =
= 0.0079
(−0.80)2
+ (−0.60 + 0.0759 · 0.94702
)2
0.94702
100 = 0.81 MW
or by using equation (8.10)
Pf (MW) = [P12 + P21]Sb = [0.8081 + (−0.80)]100 = 0.81 MW
8.1.2 Shunt capacitors and shunt reactors
As mentioned earlier in subsection 8.1.1, transmission of reactive power will increase the
line losses. An often used solution is to generate reactive power as close to the load as
possible. This is done by switching in shunt capacitors. Figure 8.2 shows a Y-connected
shunt capacitor. Figure 8.2 also shows the single-phase equivalent which can be used at
phase a
c
Three-phase connection Single-phase equivalent
phase b
phase c
c
c
c
Figure 8.2. Y -connected shunt capacitors.
symmetrical conditions. A shunt capacitor generates reactive power proportional to the bus
voltage squared U2
. In the per-unit system, we have
Qsh = BshU2
= 2πfc U2
(8.12)
An injection of reactive power into a certain bus will increase the bus voltage, see Example
8.6. The insertion of shunt capacitors in the network is also called phase compensation. This
because the phase displacement between voltage and current is reduced when the reactive
power transmission through the line is reduced.
As mentioned earlier, lines that are lightly loaded generates reactive power. The amount of
reactive power generated is proportional to the length of the line. In such situations, the
reactive power generation will be too large and it is necessary to consume the reactive power
in order to avoid overvoltages. One possible countermeasure is to connect shunt reactors.
They are connected and modeled in the same way as the shunt capacitors with the diﬀerence
that the reactors consume reactive power.

78
8.1.3 Series capacitors
By studying equation (8.5), an approximate expression of the maximum amount of power
that can be transmitted through a line, at a certain voltage level, can be written as
Pkj−max ≈ max
θkj
UkUj
X
sin θkj =
UkUj
X
(8.13)
i.e. the larger the reactance of the line is, the less amount of power can be transmitted.
One possibility to increase the maximum loadability of a line is to compensate for the series
reactance of the line by using series capacitors. In Figure 8.3, the way of connecting series
capacitors is shown as well as the single-phase equivalent of a series compensated line. The
( )kj cZ R j X X= + −
sh kjjb − sh kjjb −
Three-phase connection Single-phase equivalent
phase a
phase b
phase c
c
c
c
Figure 8.3. Series capacitors
expression for the maximum loadability of a series compensated line is
Pkj−max ≈
UkUj
X − Xc
(8.14)
It is obvious that the series compensation increases the loadability of the line.
The use of series capacitors will also reduce the voltage drop along the line, see Example
8.7.
8.2 Non-linear power flow equations
The technique of determining all bus voltages in a network is usually called load flow. When
knowing the voltage magnitude and voltage angle at all buses, the system state is completely
determined and all system properties of interest can be calculated, e.g. line loadings and
line losses.
In a power system, power can be generated and consumed at many different locations. Con-
sider now a balanced power system with N buses. Figure 8.4 schematically shows connection
of the system components to bus k.
The generator generates the current IGk, the load at the bus draws the current ILDk, and
Ikj is the currents from bus k to the neighboring buses. According to Kirchoff’s current law,

79
GkI kU
~
LDkI
1kI
2kI
kNI
Figure 8.4. Notation of bus k in a network.
the sum of all currents injected into bus k must be zero, i.e.
IGk − ILDk =
N
j=1
Ikj (8.15)
By taking the conjugate of equation (8.15) and multiply the equation with the bus voltage,
the following holds
UkI
∗
Gk − UkI
∗
LDk =
N
j=1
UkI
∗
kj (8.16)
This can be rewritten as an expression for complex power in the per-unit system as
SGk − SLDk =
N
j=1
Skj (8.17)
where
SGk = PGk + jQGk is the generated complex at bus k, complex power
SLDk = PLDk + jQLDk is the consumed complex power at bus k, complex power
Skj = Pkj + jQkj is the complex power ﬂow from bus k to bus j.
The power balance at the bus according to equation (8.17) must hold both for the active
and for the reactive part of the expression. By using PGDk and QGDk as notation for the net
generation of active and reactive power at bus k, respectively, the following expression holds
PGDk = PGk − PLDk =
N
j=1
Pkj (8.18)
QGDk = QGk − QLDk =
N
j=1
Qkj (8.19)
i.e. for any bus k in the system, the power balance must hold for both active and reactive
power.
Figure 8.5 shows a balanced power system with N = 3.

80
1 1G GP jQ+
1 1U θ∠
~
~
2 2G GP jQ+
1 1LD LDP jQ+
3 3LD LDP jQ+
3 3U θ∠
2 2U θ∠
12 12P jQ+
21 21P jQ+
23 23P jQ+
32 32P jQ+
13 13P jQ+
31 31P jQ+
1
2
3
Figure 8.5. Single-line diagram of a balanced three-bus power system.
Based on equations (8.18) and (8.19), the following system of equations can be obtained.



PG1 = P12 + P13
QG1 = Q12 + Q13
PG2 − PLD2 = P21 + P23
QG2 − QLD2 = Q21 + Q23
−PLD3 = P31 + P32
−QLD3 = Q31 + Q32
(8.20)
At each bus in Figure 8.5, four variables are of interest: net generation of active power PGDk,
net generation of reactive power QGDk, voltage magnitude Uk and voltage phase angle θk.
This gives that the total number of variables for the system are 3 · 4 = 12. The voltage
phase angles must be given as an angle in relation to a reference angle. This since the phase
angles are only relative to one another and not absolute. This reduces the number of system
variables to 12 − 1 = 11. However, there are only six equations in the system of equations
(8.20), this gives that five quantities must be known to be able to solve for the remaining
six variables. Depending on what quantities that are known at a certain bus, the buses are
mainly modeled in three different types.
PQ-bus, Load bus : For this bus, the net generated power PGDk and QGDk are assumed to
be known. The name PQ-bus is based on that assumption. On the other hand, the voltage
magnitude Uk and the voltage phase angle θk are unknown. A PQ-bus is most often a bus
with a pure load demand, as bus 3 in Figure 8.5. It represents a system bus where the power
consumption can be considered to be independent of the voltage magnitude. This model
is suitable for a load bus located on the low voltage side of a regulating transformer. The
regulating transformer keeps the load voltage constant independent of the voltage fluctua-
tions on the high voltage side of the transformer. Note that a PQ-bus can be a bus without
generation as well as load, i.e. PGDk = QGDk = 0. This holds e.g. at a bus where a line is
connected to a transformer.

81
PU-bus, Generator bus : In a PU-bus, the net active power generation PGDk as well as the
voltage magnitude Uk are assumed to be known. This gives that the net reactive power
generation QGDk and the voltage angle θk are unknown. In a PU-bus some sort of voltage
regulating device must be connected since the voltage magnitude is independent of the net
reactive power generation. For example, in a synchronous machine, the terminal voltage can
be regulated by changing the magnetizing current. In a system, voltage can be regulated
by using controllable components as controllable shunt capacitors and controllable shunt
reactors. A standard component is called SVC, Static VAr Compensator. This component
change the reactive power flow in order to regulate the bus voltage. Assume that bus
2 in Figure 8.5 is modeled as a PU-bus. This gives that the active power generation of
the generator as well as the active power consumption of the load are known. Also the
reactive power consumption of the load is known. The bus voltage is constant due to the
magnetization system of the generator. The generator may generate or consume reactive
power in such a way that the relation in equation (8.19) holds.
Uθ-bus, Slack bus : At a slack bus (only one bus in each system), The voltage magnitude
and the voltage phase angle are known and fixed. The voltage phase angle is chosen as a the
reference phase angle in the system. Normally, the phase angle θk is set to zero. Unknown
quantities are the net generation of both active and reactive power. At this bus, (as for the
PU-bus) a voltage regulating component must be present. Since the active power is allowed
to vary, a generator or an active power in-feed into the system is assumed to exist at this bus.
Since this bus also is the only bus where the active power is allowed to vary, the slack bus
will take care of the system losses since they are unknown. If the loads have been modeled
in the load flow as constant power loads and a line is tripped, the only bus which will change
the active power generation is the slack bus. If bus 1 is chosen as slack bus in Figure 8.5,
both PG1 and QG1 are unknown but the voltage U1 is given as well as the reference angle
θ1 = 0.
Assume that M of the system N buses are PU-buses. A summary of the different bus types
is given in Table 8.1. As given in equations (8.3)–(8.4), the active and reactive power flow
Bus model Number Known quantities Unknown quantities
Uθ-bus, Slack bus 1 U, θ PGD, QGD
PU-bus, Generator bus M PGD, U QGD, θ
PQ-bus, Load bus N-M-1 PGD, QGD U, θ
Table 8.1. Bus types for load flow calculations
through a line can be expressed as a function of the voltage magnitude and voltage phase
angle at both ends of the line. Assume that the power system in Figure 8.5 is modeled in
such a way that bus 1 is a slack bus, bus 2 is a PU-bus and bus 3 is a PQ-bus. By using
this bus type modeling, the system of equations (8.20) can be written as



PGD1(unknown) = P12(U1, θ1, U2, θ2) + P13(U1, θ1, U3, θ3)
QG1(unknown) = Q12(U1, θ1, U2, θ2) + Q13(U1, θ1, U3, θ3)
PGD2 = P21(U1, θ1, U2, θ2) + P23(U2, θ2, U3, θ3)
QGD2(unknown) = Q21(U1, θ1, U2, θ2) + Q23(U2, θ2, U3, θ3)
PGD3 = P31(U1, θ1, U3, θ3) + P32(U2, θ2, U3, θ3)
QGD3 = Q31(U1, θ1, U3, θ3) + Q32(U2, θ2, U3, θ3)
(8.21)

82
where also θ2, U3 and θ3 are unknown quantities whereas the others are known. As given in
equation (8.21), unknown power quantities appear only on the left hand side for buses mod-
eled as slack and PU-bus. These quantities can be easily calculated when voltage magnitudes
and angles are known. These equations are not contributing to the system of equations since
they only give one extra equation, and one extra variable which easily can be calculated. The
system of equations in (8.21) can therefore be simplified to a system of equations containing
unknown U and θ as



PGD2 = P21(U1, θ1, U2, θ2) + P23(U2, θ2, U3, θ3)
PGD3 = P31(U1, θ1, U3, θ3) + P32(U2, θ2, U3, θ3)
QGD3 = Q31(U1, θ1, U3, θ3) + Q32(U2, θ2, U3, θ3)
(8.22)
The system of equations given by (8.22) is non-linear since the expressions for power flow
through a line (equation (8.3)–(8.4)) include squared voltages as well as trigonometric ex-
pressions. This system of equations can e.g. be solved by using the Newton-Raphson’s
method.
The system of equations given by (8.22) can be generalized to a system containing N buses,
of which M have a voltage regulating device in operation. A summary of this system is given
in Table 8.2. As indicated in Table 8.2, the system of equations contains as many unknown
Bus model Number Balance equations Unknown quantities
PGDk = Pkj QGDk = Qkj Uk θk
Slack bus 1 0 st 0 st 0 st 0 st
PU-bus M M st 0 st 0 st M st
PQ-bus N-M-1 N-M-1 st N-M-1 st N-M-1 st N-M-1 st
Total N 2N-M-2 2N-M-2
Table 8.2. Summary of equations and unknown quantities at load flow calculations
quantities as the number of equations, and by that, the system is solvable.
8.3 Power flow calculations of a simple two-bus system
As shown in section 8.2, constant power loads give a non-linear system of equations, and
load flow solution for large power system requires soft-ware tools such as MATLAB which
will be used in this course. To understand the concept of load flow solution, in this section
a simple two-bus system is studied. Since for load flow solution, a bus bus must be a slack
bus, there are therefore two possible bus-type combinations, namely, slack bus + PU-bus
and slack bus + PQ-bus which can be analytically handled.
Consider the two-bus power system shown in Figure 8.6. The data given in Example 8.1 is
used for this system. Let bus 1 be a slack bus with U1 = 225 0◦
kV. Let also PLD2 = 80
MW.

83
21Z
21shjb − 21shjb −
21S1 2
2
2 2
LD
LD LD
S
P jQ
=
+
Figure 8.6. Single-line diagram of a balanced two-bus power system.
8.3.1 Slack bus + PU-bus
This combination is of interest when the voltage magnitude is known at both buses and the
net active power (i.e. PGD) is known at one of the buses. This implies that the only unknown
quantity is the voltage phase angle at the PU-bus, i.e. the bus having a known net active
power PGD.
Example 8.4 Let bus 2 be a PU-bus with U2 = 213.08 kV. Calculate the voltage phase angle
at bus 2 (the same as example 7.4). Find also Q21 in MVAr.
Solution
From equation (8.3) we have
P21 =
R
Z2
U2
2 +
U2 U1
Z
sin θ21 − arctan
R
X
⇒
θ2 = θ1 + arctan
R
X
+ arcsin
Z
U2 U1
P21 −
R
Z2
U2
2
where,
P21 = PGD2 = (0 − PLD2)/Sb = −0.8 pu, arctan
R
X
= arctan
0.0079
0.0790
= 5.71◦
Thus,
θ2 = 0 + 5.71◦
+ arcsin
0.0794
0.9470 · 1.0
−0.8 −
0.0079
0.07942
0.94702
= −3.5724◦
From equation (8.4), we have
Q21 = −bsh−21 +
X
Z2
U2
2 −
U2 U1
Z
cos θ21 − arctan
R
X
Sb
= −0.0759 +
0.0790
0.07942
0.94702
−
0.9470 1.0
0.0794
cos(−3.5724◦
− 5.71◦
) 100
= −59.9793 MVAr, from MATLAB

84
8.3.2 Slack bus + PQ-bus
This combination is of interest when the voltage magnitude is known only at one of the buses
and the net active and reactive power generation are known at the other bus.
Example 8.5 Now let bus 2 be a PQ-bus, and QLD2 = −Q21 where Q21 has been obtained
in Example 8.4. Calculate the voltage magnitude and phase angle at bus 2.
Solution
Based on equations (8.3) and (8.4), by eliminating θ21, the voltage magnitude U2 can be
analytically found, and it is given by
U2
2 = −
a4
2a3
+
(−)
a4
2a3
2
−
1
a3
(a2
1 + a2
2) (8.23)
where
a1 = −R P21 − X Q21
a2 = −X P21 + R Q21
a3 = (1 − X bsh−21)2
+ R2
b2
sh−21
a4 = 2 · a1(1 − X bsh−21) − U2
1 + 2a2R bsh−21
The voltage U2 can now be calculated as
U2 = +
(−)
U2
2 (8.24)
In our case,
a1 = −0.0079(−0.8) − 0.0790(−0.5998) = 0.0537
a2 = −0.0790(−0.8) + 0.0079(−0.6)) = 0.0585
a3 = (1 − 0.0790 · 0.0759)2
+ 0.00792
· 0.07592
= 0.9880
a4 = 2 · 0.0537(1 − 0.0790 · 0.0759) − 1.02
+
+ 2 · 0.0585 · 0.0079 · 0.0759 = −0.8931
⇒
U2
2 = 0.4520 +
(−)
0.4449 = 0.8968
⇒
U2 = +
(−)
√
0.8968 = 0.9470
⇒
U2 (kV) = 0.9470 · Ub = 213.08 kV
The voltage phase angle can now be calculated in the same way as performed in example
8.4, which results in the the same answer.

85
Example 8.6 Use the data given in Example 8.5 with PD2 = 80 MW and QD2 ≈ 60 MVAr.
Use these load levels as a base case and calculate the voltage U2 when the active and reactive
load demand are varying between 0–100 MW and 0–100 MVAr, respectively.
Solution
By using equations (8.23) and (8.24), the voltage can be calculated. The result is shown in
Figure 8.7. The base case, i.e. PD2 = 80 MW and QD2 = 60 MVAr, is marked by circles on
0 10 20 30 40 50 60 70 80 90 100
200
205
210
215
220
225
QD2=60 MVAr, PD2=0-100 MW
PD2=80 MW, QD2=0-100 MVAr
MW or MVAr
U2[kV]
Figure 8.7. The voltage U2 as a function of PD2 and QD2
both curves. As shown in the ﬁgure, the voltage drops at bus 2 as the load demand increases.
The voltage at bus 2 is much more sensitive to a change in reactive load demand compared
to a change in active demand. If a shunt capacitor generating 10 MVAr is connected at
bus 2 when having a reactive load demand of 60 MVAr, the net demand of reactive power
will decrease to 50 MVAr and the bus voltage will increase by two kV, from 213 kV to 215
kV. As discussed earlier in subsection 8.1.1, a reduced reactive power load demand will also
reduce the losses on the line.
Example 8.7 Use the base case in Example 8.6, i.e. PD2 = 80 MW and QD2 = 60 MVAr.
Calculate the voltage U2 when the series compensation of the line is varied in the interval
0–100 %.
Solution
A series compensation of 0–100 % means that 0–100 % of the line reactance is compensated
by series capacitors. 0 % means no series compensation at all and 100 % means that Xc = X.
The voltage can be calculated by using equations (8.23) and (8.24). The result is shown in
Figure 8.8. As shown in Figure 8.8, the voltage at bus 2 increases as the degree of series
compensation increases. If the degree of compensation is 40 %, the voltage at bus 2 is
increased by 4.5 kV (= 2 %) from 213.1 kV to 217.6 kV.
When having short lines or when only interested in approximate calculations, the shunt
capacitance of a line can be neglected. In these conditions, bsh−21 in equation (8.23) is

86
0 10 20 30 40 50 60 70 80 90 100
210
215
220
225
% compensation
U2[kV]
Figure 8.8. The voltage U2 as a function of degree of compensation
neglected, and the equation will be rewritten as
U2 =
U2
1 − 2a1
2
+
(−)
U2
1 − 2a1
2
2
− (a2
1 + a2
2) (8.25)
where
a1 = −R P21 − X Q21
a2 = −X P21 + R Q21
Example 8.8 Use the data given in Example 8.5. Calculate the magnitude of the voltage
by using the approximate expression given by equation (8.25).
Solution
Equation (8.25) gives that
a1 = −0.0790(−0.8) + 0.0079(−0.6) = 0.0537
a2 = 0.0079(−0.6) − 0.0790(−0.8) = 0.0585
⇒
U2 = 0.9410
⇒
U2(kV) = 0.9410 · Sb = 211.72 kV
i.e. the voltage becomes 0.6 % too low compared to the more accurate result.
Another approximation often used, is to neglect a2 in equation (8.25). That equation can
then be rewritten as
U2 ≈
U1
2
+
U2
1
4
+ R P21 + X Q21 (8.26)

87
Example 8.9 Use the same line as in example 8.5. Calculate the voltage by using the
approximate expression given by equation (8.26).
Solution
Equation (8.26) gives that
U2 = 0.9430
⇒
U2(kV ) = 0.9439 · Sb = 212.18 kV
i.e. the calculated voltage is 0.4 % too low. As indicated in this example, equation (8.26)
gives a good approximation of the voltage drop on the line. In the equation, it is also clearly
given that the active and reactive load demand have inﬂuence on the voltage drop. The
reason why the voltage drop is more sensitive to a change in reactive power compared to a
change in active power, is that the line reactance dominates the line resistance.
8.4 Newton-Raphson method
8.4.1 Theory
The Newton-Raphson method may be applied to solve for x1, x2, · · · , xn of the following
non-linear equations,
g1(x1, x2, · · · , xn) = f1(x1, x2, · · · , xn) − b1 = 0
g2(x1, x2, · · · , xn) = f2(x1, x2, · · · , xn) − b2 = 0
...
gn(x1, x2, · · · , xn) = fn(x1, x2, · · · , xn) − bn = 0
(8.27)
or in the vector form
g(x) = f(x) − b = 0 (8.28)
where
x =





x1
x2
...
xn





, g(x) =





g1(x)
g2(x)
...
gn(x)





, f(x) =





f1(x)
f2(x)
...
fn(x)





, b =





b1
b2
...
bn





x is an n × 1 vector which contains variables, b is an n × 1 vector which contains constants,
and f(x) is an n × 1 vector-valued function.
Taylor’s series expansion of (8.28) is the basis for the Newton-Raphson method of solving
(8.28) in an iterative manner. From an initial estimate (or guess) x(0)
, a sequence of gradually
better estimates x(1)
, x(2)
, x(3)
, · · · will be made that hopefully will converge to the solution
x∗
.

88
Let x∗
be the solution of (8.28), i.e. g(x∗
) = 0, and x(i)
be an estimate of x∗
. Let also
∆x(i)
= x∗
− x(i)
. Equation (8.28) can now be written as
g(x∗
) = g(x(i)
+ ∆x(i)
) = 0 (8.29)
Taylor’s series expansion of (8.29) gives
g(x(i)
+ ∆x(i)
) = g(x(i)
) + JAC(x(i))
∆x(i)
= 0 (8.30)
where
JAC(x(i))
=
∂g(x)
∂x x=x(i)
=



∂g1(x)
∂x1
· · · ∂g1(x)
∂xn
...
...
...
∂gn(x)
∂x1
· · · ∂gn(x)
∂xn



x=x(i)
(8.31)
where, JAC ia called the jacobian of g.
From (8.30), ∆x(i)
can be calculated as follows
JAC(x(i))
∆x(i)
= 0 − g(x(i)
) = ∆g(x(i)
) ⇒ (8.32)
∆x(i)
= JAC(x(i))
−1
∆g(x(i)
) (8.33)
Since g(x(i)
) = f(x(i)
) − b, ∆g(x(i)
) is given by
∆g(x(i)
) = b − f(x(i)
) = −g(x(i)
) (8.34)
Furthermore, since b is constant, JAC(x(i))
is given by
JAC(x(i))
=
∂g(x)
∂x x=x(i)
=
∂f(x)
∂x x=x(i)
=



∂f1(x)
∂x1
· · · ∂f1(x)
∂xn
...
...
...
∂fn(x)
∂x1
· · · ∂fn(x)
∂xn



x=x(i)
(8.35)
Therefore, ∆x(i)
can be calculated as follows
∆x(i)
=



∆x
(i)
1
...
∆x
(i)
n


 =



∂f1(x)
∂x1
· · · ∂f1(x)
∂xn
...
...
...
∂fn(x)
∂x1
· · · ∂fn(x)
∂xn



−1
x=x(i)



b1 − f1(x
(i)
1 , · · · , x
(i)
n )
...
bn − fn(x
(i)
1 , · · · , x
(i)
n )


 (8.36)
Finally, the following is obtained
i = i + 1
x(i)
= x(i−1)
+ ∆x(i−1)
The intention is that x(1)
will estimate the solution x∗
better than what x(0)
does. In the
same manner, x(2)
, x(3)
, · · · can be determined until a specified condition is satisfied. Thus,
we obtain an iterative method according to the flowchart in Figure 8.9.

