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AC Motors and
Generators
Student Guide
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Table of Contents
INTRODUCTION ...................................................................................................................... 1
TLO 1 TYPES OF AC GENERATORS ....................................................................................... 2
Overview ........................................................................................................................... 2
ELO 1.1 Generator Theory ............................................................................................... 3
ELO 1.2 Generator Components....................................................................................... 9
ELO 1.3 Electrical Terms ............................................................................................... 11
ELO 1.4 AC Generator Losses ....................................................................................... 15
ELO 1.5 AC Generator Ratings ...................................................................................... 17
ELO 1.6 Types of AC Generators................................................................................... 18
ELO 1.7 AC Generator Connection Schemes................................................................. 20
TLO 1 Summary ............................................................................................................. 24
TLO 2 TYPES OF AC MOTORS ............................................................................................. 25
Overview ......................................................................................................................... 25
ELO 2.1 Producing a Rotating Magnetic Field .............................................................. 26
ELO 2.2 Slip Effects on AC Induction Motor Operation ............................................... 30
ELO 2.3 Developing Torque in an AC Induction Motor ................................................ 32
ELO 2.4 Starting Current on an AC Induction Motor .................................................... 34
ELO 2.5 Developing Torque in a Single-Phase AC Motor ............................................ 38
ELO 2.6 AC Synchronous Motor Operation .................................................................. 39
TLO 2 Summary ............................................................................................................. 41
TLO 3 AC MOTOR OPERATION ........................................................................................... 42
Overview ......................................................................................................................... 42
ELO 3.1 Motor Applications .......................................................................................... 43
ELO 3.2 Indications of Motor Malfunctions .................................................................. 45
ELO 3.3 Consequences of Motor Overheating ............................................................... 46
ELO 3.4 Excessive Current in Motors ............................................................................ 48
ELO 3.5 Pump Motor Current Relationships ................................................................. 50
TLO 3 Summary ............................................................................................................. 56
TLO 4 AC GENERATOR OPERATION .................................................................................... 57
Overview ......................................................................................................................... 57
ELO 4.1 Power Factor and Generators ........................................................................... 58
ELO 4.2 Voltage Regulation and Components............................................................... 63
ELO 4.3 Paralleling Generators ...................................................................................... 66
ELO 4.4 Generator Excitation ........................................................................................ 77
TLO 4 Summary ............................................................................................................. 83
AC MOTORS AND GENERATORS SUMMARY ......................................................................... 84
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AC Motors and Generators
Revision History
Revision
Date
Version
Number
Purpose for Revision
Performed
By
11/3/2014
0
New Module
OGF Team
12/11/2014
1
Added signature of OGF
Working Group Chair
OGF Team
Introduction
AC motors and generators are vital to plant operation. The AC main
generator is the plant output, and other AC generators provide emergency
power supplies and back-ups. AC motors are the most common means to
drive the pumps, compressors and other equipment necessary to all
Rev 1
1
industrial plants, so they are common throughout the plant.
Understanding the operation and care of AC motors and generators is vital
to successful power plant operation. The plant depends on these
components, and operators must be able to properly operate and monitor
them.
Objectives
At the completion of this training session, the trainee will demonstrate
mastery of this topic by passing a written exam with a grade of 80 percent
or higher on the following Terminal Learning Objectives (TLOs):
1. Describe the construction, operating characteristics, and limitations
for an AC generator.
2. Explain the theory of operation of selected types of AC motors.
3. Explain operating characteristics and limitations on AC motors.
4. Analyze operation of generators in parallel and predict system
response to changes in frequency, voltage, and load.
TLO 1 Types of AC Generators
Overview
In this section, you will learn about the components of AC generators and
their functions, as well as the different types of AC generators and the
advantages of each.
AC Generators
AC generators are vital to the power plant. The product of the plant is the
output of the main AC generator, and each plant has additional AC
generators for emergency power supplies and back-up supplies for other
loads. Understanding the operation of each type is critical to successful
operation.
Objectives
Upon completion of this lesson, you will be able to do the following:
1. Describe the theory of operation of an AC generator.
2. State the purpose of the following components of an AC generator:
a. Field
b. Armature
c. Prime mover
d. Rotor
e. Stator
f. Slip rings
3. Define the following electrical terms:
a. Volts
b. Amps
c. VARs
Rev 1
2
4.
5.
6.
7.
d. Watts
e. Hertz
List the three losses found in an AC generator.
Describe the bases behind the kW and current ratings of an AC
generator.
Describe the following:
a. Stationary field
b. Rotating armature AC generator
c. Rotating field, stationary armature AC generator
Describe a wye-connected and delta-connected AC generator
including advantages and disadvantages of each type.
ELO 1.1 Generator Theory
Introduction
In this section, you will learn about the theory of operation of an AC
generator.
Generator Theory
A simple AC generator consists of a conductor or loop of wire in a
magnetic field produced by an electromagnet. The two ends of the loop
connect to slip rings. The slip rings are in contact with two brushes. When
the loop rotates, it cuts magnetic lines of force, first in one direction, and
then the other.
As the conductor passes through the magnetic field, the magnetic field
induces a voltage in the conductor and the slip rings conduct the induced
voltage as voltage output.
Magnitude of Generated Voltage
The magnitude of AC voltage generated by an AC generator is dependent
on the field strength and speed of the rotor. Most generators operate at a
constant speed; therefore, the generated voltage depends on field excitation,
or strength.
Developing an AC Sine Wave Voltage
As the coil rotates in a counter-clockwise direction, each side of the coil
cuts through the magnetic lines of force in opposite directions (see the
figure below). The direction (polarity) of the induced voltages depends on
the direction of movement of the coil.
Rev 1
3
Figure: Simple AC Generator
When the loop is in the vertical position, at 0 degrees, (0 degrees of rotation
in the figure below) the coils are moving parallel to the magnetic field and
do not cut magnetic lines of force. At that instant, there is no voltage
induced in the loop. As the coil rotates in a counter-clockwise direction,
each side of the coil cuts the magnetic lines of force in opposite directions.
The direction (polarity) of the induced voltages depends on the direction of
movement of the coil.
The induced voltages are additive, making slip ring X (see the previous
figure) positive (+) and slip ring Y negative (-). The potential across
resistor R causes a current to flow from Y to X through the resistor. This
current increases until it reaches a maximum value when the coil is
horizontal to the magnetic lines of force at 90 degrees. At that instant, the
horizontal coil is moving perpendicular to the magnetic field and cutting the
greatest number of magnetic lines of force.
As the coil continues to turn, the induced voltage and current decrease until
both reach zero, when the coil is again in the vertical position (180 degrees).
The next half revolution produces an equal voltage, except with reversed
polarity (270 degrees and 360 degrees). The current flow through R is now
from X to Y. The figure below shows the coil position as it rotates through
360 degrees.
Rev 1
4
Figure: Developing an AC Sine Wave Voltage
The alternating reversal of polarity results in the generation of a voltage, as
shown above. As the coil rotates through 360 degrees, voltage output in the
shape of a sine wave results.
Period and Frequency



E = voltage
I = current
P = power
When an AC generator produces a voltage, the resulting current varies in
step with the voltage. As the generator coil rotates 360 degrees, the output
voltage goes through one complete cycle. In one complete cycle, the
voltage increases from zero to Emax in one direction, decreases to zero,
increases to Emax in the opposite direction (negative Emax), and then
decreases to zero again.
The period is the time required for the generator to complete one cycle. The
frequency (measured in hertz) is the number of cycles completed per
second.
Peak Voltage and Current
In the figure below, Emax occurs at 90 degrees. This value is termed the
peak voltage. One way to quantify AC voltage or current is by peak value,
Rev 1
5
peak voltage (Ep) or peak current (Ip). Peak means the maximum voltage or
current appearing on an AC sine wave.
Peak to Peak Voltage and Current
Another commonly used term associated with AC is peak-to-peak value
(Ep-p or Ip-p). Peak to peak refers to the magnitude of voltage, or current
range, spanned by the sine wave, as denoted in the figure below.
Figure: AC Sine Wave Voltage
Effective Value of AC
A sine wave like the one shown below graphically presents the AC
generator output.
Rev 1
6
Figure: AC Voltage Sine Wave
Effective value is the value most commonly used for quantifying AC. The
effective value of AC is the amount of AC that produces the same heating
effect as an equal amount of DC. A one-ampere effective value of AC will
produce the same amount of heat in a conductor, in a given time, as one
ampere of DC.
The heating effect of a given AC current is proportional to the square of the
current. It is possible to calculate the effective value of AC by squaring all
the amplitudes of the sine wave over one period, taking the average of these
values, and then taking the square root of the average. The effective value,
because it is the root of the mean (average) square of the currents, is the
root-mean-square, or RMS value.
Phase Angle Guidelines
Phase angle is the fraction of a cycle, in degrees, that has gone by since a
voltage or current has passed through a given value. The given value is
normally zero. From the figure below, take point 1 on the sine wave as the
starting point or zero phase. The phase angle at Point 2 is 30 degrees, Point
3 is 60 degrees, Point 4 is 90 degrees, and so on, until Point 13 where the
phase angle is 360 degrees, or zero once again.
Figure: AC Voltage Sine Wave
Phase Angle Example
Phase difference is another common term for phase angle. Phase difference
describes two different voltages that have the same frequency, which pass
through zero values in the same direction, but at different times. In the
figure below, the angles along the axis indicate the phases of voltages e1
and e2. At 120 degrees, e1 passes through the zero value, which is 60
degrees ahead of e2 (e2 equals zero at 180 degrees). We describe this as
voltage e1 leads e2 by 60 electrical degrees, or voltage e2 lags e1 by 60
electrical degrees.
Rev 1
7
Figure: Phase Relationship
Phase difference can also compare two different currents or a current and a
voltage. If the phase difference between two currents, two voltages, or a
voltage and a current is zero degrees, they are termed to be in phase. If the
phase difference is any amount other than zero, they are out of phase.
Knowledge Check
An AC generator has all of the following except
______________.
A.
a commutator
B.
a magnetic field
C.
slip rings
D.
a conductor in relative motion with the magnetic field
Knowledge Check
When two AC voltages reach their peak voltage at the
same time, the voltages are said to be ______________.
Rev 1
A.
lagging
B.
out of phase
C.
in phase
D.
leading
8
ELO 1.2 Generator Components
Introduction
In this section, you will learn the major components of the AC generator
and the function of each.
Generator Components
The figure below shows components of a basic AC generator. The
following sections discuss key components of an AC generator.
Figure: Basic AC Generator
Field
The field in an AC generator consists of coils of conductors within the
generator that receive a voltage from a source (excitation) and produce a
magnetic flux. The magnetic flux in the field cuts the windings in the
armature to produce a voltage. This voltage is the output voltage of the AC
generator.
Armature
The armature is the voltage production source of an AC generator. The
armature consists of many coils of wire that are large enough to carry the
full-load current of the generator.
Prime Mover
The prime mover is the component that drives the AC generator rotor. The
prime mover may be any type of rotating machine, such as a diesel engine,
a steam turbine, or a motor.
Rev 1
9
Rotor
The rotor of an AC generator is the rotating component of the generator.
The generator's prime mover (a steam turbine, gas turbine, or diesel engine)
drives the rotor. Depending on the type of generator, the rotor may be
either the armature or the field. The rotor is the armature when it is the
source of voltage generation. The rotor is the field when it receives the
field excitation.
Stator
The stator of an AC generator is the part of the machine that is stationary.
Like the rotor, this component may be the armature or the field, depending
on the type of generator. The stator is the armature when it is the source of
voltage output generation. The stator is the field when it receives the field
excitation.
Slip Rings
Slip rings are electrical connections used to transfer power to or from the
rotor of an AC generator. A slip ring consists of a circular conducting
material mounted on the rotor shaft, as shown in the figure below. The slip
ring connects electrically to the rotor windings and insulation isolates it
electrically from the shaft. Brushes ride on and contact the slip ring as the
rotor rotates. The brushes usually connect to the output of the generator
(self-excited), by un-insulated braided copper wires.
Figure: Slip Rings and Brushes
Magnetic Field
Current flow through the field coil of the rotor produces a strong magnetic
field. Slip rings and brushes conduct excitation to the field coil in the rotor.
Rev 1
10
Continuous Connection
Brushes are spring-held in contact with the slip rings to provide a
continuous connection between the field coil and excitation circuit, on a
rotating field machine.
Conductors
Each time the rotor makes one complete revolution, one complete cycle of
AC is developed. A generator has many turns of wire wound into the slots
of the stator.
Knowledge Check
The part of an AC generator that produces the output
voltage is called the:
A.
Armature
B.
Field
C.
Stator
D.
Rotor
ELO 1.3 Electrical Terms
Introduction
In this section, you will learn about common electrical terms used in
electrical measurement and AC voltage generation.
Units of electrical measurement include the following:
Unit of Measurement
Symbol
Unit is Used to Measure:
Ampere
I or A
Electrical current
Volt
E or V
Electrical potential difference
Hertz
Hz
Frequency (f)
Ohm
R or Ω
Resistance to current flow
Siemens
S
A material’s ability to conduct
current flow (G)
Watt
W
Power (P)
Rev 1
11
Unit of Measurement
Symbol
Unit is Used to Measure:
Henry
H
Electrical inductance (L)
Farad
F
Electrical capacitance (C)
VAR
VAR
Reactive power (Q)
Voltage (Volt)
Voltage, electromotive force (EMF), or potential difference, is the pressure
or force that causes electrons to move in a conductor. In electrical formulas
and equations, you will see voltage symbolized with a capital E, while on
laboratory equipment or schematic diagrams, you will see voltage
symbolized with a capital V.
Current (Ampere or Amp)
Electron current, or amperage, is the movement of free electrons through a
conductor. In electrical formulas, you will see current symbolized with a
capital I, while in the laboratory or on schematic diagrams, you will see
current symbolized with a capital A to indicate amps or amperage (amps).
Resistance (Ohm)
Resistance is the opposition to current flow. The amount of opposition to
current flow produced by a material depends upon the amount of available
free electrons and the types of obstacles the electrons encounter as they
attempt to move through the material. The symbol (R) represents resistance
in equations.
One ohm is that amount of resistance that will limit the current in a
conductor to one ampere when the potential difference (voltage) applied to
the conductor is one volt.
The symbol for the ohm is the Greek letter capital omega (Ω).
If a voltage difference acts on a conductor, current flows. The amount of
current flow depends upon the resistance of the conductor: the lower the
resistance, the higher the current flow for a given amount of voltage; the
higher the resistance, the lower the current flow.
The relationship between these three parameters is referred to as Ohm’s
law,(𝐸 = 𝐼 × 𝑅), and will be covered in more detail later in the module.
Conductance (Siemens)
The opposite, or reciprocal, of resistance is conductance. Recall that
resistance is the opposition to current flow. Since resistance and
conductance are opposites, conductance is the ability to conduct current.
Rev 1
12
For example, if a wire has a high conductance, it will have low resistance,
and vice-versa. To calculate conductance, compute the reciprocal of the
resistance. Siemens is the unit used to specify conductance, named in honor
of German inventor Ernst Werner von Siemens. The symbol for electrical
conductance is capital letter S. When used in a formula, the letter G
represents conductance.
The equation below is the mathematical representation of conductance
obtained by relating the definition of conductance (1/R) to Ohm's Law.
𝐺=
1
𝐼
=
𝑅𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑅
For example, if a resistor (R) has five ohms, its conductance will be 0.2
siemens.
Power (Watt)
Electricity generally performs work, such as turning a motor or generating
heat. Specifically, power is the rate of performing work, or the rate of heat
generation. The unit commonly used to specify electric power is the watt.
In equations, the capital letter P denotes power, and the capital letter W
denotes watts. Power is also described as the current (I) in a circuit
multiplied by the voltage (E) across the circuit. The equation below is a
mathematical representation of this concept.
𝑃 = 𝐼 × 𝐸 or 𝑃 = 𝐼𝐸
Using Ohm’s Law for the value of voltage (E), E = I x R and using
substitution laws,
𝑃 = 𝐼 × (𝐼 × 𝑅)
Therefore, power also equals the current (I) in a circuit squared multiplied
by the resistance (R) of the circuit or as shown in the formula listed below:
𝑃 = 𝐼2𝑅
Inductance (Henry)
Inductance is the ability of a coil to store energy, induce a voltage in itself,
and oppose changes in current flowing through it. You will see a capital L
used to indicate inductance in electrical formulas and equations. Henries
are the units of measurement for inductance.
The capital letter H denotes the henry. One henry is the amount of
inductance that permits one volt to be induced when the current through the
coil changes at a rate of one ampere per second. The mathematical
representation of the rate of change in current through a coil per unit time
is:
Rev 1
13
∆𝐼
( )
∆𝑡
The equation below is the mathematical representation for the voltage VL
induced in a coil with inductance L. The negative sign indicates that
voltage induced opposes the change in current through the coil per unit
time(∆𝐼/∆𝑡).
∆𝐼
𝑉𝐿 = −𝐿 ( )
∆𝑡
A later section in this lesson presents additional detail on inductance.
Capacitance (Farad)
Capacitance is the ability to store an electric charge and is symbolized by
the capital letter C. Capacitance (C) is measured in farads, and is equal to
the amount of charge (Q) that can be stored in a device or capacitor divided
by the voltage (E) applied across the device or capacitor plates when the
charge was stored. The equation below is the mathematical representation
for capacitance.
𝐶=
𝑄
𝐸
Frequency (Hertz)
Frequency (measured in hertz) is the number of alternating voltage or
current cycles completed per second.
Volt Ampere Reactive (VAR)
VAR is the unit of reactive power; reactive power in a circuit does no useful
work. In AC circuits that are not purely resistive, voltage and current will
be out of phase with each other, power is exchanged in these circuits as
inductive fields form and collapse, and capacitors charge and discharge.
The power triangle graphically displays the relationship between reactive
power, apparent power and true power.
The power triangle, shown below, equates AC power to DC power by
showing the relationship between:



generator output (Apparent Power - S) in volt-amperes (VA),
usable power (True Power - P) in watts, and
wasted or stored power (Reactive Power - Q) in volt-amperes-reactive
(VAR).
The phase angle (θ) represents the inefficiency of the AC circuit and
corresponds to the total reactive impedance (Z) to current flow in the
circuit.
Rev 1
14
Figure: Power Triangle
The power triangle represents comparable values that can be used directly
to find the efficiency level of generated power to usable power, which is
expressed as the power factor (discussed later). You can calculate Apparent
Power, Reactive Power, and True Power by using the DC equivalent (RMS
value) of the AC voltage and current components, along with the power
factor.
Knowledge Check
The unit commonly used to specify electric power is:
A.
Volt
B.
Amp
C.
Watt
D.
Hertz or frequency
ELO 1.4 AC Generator Losses
Introduction
In this section, you will learn the different types of losses that occur during
generator operation.
Generator Losses
Three different types of physical phenomena cause generator losses:



Internal resistance and reactance
Hysteresis losses
Mechanical losses
The following sections explain each of these types.
Internal Resistance and Reactance
Rev 1
15
The load current flows through the armature in all AC generators. Like any
coil, the armature has some amount of resistance and inductive reactance.
The combination of resistance and reactance comprise the internal
resistance, which causes a loss of efficiency in an AC generator. When the
load current flows, a voltage drop develops across the internal resistance.
This voltage drop lowers the output voltage and, therefore, represents
generated voltage and power that is lost and not available to the load.
Hysteresis Losses
Hysteresis losses refer to this expenditure of energy to realign magnetic
domains in a ferromagnetic material. The figure below shows an example
of a hysteresis loop for a particular ferromagnetic material. The larger the
area enclosed by the loop, the greater the hysteresis losses associated with
the material.
Figure: Hysteresis Loop
Hysteresis losses occur when a magnetic field acts on the iron cores in an
AC generator. The number of magnetic domains held in alignment with the
field is dependent on the strength of the magnetic field. The magnetic
domains rotate, with respect to the domains not held in alignment, one
complete turn during each rotation of the rotor.
This rotation of magnetic domains in the iron causes friction and heat. The
energy associated with the heat produced is magnetic hysteresis loss. To
reduce hysteresis losses, most AC armatures are constructed of heat-treated
silicon steel, which has an inherently low hysteresis loss. After forming the
heat-treated silicon steel to the desired shape, the laminations go through an
annealing process comprised of heating the laminations to a dull red, then
allowing them to cool. This annealing process reduces hysteresis losses to a
very low value.
Mechanical Losses
Rotational or mechanical losses can be caused by bearing friction, brush
friction on the slip rings, and air friction (called windage), which is caused
Rev 1
16
by the air turbulence due to rotation of the rotor. Careful maintenance is
instrumental in keeping bearing friction to a minimum. Clean bearings and
proper lubrication are essential to the reduction of bearing friction.
Ensuring proper brush seating, correct brush selection, and maintaining
proper brush tension will reduce brush friction. A smooth and clean slip
ring also aids in the reduction of brush friction. In very large generators,
hydrogen replaces air within the generator for cooling; and being less dense
than air, causes less windage losses than air. Windage losses also are
dependent on the shape of the armature and speed of rotation.
Knowledge Check
Which of the following physical phenomena is not
considered a typical category of generator loss?
A.
Mechanical losses
B.
Hysteresis losses
C.
Power factor losses
D.
Internal resistance and reactance losses
ELO 1.5 AC Generator Ratings
Introduction
In this section, you will learn about ratings and nameplate data for AC
generators.
Generator Ratings
Typical nameplate data for an AC generator includes:










Manufacturer name and information
Serial number
Type number
Speed (rpm)
Number of poles
Frequency of output
Number of phases
Maximum supply voltage
Armature and field current per phase
Maximum temperature rise
The table below shows an example of an AC generator nameplate.
Westinghouse
AC Generator Air Cooled No. 6750616 Type ATB
3600 rpm
Rev 1
17
2 Poles 60 Hertz 3-Phase Wye-Connected for
13800 Volts
Rating 15625 kVA 12500 kW 0.80 PF Exciter
250 Volts
Armature 654 amp Field 183 amp
Guaranteed Temperature Rise Not To Exceed
60°c On Armature By Detector
80°c On Field By Resistance
Figure: AC Generator Nameplate Example
Power (KW) Ratings
The ability of the prime mover to overcome generator losses and the ability
of the machine to dissipate the internally generated heat form the basis for
AC generator power (kW) ratings.
Current Ratings
The insulation rating of the machine and the ability of the windings to
dissipate heat form the basis for AC generator current ratings.
Knowledge Check
The power rating of an AC generator is based primarily
on…
A.
the speed of the machine in rpm.
B.
the strength of the field current.
C.
the number of field pole pairs on the generator and the
speed of the machine.
D.
the ability of the prime mover to overcome generator
losses and the ability of the machine to dissipate the
internally-generated heat.
ELO 1.6 Types of AC Generators
Introduction
In this section, we will cover types and classification of AC generators.
There are two types of AC generators: the stationary field with rotating
armature; and the rotating field with stationary armature.
Rev 1
18
Stationary Field with Rotating Armature
Small AC generators usually have a stationary field and a rotating armature,
as shown in the figure below. One important disadvantage of this style
generator is that the slip ring and brush assembly is in series with the load
circuits. Dirt or wear in these components results in a reduction or
interruption in current flow to loads. For very large generators, the slip
rings and brushes become the limiting factor on the machine capacity.
Figure: Stationary Field with Rotating Armature Generator
Rotating Field with Stationary Armature
Upon connecting a DC field excitation to the rotor, the stationary coils
receive induced AC. This arrangement is a rotating field with stationary
armature AC generator, as shown in the figure below.
Figure: Rotating Field with Stationary Armature Generator
Large power generation applications typically use the rotating field,
stationary armature type AC generator. In this type of generator, a DC
source is supplied to the rotating field coils (via slip rings and brushes),
which produces a magnetic field around the rotating element.
As the prime mover turns the rotor, the field rotates across the conductors of
the stationary armature, inducing an EMF into the armature windings.
Rev 1
19
This type of AC generator has the following advantages over the stationary
field, rotating armature AC generator:

It is possible to connect a load to the armature without having moving
contacts (slip rings and brushes) in the circuit.
 It is much easier to insulate stator conductors than rotating
conductors.
 It is possible to generate very higher output voltages and currents.
Knowledge Check
The rotating armature, stationary field type is used for
most large power generators.
A.
True
B.
False
ELO 1.7 AC Generator Connection Schemes
Three-phase AC Generators
The principles of a three-phase generator are the same as that of a singlephase generator, except that there are three equally spaced windings and the
three output voltages are 120 degrees out of phase with one another.
Physically adjacent loops are separated by 60 degrees of rotation; however,
the loops are connected to the output in such a manner that there are 120
electrical degrees between phases.
The individual coils of each winding are combined, and represented as a
single coil. The top part of the figure below shows that the three-phase
generator has three separate armature windings that are 120 electrical
degrees out of phase with each other. The bottom part of the figure shows
the output voltages and the 120 electrical degree separations of the phase
voltages.
Rev 1
20
Figure: Stationary Armature Three-Phase AC Generator
As shown in the upper figure above, six leads coming from the armature of
a three-phase generator connect the generator output to an external load. In
actual practice, the windings are connected together, and only three leads
are brought out and connected to the external load.
There are two orientations available to connect the three armature windings:
a delta connection and a wye connection.
Delta (Δ) connected generator
In one type of connection, the windings connect in series, or deltaconnected (Δ) as shown in the figure below.
In a delta-connected generator, the voltage between any two phases, called
line voltage, is the same as the voltage generated in any one phase.
Rev 1
21
Figure: Delta-Connected Generator Windings
The figure below shows the electrical characteristics of the delta-connected
windings; the three-phase voltages are equal, as are the three line voltages.
The current in any line is √3 (1.73) times the phase current. A deltaconnected generator provides an increase in current, but no increase in
voltage.
Figure: Characteristics of a Delta-Connected Generator
An advantage of the delta-connected AC generator is that if one phase is
damaged or open, the remaining two phases can still deliver three-phase
power. However, the capacity of the damaged or open generator is 57.7
percent of what it was with all three-phases in operation.
Wye (Y) Connected Generator
In the other type of connection, one lead of each winding connects to a
common point, and the remaining three leads connect to an external load.
The figure below shows this type of connection, known as a wye (Y)
connection, also referred to as a Starr connection.
Figure: Wye-Connected Generator Windings
Wye (Y) Connected Generator
The voltage and current characteristics of the wye-connected AC generator
are opposite to that of the delta connection. The figure below shows the
Rev 1
22
electrical characteristics of the delta-connected windings; voltage between
any two lines in a wye connected AC generator is 1.73 times the voltage of
any one phase, while line currents are each equal to the phase currents. The
wye-connected AC generator provides an increase in voltage, but no
increase in current.
Figure: Characteristics of a Wye Connected Generator
An advantage of a wye-connected AC generator is that each phase only has
to carry 57.7 percent of line voltage and, therefore, high voltage generation
applications use this type of generator.
Knowledge Check
Select all of the following that are true.
Rev 1
A.
In a delta-connected generator, line current is equal to
phase current.
B.
In a wye-connected generator, line current is equal to
phase current.
C.
In a delta-connected generator, line voltage is equal to
phase voltage.
D.
In a wye-connected generator, line voltage is equal to
phase voltage.
23
TLO 1 Summary
In this section, you learned about generator theory, types of AC generators,
major components, ratings, and losses, and the two common generator
output connection schemes.
1. A simple AC generator consists of a conductor or loop of wire in a
magnetic field produced by an electromagnet, and relative motion
between them.
2. Major components consist of a field, armature, prime mover, rotor,
stator, and slip rings/brushes.
3. Electrical terminology associated with generators include, voltage,
current, frequency, power/watts, and volt-ampere reactive (VAR).
4. Major generator losses are caused by three different types of physical
phenomena:
 Internal resistance and reactance
 Hysteresis losses
 Mechanical losses
5. Generator ratings will normally include speed, number of poles,
frequency, output voltage and power, and maximum temperature rise.
6. AC Generators are generally classified as either stationary field with
rotating armature or rotating field with stationary armature machines.
Large commercial generators are usually rotating field with stationary
armature to mitigate the limitations of slip rings and brushes on
generator output.
7. The two common connection schemes used for AC three-phase
generators are delta-connections and wye-connections.
 Delta-connected: An advantage is that if one phase becomes
damaged or open, the remaining two phases can still deliver
three-phase power. However, the capacity of the generator is
reduced to 57.7 percent. The voltage will be constant on any two
phases, however current will be 1.73 times the single-phase
value.
 Wye-connected: The voltage and current characteristics of the
wye-connected AC generator are opposite to that of the delta
connection. Voltage between any two lines in a wye connected
AC generator is 1.73 times the voltage of any one phase, while
line currents are equal to the phase currents. An advantage of a
wye-connected AC generator is that each phase only has to carry
57.7 percent of line voltage.
Summary
Now that you have completed this lesson, you should be able to do the
following:
1. Describe the theory of operation of an AC generator.
2. State the purpose of the following components of an AC generator:
a. Field
b. Armature
c. Prime mover
Rev 1
24
3.
4.
5.
6.
7.
d. Rotor
e. Stator
f. Slip rings
Define the following electrical terms:
a. Volts
b. Amps
c. VARs
d. Watts
e. Hertz
List the three losses found in an AC generator.
Describe the bases behind the kW and current ratings of an AC
generator.
Describe the following:
a. Stationary field
b. Rotating armature AC generator
c. Rotating field, stationary armature AC generator
Describe a wye-connected and delta-connected AC generator
including advantages and disadvantages of each type.
TLO 2 Types of AC Motors
Overview
AC motors drive machinery for a wide variety of applications at the station;
to monitor and operate these motors correctly, an operator needs to have an
understanding of their design and operating characteristics.
There are two major types of AC motors used in industrial applications:
induction and synchronous motors.
Induction motors are the most commonly used AC motor in industrial
applications because of their simple, rugged construction, and relatively low
manufacturing costs. This TLO will cover the theory of operation for AC
induction motors first, followed by synchronous motors theory and
operating characteristics.
AC motors drive most of the equipment needed to make the power plant
work, including pumps, compressors, fans, and other components.
Objectives
Upon completion of this lesson, you will be able to do the following:
1.
2.
3.
4.
Describe how a rotating magnetic field is produced in an AC motor.
Define slip and explain its effect on AC induction motor operation.
Describe how torque is produced in an AC motor.
Explain differences between starting and running current for an AC
induction motor, and the operating limits used to mitigate the effects
of starting current.
5. Describe how torque is produced in a single-phase AC motor.
Rev 1
25
6. Describe how an AC synchronous motor is started and its operating
characteristics.
ELO 2.1 Producing a Rotating Magnetic Field
Introduction
The basic principle of operation for AC motors is the interaction of a
revolving magnetic field created in the stator by AC current, with an
opposing magnetic field in the rotor. The magnetic field in the rotor comes
from one of two sources: induced from the rotating stator, or provided by a
separate DC current source. The resulting interaction produces torque that
turns a shaft that can operate rotating machinery such as pumps,
compressors, fans, etc.
Rotating Magnetic Field
To understand how a rotating magnetic field causes a motor rotor to turn, it
is necessary to understand how to produce a rotating magnetic field. The
figure below illustrates a three-phase stator with three-phase AC current
applied.
The windings connect in a wye configuration. The two windings in each
phase wind in the same direction. At any instant in time, the magnetic field
generated by one particular phase will depend on the current flow through
that phase. If the current flow through that phase is zero, the resulting
magnetic field is zero. If the current flow is at a maximum value, the
resulting field is at a maximum value.
Since the currents in the three windings are 120 degrees out of phase, the
magnetic fields produced will also be 120 degrees out of phase. The stator
will produce a rotating magnetic field, which is a combination of the fields
produced by the three individual sets of windings. The stator’s magnetic
fields will act upon the rotor, inducing a current and a corresponding
magnetic field in the rotor.
In an AC induction motor, the magnetic field induced in the rotor is
opposite in polarity to the rotating magnetic field in the stator.
Consequently, as the magnetic field rotates in the stator, the rotor also
rotates in an attempt to maintain its alignment with the stator’s magnetic
field. The figure below shows each of three phases separately, and all three
combined in a three-phase motor stator.
Rev 1
26
Figure: Three-Phase Motor Stator
From one instant to the next, the magnetic fields of each stator phase
combine to produce a magnetic field whose position shifts through a certain
angle. At the end of one cycle of alternating current, the magnetic field will
have shifted through 360 degrees, or one revolution around the stator.
Since the rotor has an opposing magnetic field induced upon it, it will also
rotate through one revolution.
For the purpose of explanation, the figure below shows the rotation of the
stator’s magnetic field "stopped" at six selected positions, or instances.
These instances are marked off at 60 degrees intervals on the sine waves
representing the current flowing in the three-phases, A, B, and C.
For the following discussion, when the current flow in a particular phase is
positive, the magnetic field will develop a north pole at the poles labeled A,
B, and C. When the current flow in a particular phase is negative, the
magnetic field will develop a north pole at the poles labeled A', B', and C'.
Rev 1
27
Figure: Rotating Magnetic Field
T1
At point T1, the current in phase C (solid line on figure above) is at its
maximum positive value. At the same instance, the currents in phases A
and B are at half of the maximum negative value. The resulting magnetic
field is established vertically downward, with the maximum field strength
developed across the C phase, between pole C (north) and pole C' (south).
This magnetic field is aided by the weaker fields developed across phases A
and B, with poles A' and B' being north poles and poles A and B being
south poles.
T2
At Point T2, the current sine waves have rotated through 60 electrical
degrees. At this point, the current in phase A has increased to its maximum
negative value. The current in phase B has reversed direction. The currents
in phases B and C are at half of the maximum positive value.
Rev 1
28
The resulting magnetic field is established downward to the left, with the
maximum field strength developed across the A phase, between poles A’
(north) and A (south). This magnetic field is aided by the weaker fields
developed across phases B and C, with poles B and C being north poles and
poles B’ and C’ being south poles. Thus, the magnetic field within the
stator of the motor has physically rotated 60 degrees.
T3
At Point T3, the current sine waves have again rotated 60 electrical degrees
from the previous point for a total rotation of 120 electrical degrees. At this
point, the current in phase B has increased to its maximum positive value.
The current in phase A has decreased to half of its maximum negative
value, while the current in phase C has reversed direction and is at half of
its maximum negative value.
The resulting magnetic field is established upward to the left, with the
maximum field strength developed across phase B, between poles B (north)
and B' (south). This magnetic field is aided by the weaker fields developed
across phases A and C, with poles A' and C' being north poles and poles A
and C being south poles. The magnetic field on the stator has rotated
another 60 degrees for a total rotation of 120 degrees.
T4
At Point T4, the current sine waves have rotated 180 electrical degrees from
Point T1 so that the relationship of the phase currents is identical to Point
T1 except that the polarity has reversed. Since phase C is again at a
maximum value, the resulting magnetic field developed across phase C will
be of maximum field strength. However, reversing the current flow in
phase C establishes the magnetic field vertically upward between poles C
(north) and C (south). As can be seen, the magnetic field has now
physically rotated a total of 180 degrees from the starting point at T1.
T5
At Point T5, phase A is at its maximum positive value, which establishes a
magnetic field upward to the right. Again, the magnetic field has physically
rotated 60 degrees from the previous point for a total rotation of 240
degrees.
T6
At Point T6, phase B is at its maximum negative value, which will establish
a magnetic field downward to the right. The magnetic field has again
rotated 60 degrees from Point T5 for a total rotation of 300 degrees.
Rev 1
29
T7
Finally, at Point T7, the current returns to the same polarity and values as
that of Point T1. Therefore, the magnetic field established at this instance
will be identical to that established at Point T1.
From this discussion, for one complete cycle of the electrical sine wave
(360 degrees), the magnetic field developed in the stator of a motor will
rotate one complete revolution (360 degrees). Therefore, applying a threephase AC to three windings symmetrically spaced around a stator, produces
a rotating magnetic field in the stator.
Knowledge Check
Select all of the statements about creating a rotating
magnetic field in a three-phase induction motor that are
true.
A.
Since the windings are 120 degrees out of phase, the
magnetic field produced by the windings will also be 120
degrees out of phase.
B.
The magnetic field induced in each winding rises and
falls in a sine wave as the current through that winding
rises and falls in a sine wave.
C.
The magnetic field generated in the motor windings
makes 60 revolutions per second in a 60 Hz power
system.
D.
The magnetic field induced in the windings is
independent of how the motor is wound and connected.
ELO 2.2 Slip Effects on AC Induction Motor Operation
Introduction
In this section, you will learn how the three-phase AC induction motor
generates a magnetic field on the rotor and causes motion in the rotating
element.
Slip Effects on AC Induction Motor Operation
It is impossible for the rotor of an AC induction motor to turn at the same
speed as that of the rotating magnetic field. If the speed of the rotor were
the same as that of the stator, no relative motion between them would exist,
and there would be no induced EMF in the rotor (recall that inducing a
current requires relative motion between a conductor and a magnetic field).
Rev 1
30
Without this induced EMF, there would be no interaction of magnetic fields
to produce motion. Therefore, the rotor must rotate at some speed less than
that of the magnetic field in the stator for relative motion to exist.
Slip is the percentage difference between the speed of the rotor and the
speed of the rotating magnetic field in the stator. The smaller the
percentage difference, the closer the rotor speed is to the rotating magnetic
field speed.
The amount of torque on the rotor will change as the slip ratio changes
between the rotor and stator. The change in torque (seen in the figure
below) shows that, as slip increases from zero to about 20 percent, the
torque increases linearly. As the load and slip increase beyond full-load
torque, the torque will reach a maximum value at about 25 percent slip.
𝑆𝑙𝑖𝑝 =
𝑁𝑠 − 𝑁𝑎
= 100 𝑝𝑒𝑟𝑐𝑒𝑛𝑡
𝑁𝑠
Where Ns = Synchronous speed of the rotating field, and Na = Actual speed
or the rotor, with typical values of between 1 and 8 percent.
Figure: Torque versus Slip for an AC Induction Motor
Knowledge Check
Select all of the statements that are true regarding threephase induction motor operation.
A.
Rev 1
The rotor turns at a speed slower than the rotation of the
stator magnetic field.
31
B.
The rotor turns at the same speed as the stator magnetic
field.
C.
The difference in speed between the stator magnetic field
and the rotor is necessary to induce a magnetic field on
the rotor.
D.
The rotor has field windings supplied by an external
power supply
ELO 2.3 Developing Torque in an AC Induction Motor
Introduction
In this section, you will learn how a three-phase AC Induction Motor
develops torque.
Torque Production
When alternating current flows through the stator windings of an AC
induction motor a rotating magnetic field results, as discussed above. The
rotating magnetic field cuts the conductors of the rotor and induces a
current in them due to generator action.
This induced current will produce a magnetic field, opposite in polarity of
the field associated with the stator, around the conductors of the rotor. The
magnetic field induced on the rotor will try to line up with the magnetic
field produced by the stator. Since the stator field is rotating continuously,
the rotor cannot line up with, or lock onto, the stator field and, therefore,
must follow behind it. The figure below shows force directions of the
induced rotor magnetic field.
Figure: Induced Rotor Magnetic Field
Rev 1
32
Breakdown Torque
The maximum value of torque that an AC induction motor can produce
without stalling is the breakdown torque of the motor. If load increases
beyond this point, the motor will stall and come to a rapid stop. The typical
induction motor breakdown torque varies from 200 percent to 300 percent
of full-load torque.
Starting Torque
Starting torque is the value of torque at 100 percent slip and is normally 150
percent to 200 percent of full-load torque. As the rotor accelerates, torque
will increase to breakdown torque and then decrease to the value required to
carry the load on the motor at a constant speed, usually between 0-10
percent slip.
Developing Torque
The torque of an AC induction motor is dependent upon the strength of the
interacting rotor and stator fields and the phase relationship between them.
The equation below shows the mathematical expression for AC motor
torque:
𝑇 = 𝐾𝛷𝐼𝑅 cos 𝜃𝑅
Where:

T = torque (lb-ft or N-m)

