Download Power Factor and Generators

Operator Generic Fundamentals
Components - AC Motors and Generators
© Copyright 2016
Operator Generic Fundamentals
2
Terminal Learning 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 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 the operating characteristics and interactions between
two AC generators operating in parallel.
© Copyright 2016
Intro
Operator Generic Fundamentals
3
AC Generator
TLO 1 – Describe the construction, operating characteristics, and
limitations for an AC generator.
1.1 Describe the theory of operation of an AC generator.
1.2 State the purpose of the following components of an AC generator:
Field, Armature, Prime mover, Rotor, Stator, and Slip rings.
1.3 Define the following electrical terms: volts, amps, watts,
inductance, capacitance, VARs, and hertz.
1.4 Describe a stationary field, rotating armature AC generator and a
rotating field, stationary armature AC generator.
© Copyright 2016
TLO 1
Operator Generic Fundamentals
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AC Generator Theory
ELO 1.1 – Describe the theory of operation of an AC generator.
• A simple AC generator consists of:
– A conductor or loop of wire in a magnetic field
– The loop ends connect to slip rings that are in contact with
brushes
– When the loop rotates, it cuts magnetic lines of force
• As conductor passes through magnetic field
– A voltage is induced in the conductor and transferred through slip
rings as voltage output
Figure: Simple AC Generator
© Copyright 2016
ELO 1.1
Operator Generic Fundamentals
5
AC Generator Theory
Magnitude of Generated Voltage
• Dependent on field strength and speed of rotor
• Most generators at constant speed; generated voltage depends on
field excitation, or field strength
Developing an AC Sine Wave Voltage
• Coil rotates, each side cuts the magnetic lines of force in opposite
directions
• Direction (polarity) of induced voltages depends on direction of
movement of coil
Figure: Simple AC Generator
© Copyright 2016
ELO 1.1
Operator Generic Fundamentals
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AC Generator Theory
• Magnetic field
• Produces: Current moving
through the conductor
• Conductor
• Relative motion
Resultant
Induced
(+)Voltage
Direction of
Magnetic
Field
Direction of
Motion
( )
© Copyright 2016
ELO 1.1
Operator Generic Fundamentals
7
AC Generator Theory
Developing an AC Sine Wave Voltage
• In vertical position, 0°, coils are moving parallel to magnetic field
• As coil rotates, voltages are additive, slip ring X positive (+) and Y is
negative
• Potential across resistor R causes current flow from Y to X through
resistor
• Current increases until reaches maximum value (90°)
– perpendicular to magnetic field
Figure: Simple AC Generator
© Copyright 2016
ELO 1.1
Operator Generic Fundamentals
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AC Generator Theory
Developing an AC Sine Wave
Voltage
• As coil continues to turn,
induced voltage and current
decrease until both reach zero,
where coil is again in the
vertical position (180°)
• Next half rev produces equal
voltage, except with reversed
polarity (270° and 360°)
• Current flow through R is now
from X to Y
Figure: Developing an AC Sine Wave Voltage
© Copyright 2016
ELO 1.1
Operator Generic Fundamentals
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AC Generator Theory
Period and Frequency
• As coil rotates 360°, output voltage goes through one cycle
• Period is time required for generator to complete one cycle
• Frequency (measured in hertz) is the number of cycles per second
𝑁𝑃
𝑓=
120
• Where:
f = frequency (Hz)
P = total number of poles
N = rotor speed (rpm)
120 = conversion from minutes to seconds and from poles to pole
pairs
© Copyright 2016
ELO 1.1
Operator Generic Fundamentals
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AC Generator Theory
Peak Voltage and Current
• Emax occurs at 90°
• Value is termed peak voltage
• Can quantify AC voltage or
current as peak voltage (Ep) or
peak current (Ip)
Peak to Peak Voltage and
Current
• Another common term
associated with AC is peak-topeak value (Ep-p or Ip-p)
• Peak to peak is magnitude of
voltage, or current
© Copyright 2016
Figure: AC Sine Wave Voltage
ELO 1.1
Operator Generic Fundamentals
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Effective Value of AC
• Effective value is most commonly used for quantifying AC
• Amount of AC that produces same heating effect as equal amount of
DC
• Effective value, because it is the root of the mean (average) square of
the currents, or the root-mean-square, or RMS value
• RMS = average value X 1.73
Figure: AC Voltage Sine Wave
© Copyright 2016
ELO 1.1
Operator Generic Fundamentals
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AC Generator Theory
• Dashed line is average of I2
values, square root of that value
is RMS, or effective value
(square root of mean square
deviation of a waveform)
• Average value is ½ Imax2
• RMS value is equal to 0.707
(square root of ½)
• Effective value of voltage or
current for an AC sine wave can
be found by using: Effective
Value (RMS) = Peak Value x
0.707
© Copyright 2016
ELO 1.1
Figure: Effective Value of AC Current
Operator Generic Fundamentals
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Phase Angle Guidelines
• Phase angle is fraction of a cycle, in
degrees, that has gone by since a
voltage or current has passed
through a given value
• Phase difference is common term for
phase angle
– Describes two different voltages
that have same frequency, which
pass through zero values in
same direction at different times
• Angles along axis indicate phases of
voltages e1 and e2
• At 120°, e1 passes through zero
value 60° ahead of e2 (e2 equals zero
at 180°)
• Describes as voltage e1 leads e2 by
60 electrical degrees, or voltage e2
lags e1 by 60 electrical degrees
© Copyright 2016
ELO 1.1
Figure: Phase Relationship
Operator Generic Fundamentals
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Generator Major Components
ELO 1.2 – State the purpose of the following components of an AC
generator: field, armature, prime mover, rotor, stator, and slip rings.
Field
• Coils of conductors receive
voltage from a source
(excitation) and produce a
magnetic flux
Armature
• Voltage is produced, consists of
coils of wire large enough to
carry full-load current of
generator output
Armature
Figure: Basic AC Generator
© Copyright 2016
ELO 1.2
Operator Generic Fundamentals
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Generator Components
Prime Mover
• Component that drives the AC generator; rotating machine, such as
diesel engine, steam turbine, or a motor
Rotor
• Driven by prime mover, can be armature or field (armature when voltage
output is generated there, field when field excitation is applied)
Stator
• Stationary part of the machine, can be either armature or field
Slip Rings
• Transfer power to or from the rotor, consist of a circular conducting
material mounted on rotor shaft
• Connected electrically to rotor windings, insulated from shaft
© Copyright 2016
ELO 1.2
Operator Generic Fundamentals
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Generator Components
Magnetic Field (Rotating Field Machine)
• Current flow through field coil of rotor produces strong magnetic field
• Slip rings and brushes conduct excitation to field coil in rotor
Continuous Connection
• Brushes are spring-held in contact with slip rings to provide
continuous connection between field coil and excitation circuit
Conductors
• Each time rotor makes one complete revolution, one complete cycle
of AC is developed
• A generator has many turns of wire wound into slots of stator
© Copyright 2016
ELO 1.2
Operator Generic Fundamentals
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Common Electrical Terms
ELO 1.3 – Define the following electrical terms: volts, amps, watts,
inductance, capacitance, VARs, and hertz.