89
Yes
No
( )
Give i
x
Set 0i =
( )i
x x=
Step 1
Step 2
Step Final
( )
Calculate ( )i
g x∆
( )
Is the magnitude of the all entries of ( )
less than a small positve number
?
i
g x∆
( ) ( 1) ( 1)
1
i i i
i i
x x x− −
= +
= + ∆
Step 3
Step 4( )
Calculate i
x∆
Step 5
( )
Calculate i
JAC
Figure 8.9. Flowchart for the Newton-Raphson method.
Example 8.10 Using the Newton-Raphson method, solve for x of the equation
g(x) = k1 x + k2 cos(x − k3) − k4 = 0
Let k1 = −0.2, k2 = 1.2, k3 = −0.07, k4 = 0.4 and = 10−4
. (This equation is used in the
assignment D1.)
Solution
This equation is of the form given by (8.28), with f(x) = k1 x + k2 cos(x − k3) and b = k4.
Step 1
Set i = 0 and x(i)
= x(0)
= 0.0524 (radians), i.e. 3 (degrees).
Step 2
∆g(x(i)
) = b − f(x(i)
) = 0.4 − [(−0.2 ∗ 0.0524) + 1.2 cos(0.0524 + 0.07)] = −0.7806
Go to Step 3 since |∆g(x(i)
)| >
Step 3
JAC(x(i))
= ∂f
∂x x=x(i) = −0.2 − 1.2 sin(0.0524 + 0.07) = −0.3465
Step 4
∆x(i)
= JAC(x(i))
−1
∆g(x(i)
) = −0.7806
−0.3465
= 2.2529
Step 5

90
i = i + 1 = 0 + 1 = 1
x(i)
= x(i−1)
+ ∆x(i−1)
= 0.0524 + 2.2529 = 2.3053. Go to Step 2
After 5 iterations, i.e. i = 5, it was found that |∆g(x(i)
)| < for x(5)
= 0.9809 (rad.).
Therefore, the solution becomes x = 0.9809 (rad.) or x = 56.2000 (deg.).
MATLAB-codes for this example can be found in appendix B.
Analysis of Example 8.10
Figure 8.10 shows variations of g(x) versus x. The figure shows that the system (or equation)
has only three solutions, i.e. the points at which g(x) = 0. Due to practical issues, x∗
indicted
with (O) in the figure is the interesting solution.
−400 −300 −200 −100 0 100 200 300 400
−2.5
−2
−1.5
−1
−0.5
0
0.5
1
1.5
2
2.5
x
g(x)
x*
Figure 8.10. Variations of g(x) vs. x.
Figure 8.11 shows how the equation is solved by the Newton-Raphson method.
We first guess the initial estimate x(0)
. In this case x(0)
= 0.0524 (rad.), i.e 3 (deg.). The
tangent to g(x) through the point x(0)
, g(x(0)
) , i.e. g (x(0)
) = dg(x)
dx
x=x(0)
= JAC(x(0))
,
intersects the x–axis at point x(1)
. The equation for this tangent is given by
Y − g(x(0)
) = g (x(0)
) ∗ (x − x(0)
)
The intersection point x(1)
is obtained by setting Y = 0, i.e.
x(1)
= x(0)
−
g(x(0)
)
g (x(0))
= x(0)
− g (x(0)
)
−1
g(x(0)
)
∆x(0)
= x(1)
− x(0)
= − g (x(0)
)
−1
g(x(0)
) = JAC(x(0))
−1
∆g(x(0)
)

91
0
g(x)
x
(0)
x(1)
x
(2)
g(x
(0)
)
JAC(x
(0)
)
g(x(1)
)
JAC(x
(1)
)
Figure 8.11. Variations of g(x) vs. x.
In a similar manner, x(2)
can be obtained which is hopefully a better estimate than x(1)
. As
shown in the ﬁgure, from x(2)
we obtain x(3)
which is a better estimate of x∗
than what x(2)
does. This iterative method will be continued until |∆g(x)| < .
Example 8.11 Solve for x in Example 8.10, but let x(0)
= 0.0174 (rad.), i.e. 1 (deg.).
Solution
D.I.Y, (i.e., Do It Yourself)
8.4.2 Application to power systems
Consider a power system with N buses. The aim is to determine the voltage at all buses
in the system by applying the Newton-Raphson method. All variables are expressed in
pu.
Consider again Figure 8.1. Let
gkj + j bkj =
1
Zkj
=
1
R + j X
=
R
Z2
+ j
−X
Z2
⇒
gkj =
R
Z2
bkj = −
X
Z2
(8.37)
Based on (8.37), we rewrite (8.3) and (8.4) as follows
Pkj = gkj U2
k − Uk Uj [gkj cos(θkj) + bkj sin(θkj)] (8.38)
Qkj = U2
k (−bsh−kj − bkj) − Uk Uj [gkj sin(θkj) − bkj cos(θkj)] (8.39)

92
The current through the line, and the loss in the line can be calculated by
Ikj =
Pkj − j Qkj
U
∗
k
(8.40)
Plkj = Pkj + Pjk (8.41)
Qlkj = Qkj + Qjk (8.42)
Consider again Figure 8.4. Let Y = G + jB denote the admittance matrix of the system (or
Y-matrix), where Y is an N × N matrix, i.e. the system has N buses. The relation between
the injected currents into the buses and the voltages at the buses is given by I = Y U, see
section 5.1. Therefore, the injected current into bus k is given by Ik = N
j=1 Y kj Uj.
The injected complex power into bus k can now be calculated by
Sk = Uk I
∗
k = Uk
N
j=1
Y
∗
kjU
∗
j = Uk
N
j=1
(Gkj − jBkj) Uj(cos(θkj) + j sin(θkj))
= Uk
N
j=1
Uj [Gkj cos(θkj) + Bkj sin(θkj)] + j Uk
N
j=1
Uj [Gkj sin(θkj) − Bkj cos(θkj)]
Let Pk denote the real part of Sk, i.e. the injected active power, and Qk denote the imaginary
part of Sk, i.e. the injected reactive power, as follows:
Pk = Uk
N
j=1
Uj [Gkj cos(θkj) + Bkj sin(θkj)]
Qk = Uk
N
j=1
Uj [Gkj sin(θkj) − Bkj cos(θkj)]
(8.43)
Note that Gkj = −gkj and Bkj = −bkj for k = j. Furthermore,
Pk =
N
j=1
Pkj
Qk =
N
j=1
Qkj
Equations (8.18) and (8.19) can now be rewritten as
Pk − PGDk = 0
Qk − QGDk = 0
(8.44)

93
which are of the form given in equation (8.28), where
x =


θ
U

 =











θ1
...
θN
U1
...
UN











, f(θ, U) =


fP (θ, U)
fQ(θ, U)

 =











P1
...
PN
Q1
...
QN











, b =


bP
bQ

 =











PGD1
...
PGDN
QGD1
...
QGDN











(8.45)
the aim is to determine x = [θ U]T
by applying the Newton-Raphson method.
Assume that there are 1 slack bus and M PU-buses in the system. Therefore, θ becomes an
(N − 1) × 1 vector and U becomes an (N − 1 − M) × 1 vector, why?
Based on (8.34), we deﬁne the following:
∆Pk = PGDk − Pk k = slack bus
∆Qk = QGDk − Qk k = slack bus and PU-bus
(8.46)
Based on (8.35), the jacobian matrix is given by
JAC =


∂fP (θ,U)
∂θ
∂fP (θ,U)
∂U
∂fQ(θ,U)
∂θ
∂fQ(θ,U)
∂U

 =


H N
J L

 (8.47)
where,
H is an (N − 1) × (N − 1) matrix
N is an (N − 1) × (N − M − 1) matrix
J is an (N − M − 1) × (N − 1) matrix
L is an (N − M − 1) × (N − M − 1) matrix
The entries of these matrices are given by:
Hkj = ∂Pk
∂θj
k = slack bus j = slack bus
Nkj = ∂Pk
∂Uj
k = slack bus j = slack bus and PU-bus
Jkj = ∂Qk
∂θj
k = slack bus and PU-bus j = slack bus
Lkj = ∂Qk
∂Uj
k = slack bus and PU-bus j = slack bus and PU-bus
Based on (8.32), (8.46) and (8.47), the following is obtained
H N
J L
∆θ
∆U
=
∆P
∆Q
(8.48)
To simplify the entries of the matrices N and L , these matrices are multiplied with U.
Then, (8.48) can be rewritten as
H N
J L
∆θ
∆U
U
=
∆P
∆Q
(8.49)

94
where,
for k = j
Hkj = ∂Pk
∂θj
= Uk Uj [Gkj sin(θkj) − Bkj cos(θkj)]
Nkj = Uj Nkj = Uj
∂Pk
∂Uj
= Uk Uj [Gkj cos(θkj) + Bkj sin(θkj)]
Jkj = ∂Qk
∂θj
= −Uk Uj [Gkj cos(θkj) + Bkj sin(θkj)]
Lkj = Uj Lkj = Uj
∂Qk
∂Uj
= Uk Uj [Gkj sin(θkj) − Bkj cos(θkj)]
(8.50)
and for k = j
Hkk = ∂Pk
∂θk
= −Qk − BkkU2
k
Nkk = Uk
∂Pk
∂Uk
= Pk + GkkU2
k
Jkk = ∂Qk
∂θk
= Pk − GkkU2
k
Lkj = Uk
∂Qk
∂Uk
= Qk − BkkU2
k
(8.51)
Now based on (8.36), the following is obtained:
∆θ
∆U
U
=
H N
J L
−1
∆P
∆Q
(8.52)
Finally, U and θ will be updated as follows:
θk = θk + ∆θk k = slack bus
Uk = Uk 1 + ∆Uk
Uk
k = slack bus and PU-bus
(8.53)
8.4.3 Newton-Raphson method for solving power flow equations
Newton-Raphson method can be applied to non-linear power flow equations as follows:
• Step 1
1a) Read bus and line data. Identify slack bus (i.e. Uθ-bus), PU-buses and PQ-buses.
1b) Develop the Y-matrix and calculate the net productions, i.e. PGD = PG − PLD
and QGD = QG − QLD.
1c) Give the initial estimate of the unknown variables, i.e. U for PQ-buses and θ for
PU- and PQ-buses. It is very common to set U = Uslack and θ = θslack. However,
the flat initial estimate may also be applied, i.e. U = 1 and θ = 0.
1d) Go to Step 2.
• Step 2

95
2a) Calculate the injected power into each bus by equation (8.43).
2b) Calculate the difference between the net production and the injected power for
each bus, i.e. ∆P and ∆Q by equation (8.46).
2c) Is the magnitude of all entries of [∆P ∆Q]T
less than a specified small positive
constant ?
∗ If yes, go to Step Final.
∗ if no, go to Step 3.
• Step 3
3a) Calculate the jacobian by equations (8.50) and (8.51).
3b) Go to Step 4.
• Step 4
4a) Calculate ∆θ ∆U
U
T
by equation (8.52).
4b) Go to Step 5.
• Step 5
5a) Update U and θ by equation (8.53).
5b) Go till Step 2.
• Step Final
– Calculate the generated powers, i.e. PG (MW) and QG (MVAr) in the slack bus,
and QG (MVAr) in the PU-buses by using equation (8.44).
– Calculate the power flows (MW, MVAr) by using equations (8.38) and (8.39).
– Calculate active power losses (MW) by using equation (8.41).
– Give all the voltage magnitudes (kV) and the voltage phase angles (degrees).
– Print out the results.
Example 8.12 Consider the power system shown in Figure 8.12. Let Sb = 100 MVA, and
Ub = 220 kV.
1 2
~ ~
Figure 8.12. Single-line diagram of a balanced two-bus power system.
The following data (all in pu) is known:

96
• Line between Bus 1 and Bus 2: short line, ¯Z12 = 0.02 + j 0.2
• Bus 1: slack bus, U1 = 1, θ1 = 0, PLD1 = 0.2, QLD1 = 0.02
• Bus 2: PU-bus, U2 = 1, PG2 = 1, PLD2 = 2, QLD2 = 0.2
By applying Newton-Raphson method, calculate θ2, PG1, QG1, QG2 and the active power
losses in the system after 3 iterations.
Solution
MATLAB-codes for this example can be found in appendix C.
Step 1
1a) bus 1 is a slack bus , bus 2 is a PU-bus , U1 = 1, U2 = 1, θ1 = 0.
1b)
Y =


1
¯Z12
− 1
¯Z12
− 1
¯Z12
1
¯Z12

 =


G11 + j B11 G12 + j B12
G21 + j B21 G22 + j B22

 =
=


0.4950 − j 4.9505 −0.4950 + j 4.9505
−0.4950 + j 4.9505 0.4950 − j 4.9505

 = G + j B
PGD2 = PG2 − PLD2 = 1 − 2 = −1
No QGD since there is no PQ-bus in the system.
1c)
Since the system has only one slack bus and one PU-bus, the phase angle of the PU-bus is
the only unknown variable. As an initial value , let θ2 = 0.
Iteration 1
Step 2
2a)
P2 = U2 U1 [G21 cos(θ2 − θ1) + B21 sin(θ2 − θ1)] + U2
2 G22 =
= 1 ∗ 1 ∗ [−0.4950 ∗ cos(0 − 0) + 4.9505 ∗ sin(0 − 0)] + 12
∗ 0.4950 = 0
2b)
∆P = ∆P2 = PGD2 − P2 = −1 − 0 = −1
Step 3
Q2 = U2 U1 [G21 sin(θ2 − θ1) − B21 cos(θ2 − θ1)] − U2
2 B22 =
= 1 ∗ 1 ∗ [−0.4950 ∗ sin(0 − 0) − 4.9505 ∗ cos(0 − 0)] − 12
∗ (−4.9505) = 0
H =
∂P2
∂θ2
= −Q2 − B22U2
2 = −0 − (−4.9505 ∗ 12
) = 4.9505
JAC = H = 4.9505

97
Step 4
∆θ2 = H−1
∆P2 =
−1
4.9505
= −0.2020
Step 5
θ2 = θ2 + ∆θ2 = 0 − 0.2020 = −0.2020
Iteration 2
Step 2
2a)
2 G22 =
= 1 ∗ 1 ∗ [−0.4950 ∗ cos(−0.2020 − 0) + 4.9505 ∗ sin(−0.2020 − 0)] + 12
∗ 0.4950 = −0.9831
2b)
∆P = ∆P2 = PGD2 − P2 = −1 − (−0.9831) = −0.0169
Step 3
2 B22 =
= 1 ∗ 1 ∗ [−0.4950 ∗ sin(−0.2020 − 0) − 4.9505 ∗ cos(−0.2020 − 0)] − 12
∗ (−4.9505) = 0.2000
H =
∂P2
∂θ2
= −Q2 − B22U2
2 = −0.2000 − (−4.9505 ∗ 12
) = 4.7505
JAC = H = 4.7505
Step 4
∆θ2 = H−1
∆P2 =
−0.0169
4.7505
= −0.0035
Step 5
θ2 = θ2 + ∆θ2 = −0.2020 − 0.0035 = −0.2055
Iteration 3
Step 2
2a)
2 G22 =
= 1 ∗ 1 ∗ [−0.4950 ∗ cos(−0.2055 − 0) + 4.9505 ∗ sin(−0.2055 − 0)] + 12
∗ 0.4950 = −1.0000
2b)
∆P = ∆P2 = PGD2 − P2 = −1 − (−1.0000) ≈ 0 (in MATLAB ∆P2 = −9.3368 ∗ 10−6
)
Step 3
2 B22 =
= 1 ∗ 1 ∗ [−0.4950 ∗ sin(−0.2055 − 0) − 4.9505 ∗ cos(−0.2055 − 0)] − 12
∗ (−4.9505) = 0.2053
H =
∂P2
∂θ2
= −Q2 − B22U2
2 = −0.2053 − (−4.9505 ∗ 12
) = 4.7452
JAC = H = 4.7452

98
Step 4
∆θ2 = H−1
∆P2 =
−9.3368 ∗ 10−6
4.7452
= −1.9676 ∗ 10−6
≈ 0
Step 5
θ2 = θ2 + 0 = −0.2055 − 0 = −0.2055
Now go to Step Final
Step Final
1 G11 =
= 1 ∗ 1 ∗ [−0.4950 ∗ cos(0 + 0.2055) + 4.9505 ∗ sin(0 + 0.2055)] + 12
∗ 0.4950 = 1.0208
1 B11 =
= 1 ∗ 1 ∗ [−0.4950 ∗ sin(0 + 0.2055) − 4.9505 ∗ cos(0 + 0.2055)] − 12
∗ (−4.9505) = 0.0032
2 B22 =
= 1 ∗ 1 ∗ [−0.4950 ∗ sin(−0.2055 − 0) − 4.9505 ∗ cos(−0.2055 − 0)] − 12
∗ (−4.9505) = 0.2053
PG1 = (P1 + PLD1) ∗ Sb = (1.0208 + 0.2) ∗ 100 = 122.08 MW (in MATLAB PG1=122.0843)
QG1 = (Q1 + QLD1) ∗ Sb = (0.0032 + 0.02) ∗ 100 = 2.32 MVAr (in MATLAB QG1=2.3171)
QG2 = (Q2 + QLD2) ∗ Sb = (0.2053 + 0.2) ∗ 100 = 40.53 MVAr (in MATLAB QG2=40.5255)
g = −G , b = −B and bsh−12 = 0
P12 = g12 U2
1 − U1 U2 [g12 cos(θ1 − θ2) + b12 sin(θ1 − θ2)] ∗ Sb =
= 0.4950 ∗ 12
− 1 ∗ 1 ∗ [0.4950 ∗ cos(0 + 0.2055) − 4.9505 ∗ sin(0 + 0.2055)] ∗ 100 =
= 102.0843 MW
P21 = g21 U2
2 − U2 U1 [g21 cos(θ2 − θ1) + b21 sin(θ2 − θ1)] ∗ Sb =
= 0.4950 ∗ 12
− 1 ∗ 1 ∗ [0.4950 ∗ cos(−0.2055 − 0) − 4.9505 ∗ sin(−0.2055 − 0)] ∗ 100 =
= −100 MW
Q12 = (−bsh−12 − b12)U2
1 − U1 U2 [g12 sin(θ1 − θ2) − b12 cos(θ1 − θ2)] ∗ Sb =
= (−0 + 4.9505) ∗ 12
− 1 ∗ 1 ∗ [0.4950 ∗ sin(0 + 0.2055) + 4.9505 ∗ cos(0 + 0.2055)] ∗ 100 =
= 0.3171 MVAr
Q21 = (−bsh−12 − b21)U2
2 − U2 U1 [g21 sin(θ2 − θ1) − b21 cos(θ2 − θ1)] ∗ Sb =
= (−0 + 4.9505) ∗ 12
− 1 ∗ 1 ∗ [0.4950 ∗ sin(−0.2055 − 0) + 4.9505 ∗ cos(−0.2055 − 0)] ∗ 100
= 20.5255 MVAr
Ptot
Loss = P12 + P21 = 102.0843 − 100 = 2.0843 MW
or
Ptot
Loss = (PG1 + PG2) − (PLD1 + PLD2) = (122.0843 + 100) − (20 + 200) = 2.0843 MW
ANG = [θ1 theta2] ∗
180
π
= [0 − 0.2055] ∗
180
π
= [0 − 11.7771◦
]
VOLT = [U1 U2] ∗ Ub = [1 1] ∗ 220 = [220 220]

99
Example 8.13 Consider again the power system shown in Figure 8.12. In this example, let
bus 2 be a PQ-bus with the following data:
• Bus 2: PQ-bus, PG2 = 1, QG2 = 0.405255, PLD2 = 2, QLD2 = 0.2
Let also ε = 10−6
. By applying Newton-Raphson method, calculate θ2, PG1, QG1, QG2 and
the active power losses.
Solution
See the MATLAB-codes in appendix D.
In the next examples, it will be shown that how the ”fsolve” function in MATLAB can be
used for solving non-linear power ﬂow equations.
Example 8.14 Consider the power system shown in Figure 8.13. Let the base power be
Sb = 100 MVA, the base voltage be Ub = 220 kV. Let also, bus 1 be a slack bus.
1
2
3
~
4
System 1 System 2
Figure 8.13. Single-line diagram of a balanced four-bus power system.
The system data (in physical units, i.e. MW, MVAr, kV, Ω and S) is given as follows:
• Line between Bus 1 and Bus 2: ¯Z12 = 5 + j 65 , bsh−12 = 0.0002
• Line between Bus 3 and Bus 4: ¯Z34 = 3 + j 30, short line
• Bus 1: U1 = 220, θ1 = 0, PLD1 = 10, QLD1 = 2
• Bus 2: PLD2 = 90, QLD2 = 10
• Bus 3: PLD3 = 80, QLD3 = 10
• Bus 4: PLD4 = 50, QLD4 = 10

100
Use ”fsolve” function in MATLAB, and ﬁnd
a) the unknown voltage magnitudes and voltage phase angles,
b) the Jacobian matrix based on the obtained results in task a),
c) the generated active and reactive power at the slack bus and the PU-buses (if any),
d) the total active power losses, and the losses in System1 and System 2,
e) the changes (in % compared to the obtained results in task d)) of power losses in both
systems, for an active load increased at bus 2 with 30 MW, i.e. Pnew
LD2 = 120 MW.
f) Let PLD2 = 90 MW. Re-do task e) for a reactive load increased at bus 3 with 10 MVAr,
i.e. Qnew
LD3 = 20 MVAr.
Solution
MATLAB-codes for this example can be found in appendix E.
a) U1 = 1.0000 × Ub = 220.0000 kV, θ1 = 0◦
,
U2 = 0.9864 × Ub = 216.9990 kV, θ2 = −7.8846◦
,
U3 = 0.9794 × Ub = 215.4704 kV, θ3 = −8.7252◦
,
U4 = 0.9693 × Ub = 213.2499 kV, θ4 = −10.5585◦
,
b) Jacobian=
14.0010 -6.8457 0 0.1604 -0.4025 0
-6.8309 29.7314 -15.2054 -0.6031 1.7452 -1.0305
0 -15.1084 15.1084 0 -2.0008 1.0008
-1.9604 0.4025 0 13.8010 -6.8457 0
0.6031 -3.3452 1.0305 -6.8309 29.5314 -15.2054
0 2.0008 -2.0008 0 -15.1084 14.9084
c) Slack bus (bus 1): PG1 = 232.4938 MW, QG1 = 9.6185 MVAr
d) PSys1
Loss = 0.1715 MW , PSys2
Loss = 2.3222 MW , Ptot
Loss = 2.4937 MW
e) PSys1
Loss = 0.1729 MW ⇒ ∆PSys1
Loss = 0.8163 %
PSys2
Loss = 30.2041 %
Ptot
Loss = 3.1965 MW ⇒ ∆Ptot
Loss = 28.1830 %
f) PSys1
Loss = 1.9825 %
PSys2
Loss = 1.7526 %
Ptot
Loss = 1.7685 %

101
Example 8.15 Consider again the power system in Example 8.14. The System 1 operator
is interested in the load flow solutions when installing a controllable shunt capacitor at bus
3 to keep the voltage at its rated (or nominal) value, i.e. U3 = 220 kV. Re-do the tasks in
Example 8.14, and also find the size of the shunt capacitor Bsh in S.
Solution
In this example, bus 3 will be considered as a PU-bus with U3 = 220 kV. MATLAB-codes for
this example can be found in appendix F. Note that only the changes of the MATLAB-codes
compared to the MATLAB-codes for Example 8.14 are given in this appendix.
a) U1 = 1.0000 × Ub = 220.0000 kV, θ1 = 0◦
,
U2 = 0.9968 × Ub = 219.2882 kV, θ2 = −7.8192◦
,
U3 = 1.0000 × Ub = 220.0000 kV, θ3 = −8.6473◦
,
U4 = 0.9901 × Ub = 217.8306 kV, θ4 = −10.4051◦
,
b) Jacobian=
14.2958 -7.0632 0 0.1829 0
-7.0482 30.7646 -15.8572 -0.6208 -1.0957
0 -15.7601 15.7601 0 1.0660
-1.9829 0.4168 0 14.0958 0
0 2.0660 -2.0660 0 15.5601
c) Slack bus (bus 1): PG1 = 232.4490 MW, QG1 = −14.7469 MVAr
PU-buses (bus 3): QG3 = 22.5772 MVAr and Bsh = 109.2735 S
d) PSys1
Loss = 2.4490 MW
e) PSys1
Loss = 0 %
PSys2
Loss = 29.6595 %
Ptot
Loss = 27.6684 %
f) PSys1
Loss = 0 %
PSys2
Loss = 0 %
Ptot
Loss = 0 %
Example 8.16 Consider the power system described in Example 8.15. Now, both system
operators are interested in the load flow solutions when the generator at bus 1 has a fixed
generation with PG1 and QG1 obtained in Example 8.15, and a new generator is installed at
bus 4 to be a slack bus with U4 and θ4 obtained in Example 8.15. Re-do the tasks in Example
8.15.
Solution
In this example, bus 1 will be considered as a PQ-bus. After modifying the MATLAB-codes
for Example 8.15, the load flow simulations give the following results:

102
a) U1 = 1.0000 × Ub = 220.0000 kV, θ1 = 0◦
,
U2 = 0.9968 × Ub = 219.2882 kV, θ2 = −7.8192◦
,
U3 = 1.0000 × Ub = 220.0000 kV, θ3 = −8.6473◦
,
U4 = 0.9901 × Ub = 217.8306 kV, θ4 = −10.4051◦
,
b) Jacobian=
15.4072 -7.3870 -8.0202 3.3293 0.4415
-7.2326 14.2958 -7.0632 -1.5661 0.1829
-7.8592 -7.0482 30.7646 -1.7368 -0.6208
1.1197 -0.4415 -0.6782 15.0723 -7.3870
1.5661 -1.9829 0.4168 -7.2326 14.0958
c) Slack bus (bus 4): PG1 = 0 MW, QG1 = 0 MVAr
PU-buses (bus 3): QG3 = 22.5772 MVAr and Bsh = 109.2735 S
d) PSys1
Loss = 2.4490 MW
e) PSys1
Loss = −77.3114 %
PSys2
Loss = 1.6896 %
Ptot
Loss = −3.6137 %
f) PSys1
Loss = 0 %
PSys2
Loss = 0 %
Ptot
Loss = 0 %
Some questions regarding the obtained results:
q1: Why is ∆PSys1
Loss = 0 in Example 8.15, task e), but not in Example 8.14 and Example
8.16?
q2: Why is ∆Ptot
Loss = 0 in Example 8.15 and Example 8.16 , task f), but not in Example
8.14?
q3: Why have ∆PSys1
Loss and ∆Ptot
Loss in Example 8.16, task e), decreased?
q4: In Example 8.15, does Bsh in tasks e) and f) have the same value as that obtained in
task c)? Motivate your answer.
q5: Why are the obtained voltages in Example 8.15 and Example 8.16 identical?

