AC Motor - Rev
AC Motor - Rev
AC Motor - Rev
Lesson 12a_et332b.pptx 1
Lesson 12a_et332b.pptx 2
Learning Objectives
After this presentation you will be able to:
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Lesson 12a_et332b.pptx 3
Lesson 12a_et332b.pptx 4
120 f
ns
P
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Lesson 12a_et332b.pptx 5
Synchronous Speed
Example 12a-1: Four pole motor operating on a 60 Hz system.
What is the speed at which the magnetic field rotates
120 f P 4 poles 120 60
ns ns 1800 rpm
P f 60 Hz 4
Lesson 12a_et332b.pptx 6
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Lesson 12a_et332b.pptx 7
n sl n s n r
Where nsl = slip speed
ns = synchronous speed
nr = rotor speed
Lesson 12a_et332b.pptx 8
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Simple Construction
No brushes or other high maintenance
parts
Disadvantages
Can not easily control speed
Lesson 12a_et332b.pptx 10
Operating
Tm ( n) 100
range
50
Starting torque is developed
when n=0 rpm. In this case
approximately 100 N-m
0
0 200 400 600 800 1000 1200 1400 1600 1800
n
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Lesson 12a_et332b.pptx 12
At startup stator voltage frequency and rotor voltage frequency are equal
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Rotor reactance in
x r 2 f r Lr x r 2 s f BR Lr x r s X BR
terms of blocked
rotor inductance
Rotor current
Rotor Impedance s E BR s E BR
z r R r j x r R r j s XBR
Ir
R r j x r R r j s X BR
Lesson 12a_et332b.pptx 14
Rotor current magnitude Rotor current phase angle Rotor power factor
E BR
Ir Fpr cos(r )
r tan 1 BR
2 X
Rr Rr
X BR
2
s
s
Where r = rotor current angle
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Lesson 12a_et332b.pptx 15
Rotor Phase
Shift
Rotor
Current
Operating
Operating
range
range
Lesson 12a_et332b.pptx 16
Where E BR E BR 0 I r I r - r
Q gap E BR I r sin(r )
Pgap = active power providing shaft power, friction, windage, and rotor
resistance losses.
Qgap = reactive power that oscillates across air gap
Rotor Fp and the magnitude of the Ir determine gap active power, Pgap
EBR is assumed to be constant because it is proportional to
the flux density which is assumed to be constant
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Lesson 12a_et332b.pptx 17
Prcl 3 I r R r
2
Total 3-phase rotor losses
Lesson 12a_et332b.pptx 18
3 Ir R r
2
Total gap power Pgap
s
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3 I r R r (1 s)
2
Pmech
s
Lesson 12a_et332b.pptx 20
R r R r (1 s)
Rr
s s
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Lesson 12a_et332b.pptx 22
ns n r nr nr
s s 1- so 1 - s
ns ns ns
Substitute into the previous equation for mechanical power
3 Ir R r n r
2
Pmech
s ns
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Lesson 12a_et332b.pptx 23
180 R r E BR
2
Td N-m
2 s n s
R
2
r X BR
2
s
Lesson 12a_et332b.pptx 24
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Lesson 12a_et332b.pptx 26
Prcl
Pgap Total active power across air gap
s Prcl = rotor conductor losses
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Example 12a-2
A 3-phase 60 Hz, 75 Hp, 4 pole motor operates at a rated
terminal voltage of 230 V Under rated conditions it draws a
line current of 186 A and has an efficiency of 90%. The
following losses are measured:
Core losses = 1273 W Stator conductor losses = 2102 W
Rotor conductor losses = 1162 W
Find: a) the input power
b) the total losses
c) the air gap power
d) the shaft speed
e) the motor power factor
f) combined mechanical losses
Lesson 12a_et332b.pptx 30
Ans
Losses are the difference between the input and output powers
Ans
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Ans
Lesson 12a_et332b.pptx 32
From above
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Ans
Ans
Lesson 12a_et332b.pptx 34
Ans
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Lesson 12a_et332b.pptx 35
Example 12a-3
A 3-phase 230V, 25 HP, 60Hz, 4 pole motor rotor absorbs
20,200 W when supplying an unknown shaft load. The rotor
copper losses are measured at 975 W when supplying this
load. The friction and windage losses are known to be 250
W. Determine
a) the shaft speed;
b) mechanical power developed;
c) torque developed in the rotor;
d) shaft torque;
e) percent of rated horsepower that the motor is delivering.