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
Watt
W
Power (P)
Henry
H
Electrical inductance (L)
Farad
F
Electrical capacitance (C)
VAR
VAR
Reactive power (Q)
© Copyright 2016
ELO 1.3
Operator Generic Fundamentals
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Common Electrical Terms
Voltage (Volt)
• Potential difference that causes electrons to move in a conductor
• E in electrical formulas, V on lab equipment or schematic diagrams
Current (Ampere or Amp)
• Movement of free electrons through a conductor
• I in electrical formulas, A in lab or on schematic diagrams
Resistance (Ohm)
• One ohm is amount of resistance that limits current in conductor to
one ampere when potential difference (voltage) applied to conductor
is one volt
• R in electrical formulas, capital omega (Ω) in other instances
• Relationship between these parameters is Ohm’s law E = I × R
© Copyright 2016
ELO 1.3
Operator Generic Fundamentals
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Common Electrical Terms
Power (Watt)
• Electricity performs work
• Power (P in equations) is rate of performing work, or rate of heat
generation
• Common unit to specify electric power is watt (W in equations)
• Power also described as current (I) in a circuit times voltage (E)
across circuit
P = I ×E or P = I E
• Using Ohm’s Law for voltage, E = I × R and using substitution laws:
P = I × (I ×R)
• Therefore, power also equals current in a circuit squared multiplied
by resistance of circuit:
P = 𝐼2 × R
© Copyright 2016
ELO 1.3
Operator Generic Fundamentals
20
Common Electrical Terms
Inductance (Henry)
• Ability of a coil to store energy, induce a voltage in itself, and oppose
changes in current flowing through it
• Symbol used for inductance is L, Henries (H) are the units
• One henry is amount of inductance (L) that permits one volt to be
induced when current through coil changes at a rate of one ampere
per second
• Rate of change in current through a coil per unit time:
∆𝐼
∆𝑡
• Voltage VL induced in a coil with inductance L:
𝑉𝐿 = −𝐿
∆𝐼
∆𝑡
• Negative sign indicates voltage induced opposes change in current
through coil per unit time
© Copyright 2016
ELO 1.3
Operator Generic Fundamentals
21
Common Electrical Terms
Capacitance (Farad)
• Ability to store an electric charge (C)
• Measured in farads, equal to amount of charge (Q) that can be stored
in a device or capacitor divided by voltage (E) applied across the
device or capacitor plates when charge was stored
𝑄
𝐶=
𝐸
Frequency (Hertz)
• Measured in hertz, number of alternating voltage or current cycles
completed per second
© Copyright 2016
ELO 1.3
Operator Generic Fundamentals
22
Common Electrical Terms
Volt Ampere Reactive (VAR)
• Circulating current in a circuit that does no useful work
• In AC circuits not purely resistive, voltage and current are out of
phase
• Power is exchanged in these circuits as inductive fields form and
collapse, and capacitors charge and discharge
• Relationship between reactive power, apparent power, and true
power expressed in power triangle, which equates AC to DC power
by showing relations between:
– Generator output (Apparent Power - S) in volt-amperes (VA)
– Usable power (True Power - P) in watts
– Wasted/stored power (Reactive Power - Q) in volt-amperesreactive (VAR)
© Copyright 2016
ELO 1.3
Operator Generic Fundamentals
23
Common Electrical Terms
• Phase angle (θ) represents
inefficiency of AC circuit
• Represents comparable values
used to find efficiency level of
generated power to usable
power, expressed as power
factor
• Power factor is the ratio of True
Power/Apparent Power
Figure: Power Triangle
– “How much you get out for
what you have to put in”
© Copyright 2016
ELO 1.3
Operator Generic Fundamentals
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Types of Generators
ELO 1.4 – Describe the following types of AC generators: stationary
field/rotating armature; rotating field/stationary armature.
Stationary Field / Rotating
Armature
• Small AC generators usually
have stationary field and rotating
armature
• Disadvantage is that slip-ring
and brush assembly is in series
with load circuits
• If components are worn or dirty,
flow of current to loads may be
reduced or interrupted
• For large generators, slip rings
and brushes are limiting factor
on machine capacity
© Copyright 2016
ELO 1.4
Figure: Stationary Field with Rotating
Armature Generator
Operator Generic Fundamentals
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Types of Generators
Rotating Field / Stationary
Armature
• When DC field excitation
connected to rotor, stationary
coils have AC induced into them
• Used where large power
generation is needed
– DC source is supplied to
rotating field coils (via slip
rings and brushes), which
produces magnetic field
around rotating element
• As prime mover turns rotor, field
rotates across conductors of
stationary armature and an EMF
is induced into armature windings
© Copyright 2016
ELO 1.4
Figure: Rotating Field with Stationary
Armature Generator
Operator Generic Fundamentals
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Types of Generators
Rotating Field / Stationary
Armature
Advantages over stationary field,
rotating armature AC generator:
1. A load can be connected to
armature without having
moving contacts (slip rings
and brushes) in circuit
2. Much easier to insulate stator
conductors than rotating
conductors
3. Much higher output voltages
and currents can be
generated
© Copyright 2016
Figure: Rotating Field with Stationary
Armature Generator
ELO 1.4
Operator Generic Fundamentals
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Types of AC Motors
TLO 2 – Explain the theory of operation of selected types of AC motors.
2.1
2.2
2.3
2.4
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.
2.5 Describe how an AC synchronous motor is started and its operating
characteristics.
© Copyright 2016
TLO 2
Operator Generic Fundamentals
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Motor Rotating Field
ELO 2.1 – Describe how a rotating magnetic field is produced in an AC
motor.
• Basic principle of operation for AC motors is interaction of revolving
magnetic field created in stator by AC current with an opposing
magnetic field in rotor
• Magnetic field in rotor comes from one of two sources:
– Induced from the rotating stator
– Provided by a separate DC current source
© Copyright 2016
ELO 2.1
Operator Generic Fundamentals
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Motor Rotating Field
Rotating Magnetic Field
• Figure illustrates three-phase
stator with three-phase AC
current applied
• Windings connect in a wye
configuration
• Two windings in each phase
wind in same direction
• At any instant in time, magnetic
field generated by one phase will
depend on current flow through
that phase
• If current flow through that phase
is zero, resulting magnetic field
is zero
• If current flow is at maximum
value, resulting field is at
maximum value
© Copyright 2016
ELO 2.1
Figure: Three-Phase Motor Stator
Operator Generic Fundamentals
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Motor Rotating Field
• As magnetic field rotates in stator, rotor also rotates in an attempt
to maintain its alignment with stator’s magnetic field
• Figure on next slide shows stator’s magnetic field "stopped" at six
selected positions
• Instances are marked off at 60° intervals representing current
flowing in phases A, B, and C
© Copyright 2016
ELO 2.1
Operator Generic Fundamentals
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Motor Rotating Magnetic Field
N
N
C
B’
A
S
N
A’
N
S
S
S
B
C’
C
B’
A
A’
B
C’
N
S
N
S
T1
A
C
B’
A
N
N
N
A’
S
N
S
N
C
B’
A
A
A’
B
C’
A’
B
C’
S
N
T4
N
S
N
S
S
B’
C
A’
A
N
B
C’
S
N
S
S
T5
T6
T7
T3
C
B’
N
T3
Phase B
T2
S
N
B
C’
N
B
C’
T2
N
A’
N
S
Time 1
C
B’
S
S
S
S
Phase A
T4
T5
T6
T7
Phase C
© Copyright 2016
ELO 2.1
Operator Generic Fundamentals
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C
C’
Phase A
Phase B
Time 1
T2
T3
T4
T5
T6
T7
Phase C
© Copyright 2016
ELO 2.1
Operator Generic Fundamentals
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Slip Effects on AC Motor
ELO 2.2 – Define slip and explain its effect on AC induction motor
operation.
