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Power System Analysis

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Library of Congress Cataloging-in-Publication Data

Saadat Hadi.
t

Power system analysis / Hadi Saadat

p. cm.
lncludes bibliographical references and indexo
ISBN 0-07-012235 O
1. Electric power systems. 2. System analysis. l. Title.
TK1011. S23 1999
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 CONTENTS

PREFACE xv

1 THE POWER SYSTEM: AN OVERVIEW 1

L 1 INTRODUCTION . . . . . . . . . . . . . . . . . . . . . . . . . 1
1.2 ELECTRIC INDUSTRY STRUCTURE . . . . . . . . . . . . .. 2
1.3 MODERN POWER SYSTEM . . . . . . . . . . . . . . . . . .. 4
1.3.1 GENERATION....................... 4
1.3.2 TRANSMISSION AND SUBTRANSMISSION . . . .. 6
1.3.3 DISTRffiUTION...................... 6
1.3.4 LOADS........................... 8
1.4 SYSTEM PROTECTION . . . . . . . . . . . . . . . . . . . . .. 11
1.5 ENERGY CONTROL CENTER . . . . . . . . . . . . . . . . .. 11
1.6 COMPUTER ANALYSIS . . . . . . . . . . . . . . . . . . . . . 11

2 BASIC PRINCIPLES 14
2.1 INTRODUCTION . . . . . . . . . . . . . . . . . . . . . . . . . 14
2.2 POWER IN SINGLE-PHASE AC CIRCUITS . . . . . . . . . .. 15
2.3 COMPLEX POWER . . . . . . . . . . . . . . . . . . . . . . .. 19
2.4 THE COMPLEX POWER BALANCE. . . . . . . . . . . . . .. 21
2.5 POWER FACTOR CORRECTION. . . . . . . . . . . . . . . .. 23
2.6 COMPLEX POWER FLOW . . . . . . . . . . . . . . . . . . .. 26
2.7 BALANCED THREE-PHASE CIRCUITS. . . . . . . . . . . .. 30
2.8 Y-CONNECTED LOADS . . . . . . . . . . . . . . . . . . . . . 32
2.9 a-CONNECTED LOAD S . . . . . . . . . . . . . . . . . . . .. 34
2.10 a-y TRANSFORMATION . . . . . . . . . . . . . . . . . . .. 35
2.11 PER-PHASE ANALYSIS. . . . . . . . . . . . . . . . . . . . .. 36
2.12 BALANCEDTHREE-PHASEPOWER . . . . . . . . . . . . .. 37
vii
vüi CONTENTS

3 GENERATOR AND TRANSFORMER MODELS;

THE PER"UNIT SYSTEM 48
3.1 INTRODUCTION . . . . . . . . . . . . . . . . . . . . . . . . . 48
3.2 SYNCHRONOUS GENERATORS. . . . . . . . . . . . . . . .. 49
3.2.1 GENERATORMODEL.................. 49
3.3 STEADY-STATE CHARACTERISTICS-
CYLINDRICAL ROTOR. . . . . . . . . . . . . . . . . . . . .. 56
3.3.1 POWER FACTOR CONTROL. . . . . . . . . . . . . .. 56
3.3.2 POWER ANGLE CHARACTERISTICS . . . . . . . .. 57
3.4 SALIENT-POLE SYNCHRONOUS GENERATORS . . . . . .. 62
3.5 POWER TRANSFORMER. . . . . . . . . . . . . . . . . . . .. 64
3.6 EQUIVALENT CIRCUIT OF A TRANSFORMER . . . . . . .. 64
3.7 DETERMINATION OF EQUIVALENT
CIRCUIT PARAMETERS . . . . . . . . . . . . . . . . . . . .. 68
3.8 TRANSFORMER PERFORMANCE . . . . . . . . . . . . . .. 70
3.9 THREE-PHASE TRANSFORMER CONNECTIONS . . . . . . , 74
3.9.1 THE PER-PHASE MODEL OF
A THREE-PHASE TRANSFORMER. . . . . . . . . .. 76
3.10 AUTOTRANSFORMERS . . . . . . . . . . . . . . . . . . . .. 77
3.10.1 AUTOTRANSFORMERMODEL. . . . . . . . . . . .. 81
3.11 THREE-WINDINGTRANSFORMERS . . . . . . . . . . . . .. 81
3.11.1 THREE-WINDINGTRANSFORMERMODEL . . . .. 82
3.12 VOLTAGE CONTROL OF TRANSFORMERS . . . . . . . . .. 83
3.12.1 TAPCHANGINGTRANSFORMERS . . . . . . . . .. 83
3.12.2 REGULATING TRANSFORMERS OR BOOSTERS .. 86
3.13 THE PER-UNIT SYSTEM . . . . . . . . . . . . . . . . . . . .. 88
3.14 CHANGE OF BASE . . . . . . . . . . . . . . . . . . . . . . .. 90

4 TRANSMISSION LlNE PARAMETERS 102

4.1 INTRODUCTION . . . . . . . . . . . . . . . . . . . . . . . .. 102
4.2 OVERHEAD TRANSMISSION UNES . . . . . . . . . . . . .. 103
4.3 LINE RESISTANCE . . . . . . . . . . . . . . . . . . . . . . .. 105
4.4 INDUCTANCE OF A SINGLE CONDUCTOR . . . . . . . . .. 106
4.4.1 INTERNAL INDUCTANCE. . . . . . . . . . . . . . .. 107
4.4.2 INDUCTANCE DUE TO EXTERNAL FLUX LINKAGE 108
4.5 INDUCTANCE OF SINGLE-PHASE LINES . . . . . . . . . .. 109
4.6 FLUX LINKAGE IN TERMS OF
SELF- AND MUTUAL INDUCTANCES . . . . . . . . . ... .. 110
4.7 INDUCTANCE OF THREE-PHASE
TRANSMISSION LINES . . . . . . . . . . . . . . . . . . . . . 112
4.7.1 SYMMETRICALSPACING................ 112
 CONTENTS ix

4.7.2 ASYMMETRICAL SPACING. . . . . . . . . . . . . .. 113

4.7.3 TRANSPOSE LINE . . . . . . . . . . . . . . . . . . .. 114
4.8 INDUCTANCE OF COMPOSITE CONDUCTORS . . . . . . .. 115
4.8.1 GMROFBUNDLEDCONDUCTORS . . . . . . . . . 118
4.9 INDUCTANCE OF THREE-PHASE
DOUBLE-CIRCUITLINES . . . . . . . . . . . . . . . . . . . . 119
4.10 LINE CAPACITANCE . . . . . . . . . . . . . . . . . . . . . . . 120
4.11 CAPACITANCEOF SINGLE-PHASELINES . . . . . . . . . . . 121
4.12 POTENTIALDIFFERENCEIN A
MULTICONDUCTOR CONFIGURATION . . . . . . . . . . .. 123
4.13 CAPACITANCEOFTHREE-PHASELINES . . . . . . . . . . . 124
4.14 EFFECTOFBUNDLING . . . . . . . . . . . . . . . . . . . . . 126
4.15 CAPACITANCE OF THREE-PHASE
DOUBLE-CIRCUIT LINES . . . . . . . . . . . . . . . . . . .. 126
4.16 EFFECT OF EARTH ON THE CAPACITANCE .. . . . . . .. 127
4.17 MAGNETIC FIELD INDUCTION . . . . . . . . . . . . . . . .. 133
4.18 ELECTROSTATICINDUCTION . . . . . . . . . . . . . . . . . 135
4.19 CORONA . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 135

5 LINE MODEL AND PERFORMANCE 142

5.1 INTRODUCTION . . . . . . . . . . . . . . . . . . . . . . . . . 142
5.2 SHORT LINE MODEL . . . . . . . . . . . . . . . . . . . . . .. 143
5.3 MEDIUM LINE MODEL . . . . . . . . . . . . . . . . . . . . . 147
5.4 LONG LINEMODEL . . . . . . . . . . . . . . . . . . . . . . . 151
5.5 VOLTAGE AND CURRENT WAVES . . . . . . . . . . . . . .. 156
5.6 SURGE IMPEDANCE LOADING. . . . . . . . . . . . . . . .. 159
5.7 COMPLEX POWER FLOW
THROUGH TRANSMISSION LINES . . . . . . . . . . . . . .. 161
5.8 POWER TRANSMISSION CAPABILITY . . . . . . . . . . . . 163
5.9 LINE COMPENSATION . . . . . . . . . . . . . . . . . . . . .. 165
5.9.1 SHUNTREACTORS . . . . . . . . . . . . . . . . . . . . 165
5.9.2 SHUNT CAPACITOR COMPENSATION . . . . . . .. 168
5.9.3 SERIES CAPACITOR COMPENSATION . . . . . . . . 168
5.10 LINEPERFORMANCEPROGRAM . . . . . . . . . . . . . . . 171

6 POWER FLOW ANALYSIS 189

6.1 INTRODUCTION . . . . . . . . . . . . . . . . . . . . . . . . . 189
6.2 BUS ADMITTANCEMATRIX . . . . . . . . . . . . . . . . . . 190
6.3 SOLUTION OF NONLINEAR
ALGEBRAIC EQUATIONS . . . . . . . . . . . . . . . . . . .. 195
6.3.1 GAUSS-SEIDELMETHOD................ 195
6.3.2 NEWTON-RAPHSON METHOD . . . . . . . . . . . . , 200
4 L THE POWER SYSTEM: AN OVERVIEW

ties for power systems research and open up new opportunities to young power
engmeers.

1.3 MODERN POWER SYSTEM

The power system of today is a complex interconnected network as shown in Figure
1.1 (page 7). A power system can be subdivided into four major parts:

• Generation
• Transmission and Subtransmission
• Distribution
• Loads

1.3.1 GENERATION
Generators - One of the essential components of power systems is the three-
phase ac generator known as synchronous generator or alternator. Synchronous
generators have two synchronously rotating fields: One field is produced by the
rotor driven at synchronous speed and excited by dc current. The other field is pro-
duced in the stator windings by the three-phase armature currents. The de current
for the rotor windings is provided by excitation systems. In the older units, the ex-
citers are de generators mounted on the same shaft, providing excitation through
slip rings. Today's systems use ac generators with rotating rectifiers, known as
brushless excitation systems. The generator excitation system maintains generator
voltage and controls the reactive power fiow. Because they lack the commutator,
ac generators can generate high power at high voltage, typically 30 kV. In a power
plant, the size of generators can vary from 50 MW to 1500 MW.
The source of the mechanical power, eommonly known as the prime mover,
may be hydraulic turbines at waterfalls, steam turbines whose energy comes from
the burning of coal, gas and nuclear fuel, gas turbines, or occasionally internal com-
bustion engines burning oil. The estimated installed generation capacity in 1998 for
the United States is presented in Table 1.1.
Stearn turbines operate al relatively high speeds of 3600 or 1800 rpm. The
generators to which they are coupled are eylindrical rotor, two-pole for 3600 rpm or
four-pole for 1800 rpm operation. Hydraulic turbines, particularly those operating
with a low pressure, operate at low speed. Their generators are usually a salient
type rotor with many poles. In a power station several generators are operated in
parallel in the power grid to provide the total power needed. They are connected at
a common point called a bus.
 1.3. MODERN POWER SYSTEM 5

Today the total installed electric generating capacity is about 760,000 MW.
Assuming the United States population to be 270 million,
. . 760 X 109
Installed capaclty per caplta = 270 x 106 = 2815 W

To realize the significance of this figure, consider the average power of a

person to be approximately 50 W. Therefore, the power of 2815 W is equivalent to
2815 W
50 W = 56 (power slave)

The annual kWh consumption in the United States is about 3, 550 x 109 kWh.
The asset of the investment for investor-owned companies is about 200 billion dol-
lars and they employ close to a half million people.
With today's emphasis on environmentaI consideration and conservation of
fossil fuels, many alternate sources are considered for employing the untapped
energy sources of the sun and the earth for generation of power. Sorne of these
alternate sources which are being used to sorne extent are solar power, geothermal
power, wind power, tidal power, and biomass. The aspiration for bulk generation
of power in the future is the nuclear fusiono If nuclear fusion is harnessed economi-
cally, it would provide clean energy from an abundant source of fuel, namely water.