Chapter 9
Analysis of three-phase systems using linear
transformations
In this chapter, the possibilities of using linear transformations in order to simplify the anal-
ysis of three-phase systems, are briefly discussed. These transformations are general and are
valid under both symmetrical and un-symmetrical conditions. By generalizing the expres-
sions for a symmetric three-phase voltage given in equations (3.11) and (3.15), corresponding
expressions for an arbitrary three-phase voltage at constant frequency can be obtained as
ua(t) = UMa cos(ωt + γa)
ub(t) = UMb cos(ωt + γb)
uc(t) = UMc cos(ωt + γc)
Ua = Ua γ◦
a
Ub = Ub γ◦
b
Uc = Uc γ◦
c
(9.1)
where UMa, UMb, UMc are peak values, Ua, Ub, Uc are RMS-values and γa, γb, γc are phase
angles of the three voltages. For the un-symmetrical currents, corresponding expressions
hold as
ia(t) = IMa cos(ωt + γa − φa)
ib(t) = IMb cos(ωt + γb − φb)
ic(t) = IMc cos(ωt + γc − φc)
Ia = Ia γ◦
a − φa
Ib = Ib γ◦
b − φb
Ic = Ic γ◦
c − φc
(9.2)
where IMa, IMb, IMc are peak values, Ia, Ib, Ic are RMS-values of the three phase currents
whereas φa, φb, φc are the phase of the currents in relation to the corresponding phase
voltage.
The mean value of the total three-phase active power can be calculated as
P3 =
UMa
√
2
IMa
√
2
cos φa +
UMb
√
2
IMb
√
2
cos φb +
UMc
√
2
IMc
√
2
cos φc (9.3)
whereas the total three-phase complex power is
S3 = UaI
∗
a + UbI
∗
b + UcI
∗
c = (UaIa cos φa + UbIb cos φb + UcIc cos φc) +
+ j(UaIa sin φa + UbIb sin φb + UcIc sin φc) (9.4)
This phase representation is in many cases sufficient for a three-phase system analysis. There
are a number of important cases when the analysis can greatly be simplified by using linear
transformations.
This chapter discusses the following items. First, the advantages of using linear transfor-
mations in three-phase system analysis are generally discussed. Later on, some specific
transformations to be used in certain conditions are given. In order to really understand
the subject of transformations, the reader is referred to text books on the subject, e.g. in
electric machine theory or high power electronics. In chapter 10, one of the transformations
of interest, symmetrical components, is discussed in more detail. The purpose of chapter 9
is to show that the idea and the mathematics behind the transformations are the same. It
is only the choice of linear transformation, i.e. transformation matrix, that is different.
103

104
9.1 Linear transformations
By using transformations, components are mapped from an original space (the original space
is here the instantaneous values or the complex representation of the phase quantities) to an
image space. A linear transformation means that the components in the image space are a
linear combination of the original space. The complex values of the phase voltages can be
mapped with a linear transformation as
UA = waaUa + wabUb + wacUc
UB = wbaUa + wbbUb + wbcUc (9.5)
UC = wcaUa + wcbUb + wccUc
which in matrix form can be written as


UA
UB
UC

 =


waa wab wac
wba wbb wbc
wca wcb wcc




Ua
Ub
Uc

 (9.6)
or in a more compact notation
UABC = WUabc (9.7)
The elements in matrix W are independent of the values of the original and image space
components. In this example, the components in the original space Ua, Ub and Uc are
mapped by using the linear transformation W to the image space components UA, UB and
UC. The original space components can be calculated from the image space components by
using the inverse of matrix W (W−1
= T), i.e.
Uabc = W−1
UABC = TUABC (9.8)
The only mappings that are of interest, are those where W−1
are existing. In the following,
the matrix T or its inverse T−1
will represent the linear transformation.
9.1.1 Power invarians
A usual demand for the linear transformations in power system analysis is that it should be
possible to calculate the electric power in the image space by using the same expressions as
in the original space and that the two spaces should give the same result. A transformation
that can meet that requirement is called power invariant. Using the complex representation,
the electric power in the original space can be calculated by using equation (9.4), this gives
Sabc = UaI
∗
a + UbI
∗
b + UcI
∗
c = Ut
abc I∗
abc (9.9)
where “t” indicates the transpose.
In the image space, the corresponding expression is
SABC = UAI
∗
A + UBI
∗
B + UCI
∗
C = Ut
ABCI∗
ABC (9.10)

105
Power invarians implies that SABC = Sabc, i.e.
Ut
ABCI∗
ABC = Ut
abcI∗
abc = (TUABC)t
(TIABC)∗
= Ut
ABCTt
T∗
I∗
ABC (9.11)
This gives that the transformation matrix T must fulfill the following condition :
Tt
T∗
= ((T∗
)t
T)t
= 1 =


1 0 0
0 1 0
0 0 1

 (9.12)
which leads to
T−1
= (T∗
)t
(9.13)
If T is real, equation (9.13) implies that T is an orthogonal matrix.
9.1.2 The coefficient matrix in the original space
Consider a three-phase line between two buses. The voltage drop Uabc over the line depends
on the current Iabc flowing in the different phases. The voltage drop can be expressed as
Uabc =


Ua
Ub
Uc

 =


Zaa Zab Zac
Zba Zbb Zbc
Zca Zcb Zcc




Ia
Ib
Ic

 = ZabcIabc (9.14)
where Zabc is the coefficient matrix of the line. Note that each element in Zabc is non-zero
since a current in one phase has influence on the voltage drop in the other phases owing to
the mutual inductance, see chapter 11.
Symmetrical matrices
A matrix that is symmetrical around its diagonal is called a symmetrical matrix (or more
precisely, Hermitian if the matrix contains complex entries). For the Z-bus matrix in equation
(9.14), this implies that Zab = Zba, Zac = Zca and Zbc = Zcb, i.e.
Zabc =


Zaa Zab Zac
Zab Zbb Zbc
Zac Zbc Zcc

 = Zt
abc (9.15)
An example of a symmetrical matrix is the one representing a line (or a cable) where the
non-diagonal element are dependent on the mutual inductance, which is equal between the
phases a–b and the phases b–a, see chapter 11.
Cyclo-symmetrical matrices
The Z-bus matrix in equation (9.14) is cyclo-symmetric if Zab = Zbc = Zca, Zba = Zac = Zcb
and Zaa = Zbb = Zcc, i.e.
Zabc =


Zaa Zab Zba
Zba Zaa Zab
Zab Zba Zaa

 (9.16)

106
All normal three-phase systems are cyclo-symmetrical, i.e. if ia, ib, ic are permuted to ib,
ic, ia, the voltages ua, ub, uc will also be permuted to ub, uc, ua. This implies that ordinary
overhead lines, cables, transformers and electrical machines can be represented by cyclo-
symmetrical matrices.
9.1.3 The coefficient matrix in the image space
If both sides of equation (9.14) are multiplied with the matrix T−1
, the following is obtained
UABC = T−1
Uabc = T−1
ZabcIabc = (T−1
ZabcT)IABC = ZABCIABC (9.17)
where
ZABC = T−1
ZabcT (9.18)
ZABC is the image space mapping of the coefficient matrix Zabc. This gives that if UABC
represents the image space voltages, and IABC represents the image space currents then
ZABC will represent the impedances in the image space.
One reason of introducing a linear transformation may be to obtain a diagonal coefficient
matrix in the image space, i.e.
ZABC =


ZAA 0 0
0 ZBB 0
0 0 ZCC

 (9.19)
By having a diagonal coefficient matrix, equation (9.17) can be rewritten as
UA = ZAAIA
UB = ZBBIB (9.20)
UC = ZCCIC
i.e. the matrix equation (9.17) with mutual couplings between the phases is replaced by three
un-coupled equations. If ZABC is diagonal as in equation (9.19), both sides in equation (9.18)
can be multiplied with T and rewritten as
TZABC =

 T1 T2 T3




ZAA 0 0
0 ZBB 0
0 0 ZCC

 =

 ZAAT1 ZBBT2 ZCCT3

 =
= ZabcT = Zabc

 T1 T2 T3

 (9.21)
where T1, T2, T3 are the columns of T. Equation (9.21) can be rewritten as
ZabcT1 − ZAAT1 = 0
ZabcT2 − ZBBT2 = 0 (9.22)
ZabcT3 − ZCCT3 = 0

107
i.e. ZAA, ZBB and ZCC are the eigenvalues of matrix Zabc and the vectors T1, T2 and T3
are the corresponding eigenvectors. A transformation that maps the matrix Z to a diagonal
form should have a transformation matrix T having columns that are the eigenvectors of the
matrix Z. Note that eigenvectors can be scaled arbitrarily.
9.2 Examples of linear transformations that are used
in analysis of three-phase systems
In the following, four commonly used linear transformations will be briefly introduced. In
general, transformations can be presented in a little bit different way in different text books.
It is therefore of importance to understand the definitions used by the authors.
9.2.1 Symmetrical components
In the analysis of un-symmetrical conditions in a power system, symmetrical components
are commonly used. This complex, linear transformation uses the fact that all components
(lines, machines, etc.) in normal systems are cyclo-symmetrical, i.e. their impedances can
be modeled by equation (9.16). The power invariant transformation matrix and its inverse
for the symmetrical components are
TS =
1
√
3


1 1 1
1 α2
α
1 α α2

 T−1
S =
1
√
3


1 1 1
1 α α2
1 α2
α

 (9.23)
where α = ej120◦
. As given by the definition T−1
S = (T∗
S)t
which corresponds to the as-
sumption of power invariant according to equation (9.13). By using this transformation,
cyclo-symmetrical matrices are transformed into a diagonal form as given in equation (9.19),
i.e. the columns of matrix TS consist of the eigenvectors to a cyclo-symmetrical matrix.
This will simplify the system analysis as indicated in equation (9.20). Using the given pha-
sor voltages Ua, Ub and Uc, the power invariant symmetrical components can be calculated
as
Us =


U0
U1
U2

 = T−1
S Uabc =
1
√
3


1 1 1
1 α α2
1 α2
α




Ua
Ub
Uc

 (9.24)
The three components U0, U1 and U2 are called zero-sequence, positive-sequence and negative-
sequence, respectively. A cyclo-symmetrical impedance matrix according to equation (9.16)
can be diagonalized by using symmetrical components according to equation (9.18) as
ZS = T−1
S


Zaa Zab Zba
Zba Zaa Zab
Zab Zba Zaa

 TS =


Z0 0 0
0 Z1 0
0 0 Z2

 (9.25)

108
where
Z0 = Zaa + Zab + Zba = zero-sequence impedance
Z1 = Zaa + α2
Zab + αZba = positive-sequence impedance (9.26)
Z2 = Zaa + αZab + α2
Zba = negative-sequence impedance
The three impedances Z0, Z1 and Z2 are the eigenvalues of the cyclo-symmetrical impedance
matrix. For an impedance matrix that is both cyclo-symmetric and symmetric, i.e. Zba =
Zab, the result after a diagonalization will be that
Z0 = Zaa + 2Zab
Z1 = Zaa − Zab (9.27)
Z2 = Zaa − Zab
Transformers, overhead lines, cables and symmetrical loads (not electrical machines) can be
normally represented by impedance matrices that are both symmetrical and cyclo-symmetrical,
i.e. all diagonal elements are equal and all non-diagonal elements are equal. This gives that
the positive-sequence impedance and the negative-sequence impedance are equal.
In order to make the positive-sequence phasor voltage equal to the line-to-neutral phasor
voltage, a reference invariant form of transformation for the symmetrical components is
normally used. The reference invariant transformation matrix and its inverse are
TS =


1 1 1
1 α2
α
1 α α2

 =
√
3 · TS T−1
S =
1
3


1 1 1
1 α α2
1 α2
α

 =
1
√
3
· T−1
S (9.28)
The reference invariant transformation is not power invariant since T−1
S = 1
3
(T∗
S )t
. The
name reference invariant means that in symmetrical conditions U1 = Ua. Note that trans-
formations of coefficient matrices, according to equation (9.18), are not influenced whether
the power invariant or the reference invariant matrix is used since
ZABC(eff − inv) = T−1
S ZabcTS =
1
√
3
T−1
S Zabc
√
3TS =
= T−1
S ZabcTS = ZABC(ref − inv) (9.29)
A third variation of the transformation matrix for the symmetrical components arises when
the ordering of the sequences is changed. If the positive-sequence is given first and the
zero-sequence last, the columns of the T-matrix and the rows in the T−1
are permuted,
respectively. This results in the following reference invariant transformations matrices :
TS =


1 1 1
α2
α 1
α α2
1

 T−1
S =
1
3


1 α α2
1 α2
α
1 1 1

 (9.30)
This form of the transformation matrices will be used in chapter 10 where symmetrical
components are discussed in more detail. The only thing that happens with the coefficient
matrix in the image space is that the diagonal elements change places.
As described above, a number of different variations of the symmetrical components can
be used, all having the same fundamental purpose, to diagonalize the cyclo-symmetrical
impedance matrices.

109
9.2.2 Clarke’s components
Clarke’s components, also called α−β-components or orthogonal components, divides those
phase-quantities not having any zero-sequence into two orthogonal components. The word
zero-sequence means the same as when discussing symmetrical components, the sum of the
phase components. Components that do not have zero-sequence are those whose sum is equal
to zero. The power invariant (T−1
= (T∗
)t
, see equation (9.13)) transformation matrix and
its inverse for the Clarke’s components are
TC =
2
3



1√
2
1 0
1√
2
−1
2
√
3
2
1√
2
−1
2
−
√
3
2


 T−1
C =
2
3


1√
2
1√
2
1√
2
1 −1
2
−1
2
0
√
3
2
−
√
3
2

 (9.31)
Clarke’s components are a real orthogonal transformation that is mainly used in transfor-
mations of time quantities, e.g.


i0(t)
iα(t)
iβ(t)

 = i0αβ(t) = T−1
C iabc(t) =
2
3


1√
2
1√
2
1√
2
1 −1
2
−1
2
0
√
3
2
−
√
3
2




ia(t)
ib(t)
ic(t)

 (9.32)
where i0(t) is the zero-sequence component, iα(t) is the α-component and iβ is the β-
component for Clarke’s transformation of the phase currents ia(t), ib(t) and ic(t). The reason
why the transformation is orthogonal is given by column two and three of TC (corresponds
to the α- and β-components) since the columns are orthogonal.
For a symmetrical three-phase current given by equation (3.13)
ia(t) = IM cos(ωt − φ)
ib(t) = IM cos(ωt − 120◦
− φ) (9.33)
ic(t) = IM cos(ωt + 120◦
− φ)
the Clarke’s components are given by equation (9.32)
i0(t) =
1
√
3
(ia(t) + ib(t) + ic(t)) = 0
iα(t) =
2
3
ia(t) −
1
2
ib(t) −
1
2
ic(t) =
3
2
IM cos(ωt − φ) (9.34)
iβ(t) =
2
3
√
3
2
ib(t) −
√
3
2
ic(t) =
3
2
IM cos(ωt − φ − 90◦
)
As given above, conditions not having any zero-sequence can be fully represented by Clarke’s
α- and β-components. Conditions not having any zero-sequence are quite common and
depends, among other things, on the type of transformer connection used.
Matrices that are both symmetrical and cyclo-symmetrical can be diagonalized by using

110
Clarke’s transform as
ZC = T−1
C


Zaa Zab Zab
Zab Zaa Zab
Zab Zab Zaa

 TC =
=


Zaa + 2Zab 0 0
0 Zaa − Zab 0
0 0 Zaa − Zab

 (9.35)
For this type of matrices, the diagonalization based on Clarke’s components gives exactly
the same answer as the diagonalization based on symmetrical components, see equations
(9.25) and (9.27).
This gives that the matrix representation of transformers, overhead lines, cables and sym-
metrical loads (not electrical machines) can be diagonalized. The advantage of using Clarke’s
components is that the transformation is real which implies that the mapping of real instan-
taneous quantities are also real. The disadvantage is that electrical machines cannot be
represented by three independent variables by using Clarke’s components.
Clarke’s components are used in order to simplify the analysis of e.g. multi-phase short
circuits, transient system behavior, converter operation, etc.
9.2.3 Park’s transformation
Park’s transformation (also called dq-transformation or Blondell’s transformation) is a linear
transformation between the three physical phases and three new components. This trans-
formation is often used when analyzing synchronous machines.
In Figure 9.1, a simplified description of the internal conditions of a synchronous machine
having salient poles, is given. Two orthogonal axis are defined. One is directed along the
magnetic flux induced in the rotor. The second axis is orthogonal to the first axis. The
first axis is called the direct-axis (d-axis) and the second axis is called the quadrature-axis
(q-axis). Note that this system of coordinates follows the rotation of the rotor. The machine
given in Figure 9.1 is a two-pole machine, but Park’s transformation can be used for machines
having an arbitrary number of poles.
As indicated above, Park’s transformation is time independent since the displacement be-
tween the dq-axes and the abc-axes is changed when the rotor revolves. The Park’s trans-
formation includes not only the d- and q-components, but also the zero-sequence in order
to achieve a complete representation. The connection between phase currents ia, ib, and ic
and the dq0-components is given by the notation given in Figure 9.1
i0 =
1
√
3
(ia + ib + ic)
id =
2
3
(ia cos β + ib cos (β − 120◦
) + ic cos (β + 120◦
))
iq =
2
3
(ia sin β + ib sin (β − 120◦
) + ic sin (β + 120◦
))
(9.36)

111
a-axis
b-axis
c-axis
β
a a′
b
b′ c
c′
d-axis
q-axis
fi
a-axis
b-axis c-axis
β
d-axis q-axis
ci
f
u+f
i
bu
bi
au
+
ai
cu
n
+
+
Direction of
rotation
Figure 9.1. Deﬁnitions of quantities in Park’s transformation.
This equation can be written on matrix form as
i0dq =


i0
id
iq

 =
2
3


1√
2
1√
2
1√
2
cos β cos (β − 120◦
) cos (β + 120◦
)
sin β sin (β − 120◦
) sin (β + 120◦
)




ia
ib
ic

 = T−1
P iabc (9.37)
The matrix TP and matrix T−1
P can be transposed as
TP = (T−1
P )−1
=
2
3



1√
2
cos β sin β
1√
2
cos (β − 120◦
) sin (β − 120◦
)
1√
2
cos (β + 120◦
) sin (β + 120◦
)


 = (T−1
P )t
(9.38)
The transformation is hence power invariant according to equation (9.13). Park’s transfor-
mation is real and usable when transforming time quantities. Note that the Park’s trans-
formation is linear but the transformation matrix is time dependent. At constant frequency
β = ωt + β0.
The Park’s transformation is a frequency transformed version of Clarke’s transformation.
When β = 0 and the q-axis leads the d-axis, the transformation matrices are identical, i.e.
TC = TP(β = 0).
9.2.4 Phasor components
Phasor components are mainly used at instantaneous value analysis when a single machine
or when several machines are connected together. The power invariant (T−1
= (T∗
)t
, see

112
equation (9.13)) transformation matrix and its inverse for these components are
TR =
1
√
3


1 ejθ
e−jθ
1 α2
ejθ
αe−jθ
1 αejθ
α2
e−jθ

 (9.39)
T−1
R =
1
√
3


1 1 1
e−jθ
αe−jθ
α2
e−jθ
ejθ
α2
ejθ
αejθ


where α = ej120◦
. The phasor components of the three-phase currents ia(t), ib(t) and ic(t)
can be obtained as


i0(t)
is(t)
iz(t)

 = i0sz(t) = T−1
R iabc(t) =
1
√
3


1 1 1
e−jθ
αe−jθ
α2
e−jθ
ejθ
α2
ejθ
αejθ




ia(t)
ib(t)
ic(t)

 (9.40)
where i0(t) is the zero-sequence component and is(t) is called the field vector current, the
complex phasor of the current. The current is(t) is complex since the transformation matrix
is complex. By assuming that ia(t), ib(t) and ic(t) are real, the expression for iz(t) can be
written as
iz(t) =
1
√
3
ejθ
ia(t) + α2
ib(t) + αic(t) = (9.41)
=
1
√
3
e−jθ
ia(t) + αib(t) + α2
ic(t)
∗
= i
∗
s(t)
i.e. iz(t) is known if the field vector is(t) is known. Under conditions of no zero-sequence
components, the field vector is fully describing an arbitrary real three-phase quantity.
For a symmetrical three-phase current as given in equation (9.33), the phasor components
can be obtained according to equation (9.40)
i0(t) =
1
√
3
(ia(t) + ib(t) + ic(t)) = 0
is(t) =
e−jθ
√
3
ia(t) + αib(t) + α2
ic(t) =
√
3
2
IM ej(ωt−φ−θ)
(9.42)
iz(t) =
ejθ
√
3
ia(t) + α2
ib(t) + αic(t) =
√
3
2
IM e−j(ωt−φ−θ)
= i
∗
s(t)
Finally, for θ = ωt the following is obtained
i0(t) = 0
is(t) =
√
3
2
IM e−jφ
(9.43)
iz(t) =
√
3
2
IM ejφ
= i
∗
s(t)
i.e. the field vector current is(t) has a constant magnitude, independent of time.