Lesson 12a_et332b.pptx 36
Ans
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Ans
Ans
Lesson 12a_et332b.pptx 38
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Lesson 12a_et332b.pptx 40
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R (1 s) 2
R2 Rr a2 a2 r a Rr
s
Lesson 12a_et332b.pptx 42
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V
I1 Stator current E 2 I1 ZP Induced rotor
Zin voltage referred to
stator
Lesson 12a_et332b.pptx 44
P rcl (1 s)
Pgap Pmech P rcl
s s
Note: all power equations are for total three-phase power
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Lesson 12a_et332b.pptx 45
7.04 Pmech
TD (lb - ft) Rotor developed torque.
nr Where nr = rotor speed
7.04 Pshaft
Tshaft (lb - ft) Shaft torque
nr
Finally
3 | E 2 |2 The stator core losses are
Pcore dependent on the voltage
R fe
Lesson 12a_et332b.pptx 46
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Lesson13_et332b.pptx
ET 332b
Ac Motors, Generators and Power Systems
Lesson13_et332b.pptx
LEARNING OBJECTIVES
After this presentation you will be able to:
1
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Lesson13_et332b.pptx
R1 = 0.18 R2 = 0.20
X1 = 1.15 X2 = 1.23
XM = 40 Rfe = 317
Total mechanical power losses (friction, windage and stray) are 170 W
Find: a.) the motor slip; b.) the motor line current; c.) the apparent power
the motor draws from the system; d.) active power drawn by the motor; e.)
motor power factor; f.) total electric power losses of motor; g.) shaft power
and torque; h.) efficiency
Use per phase circuit model and circuit analysis to find these quantities
Lesson13_et332b.pptx
X1=1.15 Ω X2=1.23 Ω
R2/s=0.2/s Ω
Rfe=317 Ω
xm=40 Ω
f = 60 Hz Number of poles : P=4 nr = 1778 RPM
Pfw+Pstray = 170 W
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Lesson13_et332b.pptx
Lesson13_et332b.pptx
Zp
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Lesson13_et332b.pptx
R1
Zp
265.60
I1
14.88528.57
7
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14
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ET 332b
Ac Motors, Generators and Power Systems
15
8
Lesson 14: NEMA Designs and
Induction Motor Nameplate Data
ET 332b
Ac Motors, Generators and Power Systems
lesson14_et332b.pptx 1
Learning Objectives
After this presentation you will be able to:
lesson14_et332b.pptx 3
NEMA Motor Designs
Different motor conductor designs given different rotor resistances,
which gives different motor characteristics
Design A - Starting torque
150% rated; breakdown
torque 275% Starting current
7-10 times rated current
lesson14_et332b.pptx 4
NEMA Motor Designs
Design C - High starting torque
motors; starting torque 240% to
275% of rated; starting I < 6.4
time I rated
lesson14_et332b.pptx 6
Motor Nameplate Data
Insulation Class - specifies maximum
allowable temperature rise for motor
windings
lesson14_et332b.pptx 7
Motor Nameplate Data
Code Letter kVA Table
Code kVA/hp Code kVA/hp
Letter Letter
A 0-3.15 K 8.0-9.0
B 3.15-3.55 L 9 .0- 10.0
C 3.55-4.0 M 10-11.2
D 4.0-4.5 N 11.2-12.5
E 4.5-5.0 P 12.5-14.0
F 5.0-5.6 R 14.0-16.0
G 5.6-6.3 S 16.0 -18.0
H 6.3-7.1 T 18.0 -20.0
J 7.1-8.0 U 20-22.4
Above can be used to compute the
V >22.4
range of starting currents
lesson14_et332b.pptx 8
lesson14_et332b.pptx 9
Starting Current kVA Codes
Use kVA codes to find the range of motor starting currents
lesson14_et332b.pptx 10
Starting Current kVA Codes
Example 14-1: A NEMA design motor is rated at 150 hp at 460, 60 Hz.