Table 1.1 Installed Generation Capacity

Type Capacity, Percent Fuel

MW
Steam Plant 478,800 63 Coal, gas, petroleum
Nuclear 106,400 14 Uranium
Hydro and pumped storage 91,200 12 Water
Gas Turbine 60,800 8 Gas, petroleum
Combined cycle 15,200 2 Gas, petroleum
Internal Combustion 4,940 0.65 Gas, petroleum
Others 2,660 0.35 Geothennal, solar, wind
Total 760,000 100.00
Transformers - Another major component of a power system is the transfonner.
It transfers power with very high efficiency from one level of voltage to another
leve!. The power transferred to the secondary is almost the same as the primary,
except for losses in the transfonner, and the product VI on the secondary side is
approximately the same as the primary side. Therefore, using a step-up transfonner
of turns ratio a will reduce the secondary current by a ratio of l/a. This will re-
duce losses in the Une, which makes the transmission of power over long distances
possible.
6 l. THE POWER SYSTEM: AN OVERVIEW

The insulation requirements and other practical desígn problems limít the
generated voltage to low values, usually 30 kV. Thus, step-up transformers are
used for transmission of power. At the receiving end of the transmission lines step-
down transformers are used to reduce the voltage to suitable values for distribution
or utilízation. In a modem utilíty system, the power may undergo four or five trans-
formations between generator and ultimate user.

1.3.2 TRANSMISSION AND SUBTRANSMISSION

The purpose of an overhead transmission network is to transfer electric energy
from generating units at various locations to the distribution system which ulti-
mately supplies the load. Transmission lines also interconnect neighboring utilities
which permits not only economic dispatch of power within regions during normal
conditions, but also the transfer of power between regions during emergencies.
Standard transmission voltages are established in the United States by the
American National Standards Institute (ANSI). Transmission voltage lines operat-
ing at more than 60 kV are standardized at 69 kV, 115 kV, 138 kV, 161 kV, 230 kV,
345 kV, 500 kV, and 765 kV Hne-to-line. Transmission voltages aboye 230 kV are
usually referred to as extra-high voltage (EHV).
Figure 1.1 shows an elementary diagram of a transmission and distribution
system. High voltage transmission lines are terminated in substations, which are
called high-voltage substations, receiving substations, or primary substations. The
function of sorne substations is switching circuits in and out of service; they are
referred to as switching stations. At the primary substations, the voltage is stepped
down to a value more suitable for the next part of the joumey toward the load. Very
large industrial customers may be served from the transmission system.
The portion of the transmission system that connects the high-voltage substa-
tions through step-down transformers to the distribution substations are called the
subtransmission network. There is no c1ear delineation between transmission and
subtransmission voltage levels. Typically, the subtransmission voltage level ranges
from 69 to 138 kV. Sorne large industrial customers may be served from the sub-
transmission system. Capacitor banks and reactor banks are usually installed in the
substations for maintaining the transmission line voltage.

1.3.3 DISTRIBUTION
The distribution system is that part which connects the distribution substations to
the consumers' service-entrance equipment. The primary distribution lines are usu-
ally in the range of 4 to 34.5 kV and supply the load in a well-defined geographical
area. Sorne small industrial customers are served directly by the primary feeders.
The secondary distribution network reduces the voltage for utilization by
commercial and residential consumers. Lines and cables not exceeding a few hun-
 1.3. MODERN POWER SYSTEM 7

Thermal Nuclear Hydro

Station Station
~ Station
Fossil
T
r
a
n
s
m
i
s
s
i
o
n
--
Switching
Station
~ Very
Large
Consumers
115 -765 kV I

HV
Substation
t- - HV
Substation
HV
Substation
I
S
u
~
I
~
I
Large
b

'" /
t Consumers
r
a
n
s
m
i
s
/ 69-138 kV
s ~ Network ~
i
o
n Distribution Substations Gas
I 4-34.5 kV Turbine
I I

Medium
Consumers Distribution
I I Transformers
~ ~ Residential
Consumers
~ ~ ¡ ¡ 240/120 V

FIGURE 1.1
Basic components of a power system.
8 1. THE POWER SYSTEM: AN OVERVIEW

dred feet in length then deliver power to the individual consumers. The secondary
distribution serves most of the customers at levels of 240/120 V, single-phase,
three-wire; 208Y/120 V, three-phase, four-wire; or 480Y/277 V, three-phase, four-
wire. The power for a typical home is derived from a transformer that reduces the
primary feeder voltage to 240/120 V using a three-wire lineo
Distribution systems are both overhead and underground. The growth of un-
derground distribution has been extremely rapid and as much as 70 percent of new
residential construction is served underground.

1.3.4 LOADS
Loads of power systems are divided into industrial, commercial, and residential.
Very large industrial loads may be served from the transmission system. Large
industrial loads are served directly from the subtransmission network, and small
industrial loads are served from the primary distribution network. The industrial
loads are composite loads, and induction motors form a high proportion of these
load. These composite loads are functions of voltage and frequency and form a
major part of the system load. Commercial and residential loads consist largely
of lighting, heating, and cooling. These loads are independent of frequency and
consume negligibly small reactive power.
The real power of loads are expressed in terms of kilowatts or megawatts.
The magnitude of load varies throughout the day, and power must be available to
consumers on demando
The daily-Ioad curve of a utility is a composite of demands made by various
c1asses of users. The greatest value of load during a 24-hr period is called the peak
or maximum demando Smaller peaking generators may be commissioned to meet
the peak load that occurs for only a few hours. In order to assess the usefulness
of the generating plant the load factor is defined. The load factor is the ratio of
average load over a designated period of time to the peak load occurring in that
periodo Load factors may be given for a day, a month, or ayear. The yearly, or
annual load factor is the most useful since ayear represents a full cyc1e of time.
The daily load factor is
Dail L.F. = average load (1.1)
y peak load
Multiplying the numerator and denominator of (1.1) by a time period of 24 hr, we
have
Daily L.F. = average load x 24 hr energy consumed during 24 hr
(1.2)
peak load x 24 hr peak load x 24 hr
The annualload factor is
_ total annual energy
Annua1 L ..
F - --------'-- (1.3)
peak load x 8760 hr
 1.3. MODERN POWER SYSTEM 9

Generally there is diversity in the peak load between different c1asses ofloads,
which improves the overall system load factor. In order for a power plant to operate
economically, it must have a high system load factor. Today's typical system load
factors are in the range of 55 to 70 percent.
There are a few other factors used by utilities. Utilization factor is the ratio of
maximum demand to the installed capacity, and plant factor is the ratio of annual
energy generation to the plant capacity x 8760 hr. These factors indicate how well
the system capacity is utilized and operated.
A MATLAB function barcycle(data) is developed which obtains a plot of the
load cyc1e for a given intervalo The demand interval and the load must be defined
by the variable data in a three-column matrix. The first two columns are the de-
mand interval and the third column is the load value. The demand interval may be
minutes, hours, or months, in ascending order. Hourly intervals must be expressed
in military time.

Example 1.1
The daily load on a power system varies as shown in Table 1.2. Use the barcycle
function to obtain a plot of the daily load curve. Using the given data compute the
average load and the daily load factor (Figure 1.2).

Table 1.2 Daily System Load

Interval, hr Load,MW
12 A.M. - 2 A.M. 6
2 - 6 5
6 9 10
9 - 12 15
12 P.M. - 2 P.M. 12
2 - 4 14
4 - 6 16
6 8 18
8 -10 16
10 - 11 12
11 - 12 A.M. 6
The followmg commands

data = [ O 2 6
2 6 5
6 9 10
9 12 15
12 14 12
10 1. THE POWER SYSTEM: AN OVERVIEW

14 16 14
16 18 16
18 20 18
20 22 16
22 23 12
23 24 6];
P = data(: ,3); % Column array of load
Dt = data(:, 2) - data(:,l); % Column array of demand interval
W ,,;, P'*Dt; %Total energy, area under the curve
Pavg = W/sum(Dt) %Average load
Peak = max (P) % Peak load
LF = Pavg/Peak*100 % Percent load factor
barcycle (data) % Plots the load cycle
xlabel('Time, hr'), ylabel('P, MW')

result in

18
16
14

P, 12
MW 10

8
6

40 5 10 15 20 25
Time, hr

FIGURE 1.2
Daily load cycle for Example 1.1.

Pavg = 11.5417
Peak = 18
LF 64.12
 1.4. SYSTEM PROTECfION 11

1.4 SYSTEM PROTECTION

In addition to generators, transfonners, and transmission lines, other devices are
required for the satisfactory operation and protection of a power system. Sorne of
the protective devices directly connected to the circuits are called switchgear. They
inelude instrument transfonners, circuit breakers, disconnect switches, fuses and
lightning arresters. These devices are necessary to deenergize either for nonnal
operation or on the occurrence of faults. The associated control equipment and
protective relays are placed on switchboard in control houses.

1.5 ENERGY CONTROL CENTER

For reliable and economical operation of the power system it is necessary to mon-
itor the entire system in a control center. The modero control center of today is
called the energy control center (ECC). Energy control centers are equipped with
on-line computers perfonning all signal processing through the remote acquisition
system. Computers work in a hierarchical structure to properly coordinate different
functional requirements in nonnal as well as emergency conditions. Every energy
control center contains a control console which consists of a visual display unit
(VD U), keyboard, and light peno Computers may give alarms as advance wam-
ings to the operators (dispatchers) when deviation from the nonnal state occurs.
The dispatcher makes judgments and decisions and executes them with the aid of
a computer. Simulation tools and software packages written in high-Ievellanguage
are implemented for efficient operation and reliable control of the system. This is
referred to as SCADA, an acronym for "supervisory control and data acquisition."

1.6 COMPUTER ANALYSIS

For a power system to be practical it must be safe, reliable, and economical. Thus
many analyses must be perfonned to desig'n and operate an electrical system. How-
ever, before going into system analysis we have to model all compoIíents of elec-
trical power systems. Therefore, in this text, after reviewing the concepts of power
and three-phase circuits, we will calculate the parameters of a multí-circuit trans-
mission line. Then, we will model the transmission line and look at the perfor-
mance of the transmission lineo Since transfonners and generators are a part of
the system, we will model these devices. Design of a power system, its operation
and expansion requires much analysis. This text presents methods of power system
analysis with the aid of a personal computer and the use of MATlAB. The MAT-
IAB environment pennits a nearly direct transition from mathematical expression
12 1. THE POWER SYSTEM: AN OVERVIEW

to simulation. Sorne of the basic analysis covered in this text are:

• Evaluation of transmission line parameters

• Transmission line performance and compensatíon

• Power ftow analysis

• Economic scheduling of generation
• Synchronous machine transient analysis

• Balanced fault
• Symmetrical components and unbalanced fault

• Stability studies
• Power system control

Many MATLAB functions are developed for the aboye studies thus allowing
the student to concentrate on analysis and design of practical systems and spend
less time on programming.