• Slip Effects on AC Induction Motor Operation
– Not possible for rotor of an AC induction motor to turn at same
speed as rotating magnetic field
– If speed of rotor were same as that of stator, no relative motion
would exist and there would be no induced EMF in rotor
– Rotor must rotate at some speed less than that of magnetic field
in stator for relative motion to exist
© Copyright 2016
ELO 2.2
Operator Generic Fundamentals
34
Slip Effects on AC Motor
• Slip is percentage difference
between speed of rotor and
rotating magnetic field in stator
• Amount of torque on rotor will
change as slip ratio changes
• As slip increases from zero to
about 20%, torque increases
linearly
• As load and slip increase beyond
full-load torque, torque will reach
a maximum value at about 25%
slip
• Typical values of between 1%
and 8%
– From no-load running current
to full-load running current
© Copyright 2016
Figure: Torque versus Slip for an
AC Induction Motor
ELO 2.2
Operator Generic Fundamentals
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Developing Torque in AC Induction Motor
ELO 2.3 – Describe how torque is produced in an AC motor.
Torque Production
• When alternating current flows through stator windings of an AC
induction motor, a rotating magnetic field results
• Rotating magnetic field cuts conductors of rotor and induces a current
in them due to generator action
© Copyright 2016
ELO 2.3
Operator Generic Fundamentals
36
Developing Torque in AC Induction Motor
Torque Production
• Induced current produces a
magnetic field, opposite in
polarity of stator field, around
conductors of rotor
• Magnetic field induced in rotor
will try to line up with magnetic
field produced by stator
• Rotor unable to line up, or lock
onto, stator field; therefore,
must follow behind it
Figure: Induced Rotor Magnetic Field
• Figure shows force directions of
induced rotor magnetic field
© Copyright 2016
ELO 2.3
Operator Generic Fundamentals
37
Developing Torque in AC Induction Motor
Breakdown Torque
• Max torque motor can produce
without stalling is breakdown torque
• If load increases beyond this point,
motor will stall, come to a rapid
stop
• Typical breakdown torque varies
from 200% to 300% of full-load
torque
Starting Torque
• Value of torque at 100% slip and
normally 150% to 200% of full-load
torque
• As rotor accelerates, torque will
increase and then decrease to
value required to carry load on
motor at constant speed, between
1-8 % slip
© Copyright 2016
ELO 2.3
Figure: Torque versus Slip for an
AC Induction Motor
Operator Generic Fundamentals
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Developing Torque in AC Induction Motor
Developing Torque
• Torque of an AC induction motor dependent upon strength of
interacting rotor and stator fields and phase relationship
between them
• Equation shows 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
© Copyright 2016
ELO 2.3
Operator Generic Fundamentals
39
Developing Torque in AC Induction Motor
• During normal operation, K and cos 𝜃R are essentially constant, so
torque is directly proportional to rotor current
• Rotor current increases in almost direct proportion to slip, so
maximum current will occur at maximum slip, which is at motor
startup
© Copyright 2016
ELO 2.3
Operator Generic Fundamentals
40
Developing Torque in AC Induction Motor
• Rotating field induces a voltage
in the rotor, causing current to
flow
C
• Current flow builds rotor
magnetic field
• Rotor field chases stator
• Always turns slower
• Called “slip”
• More load = More slip
• Rotor current induces Counter
ElectroMotive Force (CEMF) in
stator
C
• CEMF controls current
© Copyright 2016
ELO 2.3
Operator Generic Fundamentals
41
Starting Current on AC Induction Motor
ELO 2.4 – Explain differences between starting and running current for
an AC induction motor and the operating limits used to mitigate the
effects of starting current.
• There are several reasons for high starting current:
– Power is required to build up rotating magnetic field in stator
– Extra energy is required to overcome inertia of rotor
– Interactions occur between rotor currents and stator’s magnetic
field, result in motor drawing high currents
– Rotor speed is too low to generate sufficient counter
electromotive force in stator
© Copyright 2016
ELO 2.4
Operator Generic Fundamentals
42
Starting Current and CEMF
• Three induced fields
C
– Stator field from electric
source
– Rotor field induced from
stator field
– Counter field from rotor back
into stator - limits current
flow in stator (CEMF)
C
© Copyright 2016
ELO 2.4
Operator Generic Fundamentals
43
Starting Current and CEMF
• Stator resistance is low at start
• When breaker is closed, stator voltage applied to rotor
– Causing high current on pump start (5 – 7 times full load)
• CEMF builds in as rotor comes to speed
– Limits stator current
• Current quickly stabilizes at no load running current
– With pump discharge valve closed or throttled
© Copyright 2016
ELO 2.4
Operator Generic Fundamentals
44
Starting Current and CEMF
• Terminal voltage applied is EMF
EMF
– (index finger of right hand)
• As rotor cuts line of force, a
current flows in the rotor
CEMF
• Rotor current generated a field
that opposes the EMF
Net Field
– Called Counter Electro
Motive Force (CEMF)
– (index finger of left hand)
• Results in lower “net” field
– No load running current
© Copyright 2016
ELO 2.4
Operator Generic Fundamentals
45
Starting Current on AC Induction Motor
Starting Current Indications
• Amps immediately increase (normally five to seven times) the normal
running current value, usually off scale high
• After a few seconds, current comes back on scale, decreases to
normal running current
• If current does not drop back into normal band within a few seconds,
this is indication that equipment not functioning properly and is
appropriate to shut it down and investigate
• Due to the large starting current and heat produced on motor startup
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.
© Copyright 2016
ELO 2.4
Operator Generic Fundamentals
46
Starting Current on AC Induction Motor
Knowledge Check – NRC Question
The starting current in an AC motor is significantly higher than the fullload running current because...
A. little counter electromotive force is induced in the rotor windings
during motor start.
B. motor torque production is highest during motor start.
C. little counter electromotive force is induced in the stator
windings during motor start.
D. work performed by the motor is highest during motor start.
Correct answer is C.
© Copyright 2016
ELO 2.4
Operator Generic Fundamentals
47
Starting Current on AC Induction Motor
Knowledge Check – NRC Question
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...
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.
D. several times the full-load current value, and then decrease to
the full-load value.
Correct answer is A.
© Copyright 2016
ELO 2.4
Operator Generic Fundamentals
48
AC Synchronous Motor Operation
ELO 2.5 – Describe how an AC synchronous motor is started and its
operating characteristics.
• Synchronous motors are like
induction motors in that they both
have stator windings that produce
rotating magnetic field
• Unlike an induction motor, an
external DC source excites the
synchronous motor rotor and
therefore requires slip rings and
brushes
Figure: AC Synchronous Motor
© Copyright 2016
ELO 2.5
Operator Generic Fundamentals
49
AC Synchronous Motor Operation
• Synchronous motors use wound
rotor
• Slip rings and brushes needed
• Rotor is much more expensive
to manufacture than squirrel
cage rotor associated with AC
induction motors
Figure: Wound Rotor
© Copyright 2016
ELO 2.5
Operator Generic Fundamentals
50
AC Synchronous Motor Operation
Starting a Synchronous Motor
• Not self-starting, torque only developed when running at synchronous
speed
• Needs some type of device to bring rotor to synchronous speed
• Two ways of starting a synchronous motor:
– Use a separate DC motor
– 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 DC motor to synchronous speed, stator windings
receive AC current, DC motor then acts as a DC generator and
supplies DC field excitation to rotor of synchronous motor
• Once operating at synchronous speed, synchronous motor is ready
for load
© Copyright 2016
ELO 2.5
Operator Generic Fundamentals
51
AC Synchronous Motor
Starting a Synchronous Motor
• More commonly, a squirrel-cage winding is embedded in face of rotor
poles to start motor
• Starts as an induction motor, speed increases to ≈95% of
synchronous speed, then direct current is applied and motor begins
to pull into synchronism
• Some plants use synchronous motors for Circ Water pumps
– Improves power factor by being able to adjust excitation
© Copyright 2016
ELO 2.5
Operator Generic Fundamentals
52
AC Motor and Generator Operation
TLO 3 – Explain operating characteristics and limitations on AC motors.