It has a rated current of 163 A and a nominal efficiency of 96.2%. The
locked rotor code is G. Find the range of starting current that can be
expected from this machine.
Vsup ply
% V 100%
Vrated
480 V
% V 100%
460 V
% V 104%
lesson14_et332b.pptx 12
Effects of Changing Voltage and Frequency on
Torque
Use the following empirical formula
V 2 s
TD k s 0.03
f
Where: TD = motor developed torque
V = motor terminal voltage
f = motor operating f
s = per unit slip
k = proportionality constant
lesson14_et332b.pptx 13
Example 14-2: A 3-phase 460 V, 20 HP, 60 Hz, 4
pole motor drives a constant torque load at rated
shaft power at rated voltage, and frequency. The
motor speed under these conditions is 1762 RPM.
A system disturbance lowers the motor voltage by
10% and the system frequency by 6%. Find: a.)
the new motor speed; b.) the new shaft power.
Assume that the mechanical losses (Pfw and Pstray)
are constant.
lesson14_et332b.pptx 14
Example 14-2 Solution (1)
Define equations
Constant
torque load so
equate torques
Solve for s2
lesson14_et332b.pptx 16
Example 14-2 Solution (3)
Compute the
value of s2
Per Unit
produces the same
result
lesson14_et332b.pptx 17
Example 14-2 Solution (4)
Need synchronous
speed for second case
Now compute
the new motor
speed
lesson14_et332b.pptx 18
Example 14-2 Solution (5)
Solve part b. P1=rated horsepower=20 hpTs1 = shaft torque state 1
Ts2= shaft torque state 2
For constant torque load Ts1=Ts2
lesson14_et332b.pptx 19
End Lesson 14: NEMA Designs and
Induction Motor Nameplate Data
ET 332b
Ac Motors, Generators and Power Systems
lesson14_et332b.pptx 20
3/24/2021
Lesson 15_et332b.pptx 1
Learning Objectives
After this presentation you will be able to:
Lesson 15_et332b.pptx 2
1
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Formulas R dc
For wye connected stator R1
V 2
R dc dc
I dc For delta connected stator R 1 1.5 R dc
Lesson 15_et332b.pptx 3
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60 Hz
Change to 60 Hz X BR 60 X BR15
15 Hz
X BR 60 x1 x 2
Divide the leakage reactances based on the NEMA design types. Use the following table.
Design A, D B C Wound
Type Rotor
x1 0.5∙XBR60 0.4∙XBR60 0.3∙XBR60 0.5∙XBR60
x2 0.5∙XBR60 0.6∙XBR60 0.7∙XBR60 0.5∙XBR60
Lesson 15_et332b.pptx 6
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Measure
PNL = No-load power losses
INL = No-load current
VNL = No-load voltage
Lesson 15_et332b.pptx 7
Q NL SNL PNL
2 2
No-load reactance is the sum of the magnetizing reactance and stator leakage
X NL x1 x M
x M X NL x1
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Lesson 15_et332b.pptx 10
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Lesson 15_et332b.pptx 12
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Lesson 15_et332b.pptx 14
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Lesson 15_et332b.pptx 15
0.1530/s Ω
Omit
Rfe
j7.58 Ω
Lesson 15_et332b.pptx 16
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ET 332b
Ac Motors, Generators and Power Systems
Lesson 15_et332b.pptx 17
9
Lesson 16: Asynchronous
Generators/Induction Generators
ET 332b
Ac Motors, Generators and Power
Systems
et332bInd.ppt 1
Learning Objectives
et332bInd.ppt 2
Induction Generators
Driving an induction motor faster than synchronous speed
when connected to the grid results in active power
generation
cogeneration
et332bInd.ppt 3
Induction Generator Starting Sequence
Existing Three Phase System
1.) Breaker open
2.) Increase prime mover
mechanical power input
until nr >ns.
Electric Breaker
Power
3.) Close Breaker
Out 4.) Adjust mechanical power
input to match electric
Induction Prime load.