PROBLEMS
1.1. The demand estimation is the starting point for planning the future electric
power supply. The consistency of demand growth over the years has led
to numerous attempts to tit mathematical curves to this trend. One of the
simplest curves is
P Poea(t-to )

where a is the average per unit growth rate, P is the demand in year t, and
Po is the given demand at year too
Assume the peak power demand in the United States in 1984 is 480 GW with
an average growth rate of 3.4 percent. Usíng MATLAB, plot the predicated
peak demand in GW from 1984 to 1999. Estímate the peak power demand
for the year 1999.
1.2. In a certain country, the energy consumption is expected to double in 10
years. Assuming a simple exponential growth given by

P = Poe at
calculate the growth rate a.
CHAPTER

2
BASIC PRINCIPLES

2.1 INTRODUCTION
The concept of power is of central importance in electrical power systems and is
the main topic of this chapter. The typical student will already have studied much
of this material, and the review here will serve to reinforce the power concepts
encountered in the electric circuit theory.
In this chapter, the ftow of energy in an ac circuit is investigated. By using
various trigonometric identities, the instantaneous power p(t) is resolved into two
components. A plot of these components is obtained using MATLAB to observe that
ac networks not only consume energy at an average rate, but also borrow and retum
energy to its sources. This leads to the basic definitions of average power P and
reactive power Q. The volt-ampere S, which is a mathematical formulation based
on the phasor forms of voltage and current, is introduced. Then the complex power
balance is demonstrated, and the transmission inefficiencies caused by loads with
low power factors are discussed and demonstrated by means of several examples.
Next, the transmission of complex power between two voltage sources is con-
sidered, and the dependency of real power on the voltage phase angle and the de-
pendency of reactive power on voltage magnitude is established. MATLAB is used
convenientIy to demonstrate this idea graphically.
Finally, the balanced three-phase circuit is examined. An important property
of a balanced three-phase system is that it delivers constant power. That is, the
14
 2.2. POWER IN SINGLE-PHASE AC ClRCUITS 15

power delivered does not fluctuate with time as in a single-phase system. For the
purpose of analysis and modeling, the per-phase equivalent circuit is developed for
the three-phase system under balanced condition.

2.2 POWER IN SINGLE-PHASE AC CIRCUITS

Figure 2.1 shows a single-phase sinusoidal voltage supplying a load.

i(t)
+

v(t)

FIGURE 2.1
Sinusoidal source supplying a load.

Let the instantaneous voltage be

v(t) = Vm cos(wt + (Jv) (2.1)

and the instantaneous current be given by

(2.2)

The instantaneous power p(t) delivered to the load is the product of voltage v(t)
and current i(t) given by

p(t) = v(t) i(t) = Vm1m cos(wt + (Jv) cos(wt + (Ji) (2.3)

In Example 2.1, MATlAB is used to plot the instantaneous power p(t), and the
result is shown in Figure 2.2. In studying Figure 2.2, we note that the frequency of
the instantaneous power is twice the source frequency. AIso, note that it is possible
for the instantaneous power to be negative for a portion of each cyc1e. In a passive
network, negative power implies that energy that has been stored in inductors or
capacitors is now being extracted.
It is informative to write (2.3) in another form using the trigonometric identity

1 1
cosAcosB = 2 cos(A - B) + '2 cos(A + B) (2.4)
16 2. BASIC PRINCIPLES

which results in
1
p(t) = '2 VmIm[cos(Ov - Oí) + cos(2wt + Ov + Oí)]
1
= '2VmIm{COS(Ov - Oí) + cos[2(wt + Ov) (Ov - Oí)]}
1
= '2 VmIm[cos(Ov - Oi) + cos 2(wt + Ov) cos(Ov - Oí)
+sin 2(wt + Ov) sin(Ov - OdJ

The root-mean-square (nns) value of v(t) is IVI = Vm/v'2 and the nns value of
i(t) is 111 = Im/v'2 . Let 0= (Ov - Oí)' The aboye equation, in tenns ofthe nns
values, is reduced to
p(t) = IVIIII cos 0[1 + cos 2(wt + Ov)] + IVIIII sin Osin 2(wt + Ov)
.... I " J
Y' v
PR(t) px(t) (2.5)
Energy ftow into Energy borrowed and
the circuit returiled by the circuit
where O is the angle between voltage and current, or the impedance angle. O is
positive ifthe load is inductive, (Le., current is lagging the voltage) and Ois negative
if the load is capacitive (i.e., current is leading the voltage).
The instantaneous power has been decomposed into two components. The
first component of (2.5) is

PR(t) = IVIIIlcosO+ IVIIIlcosOcos2(wt + Ov)] (2.6)

The second tenn in (2.6), which has a frequency twice that of the source, accounts
for the sinusoidal variation in the absorption of power by the resistive portion of
the load. Since the average value of this sinusoidal function is zero, the average
power delivered to the load is given by

P IVIIII cos O (2.7)

This is the power absorbed by the resistive component of the load and is also re-
ferred to as the active power or real power. The product of the nns voltage value
and the nns current value IVIIII is called the apparent power and is measured in
units of volt ampere. The product of the apparent power and the cosine of the angle
between voltage and current yields the real power. Because cos Oplays a k.ey role in
the detennination of the average power, it is called power factor. When the current
lags the voltage, the power factor ís considered lagging. When the current leads the
voltage, the power factor is considered leadíng.
The second component of (2.5)

px(t) = IVIIIlsinOsin2(wt + Ov) (2.8)

 2.2. POWER IN SINGLE-PHASE AC CIRCUITS 17

pulsates with twice the frequency and has an average value of zero. This compo-
nent accounts for power oscillating into and out of the load because of its reactive
element (inductive or capacitive). The amplitude of this pulsating power is called
reactive power and is designated by Q.

Q= IVIIII sin O (2.9)

Both P and Q have the same dimensiono However, in order to distinguish between
the real and the reactive power, the term "var" is used for the reactive power (var is
an acronym for the phrase "volt-ampere reactive"). For an inductive load, current is
lagging the voltage, O = (Ov - Od > O and Q is positive; whereas, for a capacitive
load, current is leading the voltage, O = (Ov - Od < O and Q is negative.
A careful study of Equations (2.6) and (2.8) reveals the following character-
istics of the instantaneous power.

• For apure resistor, the impedance angle is zero and the power factor is unity
(UPF), so that the apparent and real power are equal. The electric energy is
transformed into thermal energy.

• If the circuit is purely inductive, the current lags the voltage by 90° and the
average power is zero. Therefore, in a purely inductive circuit, there is no
transformation of energy from electrical to nonelectrical formo The instanta-
neous power at the terminal of a purely inductive circuit oscillates between
the circuit and the source. When p(t) is positive, energy is being stored in
the magnetic field associated with the inductive elements, and when p(t) is
negative, energy is being extracted from the magnetic fields of the inductive
elements.

• If the load is purely capacitive, the current leads the voltage by 90°, and the
average power is zero, so there is no transformation of energy from electri-
cal to nonelectrical formo In a purely capacitive circuit, the power oscillates
between the source and the electric field associated with the capacitive ele-
ments.

Example 2.1
The supply voltage in Figure 2.1 is given by v( t) = 100 cos wt and the load is
inductive with impedance Z = 1.25L60° n. Determine the expression for the
instantaneous current i(t) and the instantaneous power p(t). Use MATIAB to plot
i(t), v(t), p(t), PR(t), and px(t) over an interval of Oto 271'.

1 - 100LO° _ L_ °
max - 1.25L600 - 80 60 A
18 2. BASIC PRINCIPLES

v(t) = Vm cos wt, i(t) = 1m cos(wt - 60) p(t) = v(t)i(t)

100 6000 r-7"'<"-------r'<"-------,

50 4000

2000

-50 O

100 200 300 400- 2000 0 100 200 300 400

wt, degree wt, degree

Pr(t), Eq.2.6 Px(t), Eq.2.8

4000

3000 2000

2000 I - - + - - - f - - - \ - - - + - - O

1000 -2000

100 200 300 400

wt, degree wt, degree
FIGURE 2.2
Instantaneous current, voltage, power, Eqs. 2.6 and 2.8.

therefore

i(t) = 80cos(wt - 60°) A

p(t) = v(t)i(t) = 8000coswtcos(wt - 60°) W

The following statements are used to plot the aboye instantaneous quantities and
the instantaneous terms given by (2.6) and (2.8).
Vm = 100; thetav = O; %Voltage amplitude and phase angle
Z = 1.25; gama = 60; % Impedance magnitude and phase angle
thetai = thetav - gama; % Current phase angle in degree
theta = (thetav - thetai)*pi/180; %Degree to radian
1m = Vm/Z; % Current amplitude
wt = 0:.05:2*pi; %wt from O to 2*pi
v = Vm*cos(wt); % Instantaneous voltage
 2.3. COMPLEX POWER 19

i = Im*cos(wt + thetai*pi/180); % Instantaneous current

p = v.*i; % Instantaneous power
V = Vm/sqrt(2); I=Im/sqrt(2); % rms voltage and current
P = V*I*cos(theta); % Average power
Q = V*I*sin(theta); % Reactive power
S = P + j*Q % Complex power
pr = P*(l + cos(2*(wt + thetav»); % Eq. (2.6)
px = Q*sin(2*(wt + thetav»; % Eq. (2.8)
PP = P*ones(l, length(wt»;%Average power of length w for plot
xline = zeros(l, length(wt»; %generates a zero vector
wt=180/pi*wt; % converting radian to degree
subplot(2,2,1) , plot(wt, v, wt,i,wt, xline), grid
title(['v(t)=Vm coswt, i(t)=Im cos(wt+' ,num2str(thetai), ')')
xlabel('wt, degree')
subplot(2,2,2), plot(wt, p, wt, xline), grid
title('p(t)=v(t) i(t)'),xlabel('wt, degree')
subplot(2,2,3), plot(wt, pr, wt, PP,wt,xline), grid
title('pr(t) Eq. 2.6'), xlabelC'wt, degree')
subplot(2,2,4), plot(wt, px, wt, xline), grid
title('px(t) Eq. 2.8'), xlabel('wt, degree'), subplot(lll)

2.3 COMPLEX POWER

The rms voltage phasor of (2.1) and the rms current phasor of (2.2) shown in Fig-
ure 2.3 are

The term V 1* results in

V

~Q p
FIGURE 2.3
Phasor diagram and power triangle for an inductive load (lagging PF).

VI* = IVII1ILOv - Oí = IVIIIILO

20 2. BASIC PRINCIPLES

IVIIII cos e+ jlVlIII sin e

The above equation defines a complex quantity where its real part is the average
(real) power P and its imaginary part is the reactive power Q. Thus, the complex
power designated by S is given by

S = V1* P+ jQ (2.10)

The magnitude of S, ISI = V

p2 + Q2, is the apparent power; its unít is volt-
amperes and the larger units are kVA or MVA. Apparent power gives a direct indi-
cation of heating and is used as a rating unit of power equipment. Apparent power
has practical sígnificance for an electric utility company sínce a utility company
must supply both average and apparent power to consumers.
e
The reactive power Q is positive when the phase angle between voltage and
current (impedance angle) is positive (i.e., when the load impedance is inductive,
and I lags V). Q is negative when e is negative (i.e., when the load impedance is
capacitive and I leads V) as shown in Figure 2.4.
In working with Equation (2.10 ) it is convenient to think of P, Q, and S as
forming the sides of a right triangle as shown in Figures 2.3 and 2.4.

I V
P

FIGURE 2.4
Phasor diagram and power triangle for a capacitive load (leading PF).

If the load impedance is Z then

V= ZI (2.11)

substituting for V into (2.10) yields

S = V 1* = ZII* RIII2 + jXIII2 (2.12)

From (2.12) it is evident that complex power S and impedance Z have the same
angle. Because the power triangle and the impedance triangle are similar triangles,
the impedance angle is sometimes calIed the power angle.
Similarly, substituting for I from (2.11) into (2.10) yields

S = V1* = VV* (2.13)

Z*
 2.4. THE COMPLEX POWER BALANCE 21

From (2.13), the impedance of the complex power S is given by

(2.14)

2.4 TOO COMPLEX POWER BALANCE

From the conservation of energy, it is clear thal real power supplied by the source is
equal lo the sum of real powers absorbed by the load. At the sarne time, a balance
between the reactive power must be maintained. Thus the total complex power
delivered to the loads in parallel is the sum of the complex powers delivered to
each. Proof of this is as follows:

FlGURE2.S
Three loads in parallel.