113
By assuming real phase currents having no zero-sequence, they can be calculated by using
the ﬁeld vector current as


ia(t)
ib(t)
ic(t)

 = TRi0sz(t) =
1
√
3


1 ejθ
e−jθ
1 α2
ejθ
αe−jθ
1 αejθ
α2
e−jθ




0
is(t)
i
∗
s(t)

 = (9.44)
=
1
√
3


ejθ
is(t) + ejθ
is(t)
∗
α2
ejθ
is(t) + α2
ejθ
is(t)
∗
αejθ
is(t) + α2
ejθ
is(t)
∗

 =
2
√
3


Re ejθ
is(t)
Re α2
ejθ
is(t)
Re αejθ
is(t)


Phasor components are a frequency transformed form of the symmetrical components. For
θ = 0 (phasor components), the transformation matrix for the phasor components and the
symmetrical components are identical, i.e. TR(θ = 0) = TS.

Chapter 10
Symmetrical components
10.1 Deﬁnitions
Assume an arbitrary un-symmetric combination of three phases, exempliﬁed by the currents
Ia, Ib and Ic, shown in Figure 10.1 a).
b)
1aI
1cI
1bI
aI
bI
cI
a)
2aI
c) d)
2cI
2bI
0aI
0bI
0cI
1aI 2aI
0aI
1bI2bI
0bI
1cI
0cI
2cI
Figure 10.1. Unbalanced current phasors expressed as the sum of positive-, negative-,
and zero-sequence components.
Based on C. L. Fortesque’s theorem, a set of three unbalanced phasors in a three-phase
system can be resolved into the following three balanced systems of phasors (or symmetrical
components) :
A. Positive-sequence components consisting of a balanced system of three phasors with
the same amplitude, and having a phase displacement of 120 and 240◦
, respectively.
The phase sequence is abca, as shown in Figure 10.1 b).
B. Negative-sequence components consisting of a balanced system of three phasors with
the same amplitude, and having a phase displacement of 240 and 120◦
, respectively.
The phase sequence is acba, as shown in Figure 10.1 c).
C. Zero-sequence components consisting of a balanced system of three phasors with the
same amplitude and phase, as shown in Figure 10.1 d).
115

116
The three balanced systems can be symbolized with 1 (positive-sequence), 2 (negative-
sequence) and 0 (zero-sequence).
The result shown in Figure 10.1 can be mathematically expressed as :
Ia = Ia1 + Ia2 + Ia0
Ib = Ib1 + Ib2 + Ib0 (10.1)
Ic = Ic1 + Ic2 + Ic0
The three positive-sequence components can be denoted as
Ib1 = Ia1e−j120◦
(10.2)
Ic1 = Ia1ej120◦
The corresponding expressions for the negative- and zero-sequence components are as
Ib2 = Ia2ej120◦
Ic2 = Ia2e−j120◦
(10.3)
Ia0 = Ib0 = Ic0
By inserting equation (10.2) and (10.3) into equation (10.1), the following is obtained
Ia = Ia1 + Ia2 + Ia0
Ib = α2
Ia1 + αIa2 + Ia0 (10.4)
Ic = αIa1 + α2
Ia2 + Ia0
where,
α = ej120◦
= cos 120◦
+ j sin 120◦
= −
1
2
+ j
√
3
2
(10.5)
The following expressions of the symbol α are valid
α2
= ej240◦
= e−j120◦
= −
1
2
− j
√
3
2
α3
= 1
1 + α + α2
= 0 (10.6)
α∗
= α2
(α2
)∗
= α
Equation (10.4) can, by using matrix form, be written as
Iph
= TIs (10.7)
where the matrix
T =


1 1 1
α2
α 1
α α2
1

 (10.8)

117
which is called the transformation matrix for the symmetrical components. This matrix is
equal to the reference invariant matrix TS according to equation (9.30). The current vector
Iph
=


Ia
Ib
Ic

 (10.9)
represents the current phasor of each phase whereas
Is =


Ia1
Ia2
Ia0

 or just Is =


I−1
I−2
I−0

 (10.10)
represents the symmetrical components of the phase currents.
By using equation (10.7), the symmetrical components as a function of the phase currents
can be obtained :
Is = T−1
Iph
(10.11)
where
T−1
=
1
3


1 α α2
1 α2
α
1 1 1

 (10.12)
which is equal to T−1
S according to equation (9.30). Of course, the symmetrical components
can also be applied to voltages. Using the vectors
Uph
=


Ua
Ub
Uc

 and Us =


Ua1
Ua2
Ua0

 or just Us =


U−1
U−2
U−0

 (10.13)
for representing line-to-neutral voltage phasors and symmetrical components, respectively,
the relation between them can be written as
Uph
= TUs (10.14)
Us = T−1
Uph
(10.15)
Example 10.1 Calculate the symmetrical components for the following symmetrical voltages
Uph
=


Ua
Ub
Uc

 =


277 0◦
277 − 120◦
277 + 120◦

 V (10.16)
Solution
By using equation (10.12) and (10.15), the symmetrical components of the voltage Uph
can

118
be calculated as


U−1
U−2
U−0

 = Us = T−1
Uph
=
1
3


1 α α2
1 α2
α
1 1 1




Ua
Ub
Uc

 = (10.17)
=
1
3


1 · 277 + 1 120◦
· 277 − 120◦
+ 1 240◦
· 277 + 120◦
1 · 277 + 1 240◦
· 277 − 120◦
+ 1 120◦
· 277 + 120◦
1 · 277 + 1 · 277 − 120◦
+ 1 · 277 + 120◦

 =
=


277 0◦
0
0


As given in the example, a symmetric three-phase system with a phase sequence of abc gives
rise to a positive-sequence voltage only, having the same amplitude and angle as the voltage
in phase a.
Example 10.2 For a Y-connected three-phase load with zero conductor, phase b is at one
occasion disconnected. The load currents at that occasion is :
Iph
=


Ia
Ib
Ic

 =


10 0◦
0
10 + 120◦

 A (10.18)
Calculate the symmetrical components of the load current as well as the current in the zero
conductor, In.
Solution


I−1
I−2
I−0

 =
1
3


1 · 10 0◦
+ 1 120◦
· 0 + 1 240◦
· 10 + 120◦
1 · 10 0◦
+ 1 240◦
· 0 + 1 120◦
· 10 + 120◦
1 · 10 0◦
+ 1 · 0 + 1 · 10 + 120◦

 = (10.19)
=


6.667 0◦
3.333 − 60◦
3.333 60◦


In = Ia + Ib + Ic = 10 0◦
+ 0 + 10 + 120◦
= 10 60◦
= 3I−0 (10.20)
As given in the example, the current in the zero conductor is three times as large as the
zero-sequence current.
10.2 Power calculations under unbalanced conditions
Based on the voltage and current phasors of each phase, the three-phase complex power can
be calculated as
S = P + jQ = UaI
∗
a + UbI
∗
b + UcI
∗
c = Ut
ph
I∗
ph
(10.21)

119
By introducing symmetrical components, the expression above can be converted to
S = Ut
ph
I∗
ph
= (TUs)t
(TIs)∗
= Ut
sTt
T∗
I∗
s (10.22)
The expression Tt
T∗
can be written as
Tt
T∗
=


1 α2
α
1 α α2
1 1 1




1 1 1
α α2
1
α2
α 1

 = 3


1 0 0
0 1 0
0 0 1

 (10.23)
i.e. the transformation is not power invariant, see subsection 9.1.1, equation (9.12). Equation
(10.22) can be rewritten as
S = 3 Ut
s I∗
s = 3 U−1 I
∗
−1 + 3 U−2 I
∗
−2 + 3 U−0 I
∗
−0 (10.24)
Since the magnitude of the line-to-line voltages are
√
3 times the line-to-neutral voltages and
Sb =
√
3 · Ub · Ib, the introduction of the per-unit system gives that equation (10.24) can be
rewritten as
Spu =
√
3(
√
3Us)t
· I∗
s
√
3 · Ub · Ib
= Upu−1 I
∗
pu−1 + Upu−2 I
∗
pu−2 + Upu−0 I
∗
pu−0 p.u (10.25)
This implies that the total power (in per-unit value) in an unbalanced system can be ex-
pressed by the sum of the symmetrical components of power. The total power in physical
unit can be obtained by multiplying Spu with Sb, i.e.
S = Spu Sb = Upu−1 I
∗
pu−1 Sb + Upu−2 I
∗
pu−2 Sb + Upu−0 I
∗
pu−0 Sb MVA (10.26)

K Chapter 11
Sequence circuits of transmission lines
11.1 Series impedance of single-phase overhead line
The theory of having an overhead line using the ground as a return conductor was discussed
by Carson in 1923. Carson considered a single conductor a of unity length (e.g. one meter)
running in parallel with the ground, see Figure 11.1.
a
+
−
aU
+
−
Local earth
Ref.
Surface of
remote earth
0dU =
d d′
Fictitious ground
return conductor
a′
aaZ
ddZ
aI
d aI I= −
adD





adZ
1 Unit
Figure 11.1. Carson’s single-phase overhead line using the ground as return path
The current Ia flows in the conductor using the ground between d − d as return path. The
ground is assumed to have an uniform resistance and an infinite extension. The current Id
(=−Ia) is distributed over a large area, flowing along the ways of least resistance. Kirchhoff’s
law about the same voltage drop along each path is fulfilled. It has been shown that these
distributed return paths may, in the analysis, be replaced by a single return conductor having
a radius = rd located at a distance Dad from the overhead line according to Figure 11.1. The
distance Dad is a function of the resistivity of the ground ρ. The distance Dad increases as
the resistivity ρ increases.
The inductance of this circuit can be calculated as
La =
µ
2π
ln
1
Da
Laa
+
µ
2π
ln
1
Dd
Ldd
−2
µ
2π
ln
1
Dad
Lad
=
µ
2π
ln
Dad
Da
+ ln
Dad
Dd
(11.1)
where
µ = the permeability of the conductor
Da = e−1/4
ra for a single conductor with radius ra
Dd = e−1/4
rd for a return conductor in ground with radius rd
The inductance can according to equation (11.1) be divided into three parts, two apparent self
inductances (Laa, Ldd) and one apparent mutual inductance (Lad). Note that these quantities
121

122
are only mathematical quantities without any physical meaning. For instance, they have
wrong dimension inside the ln-sign. It is only after the summation they achieve a physical
meaning. Hopefully, the diﬀerent part expressions will simplify the understanding of the
behavior of a three-phase line. The total series reactance of this single-phase conductor is
Xa = ωLa = ω(Laa + Ldd − 2Lad) (11.2)
By using this line model, having apparent inductances, the voltage drop for a single-phase
line can be calculated as
Uaa
Udd
=
Ua − Ua
Ud − Ud
=
zaa zad
zad zdd
Ia
−Ia
V/length unit (11.3)
where Ua, Ua , Ud and Ud are given in proportion to the same reference. Since Ud = 0 and
Ua − Ud = 0, Ua can be obtained by subtracting the two equations from each other :
Ua = (zaa + zdd − 2zad)Ia = ZaIa (11.4)
By deﬁnition
Za ≡ zaa + zdd − 2zad Ω/length unit (11.5)
The impedances in this equation can be calculated as
zaa = ra + jxaa = ra + jωLaa Ω/length unit
zdd = rd + jxdd = rd + jωLdd Ω/length unit (11.6)
zad = jxad = jωLad Ω/length unit
Za = ra + rd + jXa Ω/length unit
where
ra = conductor resistance per length unit
rd = ground resistance per length unit
11.2 Series impedance of a three-phase overhead line
In order to obtain the series impedance of a three-phase line, the calculations are performed
in the same way as for the single-phase line. In Figure 11.2, the impedances, voltages and
currents of the line are given.
Since all conductors are grounded at a , b , c , the following are valid
Ua − Ud = 0 , Ub − Ud = 0 , Uc − Ud = 0
Id = − (Ia + Ib + Ic)
(11.7)
The voltage drop over the conductors can be calculated as




Uaa
Ubb
Ucc
Udd



 =




Ua − Ua
Ub − Ub
Uc − Uc
Ud − Ud



 =




zaa zab zac zad
zab zbb zbc zbd
zac zbc zcc zcd
zad zbd zcd zdd








Ia
Ib
Ic
Id



 V/length unit (11.8)

123
All wires grounded
here to local earth
potential
adZ

















ddZdI
a aI
d
+
−
Ref.
0dU =
ccZ
c cI
aaZ
bbZ
b bI
d′
c′
a′
b′
1 Unit
cU
aU
bU
bdZ













cdZ








abZ





bcZ





acZ









Figure 11.2. Three-phase overhead line with ground as return path
In the same way as for the single-phase conductor, the impedances in equation (11.8) are
apparent without any physical relevance. With Ud = 0 and by using equation (11.7), row
four can be subtracted from row one in equation (11.8) which gives
Ua − (Ua − Ud ) = (zaa − 2zad + zdd)Ia + (zab − zad − zbd + zdd)Ib +
+ (zac − zad − zcd + zdd)Ic (11.9)
This can be simplified to Ua = ZaaIa + ZabIb + ZacIc. The impedances Zaa, Zab and Zac are
defined below. Note that when Ib = Ic = 0, the impedance Zaa is exactly the impedance
of a single-phase line using the ground as return path as described in section 11.1. If the
calculations above are repeated for the phases b and c, the following can be obtained


Ua
Ub
Uc

 =


Zaa Zab Zac
Zab Zbb Zbc
Zac Zbc Zcc




Ia
Ib
Ic

 V/length unit (11.10)
where
Zaa = zaa − 2zad + zdd Ω/length unit
Zbb = zbb − 2zbd + zdd Ω/length unit
Zcc = zcc − 2zcd + zdd Ω/length unit (11.11)
Zab = Zba = zab − zad − zbd + zdd Ω/length unit
Zbc = Zcb = zbc − zbd − zcd + zdd Ω/length unit
Zac = Zca = zac − zad − zcd + zdd Ω/length unit
The magnitude of the impedances can be calculated as the impedances in section 11.1,
equation (11.1) and 11.6. It is important to note the coupling between the phases. A
current flowing in one phase will influence the voltage drop in other phases. The replacing of
a three-phase line with three parallel impedances, is an approximation which gives that all

124
non-diagonal element of the Z-bus matrix in equation (11.10) are neglected. In other words,
the mutual inductance between the conductors are neglected. The error this simpliﬁcation
gives is dependent on several things, e.g. the distance between the conductors, the length of
the conductors and the magnitude of the currents in the conductors.
11.2.1 Symmetrical components of the series impedance of a three-
phase line
Symmetrical components are often used in the analysis of power systems having three-phase
lines, in order to simplify the complicated cross-couplings that exist between the phases.
The quantities in equation (11.10) can be deﬁned as :


Ua
Ub
Uc

 = Uph
= Zph
Iph
=


Zaa Zab Zac
Zab Zbb Zbc
Zac Zbc Zcc




Ia
Ib
Ic

 (11.12)
The voltage vector (Uph
) and current vector (Iph
) can be replaced by the corresponding
symmetrical component multiplied with matrix T according to the section on symmetrical
components :
Uph
= TUs = Zph
TIs = Zph
Iph
(11.13)
This equation can be rewritten as
Us = T−1
Zph
TIs = ZsIs (11.14)
If a symmetrical overhead line (or cable) is assumed, i.e. Zaa = Zbb = Zcc and Zab = Zbc =
Zac, the following is obtained
Zs = T−1
Zph
T =
1
3


1 α α2
1 α2
α
1 1 1




Zaa Zab Zac
Zab Zbb Zbc
Zac Zbc Zcc




1 1 1
α2
α 1
α α2
1

 =
=


Zaa − Zab 0 0
0 Zaa − Zab 0
0 0 Zaa + 2Zab

 (11.15)
Equation (11.14) can be rewritten as


U−1
U−2
U−0

 = Us = Zs Is = (11.16)
=


Zaa − Zab 0 0
0 Zaa − Zab 0
0 0 Zaa + 2Zab




I−1
I−2
I−0

 =


Z−1 I−1
Z−2 I−2
Z−0 I−0


where
Z−1 = Zaa − Zab = positive-sequence impedance
Z−2 = Zaa − Zab = negative-sequence impedance (11.17)
Z−0 = Zaa + 2Zab = zero-sequence impedance

125
By inserting the expressions used in equation (11.11) into equation (11.17), the following
can be obtained
Z−1 = Z−2 = zaa − zab (11.18)
Z−0 = zaa + 2zab − 6zad + 3zdd
Note that the coupling to ground are not present in the expressions for the positive- and
negative-sequence impedances, i.e. the elements having index d in the Z-bus matrix in
equation (11.8) are not included. This means that the zero-sequence current is zero in the
positive- and negative-sequence reference frame, which is quite logical. All couplings to
ground are represented in the zero-sequence impedance. As indicated above, a line by using
this model, can be represented as three non-coupled components : positive-, negative-, and
zero-sequence components. It should be pointed out that some loss of information will occur
when using this model. For example, if only positive-, negative-, and zero-sequence data are
given, the potential of the ground, Ud in Figure 11.2, cannot be calculated. To calculate
that potential, more detailed data are needed. The line model introduced in section 7.1.2,
is based on positive-sequence data only, since symmetrical conditions are assumed.
11.2.2 Equivalent diagram of the series impedance of a line
As given above, for a symmetrical line Z−1 = Z−2. Assume that this line can be replaced by
an equivalent circuit according to Figure 11.3, i.e. three phase impedances Zα and one return
aI aU′
bU′
cU′
0U′
aU
bU
cU
0U
bI
cI
0I Zβ
Zα
Zα
Zα
Figure 11.3. Equivalent diagram of the series impedance of a line
impedance Zβ where the mutual inductance between the phases is assumed zero. With three
phases and one return path, as given by the equivalent in Figure 11.3, the following is valid
I0 = Ia + Ib + Ic (11.19)
By using equation (11.19), the voltage drop between the phases and the return conductor
can be calculated as
Ua − U0 = Ua − U0 − Ia · Zα − (Ia + Ib + Ic)Zβ
Ub − U0 = Ub − U0 − Ib · Zα − (Ia + Ib + Ic)Zβ (11.20)
Uc − U0 = Uc − U0 − Ic · Zα − (Ia + Ib + Ic)Zβ

126
which can be rewritten to matrix form



Ua − U0
Ub − U0
Uc − U0


 =


Ua − U0
Ub − U0
Uc − U0

 −


Zα + Zβ Zβ Zβ
Zβ Zα + Zβ Zβ
Zβ Zβ Zα + Zβ




Ia
Ib
Ic

 (11.21)
or
Uph
= Uph
− ZαβIph
(11.22)
Since the matrix Zαβ is both symmetric and cyclo-symmetric, it can represent a line according
to the assumption made. The matrix Zαβ can be converted to symmetrical components by
using equation (11.15) :
Zsαβ = T−1
ZαβT =


Zα 0 0
0 Zα 0
0 0 Zα + 3Zβ

 (11.23)
When the symmetrical components Z1 = Z2 and Z0 for the line are known, the following is
obtained
Zα = Z−1
Zβ = Z−0−Z−1
3
(11.24)
With these values of Zα and Zβ, the equivalent in Figure 11.3 can be used, together with
equation (11.21), to calculate the voltage drop between the phases and the return conductor
(= Uph
− Uph
) as a function of the phase currents (= Iph
).
Note that the equivalent cannot be used to calculate e.g. U0 − U0 or Ua − Ua but only e.g.
(Ua − U0) − (Ua − U0).
Example 11.1 Solve example 3.5 by using symmetrical components.
LZ
aI aU′
bU′
cU′
0U′
aU
bU
cU
0U
bI
cI
0I
LZ
0LZ
aZ
bZ
cZ
LZ
Figure 11.4. Network diagram for example 11.1
Solution
According to the solutions in example 3.5, the impedances of interest are ZL = 2.3 + j0.16
Ω, ZL0 = 2.3 + j0.03 Ω, Za = 47.9 + j4.81 Ω, Zb = 15.97 + j1.60 Ω, Zc = 23.96+j2.40 Ω.

127
The symmetrical components of the line will ﬁrst be calculated. Note that the line in the
example is given in the same way as the equivalent. The symmetrical components can be
calculated by using equation (11.23) :
Z−1 = Z−2 = ZL = 2.3 + j0.16 Ω
Z−0 = ZL + 3ZL0 = 9.2 + j0.25 Ω
(11.25)
which gives that
Zs =


Z−1 0 0
0 Z−2 0
0 0 Z−0

 =


2.3 + j0.16 0 0
0 2.3 + j0.16 0
0 0 9.2 + j0.25

 (11.26)
The symmetrical components for the load can be calculated by using equation (11.15)
ZLDs =
1
3


1 α α2
1 α2
α
1 1 1




Za 0 0
0 Zb 0
0 0 Zc




1 1 1
α2
α 1
α α2
1

 =
=


29.28 + j2.94 9.09 + j3.24 9.55 − j1.37
9.55 − j1.37 29.28 + j2.94 9.09 + j3.24
9.09 + j3.24 9.55 − j1.37 29.28 + j2.94

 Ω (11.27)
The applied voltage is symmetric, i.e. it has only one sequence, the positive one :
Us = T−1
Uph
=


220 0◦
0
0

 V (11.28)
The equation for this un-symmetric three-phase network can be described as
Us = (Zs + ZLDs)Is (11.29)
which can be rewritten as
Is = (Zs + ZLDs)−1
Us =


8.11 − 5.51◦
2.22 149.09◦
1.75 − 155.89◦

 A (11.30)
The symmetrical components for the voltage at the load can be calculated as
ULDs = ZLDsIs =


201.32 − 0.14◦
5.13 − 26.93◦
16.10 25.67◦

 V (11.31)
The power obtained in the radiators can be calculated by using equation (10.24)
S = 3Ut
LDsI∗
s = 4754 + j477 VA (11.32)
i.e. the thermal power is 4754 W.