3.1 State the applications of the following types of AC motors:
a. Induction
b. Single-phase
c. Synchronous
3.2 Describe the indications resulting from a locked motor rotor, sheared shaft,
or miswired phases.
3.3 Describe the consequences of overheated motor windings or motor
bearings.
3.4 Describe the causes of excessive current in motors and generators resulting
from conditions such as low voltage, overloading, and mechanical binding.
3.5 Describe the relationship between pump motor current and the following
parameters:
a. Pump flow
b. Pressure
c. Speed
d. Motor stator temperature
© Copyright 2016
TLO 3
Operator Generic Fundamentals
53
Motor Applications
ELO 3.1 – State the applications of the following types of AC motors:
induction, single-phase, and synchronous.
Induction Motors
• Most commonly used motor in industrial applications
– less expensive to manufacture and maintain and operates reliably
• Pumps, fans, and compressors are examples of induction motors
© Copyright 2016
ELO 3.1
Operator Generic Fundamentals
54
Motor Applications
Induction Motors
• Named from rotating magnetic
field of stator
– induces a voltage into the
rotor
• Most commonly used AC motor
in industrial applications
– simplicity, rugged
construction, and relatively
low manufacturing costs
– no external connections
• Most AC induction motors use a
“squirrel-cage” rotor
© Copyright 2016
ELO 3.1
Figure: Squirrel Cage Rotor
Operator Generic Fundamentals
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Motor Applications
Squirrel-Cage Rotor
• Made of a laminated cylinder
with slots in its surface
• Made of heavy copper bars that
connect at each end by a metal
ring made of copper or brass
• No insulation required because
of low voltages induced into
rotor bars
• For maximum field strength,
size of air gap between rotor
bars and stator windings is
small
© Copyright 2016
Figure: Squirrel-Cage Induction Rotor
ELO 3.1
Operator Generic Fundamentals
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Motor Applications
Single-Phase Motors
• Very small commercial applications
• Mostly used in household appliances
Synchronous Motors
• More complex design and higher manufacturing cost
• Rarely used in commercial applications
• Sometimes used for Circ Water pumps
– Excitation can be adjusted to minimize impact of inductive loading
© Copyright 2016
ELO 3.1
Operator Generic Fundamentals
57
Indications of Motor Malfunctions
ELO 3.2 – Describe the indications resulting from a locked motor rotor or
shaft, sheared shaft, or miswired phases.
Indications of Motor Malfunctions
• Motors or their connected loads
are subject to several different
types of mechanical failures,
including:
– Locked (seized) rotor
– Sheared rotor
– Miswired phases
Figure: Damage From Locked Rotor
© Copyright 2016
ELO 3.2
Operator Generic Fundamentals
58
Indications of Motor Malfunctions
Locked (Seized) Rotor
• Typical indications of a locked rotor:
– Rotor speed decreases
– Immediate increase in current to essentially starting current
o No rotor speed, no CEMF
– Immediate reduction in system flow rate
– Immediate reduction in component discharge pressure
– Immediate rise in motor winding temperatures resulting from
higher current flow
– Eventual timed delay tripping of circuit breaker due to high current
o Therefore, current drops to ZERO after breaker trip
© Copyright 2016
ELO 3.2
Operator Generic Fundamentals
59
Indications of Motor Malfunctions
Sheared Rotor
• Allows motor to operate freely (spinning) with no load mechanically
attached
• Following indications are typical of a sheared rotor:
– Rotor speed initially increases due to loss of load (pump impeller)
– Load has no (low) running current (as indicated on ampere meter)
– Immediate reduction in system flow rate
– Immediate reduction in component discharge pressure
• Reduction in flow and pressure similar to locked rotor indication
• Current indication differs from locked rotor
– Sheared shaft goes to no load running current
– Locked rotor goes to starting current then 0 after breaker trip
© Copyright 2016
ELO 3.2
Operator Generic Fundamentals
60
Indications of Motor Malfunctions
Miswired Phases
• Phases can be miswired after work in which motor leads were
disconnected
• If two motor leads (phases) reversed, motor will turn backwards
• Motor will work fine, but component it is driving generally will not
© Copyright 2016
ELO 3.2
Operator Generic Fundamentals
61
Indications of Motor Malfunctions
Knowledge Check – NRC Question
A motor-driven cooling water pump is operating normally. How will
pump motor current respond if the pump experiences a locked rotor?
A. Decreases immediately to zero due to breaker trip.
B. Decreases immediately to no-load motor amps.
C. Increases immediately to many times running current, then
decreases to no-load motor amps.
D. Increases immediately to many times running current, then
decreases to zero upon breaker trip.
Correct answer is D.
© Copyright 2016
ELO 3.2
Operator Generic Fundamentals
62
Consequences of Motor Overheating
ELO 3.3 – Describe the consequences of overheated motor windings or
bearings.
Loss of Motor Cooling
• Continuous operation of a motor at rated load with a loss of required
cooling to motor windings will result in breakdown of motor insulation
due to overheating
• Causes increased temperature and current flow due to a decrease in
insulation resistance
– Possible short circuit
• Thermal overload device protects many large motors from high
current
© Copyright 2016
ELO 3.3
Operator Generic Fundamentals
63
Consequences of Motor Overheating
Loss of Bearing Cooling or Lubrication
• Various means of cooling and lubricating motor bearings, depending
on size and application of motor
• Two needs are common to all bearings:
– Must be properly lubricated to minimize friction
– Heat generated in the bearing must be removed
• High bearing friction results in additional load that causes motor
current and motor winding temperature to rise
• Bearing wear can increase runout, permitting destructive contact
between rotor and stator
© Copyright 2016
ELO 3.3
Operator Generic Fundamentals
64
Consequences of Motor Overheating
Loss of Bearing Cooling or Lubrication
• Increase in bearing friction will cause bearing to degrade until it fails,
resulting in component trip and possible damage to component
– If increase is significant, can lead to motor damage
• If a bearing is running with higher temperature, but friction is not
excessive, it may run for some time without causing damage
– Most bearing lubricants are effective within a specified
temperature range
© Copyright 2016
ELO 3.3
Operator Generic Fundamentals
65
Consequences of Motor Overheating
Knowledge Check – NRC Question
Which one of the following will result from prolonged operation of an AC
induction motor with excessively high stator temperatures?
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
Correct answer is B.
© Copyright 2016
ELO 3.3
Operator Generic Fundamentals
66
Excessive Current in Motors
ELO 3.4 – Describe the causes of excessive current in motors resulting
from conditions such as low voltage, overloading, and mechanical
binding.
Motor Overloading
• Overloading or just increased loading can result in increased current
• Overloading can be caused by:
– Gradual bearing failure
– Undervoltage
– Locked rotor
– Shorted windings
– One open phase
© Copyright 2016
ELO 3.4
Operator Generic Fundamentals
67
Excessive Current in Motors
Malfunction
Likely Result
Gradual motor
bearing failure
Increased friction, current and temperature could be
high enough to cause a thermal overload trip
Packing on a
motor-operated
valve tightened
excessively
Motor current increases due to increased torque on
motor caused by additional friction associated with
packing
© Copyright 2016
High currents can result in excessive heat within
motor, causing breakdown of motor winding
insulation and damage to motor (grounds)
ELO 3.4
Operator Generic Fundamentals
68
Excessive Current in Motors
Malfunction
Likely Result
Locked or seized Overcurrent trip of the supply circuit breaker
rotor of motor or
component
driven (pump,
etc.)