For the three loads shown in Figure 2.5, the total complex power is given by

S = VI* = V[l¡ + I 2 + I3]* = VIi + VI'; + VI; (2.15)

Example2.2
In the aboye circuit V = 1200LO° V, Z1 = 60 + jO n , Z2 = 6 + j12 n and
Z3 30 j30 n. Find the power absorbed by each load and the total complex
power.
22 2. BASIC PRINCIPLES

SI = V 1; = 1200LOO(20 jO) = 24,000 W + jO var

S2 = V li = 1200LOO(40 + j80) 48,000 W + j96, 000 var
S3 = V 1; = 1200LOO(20 j20) 24,000 W - j24, 000 var
The total load complex power adds up to
S = SI + S2 + S3 = 96,000 W + j72, 000 var
Altematively, the sum of complex power delivered to the load can be obtained by
first finding the total current.
1 = l¡ + h + la = (20 + jO) + (40 j80) + (20 + j20)
= 80 - j60 = 100L -36.87° A
and
S = V 1* = (1200LOO)(100L36.87°) = 120, 000L36.87° VA
= 96,000 W + j72, 000 var
A final insight is contained in Figure 2.6, which shows the current phasor diagram
and the complex power vector representation.
S2
S

FIGURE 2.6
Current phasor diagram and power plane diagram.

The complex powers may also be obtained directly from (2.14)

1V12 (1200)2 .
SI = Zi = 60 = 24, 000 W + J O

S2 = I~t = ~1:0;{~ = 48,000 W + j96, 000 var

1V12 (1200)2 .
S3 = Z; = 30 + j30 = 24,000 W - J24, 000 var
 2.5. POWER FACTOR CORRECTION 23

2.5 POWER FACTOR CORRECTION

It can be seen from (2.7) that the apparent power will be larger than P if the power
factor is less than 1. Thus the current 1 that must be supplied will be larger for
PF < 1 than it would be for P F 1, even though the average power P supplied
is the same in either case. A larger current cannot be supplied without additional
cost to the utility company. Thus, it is in the power company's (and its customer's)
best interest that major load s on the system have power factors as c10se to 1 as
possible. In order to maintain the power factor close to unity, power companies
install banks of capacitors throughout the network as needed. They also impose an
additional charge to industrial consumers who operate at 10w power factors. Since
industrialloads are inductive and have low lagging power factors, it is beneficial to
install capacitors to improve the power factor. This consideration is not important
for residential and small commercial customers because their power factors are
c10se to unity.

Example2.3
Two 10ads Zl 100 + jO O and Z2 10 + j20 O are connected across a 200-V
nns, 60-Hz source as shown in Figure 2.7.

(a) Find the total real and reactive power, the power factor at the source, and the
total current.

Q
1 l¡ h le
100 I

200 V 100 O ---Le

-,-
j20 O

Qc

FIGURE 2.7
Circuit for Example 2.3 and the power triangle.

1 = 200'::0° = UO° A
1 100
1 - 200'::0° - _·S A
2 - 10 + j20 - 4 J
SI = VI; 200'::0°(2 - jO) = 400 W + jO var
S2 = V 12 200'::0°(4 + jS) = SOO W + j1600 var
24 2. BASIC PRINCIPLES

Total apparent power and current are

s = P + jQ 1200 + j1600 = 2000L53.13° VA

1 = S* 2000L-53.13° = lOL-53.13° A
V* 200LO°
Power factor at the source is

PF = cos(53.13) = 0.6 lagging

(b) Find the capacitance of the capacitor connected across the loads to improve the
overall power factor to 0.8 lagging.

Total real power P = 1200 W at the new power factor 0.8 lagging. Therefore

e' cos-1(0.8) = 36.87°

Q' Ptane' = 1200tan(36.87°) = 900 var
Qc = 1600 -
900 = 700 var

Zc
1V1 2(200)2
-j57.14 n
S~ j700
106
e = 27f(60)(57.14) 46.42 f.LF
The total power and the new current are

s' = 1200 + j900 1500L36.87°

l' - S'* _ 1500L-36.87° = 7.5L-36.8~
- V* - 200LO°
Note the reduction in the supply current from 10 A to 7.5 A.

Example2.4
Three loads are connected in parallel across a 1400-V rms, 60-Hz single-phase
supply as shown in Figure 2.8.

Load 1: Inductive load, 125 kVA at 0.28 power factor.

Load 2: Capacitive load, 10 kW and 40 kvar.

Load 3: Resistive load of 15 kW.

(a) Find the total kW, kvar, kVA, and the supply power factor.
 2.5. POWER FACTOR CORRECTION 25

J JI

1400 V 1 2

FIGURE 2.8
Circuit for Example 2.4.

An inductive load has a lagging power factor, the capacitive load has a lead-
ing power factor, and the resistive load has a unity power factor.

ForLoad 1:

01 = cos- l (0.28) 73.74° lagging

The load complex powers are

8 1 = 125L73.74 kVA = 35 kW + j120 kvar

82 = 10 kW - j40 kvar
83 = 15 kW + JO kvar
The total apparent power is

8= P + jQ = 8 1 + 82 + 83
= (35 + j120) + (10 j40) + (15 + JO)
= 60 kW + j80 kvar 100L53.13 kVA

The total current is

8* 100, OOOL -53.13° = 71.43L -53.13° A
J=-
V* 1400LO°
The supply power factor is

PF = cos(53.13) = 0.6 lagging

(b) A capacitor of negligible resistance is connected in parallel with the aboye loads
to improve the power factor to 0.8 lagging. Determine the kvar rating of this ca-
pacitor and the capacitance in ¡.¡,F.
26 2. BASIC PRINCIPLES

Total real power P = 60 kW at the new power factor of 0.8 lagging results in the
new reactive power Q'.
e' = cos- 1 (0.8) = 36.87°
Q' = 60tan(36.87°) = 45 kvar
Therefore, the required capacitor kvar is
Qc = 80 - 45 = 35 kvar
and
IVI 2 14002 _ _ . n
Xc 8*c = j35 , 000 - )56
106
e= 211"(60)(56) = 47.37 ¡.tF

and the new current is

l' = 8'* 60,000 - 000 = 53.57 L -36.87° A

V* 1400 LO°
Note the reduction in the supply current from 71.43 Ato 53.57 A.

2.6 COMPLEX POWER FLOW

Consíder two ideal voltage sources connected by a line of impedance Z = R +
jX nas shown in Figure 2.9.

FIGURE 2.9
1\vo interconnected voltage sources.

Let the phasor voltage be VI = IVl1Ló 1 and V2 = 1V21Ló2. For the assumed direc-
tion of current
1 - IVdLó1 -1V21Ló2 _ IVI Ia _ _ 1"V21 LO _
12 - IZIL¡ - IZI 1 ¡ IZI 2 ¡
 2.6. COMPLEX POWER FLOW 27

The complex power 8 12 is given by

Thus, the real and reactive power at the sending end are
2
P12 = lZf cos 1- !VIIZI1!V21 cos(¡. + 61 -
!VI 1 62 ) (2.16)

!V112. !VI 1!V2 1 .

Q12 = lZf sm 1- IZI sm(¡ + 61 - 62 ) (2.17)

Power system transmission lines have small resistance compared to the reactance.
Assuming R = O (i.e., Z = X L90° ), the aboye equations become

(2.18)

(2.19)

Since R = O, there are no transmission line losses and the real power sent equals
the real power received.
From the aboye results, for a typical power system with small R/ X ratio, the
following important observations are made :

1. Equation (2.18) shows that small changes in 61 or 62 will have a significant

effect on the real power flow, while small changes in voltage magnitudes will
not have appreciable effect on the real power flow. Therefore, the flow of real
power on a transmission line is govemed mainly by the angle difference of
the terminal voltages (i.e., P 12 ex: sin 6 ), where 6 = 61 - 62 . If VI leads V2,
6 is positive and the real power flows from node 1 to node 2. If VI lags V2, 6
is negative and power flows from node 2 to node 1.

2. Assuming R = O, the theoretical maximum power (static transmission ca-

pacity) occurs when 6 = 90 0 and the maximum power transfer is given by

(2.20)

In Chapter 3 we learn that increasing 6 beyond the static transmission capac-

ity will result in loss of synchronism between the two machines.
 28 2. BASIC PRINCIPLES

3. For maintaining transient stability, the power system is usualIy operated with
smallload angle Ó. Also, from (2.19) the reactive power flow is determined
by the magnitude difference of terminal voltages, (Le., Q ex: !VII - 1V21).

Example2.S
Two voltage sources VI = 120L-5 V and V2 = 100LO V are connected by a short
line of impedance Z = 1 + j7 n as shown in Figure 2.9. Determine the real and
reactive power supplied or received by each source and the power los s in the lineo

112 = 120L-::~;OOLO° = 3.135L 110.02° A

100LO° - 120L 5°
121 = 3.135L69.98° A
1 + j7
1:
8 12 = V 1 2 376.2L105.02° -97.5 W + j363.3 var
8 21 = V2 121 = 313.5L -69.98° = 107.3 W - j294.5 var

Line loss is given by

8L 8 1 + 82 = 9.8 W + j68.8 var

From the aboye results, since P¡ is negative and P2 is positive, source 1 receives
97.5 W, and source 2 generates 107.3 W and the real power 10ss in the line is 9.8
W. The real power loss in the line can be checked by

Also, since Ql is positive and Q2 is negative, source 1 delivers 363.3 var and source
• 2 receives 294.5 var, and the reactive power los s in the line is 68.6 varo The reactive
power 10ss in the line can be ~hecked by

2
QL = Xlh21 = (7)(3.135)2 = 68.8 var

Example2.6
This example concems the direction of power flow between two voltage sources.
Write a MATIAB prograrn for the system of Example 2.5 such that the phase an-
gle of source 1 is changed from its initial value by ±30° in steps of 5°. Voltage
magnitudes of the two sources and the voltage phase angle of source 2 is to be kept
constant. Compute the complex power for each source and the line loss. Tabulate
the real power and plot PI , P2, and PL versus voltage phase angle ó. The following
commands
 2.6. COMPLEX POWBR FLOW 29

El = input('Source # 1 Voltage Mag. = ');

al = input('Source# 1 Phase Angle = ');
E2 ;: input('Source # 2 Voltage Mag. = ');
a2 = input('Source # 2 Phase Angle = ');
R = input('Line Resistance = ');
x = input('Line Reactance = ');
Z R + j*X; % Line impedance
al = (-30+al:5:30+al)'; % Change al by +/- 30, col. array
alr = al*pi/180; % Convert degree to radian
k = length(al);
a2 = ones(k,1)*a2; %Create col. array of same length for a2
a2r = a2*pi/180; %Convert degree to radian
Vl = El.*cos(alr) + j*El.*sin(alr);
V2 = E2.*cos(a2r) + j*E2.*sin(a2r);
112 = (Vl - V2)./Z; 121=-I12;
Sl = Vl.*conj(I12); Pl = real(Sl); Ql ;: imag(Sl);
S2 ;: V2.*conj(I21); P2 = real(S2); Q2 = imag(S2);
SL = Sl+S2; PL = real(SL); QL = imag(SL);
Resultl = [al, Pl, P2, PL];
disp(' Delta 1 P-l P-2 P-L ')
disp(Resultl)
plot(al, Pl, al, P2, al,PL)
xlabel('Source #1 Voltage Phase Angle')
ylabel(' P, Watts'),
text(-26, -550, 'Pl'), text(-26, 600,'P2'),
text(-26, lOO, 'PL')

result in

Source # 1 Voltage Mag. = 120

Source # 1 Phase Angle ;: -5
Source # 2 Voltage Mag. = 100
Source # 2 Phase Angle = O
Line Resistance = 1
Line Reactance = 7

Delta 1 P-l P-2 P-L

-35.0000 -872.2049 967.0119 94.8070
-30.0000 -759.8461 832.1539 72.3078
-25.0000 -639.5125 692.4848 52.9723
-20.0000 -512.1201 549.0676 36.9475
-15.0000 -378.6382 402.9938 24.3556
-10.0000 -240.0828 255.3751 15.2923
-5.0000 -97.5084 107.3349 9.8265
O 48.0000 -40.0000 8.0000
30 2. BASIC PRINCIPLES

5.0000 195.3349 -185.5084 9.8265

10.0000 343.3751 -328.0828 15.2923
15.0000 490.9938 -466.6382 24.3556
20.0000 637.0676 -600.1201 36.9475
25.0000 780.4848 -727.5125 52.9723

1000
800
600
400
200
JaUS O
-200
-400
-600
-800
-1002.
40 -30 -20 -10 O 10 20 30
Source # 1 Voltage Phase Angle
FIGURE 2.10
Real power versus voltage phase angle o.