128
As given above, only the voltage drop at the load and the load currents can be calculated by
using the symmetrical components. The ground potential at the load cannot be calculated,
but that is usually of no interest.
Previously, in example 3.5, 5.1 and 5.2, the ground potential at the load has been calculated
by using other types of circuit analyses. It should be pointed out that the value of the ground
potential has no physical interpretation if the value of ZL and ZL0 has been obtained by
using the symmetrical components of the line according to equation (11.24). As given by
the solutions, the load demand, phase voltages at the load and the currents at the load are
physically correct by using either one of the four methods of solution.
11.3 Shunt capacitance of a three-phase line
The line resistance and inductance are components that together form the series impedance
of the line. The capacitance that is of interest in this section, forms the shunt component.
The series component, usually the inductance, gives a limit on the maximum amount of the
current that can be transmitted over the line, and by that also the maximum power limit.
The capacitive shunt component behaves as a reactive power source. The reactive power
generated, is proportional to the voltage squared, which implies that the importance of the
shunt capacitance increases with the voltage level. For lines having a nominal voltage of
300–500 kV and a length of more than 200 km, these capacitances are of great importance.
In high voltage cables where the conductors are more close to one another, the capacitance
is up to 20–40 times larger than for overhead lines. The reactive power generation can be a
problem in cables having a length of only 10 km.
There is a fundamental law about electric fields saying that the electric potential v at a
certain point on the distance r from a point charge q, can be calculated as :
v =
q
4π 0r
V (11.33)
where 0 = 8.854 × 10−12
F/m, permittivity of vacuum. This law gives that there is a direct
relationship between the difference in potential and accumulation of charges. If two long,
parallel conductors are of interest, and if there is a voltage difference, v1 − v2, between the
lines, an accumulation of charges with different sign, +Q and −Q, will take place. The
magnitude of the total charge Q depends mainly on the distance between the lines but also
on the design of the lines. For cables, the material between the conductors will also have an
influence on the charge accumulation. The capacitance between the two conductors is equal
to the quotient between the charge Q and the difference in potential :
C ≡
Q
v1 − v2
(11.34)
For a three-phase line, the corresponding capacitance is located between all conductors.
When having a difference in potential between a conductor and ground, an accumulation of
charges will also occur in relation to the magnitude of the capacitance. In Figure 11.5, the
different capacitances of a three-phase overhead line are given. A line is normally constructed

129
b
bcc
abc
acc
agc bgc cgc
c
a
Figure 11.5. Capacitances of a three-phase overhead line without earth wires
in a symmetrical way, i.e. the mean distance between the phases are equal. Also, the
mean distance between a phase and ground is the same for all phases. In Figure 11.5, this
corresponds to the case that cab = cbc = cac and cag = cbg = ccg when the entire line is of
interest.
In the same way as given earlier for the series impedances, the positive-, negative- and
zero-sequence capacitances can be calculated. Only the results from the calculations will be
presented here.
C−1 = C−2 =
2π 0
ln 2Ha
Ar
F/m (11.35)
C−0 =
2π 0
ln 2HA2
ra2
F/m (11.36)
where, according to Figure 11.6
C−1 = positive-sequence capacitance
C−2 = negative-sequence capacitance
C−0 = zero-sequence capacitance
H = 3
H1H2H3
A = 3
A1A2A3
a = 3
√
a12a13a23
r = the equivalent radius of the line = e−1/4
· real radius of the line
Note that C1 is equal to C2, but C0 has a diﬀerent value. When having a closer look
at the equations for C1, it can be seen that 2H/A ≈ 1 according to Figure 11.6, which
means that the distance to ground has a relatively small inﬂuence. If the conductors are
located close to one another, then 2H = A. The line model described in section 7.1.2, uses
only the positive-sequence capacitance C1 for the line. In principle, this can be regarded
as a ∆-Y-transformation of the capacitances between the phases since they are the main
contributors to the positive- and negative-sequence capacitances. The coupling to ground is
of less importance. In cables, the positive- and negative-sequence capacitances are usually
higher owing to the short distance between the phases.

130
Ground level
H1
H2 H3
a12
a23
a13
A1
A2
A3
Figure 11.6. Geometrical quantities of a line in the calculation of capacitance
For C−0, the coupling to ground is very important. When calculating C−0, all phases have
the same potential by the deﬁnition of zero-sequence. This implies that the capacitances
between the phases cab, cbc, cac are not of interest. However, the electric ﬁeld is changed
since all three conductors have the same potential. As given in the equation, the distance
to ground is very important (power of three inside the ln-sign) in the calculations of the
zero-sequence capacitance.

Chapter 12
Sequence circuits of transformers
In the analysis of three-phase circuits under unbalanced conditions, the transformer is rep-
resented by its positive-, negative- and zero-sequence impedances. These can be determined
by analyzing the three-phase transformer, e.g. the Y0-∆ connected shown in Figure 12.1.
aIa
b
c
n
A
C
B
bI
cI
eZ
eZ
eZ
nZ nI
Figure 12.1. Y0-∆ connected transformer with neutral point grounded through an
impedance Zn.
The impedance Ze represents the equivalent impedance of each phase and consists of both
leakage reactance of the primary and secondary windings as well as the resistance of the
windings, i.e. the windings shown in the ﬁgure are considered as ideal windings. The
magnetizing current of the transformer can be neglected, i.e. the magnetizing impedance is
assumed to be inﬁnitely large.
By using the direction of currents as shown in Figure 12.1, the following expressions for the
three phases of the transformer can be held
∆Ua = IaZe + InZn
∆Ub = IbZe + InZn (12.1)
∆Uc = IcZe + InZn
Since In = Ia + Ib + Ic, this can be rewritten as
∆Ua = Ia Ze + Zn + IbZn + IcZn
∆Ub = IaZn + Ib Ze + Zn + IcZn (12.2)
∆Uc = IaZn + IbZn + Ic Ze + Zn
which can be written on matrix form
∆U =


∆Ua
∆Ub
∆Uc

 =


Ze + Zn Zn Zn
Zn Ze + Zn Zn
Zn Zn Ze + Zn




Ia
Ib
Ic

 = ZtrIph
(12.3)
131

132
By diagonalizing the matrix Ztr, the symmetrical components of the transformer can be
obtained as the eigenvalues of Ztr
Ztrs = T−1
ZtrT =


Ze 0 0
0 Ze 0
0 0 Ze + 3Zn

 (12.4)
i.e.
Zt−1 = Ze = positive-sequence impedance
Zt−2 = Ze = negative-sequence impedance (12.5)
Zt−0 = Ze + 3Zn = zero-sequence impedance
As given above, the positive- and negative-sequence impedances are the same and equal to
the leakage reactance of each phase. That Z1 = Z2 is not surprising since the transformer
impedance does not change if the phase ordering is changed from abc (positive-sequence) to
acb (negative-sequence).
The zero-sequence impedance includes the leakage reactance but a factor of 3Zn is added
where Zn is the impedance connected between the transformer neutral and the ground. If
Zn = 0, the zero-sequence impedance will be equal to the leakage reactance of the trans-
former.
Note that to obtain a zero-sequence current, it must be a connection between the trans-
former neutral and the ground. Figure 12.2 shows the zero-sequence equivalent circuits of
transformers with different winding connections. In the figure, P stands for Primary side
and S for Secondary side of transformer.
For the transformer shown in Figure 12.1, the neutral of the primary side (P) is grounded,
It should be pointed out that this analysis of the zero-sequence impedance is dependent on
the Y0-∆ connected according to Figure 12.1. To obtain a zero-sequence current, it must be
a connection between the transformer neutral and the ground. On the secondary side, the
∆-connected side in Figure 12.1, there is no such connection, i.e. seen from the secondary
side, Zt−0 = ∞ and a zero-sequence current cannot flow.
Whereas the positive- and negative-sequence impedances of the transformer are independent
on from which side of the transformer the analysis is performed, the zero-sequence impedance
can vary with a large amount. Figure 12.2 a) shows a Y0-Y0 connected through which a
zero-sequence current can flow. This gives that the zero-sequence impedance is equal to the
leakage reactance as discussed above. However, Figure 12.2 b)-c) show a Y-Y connected and
a Y0-Y connected through which a zero-sequence current cannot flow.
Figure 12.2 d) shows a Y0-∆ connected which is allows a zero-sequence current to flow only
on the Y0-side since a return conductor exists and mmf-balance can be obtained owing to
the ∆-winding. On the ∆-side, no zero-sequence current can flow. In Figure 12.2 e)–f), no
zero-sequence current can flow on either side due to the connection types.

133
n nZ nI
n nI
NNZNI
Winding connection
a)
Y0-Y0
Zero-sequence
equivalent circuit
Ref.
SP
3 NZ3 nZeZ
Y0-∆
d)
n Nb)
Y-Y
SP
eZ
n Nc)
Y0-Y
Ref.
SP
eZ
nI
Ref.
SP
eZ
n
Y-∆
e)
Ref.
SP
eZ
-∆ ∆
f)
0PI − 0SI −
Ref.
0 0PI − = 0 0SI − =
0 0PI − = 0 0SI − =
0 0SI − =0PI −
0 0SI − =0 0PI − =
Ref.
SP
eZ 0 0SI − =0 0PI − =
Figure 12.2. Zero-sequence equivalent circuits of transformers with diﬀerent winding
connections.

Chapter 13
Analysis of unbalanced three-phase systems
As discussed in chapters 11 and 12, lines and transformers can be represented by their
positive-, negative- and zero-sequence impedances. These sequences are decoupled which
implies that for instance a certain zero-sequence current will only cause a zero-sequence
voltage drop whereas positive- and negative-sequence voltages will be unchanged. Also three-
phase generators can in an equivalent way be described by decoupled positive-, negative- and
zero-sequence systems.
This property implies that the entire system including generators, lines and transformers
can be represented by three decoupled systems.
13.1 Symmetrical components of impedance loads
A three-phase impedance load is normally Y - or ∆-connected as shown in Figure 13.1.
a) Y0-connected load
nZ
aZaI
aU
bZbI
bU
cZcI
cU
a b cI I I+ +
b) -connected load
abZ
aI
aU
bcZ
bI
bU acZ
cI
cU
Figure 13.1. Two possible load conﬁgurations
The neutral of a Y -connected load may be grounded with or without an impedance. Then
it is termed Y0-connected. For the Y 0-connected load shown in Figure 13.1 a), we have:
Uph
=


Ua
Ub
Uc

 =


Za + Zn Zn Zn
Zn Zb + Zn Zn
Zn Zn Zc + Zn




Ia
Ib
Ic

 = ZLDph
Iph
(13.1)
This equation can be transformed to symmetrical components as follows:
Uph
= TUs = ZLDph
Iph
= ZLDph
T Is
⇒
Us = T−1
ZLDph
T Is = ZLDsIs (13.2)
135

136
where
ZLDs ≡ T−1
ZLDph
T = (13.3)
=
1
3


Za + Zb + Zc Za + α2
Zb + αZc Za + αZb + α2
Zc
Za + αZb + α2
Zc Za + Zb + Zc Za + α2
Zb + αZc
Za + α2
Zb + αZc Za + αZb + α2
Zc Za + Zb + Zc + 9Zn


As indicated in this matrix, there are non-diagonal elements that are nonzero, i.e. there
exists couplings between the positive-, negative- and zero-sequences. A special case is when
Za = Zb = Zc. In this special case, ZLDs can be written as
ZLDs =


Za 0 0
0 Za 0
0 0 Za + 3Zn

 (13.4)
For a symmetric Y 0-connected load ZLD−1 = ZLD−2 = Za and ZLD−0 = Za + 3Zn. If
Zn = 0 then ZLD−1 = ZLD−2 = ZLD−0. However, if the neutral of the load is not grounded,
i.e a Y -connected load, then Zn = ∞ = ZLD−0 which means that no zero-sequence current
can flow.
For the ∆-connected load shown in Figure 13.1 b), the impedance can first be ∆-Y trans-
formed which results in a Y -connected load:
Za =
ZabZac
Zab + Zac + Zbc
(13.5)
Zb =
ZabZbc
Zab + Zac + Zbc
(13.6)
Zc =
ZacZbc
Zab + Zac + Zbc
(13.7)
Zn = ∞ (13.8)
For a symmetric ∆-connected load, i.e. Zab = Zbc = Zac, the symmetrical components can
be calculated by using equations (13.4) to (13.8) :
ZLD−1 = Zab/3 (13.9)
ZLD−2 = Zab/3 (13.10)
ZLD−0 = ∞ (13.11)
13.2 Connection to a system under unbalanced condi-
tions
In subsection 7.1.3, the connection to a network under symmetrical (or balanced) conditions
was discussed. It has been shown that by applying the Thévenin theorem the entire linear
balanced system (as seen from a selected point) can be represented by a voltage source

137
behind an impedance. The value of the impedance can be calculated when knowing the
three-phase short circuit current at the selected point.
A balanced power system as seen from a selected point p can be described by three de-
coupled single-line sequence systems (or networks) termed as positive-, negative and zero-
sequence systems. The model of the positive-sequence system is indeed the single-line system
of a balanced three-phase system that has been studied from Chapter 6 to Chapter 8, i.e.
in these chapters we have studied the positive-sequence system of a balanced three-phase
system.
Assuming a linear balanced three-phase power system, the sequence systems can be repre-
sented by their Th´evenin equivalents as shown in Figure 13.2. Note that there are no voltage
sources in the network for the negative- and zero-sequence systems. Thus, the negative- and
zero-sequence systems only consist of impedances.
~
1ThpZ −
ThpU
1pI −
a) Positive-sequence
2ThpZ −
2pI −
b) Negative-sequence
0ThpZ −
0pI −
c) Zero-sequence
1pU − 2pU − 0pU −
p p p
Figure 13.2. Th´evenin equivalents as seen from a selected point p in the system.
From Figure 13.2, the following can be obtained in p.u:
Up−1 = UThp − ZThp−1 Ip−1
Up−2 = 0 − ZThp−2 Ip−2 (13.12)
Up−0 = 0 − ZThp−0 Ip−0
13.3 Single line-to-ground fault
Assume that a single line-to-ground through an impedance Zf occurs at a point p in the
system, as shown in Figure 13.3.
Based on equation equation (10.11) the following is obtained:
Is =


Ip−1
Ip−2
Ip−0

 =
1
3


1 α α2
1 α2
α
1 1 1




Ifa
0
0

 =
1
3


Ifa
Ifa
Ifa


⇒ Ip−1 = Ip−2 = Ip−0 =
1
3
Ifa (13.13)
From Figure 13.3, we have Upa = Zf Ifa.

138
faI
p
fZ
Phase a
0fbI =
0fcI =
Phase b
Phase c
Three-phase power system
Figure 13.3. Single-phase short circuit in phase a
Using the first row in equation (10.14) and equations (13.12)-(13.13), the following is obtained
in p.u:
Upa = Up−1 + Up−2 + Up−0 = Zf Ifa = 3Zf Ip−1 ⇒
Upa = UThp − ZThp−1 Ip−1 − ZThp−2 Ip−1 − ZThp−0 Ip−1 = 3Zf Ip−1 p.u
(13.14)
Thus,
Ifa = 3 Ip−1 =
3 UThp
ZThp−1 + ZThp−2 + ZThp−0 + 3 Zf
p.u (13.15)
If the quantities are expressed in their physical units (i.e. kV, kA and Ω), then
Upa =
UThp
√
3
− ZThp−1 Ip−1 − ZThp−2 Ip−1 − ZThp−0 Ip−1 = 3Zf Ip−1 kV
Ifa = 3 Ip−1 =
3 UT hp
√
3
ZThp−1 + ZThp−2 + ZThp−0 + 3 Zf
kA
(13.16)
If the equivalent diagram of a line shown in Figure 11.3 is used, and if Ua in the figure is
connected to U0 through an impedance Zf , then by virtue of equation (11.24) the current
Ia can be obtained as
Ia =
Ua − U0
Zα + Zβ + Zf
=
Ua − U0
Z−1 + Z−0−Z−1
3
+ Zf
=
3(Ua − U0)
2Z−1 + Z−0 + 3Zf
(13.17)
which is similar to equation (13.16). To calculate the current of a single line-to ground fault,
the equivalent in Figure 11.3 can be used if ZThp−1 = ZThp−2.
Example 13.1 At a 400 kV bus, a solid three-phase short circuit occurs, giving a fault
current of 20 kA per phase. If a solid single line-to-ground fault occurs at the same bus, the
fault current will be 15 kA in the faulted phase. The Thévenin impedances in the positive- and

139
negative-sequence systems at the bus can be assumed to be purely reactive and equal. (This
is normal for high voltage systems since the dominating impedances origin from lines and
transformers which have dominating reactive characteristics, equal for positive- and negative-
sequences). Also the zero-sequence impedance can be assumed to be purely reactive. Calculate
the Th´evenin equivalents for the positive-, negative- and zero-sequences at the fault.
Solution
Solid short circuits means that Zf = 0. Since all impedances are purely inductive, the fault
currents will also be inductive, i.e.
Isc3Φ
= −j20 kA Isc1Φ
= −j15 kA (13.18)
Three-phase fault :
Based on equation (7.25),
ZTh−1 = ZTh−2 =
UTh
√
3 Isc3Φ
=
400
√
3 · (−j20)
= j11.55 Ω (13.19)
Single-phase fault :
From equation (13.16),
ZTh−0 =
3 UTh
√
3 Isc1Φ
−ZTh−1 −ZTh−2 −3 Zf =
3 · 400
√
3 · (−j15)
−2·(j11.55) = j23.09 Ω (13.20)
13.4 Analysis of a linear three-phase system with one
unbalanced load
As discussed in section 13.1, the three sequence systems are decoupled when having a bal-
anced system. However, in the case having unsymmetrical loads, these three sequence sys-
tems will not be decoupled.
Assume a linear power system with an unbalanced load. The system is composed of a voltage
source, lines and transformers. This system can be analyzed as follows:
1. Draw the impedance diagrams of the positive-, negative and zero-sequence systems,
for the entire network excluding the unbalanced load.
2. Find the Th´evenin equivalents of the positive-, negative- and zero-sequence systems as
seen from the point the unsymmetrical load is located.
3. Calculate the positive-, negative- and zero-sequence currents through the unbalanced
load.
4. Calculate the positive-, negative- and zero-sequence voltages at the points of interest.

140
5. Calculate the positive-, negative- and zero-sequence currents through components that
are of interest.
6. Transform those symmetrical components to the phase quantities that are asked for.
The items above can be treated in diﬀerent ways which will be shown in the following
example.
Example 13.2 Consider again the system described in Example 7.2. The following addi-
tional data is also given:
• Transformer is ∆-Y0 connected with Y 0 on the 10 kV-side, and Zn = 0.
• The zero-sequence impedance of the line is 3 times the positive-sequence impedance,
i.e. Z21−0 = 3 Z21−1.
• The zero-sequence shunt admittance of the line is 0.5 times the positive-sequence shunt
admittance, i.e. ysh−21−0 = 0.5 ysh−21−1.
• When the transformer is disconnected from bus 3, a solid (i.e. Zf = 0) single line-to-
ground applied to this bus results in a pure inductive fault current of 0.2 kA.
• The positive- and negative-sequence Th´evenin impedances of the power system are iden-
tical, i.e. ZTh−1 = ZTh−2.
• The load is Y0-connected with Zn = 0. Furthermore, half of the normal load connected
to phase a is disconnected while the other phases are loaded as normal, i.e. it is an
unsymmetrical load.
Calculate the voltage at the industry as well as the power fed by the transformer into the
line.
Solution
1) Start with the building of the impedance diagram of the positive-, negative- and zero-
sequence for the whole system excluding the unsymmetrical load, see Figure 13.4.
Positive- and negative-sequence components in per-unit values (from the
solution to Example 7.2):

141
1ThZ −
ThU
~
21 1Z −1tZ −
123
4
power system
21 1shy − − 21 1shy − −
a) Positive-sequence system
2ThZ − 21 2Z −2tZ −
123
power system
21 2shy − − 21 2shy − −
b) Negative-sequence system
0ThZ − 21 0Z −0tZ −
12
power system
21 0shy − − 21 0shy − −
c) Zero-sequence system
4
4
3
Figure 13.4. Positive-, negative- and zero-sequence systems.
UTh = UThpu = 1 0◦
ZTh−1 = ZTh−2 = ZThpu = j 0.0137
Zt−1 = Zt−2 = Ztpu = j 0.004
Z21−1 = Z21−2 = Z21pu = 0.0225 + j 0.0075
ysh−21−1 = ysh−21−2 = ysh−21pu =
j 0.003
2
AL−1 = AL−2 = AL , BL−1 = BL−2 = BL
CL−1 = CL−2 = CL , DL−1 = DL−2 = DL
A−1 = A−2 = A , B−1 = B−2 = B
C−1 = C−2 = C , D−1 = D−2 = D
(13.21)

142
Zero-sequence components in per-unit values:
Isc1Φ
=
0.2 − 90◦
Ib70
=
0.2 − 90◦
0.00412
= 48.5437 − 90◦
ZTh−0 =
3 UTh
Isc1Φ
− 2 ZTh−1 − 0 = j 0.0344
Zt−0 = Zt−1 = j 0.004 , since Zn = 0
Z21−0 = 3 Z21−1 = 0.0675 + j 0.0225
ysh−21−0 = 0.5 ysh−21−1 =
j 0.003
4
AL−0 = 1 + ysh−21−0 · Z21−0 = 1.0000 + j 0.0001
BL−0 = Z21−0 = 0.0675 + j 0.0225
CL−0 = ysh−21−0(2 + ysh−21−0 · Z21−0) = 0.0000 + j 0.0015
DL−0 = AL−0 = 1.0000 + j 0.0001
(13.22)
2) Next step is to replace the networks with Thévenin equivalents as seen from the in-
dustry connection point (bus 1), i.e. UThbus1, ZThbus1−1, ZThbus1−2 and ZThbus1−0, see
Figure 13.2.
The twoport of the entire positive-sequence network between bus 4 (which represents
the voltage source) and bus 1 (the industry connection point) is given by (see also
Example 7.2):
UTh
Ibus4−1
=
A−1 B−1
C−1 D−1
Ubus1−1
Ibus1−1
= (13.23)
=
0.9999 + j0.0000 0.0225 + j0.0252
0.0000 + j0.0030 1.0000 + j0.0000
Ubus1−1
Ibus1−1
Based on Figure 13.2 a), UThbus1−1 is obtained by setting Ibus1−1 = 0 as follows:
UTh = A−1 UThbus1−1 + B−1 · 0 ⇒
UThbus1−1 =
UTh
A−1
= 1.0001 − 0.0019◦
(13.24)
The Thévenin impedance ZThbus1−1 is obtained by setting UTh = 0 as follows:
0 = A−1 Ubus1−1 + B−1 Ibus1−1 ⇒
ZThbus1−1 = −
Ubus1−1
Ibus1−1
=
B−1
A−1
= 0.0225 + j0.0252
(13.25)
To set UTh = 0, it means that bus 4 is short circuited. Therefore in this case, the
positive-sequence system will have a configuration similar to the negative-sequence
system as shown in Figure 13.2 b). Furthermore, since Zt−1 = Zt−2, Z12−1 = Z12−2 and
ysh−21−1 = ysh−21−2, they imply that A−1 = A−2 and B−1 = B−2. Thus, in a similar