Reduced voltage Motor supplies power needed to drive the load,
supplied to motor when voltage drops, current increases
Operating for extended periods with reduced
voltage can lead to winding damage
Sheared shaft or Abnormally low or no running currents as indicated
rotor
on an amp meter, with low or no flow/pressure
indicated
© Copyright 2016
ELO 3.4
Operator Generic Fundamentals
69
Excessive Current in Motors
Failed Bearings
• Bearing failure can result from a number of circumstances, including:
– Insufficient lubrication
– Poor bearing maintenance practices
– Improper loading of component
– Motor undervoltage situation
• Any condition leading to overheating of bearings can cause bearing
failure
• Excessive work can cause heat buildup in machine windings
© Copyright 2016
ELO 3.4
Operator Generic Fundamentals
70
Excessive Current In Motors
Knowledge Check
Which of the following will cause an increase in motor current? (Select
all that are true.)
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
Correct answers are A, B, and D.
© Copyright 2016
ELO 3.4
Operator Generic Fundamentals
71
Excessive Current In Motors
Knowledge Check – NRC Question
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
Correct answer is D.
© Copyright 2016
ELO 3.4
Operator Generic Fundamentals
72
Pump Motor Current Relationships
ELO 3.5 – Describe the relationship between pump motor current and the
following parameters: pump flow, pressure, speed, and motor stator
temperature.
Operating Limits to Mitigate Effect of Starting Current
• Recall from the pumps module that centrifugal pumps operate within
a known set of laws, known as pump laws
• Applying pump laws, we can determine how changes in pump
parameters will affect motor power and current
© Copyright 2016
ELO 3.5
Operator Generic Fundamentals
73
Pump Motor Current Relationships
Pump Law Review
• Flow rate of a pump is directly proportional to pump speed
𝑉∝𝑛
• Head of a pump is directly proportional to pump speed squared
𝐻𝑝 ∝ 𝑛2
• Power of a pump is directly proportional to pump speed cubed
𝑝 ∝ 𝑛3
• Where:
n = speed of pump impeller (rpm)
𝑉 = volumetric flow rate of pump (gpm or ft3/hr)
Hp = head developed by pump (feet)
p = pump power (kW)
© Copyright 2016
ELO 3.5
Operator Generic Fundamentals
74
Pump Motor Current Relationships
• These relationships are 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 will draw.
Action
Equation
Determine change in flow by using this relationship.
𝑉2 = 𝑉1
𝑛2
𝑛1
Determine change in head by using this relationship.
Notice that head changes by the square of change in
speed.
𝐻𝑃2 = 𝐻𝑃1
Determine change in power by using this relationship.
Notice that power changes by the cube of change in
speed.
𝑃2 = 𝑃1
© Copyright 2016
ELO 3.5
𝑛2
𝑛1
𝑛2
𝑛1
2
3
Operator Generic Fundamentals
75
Pump Motor Current Relationships
Flow Example
• 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:
𝑉2 = 𝑉1
𝑛2
𝑛1
𝑉2 = 400 𝑔𝑝𝑚
3,600 𝑟𝑝𝑚
1,800 𝑟𝑝𝑚
𝑉2 = 800 𝑔𝑝𝑚
© Copyright 2016
ELO 3.5
Operator Generic Fundamentals
76
Pump Motor Current Relationships
Head Example
• 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
𝐻𝑃2
𝑛2
𝑛1
2
3,600 𝑟𝑝𝑚
= 48 𝑓𝑡
1,800 𝑟𝑝𝑚
2
𝐻𝑃2 = 192 𝑓𝑡
© Copyright 2016
ELO 3.5
Operator Generic Fundamentals
77
Pump Motor Current Relationships
Power Example
• 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
𝑛1
3
3,600 𝑟𝑝𝑚
𝑃2 = 45 𝑘𝑊
1,800 𝑟𝑝𝑚
3
𝑃2 = 360 𝑘𝑊
© Copyright 2016
ELO 3.5
Operator Generic Fundamentals
78
Pump Motor Current Relationships
• Relationship of these
parameters can be calculated
and plotted on a pump
performance curve for specific
pumps and operating conditions
(see graph)
• NRC GFE questions use these
curves frequently
Figure: Pump Flow, Head, and Brake Horsepower
© Copyright 2016
ELO 3.5
Operator Generic Fundamentals
79
Pump Motor Current Relationships
• Brake horsepower is input power delivered to pump by motor (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 are proportional to changes in motor
amps
• Increases in motor amps will result in increases in motor winding
temperature
© Copyright 2016
ELO 3.5
Operator Generic Fundamentals
80
Pump Motor Current Relationships
Example: 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
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
© Copyright 2016
Figure: Pump Flow, Head, and Brake Horsepower
ELO 3.5
Operator Generic Fundamentals
81
Pump Motor Current Relationships
• Using the pump curve, 400 gpm
is equivalent to ~ 220 BHP;
– 800 gpm is ~ 270 BHP
– 220/270 = 50 amps/x
• Solving for x:
– x = 50/(220/270)
– x = 50/0.82 = ~ 61 amps,
• 61 amps is less than 100 amps,
so the correct answer is A
© Copyright 2016
Figure: Pump Flow, Head, and Brake Horsepower
ELO 3.5
Operator Generic Fundamentals
82
Pump Motor Current Relationships
Knowledge Check – NRC Question
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 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
Correct answer is C.
© Copyright 2016
ELO 3.5
Operator Generic Fundamentals
83
AC Generator Operation
TLO 4 – Analyze operation of generators in parallel and predict system
response to changes in frequency, voltage, and load.
4.1 Define apparent, true, and reactive power and power factor using a
power triangle and their effect on generator operation.
4.2 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
4.3 Describe the conditions that must be met prior to paralleling two
generators, including consequences of not meeting these conditions.
4.4 Describe the consequences of over-excitation and under-excitation
when load sharing.
© Copyright 2016
TLO 4
Operator Generic Fundamentals
84
Power Factor and Generators
ELO 4.1 – Define apparent, true, and reactive power and power factor
using a power triangle and their effect on generator operation.