Examination of Figure 2.10 shows that the ftow of real power along the intercon-
nection is detennined by the angle difference of the tenninal voltages. Problem 2.9
requires the development of a similar program for demonstrating the dependency
of reactive power on the magnitude difference of tenninal voltages.

2.7 BALANCED THREE-PHASE CIRCUITS

The generation, transmission and distribution of electric power is accomplished by
means of three-phase circuits. At the generating station, three sinusoidal voltages
are generated having the same amplitude but displaced in phase by 1200 • This is
called a balanced source. If the generated voltages reach their peak values in the
sequential order ABC, the generator is said to have a positive phase sequence,
shown in Figure 2. 11 (a). If the phase order is ACB, the generator is said to have a
negative phase sequence, as shown in Figure 2.11 (b).
 2.7. BALANCEDTHREE-PHASECIRCUrrS 31

ECn

EBn (a) ECn (b)

FIGURE 2.11
(a) Positive, or ABC, phase sequence. (b) Negative. or ACB, phase sequence.

In a three-phase system, the instantaneous power delivered to the external

Ioads is constant rather than puIsating as it is in a single-phase circuito AIso, three-
phase motors, having constant torque, start and ron much better than single-phase
motors. This feature of three-phase power, coupled with the inherent efficiency of
its transmission eompared to single-phase (Iess wire for the same delivered power),
accounts for its universal use.
A power system has Y-conneeted generators and usually ineludes both ~­
and Y-connected Ioads. Generators are rarely ~-connected, because if the voltages
are not perfeetly balanced, there will be a net voltage, and eonsequently a circulat-
ing eurrent, around the ~. AIso, the phase voltages are Iower in the Y-eonnected
generator, and thus less insulation is required. Figure 2.12 shows a Y-eonnected
generator supplying balaneed Y-eonneeted loads through a three-phase lineo As-
suming a positive phase sequenee (phase order ABC) the generated voltages are:

EAn = IEpILO°
EBn = IEpIL-120° (2.21)
ECn = IEpIL-240°
In power systems, great eare is taken to ensure that the loads of transmission lines
are balanced. For balaneed loads, the terminal voltages of the generator VAn, VBn
and VCn and the phase voltages Van, Vbn and Ven at the load terminals are balanced.
For "phase A," these are given by

VAn = EAn Za1a (2.22)

Van = VAn - ZL1a (2.23)
32 2. BASIC PRINCIPLES

FIGURE 2.12
A Y-connected generator supplying a Y-connected load.

2.8 Y-CONNECTED LOADS

To find the relationship between the line voltages (line-to-line voltages) and the
phase voltages (line-to-neutral voltages), we assume a positive, or ABe, sequence.
We arbitrarily choose the line-to-neutral voltage of the a-phase as the reference,
thus

Van = ¡VpILO°
Vbn = ¡VpIL-120° (2.24)
Ven IVpIL-240°

where ¡Vpl represents the magnitude ofthe phase voltage (line-to-neutral voltage).
The line voltages al the load terminals in terms of the phase voltages are found
by the application of Kirchhoff's voltage law

Vab = Van - Vbn = ¡Vpl(lLO° 1L-1200) = v'3IVpIL30°

\!be = Vbn - Ven = ¡Vpl(lL-120° - 1L-2400) = v'3¡VpIL-90° (2.25)
Vea = Ven - Van = ¡Vpl(lL-240° -110°) v'3¡VplL150°

The voltage phasor diagram of the Y-connected loads of Figure 2.12 is shown
in Figure 2.13. The relationship between the line voltages and phase voltages is
demonstrated graphical1y.
 2.8. Y-CONNECTED LOADS 33

FIGURE 2.13
Phasor diagram showing phase and line voltages.

If the rms value of any of the line voltages is denoted by VL, then one of the
important characteristics of the Y-connected three-phase load may be expressed as

(2.26)

Thus in the case of Y-connected loads, the magnitude of the line voltage is
v'3 times the magnitude of the phase voltage, and for a positive phase sequence,
the set of line voltages leads the set of phase voltages by 30°.
The three-phase currents in Figure 2.12 also possess three-phase symmetry
and are given by

(2.27)

where () is the impedance phase angle.

The currents in lines are also the phase currents (the current carried by the
phase impedances). Thus

(2.28)
34 2. BASIC PRINCIPLES

2.9 .6.-CONNECTED LOADS

A baIanced Ll-connected load (with equaI phase impedances) is shown in Fig-
ure 2.14.

a la

FIGURE 2.14
A ~-connecled load.

It is clear from the inspection of the circuit that the line voltages are the same
as phase voltages.
(2.29)
Consider the phasor diagram shown in Figure 2.15, where the phase current lab is
arbitrarily chosen as reference. we have

lab IlpILO°
lbe = IlpIL-120° (2.30)
lea = IlpIL-240°

where Ilpl represents the magnitude of the phase current.

lbe
FIGURE 2.15
Phasor diagram showing phase and line currenls.
 2.10. .6.-YTRANSFORMATION 35

The relationship between phase and Une currents can be obtained by applying
Kirchhoff's current law at the comers of ~.
la = lab -lea = Ilpl(L:::O° -
lL-2400) = J31 1pIL-30°
h = lbe -lab = Ilpl(lL-120° -lLOO) = J311pIL-150° (2.31)
le = lea - he = Ilpl(lL-240° -lL-1200) = J311pIL90°
The relationship between the Une currents and phase currents is demonstrated
graphically in Figure 2.15.
If the rms of any of the Une currents is denoted by h, then one of the impor-
tant characteristics of the ~-connected three-phase load may be expressed as
(2.32)

Thus in the case of ~-connected loads, the magnitude of the line current is J3
times the magnitude of the phase current, and with positive phase sequence, the set
of line currents lags the set of phase currents by 30°.

2.10 ,6.- y TRANSFORMATION

For analyzing network problems, it is convenient to replace the ~-connected cir-

cuit with an equivalent Y-connected circuit. Consider the fictitious Y-connected
circuit of Zy nJphase which is equivalent to a balanced ~-connected circuir of
Zf!,. nJphase, as shown in Figure 2.16.
a a

h b e
(a) (b)
FIGURE 2.16
(a) t. to (b) Y-connection.

For the ~-connected circuit, the phase current la is given by

la = Vab + Vae = Vab + Vae (2.33)

Zf!,. Zf!,. Zf!,.
36 2. BASrC PRINCIPLES

FIGURE 2.17
Phasor diagram showing phase and Une voltages.

The phasor diagram in Figure 2.17 shows the relationship between balanced phase
and line-to-line voltages. From this phasor diagram, we find

Vab + Yac = J3lVanIL30° + J3lVanI L- 30° (2.34)

= 3Van (2.35)

Substituting in (2.33), we get

1 _ 3Van
a - Zt1
or

Zt1
Van = ala (2.36)

Now, for the Y-connected circuit, we have

Van = Zyla (2.37)

Thus, from (2.36) and (2.37), we find that

Zt1
ZY=a (2.38)

2.11 PERMPHASE ANALYSIS

The current in the neutral of the balanced Y-connected 10ads shown in Figure 2.12
is given by
(2.39)
 2.12. BALANCED THREE-PHASE POWER 37

Since the neutral carnes no current, a neutral wire of any impedance may be re-
placed by any other impedance, including a short circuit and an open circuito The
retum line may not actually exist, but regardless, a line of zero impedance is in-
c1uded between the two neutral points. The balanced power system problems are
then solved on a "per-phase" basis. It is understood that the other two phases carry
identical currents except for the phase shift.
We may then look at only one phase, say "phase A," consisting of the source
VAn in series with Z L and Zp, as shown in Figure 2.18. The neutral is taken as
datum and usually a single-subscript notation is used for phase voltages.

FIGURE 2.18
Single-phase circuit for per-phase analysis.

If the load in a three-phase circuit is connected in a Ll, it can be transformed

into a Y by using the Ll-to-Y transformation. When the load is balanced, the
impedance of each leg of the Y is one-third the impedance of each leg of the Ll, as
given by (2.38), and the circuit is modeled by the single-phase equivalent circuit.

2.12 BALANCED THREE-PHASE POWER

Consider a balanced three-phase source supplying a balanced Y- or Ll- connected

load with the following instantaneous voltages
Van = V2¡Vpl cos(wt + (}v)
v'm = V2¡Vpl cos(wt + (}v - 120°) (2.40)
Ven = V2¡Vpl cos(wt + (}v - 240°)
For a balanced load the phase currents are
ia V2I I pl cos(wt + (}í)
ib = V2IIpl cos(wt + (}í - 120°) (2.41)
ic V2IIpl cos(wt + (}í - 240°)
38 2. BASIC PRINCIPLES

where IVpl and ¡Ipl are the magnitudes of the nns phase voltage and current, re-
spectively. The total instantaneous power is the sum of the instantaneous power of
each phase, given by
(2.42)
Substituting for the instantaneous voltages and currents from (2.40) and (2.41) into
(2.42)

P3</> = 21Vpll I pl cos(wt + (1)) cos(wt + Oi)

+2¡VpIlIpl cos(wt + (1) - 120°) cos(wt + Oi - 120°)
+2¡VpIIIpl cos(wt + (1) - 240°) cos(wt + Oi - 240°)

Using the trigonometric identity (2.4)

P3</> = IVpIlIpl[cos(01) - lh)+ cos(2wt + (1) + Oi)]

+lVpIlIpl[cos(Ov Oi) + cos(2wt + Ov + Oi - 240°)] (2.43)
+lVpIIIpl[cos(Ov - Oi) + cos(2wt + (1) + Oi - 480°)]

The three double frequency cosine tenns in (2.43) are out of phase with each other
by 120° and add up to zero, and the three-phase instantaneous power is

(2.44)

o= Ov - Oi is the angle between phase voltage and phase current or the impedance
angle.
Note that although the power in each phase is pulsating, the total instanta-
neous power is constant and equal to three times the real power in each phase. In-
deed, this constant power is the main advantage of the three-phase system over the
single-phase system. Since the power in each phase is pulsating, the power, then,
is made up of the real power and the reactive power. In order to obtain fonnula
symmetry between real and reactive powers, the concept of complex or apparent
power (8) is extended to three-phase systems by defining the three-phase reactive
power as
Q3</> = 3 IVpIIIpI sin O (2.45)
Thus, the complex three-phase power is

(2.46)

or
(2.47)
Equations (2.44) and (2.45) are sometimes expressed in tenns of the nns
magnitude of the line voltage and the nns magnitude of the line current. In a Y-
connected load the phase voltage IVpl = IVLI/V3 and the phase current Ip = h.
 2.12. BALANCED THREE-PHASE POWER 39

In the 6.-connection Vp = VL and IIpl = Ihl/J3. Substituting for the phase volt-
age and phase currents in (2.44) and (2.45), the real and reactive powers for either
connection are given by
(2.48)

and
(2.49)
A comparison of the last two expressions with (2.44) and (2.45) shows that the
equation for the power in a three-phase system is the same for either a Y or a 6.
connection when the power is expressed in terms of line quantities.
When using (2.48) and (2.49) to calculate the total real and reactive power,
remember that eis the phase angle between the phase voltage and the phase current.
As in the case of single-phase systems for the computation of power, it is best to
use the complex power expression in terms of phase quantities given by (2.47).
The rated power is customarily given for the three-phase and rated voltage is the
line-to-line voltage. Thus, in using the per-phase equivalent circuit, care must be
taken to use per-phase voltage by dividing the rated voltage by J3.