143
way as shown in equation (13.25), the Thévenin impedance ZThbus1−2 is obtained as
follows:
0 = A−2 Ubus1−2 + B−2 Ibus1−2 ⇒
ZThbus1−2 = −
Ubus1−2
Ibus1−2
=
B−2
A−2
=
B−1
A−1
= ZThbus1−1 = 0.0225 + j0.0252
(13.26)
Based on Figure 12.2 d), the zero-sequence of a ∆-Y0-transformer should be modeled
as an impedance to ground on the Y 0-side, as shown in Figure 13.4 c). As seen in the
figure, the feeding network (i.e. power system) is not connected to the industry load
from a zero-sequence point of view. The twoport of the network from the transformer
(bus 3) to the connection point of the industry (bus 1) is given by
Ubus3−0
Ibus3−0
=
0
Ibus3−0
=
1 Zt−0
0 1
AL−0 BL−0
CL−0 DL−0
Ubus1−0
Ibus1−0
=
=
A−0 B−0
C−0 D−0
Ubus1−0
Ibus1−0
= (13.27)
=
1.0000 + j0.0001 0.0675 + j0.0265
0.0000 + j0.0015 1.0000 + j0.0001
Ubus1−0
Ibus1−0
The Thévenin impedance ZThbus1−0 is obtained as follows:
0 = A−0 Ubus1−0 + B−0 Ibus1−0 ⇒
ZThbus1−0 = −
Ubus1−0
Ibus1−0
=
B−0
A−0
= 0.0675 + j0.0265
(13.28)
3) From Example 7.2 the the per-unit value of the load (i.e. ZLD) is known. At the
half load in phase a (i.e. ZLDa = 2ZLD), the impedance matrix of the symmetrical
components is calculated based on equation (13.3) :
ZLDs = T−1
ZLDph
T = T−1


2ZLD 0 0
0 ZLD 0
0 0 ZLD

 T
=


1.0667 + j0.8000 0.2667 + j0.2000 0.2667 + j0.2000
0.2667 + j0.2000 1.0667 + j0.8000 0.2667 + j0.2000
0.2667 + j0.2000 0.2667 + j0.2000 1.0667 + j0.8000


(13.29)
The equation of the entire system is now given by
UTh =


UThbus1
0
0

 = (Zs + ZLDs) Is (13.30)
where
Zs =


ZThbus1−1 0 0
0 ZThbus1−2 0
0 0 ZThbus1−0

 and Is =


Ibus1−1
Ibus1−2
Ibus1−0



144
The symmetrical components of the currents through the load can be calculated as:
Is = (Zs + ZLDs)−1
UTh =


0.8084 − 37.1616◦
0.1596 142.3433◦
0.1541 143.7484◦

 (13.31)
4-5) The symmetrical components of the voltage at the industry (bus 1) are given by
Ubus1s =


Ubus1−1
Ubus1−2
Ubus1−0

 = ZLDs Is =


0.9733 − 0.3126◦
0.0054 10.6340◦
0.0112 − 14.8199◦

 (13.32)
The positive-, negative- and zero-sequence voltages and currents (in per-unit values)
at the transformer connection to the line (bus 2) are given by
Ubus2−1
Ibus2−1
=
AL−1 BL−1
CL−1 DL−1
Ubus1−1
Ibus1−1
=
0.9915 − 0.6607◦
0.8066 − 36.9937◦
Ubus2−2
Ibus2−2
=
AL−2 BL−2
CL−2 DL−2
Ubus1−2
Ibus1−2
=
0.0028 52.3414◦
0.1596 142.3414◦
Ubus2−0
Ibus2−0
=
AL−0 BL−0
CL−0 DL−0
Ubus1−0
Ibus1−0
=
0.0004 0.0005◦
0.1541 143.7455◦
(13.33)
The symmetrical components of power (in physical units) fed by the transformer into
the line can be expressed by
Sbus2−1 = Ubus2−1 I
∗
bus2−1 Sb = 0.3221 + j 0.2369 MVA
∗
bus2−2 Sb = 0 − j 0.0002 MVA
∗
bus2−0 Sb = 0 − j 0.00005 MVA
(13.34)
6) Based on equation (10.14), the line-to-neutral voltages can be obtained. To express
these quantities in physical units, they must be multiplied with the corresponding base
voltage, then divided by
√
3, since the base voltage is based on line-to-line voltage.
Thus, 

Ubus1a
Ubus1b
Ubus1c

 = T Ubus1s ·
Ub10
√
3
=


5.7122 − 0.4154◦
5.5918 − 119.9774◦
5.5535 119.4555◦

 (13.35)
Based on equation (10.26), the total power fed by the transformer into the line is given
by
S = Sbus2−1 + Sbus2−2 + Sbus2−0 = 0.3221 + j0.2366 MVA (13.36)
13.5 A general method for analysis of linear three-
phase systems with one unbalanced load
In larger unsymmetrical systems, it is necessary to use a systematic approach to analyze
system voltages and currents. In this section, all system components except one load, are

145
symmetrical. In the demonstration below, a small system is analyzed in the same way as
can be performed for a large system. The example given below is identical to the one in
section 7.2 but with the diﬀerence that the load which will be connected to bus 2 is assumed
unbalanced. The voltage source is represented by bus 3. All quantities are expressed in
per-unit values.
Consider the simple balanced system shown in Figure 13.5. For a balanced system, all system
quantities and components can be represented only by their positive-sequence components.
The only diﬀerence between this system and the system studied in section 7.2 is that here
we use index -1 which has been omitted in section 7.2.
~
1 1LDZ −
21 1Z −1tZ −
123
3 1busI −
2 1busI − 1 1busI −
Figure 13.5. Impedance diagram of a symmetrical system
The Y-bus matrix of the positive-sequence system is identical with Y in in section 7.2, i.e.


Ibus1−1
Ibus2−1
Ibus3−1

 = I1 = Y1 U1 =
=



1
ZLD1−1
+ 1
Z12−1
− 1
Z12−1
0
− 1
Z21−1
1
Z21−1
+ 1
Zt−1
− 1
Zt−1
0 − 1
Zt−1
1
Zt−1





Ubus1−1
Ubus2−1
Ubus3−1


(13.37)
This Y-bus matrix can be inverted which results in the corresponding Z-bus matrix :
U1 = Y−1
1 I1 = Z1 I1 (13.38)
Since Ibus1−1 = Ibus2−1 = 0, the third row in equation (13.38) can be written as
Ubus3−1 = Z1(3, 3) · Ibus3−1 ⇒
Ibus3−1 =
Ubus3−1
Z1(3, 3)
(13.39)
where Z1(3, 3) is an element in the Z-bus matrix. With this value of the current inserted
into equation (13.38) all voltages are obtained.
Ubus1−1 = Z1(1, 3) · Ibus3−1 (13.40)
Ubus2−1 = Z1(2, 3) · Ibus3−1 (13.41)

146
So far, all calculations are identical to those in section 7.2. Corresponding calculations can
be performed for an arbitrary large system having impedance loads and one voltage
source. Assume a system with a voltage source at bus i. The current at bus i and the
voltage at another bus r can then be calculated as:
Ibusi−1 =
Ubusi−1
Z1(i, i)
(13.42)
Ubusr−1 = UThbusr = Z1(r, i) · Ibusi−1 (13.43)
Now assume that an unsymmetrical impedance load is connected to bus 2 which apparently
leads to the changes of the system voltages and currents. The actual voltages can be obtained
by using the theorem of superposition, i.e. as the sum of the voltages before the connection of
the load and with the voltage change obtained by the load connection. This can be expressed
by using symmetrical components as:
U1 = Upre1 + U∆1
U2 = Upre2 + U∆2 (13.44)
U0 = Upre0 + U∆0
where, (below with all buses it means all buses in the system excluding the bus
connected to the voltage source)
U1 is a vector containing the positive-sequence voltages at all buses (i.e. bus 1 and bus 2
in this example) due to the connection of the unsymmetrical load.
U2 is a vector containing the negative-sequence voltages at all buses due to the connection
of the unsymmetrical load.
U0 is a vector containing the zero-sequence voltages at all buses due to the connection of
the unsymmetrical load.
Upre1 is a vector containing the positive-sequence voltages at all buses prior to the connec-
tion of the unsymmetrical load.
Upre2 is a vector containing the negative-sequence voltages at all buses prior to the connec-
tion of the unsymmetrical load. All elements of this vector are zero, since the system
is under balanced conditions prior to the connection of the unsymmetrical load.
Upre0 is a vector containing the zero-sequence voltages at all buses prior to the connection
of the unsymmetrical load. All elements of this vector are zero, since the system is
under balanced conditions prior to the connection of the unsymmetrical load.
U∆1 is a vector containing the changes in the positive-sequence voltages at all buses due to
the connection of the unsymmetrical load.
U∆2 is a vector containing the changes in the negative-sequence voltages at all buses due
to the connection of the unsymmetrical load.
U∆0 is a vector containing the changes in the zero-sequence voltages at all buses due to the
connection of the unsymmetrical load.

147
Equation (13.44) can be rewritten by expressing the voltage changes by a Z-bus matrix
multiplied with the current changes injected into the buses as follows:
U1 = Upre1 + Z∆1 I∆1
U2 = 0 + Z∆2 I∆2 (13.45)
U0 = 0 + Z∆0 I∆0
where
Z∆1 is the Z-bus matrix of the positive-sequence system with the shortened voltage source.
Z∆2 is the Z-bus matrix of the negative-sequence system.
Z∆0 is the Z-bus matrix of the zero-sequence system.
I∆1 is a vector containing the injected positive-sequence current changes into the buses. In
this example only I∆1(2) = 0, since the load is connected to bus 2.
I∆2 is a vector containing the injected negative-sequence current changes into the buses. In
this example only I∆2(2) = 0.
I∆0 is a vector containing the injected zero-sequence current changes into the buses. In this
example only I∆0(2) = 0.
Figure 13.6 shows the positive-, negative- and zero-sequence systems which will be used to
calculate the voltage changes due to connection of the unsymmetrical load at bus 2. The
diﬀerence between the positive-sequence system in Figure 13.6 and the ∆-system used in
section 7.2 is that the load is now represented by the currents injected into the buses. The
inﬁnite bus (bus 1) is assumed to be directly connected to ground and the transformer is
Y0-Y0 connected. The admittance matrices of the sequence networks shown in Figure 13.6
can be formed as
Y∆1 =
1
ZLD1−1
+ 1
Z21−1
− 1
Z21−1
− 1
Z21−1
1
Z21−1
+ 1
Zt−1
Y∆2 =
1
ZLD1−2
+ 1
Z21−2
− 1
Z21−2
− 1
Z21−2
1
Z21−2
+ 1
ZT −2
(13.46)
Y∆0 =
1
ZLD1−0
+ 1
Z21−0
− 1
Z21−0
− 1
Z21−0
1
Z21−0
+ 1
Zt−0
Note that Y∆1 = Y1(1 : 2, 1 : 2), i.e. the row and column corresponding to the bus
connected to the voltage source (bus 1 in this example) are removed, see also Y∆ in section
7.2. Furthermore, Y∆2 = Y∆1 for a system that is only composed of lines, transformers and
symmetrical impedance loads, since their positive- and negative-sequence components are
identical.
From the above Y-bus matrices the corresponding Z-bus matrices can be calculated as
Z∆1 = Y−1
∆1
Z∆2 = Y−1
∆2 (13.47)
Z∆0 = Y−1
∆0

148
1 1LDZ −
21 1Z −1tZ −
123
1(2)I∆ 1(1) 0I∆ =
Positive-sequence
with shortened
voltage source
1 2LDZ −
21 2Z −2tZ −
123
2(2)I∆ 2(1) 0I∆ =
Negative-sequence
1 0LDZ −
21 0Z −0tZ −
123
0(2)I∆ 0(1) 0I∆ =
Zero-sequence
Figure 13.6. Positive-, negative- and zero-sequence diagrams for calculations of the
voltage changes.
Since only the sequence components of the injected currents (i.e. I∆) into the bus to which
the unsymmetrical load is connected (bus 2 in this example) are nonzero, these currents are
of interest and will be calculated as follows. Based on equation (13.45), we have
Ubus2−1 = U1(2) = Ubus2−1
from eq. (13.43)
+ Z∆1(2, 2) I∆1(2)
Ubus2−2 = U2(2) = 0 + Z∆2(2, 2) I∆2(2) (13.48)
Ubus2−0 = U0(2) = 0 + Z∆0(2, 2) I∆0(2)
Assuming that the unsymmetrical load is connected to bus r (r = 2 in this example), the
equations can be summarized as
U (r) =



Ubusr−1
Ubusr−2
Ubusr−0


 = Upre(r) + Z∆(r, r) I∆(r) = (13.49)
≡


UThbusr
0
0

 +


Z∆1(r, r) 0 0
0 Z∆2(r, r) 0
0 0 Z∆0(r, r)




I∆1(r)
I∆2(r)
I∆0(r)


It should be pointed out that equation (13.49) indeed describes the Th´evenin equivalents
as seen from bus r, where the voltage behind the positive-sequence impedance is UThbusr =
Ubus2−1 and the three Th´evenin impedances are Z∆1(r, r), Z∆2(r, r) and Z∆0(r, r).
Assume that the unsymmetrical load is Y0-connected with ZLDbusra in phase a, ZLDbusrb in
phase b and ZLDbusrc in phase c. The voltage drop over the load is

149
ULDbusrph
=


ULDbusra
ULDbusrb
ULDbusrc

 =


ZLDbusra 0 0
0 ZLDbusrb 0
0 0 ZLDbusrc




ILDbusra
ILDbusrb
ILDbusrc

 =
= ZLDbusrph
ILDbusrph
(13.50)
By introducing symmetrical components, this can be converted to
U (r) =



Ubusr−1
Ubusr−2
Ubusr−0


 = T−1
ULDbusrph
= T−1
ZLDbusrph
ILDbusrph
=
= T−1
ZLDbusrph
T
=ZLDbusrs
ILDbusrs = −ZLDbusrs I∆(r) (13.51)
Note that ILDbusrs is injected into the load, however I∆(r) is injected into the bus. Therefore,
ILDbusrs = −I∆(r).
Next based on equations (13.49) and (13.51), the current I∆(r) can be expressed by
I∆(r) = − [Z∆(r, r) + ZLDbusrs]−1
Upre(r) (13.52)
These values of the symmetrical components of the current at bus r can then be inserted
into equation (13.45) where the symmetrical components of all voltages can be calculated.
The voltage at bus k can then be calculated as :
U1(k) = Upre1(k) + Z∆1(k, r) I∆1(r)
U2(k) = Z∆2(k, r) I∆2(r) (13.53)
U0(k) = Z∆0(k, r) I∆0(r)
Example 13.3 Consider again the system described in Example 7.3. The following addi-
tional data is also given:
• The inﬁnite bus (i.e. bus 1) has and a grounded zero connection point.
• Transformer T1 is ∆-Y0 connected with Y 0 on the 10 kV-side, and Zn = 0.
• Transformer T2 is Y0-Y0 connected with Zn = 0.
• The zero-sequence impedances of the lines are 3 times the positive-sequence impedances,
and the zero-sequence shunt admittances of the lines are 0.5 times the positive-sequence
shunt admittances.
• The load LD1 is ∆-connected.
• The load LD2 is Y0-connected with Zn = 0. Furthermore, half of the normal load
connected to phase a is disconnected while the other phases are loaded as normal, i.e.
LD2 is an unsymmetrical load.

150
Calculate the eﬃciency of the internal network operating in this unbalanced condition.
Solution
1) Start with the building of the impedance diagram of the positive-, negative- and zero-
sequence networks for the entire system excluding the unsymmetrical load, see Figure
13.7.
1
2 3
4 5
1 1tZ −
23 1Z −
23 1shy − − 1 1LDZ −
24 1Z − 2 1tZ −
24 1shy − −
ThU
a) Positive-sequence system
1
2 3
4 5
1 2tZ −
23 2Z −
23 2shy − − 1 2LDZ −
24 2Z − 2 2tZ −
24 2shy − −
b) Negative-sequence system
1
2 3
4 5
1 0tZ −
23 0Z −
23 0shy − −
24 0Z − 2 0tZ −
24 0shy − −
c) Zero-sequence system
Figure 13.7. The sequence networks of the system in example 13.3.

151
Positive- and negative-sequence components in per-unit values (from the
solution to Example 7.3):
UTh = U1 = 1 0◦
Zt1−1 = Zt1−2 = Zt1pu = j 0.0438
Zt2−1 = Zt2−2 = Zt2pu = j 0.1333
Z23−1 = Z23−2 = Z23pu = 0.0017 + j 0.003
ysh−23−1 = ysh−23−2 = ysh−23pu =
j 0.0013
2
Z24−1 = Z24−2 = Z24pu = 0.0009 + j 0.0015
ysh−24−1 = ysh−24−2 = ysh−24pu =
j 0.00064
2
ZLD1−1 = ZLD1−2 = ZLD1pu = 0.64 + j 0.48
(13.54)
Zero-sequence components in per-unit values:
Zt1−0 = Zt1−1 = j 0.0438 , since Zn = 0
Zt2−0 = Zt2−1 = j 0.1333 , since Zn = 0
Z23−0 = 3 Z23−1 = 0.0051 + j 0.009
ysh−23−0 = 0.5 ysh−23−1 =
j 0.0013
4
Z24−0 = 3 Z24−1 = 0.0026 + j 0.0045
ysh−24−0 = 0.5 ysh−24−1 =
j 0.00064
4
ZLD1−0 = ∞ , since ∆-connected
(13.55)
Next, the admittance matrix of the positive-sequence network (i.e. Y1) will be formed
in a manner described in section 13.5. Bus 1 is included in order to determine the
voltage at all buses prior to the connection of the unsymmetrical load. However, the
load LD2 is not included in the Y-bus matrix since it is unsymmetrical. It implies
that Y1 is identical with the admittance matrix Y in Example 7.3 with the exception
of the ﬁfth diagonal element in which the unsymmetrical load LD2 is not included, i.e.
Y1 = Y with Y 55 = 1
Zt2−1
, see equation (7.49).
2) Next step is to replace the sequence networks with Th´evenin equivalents as seen from
bus 5, i.e. UThbus5, ZThbus5−1, ZThbus5−2 and ZThbus5−0, see Figure 13.2.
First, the Z-bus matrix is calculated as follows:
Z1 = Y−1
1 =






0.6429+j0.5264 0.6429+j0.4827 0.6412+j0.4797 0.6429+j0.4827 0.6429+j0.4827
0.6429+j0.4827 0.6429+j0.4827 0.6412+j0.4797 0.6429+j0.4827 0.6429+j0.4827
0.6412+j0.4797 0.6412+j0.4797 0.6412+j0.4797 0.6412+j0.4797 0.6412+j0.4797
0.6429+j0.4827 0.6429+j0.4827 0.6412+j0.4797 0.6437+j0.4842 0.6437+j0.4842
0.6429+j0.4827 0.6429+j0.4827 0.6412+j0.4797 0.6437+j0.4842 0.6437+j0.6175







152
Then, based on equation (13.43), the Thévenin voltage UThbus5 can be obtained as
UThbus5 = Z1(5, 1)Ibus1−1 = Z1(5, 1)
UTh
Z1(1, 1)
= 0.968 − 2.413◦
(13.56)
Based on equations (13.46)-(13.47), the positive-sequence impedance of the Thévenin
equivalent can be obtained from the Z-bus matrix of the positive-sequence system
with voltage source shortened. This implies that in forming Y∆1, the row and column
corresponding to the bus connected to voltage source (i.e. bus 1) in matrix Y1 are
removed. Thus,
Y∆1 =





Y 22−1 − 1
Z23−1
− 1
Z24−1
0
− 1
Z23−1
Y 33−1 0 0
− 1
Z24−1
0 Y 44−1 − 1
Zt2−1
0 0 − 1
Zt2−1
1
Zt2−1





(13.57)
The matrix Z∆1 can now be obtained as follows:
Z∆1 = Y−1
∆1 =




0.0018+j0.0423 0.0018+j0.0421 0.0018+j0.0423 0.0018+j0.0423
0.0018+j0.0421 0.0036+j0.0449 0.0018+j0.0421 0.0018+j0.0421
0.0018+j0.0423 0.0018+j0.0421 0.0026+j0.0438 0.0026+j0.0438
0.0018+j0.0423 0.0018+j0.0421 0.0026+j0.0438 0.0026+j0.1771



 (13.58)
Note that the element (4,4) corresponds to bus 5 since the row and column correspond-
ing to bus 1 is removed. This implies that
ZThbus5−1 = 0.0026 + j0.1771 (13.59)
The Thévenin impedance of the negative-sequences ZThbus5−2 can be calculated using
the corresponding matrix of the negative-sequence. The only difference between the
positive- and negative-sequence networks is that there is no voltage source in the
negative-sequence system.
Since all impedances (and thereby all admittances) in positive- and negative-sequence
networks are identical, the following is valid
Y∆2 = Y∆1 (13.60)
Thus,
ZThbus5−2 = ZThbus5−1 = 0.0026 + j0.1771 (13.61)
The Y-bus matrix of the zero-sequence is different compared with the other sequences,
both owing to different numerical values but also because of the zero-sequence connec-
tions in transformers and loads.
Y∆0 =





Y 22−0 − 1
Z23−0
− 1
Z24−0
0
− 1
Z23−0
Y 33−0 0 0
− 1
Z24−0
0 Y 44−0 − 1
Zt2−0
0 0 − 1
Zt2−0
1
Zt2−0





(13.62)

153
where
Y 22−0 =
1
Zt1−0
+
1
Z23−0
+ ysh−23−0 +
1
Z24−0
+ ysh−24−0
Y 33−0 =
1
Z23−0
+ ysh−23−0
Y 44−0 =
1
Z24−0
+ ysh−24−0 +
1
Zt2−0
Corresponding Z-bus matrix is obtained as the inverse
Z∆0 = Y−1
∆0 =




0.0000+j0.0438 0.0000+j0.0438 0.0000+j0.0438 0.0000+j0.0438
0.0000+j0.0438 0.0051+j0.0528 0.0000+j0.0438 0.0000+j0.0438
0.0000+j0.0438 0.0000+j0.0438 0.0026+j0.0483 0.0026+j0.0483
0.0000+j0.0438 0.0000+j0.0438 0.0026+j0.0483 0.0026+j0.1816



 (13.63)
Note that element (4,4) corresponds to bus 5. Thus,
ZThbus5−0 = 0.0026 + j0.1816 (13.64)
Next, since all Th´evenin equivalents as seen from bus 5 are identiﬁed, the voltage vector
and impedance matrix expressed in equation (13.49) can now be obtained as follows:
Upre(5) =


UThbus5
0
0

 =


0.968 − 2.413◦
0
0


Z∆(5, 5) =


ZThbus5−1 0 0
0 ZThbus5−2 0
0 0 ZThbus5−0

 = (13.65)
=


0.0026 + j0.1771 0 0
0 0.0026 + j0.1771 0
0 0 0.0026 + j0.1816


3) Determine the symmetrical components of the unsymmetrical load by using equation
(13.51) and ZLD2pu from the solution to Example 7.3.
ZLDbus5s = T−1
ZLDbus5ph
T = T−1


2ZLD2pu 0 0
0 ZLD2pu 0
0 0 ZLD2pu

 T =
=


3.0083 + j0.9888 0.7521 + j0.2472 0.7521 + j0.2472
0.7521 + j0.2472 3.0083 + j0.9888 0.7521 + j0.2472
0.7521 + j0.2472 0.7521 + j0.2472 3.0083 + j0.9888

 (13.66)
The symmetrical components of the currents through the load can now be determined
by using equations (13.52) and (13.65).
I∆(5) = − [Z∆(5, 5) + ZLDbus5s]−1
U(5) =