Power Triangle
• In AC circuits, current and voltage are normally out of phase due to
effects of inductive and capacitive reactance
• As a result, not all 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
© Copyright 2016
ELO 4.1
Operator Generic Fundamentals
85
Power Triangle
• Can be used to determine a
generator’s efficiency
• The power output of a generator is
referred to as apparent power (S)
‒ Measured in volt-amperes (VA)
• True power (P) is the amount of
usable power
‒ Measured in watts
• Wasted or stored power is referred to
as reactive power (Q)
‒ Measured in volt-amperesreactive (VAR)
• Phase angle (θ) represents
inefficiency of circuit
‒ Corresponds total reactive
impedance (Z) to current flow
© Copyright 2016
ELO 4.1
Figure: Power Triangle
Operator Generic Fundamentals
86
Power Factor and Generators
• Apparent power is power delivered to an electrical circuit
𝑆 = 𝐼2 𝑍 = 𝐼𝐸
• Where:
S = apparent power (VA)
I = RMS current (A)
E = RMS voltage (V)
Z = impedance (Ω)
• True power is power consumed by resistive loads in electrical circuit
𝑃 = 𝐼2 𝑅 = 𝐸𝐼 cos 𝜃
• Where:
P = true power (watts)
I = RMS current (A)
E = RMS voltage (V)
R = resistance (Ω)
θ = angle between E and I sine waves
© Copyright 2016
ELO 4.1
Operator Generic Fundamentals
87
Power Factor and Generators
• Reactive power is power component in AC circuit (inductive) and
(capacitive) fields, units are volt-amperes-reactive (VAR)
𝑄 = 𝐼 2 𝑋 = 𝐸𝐼 sin 𝜃
• Where:
Q = reactive power (VAR)
I = RMS current (A)
X = net reactance (Ω)
E = RMS voltage (V)
θ = angle between the E and I sine waves
• Reactive power is unusable power because it is stored in circuit by:
– Inductors, expand and collapse magnetic fields, try to keep current
constant
– Capacitors, charge and discharge in an attempt to keep voltage
constant
– Inductance and capacitance alternately exchange reactive power
(consume and give back) with circuit’s AC source
© Copyright 2016
ELO 4.1
Operator Generic Fundamentals
88
Power Factor and Generators
• Total power delivered by the source, same as apparent power
• Part of this apparent power, called true power, dissipates by circuit
resistance in form of heat
• Rest of apparent power returns to source by circuit inductance and
capacitance (reactive power)
© Copyright 2016
ELO 4.1
Operator Generic Fundamentals
89
Power Factor and Generators
• Power factor (PF) is represented by
cos θ in an AC circuit
• Ratio of true power to apparent
power, where θ is phase angle
between applied voltage and
current sine waves and is angle
between P and S on a power
triangle
𝑃
cos 𝜃 =
𝑆
• Where:
cos 𝜃 = power factor (pf)
P = true power (watts)
S = apparent power (VA)
Figure: Power Triangle
NOTE: The longer the hypotenuse,
the greater the generator current, and
higher the heat from that current.
© Copyright 2016
ELO 4.1
Operator Generic Fundamentals
90
Power Factor and Generators
Lagging Power Factor
• Power factor determines what part of apparent power is true power
• Can vary from 1, when the phase angle is 0°, to 0, when the phase
angle is 90°
• Inductive circuit, current lags voltage
• This type of circuit has a lagging power factor
ELI the ICE Man
• ELI refers to an inductive circuit (L)
– Where, current (I) lags voltage (E)
© Copyright 2016
ELO 4.1
Operator Generic Fundamentals
91
Power Factor and Generators
Leading Power Factor
• In capacitive circuit, current leads voltage
• Circuits with loads like motors will have lagging power factor
• Circuits with loads like fluorescent lighting will have leading power
factor
• Most industrial electrical systems exhibit lagging power factor
because inductive loads account for larger percentage of reactance
in these circuits
ELI the ICE Man
• ICE refers to a capacitive circuit (C) where current (I) leads voltage
(E)
© Copyright 2016
ELO 4.1
Operator Generic Fundamentals
92
Sample Generator Capability Curve
• Megawatts is the solid horizontal
line (also, 0 VARs)
• In a purely resistive circuit
– Power triangle is a straight line
– Power Factor = 1
– True Power = Apparent Power
• As you raise (or lower) excitation
(voltage), you become more (or
less) “overexcited”
– Move “up (down)” the Reactive
KVA axis
– Phase angle increases as you
move away from “Megawatts”
line
o Generator current increases
© Copyright 2016
Figure: Sample Generator Capability Curve
ELO 4.1
Operator Generic Fundamentals
93
VARs Terminology
• Consider the Generator Capability Curve:
Increase
excitation
• Lagging VARs:
– Overexcited
Increase
generator
amps
– Positive VARs
– VARs OUT
– Supplying VARs
• Leading VARs:
– Underexcited
Increase
generator
amps
– Negative VARs
Decrease
excitation
© Copyright 2016
ELO 4.1
– VARs IN
Operator Generic Fundamentals
94
VARs Terminology
• Consider an increase in excitation from an initial point of some
“leading VARs”:
New
Power
Triangle
Still LEADING VARs, just less
leading. Power Factor closer to 1.
(smaller phase angle)
Increase
excitation
© Copyright 2016
ELO 4.1
Operator Generic Fundamentals
95
Voltage Regulation and Components
ELO 4.2 – Describe the purpose of a voltage regulator and function of
each of the following typical voltage regulator components: sensing
circuit, reference circuit, comparison circuit, amplification circuit(s), signal
output circuit, and feedback circuit.
• Output voltage of an AC generator varies with both changes in load
(i.e., current) and changes in power factor
• Due to output variation, AC generators require voltage regulation
• Voltage regulator compares output voltage of machine to a desired
value
– Varies excitation applied to generator’s field to maintain generator
output voltage constant
© Copyright 2016
ELO 4.2
Operator Generic Fundamentals
96
Voltage Regulation and Components
Voltage Regulator Components
• Below is a typical block diagram of a voltage regulator, which consists
of six basic circuits that act together to regulate output voltage from
no-load to full-load
Figure: Voltage Regulator Block Diagram
© Copyright 2016
ELO 4.2
Operator Generic Fundamentals
97
Voltage Regulation and Components
Sensing Circuit
• Senses output voltage of AC generator
• As generator is loaded or unloaded, output voltage changes and sensing
circuit provides a signal of voltage changes
• Signal proportional to output voltage, sends signal to comparison circuit
Reference Circuit
• Maintains a constant output for reference, represents desired voltage
output
Comparison Circuit
• Electrically compares reference voltage to received sensed voltage and
provides error signal
• Sends error signal to amplification circuit
© Copyright 2016
ELO 4.2
Operator Generic Fundamentals
98
Voltage Regulation and Components
Amplification Circuit
• Magnetic amplifier or transistor amplifier, takes error signal from
comparison circuit and amplifies milliamp input to an amp output then
sends amplified signal to signal output or field circuit
Signal Output Circuit
• Controls field excitation of AC generator, increases or decreases field
excitation to either raise or lower AC output voltage
Feedback Circuit
• Takes some output of signal output circuit and feeds it back to
amplification circuit, prevents overshooting or undershooting desired
voltage by slowing down circuit response
Changing Output Voltage
• As generator load changes, output voltage will change if excitation
remains constant; voltage regulator senses change in output voltage and
changes generator excitation to return output voltage to desired value
© Copyright 2016
ELO 4.2
Operator Generic Fundamentals
99
Voltage Regulation and Components
Generator Voltage Control (most plants are similar to this)
• Steam applied to rotor of turbine to get it up to 1800 rpm
• Permanent Magnet Generator (PMG) generates AC and sends to
voltage regulator
– Sometimes called Field Flashing
• Voltage regulator rectifies the AC to DC then sends DC to the exciter
– This is the “excitation” discussed earlier to change VARs
• Exciter generates AC and sends it to the rectifier banks
• Rectifier banks rectify AC to DC
• DC applied to the rotor of the main generator and generates AC in
the stator (generator output voltage)
• This voltage also sent to voltage regulator for voltage control
© Copyright 2016
ELO 4.2
Operator Generic Fundamentals
100
Voltage Regulation and Components
Increasing Generator Load (Output Voltage Drop)
• Increase in generator load results in a drop in generator output voltage
– voltage regulator will respond to raise output voltage
• Sensing circuit senses decrease in output voltage as compared to
reference and lowers its input to comparison circuit
• Comparison circuit develops an error signal
• Error signal developed will be a positive value with magnitude of signal
dependent on difference between sensed and reference voltage
• Amplifier circuit amplifies this output and sends amplified signal to signal
output circuit
• Signal output circuit increases field excitation to AC generator, causing
generated voltage to increase to desired output
© Copyright 2016
ELO 4.2
Operator Generic Fundamentals
101
Voltage Regulation and Components
Decreasing Generator Load (Output Voltage Increase)
• If load on generator decreases, voltage output of machine would rise
• Actions of voltage regulator for an increase in output voltage would
be opposite of those for a lowering output voltage
• Comparison circuit will develop a negative error signal whose
magnitude is again dependent on difference between sensed voltage
and reference voltage
• Signal output circuit will decrease field excitation to AC generator,
causing generated voltage to decrease to desired output
© Copyright 2016
ELO 4.2
Operator Generic Fundamentals
102
Paralleling Generators
ELO 4.3 – Describe the conditions that must be met prior to paralleling
two generators, including consequences of not meeting these conditions.