Example 2.7
A three-phase line has an impedance of 2 + j4 O as shown in Figure 2.19.

2+j40

IVLI = 207.85 V 600

b ____~~~~-T--__+b~
-j45 O

FIGURE 2.19
Three-phase circuit diagram for Example 2.7.

The Une feeds two balanced three-phase loads that are connected in parallel. The
first load is Y-connected and has an impedance of 30+ j40 O per phase. The second
load is 6.-connected and has an impedance of 60 j45 O. The line is energized
at the sending end from a three-phase balanced supply of line voltage 207.85 V.
Taking the phase voltage Va as reference, determine:
(a) The current, real power, and reactive power drawn from the supply.
40 2. BASIC PRINCIPLES

(b) The line voltage at the combined loads.

(c) The current per phase in each load.
(d) The total real and reactive powers in each load and the lineo

(a) The ~-connected load is transformed into an equivalent Y. The impedance per
phase of the equivalent Y is

Z2 = 60 j45 = 20 - j15 O
3
The phase voltage is

VI = 207.85 = 120 Y
J3
The single-phase equivalent circuit is shown in Figure 2.20.

1 2+j40
a
+ l¡ 12
300 200
VI = 120LOoy V2
j400 -j150
n~----------------------~------~

FIGURE 2.20
Single-phase equivalent circuit for Example 2.7.

The total impedance is

. (30 + j40)(20 - j15)
Z = 2 + )4 + (30 + j40) + (20 - j15)
= 2 + j4 + 22 - j4 = 24 O
With the phase voltage Van as reference, the current in phase a is

1 = VI = 120LO° = 5 A
Z 24
The three-phase power supplied is

(b) The phase voltage at the load terminal is

V2 = 120LO° - (2 + j4)(5LOO) = 110 - j20

= 111.8L-1O.3° Y
 2.12. BALANCED THREB-PHASE POWER 41

The line voltage at the load terminal is

(c) The current per phase in the Y-connected load and in the equivalent Y of the Ll
load is

V2 110 . °
l¡ = Zl = 30 + j40 = 1-)2 = 2.236L - 63.4 A

1 = V2 = 110 j20 =4+ '2=4472L2656° A

2 Z2 20 j15 ) . .

The phase current in the original Ll-connected load, Le., 1ab is given by

1ab = 12 = 4.472L26.56° = 2.582L56.56° A

V3L-30° V3L-30°
(d) The three-phase power absorbed by each load is

SI = 3V21i = 3(111.8L - 1O.3°)(2.236L63.4°) = 450 W + j600var

S2 = 3V212 = 3(111.8L -1O.3°)(4.472L-26.56°) = 1200 W - j900var

The three-phase power absorbed by the line is

SL = 3(RL + jXdl1I2 = 3(2 + j4)(5)2 = 150 W + j300 var

It is clear that the sum of load powers and line losses is equal to the power delivered
from the supply, Le.,

SI + S2 + SL = (450 + j600) + (1200 - j900) + (150 + j300)

= 1800 W + jO var

Example2.8
A three-phase line has an impedance of 0.4 + j2. 7 n per phase. The line feeds two
balanced three-phase loads that are connected in parallel. The first load is absorb-
ing 560.1 kVA at 0.707 power factor lagging. The second load absorbs 132 kWat
unity power factor. The line-to-line voltage at the load end of the line is 3810.5 V.
Determine:

(a) The magnitude of the line voltage at the source end of the lineo
(b) Total real and reactive power loss in the lineo
(c) Real power and reactive power supplied at the sending end of the lineo
42 2. BASIC PRINCIPLES

1 0.4 + j2.7D.
a~~--~~rn~----~~------~
+ l¡

n~------------------~------~

FIGURE 2.21
Single-phase equivalent diagram for Example 2.8.

(a) The phase voltage at the load tenninals is

V2 = 3810.5 = 2200 V
V3
The single-phase equivalent circuit is shown in Figure 2.21.
The total complex power is

8 R (31/» = 560.1(0.707 + jO.707) + 132 = 528 + j396

= 660L36.87° kVA

With the phase voltage V2 as reference, the current in the line is

1 = 8 R(31/» = 660, OOOL -36.87° = 100L -36.87° A

3V2* 3(2200LOO)
The phase voltage at the sending end is

VI = 2200LO° + (0.4 + j2.7)100L-36.87° = 2401.7 L4.58° V

The magnitude of the line voltage at the sending end of the line is

!VIL I = V3!VII = V3(2401.7) = 4160 V

(b) The three-phase power los s in the line is

8 L (31/» = 3RIII2 + j3XIII2 = 3(0.4)(100)2 + j3(2.7)(100)2

= 12 kW + j81 kvar

(e) The three-phase sending power is

8 S(31/» = 3VI I* = 3(2401.7 L4.58°)(100L36.87°) = 540 kW + j477 kvar

lt is clear that the sum of load powers and the line losses is equal to the power
delivered from the supply, i.e.,

8 S(31/» = 8 R (31/» + 8 L (31/» = (528 + j396) + (12 + j81) = 540 kW + j477 kvar
 2.12. BALANCED THREE-PHASE POWER 43

PROBLEMS
2.1. Modify the program in Example 2.1 such that the following quantities can
be entered by the user:
The peak: amplitude Vm , and the phase angle B'IJ of the sinusoidal supply
v(t) = Vm cos(wt + B'IJ)' The impedance magnitude Z, and the phase angle
'Y of the load.
The program should produce plots for i(t), v(t), p(t), Pr(t) and Px(t), sim-
ilar to Example 2.1. Run the program for Vm = 100 V, B'IJ O and the
following loads:

An inductive load, Z = 1.25L6000

A capacitive load, Z = 2.0L -30°0
A resistive load, Z 2.5LOoO

(a) From Pr(t) and Px(t) plots, estimate the real and reactive power for each
load. Draw a conclusion regarding the sign of reactive power for inductive
and capacitive loads.
(b) Using phasor values of current and voltage, calculate the real and reactive
power for each load and compare with the results obtained from the curves.
(c) Ifthe above loads are all connected across the same power supply, deter-
mine the total real and reactive power taken from the supply.

2.2. A single-phase load is supplied with a sinusoidal voltage

v(t) = 200cos(377t)
The resulting instantaneous power is

p(t) = 800 + 1000 cos(754t 36.87°)

(a) Find the complex power supplied to the load.
(b) Find the instantaneous current i (t) and the rms value of the current sup-
plied to the load.
(c) Find the load impedance.
(d) Use MATIAB to plot v(t), p(t), and i(t) = p(t)/v(t) over a range of
Oto 16.67 ms in steps of 0.1 ms. From the current plot, estimate the peak:
amplitude, phase angle and the angular frequency of the current, and verify
the results obtained in part (b). Note in MATLAB the command for array or
element-by-element division is ./.

l.3. An inductive load consisting of R and X in series feeding from a 2400-V

rms supply absorbs 288 kW at a lagging power factor of 0.8. Determine R
andX.
44 2. BASIC PRINClPLES

2.4. An inductive load consisting of R and X in parallel feeding from a 2400-V

rms supply absorbs 288 kW at a lagging power factor of 0.8. Determine R
andX.

2.5. Two loads connected in parallel are supplied from a single-phase 240-V rms
source. The two loads draw a total real power of 400 kW at a power factor
of 0.8 lagging. One of the loads draws 120 kW at a power factor of 0.96
leading. Find the complex power of the other load.

2.6. The load shown in Figure 2.22 consists of a resistance R in parallel with a
capacitor of reactance X. The load is fed from a single-phase supply through
a line of impedance 8.4 + jl1.2 O. The rms voltage at the load terminal is
1200LO° V rms, and the load is taking 30 kVA at 0.8 power factor leading.
(a) Find the values of R and X.
(b) Determine the supply voltage V.

8.4 + jl1.2 O

V -jX

FIGURE 2.22
Circuit for Problem 2.6.

2.7. 1\vo impedances, Z¡ = 0.8+ j5.6 O and Z2 = 8- j16 O, and a single-phase

motor are connected in parallel across a 200-V rms, 60-Hz supply as shown
in Figure 2.23. The motor draws 5 kVA at 0.8 power factor lagging.

+ 1
0.8
8 3 = 5kVA
200LO° V
atO.8 PF lag
-j16

FlGURE2.23
Circuit for Problem 2.7.
 2.12. BALANCED THREB-PHASB POWER 45

(a) Find the complex powers SI, S2 for the two impedances, and S3 for the
motor.
(b) Determine the total power taken from the supply, the supply current, and
the overall power factor.
(c) A capacitor is connected in parallel with the loads. Find the kvar and the
capacitance in JLF to improve the overall power factor to unity. What is the
new line current?

2.8. Two single-phase ideal voltage sources are connected by atine of impedance
of 0.7 + j2.4 nas shown in Figure 2.24. VI = 500L16.26° V and V2 =
585 L0° V. Find the complex power for each machine and determine whether
they are delivering or receiving real and reactive power. Also, find the real
and the reactive power loss in the lineo

0.7 + j2.4 n

500L16.26° V 585LO° V

FIGURE 2.24
Circuit for Problem 2.8.

2.9. Write a MATlAB program for the system of Example 2.6 such that the volt-
age magnitude of source 1 is changed from 75 percent to 100 percent of
the given value in steps of 1 V. The voltage magnitude of source 2 and the
phase angles of the two sources is to be kept constant. Compute the complex
power for each source and the tine loss. Tabulate the reactive powers and
plot Q 1> Q2, and Q L versus voltage magnitude !VII. From the results, show
that the flow of reactive power along the interconnection is determined by
the magnitude difference of the terminal voltages.

2.10. A balanced three-phase source with the following instantaneous phase volt-
ages

Van = 2500cos(wt)
Vbn 2500cos(wt - 120°)
Ven = 2500cos(wt - 240°)
46 2. BASIC PRINCIPLES

supplies a balanced Y-connected load of impedance Z = 250L36.87° n per

phase.
(a) Using MATLAB, plot the instantaneous powers Pa, Pb, Pe and their sum
versus wt over a range of 0:0.05: 27r on the same graph. Comment on the
nature of the instantaneous power in each phase and the total three-phase
real power.
(b) Use (2.44) to verify the total power obtained in part (a).

2.11. A 4157-V rms, three-phase supply is applied to a balanced Y-connected

three-phase load consisting of three identical impedances of 48L36.87°n.
Taking the phase to neutral voltage Van as reference, calculate
(a) The phasor currents in each lineo
(b) The total active and reactive power supplied to the load.

2.12. Repeat Problem 2.11 with the same three-phase impedances arranged in a .6.
connection. Take Vab as reference.

2.13. A balanced delta connected load of 15 + j18 n per phase is connected at

the end of a three-phase line as shown in Figure 2.25. The line ímpedance is
1 + j2 n per phase. The line is supplied from a three-phase source with a
line-to-line voltage of 207.85 V rms. Taking Van as reference, determine the
following:
(a) Current in phase a.
(b) Total complex power supplied from the source.
(c) Magnitude of the line-to-line voltage at the load terminal.

1 + j2n
ao-------'\N\

IVLI = 207.85 V
b
bo-----~~ ~nn~---- ____< 15 + j18n

Co-----'V\¡'V\t-'

FIGURE 2.25
Circuit for Problem 2.13.

2.14. Three parallel three-phase Ioads are supplied from a 207.85-V rms, 60-Hz
three-phase supply. The loads are as follows:

Load 1: A 15 hp motor operating at fuIl-Ioad, 93.25 percent efficiency, and

0.6 Iagging power factor.
 2.12. BALANCED THREE-PHASE POWER 47

Load 2: A balanced resistive load that draws a total of 6 kW.