Generator Mover
Pmech=Pe+Ploss
nr >ns
Generator
Losses Induction generator can
Mechanical not vary terminal voltage
Power In
or frequency. Set by
system.
et332bInd.ppt 4
Induction Generator Speed Power Curves
Induction Machine Speed-Power Curve
100
ns
1.1ns
Generator
Air Gap Power (kW)
Operation
50
-50
Pushover
Power
-100
0 500 1000 1500 2000 2500 3000 3500 4000
Rotor Speed (rpm)
et332bInd.ppt 5
Limitations of Induction Generations
• Require existing power grid for synchronous
operation.
– Can not control frequency or voltage independently
• Can not operate above pushover speed
• Require a source of reactive power to operate
– When connect to grid, system supplies reactive power
to operate generator
• When operating without grid connection
frequency varies with power output.
– Parallel capacitors supply reactive power
et332bInd.ppt 6
Induction Generator Example
Find the active power delivered by the generator and the reactive
power it requires from the system to operate.
et332bInd.ppt 7
Example Solution
et332bInd.ppt 8
Example Solution
et332bInd.ppt 9
Example Solution
et332bInd.ppt 10
Example Solution
et332bInd.ppt 11
Isolated Operation of Induction Generators
Isolated Three Phase System
Mechanical
Power In
et332bInd.ppt 12
Voltage Build-up in Isolated Induction Generators
jX1
R1
R2/s
External Capacitor provides
-jXc jXm
Vin Rfe
jX2
Reactive power for operation
Xc0 Vop
Xc
Voltage
Xc1
Iop
Iop 1
Current
Iop C
Vop 2f
et332bInd.ppt 13
Voltage Build-up in Isolated Induction Generators
Lab measurements determine the magnetization curve
Three Phase Induction Motor Magnetization Curve
140
120
Stator Voltage (V)
100
80
60
Iop 1 1.18 1
40 C
Vop 2f 110 260
Inductance
change
20
C 2.845 10 5 F or 28.45 F
due to rotor 0
motion 0.00 0.25 0.50 0.75 1.00 1.25 1.50
Stator Current (A)
et332bInd.ppt 14
Voltage Build-up in Isolated Induction Generators
Single phase motor magnetization curve
140.00
120.00
100.00
80.00
Voltage (V)
60.00
I op 1 5.74 1
40.00
C
Vop 2f 120 260
20.00 C 1.27 10 4 F or 127 F
0.00
0.00 1.00 2.00 3.00 4.00 5.00 6.00 7.00
Current (A)
et332bInd.ppt 15
ET 332b
Ac Motors, Generators and Power Systems
Lesson 17_et332b.pptx 1
Learning Objectives
After this presentation you will be able to:
Lesson 17_et332b.pptx 2
Synchronous Machines
Synchronous Motors – Convert electrical power into
mechanical power. They operate at a constant speed
Lesson 17_et332b.pptx 3
Basic Construction of Synchronous
Machines
Stationary part of machine
called the armature. Similar
to induction motor stator
Lesson 17_et332b.pptx 4
Operational Theory for Synchronous
Machines
Motor Operation
1.) Armature is connected to 3-phase source - rotating magnetic
field formed in armature windings
2.) Rotor (field) is energized by dc source. Creates a magnetic
dipole
3.) Motor has no starting torque, must be started as induction
motor. Damper windings engaged (induction motor action).
Motor accelerates to almost synchronous speed.
4.) Damper winding disengaged, motor stays locked to rotating
magnetic field and produces torque
Lesson 17_et332b.pptx 5
Operational Theory for Synchronous
Machines
Alternator Operation
Lesson 17_et332b.pptx 6
Synchronous Motor Operation
Motor speed constant at synchronous speed, ns
120 f
ns =
P
Where f = power system frequency (Hz)
P = number of motor poles
Speed can only be changed by changing system f or number of poles in
machine
Lesson 17_et332b.pptx 7
Motor Reaction to Application Load
Application
1.) Motor rotor is in phase with rotating armature (stator) field
Lesson 17_et332b.pptx 8
Synchronous Motor Operation
Torque Angle
Lesson 17_et332b.pptx 9
Counter-EMF Components in
Synchronous Machines
Field flux induces voltage into armature (stator) due to rotation (Ef).