• Most power grids and distribution systems have more than one AC
generator operating at one time
• Three required conditions for paralleling (or synchronizing) AC
generators:
– Terminal voltages must be essentially equal
o Minimizes VAR loading
– Frequency of incoming generator slightly > grid frequency
o Ensures generator picks up load instead of becoming a load
(motorizing)
– Output voltages must be in phase
o Minimize current through generator output breaker
© Copyright 2016
ELO 4.3
Operator Generic Fundamentals
103
Paralleling Generators
• During paralleling operations, voltmeters indicate voltages of
generators
• Output frequency meters facilitate frequency matching
• A synchroscope, a device that senses two frequencies and indicates
phase differences between generators, allows phase matching of two
generators
© Copyright 2016
ELO 4.3
Operator Generic Fundamentals
104
Paralleling Generators
• With two generators in parallel,
real load carried by each
generator can be determined by
examining frequency-power
characteristics for each
generator
• The “house curves” to the right
graphically describes load
sharing between two power
sources
• Note that power values for
generators G1 increases to the
left and G2 increases to the
right
© Copyright 2016
Figure: Two AC Generators Operating in Parallel
ELO 4.3
Operator Generic Fundamentals
105
Paralleling Generators
• Gen G1 – power output to left of
the solid vertical centerline
(frequency axis)
• Gen G2 - power output to right
of solid vertical centerline
(frequency axis)
• Since parallel generators must
operate at same frequency, we
can determine load carried by
each generator by noting the
intersection of generator
characteristic for particular
speed regulator setting in use
and operating frequency
© Copyright 2016
Figure: Two AC Generators Operating in Parallel
ELO 4.3
Operator Generic Fundamentals
106
Paralleling Generators
Real Power Sharing
• For settings in figure, at 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 frequency
• Total length of line A-B represents
total load (400 kW)
• If total load increased to 500kW
with no changes in speed
regulator settings, frequency
would drop to 59.95 Hz and G1
would carry 400 kW and G2
100kW, line C-D
© Copyright 2016
ELO 4.3
Figure: Two AC Generators Operating in Parallel
Operator Generic Fundamentals
107
Paralleling Generators
Real Load Sharing
• Load initially unbalanced
between generators and proper
distribution of load between
generators
• Dashed lines show original
condition (freq. = 60Hz, G1 load
= 350 kW, and G2 load = 50
kW)
• Increasing speed setting on G2
and decreasing speed setting
on G1 until no-load frequencies
are same, load on G2 increases
from 50 kW to 200 kW
• Load on G1 decreases from
350 kW to 200 kW
© Copyright 2016
ELO 4.3
Figure: Kilowatt Load Balanced by Adjusting
Speed Regulator
Operator Generic Fundamentals
108
Paralleling Generators
Reactive Load Sharing
• Consider two generators
operating in parallel, supplying a
constant real power
• The reactive loads carried by
each generator determined by
terminal voltage
• Different curves represent
different values of field excitation
• Fixed value of field excitation,
terminal voltage drops off as
reactive load increases,
exhibiting a drooping
characteristic
© Copyright 2016
Figure: Reactive Load Balance by
Adjusting Voltage Regulator
ELO 4.3
Operator Generic Fundamentals
109
Paralleling Generators
Reactive Load Sharing
• Generators typically use
automatic voltage regulators to
correct terminal voltage-reactive
power characteristic to a linear
droop
• Parallel generators operate at
same voltage
• Reactive load determined by
intersection of generator
characteristic curve for voltage
regulator setting and operating
voltage
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Figure: Reactive Load Balance by
Adjusting Voltage Regulator
ELO 4.3
Operator Generic Fundamentals
110
Paralleling Generators
Reactive Load Sharing
• For example, using figure, at an
operating voltage of 450KV, G1
initially carries 600 KVAR and
G2 carries 200 KVAR, shown by
dotted lines; to balance reactive
load while maintaining terminal
voltage at 450KV, decrease noload voltage regulator setting on
G1 to 460 KV while increasing
voltage regulator setting on G2
to 460 KV (desirable to
distribute load in proportion to
generator ratings)
© Copyright 2016
ELO 4.3
Figure: Reactive Load Balance by
Adjusting Voltage Regulator
Operator Generic Fundamentals
111
Paralleling Generators
Operation in Parallel with a Large
Grid
• When paralleling a generator with
a large power system (grid),
output of generator is minimal in
comparison
• House curve shows system as a
constant frequency, constant
voltage power source
• Power grid automatically supplies
remainder of required load; when
single generator supplies
negative reactive power –
decreases total reactive load,
improves power factor, and
requires less current for same
total real power
© Copyright 2016
ELO 4.3
Figure: Parallel Operation of a
Single Generator with the Grid
Operator Generic Fundamentals
112
Paralleling Generators Example
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.
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. Reverse power
B. Under-frequency
C. Under-voltage
D. Over-speed
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.
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ELO 4.3
Operator Generic Fundamentals
113
Paralleling Generators
Re-Energizing a Dead Bus
• Consider the process of re-energizing a dead bus with a generator
– Could occur when using plant emergency diesel generators
• Assuming none of the loads on the bus are running
• If you connect generator onto bus without first stripping loads, all of
the bus loads will start at once when generator energizes bus
• Starting current is several times higher than running current, if all
start at once, the bus will draw several times its normal full load
current
• Generator rating is sufficient to carry all bus loads
• Not sufficient to carry all starting currents simultaneously
• Starting all loads together will result in an over-current condition,
which may trip the generator
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ELO 4.3
Operator Generic Fundamentals
114
Paralleling Generators
Typical NRC Exam Question
Closing the output breaker of a three-phase generator onto a deenergized bus can result in...
A. An over-voltage condition on the bus
B. An over-current condition on the generator if the bus was not
first unloaded
C. A reverse power trip of the generator circuit breaker if generator
frequency is low
D. A large reactive current in the generator
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.
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ELO 4.3
Operator Generic Fundamentals
115
Paralleling Generators
Explanation for NRC Exam Question
Closing the output breaker of a three-phase generator onto a de-energized bus
can result in...
A. An over-voltage condition on the bus. Incorrect. If voltage was set
correctly prior to energizing dead bus, voltage would tend to drop slightly
when loaded. Voltage regulator would compensate and keep it within
the band, but it would certainly not result in voltage going too high.
B. An over-current condition on the generator if the bus was not first
unloaded. Correct. All 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 de-energize the bus again.
C. A reverse power trip of the generator circuit breaker if generator
frequency is low. Incorrect. Since the generator is 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. Incorrect. Reactive power
should not be higher than normal reactive load for running this
bus. Real load from starting currents will be the problem in this
circumstance.
© Copyright 2016
ELO 4.3
Operator Generic Fundamentals
116
Paralleling Generators Example
A main generator is about to be connected to an infinite power grid. Closing the
generator output breaker with generator voltage slightly lower than grid voltage
and generator frequency slightly higher than grid frequency will initially result
in… (assume no generator breaker protective trip occurs)
A.
The generator supplying reactive power to the grid
B.