Load 3: A Y-connected capacitor bank with a total rating of 16 kvar.

(a) What is the total system kW, kvar, power factor, and the supply current
per phase?
(b) What is the system power factor and the supply current per phase when
the resistive load and induction motor are operating but the capacitor bank is
switched off?

2.15. Three loads are connected in parallel across a 12.47 kV three-phase supply.
Load 1: Inductive load, 60 kW and 660 kvar.
Load 2: Capacitive load, 240 kW at 0.8 power factor.
Load 3: Resistive load of 60 kW.
(a) Find the total complex power, power factor, and the supply current.
(b) A Y-connected capacitor bank is connected in parallel with the loads.
Find the total kvar and the capacitance per phase in ¡.tF to improve the overall
power factor to 0.8 lagging. What is the new line current?

2.16. A balanced Ll.-connected load consisting of apure resistances of 18 O per

phase is in parallel with a purely resistive balanced Y-connected load of 12 O
per phase as shown in Figure 2.26. The combination is connected to a three-
phase balanced supply of 346.41-V rms (line-to-line) vía a three-phase line
having an inductive reactance of j3 O per phase. Taking the phase voltage
Van as reference, determine
(a) The current, real power, and reactive power drawn from the supply.
(b) The line-to-neutral and the line-to-line voltage of phase a at the combined
load terminals.
j30
~-----J~~---------4r-------~a

IVLI = 346.41 V
b 180

120

FIGURE 2.26
Circuit for Prob1em 2.16.
CHAPTER

3
GENERATOR AND
TRANSFORMER MODELS;
THE PER-UNIT SYSTEM

3.1 INTRODUCTION
Before the power systems network can be solved, it must first be modeled. The
three-phase balanced system is represented on a per-phase basis, which was de-
scribed in Section 2.10. The single-phase representation is also used for unbalanced
systems by means of symmetrical components which is treated in a later chapter.
In this chapter we deal with the balanced system, where transmission lines are rep-
resented by the 7r model as described in Chapter 4. Other essential components
of a power system are generators and transformers; their theory and construction
are discussed in standard electric machine textbooks. In this chapter, we represent
simple models of generators and transformers for steady-state balanced operation.
Next we review the one-line diagram of a power system showing generators,
transformers, transmission lines, capacitors, reactors, and loads. The diagram is
usually limited to major transmission systems. As a rule, distribution circuits and
small loads are not shown in detail but are taken into account merely as lumped
loads on substation busses.
48
 3.2. SYNCHRONOUS GENERATORS 49

In the analysis of power systems, it is frequentIy convenient to use the per-

unit system. The advantage of this method is the elimination of transformers by
simple impedances. The per-unit system is presented, followed by the impedance
diagram ofthe network, expressed to a common MVA base.

3.2 SYNCHRONOUS GENERATORS

Large-scale power is generated by three-phase synchronous generators, known as
altemators, driven either by stearn turbines, hydroturbines, or gas turbines. The
armature windings are placed on the stationary part called stator. The annature
windings are designed for generation of balanced three-phase voltages and are ar-
ranged 10 develop the sarne number of magnetie poles as the field winding that is
on the rotor. The field which requires a relatively small power (0.2-3 percent of the
machine rating) for its excitation is plaeed on the rotor. The rotor is also equipped
with one or more short-cireuited windings known as damper windings. The rotor is
driven by a prime mover at constant speed and its field cireuit is excited by direct
current. The excitation may be provided through slip rings and brushes by means of
de generators (referred to as exciters) mounted on the sarne shaft as the rotor of the
synchronous machine. However, modern excitation systems usually use ae gener-
ators with rotating rectifiers, and are known as brushless excitation. The generator
excitation system maintains generator voltage and eontrols the reactive power flow.
The rotor of the synehronous machine may be of cylindrical or salient con-
struction. The eylindrieal type of rotor, also called round rotor, has one distributed
winding and a uniform air gap. These generators are driven by stearn turbines and
are designed for high speed 3600 or 1800 rpm (two- and four-pole machines, re-
spectively) operation. The rotor of these generators has a relatively large axial
length and small diarneter to limit the centrifugal forces. Roughly 70 pereent of
large synchronous generators are cylindrical rotor type ranging from about 150 to
1500 MVA. The salient type of rotor has eoncentrated windings on the poles and
nonuniform air gaps. It has a relatively large number of poles, short axial length,
and large diarneter. The generators in hydroelectric power stations are driven by
hydraulic turbines, and they have salient-pole rotor construction.

3.2.1 GENERATOR MODEL

An elementary two-pole three-phase generator is illustrated in Figure 3.1. The sta-
tor contains three coils, aa', bb', and c¿, displaced from eaeh other by 120 elec-
trica1 degrees. The concentrated full-piteh eoils shown here may be eonsidered to
represent distributed windings producing sinusoidal mmf waves coneentrated on
the magnetic axes of the respective phases. When the rotor is excited to produce
SO 3. GENERATOR AND TRANSFORMER MODELS; THE PER-UNlT SYSTEM

,,

FIGURE 3.1
Elementary two-pole three-phase synchronous generator.

an air gap flux of 4> per pole and is revolving at constant angular velocity w, the
flux linkage of the coil varíes with the position of the rotor mmfaxis wt, where
wt is measured in electrical radians from coil aa' magnetic axis. The flux linkage
for an N-tum concentrated coil aa' will be maximum (N4)) at wt = O and zero
at wt 1f' /2. Assuming distributed winding, the flux linkage Aa will vary as the
cosine of the angle wt. Thus, the flux linkage with coil a is

Aa = N4>coswt (3.1)

The voltage induced in coil aa' is obtained from Faraday's law as

ea = - dA
dt = wNA.'
'f'smwt
= Emax sin wt (3.2)
1f'
= Emax cos(wt - 2')
where

Emax = wN4> = 21f'j N4>

 3.2. SYNCHRONOUS GENERATORS 51

Therefore, the nns value of the generated voltage is

E = 4.44fN<jJ (3.3)

where f is the frequency in hertz. In actual ac machine windings, the annature

coil of each phase is distributed in a number of slots. Since the emfs induced in
different slots are not in phase, their phasor sum is less than their numerical sumo
Thus, a reduction factor Kw, called the winding factor, must be applied. For most
three-phase windings Kw is about 0.85 ro 0.95. Therefore, for a distributed phase
winding, the nns value of the generated voltage is

E = 4.44Kw fN<jJ (3.4)

The magnetic field of the rotor revolving at constant speed induces three-phase
sinusoidal voltages in the annature, displaced by 27r /3 radians. The frequency of
the induced annature voltages depends on the speed at which the rotor mns and
on the number of poles for which the machine is wound. The frequency of the
annature voltage is given by

Pn
f 260
(3.5)

where n is the rotor speed in rpm, referred to as synchronous speed. During nonnal
conditions, the generator operates synchronously with the power grid. This results
in three-phase balanced currents in the annature. Assuming current in phase a is
lagging the generated emf ea by an angle '!/J, which is indicated by tine mn in
Figure 3.1, the instantaneous annature currents are

ia = I max sin(wt
ib = I max sin(wt (3.6)

ie = I max sin(wt

According to (3.2) the generated emf ea is maximum when the rotor magnetic axis
is under phase a. Since ia is lagging ea by an angle '!/J. when line mn reaches
the axis of coil aal , current in phase a reaches its maximum value. At any instant
of time, each phase winding produces a sinusoidally distributed mmf wave with
íts peak along the axis of the phase winding. These sinusoidally distributed fields
can be represented by vectors referred to as space phasors. The amplitude of the
sinusoidalIy distributed mmf fa((}) is represented by the vector Fa along the axis of
phase a. Similarly, the amplitude ofthe mmfs fb((}) and fe((}) are shown by vectors
Fb and Fe along their respective axis. The mmf amplitudes are proportionaI to the
52 3. GENERATOR AND TRANSFORMER MODELS; THB PBR-UNIT SYSTBM

instantaneous value of the phase current, Le.,

Fa Kia = Klmaxsin(wt - 'I/J) = Fmsin(wt 'I/J)
Fb = Kib Klmaxsin(wt - 'I/J - 2;) = Fmsin(wt 'I/J 2; (3.7)

Fe = Kie = Klmaxsin(wt - 'I/J - ~) = Fmsin(wt - 'I/J - ~)

where K is proportional to the number of annature tums per phase and is a function
of the winding type. The resultant annature mmf is the vector sum of the above
mmfs. A suitable method for finding the resultant mmf is to project these mmfs on
line mn and obtain the resultant in-phase and quadrature-phase components. The
resultant in-phase components are

FI = Fm sin(wt 'I/J) cos(wt - 'I/J) + Fm sin(wt - 'I/J _ 2;)

211" 411" 411"
cos(wt - 'I/J - 3 + Fm sin(wt - 'I/J - 3) cos(wt - 'I/J - 3

Using the trigonometric identity sin a cos a = (1/2) sin 2a, the above expression
becomes

F I = ~m [sin 2(wt 'I/J) + SIn 2(wt - 'I/J - 3 )

. 211"

+sin2(wt 411" )J
3
The above expression is the sum of three sinusoidal functions displaced from each
other by 211"/3 radians, which adds up to zero, i.e., FI = O.
The sum of quadrature components results in

F2 = Fm sin(wt - 'I/J) sin(wt - 'I/J) + Fm sin(wt 'I/J 2;) sin(wt - 'I/J 211")
3
+Fm sin(wt - 'I/J - ~) sin(wt - 'I/J
Using the trigonometric identity sin 2 a = (1/2)(1 cos 2a), the above expression
becomes
Fm [3 - cos 2(wt - 'I/J) + cos 2(wt2- 1
F2 = 2 'I/J 1 "
- 3)
411"
+cos2(wt-'l/J- 3)J
The sinusoidal terms of the above expression are displaced from each other by
211" /3 radians and add up to zero, with F2 = 3/2Fm. Thus, the amplitude of the
resultant annature mmf or stator mmf becomes
3
Fs = '2 Fm (3.8)
 3.2. SYNCHRONOUS GENERATORS S3

\
\
\

/
/
/
/
/
/
/
I /
/
I

~--~~~-------------'E
/
/
/
/
/
/'
m

FIGURE 3.2
Combined phasor/vector diagram for one phase of a cylindrical rotor generator.