Ef depends on If (dc field current) and ns The parameter Ef , called
excitation voltage.
Lesson 17_et332b.pptx 10
Per Phase Circuit Model
Lesson 17_et332b.pptx 11
Simplified Synchronous Machine
Model
Let Xs = Xl + Xar = synchronous reactance
+ Ra << Xs so it is usually ignored
Ia +
Ef also called voltage behind
synchronous reactance.
Lesson 17_et332b.pptx 12
Power Transfer in Synchronous
Machines
d Ia X
E f sin(d)
Ef qi
I a X s cos(qi )
From phasor diagram above E sin(d) = I X cos(q )
f a s i
Lesson 17_et332b.pptx 13
Power Transfer In Synchronous
Machines
Multiplying the equation from the phasor diagram by VT and dividing
both sides by Xs gives power/phase.
VT Ef
sin(d) = VT I a cos(qi )
Xs
Lesson 17_et332b.pptx 14
Maximum Power Transfer in
Synchronous Machines
There is a maximum power that a synchronous machine can develop
before it loses synchronism
Pmax Generator action
(+90 degrees)
12 10
− 90 90
Lagging angle-motor action
P(d ) 0
Leading angle-generator
action
10
− 12
200 150 100 50 0 50 100 150 200
− 180 d 180
deg
Pmax Motor action
(-90 degrees)
Lesson 17_et332b.pptx 15
Example 17-1: A 100 hp 460 volt 60 Hz 4-pole synchronous
motor is operating at rated conditions and a power factor
of 80% leading. The motor efficiency is 96% and the
synchronous reactance is 2.72 ohms/phase. Find:
a.) developed torque;
b.) armature current;
c.) excitation voltage (Ef);
d.) power angle;
e.) the maximum torque the motor can develop without
loss of synchronization. (pull-out torque).
Lesson 17_et332b.pptx 16
Example 17-1 Solution (1)
Find the input power
No losses in
Armature so
Motor operates
at synchronous
speed
Lesson 17_et332b.pptx 17
Example 17-1 Solution (2)
Lesson 17_et332b.pptx 18
Example 17-1 Solution (3)
Find voltage behind synchronous reactance
Lesson 17_et332b.pptx 19
Example 17-1 Solution (4)
Now find the pullout torque
Lesson 17_et332b.pptx 20
ET 332b
Ac Motors, Generators and Power Systems
END LESSON 17
Lesson 17_et332b.pptx 21
Lesson 18_et332b.pptx
LESSON 18 SYNCHRONOUS MOTOR
OPERATION AND APPLICATIONS
1 ET 332b
Ac Motors, Generators and Power Systems
LEARNING OBJECTIVES
Lesson 18_et332b.pptx
➢ Interpret a synchronous motor phasor diagram
➢ Explain how increasing mechanical load affects
synchronous motor electrical characteristics
➢ Explain the difference between electrical and
mechanical degrees in synchronous motor operation
➢ Compute the motor developed power and torque
2
LOAD CHANGE ON SYNCHRONOUS Mcurrent
Armature OTORS increases
and power factor angle
decreases.