The generator attaining a leading power factor
C. The generator acting as a real load to the grid
D. Motoring of the generator
Requires you to consider both real load and reactive load. Since generator
frequency is slightly higher than grid frequency (the synchroscope will be
rotating slowly clockwise as it should be with parallel generators), the generator
will pick up real load from 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 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.
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ELO 4.3
Operator Generic Fundamentals
117
Paralleling Generators
Knowledge Check – NRC Question
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-of-phase
voltages.
D. excessive arcing within the generator output breaker due to outof-phase voltages.
Correct answer is D.
© Copyright 2016
ELO 4.3
Operator Generic Fundamentals
118
Paralleling Generators
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...
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.
Correct answer is C.
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ELO 4.3
Operator Generic Fundamentals
119
Paralleling Generators
Knowledge Check – Explanation
• Generator Capability Curve only shown with “positive” MW loading
– But “negative” MW loading can be shown for training purposes
• Since generator frequency < grid frequency and generator voltage >
grid voltage
• Power triangle would actually look like this:
VARs (Supplying/Acquiring)
MW (Becoming)
(Reverse Power)
MW (Supplying/Acquiring)
Therefore, correct answer is:
C. become a real load to the grid,
but acquire reactive load.
VARs (Becoming, for lack of better word)
© Copyright 2016
ELO 4.3
Operator Generic Fundamentals
120
Paralleling Generators
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 over-speed conditions.
Correct answer is C.
© Copyright 2016
ELO 4.3
Operator Generic Fundamentals
121
Generator Excitation
ELO 4.4 – Describe the consequences of over-excitation and underexcitation when load sharing.
Generator Excitation Guidelines
• Terms commonly used to explain changes in generator field current:
– over-excitation
– under-excitation
© Copyright 2016
ELO 4.4
Operator Generic Fundamentals
122
Generator Excitation Guidelines
• If generator in under-excited condition, power factor angle is leading
– Machine is absorbing reactive power from system
• If generator in over-excited condition, power factor angle is lagging
– Generator is supplying reactive power to system
• Based on grid conditions and loads, power control/load dispatch may
request change in generator excitation to maintain grid voltage
stability
• Generator will react differently to governor speed and field changes
when operating un-paralleled, paralleled with a generator, or
supplying power grid
• During parallel operation, generator is a small power supply in
comparison to entire grid network, not able to change grid voltage or
frequency significantly
© Copyright 2016
ELO 4.4
Operator Generic Fundamentals
123
Generator Excitation Guideline Terms
• Unity power factor – (VARs at lowest point or zero) for every
operating condition, one value of field current causes generator to
deliver only real power
• Under-excitation – when current falls below value that yields
operation at unity power factor, the generator absorbs reactive power
(VARs increase)
– Generator acts as inductive coil
– Not desired mode of operation, since generator is a reactive load
• Over-excitation – increasing current causes generator to supply
increased reactive power
– Generator acts as capacitor (delivery of reactive power)
– Generators normally provide VARs together with watts and nearly
always operate in overexcited conditions
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ELO 4.4
Operator Generic Fundamentals
124
Generator Excitation
When generator is under-excited:
• Reducing generator excitation 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 underexcited
• It will draw less reactive load and move closer to unity power factor
• In under-excited condition, generator has a leading power factor
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ELO 4.4
Operator Generic Fundamentals
125
Generator Excitation
When generator is over-excited:
• Increasing generator voltage will make the generator more overexcited
– will supply more reactive load to the grid
• In over-excited condition, generator has a lagging power factor
• Power plants generally operate in 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).
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ELO 4.4
Operator Generic Fundamentals
126
Infinite Versus Isolated Bus
Infinite Bus
• Nuclear power plants have turbine generators that are tied to an
infinite bus
– Speed and voltage changes do not effect grid frequency or
voltage, just real and reactive loading
Isolated Bus
• Large factories might supply their own power
• This is considered an isolated or “finite” bus
• There is a finite amount of real and reactive load being supplied
• If two generators are operated in parallel
– changes made by one or the other will effect frequency or voltage
• Several bank questions in this chapter and 191008 test this concept
© Copyright 2016
Operator Generic Fundamentals
127
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 ______________.
In order to evaluate this item, we need to consider real load and
reactive load separately.
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ELO 4.4
Operator Generic Fundamentals
128
Generator Excitation Example Answer
• Main generator voltage excitation controls reactive load
• Main generator real load is controlled by changing speed control
– 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
• The generator begins at 0 MVARS or no reactive load
– An adjustment in either direction will increase current
• Since excitation is “reduced slightly”, generator amps will increase
• Therefore, in answer to the question, amps will increase and MW will
remain the same
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ELO 4.4
Operator Generic Fundamentals
129
Generator Excitation
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
22.5 kV
60.2 Hertz
750 MW
25 MVAR (out)
© Copyright 2016
Generator B
22.5 kV
60.2 Hertz
750 MW
50 MVAR (out)
ELO 4.3
Operator Generic Fundamentals
130
Generator Excitation
Knowledge Check (continued)
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?
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.
Correct answer is D.
© Copyright 2016
ELO 4.3
Operator Generic Fundamentals
131
Generator Excitation - Explanation
• Stem states it is an “isolated” bus (finite amount of load)
– Total of: 1500 MW and 75 MVAR (Out)
• Power triangles for both generators shown below (not to scale)
• B regulator starts to fail “high” (increases excitation)
– When “B” raises to 75 MVAR, “A” lowers to 0 MVAR
– “B” generator current increases, “A” generator current decreases
VARs (Supplying/Acquiring)
MW (Becoming)
(Reverse Power)
B
B
A
A
MW (Supplying/Acquiring)
VARs (Absorbing, for lack of better word)
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ELO 4.4
Operator Generic Fundamentals
132
Generator Excitation - Explanation
• As “B” continues to fail high (“B” to 100, “A” to -25)
– “B” current continues to increase, while “A” now increases
– “B” will reach an “upper” limit before “A” reaches a “lower” limit
– Therefore, since the question asks what happens to “A”:
o Current will initially decrease, then increase until the output
breaker for generator B trips on overcurrent
VARs (Supplying/Acquiring)
MW (Becoming)
(Reverse Power)
B
B
A
A
MW (Supplying/Acquiring)
VARs (Becoming, for lack of better word)
© Copyright 2016
ELO 4.4
Operator Generic Fundamentals
NRC KA to ELO Tie
KA #
KA Statement
RO SRO ELO
K1.01 Indication of a locked rotor
2.8
3.1
3.2
K1.02 Potential consequences of overheating insulation or bearings
Causes of excessive current in motors and generators, such as low voltage, overloading,
K1.03 and mechanical binding
Relationship between pump motor current (ammeter reading) and the following: pump fluid
K1.04 flow, head, speed, and stator temperature
2.8
2.9
3.3
2.7
2.8
3.4
2.7
2.8
3.5
K1.05 Explain the difference between starting current and operating (running) current in a motor
2.8
2.7
2.4
K1.06 Reason for limiting the number of motor starts in a given time period
3.0
3.1
2.4
K1.07 Electrical units: Volts, Amps, VARs, Watts, and Hertz
2.1
2.3
1.3
K1.08 Consequences of overexcited/under excited
2.1
2.3
K1.09 Interrelations of the following: VARs, Watts, Amps, Volts, Power factor
1.9
2.1
4.4
1.3,
4.1
K1.10 Load sharing with parallel generators
2.3
2.4
4.3
K1.11 Motor and generator protective devices
* Covered in greater detail in 191008 - Breakers, Relays, and Disconnects for KA's
K1.05 and K1.08
2.3
2.4
2.4*
© Copyright 2016
Operator Generic Fundamentals