We thus conclude that the resultant annature mmf has a constant amplitude
perpendicular to line mn, and rotates at a constant speed and in synchronism
with the field mmf Fr. To see a demonstration of the rotating magnetic field, type
rotfield at the MATlAB prompt.
A typical synchronous machine field alignment for operation as a generator is
shown in Figure 3.2, using space vectors to represent the various fields. When the
rotor is revolving at synchronous speed and the annature current is zero, the field
rnmf Fr produces the no-load generated emf E in each phase. The no-load gen-
erated voltage which is proportional to the field current is known as the excitatíon
voltage. The phasor voltage for phase a, which is lagging Fr by 90°, is combined
with the mmf vector diagram as shown in Figure 3.2. This combined phasor/vector
diagram leads to a circuit model for the synchronous machine. Jt must be empha-
sized that in Figure 3.2 rnmfs are space vectors, whereas the emfs are time phasors.
When the annature is carrying balanced three-phase currents, Fs is produced per-
pendicular to line mn. The interaction of annature mmf and the field mmf, known
as armature reaction, gives rise to the resultant air gap mmf Fsr . The resultant mmf
Fsr is the vector sum of the field mmf Fr and the annature mmf Fs. The resultant
mmf is responsible for the resultant air gap flux 4>sr that induces the generated emf
on-load, shown by E sr . The annature mmf Fa induces the emf E ar , known as the
armature reaction voltage, which is perpendicular to FIJ. The voltage Ear leads
S4 3. GENERATOR AND TRANSFORMER MODELS; THE PER-UNIT SYSTEM

la by 900 and thus can be represented by a voltage drop across a reactance X ar

due to the current la. X ar is known as the reactance o/the armature reaction. The
phasor sum of E and Ear is shown by E sr perpendicular to Fsr , which represents
the on-Ioad generated emf.
(3.9)
The terminal voltage V is less than E sr by the amount of resistive voltage drop
Rala and leakage reactance voltage drop X/la. Thus
E= V + [Ra + j(Xl + Xar)]Ia (3.10)
or
E = V + [Ra + jXs]Ia (3.11)

where X s = (Xl + X ar ) is known as the synchronous reactance. The cosine ofthe

angle between I and V, i.e., cos (J represents the power factor at the generator ter-
minals. The angle between E and E sr is equal to the angle between the rotor mmf
Fr and the air gap mmf Fsr , shown by dr . The power developed by the machine
is proportional to the product of Fr, Fsr and sin dr • The relative positions of these
mmfs dictates the action of the synchronous machine. When Fr is ahead of Fsr by
an angle dr. the machine is operating as a generator and when Fr falls behind Fsr ,
the machine will act as a motor. Since E and E sr are proportional to Fr and Fsr ,
respectively, the power developed by the machine is proportional to the products of
E, E sr , and sin dr • The angle dr is thus known as the power angle. This is a very
important result because it relates the time angle between the phasor emfs with
the space angle between the magnetic fields in the machine. Usually the developed
power is expressed in terms of the excitation voltage E, the terminal voltage V,
and sin d. The angle d is approximately equal to dr because the leakage impedance
is very small compared to the magnetization reactance.
Due to the nonlinearity of the machine magnetization curve, the synchronous
reactance is not constant. The unsaturated synchronous reactance can be found
from the open- and short-circuit data. For operation at or near rated terminal volt-
age, it is usually assumed that the machine is equivalent to an unsaturated one
whose magnetization curve is a straight line through the origin and the rated volt-
age point on the open-circuit characteristic. For steady-state analysis, a constant
value known as the saturated value o/ the synchronous reactance corresponding to
the rated voltage is used. A simple per-phase model for a cylindrical rotor genera-
tor based on (3.11) is obtained as shown in Figure 3.3. The armature resistance is
generally much smaller than the synchronous reactance and is often neglected. The
equivalent circuit connected to an infinite bus becomes that shown in Figure 3.4,
and (3.11) reduces to
(3.12)
 3.2. SYNCHRONOUS GENERATORS SS

FIGURE 3.3
Synchronous machine equivalent circuito

E v
rt%~_~{a_---11
01
FIGURE 3.4
Synchronous machíne connected to an ínfinite bus.

Figure 3.5 shows the phasor diagram of the generator with tenninal voltage
as referenee for exeitations eorresponding to lagging, unity, and leading power fac-
torso The voltage regulation of an alternator is a figure of merit used for eompari-

E E

~ M3>
E

la V V

(a) Lagging pf load (b) Upfload (e) Leading pf load

FIGURE 3.5
Synchronous generator phasor diagram.

son with other machines. It is defined as the pereentage ehange in terminal voltage
from no-load to rated load. This gives an indication of the ehange in field eurrent
required to rnaintain system voltage when going frorn no-load to rated load at sorne
specifie power factor.

VR IVnll - IVrated I x 100 IEI - IVrated I x 100 (3.13)

IVratedl IVrated I
The no-load voltage for a specifie power factor rnay be detennined by operating
the maehine at rated load eonditions and then rernoving the load and observing
56 3. GENERATOR ANO TRANSFORMER MODELS; THE PER-UNIT SYSTEM

the no-load voltage. Since this is not a practical method for very large machines,
an accurate analytical method recommended by IEEE as given in reference [43]
may be used. An approximate method that provides reasonable results is to con-
sider a hypothetical linearized magnetization curve drawn to intersect the actual
magnetization curve at rated voltage. The value of E calculated from (3.12) is then
used to find the field current from the linearized curve. Finally, the no-load voltage
corresponding to this field current is found from the actual magnetization curve.

3.3 STEADY-STATE CHARACTERISTICS-

CYLINDRICAL ROTOR
3.3.1 POWER FACTOR CONTROL
Most synchronous machines are connected to large interconnected electric power
networks. These networks have the important characteristic that the system voltage
at the point of connection is constant in magnitude, phase angle, and frequency.
Such a point in a power system is referred to as an infinite bus. That is, the voltage
at the generator bus will not be altered by changes in the generator's operating
condition.
The ability to vary the rotor excitation is an important feature of the syn-
chronous machine, and we now consider the effect of such a variation when the
machine operates as a generator with constant mechanical input power. The per-
phase equivalent circuit of a synchronous generator connected to an infinite bus is
shown in Figure 3.4. Neglecting the armature resistance, the output power is equal
to the power developed, which is assumed to remain constant given by

(3.14)

where V is the phase-to-neutral terminal voltage assumed to remain constant. From

(3.14) we see that for constant developed power at a fixed terminal voltage V,
la cos (J must be constant. Thus, the tip of the armature current phasor must fall on
a verticalline as the power factor is varied by varying the field current as shown in
Figure 3.6. From this diagram we have

cd = El sin 81 = Xslal cos (JI (3.15)

Thus El sin 81 is a constant, and the locus of El is on the line ej. In Figure 3.6,
phasor diagrams are drawn for three armature currents. Application of (3.12) for a
lagging power factor armature current lal results in El. If (J is zero, the generator
operates at unity power factor and armature current has a minimum value, shown
by la2, which results in E2. Similarly, Es is obtained corresponding to laS at a
leading power factor. Figure 3.6 shows that the generation of reactive power can
r
3.3. STEADY-STATE CHARACTERISTICS- CYLINDRICAL ROTOR 57

FIGURE 3.6
Variation of field current at constant power.

be controlled by means of the rotor excitation while maintaining a constant real

power output. The varíation in the magnitude of armature current as the excitation
voltage is varíed is best shown by a curve. Usually the field current is used as the
abscissa instead of excitation voltage because the field current is readily measured.
The curve of the armature current as the function of the field current resembles the
letter V and is often referred to as the V curve of synchronous machines. These
curves constitute one of the generator's most important characteristics. There is, of
course, a limit beyond which the excitation cannot be reduced. This limit is reached
when 6 = 900 • Any reduction in excitation below the stability limit for a particular
load will cause the rotor to puli out of synchronism. The V curve is illustrated in
Figure 3.7 (page 62) for the machine in Example 3.3.

3.3.2 POWER ANGLE CHARACTERISTICS

Consider the per-phase equivalent circuit shown in Figure 3.4. The three-phase
complex power at the generator terminal is

(3.16)

Expressing the phasor voltages in polar form, the armature current is

la = IEIL6 -IVILO (3.17)

IZsIL,
Substituting for I~ in (3.16) results in

S3.!. 31EIIVI L'V - 6 (3.18)

'1' IZsl I
58 3. GENERATOR AND TRANSFORMER MODELS; THE PER-UNIT SYSTEM

Thus, the real power P3</> and reactive power Q3</> are

IEIIVI
P3</> = 3--¡z:r cosb' - ó) - 3
1V12 cos')' (3.19)
1Zs1
IEIIVI .
Q3</> = 3--¡z:r smb' - ó) - 3
1V12 sm')'
. (3.20)
1Zs1
If Ra is neglected, then Zs = jXs and')' 90°. Equations (3.19) and (3.20) reduce
to

P3</> = 3IE~~1 sin ó (3.21)

Q3</> = 3~sl(IElcosó IVD (3.22)

Equation (3.21) shows that if IEI and IVI are held fixed and the power angle ó
is changed by varying the mechanical driving torque, the power transfer varies
sinusoidally with the angle ó. From (3.21), the theoretical maximum power occurs
when ó = 90°

IEIIVI
Pmax (3</» = 3~ (3.23)

The behavior of the synchronous machine can be described as follows. If we start

with ó = 0° and increase the driving torque, the machine accelerates, and the rotor
mmf Fr advances with respect to the resultant mmf Fsr ' This results in an increase
in ó, causing the machine to deliver electric power. At sorne value of ó the machine
reaches equilibrium where the electric power output balances the increased me-
chanícal power owing to the increased driving torque. It is clear that if an attempt
were made to advance ó further than 90° by increasing the driving torque, the
electric power output would decrease from the Pmax point. Therefore, the excess
driving torque continues to accelerate the machine, and the mmfs will no longer be
magnetically coupled. The machine loses synchronism and automatic equipment
disconnects it from the system. The value Pmax is called the steady-state stabil-
ity limit or static stability limito In general, stability considerations dictate that a
synchronous machine achieve steady-state operation for a power angle at consid-
erably less than 90°. The control of real power ftow is maintained by the generator
govemor through the frequency-power control channel.
Equation (3.22) shows that for small ó, cos ó is nearly unity and the reactive
power can be approximated to

(3.24)
204 6. POWER FLOW ANALYSIS

or in matrix form
CI - (f¡)(0) b.. (O)
(~) (O) (~)(O) Xl

C2 - (12)(0)
=
(!!h.) (O)
8Xl
(~ )(0)
8x n
b.. x (O)
2

(~) (O) b.. Xn(O)

In short form, it can be written as

b..C(k) = J(k) b..X(k)

or
b..X(k) = [J(k)¡-l b..C(k) (6.19)
and the Newton-Raphson algorithm for the n-dimensional case becomes
X(k+l) = X(k) + b..X(k) (6.20)

where
b.. (k)
Xl
b.. X (k)
2
[ el -
C2 -
(Jl)(k)
(12)(k)
1
b..X(k) = and b..C(k) = (6.21)
..
b.. (k)
Xn
en - (Jn)(k)

(~)(k) (~)(k) (~)(k)

(~)(k) (~)(k) (~)(k)
J(k) = (6.22)

(Q.fu ) (k) (Q.fu ) (k) ( !Un. ) (k)

8Xl 8X2 8x n

J(k) is called the Jacobian matrix. Elements of this matrix are the partial
derivatives evaluated at X(k). It is assumed that J(k) has an inverse during each
iteration. Newton's method, as applied to a set of nonlinear equations, reduces the
problem to solving a set of linear equations in order to determine the values that
improve the accuracy of the estimates.
The solution of (6.19) by inversion is very inefficient. It is not necessary
to obtain the inverse of J(k). Instead, a direct solution is obtained by optimally
ordered triangular factorization. In MATLAB, the solution of linear simultaneous
equations b..C = J b..X is obtained by using the matrix division operator \ (i.e.,
b..X = J \ b..C) which is based on the triangular factorization and Gaussian elim-
ination.
486 11. STABILITY

A = [O 1 ; -37.705 -2.617] ;
r = eig(A)
result in

r =
-1.3 + 6.00i
-1.3 + 6.00i

The linearized model for small disturbances is very useful when the system
is extended to include the govemor action and the effect of automatic voltage reg-
ulators in a multimachine system. The linearized model allows the application of
the linear control system analysis and compensation, which will be dealt with in
Chapter 12.

11.5 TRANSIENT STABILITY -

EQUAL-AREA CRITERION

The transient stability studies involve the determination of whether or not synchro-
nism is maintained after the machine has been subjected to severe disturbance.
This may be sudden application of load, loss of generation, los s of large load, or a
fault on the system. In most disturbances, oscillations are of such magnitude that
linearization is not permissible and the nonlinear swing equation must be solved.
A method known as the equal-area criterion can be used for a quick predic-
tion of stability. This method is based on the graphical interpretation of the energy
stored in the rotating mass as an aid to determine if the machine maintains its sta-
bility after a disturbance. The method is only applicable to a one-machine system
connected to an infinite bus or a two-machine system. Because it provides physi-
cal insight to the dynamic behavior of the machine, application of the method to
analysis of a single machine connected to a large system is considered here.
Consider a synchronous machine connected to an infinite bus. The swing
equation with damping neglected as given by (11.21) is

where Pa is the accelerating power. From the aboye equation, we have

2
d6 = 'Ir Jo (P. _ P. )
dt2 H m e