K = constant

Φ = stator magnetic flux

IR = rotor current (A)

cos 𝜃𝑅 = power factor of rotor
During normal operation, K, Φ, and cos 𝜃𝑅 are essentially constant, so
torque is directly proportional to the rotor current. Rotor current increases
in almost direct proportion to slip, so maximum current will occur at
maximum slip, which is at motor start-up.
Rev 1
33
Note
100 percent slip occurs upon motor startup. The
magnetic field on the stator is rotating as soon as current
is applied to the stator upon starting the motor. Initially,
there is no magnetic field on the rotor. It is necessary to
induce the magnetic field on the rotor; the induced field
must always lag the magnetic field on the stator. Once
there is relative motion between the stator and the rotor,
the rotor will develop a magnetic field, which will try to
lineup with the rotating magnetic field of the stator.
Knowledge Check
The maximum value of torque that an AC induction
motor can produce without stalling is known as:
A.
Starting torque
B.
Breakdown torque
C.
Stalling torque
D.
Critical torque
ELO 2.4 Starting Current on an AC Induction Motor
Introduction
In this section, you will learn the causes and effects of motor starting
current.
Starting Current
Starting current in an AC motor is a phenomenon that the operator must
understand in order to properly monitor and operate large loads in the
station.
Starting current in an AC motor is greater than the motor’s normal running
current. There are several reasons for this high starting current:

Power is required to build up the rotating magnetic field in the stator
in its initial stationary position.
 Extra energy is required to overcome the inertia of the rotor in order
to place it in motion.
 Interactions occur between the rotor currents and the stator’s magnetic
field, which result in the motor drawing high currents.
 Rotor speed is too low to generate sufficient counter electromotive
force in the stator.
Rev 1
34
At motor start, the rotating magnetic field from the stator cuts the
conductors in the rotor inducing in EMF in the rotor. The rotor’s
conductors short to one another, causing current flow and a resulting
magnetic field around the rotor. Upon energizing the motor, at the instant
of energizing, the rotating magnetic field is produced in the stator, but the
rotor is not yet rotating. This results in high voltages and induced currents
in the rotor.
Since the rotor requires time to develop a magnetic field, the counter EMF
takes some time to build up. The lack of counter EMF during starting
combined with the very low resistance of the motor windings cause the very
high currents on startup of a motor.
Starting current is typically five to seven times the amount of motor full
load current. The motor full load current is the amount of current that a
motor draws from the power source when the motor is at full capacity. The
motor’s nameplate data (ratings) always shows full load current for the
motor.
An ammeter (current meter) shows the high starting current for a large
motor upon startup. In many cases, the ammeter’s needle will “peg high”
on motor start, and then drop down to the normal operating current value.
This is a normal indication. If starting current is not observed, it should be
considered an abnormal indication that requires investigation.
Running Current
Once an AC motor passes beyond startup, the amount of current it draws
while running is directly proportional to the load placed on the motor, a
motor attached to a pump provides a common example. As flow through
the pump increases (such as when personnel open the throttle valve), the
load on the motor increases, causing the pump to slow down slightly under
the increased load.
When the pump slows, slip increases and the magnitude of the rotor
magnetic field and torque increase. The increase in rotor field strength
allows the motor to carry more load, but the slower rotor causes counter
EMF to drop slightly, increasing motor current. If the throttle valve closes,
flow through the pump decreases, load on the motor decreases, and motor
current decreases.
Starting Current Discussion
To better understand the effect of starting current, the following example
considers the reason why starting current is higher than running current.
Resistance in the windings and voltage across the windings determines the
current drawn by a three-phase induction motor (stator current). The
resistance does not change significantly, and the voltage applied does not
change significantly.
Rev 1
35
However, the counter EMF caused by the rotating magnetic field of the
rotor does change.
Upon motor startup, the rotor is not moving, so the stator generates no
counter EMF. As the rotor comes up to speed, the counter EMF that it
induces in the stator increases, until it reaches a balance point.
When the load on the motor balances with the torque it is producing, rotor
speed will stabilize and the counter EMF induced by the rotor stabilizes.
This will cause the stator current to stabilize at the running current.
Starting Current Indications
The indications expected when starting a large AC induction motor are:
Amps immediately increase to many times (normally stated as five to seven
times) the normal running current value. Usually this will cause the current
indication to go off scale high, since monitoring running current is the
normal basis for meter scale selection.
After a few seconds, current will come back in scale, and decrease to
normal running current for the load applied. Normally, startup loads for
large motors are no load or very low load, so the current should drop to the
low end of the running current band after startup.
If the current does not drop back into the normal band for running current
within a few seconds, this is an indication that the equipment is not
functioning properly, and it is appropriate to shut it down and investigate
the reason for the malfunction.
Due to the large starting current and heat produced on motor start-up on
large pumps, plant operating procedures will often limit the number of
pump starts allowed within a given time period to prevent overheating the
motor windings.
Knowledge Check
If the discharge valve of a large motor-driven centrifugal
pump remains closed during a normal pump start, the
current indication for the AC induction motor will rise
to...
Rev 1
A.
several times the full-load current value, and then
decrease to the no-load current value.
B.
approximately the full-load current value, and then
decrease to the no-load current value.
C.
approximately the full-load current value, and then
stabilize at the full-load current value.
36
D.
several times the full-load current value, and then
decrease to the full-load value.
Knowledge Check
Which one of the following is a characteristic of a typical
AC induction motor that causes starting current to be
greater than running current?
A.
After the motor starts, resistors are added to the electrical
circuit to limit the running current.
B.
A large amount of starting current is required to initially
establish the rotating magnetic field.
C.
The rotor does not develop maximum induced current
flow until it has achieved synchronous speed.
D.
The rotor magnetic field induces an opposing voltage in
the stator that is proportional to rotor speed.
Knowledge Check – NRC Bank
Two identical AC induction motors are connected to
identical radial-flow centrifugal pumps in identical but
separate cooling water systems. Each motor is rated at
200 hp. The discharge valve for pump A is fully open
and the discharge valve for pump B is fully closed. Each
pump is currently off.
If the pumps are started under these conditions, the
shorter time period required to reach a stable running
current will be experienced by the motor for pump _____
and the higher stable running current will be experienced
by the motor for pump _____.
Rev 1
A.
A; A
B.
A; B
C.
B; A
D.
B; B
37
ELO 2.5 Developing Torque in a Single-Phase AC Motor
Introduction
In this section, you will learn about single-phase motor operation.
The rotating magnetic field associated with AC motors described above was
for a three-phase electrical system. A single-phase electrical system must
use a specially designed motor to obtain a “rotating” magnetic field with
sufficient strength to produce starting torque. These types of motors are
split-phase induction motors.
Single-Phase AC Motors
If two stator windings of unequal impedance are spaced 90 electrical
degrees apart and connected in parallel to a single-phase source, the field
produced when current is applied will appear to rotate. This phenomenon is
“phase splitting”.
A split-phase motor uses a starting winding. This winding has a higher
resistance and lower reactance than the main winding. When the starting
and main windings receive the same voltage (Vt), the current in the main
winding (Im) lags behind the current of the starting winding (Is). The angle
between the two windings (θ) is enough of a phase difference to provide a
rotating magnetic field strong enough to produce a starting torque. The
figure below shows a schematic diagram and the phase relationships in a
split-phase motor.
Figure: Split-Phase Induction Motor
After starting, when the motor reaches 70 percent to 80 percent of
synchronous speed, a centrifugal switch on the motor shaft opens and
disconnects the starting winding from the circuit. Once the motor is up to
normal operating speed, the rotating magnetic field created by the singlephase windings is sufficient to keep the motor running and the starting
winding is no longer required.
Very small commercial applications such as household appliances and floor
buffers commonly use single-phase motors.
Rev 1
38
Knowledge Check
Select all of the statements about single-phase AC
motors that are true.
A.
The starting winding in a single-phase AC motor is
always energized while the motor is running.
B.
Single-phase AC motors have a starting winding that is
out of phase with the main winding.
C.
The resistance and reactance of the starting winding must
be the same as the main winding.
D.
Single-phase AC motors are used for small load
applications.
ELO 2.6 AC Synchronous Motor Operation
Introduction
In this section, you will learn about the operating characteristics of a
synchronous motor.
AC Synchronous Motor Operation
The second major type of AC motor used in industrial applications is the
synchronous motor. Synchronous motors are like induction motors in that
they both have stator windings that produce a rotating magnetic field.
Unlike an induction motor, an external DC source excites the synchronous
motor rotor, and therefore, a synchronous motor requires slip rings and
brushes to provide current to the rotor.
In the synchronous motor, the rotor locks into step with the rotating
magnetic field of the stator and rotates at synchronous speed. If the
synchronous motor is loaded to the point where it pulls the rotor out of step
with the rotating magnetic field, no torque is developed, and the motor will
stop.
AC Synchronous Motor Operation
Synchronous motors use a wound rotor, shown in the figure below. This
type of rotor contains coils of wire placed in the rotor slots. Slip rings and
brushes convey current to the rotor. This type of rotor is much more
expensive to manufacture than the squirrel-cage rotor associated with AC
induction motors.
Rev 1
39
Figure: Wound Rotor
Starting a Synchronous Motor
A synchronous motor is not a self-starting motor because it develops torque
only when it is running at synchronous speed; therefore, the motor needs
some type of device to bring the rotor up to synchronous speed.
There are two ways of starting a synchronous motor:
1. Use a separate DC motor
2. Embed squirrel-cage windings on the face of the rotor
If starting a synchronous motor with a separate DC motor, both motors may
share a common shaft. Upon bringing the DC motor to synchronous speed,
the stator windings receive AC current. The DC motor then acts as a DC
generator and supplies DC field excitation to the rotor of the synchronous
motor. Once operating at synchronous speed, the synchronous motor is
ready for load.
More often, a squirrel-cage winding embedded in the face of the rotor poles
starts synchronous motors. The motor starts as an induction motor, speed
increases to about 95 percent of synchronous speed, then direct current is
applied, and the motor begins to pull into synchronism. Pull-in torque is the
term for the torque required to pull the motor into synchronism.
Because of their more complex nature and higher manufacturing cost,
designers rarely specify synchronous motors as their first choice. However,
large industrial applications sometimes use synchronous motors to
accommodate large loads or to improve power factor of transformers in
Rev 1
40
large industrial applications, and in applications where a constant speed is
important.
Knowledge Check
Select all the statements that are true.
A.
Synchronous motors are the most commonly used motors
in large industrial applications.
B.
Synchronous motors develop starting torque in the same
manner as induction motors.
C.
Synchronous motors are not often used due to their more
complex nature and higher manufacturing costs.
D.
Synchronous motors can continue to operate with more
slip than induction motors, thus producing more torque.
TLO 2 Summary
Types of AC Motors
In this section, you learned how AC motors work, including three-phase
induction motors, single-phase (split-phase) induction motors, and
synchronous motors.
1. AC current in the stator induces a rotating magnetic field in a threephase motor. An AC induction motor is the most commonly used
type of AC motor in industrial use.
 The rotating magnetic fields in the stator induce an opposing
rotating field in the rotor that will induce a CEMF in the rotor.
2. The rotor does not rotate at the same speed as the stator, the
difference in speed is slip, or ratio of stator speed to rotor speed, the
rotor will operate at about 5 to 10 percent less than the speed of the
stator.
 The magnetic field induced on the rotor will try to line up with the
magnetic field produced by the stator. Since the stator field is
rotating continuously, the rotor cannot line up with, or lock onto,
the stator field and, therefore, must follow behind it.
3. The torque of an AC induction motor is dependent upon the strength
of the interacting rotor and stator fields and the phase relationship
between them. Torque for an AC motor can be expressed
mathematically as 𝑇 = 𝐾𝛷𝐼𝑅 cos 𝜃𝑅
4. Starting current in an AC motor is greater than the normal running
current that is drawn by the motor, due to:
 Power is required to build up the rotating magnetic field in the
stator in its initial stationary position.
 Extra energy is required to overcome the inertia of the rotor.
Rev 1
41
 Interactions occur between the rotor currents and the stator’s
magnetic field, resulting in the motor drawing high currents.
 Rotor speed is too low to generate sufficient counter
electromotive force in the stator.
5. A split-phase motor uses a starting winding. This winding has a
higher resistance and lower reactance than the main winding. The
angle between the two windings is enough of a phase difference to
provide a rotating magnetic field strong enough to produce a starting
torque. A centrifugal switch removes the starting winding from the
circuit at about 75 percent of synchronous speed.
 An AC synchronous motor rotates at the same speed as the
rotating magnetic field (frequency) of the incoming voltage.
6. A synchronous motor is not a self-starting motor because it develops
torque only when it is running at synchronous speed; therefore, the
motor needs some type of device to bring the rotor up to synchronous
speed. There are two ways of starting a synchronous motor:
 Use a separate DC motor
 Embed squirrel-cage windings on the face of the rotor
Summary
Now that you have completed this lesson, you should be able to do the
following:
1.
2.
3.
4.
Describe how an AC motor produces a rotating magnetic field.
Define slip and explain its effect on AC induction motor operation.
Describe how an AC motor produces torque.
Explain differences between starting and running current for an AC
induction motor, and the operating limits used to mitigate the effects
of starting current.
5. Describe how a single-phase AC motor produces torque.
6. Describe how an AC synchronous motor is started, and describe its
operating characteristics.
TLO 3 AC Motor Operation
Overview
In this section, you will learn about applications for different AC motor
types, indications, and consequences of motor malfunctions, methods to
limit starting current, and relationships of motor power to pump head and
flow conditions.
AC motors power most of the equipment required to operate the power
plant, including pumps, compressors and other auxiliary equipment.
Understanding their operation, how to monitor them, and care for them is
vital to the operator role.
Objectives
Upon completion of this lesson, you will be able to do the following:
Rev 1
42
1. State the applications of the following types of AC motors:
a. Induction
b. Single-phase
c. Synchronous
2. Describe the indications resulting from a locked motor rotor, sheared
shaft, or miswired phases.
3. Describe the consequences of overheated motor windings or motor
bearings.
4. Describe the causes of excessive current in motors and generators
resulting from conditions such as low voltage, overloading, and
mechanical binding.
5. Describe the relationship between pump motor current and the
following parameters:
a. Pump flow
b. Pressure
c. Speed
d. Motor stator temperature
ELO 3.1 Motor Applications
Introduction
Specific applications require AC motors with varying attributes. Best
equipment performance results when the application requirements for a
motor match the motor attributes. The sections below describe some of the
more common types of AC motors used in industry.
Induction Motors
The induction motor is the most commonly used motor in industrial
applications because it is less expensive to manufacture and maintain, and
operates reliably. Pumps, fans, compressors and other equipment used
throughout the station commonly use induction motors.
The induction motor derives its name from the fact that the rotating
magnetic field of the stator induces AC currents into the rotor. It is the
most commonly used AC motor in industrial applications because of its
simplicity, rugged construction, and relatively low manufacturing costs.
This mainly attributed to the construction of the self-contained rotor, with
no external connections. Most AC induction motors use a particular type of
rotor, known as a “squirrel-cage” rotor.
Rev 1
43
Figure: Squirrel-Cage Rotor
Squirrel-Cage Rotor
The induction motor rotor (shown in the figure below) is made of a
laminated cylinder with slots in its surface. The name squirrel-cage comes
from the idea that the rotor resembles a cage that could contain a small
animal, such as a squirrel. This rotor is made of heavy copper bars that
connect at each end by a metal ring made of copper or brass. No insulation
is required between the core and the bars because of the low voltages
induced into the rotor bars. To obtain maximum field strength, the size of
the air gap between the rotor bars and stator windings is small.
Figure: Squirrel-Cage Induction Rotor
Single-phase Motors
Very small commercial applications such as household appliances and floor
buffers typically use single-phase motors.
Rev 1
44
Synchronous Motors
Because of their more complex design and higher manufacturing cost,
designers rarely specify synchronous motors as their first choice. However,
large industrial applications sometimes use synchronous motors to
accommodate large loads and to improve the power factor of transformers
in large industrial complexes.
Knowledge Check
The least often used motor in station applications is the
____________.
A.
three-phase induction motor
B.
single-phase (split-phase) motor
C.
synchronous motor
D.
squirrel-cage motor
ELO 3.2 Indications of Motor Malfunctions
Introduction
In this section, you will learn about some of the more common motor
malfunctions and indications associated with these malfunctions.
Indications of Motor Malfunctions
Motors or their connected loads are subject to several different types of
mechanical failures. Some of the most common are described below.
Locked (Seized) Rotor
One type of mechanical problem results from binding of the mechanical
load component; this binding causes shaft bending, which results in motor
rotor (shaft) binding along with the associated motor or component bearing.
At the extreme, this situation may cause a locked (seized) rotor. The
following indications are typical of a locked rotor:





Tripping of component circuit breaker
Immediate reduction in system flow rate
Immediate reduction in component discharge pressure
Immediate rise in component current to supply the needed torque
Immediate rise in motor winding temperatures resulting from
higher current flow
Sheared Rotor
Rev 1
45
A second mechanical problem that can occur is a sheared rotor (shaft). This
allows the motor to operate freely (spinning) with no load mechanically
attached. The following indications are typical of a sheared rotor:
 Load has no (low) running current (as indicated on ampere meter)
 Immediate reduction in system flow rate
 Immediate reduction in component discharge pressure
Miswired Phases
Phases can be miswired after work in which the motor leads were
disconnected. Normally, personnel verify the phase rotation following this
type of work to ensure that it is correct. If the motor leads (phases) are
miswired and the wires are reversed, the motor will turn in the wrong
direction. While the motor will work fine turning backward, the component
it is driving generally will not.
The consequence to the component can vary from the component not
functioning appropriately to the component damaging itself. If the
component encounters resistance because of turning in the wrong direction,
it can cause more resistance to the motor and damage the motor.
Knowledge Check
If a reactor coolant pump (RCP) rotor seizes, RCP motor
current will __________; if the rotor shears, RCP motor
speed will __________.
A.
increase; increase
B.
increase; decrease
C.
decrease; increase
D.
decrease; decrease
ELO 3.3 Consequences of Motor Overheating
Introduction
In this section, you will learn about the consequences of overheating motor
windings and bearings.
Loss of Motor Cooling
Continuous operation of a motor at rated load with a loss of required
cooling to the motor windings would eventually result in breakdown of the
motor insulation due to overheating. This will cause increased temperature
and current flow due to a decrease in the insulation resistance and, if severe,
will result in a short circuit.
Rev 1
46
Overheating, caused by loss of cooling or by any other mechanism will
result in this insulation breakdown and eventually lead to motor damage,
and need for motor rewind or replacement. A thermal overload device
protects many large motors from high temperature by tripping the motor off
upon exceeding temperature limits.
Loss of Bearing Cooling or Lubrication
There are various means of cooling and lubricating motor bearings,
depending on the size and application of the motor, however two needs are
common to all bearings:
1. The bearing must be properly lubricated to minimize friction
2. The heat generated in the bearing must be removed
High bearing friction manifests as an additional load to the motor that
causes both motor current and motor winding temperature to rise. An
increase in bearing friction will cause the bearing to degrade until it fails,
resulting in component trip and possible damage to the component, as well.
If the increase in friction is significant, it can lead to motor damage.
If a bearing is running with higher than normal temperature, but the friction
is not excessive, it may run for some time without causing damage. Most
bearing lubricants are effective within a specified temperature range; if
lubricant temperature is outside that range, there is a reduction in
lubricating properties. This can cause increased friction and lead to the
difficulties already discussed.
Important operator tasks include maintaining bearing lubrication and
temperature within the design limits, and maintaining winding temperatures
below design limits.
Knowledge Check
Which one of the following will result from prolonged
operation of an AC induction motor with excessively
high stator temperatures?
Rev 1
A.
Decreased electrical current demand due to reduced
counter electromotive force.
B.
Decreased electrical resistance to ground due to
breakdown of winding insulation.
C.
Increased electrical current demand due to reduced
counter electromotive force.
D.
Increased electrical resistance to ground due to
breakdown of winding insulation.
47
Knowledge Check
Continuous operation of a motor at rated load with a loss
of required cooling to the motor windings will eventually
result in ____________.
A.
cavitation of the pumped fluid
B.
failure of the motor overcurrent protection devices
C.
breakdown of the motor insulation and electrical grounds
D.
phase current imbalance in the motor and overspeed trip
actuation
ELO 3.4 Excessive Current in Motors
Introduction
In this section, you will learn what about causes of excessive current and
overload conditions in a motor.
Motor Overloading
Overloading or just increased loading can result in increased current draw
by the motor. Operators should investigate abnormally high running
currents promptly, as these high currents may indicate a problem such as
mechanical binding of the motor or of the associated component.
The table below summarizes overcurrent and overload malfunctions and the
indicators with likely results:
Malfunction
Indicator(s) with Likely Result
Gradual motor bearing
failure
Increased friction, current, and temperature
could be high enough to cause a thermal
overload trip.
Locked or seized rotor
(shaft) of the motor or
component driven (pump,
valve, etc.)
Overcurrent trip of the supply circuit breaker
Packing on a motoroperated valve is
tightened excessively
Motor current would increase due to
increased torque on motor caused by
additional friction associated with the
packing. These high currents can result in
excessive heat within the motor, causing
Rev 1
48
Malfunction
Indicator(s) with Likely Result
break down of the motor winding insulation
and damage to the motor (grounds).
Reduced voltage supplied
to the motor
The motor supplies power needed to drive the
load, and when voltage drops, the current
increases. Operating for extended periods
with reduced voltage can lead to winding
damage.
Sheared shaft or rotor
Abnormally low or no running currents as
indicated on an amp meter, with low or no
flow/pressure indicated.
Failed Bearings
Failed bearings are another mechanical problem that can affect electric
motor operation. Bearing failure can result from a number of
circumstances, including:




Insufficient lubrication
Poor bearing maintenance practices
Improper loading of the component
Motor under-voltage situation
An under-voltage condition may result in bearing failure due to the
development of excessive torque on the motor. Any condition leading to
overheating of the bearings can cause bearing failure. The excessive work
can cause heat buildup in the machine windings.
Knowledge Check
Which of the following will cause an increase in motor
current? (Select all that are true.)
Rev 1
A.
An increase in bearing friction due to inadequate
maintenance
B.
Increasing the load on the component driven, such as
opening a pump discharge valve to provide more flow
C.
Increased voltage to the motor
D.
Reduced voltage to the motor
49
Knowledge Check
Excessive current will be drawn by an AC induction
motor that is operating ____________.
A.
completely unloaded
B.
at full load
C.
with open circuited stator windings
D.
with short circuited stator windings
ELO 3.5 Pump Motor Current Relationships
Introduction
In this section, you will learn about the operating limits imposed to mitigate
the effects of starting current.
Operating Limits to Mitigate the Effect of Starting Current
Recall from the pumps module that centrifugal pumps operate within a
known set of laws, known as pump laws. It is possible to determine how
changes in pump parameters will affect motor power and current by
applying these laws.
Pump Law Review
The pump laws state the relationships between pump speed and the
resulting flow, pump head, and power. The flow rate of a pump is directly
proportional to pump speed. The discharge head or pressure is directly
proportional to the square of the speed. The power required by the pump
driver is directly proportional to the cube of pump speed. The following
equations express these relationships:
a. The flow rate of a pump is directly proportional to pump speed:
𝑉̇ ∝ 𝑛
b. The head of a pump is directly proportional to pump speed squared:
𝐻𝑝 ∝ 𝑛2
c. The power of a pump is directly proportional to pump speed cubed:
𝑝 ∝ 𝑛3
Where:
Rev 1
50
n = speed of pump impeller (rpm)
𝑉̇ = volumetric flow rate of pump (gpm or ft3/hr)
Hp = head developed by pump (feet)
p = pump power (kW)
Determining Changes in Flow, Head, and Power
Knowing a change in pump speed allows you to calculate the resultant
change in flow, head, and power, using the following relationships:
Rev 1
51
Action
Determine the change in flow by
using this relationship.
Determine the change in head by
using this relationship. Notice that
the head changes by the square of
the change in speed.
Determine the change in power by
using this relationship. Notice that
the power changes by the cube of
the change in speed.
Equation
𝑉̇2 = 𝑉1̇ (
𝑛2
)
𝑛1
𝐻𝑃2 = 𝐻𝑃1 (
𝑛2 2
)
𝑛1
𝑛2 3
𝑃2 = 𝑃1 ( )
𝑛1
These relationships are very useful in determining the effects on pump
performance, when flow is increased or decreased. Changes in power can
affect the amount of current an electric motor (prime mover) will draw.
Flow
A cooling water pump is operating at a speed of 1,800 rpm. Its flow rate is
400 gpm at a head of 48 ft. The pump’s motor is drawing 45 kW.
Determine the new pump flow rate if the pump speed increases to 3,600
rpm.
Solution:
𝑉̇2 = 𝑉1̇ (
𝑛2
)
𝑛1
3,600 𝑟𝑝𝑚
𝑉̇2 = (400 𝑔𝑝𝑚) (
)
1,800 𝑟𝑝𝑚
𝑉̇2 = 800 𝑔𝑝𝑚
Head
A cooling water pump is operating at a speed of 1,800 rpm. Its flow rate is
400 gpm at a head of 48 ft. The pump's motor is drawing 45 kW.
Determine the new pump head if the pump speed increases to 3,600 rpm.
Solution:
𝐻𝑃2 = 𝐻𝑃1 (
Rev 1
𝑛2 2
)
𝑛1
52
𝐻𝑃2
3,600 𝑟𝑝𝑚 2
= 48 𝑓𝑡 (
)
1,800 𝑟𝑝𝑚
𝐻𝑃2 = 192 𝑓𝑡
Power
A cooling water pump is operating at a speed of 1,800 rpm. Its flow rate is
400 gpm at a head of 48 ft. The pump's motor is drawing 45 kW.
Determine the new pump power requirements if the pump speed increases
to 3,600 rpm.
Solution:
𝑃2 = 𝑃1 (
𝑛2 3
)
𝑛1
𝑃2 = 45 𝑘𝑊 (
3,600 𝑟𝑝𝑚 3
)
1,800 𝑟𝑝𝑚
𝑃2 = 360 𝑘𝑊
Relationship between Pump Head, Brake Horsepower, and
Pump Flow
The relationship of these parameters can be calculated and plotted on a
pump performance curve for specific pumps and operating conditions; the
figure below shows these curves.
Rev 1
53
Figure: Pump Flow, Head, and BHP
There are some important characteristics to remember when using this type
of curves to solve problems; the NRC GFE questions use these curves
frequently.
Brake horsepower is the input power delivered to the pump by the motor,
the SI system units are kilowatts (kW).
If flow rate is increased, brake horsepower will increase, and pump head
will decrease. If flow rate decreases, brake horsepower decreases and pump
head will increase.
Changes in brake horsepower will be proportional to changes in motor amps
(voltage does not change), so if horsepower changes by 0.80, then the
resultant change in motor amps will change by the same ratio. Increases in
motor amps will result in increases in motor winding temperature. (Heat
generation is proportional to the square of the current.)
Example: Using the pump curves above, solve the following problem:
Consider a pump driven by a single-speed AC induction motor. A throttled
discharge flow control valve controls the pump flow rate.
The following initial pump conditions exist:
Pump motor current = 50 amps
Pump flow rate = 400 gpm
Rev 1
54
What will be the approximate value of pump motor current if the flow
control valve is repositioned such that pump flow rate is 800 gpm?
A. Less than 100 amps
B. 200 amps
C. 400 amps
D. More than 500 amps
Solution
Note
Using the pump curve, 400 gpm is equivalent to ~ 220
BHP; 800 gpm is~ 270 BHP
220
= 50 𝑎𝑚𝑝𝑠/𝑥
270
0.82𝑥 = 50 𝑎𝑚𝑝𝑠
𝑥=
50
= ~ 61 𝑎𝑚𝑝𝑠
0.82
61 amps is less than 100 amps, so the correct answer is A.
Knowledge Check
A centrifugal pump is operating with the initial
conditions as listed:
100 kW
500 gpm
150 ft
The operator notices that pump power decreased 50 kW.
What is the corresponding change in flow?
A.
350 gpm
B.
300 gpm
C.
396 gpm
D.
250 gpm
Knowledge Check
A multispeed centrifugal pump is operating with a flow
rate of 1,800 gpm at a speed of 3,600 rpm.
Which one of the following approximates the new flow
Rev 1
55
rate if the pump speed is decreased to 2,400 rpm?
A.
900 gpm
B.
1,050 gpm
C.
1,200 gpm
D.
1,350 gpm
TLO 3 Summary
AC Motor Operation
In this section, you learned the about applications for the different types of
motors, the indications and consequences of common equipment problems,
and the types of operating limits used to mitigate the effects of starting
currents.
1. Most common types of AC motors are:
a. Induction – most common used motor for industrial use
b. Single-phase – used for small commercial and household
applications
c. Synchronous – used for constant speed applications and where
large loads are needed to improve transformer power factor, more
complex design and cost
2. Indications form motor malfunctions are:
a. Locked rotor – high current and trip of circuit breaker, if on a
pump, both flow and pressure immediately go to zero
b. Sheared rotor – motor spins freely with no load, decrease in
current, and if connected to a pump, reduction in flow and pressure.
c. Miswired motor – if two leads are reversed during motor reassembly, the motor will turn backwards from proper rotation
3. Motor overheating
a. Loss of motor cooling - this will cause increased temperature and
current flow due to a decrease in the insulation resistance, and if
severe will result in a short circuit.
b. Loss of bearing cooling or lubrication – this will result in increased
friction and load, increased motor current and temperature, and
eventually bearing failure, and/or motor damage.
4. Excessive motor current can be caused from:
a. Overloading – multiple sources, likely causes are loss of cooling or
lubrication to bearings, and loss of motor cooling
b. Mechanical binding – such as valve packing over tightened
resulting in excessive torque and current to move valve
c. Low voltage supplied – results in higher current to produce same
amount of work, this leads to overheating and insulation break down.
Rev 1
56
5. Pump motor current follows power changes with respect pump curves
and laws; and changes proportionally with brake horsepower with
respect to pump flow changes.
Summary
Now that you have completed this lesson, you should be able to do the
following:
1. State the applications of the following types of AC motors:
a. Induction
b. Single-phase
c. Synchronous
2. Describe the indications resulting from a locked motor rotor, sheared
shaft, or miswired phases.
3. Describe the consequences of overheated motor windings or motor
bearings.
4. Describe the causes of excessive current in motors and generators
resulting from conditions such as low voltage, overloading, and
mechanical binding.
5. Describe the relationship between pump motor current and the
following parameters:
a. Pump flow
b. Pressure
c. Speed
d. Motor stator temperature
TLO 4 AC Generator Operation
Overview
On completion of this section, the student will be able to analyze operation
of generators in parallel and predict system response to changes in
frequency, voltage, and load.
The power plant must operate in parallel with the grid to supply power.
Knowledge of how to parallel, and control generator parameters is an
essential job activity for plant operators.
Objectives
Upon completion of this lesson, you will be able to do the following:
1. Define apparent, true, and reactive power and power factor using a
power triangle, and their effect on generator operation.
2. Describe the purpose of a voltage regulator and function of each of
the following typical components:
a. Sensing circuit
b. Reference circuit
c. Comparison circuit
d. Amplification circuit(s)
Rev 1
57
e. Signal output circuit
f. Feedback circuit
3. Describe the conditions that must be met prior to paralleling two
generators including consequences of not meeting these conditions.
4. Describe the consequences of over-excitation and under-excitation
when load sharing.
ELO 4.1 Power Factor and Generators
Introduction
In this section, you will learn about the relationship between apparent, true,
and reactive power, and the effect that leading and lagging power factors
have on generator operation.
Power Triangle
In AC circuits, current and voltage are normally out of phase due to the
effects of inductive and capacitive reactance. As a result, not all the power
produced by a generator in an AC application is available to accomplish
work.
Similarly, power calculations in AC circuits vary from those in DC circuits.
The power triangle, shown in the figure below, equates AC power to DC
power by showing the relationship between generator output (called
apparent power - S) in volt-amperes (VA), usable power (called true power
- P) in watts, and wasted or stored power (called reactive power - Q) in voltamperes reactive (VAR). The phase angle (θ) represents the inefficiency of
the AC circuit and corresponds to the total reactive impedance (Z) to current
flow in the circuit.
Figure: Power Triangle
The power triangle represents comparable values that can be used directly
to find the efficiency level of generated power to usable power, which is
expressed as the power factor (discussed later). With the DC equivalent
(RMS value) of the AC voltage, current components, and the power factor,
Rev 1
58
we can calculate apparent power, reactive power, and true power using the
power triangle and trigonometric laws. The following sections demonstrate
these calculations.
Apparent Power
Apparent power (S) is the power delivered to an electrical circuit, as shown
below in a mathematical representation of apparent power. The
measurement of apparent power is in volt-amperes (VA).
𝑆 = 𝐼 2 𝑍 = 𝐼𝐸
Where:

S = Apparent Power (VA)

I = RMS current (A)

E = RMS voltage (V)

Z = Impedance (Ω)
True Power
True power (P) is the power consumed by the resistive loads in an electrical
circuit, measured in watts. The following equation is a mathematical
representation of true power:
𝑃 = 𝐼 2 𝑅 = 𝐸𝐼 cos 𝜃
Where:

P = True Power (watts)

I = RMS current (A)

R = Resistance (Ω)

E = RMS voltage (V)

θ = Angle between E and I sine waves
Reactive Power
Reactive power (Q) is the power component in an AC circuit necessary for
the expansion and collapse of magnetic (inductive) and electrostatic
(capacitive) fields, expressed in volt-amperes-reactive (VAR). The
following equation is a mathematical representation for reactive power:
𝑄 = 𝐼 2 𝑋 = 𝐸𝐼 sin 𝜃
Where:
Rev 1
59

Q = Reactive Power (VAR)

I = RMS current (A)

X = Net reactance (Ω)

E = RMS voltage (V)

θ = Angle between E and I sine waves
Unlike true power, reactive power is unavailable power because it is stored
in the circuit itself. This power is stored by inductors, because they expand
and collapse their magnetic fields in an attempt to keep current constant,
and by capacitors, because they charge and discharge in an attempt to keep
voltage constant. In AC electrical circuits, the circuit’s inductance and
capacitance consume and give back (exchange) reactive power with the
circuit’s AC power source.
Reactive power is a function of a system’s amperage. The power delivered
to the inductance is stored in the magnetic field when the field expands and
returns to the source when the field collapses. The power delivered to the
capacitance is stored in the electrostatic field when the capacitor charges
and returns to the source when the capacitor discharges.
There is no consumption of the reactive power delivered to the circuit by
the source; circuit components use reactive power as needed, then return it
to the source. Therefore, these reactive loads consume no true power to
maintain their magnetic and electrostatic fields. Alternating current
constantly changes; thus, the cycle of expansion and collapse of the
magnetic and electrostatic fields constantly occurs.
Circulating current describes the current that is constantly flowing between
the source and the inductive and capacitive loads in an AC circuit.
Circulating currents account for no real work in the circuit. Circulating
currents are measurable and can cause protective relaying to actuate, when
significant amounts are present.
Total Power
The total power delivered by the source is the same as apparent power. Part
of this apparent power, is dissipated by the circuit resistance in the form of
heat. This power dissipated in heat is true power. The circuit inductance
and capacitance return the rest of the apparent power to the source (reactive
power).
𝑇𝑜𝑡𝑎𝑙 𝑃𝑜𝑤𝑒𝑟 = 𝐴𝑝𝑝𝑎𝑟𝑒𝑛𝑡 𝑃𝑜𝑤𝑒𝑟
= 𝑃𝑜𝑤𝑒𝑟 𝑑𝑒𝑙𝑖𝑣𝑒𝑟𝑒𝑑 𝑡𝑜 𝑎𝑛 𝑒𝑙𝑒𝑐𝑡𝑟𝑖𝑐𝑎𝑙 𝑐𝑖𝑟𝑐𝑢𝑖𝑡
Rev 1
60
Power Factor
Power factor (pf) is the ratio between true power and apparent power. True
power is the power consumed by an AC circuit, whereas apparent power is
a representation of the total power delivered to an AC circuit. Reactive
power comprises a portion of the apparent power; reactive power is power
that is stored in an AC circuit and accomplishes no real work in the circuit.
Power factor is represented by cos θ in an AC circuit. It is the ratio of true
power to apparent power, where θ is the phase angle between the applied
voltage and current sine waves; the power triangle (see the figure above)
shows that θ is the angle between P and S. The following equation is a
mathematical definition of power factor:
cos 𝜃 = 𝑃𝑜𝑤𝑒𝑟 𝐹𝑎𝑐𝑡𝑜𝑟(𝑝𝑓) =
𝑃
𝑇𝑟𝑢𝑒 𝑃𝑜𝑤𝑒𝑟
=
𝑆 𝐴𝑝𝑝𝑎𝑟𝑒𝑛𝑡 𝑃𝑜𝑤𝑒𝑟
Where:


P = true power (watts)
S = apparent power (VA)
Power factor can be either lagging (current lags voltage) or leading (current
leads voltage). The make-up of the circuit components determines if the
power factor is leading or lagging. Capacitive loads are leading, and
inductive loads are lagging. Resistive loads are real loads with a power
factor of one. The combination of all loads on the system will determine
whether the overall system is leading, lagging, or neutral.
Most industrial power grids are lagging (inductive) because more of the
loads, (motors) are inductive than capacitive.
It is important to know whether the power factor of the system is inductive
or capacitive, because it allows the system operator to minimize the impact
of reactive power on the grid, to allow as much generation as possible from
each operating generator. The section on generator excitation covers this
topic in more detail.
Lagging Power Factor
In an inductive circuit, the current lags the voltage. This type of circuit has
a lagging power factor, as shown in the following figure:
Figure: Lagging Power Factor
Rev 1
61
Power factor determines what part of the apparent power is true power. It
can vary from 1, when the phase angle is 0 degrees, to 0, when the phase
angle is 90 degrees.
Leading Power Factor
In a capacitive circuit, the current leads the voltage. This type of circuit has
a leading power factor, as shown in the following figure:
Figure: Leading Power Factor
An electrical circuit that supplies power to loads such as motors, will
exhibit a lagging power factor. An electrical circuit that supplies power to
loads such as fluorescent lighting will exhibit a leading power factor. Most
industrial electrical distribution systems exhibit a lagging power factor
because inductive loads normally account for a larger percentage of the
reactance seen in these types of circuits.
ELI the ICE Man
You may choose to use a mnemonic memory device for voltage and current
phase relationships, "ELI the ICE man," to remember the voltage/current
relationship in AC circuits. ELI refers to an inductive circuit (L) where
current (I) lags voltage (E). ICE refers to a capacitive circuit (C) where
current (I) leads voltage (E).
Knowledge Check
Select all that are true.
Rev 1
A.
Reactive power is imaginary and the currents caused by
reactive power do not contribute to heat loads on the
generator or current detected by protective relays.
B.
True power is the power consumed by resistive loads on
the system.
C.
Total power is the power available to do work on the
system.
D.
Reactive power is held and given back by inductive and
capacitive loads on the system.
62
Knowledge Check
Select all that are true.
A.
Power factor is the ratio of true power to apparent power
and is always greater than 1.
B.
Power factor is the ratio of reactive power to real power
and is always greater than 1.
C.
Power factor is the ratio of true power to apparent power
and can vary from 0 to 1.
D.
Power factor is the ratio of true power to reactive power
and is always less than 1.
ELO 4.2 Voltage Regulation and Components
Introduction
In this section, you will learn the purpose, basic components, and method of
operation of voltage regulators.
Voltage Regulation
The purpose of a voltage regulator is to maintain the output voltage of a
generator at a desired value. As load on an AC generator changes, the
output voltage will also change. The main reason for this change is the
change in the voltage drop across the armature winding of the generator
caused by a change in load current.
In an AC generator, there are losses caused by the current flowing through
the resistance and inductance of the armature windings. The losses are due
to resistance to current flow (IR losses) and inductive reactance to current
flow (IXL losses). The IR drop is dependent on the amount of the electrical
load change only.
The IXL drop is dependent on not only the load change, but also on the
power factor of the circuit to which the generator is connected. Therefore,
the output voltage of an AC generator varies with both changes in load (i.e.,
current) and changes in power factor. Because of this output variation, AC
generators require some method of regulating output voltage.
An AC generator’s voltage regulator continuously compares the voltage
output of the machine to a desired value. The voltage regulator varies the
DC (rectified AC) excitation applied to the generator’s field in order to
maintain generator output voltage constant.
Rev 1
63
Voltage Regulator Components
The figure below shows a typical block diagram of an AC generator voltage
regulator. This regulator consists of six basic circuits that act together to
regulate the output voltage of an AC generator from no-load to full-load.
Figure: Voltage Regulator Block Diagram
Voltage Regulator Components
Voltage regulators have the following components to monitor and control
generator output voltage:
Sensing Circuit
The sensing circuit senses output voltage of the AC generator. As the
generator is loaded or unloaded, the output voltage changes, and the sensing
circuit provides a signal of these voltage changes. This signal is
proportional to output voltage the sensing circuit sends the signal to the
comparison circuit.
Reference Circuit
The reference circuit maintains a constant output for reference. This
reference signal represents the desired voltage output of the AC generator.
Comparison Circuit
The comparison circuit electrically compares the reference voltage to the
received sensed voltage and provides an error signal. This error signal
represents an increase or decrease in output voltage. The comparison circuit
sends the error signal to the amplification circuit.
Amplification Circuit
The amplification circuit, which can be a magnetic amplifier or transistor
amplifier, takes the error signal from the comparison circuit, and amplifies
the milliamp input to an amp output, then, sends the amplified signal to the
signal output, or field circuit.
Signal Output Circuit
Rev 1
64
The signal output circuit, which controls field excitation of the AC
generator, increases or decreases field excitation to either raise or lower the
AC output voltage, based on the received, amplified error signal.
Feedback Circuit
The feedback circuit takes some of the output of the signal output circuit
and feeds it back to the amplification circuit. It does this to prevent
overshooting or undershooting of the desired voltage by slowing down the
circuit response.
Changing Output Voltage
As generator load changes, output voltage will change, if the generator
excitation remains constant. This occurs because load changes cause
increases or decreases in the current losses in the generator. The voltage
regulator senses the change in output voltage, and changes the generator
excitation to return the generator output voltage to the desired value.
Increasing Generator Load (Output Voltage Drop)
An increase in generator load results in a drop in generator output voltage.
The voltage regulator will respond to raise output voltage. First, the sensing
circuit senses the decrease in output voltage as compared to the reference
and lowers its input to the comparison circuit.
Since the reference circuit is always a constant, the comparison circuit will
develop an error signal due to the difference between the sensed voltage and
the reference voltage. The error signal developed will be of a positive value
with the magnitude of the signal dependent on the difference between the
sensed voltage and the reference voltage.
The amplifier circuit amplifies this output from the comparison circuit and
sends the amplified signal to the signal output circuit. The signal output
circuit then increases field excitation to the AC generator. This increase in
field excitation causes generated voltage to increase to the desired output.
Decreasing Generator Load (Output Voltage Increase)
If the load on the generator decreases, the voltage output of the machine
would rise. The actions of the voltage regulator for an increase in output
voltage would be the opposite of those for a lowering output voltage. In
this case, the comparison circuit will develop a negative error signal whose
magnitude is again dependent on the difference between the sensed voltage
and the reference voltage. As a result, the signal output circuit will decrease
field excitation to the AC generator, causing the generated voltage to
decrease to the desired output.
Knowledge Check
Select all that are true.
Rev 1
65
A.
Generator output voltage is affected by the current drawn
by the system the generator supplies.
B.
Generator output voltage is affected by the power factor
of the system the generator supplies.
C.
Generator output voltage is not affected by the system
the generator supplies at all.
D.
Generator output voltage is a function of the number of
generator windings and the output voltage variance is
insignificant.
Knowledge Check
The __________ prevents overshooting or undershooting
of the desired voltage by slowing down the circuit
response.
A.
sensing circuit
B.
comparison circuit
C.
feedback circuit
D.
amplification circuit
ELO 4.3 Paralleling Generators
Introduction
In this section, you will learn about the required conditions prior to
paralleling two generators including consequences of failing to meet these
conditions.
Paralleling AC Generators
Most electrical power grids and distribution systems have more than one
AC generator operating at one time. Normally, systems operate two or
more generators in parallel in order to increase the available power.
There are three required conditions prior to paralleling (or synchronizing)
AC generators:
1. Their terminal voltages must be essentially equal. If the voltages of
the two AC generators are not close to equal when the breaker is
closed, one of the AC generators will act as a reactive load to the
Rev 1
66
other AC generator. This causes high current exchange between the
two machines, possibly causing generator or distribution system
damage. Reactive load (VAR) will transfer from the generator with a
higher voltage (lagging power factor) to the generator with a lower
voltage (leading power factor).
2. Their frequencies must be almost equal. A mismatch in frequencies
of the two AC generators will cause the generator with the lower
frequency to transfer its real load (watts) to the generator operating at
the higher frequency. If the frequency mismatch is significant, the
lower frequency machine will act as a load on the other generator (a
condition referred to as "motoring or reverse powering") and
potentially trip the generator off line. The amount of real load (watts)
transfer is relative to the frequency difference between the generators.
This can cause an overload in the generators and the distribution
system.
3. Their output voltages must be in phase. A mismatch in the phases
will cause development of large opposing voltages in the two sources.
The worst-case mismatch is 180 degrees out of phase, resulting in an
opposing voltage between the two generators of twice the output
voltage. This high voltage can cause damage to the distribution
system due to extremely high currents and large mechanical torque
exerted on both of the generators. The greater the phase mismatch,
the greater the damage that will occur to the generator output breaker
due to excessive arcing when the circuit breaker is closed.
During paralleling operations, voltmeters indicate the voltages of the two
generators. Output frequency meters facilitate frequency matching. A
synchroscope, a device that senses the two frequencies and indicates the
phase differences between the generators allows phase matching of the two
generators.
AC Generator Real Power Sharing
With two generators operating in parallel, the real load carried by each
generator can be determined by examining the frequency-power
characteristics for each generator, as shown in the figure below. This type
of figure, which graphically describes load sharing between two power
sources, is termed a “House Curve”. Note that the power values for
generators G1 and G2 increase both to the left and to the right of the graph
vertical centerline.
Rev 1
67
Figure: Two AC Generators Operating in Parallel
Those values to the left of the vertical centerline (frequency axis) are the
power outputs of generator 1 (G1), while those on the right side of the
vertical centerline (frequency axis) are the power outputs of generator 2
(G2). Since parallel generators must operate at the same frequency, we can
determine the load carried by each generator by noting the intersection of
the generator characteristic for the particular speed regulator setting in use
and the operating frequency.
For example, for the settings shown in the figure above, at an operating
frequency of 60 Hz, G1 would carry 350 kW and G2 would carry 50 kW.
Points A (generator G1) and B (generator G2) are the points where each
generator curve intersects with the operating frequency. The total length of
the line A-B represents the total load (400 kW).
The figure also shows that if the total load increased to 500kW with no
changes in the speed regulator settings, the system frequency would drop to
59.95 Hz and G1 would carry 400 kW and G2 100kW, as indicated by line
C-D.
AC Generator Real Load Sharing
The figure below depicts a load initially unbalanced between two
generators, and the proper distribution of the real load between the two
generators. The dashed lines in the figure below show the original
condition (frequency = 60Hz, G1 load = 350 kW, and G2 load = 50 kW).
The solid lines in the figure below show the proper load distribution
condition (frequency = 60Hz, G1 load = 200 kW, and G2 load = 200 kW).
Rev 1
68
Figure: Kilowatt Load Balanced by Adjusting Speed Regulator
By increasing the speed regulator setting on G2 and decreasing the speed
regulator setting on G1 until the no-load frequency of the generators are the
same, the load on G2 increases from 50 kW to 200 kW. At the same time,
the load on G1 decreases from 350 kW to 200 kW.
Note that the total load (400 kW) remains unchanged, as does the system
frequency (60 Hz). Also, note that raising or lowering both regulator
settings together results in a change in system frequency without altering
the generator load distribution. For generators with different ratings, the
best operating practice is to distribute the kW load (real load) in proportion
to the generator ratings.
Consider an example 1,000 kW load distributed between two generators;
generator G1 carries a 500 kW rating and G2 carries a 1,500 kW rating.
Distributing the load between the generators in proportion to their ratings
results in G1 carrying 250 kW and G2 carrying 750 kW, combined they
carry the 1,000 kW load. When removing a generator from service, first
minimize the load on that generator by lowering its speed regulator setting
while simultaneously increasing the speed regulator setting on the other
generator to maintain a constant system frequency.
AC Generator Reactive Load Sharing
After balancing the real load, use a similar procedure to balance the reactive
load. With two generators operating in parallel and supplying a constant
real power, the reactive loads carried by each of the generators can be
determined by examining the terminal voltage versus reactive power
characteristics. Different curves on a terminal voltage-reactive power
characteristic represent different values of field excitation.
Rev 1
69
For a fixed value of field excitation, the terminal voltage drops off as the
reactive load increases, thereby exhibiting a drooping characteristic.
Practical generators incorporate automatic voltage regulators, which act on
the generator’s field circuitry to correct the terminal voltage-reactive power
characteristic to a linear droop. The figure below depicts the redistribution
of the reactive load.
Figure: Reactive Load Balance by Adjusting Voltage Regulator
As with the frequency-power characteristic, values measured along the
horizontal axis in both directions from the graph centerline are positive.
Since parallel generators must operate at the same voltage, determine the
reactive load carried by each generator by finding the intersection of the
generator characteristic curve for the particular voltage regulator setting in
use and the actual operating voltage.
For example, for the figure above, at an operating voltage of 450V, G1
initially carries 600 KVAR and G2 carries 200 KVAR, as shown by the
dotted lines. To balance the reactive load while maintaining the terminal
voltage at 450V, decrease the voltage regulator setting on G1 so that G1
carries 400 KVAR while increasing the voltage regulator setting on G2 so
that G2 carries 400 KVAR. As in the real load case, it is desirable to
distribute the load in proportion to generator ratings.
AC Generator Operation in Parallel with a Large Grid
Paralleling a generator with the grid
When paralleling a single generator with a large power system (grid), the
output of the single oncoming generator is minimal compared with the
power output of the entire grid. Hence, the House Curve below shows the
system as a constant frequency, constant voltage power source.
Rev 1
70
Figure: Parallel Operation of a Single Generator with the Grid
The figure above illustrates the distribution of real and reactive load of a
single generator in parallel with a large power grid. An operator
independently adjusts the appropriate regulators to set the load distributions
on the single generator G1. The power grid will automatically supply the
remainder of the required load. It is possible to have a single generator
supply negative reactive power, thereby decreasing the total reactive load,
improving the power factor, and requiring less current for the same total
real power.
Paralleling Generators Example
Consider the following set of conditions:
An operator is connecting a main generator to an infinite power grid that is
operating at 60 Hz. Generator output voltage is equal to the grid voltage but
generator frequency is at 57 Hz.
Rev 1
71
Which one of the following generator conditions is most likely to occur if
the operator closes the generator output breaker with the voltages in phase
(synchronized) but with the existing frequency difference? (Assume no
generator breaker protective trip occurs.)
A.
B.
C.
D.
Reverse power
Underfrequency
Undervoltage
Overspeed
In evaluating this scenario, first remember that the operating is paralleling
the generator to an assumed infinite grid. The grid is not infinite, but it is so
much larger than any one generator that it can be considered as infinite
when determining the expected response for your power plant. When
paralleling to an infinite grid, the actions you take with respect to your
generator will not change either the frequency or voltage of the grid. They
will determine the amount of real and reactive load your generator carries.
When paralleling the generator to the grid, the grid will govern the
generator’s frequency, voltage, and speed. Therefore, underfrequency,
undervoltage, and overspeed trips are not possible answers.
By synchronizing your generator to the grid with a frequency much lower
than the grid, you have forced your generator to speed up when connected.
The additional work to speed up your generator came from the grid; the grid
like a motor (motorized) is now driving your generator. The real power is
flowing into your generator rather than out, and a reverse power trip is
plausible. Therefore, A is the only correct answer.
Note that as an operating practice, you synchronize with the generator
running slightly faster than the grid (synchroscope moving slowly
clockwise), so that when you close the breaker, the generator picks up a
small amount of load, rather than being a load on the grid. This prevents
motorizing or a reverse power condition being established.
Of the choices given, answer A. is the only credible choice. However, in
practice, the generator would certainly be reverse powered, and with a
frequency difference this large it would almost certainly trip on reverse
power if the synchronizing circuit allowed the breaker to close at all.
Paralleling Generators Example
Consider the process of re-energizing a dead bus with a generator. This
could occur when using your plant emergency diesel generators. In this
circumstance, none of the loads on the bus are running. If you parallel the
generator onto the bus without first stripping the loads, all of the bus loads
will start at once, when the generator energizes the bus.
Recall that electric motors have a starting current equal to several times
higher than running current, and if they all start at once, the bus will draw
several times its normal full load current. While the generator rating is
Rev 1
72
sufficient to carry all of the bus loads, it is insufficient to carry all of the
starting currents simultaneously. Starting all the loads together will result
in an overcurrent condition, which will trip the generator.
Rev 1
73
A typical exam question could be:
Closing the output breaker of a three-phase generator onto a deenergized
bus can result in...
A. an overvoltage condition on the bus. - This answer is
incorrect. If the voltage was set correctly prior to energizing
the dead bus, voltage would tend to drop slightly when
loaded. The voltage regulator would compensate and keep it
within the band, but it would certainly not result in voltage
going too high.
B. an overcurrent condition on the generator if the bus was not
first unloaded. This answer is correct. All of the motors
supplied by this bus would be accelerating to speed and
drawing starting current at the same time. This will result in
an overcurrent condition that will likely trip the generator
and deenergize the bus again.
C. a reverse power trip of the generator circuit breaker if
generator frequency is low. - Since the generator is the only
source of power to this bus, it cannot receive power from any
other source- reverse power is not plausible.
D. a large reactive current in the generator. - Reactive power
should not be higher than the normal reactive load for
running this bus. Real load from starting currents will be the
problem in this circumstance.
In this instance, the only correct choice is B. This is a real operating
concern. If a dead bus is re-energized without stripping loads first, it will
result in an overcurrent condition, possibly tripping the bus again, and
possibly damaging either the generator or the feeder breaker.
Paralleling Generators Example
Consider the following situation.
A main generator is about to be connected to an infinite power grid.
Closing the generator output breaker with the generator voltage slightly
lower than grid voltage and with generator frequency slightly higher than
grid frequency will initially result in: (Assume no generator breaker
protective trip occurs).
A.
B.
C.
D.
the generator supplying reactive power to the grid.
the generator attaining a leading power factor.
the generator acting as a real load to the grid.
motoring of the generator.
This circumstance requires you to consider both real load and reactive load.
Since the generator frequency is slightly higher than grid frequency (the
synchroscope will be rotating slowly in the clockwise direction, as it should
be when you parallel generators), the generator will pick up real load from
Rev 1
74
the grid. This eliminates options C and D, since both of them imply the
generator becomes a load on the grid.
You should evaluate reactive load next. Since the generator voltage is
slightly lower than grid voltage, it will not supply reactive power to the
grid, eliminating option A. It will have a leading power factor, and appear
as a capacitive load to the rest of the grid; we have eliminated all possible
answers except option B.
Knowledge Check
A main generator is being paralleled to the power grid.
Generator voltage has been properly adjusted and the
synchroscope is rotating slowly clockwise. The generator
breaker must be closed just as the synchroscope pointer
reaches the 12 o'clock position to prevent...
A motoring of the generator due to unequal frequencies.
B excessive MW load transfer to the generator due to
unequal frequencies.
C excessive MW load transfer to the generator due to out-ofphase voltages.
D excessive arcing within the generator output breaker due
to out-of-phase voltages.
Knowledge Check
A main generator is about to be connected to an infinite
power grid with the following conditions:
Generator frequency: 59.5 Hz
Grid frequency: 59.8 Hz
Generator voltage: 115.1 kV
Grid voltage: 114.8 kV
When the generator output breaker is closed, the
generator will...
Rev 1
A.
acquire real load and reactive load.
B.
acquire real load, but become a reactive load to the grid.
C.
become a real load to the grid, but acquire reactive load.
D.
become a real load and a reactive load to the grid.
75
Knowledge Check
A main generator is about to be connected to an infinite
power grid. Closing the generator output breaker with
generator and grid voltages matched, but with generator
frequency lower than grid frequency will initially result
in the generator...
A.
picking up a portion of the grid real load.
B.
picking up a portion of the grid reactive load.
C.
experiencing reverse power conditions.
D.
experiencing overspeed conditions.
Knowledge Check
Which one of the following evolutions will draw the
highest current from the main generator during operation
of the output breaker?
Rev 1
A.
Closing the output breaker with voltages in phase
B.
Opening the output breaker under full-load conditions
C.
Opening the output breaker under no-load conditions
D.
Closing the output breaker with voltages out of phase
76
Knowledge Check - NRC GFE Question
Two identical 1,000 MW generators are operating in
parallel supplying the same isolated electrical bus. The
generator output breakers provide identical protection for
the generators. Generator A and B output indications are
as follows:
Generator A Generator B
22 kV
22 kV
60.2 Hertz
60.2 Hertz
200 MW
200 MW
25 MVAR
(out)
50 MVAR
(out)
A malfunction causes the voltage regulator setpoint for
generator B to slowly and continuously increase. If no
operator action is taken, generator A output current will...
A.
increase continuously until the output breaker for
generator A trips on overcurrent.
B.
decrease continuously until the output breaker for
generator B trips on overcurrent.
C.
initially decrease, and then increase until the output
breaker for generator A trips on overcurrent.
D.
initially decrease, and then increase until the output
breaker for generator B trips on overcurrent.
ELO 4.4 Generator Excitation
Introduction
An understanding of generator response to over and under excitations
conditions and governor speed changes is important to proper generator
monitoring and operation.
Generator Excitation Guidelines
There are two terms commonly used to explain changes in a generator’s airgap flux: over-excitation and under-excitation. These terms describe the
changes in a generator's air-gap flux (magnetic field) due to the internally
generated voltage (emf) during paralleled generators or generator-grid
Rev 1
77
operation. If the generator is initially in an under excited condition, the
power factor angle is lagging. This means that the machine is absorbing
reactive power from the system (acting as a reactive load on the system).
Conversely, if the generator is in an over-excited condition, the power
factor angle is leading. This means that the generator is supplying reactive
power to the system. Based on grid conditions and supplied loads, power
control/load dispatch may request a change in generator excitation to help
maintain grid voltage stability.
A generator will react differently to governor speed and field changes,
dependent on whether it is operating un-paralleled, paralleled with another
generator, or supplying the power grid.
During parallel operation to the grid, the generator is a relative small power
supply in comparison to the entire grid network, and therefore not able to
change grid voltage or frequency significantly.
Operation with generator not paralleled:
 Changes to the governor control (prime mover) result in changes to
generator output frequency. If you increase the speed of the
generator, the frequency will increase (recall that frequency =
NP/120, which is rotor speed times number of poles, divided by 120);
and if you decrease the generator speed, the frequency will decrease.
 Changes to the generator field will result in changes in the generator
output voltage (recall that the EMF induced in the armature is
dependent on the field strength of the rotating magnetic field). If you
increase current flow to the field, that will increase field strength,
which will increase EMF and generator output voltage.
Generator Excitation Guidelines (parallel operation)
Some terms associated with generator over-excitation and under-excitation
with a generator paralleled with another generator or the grid include the
following:
Unity power factor – (VARs at lowest point or zero) for every
operating condition, there is one value of field current that will cause
the generator to deliver only real power. It is impractical to operate
an AC generator at a unity power factor because all AC systems
require some reactive power for the generation of magnetic fields, etc.
 Under-excitation - when field current falls below the value that yields
operation at a unity power factor, the generator will absorb reactive
power (VARs will increase). The generator acts as inductive coil
(consumption of inductive reactive power). This condition is usually
not the desired mode of operating an AC generator, since in this
condition, the generator is acting as a reactive load on the system.
 Over-excitation - increasing field current causes the generator to
supply increased reactive power. The generator acts as capacitor
(delivery of reactive power). Normally, we require generators to
provide VARs together with watts; they nearly always operate in the