0.3315 155.8442◦
0.0653 − 26.5244◦
0.0653 − 26.6221◦

 =


I∆1(5)
I∆2(5)
I∆0(5)


(13.67)

154
4) The symmetrical components of all voltages can be calculated by using equations
(13.43) and (13.53).
U1(2) = Z1(2, 1) · Ibus1−1 + Z∆1(2, 5)I∆1(5) = 0.9618 − 3.1760◦
U2(2) = Z∆2(2, 5)I∆2(5) = 0.0028 61.0623◦
U0(2) = Z∆0(2, 5)I∆0(5) = 0.0029 63.3779◦
U1(3) = Z1(3, 1) · Ibus1−1 + Z∆1(3, 5)I∆1(5) = 0.9581 − 3.2745◦
U2(3) = Z∆2(3, 5)I∆2(5) = 0.0028 60.9638◦
U0(3) = Z∆0(3, 5)I∆0(5) = 0.0029 63.3778◦
(13.68)
U1(4) = Z1(4, 1) · Ibus1−1 + Z∆1(4, 5)I∆1(5) = 0.9614 − 3.1976◦
U2(4) = Z∆2(4, 5)I∆2(5) = 0.0029 60.0356◦
U0(4) = Z∆0(4, 5)I∆0(5) = 0.0032 60.3527◦
U1(5) = Z1(5, 1) · Ibus1−1 + Z∆1(5, 5)I∆1(5) = 0.9465 − 5.6970◦
U2(5) = Z∆2(5, 5)I∆2(5) = 0.0116 62.6242◦
U0(5) = Z∆0(5, 5)I∆0(5) = 0.0119 62.5733◦
Note that the element numbers given in the above equations are the bus numbers.
However, for Z∆ matrices, since the row and column corresponding to the bus con-
nected to the voltage source (in this example bus 1) are removed, Z∆(k, r) corresponds
to the element Z∆(k − 1, r − 1), i.e. with Z∆(2, 5) it means the element Z∆(1, 4).
Since ¯Zt and ¯ysh are lossless, the system losses are in the lines, i.e. ¯Z23 and ¯Z24. The
positive-, negative- and zero-sequence currents through these impedances are expressed
by
IZ23−1 =
U1(2) − U1(3)
Z23−1
= 1.1972 − 401209◦
IZ23−2 =
U2(2) − U2(3)
Z23−2
= 0.0034 − 24.1174◦
IZ23−0 =
U0(2) − U0(3)
Z23−0
= 0.0000 153.3778◦
(13.69)
IZ24−1 =
U1(2) − U1(4)
Z24−1
= 0.3314 − 24.1061◦
IZ24−2 =
U2(2) − U2(4)
Z24−2
= 0.0653 153.4756◦
IZ24−0 =
U0(2) − U0(4)
Z24−0
= 0.0653 153.3779◦
Positive-, negative- and zero-sequence powers injected into the line with impdance ¯Z23
are given by
SZ23−1 = U1(2)I
∗
Z23−1 = 0.9203 + j0.6921
SZ23−2 = U2(2)I
∗
Z23−2 = (7.60 + j5.72) × 10−6
(13.70)
SZ23−0 = U0(2)I
∗
Z23−0 = 4.24 × 10−15
+ j2.61 × 10−15

155
By using equation (10.25), the total power flowing through ¯Z23 is given by
SZ23in
= SZ23−1 + SZ23−2 + SZ23−0 = 0.4602 + j0.3461 (13.71)
In a similar manner, the following can also be obtained:
SZ24in
= U1(2)I
∗
Z24−1 + U2(2)I
∗
Z24−2 + U0(2)I
∗
Z24−0 = 0.1489 + j0.0567
SZ23out = U1(3)I
∗
Z23−1 + U2(3)I
∗
Z23−2 + U0(3)I
∗
Z23−0 = 0.4589 + j0.3461
SZ24out = U1(4)I
∗
Z24−1 + U2(4)I
∗
Z24−2 + U0(4)I
∗
Z24−0 = 0.1488 + j0.0567
The efficiency can now be obtained as:
η = 100 ·
Real(SZ23out ) + Real(SZ24out )
Real(SZ23in
) + Real(SZ24in
)
= 99.7911% (13.72)
The efficiency can also be calculated as follows:
Sinj = UTh
UTh − U1(2)
Zt1−1
∗
+ U2(2)
0 − U2(2)
Zt1−2
∗
+ U0(2)
0 − U0(2)
Zt1−0
∗
where, Sinj is the total power injected into the system by the infinite bus. Next, the
total load is calculated as follows:
SLDtot =
|U1(3)|2
ZLD1−1
+
|U2(3)|2
ZLD1−2
+0+U1(5)[−I∆1(5)]∗
+U2(5)[−I∆2(5)]∗
+U0(5)[−I∆0(5)]∗
Thus,
η = 100 ·
Real(SLDtot )
Real(Sinj)
% (13.73)
Note that (13.72) is valid if the losses are only in the lines. However, (13.73) is a
general expression regardless of where the losses are.

Chapter 14
Power system harmonics
Currents and voltages that are not pure sinusoidal occur when non-linear components are
connected to the power system. Non-linear components gives for example that a complex
notation of the voltage drop
U = ZI (14.1)
over a component cannot be used directly since the component itself cannot be describes
as an impedance. Examples of non-linear components are power electronic devices and
transformer operating with saturation.
The influence non-linear components may have on a power system will be illustrated by an
example. Assume that a symmetric three-phase voltage, UM = 400 V , f = 50 Hz, feeds a
resistive, Y-connected three-phase load, R = 400 Ω, via anti-parallel connected thyristors as
given in Figure 14.1.
ua(t)
ub(t)
uc(t)
ia(t)
ib(t)
ic(t)
i0(t)
u0=0
T1
T4
T2
T3
T6
R
R
R
T5
Figure 14.1. Three-phase load fed via thyristors all having the same firing angle α
Thyristors are used in power systems to control the power flow. A thyristor is a controllable
diode where the firing angle α denotes the delay after the voltage zero crossing that the
thyristor starts to behave as a conductor. Anti-parallel connected thyristors are used in e.g.
dimmers that are used to control the light from a lamp. In Figure 14.2 the voltage in phase
a in Figure 14.1 is shown as well as how the current changes when the firing angle of the
thyristors T1 and T2 is changed. As indicated in Figure 14.2, ia decreases as the firing
angle α increases. The firing angle α can be increased to a maximum of 180◦
, i.e. half cycle.
The firing angle α = 45◦
means that the thyristor is ignited and behaves as a conductor
after 45
360
= 1
8
cycle, i.e. after 0.0025 s. When i2
adt decreases, the mean power delivered is
also decreased, i.e. the thyristors controls the power flow. The figure also shows that when
the firing angle α = 0◦
, the current is not purely sinusoidal.
If all thyristors in Figure 14.1 have the firing angle α = 0◦
, the phase currents ia, ib and
ic will be sinusoidal and thereby the load will draw a sinusoidal current. The current i0 is
equal to zero.
157

158
Assume that all thyristors instead have a firing angle of 45◦
. The three phase currents as
well as the current in the neutral will have the shape shown in Figure 14.3.
As given in Figure 14.3, the occurrence of thyristors gives a current in the neutral conductor.
If the firing angle increases in the circuit, Figure 14.1, the shape of the currents will also
change. In Figure 14.4 the currents when having a firing angle of 135◦
at all thyristors are
shown.
0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04
-500
0
500
ua
0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04
-1
0
1
ia_0
0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04
-1
0
1
ia_45
0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04
-1
0
1
ia_135
Figure 14.2. The voltage ua(t) and the current ia at a firing angle of 0◦, 45◦ and
135◦, respectively, of the thyristors T1 and T2
0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04
-1
0
1
T1
T2
T1
T2
ia(t)
0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04
-1
0
1
T4
T3 T3
ib(t)
T4
0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04
-1
0
1
T6
T5
T6
T5
ic(t)
0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04
-1
0
1
i0(t)
Figure 14.3. Currents in the circuit given in figure 14.1 when the firing angle is 45◦
In the analysis of not purely sinusoidal quantities in the power system, Fourier analysis is
often used. The quantity of interest is then studied as the sum of sinusoidal quantities with

159
a fundamental frequency and multiples of it, so called harmonics. The current in e.g. phase
a can then be calculated as
ia(t) = a0 +
∞
n=1
an cos(nωt + γn) (14.2)
The amplitudes an and phase angles γn of the fundamental frequency as well as harmonics
of the current, can be calculated by using Fast Fourier Transform, FFT, using samples of
the waveform of the current as given in Figure 14.3.
0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04
-1
0
1
ia(t)
0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04
-1
0
1
ib(t)
0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04
-1
0
1
ic(t)
0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04
-1
0
1
i0(t)
Figure 14.4. Currents in the circuit according to figure 14.1 when the firing angle is 135◦
0 500 1000 1500
0
0.5
1
Phasea
0 500 1000 1500
0
0.5
1
Phaseb
0 500 1000 1500
0
0.5
1
Phasec
0 500 1000 1500
0
0.5
1
Return
Figure 14.5. Frequency spectra of the currents shown in figure 14.3
In Figure 14.5,the result of a Fourier analysis of the currents given in Figure 14.3 is shown.
As shown in the figure, the harmonics in each phase are the same since the waveform of

160
the currents are the same in the three phases. For the current i0, the rate of harmonics is
different. The rate of harmonics is also given in Table 14.1 which also includes an analysis
of the system condition at a firing angle of 135◦
.
α = 45◦
α = 135◦
ia i0 ia i0
Order Hz A % of 50 Hz A % of 50 Hz A % of 50 Hz A % of 50 Hz
1 50 0.921 100.0 0.004 0.9 0.187 100.0 0.004 0.9
2 100 0.000 0.0 0.000 0.0 0.000 0.0 0.000 0.0
3 150 0.162 17.6 0.480 100.0 0.162 86.6 0.480 100.0
4 200 0.000 0.0 0.000 0.0 0.000 0.0 0.000 0.0
5 250 0.120 13.0 0.004 0.8 0.120 64.2 0.004 0.8
6 300 0.000 0.0 0.000 0.0 0.000 0.0 0.000 0.0
7 350 0.075 8.1 0.004 0.9 0.075 40.2 0.004 0.9
8 400 0.000 0.0 0.000 0.0 0.000 0.0 0.000 0.0
9 450 0.045 4.9 0.135 28.1 0.045 24.1 0.135 28.1
10 500 0.000 0.0 0.000 0.0 0.000 0.0 0.000 0.0
11 550 0.039 4.2 0.004 0.8 0.039 20.9 0.004 0.8
12 600 0.000 0.0 0.000 0.0 0.000 0.0 0.000 0.0
13 650 0.038 4.2 0.004 0.9 0.038 20.6 0.004 0.9
14 700 0.000 0.0 0.000 0.0 0.000 0.0 0.000 0.0
15 750 0.032 3.5 0.096 20.1 0.032 17.3 0.096 20.1
16 800 0.000 0.0 0.000 0.0 0.000 0.0 0.000 0.0
17 850 0.025 2.7 0.004 0.8 0.025 13.4 0.004 0.8
18 900 0.000 0.0 0.000 0.0 0.000 0.0 0.000 0.0
19 950 0.023 2.5 0.004 0.9 0.023 12.4 0.004 0.9
20 1000 0.000 0.0 0.000 0.0 0.000 0.0 0.000 0.0
THD 100-∞ 0.171 25.4 0.371 100.0 0.171 79.2 0.371 100.0
RMS 50-∞ 0.673 100.0 0.371 100.0 0.216 100.0 0.371 100.0
Table 14.1. Harmonics in the phase currents in the circuit given in Figure 14.1 at
α = 45◦ and α = 135◦
As given in table 14.1, no even harmonics exist. This owing to the anti-parallel connected
thyristors in the three phases that all have the same firing angle giving that the positive
and the negative part of the current have the same waveform, i.e. a mirror image along the
time-axis, see Figure 14.3 and 14.4. In Table 14.1, the Total Harmonic Distortion, THD, is
given which is an RMS-value of the total harmonics.
THD =
1
2
∞
n=2
a2
n (14.3)
As given in Table 14.1, the harmonics in the neutral conductor is (mainly) the odd multiples
of three times the fundamental frequency, i.e. 3, 9, 15, etc. (The low percentage levels
given for other harmonics in the table are due to limitations in the calculation routine used.)
Also the amplitude of the harmonics are higher in the neutral conductor compared with the
phase conductors. The reason for this is that the harmonics in the three phases have the
same phase position since the phase displacement of 120◦
is an even number of cycles of the

161
frequency, because
i0−3n = ia−3n + ib−3n + ic−3n =
= an cos(3nωt + γn) + an cos(3nωt + γn + 120◦
) + (14.4)
+ an cos(3nωt + γn − 120◦
) = 3an cos(3nωt + γn)
For the other harmonics, the current in the different phases are cancelling one another in
the neutral conductor since each of them represents a symmetrical phase sequence
0 20 40 60 80 100 120 140 160 180
0
0.5
1
1.5
2
RMS(ia) RMS(i0)
RMS(i0) / RMS(ia)
alfa [degrees]
Figure 14.6. RMS(ia) (’—’), RMS(i0) (’- - -’), and the quotient RMS(i0)/RMS(ia) (’...’)
i0−(3n±1) = ia−(3n±1) + ib−(3n±1) + ic−(3n±1) =
= an cos(3nωt + γn) + an cos(3nωt + γn ± 120◦
) + (14.5)
+ an cos(3nωt + γn 120◦
) = 0
Note that if no neutral conductor existed, no harmonics having a multiple of three times the
fundamental frequency could occur in the phase conductors either. Since the harmonics of
multiples of three times the fundamental frequency has the same phase position, they will
appear as zero-sequence components which implies that they cannot pass ∆-Y-connected
transformers.
The firing angle of the thyristors will of course influence the currents in the phases as well as
in the neutral conductor. In Figure 14.6, the change in the RMS-value of the current when
the firing angle is increased is shown.
As shown in Figure 14.6, the i0-current is relatively small for firing angles up to approximately
20◦
. When the firing angle has increased to about 70◦
, the RMS-values of the current in the
neutral conductor is as large as the RMS-value of the phase current. The RMS-value of the
current in the neutral conductor reach a maximum value when the firing angle is α = 90◦
and for even higher firing angles, the current in the neutral conductor is
√
3 times the phase
current.

Appendix A
MATLAB-codes for Examples 7.2, 7.3, 13.2
and 13.3
%--- Example 7.2
% File name: Ex7_2.m
clear%
deg=180/pi;%
rad=1/deg;
% Choose the base values
Sb=0.5; Ub10=10; Ib10=Sb/Ub10/sqrt(3);Zb10=Ub10^2/Sb;%
Ub70=70; Ib70=Sb/Ub70/sqrt(3);
%Calculate the per-unit values of the Th´evenin equivalent of the system
UTh=70*exp(j*0*rad); Isc=0.3*exp(j*-90*rad);%
UThpu =UTh/Ub70; Iscpu =Isc/Ib70; ZThpu =UThpu/Iscpu;
% Calculate the per-unit values of the transformer
Zt=j*4/100;Snt=5; Ztpu=Zt*Sb/Snt;
%Calculate the per-unit values of the line
Z21pu=5*(0.9+j*0.3)/Zb10; ysh21pu=5*(j*3*1E-6)*Zb10/2;
%Calculate the per-unit values of the industry impedance
cosphi=0.8; sinphi=sqrt(1-cosphi^2);%
Un=Ub10; PLD=0.4;absSLD=PLD/cosphi; SLD=absSLD*(cosphi+j*sinphi);%
ZLDpu=Un^2/conj(SLD)/Zb10
% The twoport of the system
AL=1+ysh21pu*Z21pu; BL=Z21pu; CL=ysh21pu*(2+ysh21pu*Z21pu); DL=AL;
F_L=[AL BL ; CL DL]; F_Th_tr=[1 ZThpu+Ztpu ; 0 1]; F_tot=F_Th_tr*F_L;
%The impedance of the entire system
Ztotpu=(F_tot(1,1)*ZLDpu+F_tot(1,2))/(F_tot(2,1)*ZLDpu+F_tot(2,2));%
I4pu = UThpu/Ztotpu;
%The power fed by the transformer into the line
U2pu_I2pu=inv(F_Th_tr)*[UThpu;I4pu];%
S2=U2pu_I2pu(1,1)*conj(U2pu_I2pu(2,1))*Sb
%The voltage at the industry
U1pu_I1pu=inv(F_tot)*[UThpu;I4pu];%
163

164
U1=abs(U1pu_I1pu(1,1))*Ub10,
%--- Example 7.3
clear%
deg=180/pi;%
rad=1/deg;
% Choose the base values
Sb=0.5; Ub70=70; Ib70=Sb/Ub70/sqrt(3);%
Ub10=10; Ib10=Sb/Ub10/sqrt(3);Zb10=Ub10^2/Sb;%
Ub04=04; Ib04=Sb/Ub04/sqrt(3);Zb04=Ub04^2/Sb;
%Calculate the per-unit values of the infinite bus
U1=70/Ub70;
%Calculate the per-unit values of the transformer T1 and T2
Zt1=j*7/100;Snt1=0.8; Zt1pu=Zt1*Sb/Snt1;%
Zt2=j*8/100;Snt2=0.3; Zt2pu=Zt2*Sb/Snt2;
%Calculate the per-unit values of Line1and Line2
Z23pu=2*[0.17+j*0.3]/Zb10; ysh23pu=2*(3.2*1E-6)*Zb10/2;%
Z24pu=1*[0.17+j*0.3]/Zb10; ysh24pu=1*(3.2*1E-6)*Zb10/2;
%Calculate the per-unit values of the impedance LD1 and LD2
cosphiLD1=0.8; sinphiLD1=sqrt(1-cosphiLD1^2);%
UnLD1=Ub10; PLD1=0.5;absSLD1=PLD1/cosphiLD1; SLD1=absSLD1*(cosphiLD1+j*sinphiLD1);%
ZLD1pu=UnLD1^2/conj(SLD1)/Zb10
cosphiLD2=0.95; sinphiLD2=sqrt(1-cosphiLD2^2);%
UnLD2=Ub04;PLD2=0.2;absSLD2=PLD2/cosphiLD2; SLD2=absSLD2*(cosphiLD2+j*sinphiLD2);%
ZLD2pu=UnLD2^2/conj(SLD2)/Zb04
%Y-BUS
Y22=1/Zt1pu+1/Z23pu+ysh23pu+1/Z24pu+ysh24pu;%
Y33=1/Z23pu+ysh23pu+1/ZLD1pu;%
Y44=1/Z24pu+ysh24pu+1/Zt2pu;
Ybus=[ 1/Zt1pu -1/Zt1pu 0 0 0;
-1/Zt1pu Y22 -1/Z23pu -1/Z24pu 0;
0 -1/Z23pu Y33 0 0;
0 -1/Z24pu 0 Y44 -1/Zt2pu;
0 0 0 -1/Zt2pu 1/Zt2pu+1/ZLD2pu];
Zbus=inv(Ybus);
%Calculate the efficiency
I1=U1/Zbus(1,1);%

165
U2=Zbus(2,1)*I1;%
U3=Zbus(3,1)*I1;%
U4=Zbus(4,1)*I1;%
U5=Zbus(5,1)*I1;
S1=U1*conj(I1)*Sb;%
IZ23=(U2-U3)/Z23pu;%
IZ24=(U2-U4)/Z24pu;%
PfLine1=real(Z23pu)*abs(IZ23)^2*Sb;%
PfLine2=real(Z24pu)*abs(IZ24)^2*Sb;
eta=(real(S1)-PfLine1-PfLine2)/real(S1);
Zf4=0; YD=Ybus(2:5,2:5); ZD=inv(YD);%
Isc4=U4*Ib10/(Zf4+ZD(4-1,4-1));%
absIsc4=abs(Isc4)%
angIsc4=angle(Isc4)*deg
%--- Example 13.2
clear
% 1)
Ex7_2, % run Example 7.2
%Positive- and negative-sequence components in per-unit values (from the
%solution to Example 7.2):
UTh=UThpu; ZTh_1=ZThpu; ZTh_2=ZTh_1;%
Zt_1=Ztpu; Z21_1=Z21pu; Z21_2=Z21_1;%
ysh21_1=ysh21pu; ysh21_2=ysh21_1;
AL_1=AL ; AL_2=AL_1; BL_1=BL ; BL_2=BL_1;%
CL_1=CL ; CL_2=CL_1; DL_1=DL ; DL_2=DL_1;
A_1=F_tot(1,1) ; A_2=A_1; B_1=F_tot(1,2) ; B_2=B_1;%
C_1=F_tot(2,1) ; C_2=C_1; D_1=F_tot(2,2) ; D_2=D_1;
%Zero-sequence components in per-unit values
Isc1phi=0.2*exp(-j*90*rad)/Ib70; ZTh_0=3*UTh/Isc1phi-2*ZTh_1;%
Zt_0=Zt_1; Z21_0=3*Z21_1; ysh21_0=0.5*ysh21_1;
AL_0=1+ysh21_0*Z21_0; BL_0=Z21_0;%
CL_0=ysh21_0*(2+ysh21_0*Z21_0); DL_0=AL_0;

166
% 2)
UThbus1=UTh/A_1; ZThbus1_1=B_1/A_1; ZThbus1_2=ZThbus1_1;
F_tot_0=[ 1 Zt_0 ; 0 1]*[AL_0 BL_0 ; CL_0 DL_0];%
ZThbus1_0=F_tot_0(1,2)/F_tot_0(1,1);
% 3)
alfa=exp(j*120*rad); TT=[1 1 1 ; alfa^2 alfa 1 ; alfa alfa^2 1];%
ZLDs=inv(TT)*[2*ZLDpu 0 0 ; 0 ZLDpu 0 ; 0 0 ZLDpu]*TT;%
UTH=[UThbus1 ; 0 ; 0];%
Zs=[ZThbus1_1 0 0 ; 0 ZThbus1_2 0 ; 0 0 ZThbus1_0];%
Is=inv(Zs+ZLDs)*UTH;
% 4-5)
Ubus1s=ZLDs*Is;%
Ubus2_Ibus2_1=[AL_1 BL_1 ; CL_1 DL_1]*[Ubus1s(1) ; Is(1)];%
Sbus2_1=Ubus2_Ibus2_1(1,1)*conj(Ubus2_Ibus2_1(2,1))*Sb;%
Sbus2_2=Ubus2_Ibus2_2(1,1)*conj(Ubus2_Ibus2_2(2,1))*Sb;%
Sbus2_0=Ubus2_Ibus2_0(1,1)*conj(Ubus2_Ibus2_0(2,1))*Sb;
% 6)
Ubus1_Ph=[abs(TT*Ubus1s*Ub10/sqrt(3)) angle(TT*Ubus1s*Ub10/sqrt(3))*deg]%
Stot=Sbus2_1+Sbus2_2+Sbus2_0
%--- Example 13.3
% File name: Ex13_3
clear
Ex7_3, %Run Example 7.3
% 1)
% Positive- and negative-sequence components in per-unit values (from the
% solution to Example 7.3)
UTh=U1;
Zt1_1=Zt1pu ; Zt1_2=Zt1_1;%
Zt2_1=Zt2pu ; Zt2_2=Zt2_1;%
Z23_1=Z23pu ; Z23_2=Z23_1;%
ysh23_1=ysh23pu ; ysh23_2=ysh23_1;%
Z24_1=Z24pu ; Z24_2=Z24_1;%
ysh24_1=ysh24pu ; ysh24_2=ysh24_1;%
ZLD1_1=ZLD1pu ; ZLD1_2=ZLD1_1;%
%Zero-sequence components in per-unit values
Zt1_0=Zt1_1 ; Zt2_0=Zt2_1;%
Z23_0=3*Z23_1 ; ysh23_0=0.5*ysh23_1;%