Lesson 18_et332b.pptx
Increasing torque
increases d
More active power enters the motor to maintain speed and develop the new 3
value of torque
ELECTRICAL DEGREES VS MECHANICAL DEGREES IN
SYNCHRONOUS MACHINES
Electrical power/torque angle measurement is related to the rotor phase
shift by the following formula
Lesson 18_et332b.pptx
P
de = d m
2
Where: de = phase shift in electrical degrees
dm = mechanical phase shift in the rotating magnetic
field of the rotor
4
ELECTRICAL DEGREES VS MECHANICAL DEGREES IN
SYNCHRONOUS MACHINES
Example 18-1: A 100 hp, 460 V, 4-pole, wye-connected,
cylindrical rotor synchronous motor is operating with a
mechanical power angle of 5.5 degrees. Determine the
Lesson 18_et332b.pptx
electrical torque angle
5.5
5
Example 18-2: A 2-pole, 1000 hp, 6000 V, 3-phase wye-
connected synchronous motor is operating at rated load and
0.8 leading power factor. At this operating point, the machine
efficiency is 92%. The motor synchronous reactance is 4.2
ohms/phase Find:
Lesson 18_et332b.pptx
a.) the motor power angle
b.) the power developed in the rotor
c.) the torque developed in the rotor (lb-ft)
6
EXAMPLE 18-2 SOLUTION (1)
Need to find angle d. Start by finding the armature current
Lesson 18_et332b.pptx
Use power factor to find current phase angle
7
EXAMPLE 18-2 SOLUTION (2)
Find Eaf Wye connect motor. Need phase voltage
Lesson 18_et332b.pptx
8
EXAMPLE 18-2 SOLUTION (3)
b.) Find the developed power
Lesson 18_et332b.pptx
Compute
value
9
EXAMPLE 18-2 SOLUTION (4)
c.) Use the developed power and motor speed to find torque
Lesson 18_et332b.pptx
10
SYNCHRONOUS MACHINE OPERATION AND EXCITATION
Motor Operation: Increasing excitation increases magnetic field
strength. Decreases power angle for a fixed load. Increases Max
power.
Lesson 18_et332b.pptx
− 3 VT Ef
Pin = sin(d)
Xs
11
MACHINE EXCITATION AND REACTIVE POWER TRANSFER
Increases in excitation also control reactive power transfer.
3 V T E f cos(d) − V T
2
Lesson 18_et332b.pptx
Q=
Xs
12
TYPES OF MACHINE EXCITATION
Ef = VT Called normal excitation. Synchronous motor
supplies all magnetizing current to transfer power
across air gap.
Lesson 18_et332b.pptx
Ef > VT Called overexcited. Synchronous motor has
surplus of reactive power. Supplies vars to power
system. (leading Fp)
Lesson 19_et332b.pptx
Explain how a synchronous motor can provide
reactive power to other electric loads
Use circuit model and power formulas to find the
amount of reactive power a synchronous motor
must deliver to produce a given power factor
Find the motor excitation voltage required to give a
desired reactive power
2
POWER FACTOR CORRECTION USING
SYNCHRONOUS MOTORS
Overexcited motor supplied mechanical power to load and
reactive power to the inductive loads of system
Lesson 19_et332b.pptx
Synchronous Condenser
Synchronous motor
designed for power factor
correction. No shaft.
3
SYNCHRONOUS CONDENSER EXAMPLE
Example 19-1: 3-phase, 460 V 60 Hz system
Lesson 19_et332b.pptx
Load 2: 75 kW delta connected resistance heater
Load 3: 300 hp 60 Hz 4-pole Y-connected cylindrical rotor
synchronous motor operating at 50% rated torque angle of
-16.4 degrees Efficiency = 95%Xs = 0.667 ohms/phase
Lesson 19_et332b.pptx
5
EXAMPLE SOLUTION CONTINUED
Find the total input power
Lesson 19_et332b.pptx
Answer Part a
6
EXAMPLE SOLUTION CONTINUED
Lesson 19_et332b.pptx
7
EXAMPLE SOLUTION CONTINUED
Compute the value of armature current
Lesson 19_et332b.pptx
8
EXAMPLE SOLUTION CONTINUED
Part c. Find system power factor. Construct power triangles for all
loads
Lesson 19_et332b.pptx
9
EXAMPLE SOLUTION CONTINUED
Find the reactive power produced by the synchronous motor