Rev 1
78
overexcited condition. While there are protective devices that guard
against voltage regulator failure, none protect from over-excitation or
under-excitation.
Generator Excitation Guidelines
When the generator is under-excited:
Reducing the generator voltage signal will make the generator more underexcited. It will look like a larger inductive load to the grid, and draw more
reactive load from the grid.
Increasing generator voltage will make the generator less under-excited. It
will draw less reactive load and move closer to unity power factor. Upon
increasing voltage more, the generator will pass through unity power factor
and become over-excited.
In the under-excited condition, the generator has a lagging power factor.
When the generator is over-excited:
Reducing generator voltage will make the generator less over-excited,
moving toward unity power factor. It will supply less reactive load to the
grid and less current (VARs).
Increasing generator voltage will make the generator more over-excited, it
will supply more reactive load to the grid.
In the over-excited condition, the generator has a leading power factor.
Power plants generally operate in the over-excited condition because the
grid has more inductive than capacitive loads.
Note:
Generator voltage changes will only affect reactive load (VAR) and not real
load (kW).
Generator Excitation Example
A main generator that connects to an infinite power grid has the following
initial indications:
100 MW
0 MVAR
2,900 Amps
20,000 VAC
If we reduce the main generator excitation slightly, amps will
_________________ and MW will ______________.
Rev 1
79
In order to evaluate this item, we need to consider real load and reactive
load separately.
Main generator excitation controls reactive load, and generator real load is
controlled by raising speed. Speed will actually be unchanged due to the
connection to an infinite grid, but the generator will take load from the grid
in its attempt to raise speed. Since speed control has not been touched, real
load will remain unchanged.
Next, we must consider the effect of reducing generator excitation. The
generator begins at 0 MVARS or no reactive load at all. Therefore, an
adjustment in either direction will increase current. In this case, the
generator will go from unity power factor to under-excited. It will appear
as an inductive load to the grid.
Therefore, in answer to the question above:
If we reduce the main generator excitation slightly, amps will increase and
MW will remain the same.
Knowledge Check
Two identical 1,000 MW electrical generators are
operating in parallel, supplying the same isolated
electrical bus. The generator output breakers also
provide identical protection for the generators.
Generator A and B output indications are as follows:
Generator A
Generator B
22.5 kV
22.5 kV
60.2 Hertz
60.2 Hertz
750 MW
750 MW
25 MVAR (out)
50 MVAR (out)
A malfunction causes the voltage regulator for generator
B to slowly and continuously increase the terminal
voltage for generator B.
If the operator takes no action, which one of the
following describes the electrical current indications for
generator A?
Rev 1
80
A.
Current will decrease continuously, until the output
breaker for generator A trips on reverse power.
B.
Current will decrease continuously, until the output
breaker for generator B trips on reverse power.
C.
Current will initially decrease, then increase until the
output breaker for generator A trips on overcurrent.
D.
Current will initially decrease, then increase until the
output breaker for generator B trips on overcurrent.
Knowledge Check
Two identical 1,000 MW AC electrical generators are
operating in parallel, supplying all the loads on a
common electrical bus. The generator output breakers
provide identical protection for the generators.
Generator A and B output indications are as follows:
Generator A
Generator B
28 kV
28 kV
60 Hertz
60 Hertz
150 MW
100 MW
25 MVAR (out)
50 MVAR (out)
A malfunction causes the voltage regulator setpoint for
generator B to slowly and continuously decrease. If no
operator action is taken, the electrical current indication
for generator B will...
Rev 1
81
A.
initially decrease, then increase until the output breaker
for generator A trips on overcurrent.
B.
initially decrease, then increase until the output breaker
for generator B trips on overcurrent.
C.
decrease continuously until the output breaker for
generator A trips on overcurrent.
D.
decrease continuously until the output breaker for
generator B trips on reverse power.
Knowledge Check
A main generator is about to be connected to an infinite
power grid. Generator voltage is slightly higher than
grid voltage and the synchroscope is rotating slowly in
the clockwise direction. The generator breaker is closed
just as the synchroscope pointer reaches the 3 o'clock
position.
Which one of the following will occur after the breaker is
closed?
Rev 1
A.
The breaker will remain closed and the generator will
supply only MW to the grid.
B.
The breaker will remain closed and the generator will
supply both MW and MVAR to the grid.
C.
The breaker will open due to overcurrent.
D.
The breaker will open due to reverse power.
82
TLO 4 Summary
Parallel Operation of Generators
In this section, you learned how generators operate in parallel with other
generators of similar size, and in parallel with an infinite grid. You learned
how to control real and reactive load and maintain the generator in a stable
operating condition in all normal operating circumstances.
1. In AC circuits, current and voltage are normally out of phase due to
the effects of inductive and capacitive reactance. As a result, not all
the power produced by a generator in an AC application can
accomplish work.
2. A power triangle shows apparent, true, and reactive power, and the
relationships between them.
3. The power triangle equates AC power to DC power by showing the
relationship between generator output (Apparent Power - S) in voltamperes (VA), usable power (True Power - P) in watts, and wasted or
stored power (Reactive Power - Q) in volt-amperes-reactive (VAR).
The phase angle (θ) represents the inefficiency of the AC circuit and
corresponds to the total reactive impedance (Z) to current flow in the
circuit.
4. Power factor (pf) is the ratio between True Power and Apparent
Power. True Power is the power consumed by an AC circuit, whereas
Apparent Power is a representation of the total power delivered to an
AC circuit.
5. The purpose of a voltage regulator is to maintain the output voltage of
a generator at a desired value.
6. An AC generator’s voltage regulator continuously compares the
voltage output of the machine to a desired value. The voltage
regulator varies the DC (rectified AC) excitation applied to the
generator’s field in order to maintain generator output voltage
constant.
7. To parallel AC generators, three conditions must be met:
 Their terminal voltages must be essentially equal.
 Their frequencies must be almost equal.
 Their output voltages must be in phase.
8. Generator Over and Under-Excitation
 Unity power factor – (VARs at lowest point or zero) for every
operating condition, there is one value of field current that will
cause the generator to deliver only real power.
 Under-excitation - when field current falls below the value for
unity power factor operation, the generator will absorb reactive
power (VARs will increase). Generator acts as inductive coil
(consumption of inductive reactive power). This condition is
usually not the desired mode of operating an AC generator, since
in this condition the generator is acting as a reactive load on the
system.
 Over-excitation - increasing field current causes generator to
supply increasing reactive power. Generator acts as capacitor
Rev 1
83
(delivery of reactive power). Generators normally provide VARs
together with watts; they usually operate in the over-excited
condition.
Summary
Now that you have completed this lesson, you should be able to do the
following:
1. Define apparent, true, and reactive power and power factor using a
power triangle, and their effect on generator operation.
2. Describe the purpose of a voltage regulator and function of each of
the following typical components:
a. Sensing circuit
b. Reference circuit
c. Comparison circuit
d. Amplification circuit(s)
e. Signal output circuit
f. Feedback circuit
3. Describe the conditions that must be met prior to paralleling two
generators including consequences of not meeting these conditions.
4. Describe the consequences of over-excitation and under-excitation
when load sharing.
AC Motors and Generators Summary
This module presented the types of AC motors and generators, their
operating characteristics and applications, and the means to control them in
conjunction with the power system.
Now that you have completed this module, you should be able to
demonstrate mastery of this topic by passing a written exam with a grade of
80 percent or higher on the following TLOs:
1. Describe the construction, operating characteristics and limitations for
an AC generator.
2. Explain the theory of operation of selected types of AC motors.
3. Explain operating characteristics and limitations on AC motors and
generators.
4. Analyze operation of generators in parallel and predict system
response to changes in frequency, voltage, and load.
Rev 1
84