167
Z24_0=3*Z24_1 ; ysh24_0=0.5*ysh24_1;%
% 2)
Y1=Ybus; Y1(5,5)=1/Zt2_1; Z1=inv(Y1);%
Ibus1_1=UTh/Z1(1,1); UThbus5=Z1(5,1)*Ibus1_1;%
YD1=Y1(2:5,2:5); ZD1=inv(YD1); ZD2=ZD1;
% 3)
ZThbus5_1=ZD1(5-1,5-1); ZThbus5_2=ZThbus5_1;
Y22_0=1/Zt1_0+1/Z23_0+ysh23_0+1/Z24_0+ysh24_0;%
Y33_0=1/Z23_0+ysh23_0;%
Y44_0=1/Z24_0+ysh24_0+1/Zt2_0;
YD0=[ Y22_0 -1/Z23_0 -1/Z24_0 0;
-1/Z23_0 Y33_0 0 0;
-1/Z24_0 0 Y44_0 -1/Zt2_0;
0 0 -1/Zt2_0 1/Zt2_0];
ZD0=inv(YD0); ZThbus5_0=ZD0(5-1,5-1);
UPre_5=[UThbus5 ; 0 ; 0];%
ZD_5=[ZThbus5_1 0 0 ; 0 ZThbus5_2 0 ; 0 0 ZThbus5_0];%
alfa=exp(j*120*rad);%
TT=[1 1 1 ; alfa^2 alfa 1 ; alfa alfa^2 1];%
ZLDbus5s=inv(TT)*[2*ZLD2pu 0 0 ; 0 ZLD2pu 0 ; 0 0 ZLD2pu]*TT;%
ID_5=-inv(ZD_5+ZLDbus5s)*UPre_5;
% 4)
Up2_1=Z1(2,1)*Ibus1_1+ ZD1(2-1,5-1)*ID_5(1);%
Up2_2= 0 + ZD2(2-1,5-1)*ID_5(2);%
Up2_0= 0 + ZD0(2-1,5-1)*ID_5(3);%
Up3_1=Z1(3,1)*Ibus1_1+ ZD1(3-1,5-1)*ID_5(1);%
Up3_2= 0 + ZD2(3-1,5-1)*ID_5(2);%
Up3_0= 0 + ZD0(3-1,5-1)*ID_5(3);%
Up4_1=Z1(4,1)*Ibus1_1+ ZD1(4-1,5-1)*ID_5(1);%
Up4_2= 0 + ZD2(4-1,5-1)*ID_5(2);%
Up4_0= 0 + ZD0(4-1,5-1)*ID_5(3);%
Up5_1=Z1(5,1)*Ibus1_1+ ZD1(5-1,5-1)*ID_5(1);%
Up5_2= 0 + ZD2(5-1,5-1)*ID_5(2);%
Up5_0= 0 + ZD0(5-1,5-1)*ID_5(3);
IZ23_1=(Up2_1-Up3_1)/Z23_1; IZ23_2=(Up2_2-Up3_2)/Z23_2; IZ23_0=(Up2_0-Up3_0)/Z23_0;%
IZ24_1=(Up2_1-Up4_1)/Z24_1; IZ24_2=(Up2_2-Up4_2)/Z24_2; IZ24_0=(Up2_0-Up4_0)/Z24_0;
SZ23_1=Up2_1*conj(IZ23_1); SZ23_2=Up2_2*conj(IZ23_2); SZ23_0=Up2_0*conj(IZ23_0);%
SZ23_in=SZ23_1+SZ23_2+SZ23_0;%
SZ24_in=Up2_1*conj(IZ24_1)+Up2_2*conj(IZ24_2)+Up2_0*conj(IZ24_0);

168
SZ23_out=Up3_1*conj(IZ23_1)+Up3_2*conj(IZ23_2)+Up3_0*conj(IZ23_0);%
SZ24_out=Up4_1*conj(IZ24_1)+Up4_2*conj(IZ24_2)+Up4_0*conj(IZ24_0);
eta=100*(real(SZ23_out)+real(SZ24_out))/(real(SZ23_in)+real(SZ24_in))
Sinj=UTh*conj((UTh-Up2_1)/Zt1_1)+...
Up2_2*conj((0-Up2_2)/Zt1_2)+...
Up2_0*conj((0-Up2_0)/Zt1_0);
SLDtot=abs(Up3_1)^2/ZLD1_1+abs(Up3_2)^2/ZLD1_2+0-...
Up5_1*conj(ID_5(1))-Up5_2*conj(ID_5(2))-Up5_0*conj(ID_5(3));
eta=100*real(SLDtot)/real(Sinj)

Appendix B
Matlab-codes for Example 8.10
% Start of file
clear,
clear global
deg=180/pi;
maxiter=10;
EPS=1e-4;
k1=-0.2; k2=1.2; k3=-0.07; k4=0.4;
% Step 1
converged=0; iter=0; x=3/deg;
while ~converged & iter < maxiter,
% Step 2
delta_gx=k4-(k1*x+k2*cos(x-k3));
% Step Final
if all(abs(delta_gx)< EPS),
converged=1;
iter=iter,
xdeg=x*deg
else
% Step 3
Jac=k1-k2*sin(x-k3); %Jac=dfx/dx;
% Step 4
delta_x=inv(Jac)*delta_gx;
% Step 5
x=x+delta_x;
iter=iter+1;
end, % if all
if iter==maxiter,
iter=iter,
disp(’The equation has no solutions’)
disp(’or’)
disp(’bad initial value, try with another initial value’)
end, % iter
end, % while
% End of file
169

Appendix C
MATLAB-codes for Example 8.12
% Start of file
clear,%
clear global
Sb=100; Ub=220; deg=180/pi;
%Step 1
% 1a)
U1=1;theta1=0;PLD1=0.2;QLD1=0.02;%
U2=1;PG2=1;PLD2=2;QLD2=0.2;
%1b)
Z12=0.02+j*0.2;%
Y=[1/Z12 -1/Z12 ; -1/Z12 1/Z12];%
G=real(Y); B=imag(Y);%
PGD2=PG2-PLD2;
%1c)
theta2=0;
iter=0;%
while iter < 3,
iter=iter+1;
%Step 2
%2a)
P2=U1*U2*(G(2,1)*cos(theta2-theta1)+B(2,1)*sin(theta2-theta1))+U2^2*G(2,2);
%2b
deltaP=PGD2-P2;
%Step 3
Q2=U2*U1*(G(2,1)*sin(theta2-theta1)-B(2,1)*cos(theta2-theta1))-U2^2*B(2,2);
H=-Q2-U2^2*B(2,2);
JAC=[H];
%Step 4
DX=inv(JAC)*[deltaP];
delta_theta2=DX;
%Step 5
theta2=theta2+delta_theta2;
end,
%Step final
P1=U1*U2*(G(1,2)*cos(theta1-theta2)+B(1,2)*sin(theta1-theta2))+U1^2*G(1,1);%
171

172
PG1=(P1+PLD1)*Sb; QG1=(Q1+QLD1)*Sb; QG2=(Q2+QLD2)*Sb;
g=-G;b=-B;bsh_12=0;
P_12=(U1^2*g(1,2)-U1*U2*(g(1,2)*cos(theta1-theta2)+b(1,2)*sin(theta1-theta2)))*Sb;
Q_12=((-bsh_12-b(1,2))*U1^2-...
U1*U2*(g(1,2)*sin(theta1-theta2)-b(1,2)*cos(theta1-theta2)))*Sb;
Q_21=((-bsh_12-b(2,1))*U2^2-...
PLoss=P_12+P_21; %or PLoss=(PG1+PG2)-(PLD1+PLD2)*Sb;
ANG=[theta1 theta2]’*deg;
VOLT=[U1 U2]*Ub;
% End of file

Appendix D
% Start of file
clear,%
clear global
tole=1e-6;
Sb=100; Ub=220; deg=180/pi;
%Step 1
% 1a)
U1=1;theta1=0;PLD1=0.2;QLD1=0.02;
PG2=1;QG2=0.405255;PLD2=2;QLD2=0.2;
%1b)
Z12=0.02+j*0.2;
Y=[1/Z12 -1/Z12 ; -1/Z12 1/Z12];
G=real(Y); B=imag(Y);
PGD2=PG2-PLD2;
QGD2=QG2-QLD2;
%1c)
theta2=0;
U2=1;
deltaP=PGD2-P2;
deltaQ=QGD2-Q2;
while all(abs([deltaP;deltaQ])> tole),
%Step 2
%2a)
%2b
deltaP=PGD2-P2,
deltaQ=QGD2-Q2,
%Step 3
H=-Q2-B(2,2)*U2^2;
N=P2+G(2,2)*U2^2;
J=P2-G(2,2)*U2^2;
L=Q2-B(2,2)*U2^2;
JAC=[H N ; J L];
%Step 4
DX=inv(JAC)*[deltaP;deltaQ];
173

174
delta_theta2=DX(1);
delta_U2=DX(2);
%Step 5
theta2=theta2+delta_theta2;
U2=U2*(1+delta_U2/U2);
end, %while
%Step final
P1=U1*U2*(G(1,2)*cos(theta1-theta2)+B(1,2)*sin(theta1-theta2))+U1^2*G(1,1);%
PG1=(P1+PLD1)*Sb; QG1=(Q1+QLD1)*Sb; QG2=(Q2+QLD2)*Sb;
g=-G;b=-B;bsh_12=0;
Q_12=((-bsh_12-b(1,2))*U1^2-...
Q_21=((-bsh_12-b(2,1))*U2^2-...
PLoss=P_12+P_21; %or PLoss=(PG1+PG2)-(PLD1+PLD2)*Sb;
ANG=[theta1 theta2]’*deg;
VOLT=[U1 U2]*Ub;
% End of file

Appendix E
% To run Load Flow (LF), two MATLAB-files are used, namely
% (run_LF.m) and (solve_lf.m)
% Start of file (run_LF.m)
clear%
clear global
%%%%%%%%%%%%%%%%%%%%%%%%%%
tole=1e-9; deg=180/pi; rad=1/deg ;
%%%%%%%%%%%%%%
% Base values
%%%%%%%%%%%%%%
Sb=100; Ub=220; Zb=Ub^2/Sb;
%%%%%%%%%%%%
% Bus data
%%%%%%%%%%%%
% Number of buses
nbus=4;
%Bus 1, slack bus
U1=220/Ub; theta1=0*rad; PLD1=10/Sb; QLD1=2/Sb;
%Bus 2, PQ-bus
PG2=0/Sb; QG2=0/Sb; PLD2=90/Sb; QLD2=10/Sb;
%Bus 3, PQ-bus
%Bus 4, PQ-bus
%%%%%%%%%%%%
% Line data
%%%%%%%%%%%%
Z12=(5+j*65)/Zb;bsh12=0.0002*Zb;%
Z13=(4+j*60)/Zb;bsh13=0.0002*Zb;%
Z23=(5+j*68)/Zb;bsh23=0.0002*Zb;%
Z34=(3+j*30)/Zb;bsh34=0;
%%%%%%%%%%%%%%
% YBUS matrix
%%%%%%%%%%%%%%
y11=1/Z12+1/Z13+j*bsh12+j*bsh13; y12=-1/Z12; y13=-1/Z13; y14=0;%
y21=-1/Z12; y22=1/Z12+1/Z23+j*bsh12+j*bsh23; y23=-1/Z23; y24=0;%
y31=-1/Z13; y32=-1/Z23; y33=1/Z13+1/Z23+1/Z34+j*bsh13+j*bsh23; y34=-1/Z34;%
y41=0; y42=0; y43=-1/Z34; y44=1/Z34;%
YBUS=[y11 y12 y13 y14; y21 y22 y23 y24; y31 y32 y33 y34; y41 y42 y43 y44];%
G=real(YBUS); B=imag(YBUS);
175

176
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Define PGD for PU- and PQ-buses
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
PGD2=PG2-PLD2; % for bus 2 (PQ-bus)
PGD=[PGD2 ; PGD3 ; PGD4];
%%%%%%%%%%%%%%%%%%%%%%%%%%
% Define QGD for PQ-buses
%%%%%%%%%%%%%%%%%%%%%%%%%%
QGD2=QG2-QLD2; % for bus 2 (PQ-bus)
QGD=[QGD2 ; QGD3 ; QGD4];
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Use fsolve function in MATLAB to run load flow
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Unknown variables [theta2 theta3 theta4 U2 U3 U4]’;
% Define the initial values of the unknown variables
X0=[0 0 0 1 1 1]’; % Flat initial values
s_z=size(X0);
nx=s_z(1,1); % number of unknown variables
% The function below is used for fsolve (type "help fsolve" in MATLAB)
options_solve=optimset(’Display’,’off’,’TolX’,tole,’TolFun’,tole);
% Parameters used for fsolve
PAR=[nx ; nbus ; U1 ; theta1];%
[X_X]=fsolve(’solve_lf’,X0,options_solve,G,B,PGD,QGD,PAR);
% Solved variables X_X=[theta2 theta3 theta4 U2 U3 U4]’;
ANG=[theta1 X_X(1) X_X(2) X_X(3)]’;%
VOLT=[U1 X_X(4) X_X(5) X_X(6)]’,
ANG_deg=ANG*deg;%
VOLT_kV=VOLT*Ub;
%%%%%%%%%%%%%%%%%
%Jacobian matrix
%%%%%%%%%%%%%%%%%
for m=1:nbus
for n=1:nbus
PP(m,n)=VOLT(m)*VOLT(n)*(G(m,n)*cos(ANG(m)-ANG(n))+B(m,n)*sin(ANG(m)-ANG(n)));
QQ(m,n)=VOLT(m)*VOLT(n)*(G(m,n)*sin(ANG(m)-ANG(n))-B(m,n)*cos(ANG(m)-ANG(n)));
end, %for n
end, % for m
P=sum(PP’); Q=sum(QQ’);

177
for m=1:nbus
for n=1:nbus
if m==n,
H(m,m)=-Q(m)-B(m,m)*VOLT(m)*VOLT(m);
N(m,m)= P(m)+G(m,m)*VOLT(m)*VOLT(m);
J(m,m)= P(m)-G(m,m)*VOLT(m)*VOLT(m);
L(m,m)= Q(m)-B(m,m)*VOLT(m)*VOLT(m);
else
H(m,n)= VOLT(m)*VOLT(n)*(G(m,n)*sin(ANG(m)-ANG(n))-B(m,n)*cos(ANG(m)-ANG(n)));
N(m,n)= VOLT(m)*VOLT(n)*(G(m,n)*cos(ANG(m)-ANG(n))+B(m,n)*sin(ANG(m)-ANG(n)));
J(m,n)=-VOLT(m)*VOLT(n)*(G(m,n)*cos(ANG(m)-ANG(n))+B(m,n)*sin(ANG(m)-ANG(n)));
L(m,n)= VOLT(m)*VOLT(n)*(G(m,n)*sin(ANG(m)-ANG(n))-B(m,n)*cos(ANG(m)-ANG(n)));
end, %if
end, %for n
end, % for m
slack_bus=1; %bus 1
PU_bus=[ ]; % No buses
% Remove row and column corresponding to slack bus
H(slack_bus,:)=[ ]; H(:,slack_bus)=[ ];
% Remove row corresponding to slack bus
N(slack_bus,:)=[ ];
% Remove columns corresponding to slack bus and PU-buses
N(:,sort([slack_bus PU_bus]))=[ ];
% Remove rows corresponding to slack bus and PU-buses
J(sort([slack_bus PU_bus]),:)=[ ];
% Remove column corresponding to slack bus
J(:,slack_bus)=[ ];
%Remove rows and columns corresponding to slack bus and PU-buses
L(sort([slack_bus PU_bus]),:)=[ ]; L(:,sort([slack_bus PU_bus]))=[ ];
JAC=[H N ; J L],
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% The generated active and reactive power at slack bus and PU-buses
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
g=-G;b=-B;
% At slack bus, bus 1
P12=g(1,2)*VOLT(1)^2-...
VOLT(1)*VOLT(2)*(g(1,2)*cos(ANG(1)-ANG(2))+b(1,2)*sin(ANG(1)-ANG(2)));
P13=g(1,3)*VOLT(1)^2-...
Q12=(-bsh12-b(1,2))*VOLT(1)^2-...
VOLT(1)*VOLT(2)*(g(1,2)*sin(ANG(1)-ANG(2))-b(1,2)*cos(ANG(1)-ANG(2)));
Q13=(-bsh13-b(1,3))*VOLT(1)^2-...

178
VOLT(1)*VOLT(3)*(g(1,3)*sin(ANG(1)-ANG(3))-b(1,3)*cos(ANG(1)-ANG(3)));%
PG1=P12+P13+PLD1;%
QG1=Q12+Q13+QLD1;%
PG1_MW=PG1*Sb;%
QG1_MVAr=QG1*Sb;
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% or
%PG_slackbus_MW=(P(slack_bus)+PLD1)*Sb,
%QG_slackbus_MW=(Q(slack_bus)+QLD1)*Sb,
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%
% Losses
%%%%%%%%%
PLoss_tot=(PG1+PG2+PG3+PG4)-(PLD1+PLD2+PLD3+PLD4);%
PLoss_tot_MW=PLoss_tot*Sb;
P34=g(3,4)*VOLT(3)^2-...
P43=g(4,3)*VOLT(4)^2-...
VOLT(4)*VOLT(3)*(g(4,3)*cos(ANG(4)-ANG(3))+b(4,3)*sin(ANG(4)-ANG(3)));%
PLoss_Sys1=(P34+P43);%
PLoss_Sys1_MW=(P34+P43)*Sb;
PLoss_Sys2_MW=PLoss_tot_MW-PLoss_Sys1_MW;
% End of file (run_LF.m)
%%%%%%%%%%%%%%
% Second file
%%%%%%%%%%%%%%
% Start of file (solve_lf.m)
% This function solves g(x)=0 for x.
function [g_x]=solve_lf(X,G,B,PGD,QGD,PAR);
nx=PAR(1); nbus=PAR(2); U1=PAR(3); theta1=PAR(4);
PGD2=PGD(1); PGD3=PGD(2); PGD4=PGD(3); QGD2=QGD(1); QGD3=QGD(2); QGD4=QGD(3);
theta2=X(1); theta3=X(2); theta4=X(3); U2=X(4); U3=X(5); U4=X(6);
ANG=[theta1 theta2 theta3 theta4]’ ; VOLT=[U1 U2 U3 U4]’;

179
% We have nx unknown variables, therefore the
% size of g(x) is nx by 1.
g_x=zeros(nx,1);
for m=1:nbus
for n=1:nbus
end, %for n
end, % for m
P=sum(PP’)’;
Q=sum(QQ’)’;
% Active power mismatch (PU- and PQ-buses)
% Bus 2
g_x(1)=P(2)-PGD2;
% Bus 3
g_x(2)=P(3)-PGD3;
% Bus 4
g_x(3)=P(4)-PGD4;
% Reactive power mismatch (PQ-buses)
% Bus 2
g_x(4)=Q(2)-QGD2;
% Bus 3
g_x(5)=Q(3)-QGD3;
% Bus 4
g_x(6)=Q(4)-QGD4;
% End of file (solve_lf.m)

Appendix F
Note that only the changes of the MATLAB-codes compared to the MATLAB-codes for
Example 8.14 are given in this appendix.
% Changes in file run_LF.m
%%%%%%%%%%%%
% Bus data
%%%%%%%%%%%%
% Number of buses
nbus=4;
%Bus 3, PU-bus
PG3=0/Sb; U3=220/Ub; PLD3=80/Sb; QLD3=10/Sb;
%%%%%%%%%%%%
% Line data
%%%%%%%%%%%%
%%%%%%%%%%%%%%
% YBUS matrix
%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Define PGD for PU- and PQ-buses
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%
% Define QGD for PQ-buses
%%%%%%%%%%%%%%%%%%%%%%%%%%
QGD=[QGD2 ; QGD4];
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Use fsolve function in MATLAB to run load flow
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Unknown variables [theta2 theta3 theta4 U2 U4]’;
% Define the initial values of the unknown variables
X0=[0 0 0 1 1]’; % Flat initial values
% Parameters used for fsolve
PAR=[nx ; nbus ; U1 ; U3 ; theta1];%
181

182
% Solved variables X_X=[theta1 theta2 theta4 U2 U4]’;
ANG=[theta1 X_X(1) X_X(2) X_X(3)]’;%
VOLT=[U1 X_X(4) U3 X_X(5)]’;
%%%%%%%%%%%%%%%%%
%Jacobian matrix
%%%%%%%%%%%%%%%%%
slack_bus=1; %bus 1
PU_bus=[3]; %bus 3
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% The generated active and reactive power at slack bus and PU-buses
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
g=-G;b=-B;
% At slack bus, bus 1
% At PU-buses, bus 3
Q31=(-bsh13-b(3,1))*VOLT(3)^2-...
Q32=(-bsh23-b(3,2))*VOLT(3)^2-...
Q34=(-bsh34-b(3,4))*VOLT(3)^2-...
VOLT(3)*VOLT(4)*(g(3,4)*sin(ANG(3)-ANG(4))-b(3,4)*cos(ANG(3)-ANG(4)));%
QG3=Q31+Q32+Q34+QLD3;%
QG3_MVAr=QG3*Sb;
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% or
%QLD_pubus=[QLD3];
%QG_pubus_MVAr=(Q(PU_bus’)+QLD_pubus’)*Sb,
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%
% Losses
%%%%%%%%%
% End of file (run_LF.m)

183
%%%%%%%%%%%%%%
% Second file
%%%%%%%%%%%%%%
% Start of file (solve_lf.m)
function [g_x]=solve_lf(X,G,B,PGD,QGD,PAR);
nx=PAR(1); nbus=PAR(2); U1=PAR(3); U3=PAR(4); theta1=PAR(5);
PGD2=PGD(1); PGD3=PGD(2); PGD4=PGD(3); QGD2=QGD(1); QGD4=QGD(2);
theta2=X(1); theta3=X(2); theta4=X(3); U2=X(4); U4=X(5);
ANG=[theta1 theta2 theta3 theta4]’; VOLT=[U1 U2 U3 U4]’;
% We have nx unknown variables, therefore the
% size of g(x) is nx by 1.
g_x=zeros(nx,1);
for m=1:nbus
for n=1:nbus
end, %for n
end, % for m
P=sum(PP’)’;
Q=sum(QQ’)’;
% Active power mismatch (PU- and PQ-buses)
% Bus 2
g_x(1)=P(2)-PGD2;
% Bus 3
g_x(2)=P(3)-PGD3;
% Bus 4
g_x(3)=P(4)-PGD4;
% Reactive power mismatch (PQ-buses)
% Bus 2
g_x(4)=Q(2)-QGD2;
% Bus 4
g_x(5)=Q(4)-QGD4;
% End of file (solve_lf.m)

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