Lesson 19_et332b.pptx
10
EXAMPLE SOLUTION CONTINUED
Construct power triangles
Load 1
173,571 VAR
Load 3
Lesson 19_et332b.pptx
Load 2
117,790 W
248,667 W 75,000 W
-82,830 VAR
11
EXAMPLE SOLUTION CONTINUED
Find the apparent power of the total system
Lesson 19_et332b.pptx
Part d. Find excitation of synchronous motor for system unity Fp
12
EXAMPLE SOLUTION CONTINUED
Determine the required motor reactive power
Lesson 19_et332b.pptx
Find power/phase
13
EXAMPLE SOLUTION CONTINUED
Compute the armature current
Lesson 19_et332b.pptx
Ef1 = voltage at the original operating point and Fp
14
Lesson 19_et332b.pptx
END LESSON 19
15 ET 332b
Ac Motors, Generators and Power Systems
Lesson 20_et332b.pptx
ET 332b
Ac Motors, Generators and Power Systems
LESSON 20 ALTERNATOR
OPERATION OF SYNCHRONOUS
MACHINES 1
Lesson 20_et332b.pptx
LEARNING OBJECTIVES
2
Lesson 20_et332b.pptx
ALTERNATOR OPERATION
Synchronous machines can convert from motor to generator
operation by having the shaft driven by a source of mechanical
power
Lagging Ia
4
Lesson 20_et332b.pptx
Ef1
d=0
5
Lesson 20_et332b.pptx
6
Lesson 20_et332b.pptx
GENERATOR LOADING
Excess field
Mechanical Exciter flux converted
power to reactive power
converted to
active power Idc
Pmech
Pe
ELECTR
Prime
Generator IC
Mover
LOADS
Governo Q
Prime r Electrical load
Mover produces counter-
Torque torque
peed Governor – device that regulates
Increased electrical load produces counter
eed to match electric load torque that prime mover must overcome or
prime mover slows down 8
Lesson 20_et332b.pptx
11
Lesson 20_et332b.pptx
Solve for Ef
12
Lesson 20_et332b.pptx
13
Lesson 20_et332b.pptx
14
Lesson 20_et332b.pptx
15
Lesson 20_et332b.pptx
16
Lesson 20_et332b.pptx
Mathematically,
phase shift is Where: f = phase shift
integral of fs = system frequency
frequency change fg(t) = generator frequency
17
Lesson 20_et332b.pptx
SYNCH EXAMPLE
Generator voltage lagging system voltage - crosses zero later
in time
Also cause
rotor
oscillations
which cause
the
generator
to be
unstable Df
SYNCHRONIZING GENERATOR TO
SYSTEM
Increasing speed of prime mover increases frequency which
reduces phase difference
19
Lesson 20_et332b.pptx
SYNCHRONIZING GENERATOR TO
SYSTEM
Phase shift near zero at ts. Prime
mover is accelerating the rotor ahead
of the system voltage after this point.
(Vg leads Vs)
Close generator
breaker when phase
is slightly leading
system
STIFFNESS OF SYNCHRONOUS
MACHINES
Stiffness -Ability of a synchronous machine to resist forces that
pull it out of synchronism. Slope of the power-angle curve around
a given operating point
STIFFNESS OF SYNCHRONOUS
MACHINES
Remember power equation
22
Lesson 20_et332b.pptx
STIFFNESS OF SYNCHRONOUS
MACHINES
Example 20-2: A 3-phase 13.2 kV 60 Hz, 50 MVA
wye-connected cylindrical rotor synchronous
generator has Xs = 2.49 ohms/phase and an
internally generated voltage at the operating point of
15,767 VLL with a power angle of 11.1 degrees. The
machine has 8 poles. Determine:
23
Lesson 20_et332b.pptx
24
Lesson 20_et332b.pptx
25
Lesson 20_et332b.pptx
ET 332b
Ac Motors, Generators and Power Systems
END LESSON 20
ALTERNATOR OPERATION OF
SYNCHRONOUS MACHINES
26
Lesson 21: Alternator
Capabilities and
Mechanical Power Control
ET 332b
Ac Motors, Generators and Power Systems
Lesson 21_et332b.pptx 1
Learning Objectives
After this presentation you will be able to:
Lesson 21_et332b.pptx 2
Power Balance in Alternators
Alternator Ratings
Alternators rated in kVA
or MVA at a specified
power factor
Generally
Pout Fp Srated
Lesson 21_et332b.pptx 4
Alternator Capability Curve
Alternator Capability Curve
150
Operating
Reactive power output
100 Range limited by machine
excitation system
50 In this diagram
Reactive Power (kVAR)
50
100
80% lead
150
0 20 40 60 80 100 120
Active Power Into System (kW)
kVAR out 80% lag
kVAR in
0.8 Leading
0.8 Lagging
Lesson 21_et332b.pptx 5
Mechanical Power Input
Control
Governor - electromechanical speed control used to
maintain constant speed as machine power load
changes
Power transfer between parallel alternators is controlled
by change in prime-mover power input and speed. If
speed remains constant, then torque increases as
developed power increases
Mechanical Power In
n= constant
Speed
Increasing Torque
Lesson 21_et332b.pptx 6
Governor Characteristic and Speed Regulation
Lesson 21_et332b.pptx 7
Governor Characteristic Curve
fnl and nnl
61.2 Hz
3672 rpm
rated f and n
60 Hz GD
3600
rpm
f (Hz)
n (rpm) rated P out
0
0 500 kW
P (kW)
Lesson 21_et332b.pptx 8
Governor Characteristic Curve Example
Example 21-1: Find the speed
61.2 Hz regulation and the governor droop
3672 rpm for the machine that has the curve
GD
shown
60 Hz
3600
rpm
f (Hz)
n (rpm)
0
0 500 kW
P (kW)
f 61.2 60 Hz
GD 0.0024 Hz/kW
P 500 kW
Lesson 21_et332b.pptx 9
End Lesson 21: Alternator Capabilities
and Mechanical Power Control
ET 332b
Ac Motors, Generators and Power Systems
Lesson 21_et332b.pptx 10
Lesson 22_et332b.pptx
LESSON 22: ACTIVE POWER
DIVISION BETWEEN ALTERNATORS
1 ET 332b
Ac Motors, Generators and Power Systems
LEARNING OBJECTIVES
Lesson 22_et332b.pptx
Explain how active power load divides between
parallel alternators with equal governor droop
Explain how active power load divides between
parallel alternators with unequal governor droop
Define isochronous governor operation
Compute the load division and system frequency
of a load increase on parallel generators.
2
ACTIVE POWER DIVISION BETWEEN
ALTERNATORS
Governor droop determines the active power division between
parallel alternators
Lesson 22_et332b.pptx
Load shifting requires changes in the To grid
mechanical power of each machine’s
prime-mover
Pin Pin
ACTIVE POWER DIVISION BETWEEN
ALTERNATORS
Power transfer between alternators accomplished by adjusting their
governor’s no-load speed settings
Lesson 22_et332b.pptx
Example 22-1: Two alternators serve a 150 MW load in an isolated power
system and have identical governor characteristics
A1 B1
A B 60 Hz
B
PA =0 MW PB =150 MW
A
To load 0 150 MW
4
Frequency is constant in power system. Change
Initial conditions: load by shifting governor characteristics.
ACTIVE POWER DIVISION BETWEEN
ALTERNATORS
To shift load: raise the no-load speed of Gen A raising the whole characteristic.
Lesson 22_et332b.pptx
If 50 MW change
required, pick up 25
MW on Gen A
A2 B2
f new
60 Hz
B1 B
A1
A
0 150 MW
25 MW 125 MW
Lesson 22_et332b.pptx
Gen B lowers characteristic to
A2 B2 reduce load by 25 MW
f new
60 Hz A
A3 B3 B
0 50 MW
100 MW
Lesson 22_et332b.pptx
B system with similar
governor droops
A
60 Hz
DPL
59.5 Hz
DGA DGB DGA = DGB
0
P (MW)
Automatic control of governor no-load speed corrects for change in load
by raising or lowering the characteristics of the machines. This causes
the development of more or less mechanical power.
Steam Turbines – open turbine control valve more 7
Diesel Engine – open throttle more
Hydropower – open water control valves
ISOCHRONOUS GOVERNORS AND FREQUENCY
CONTROL
Isochronous governors - This type of governor can maintain
constant speed for any level of output power.
Lesson 22_et332b.pptx
Df
For isochronous governors 0
DP
Gen B is isochronous
Lesson 22_et332b.pptx
From Similar Triangles
f nl f nl f rated
GD
frequency
Dfrated Prated
Df So
f rated f new
f nl f rated Df
DPrated DP Prated DP
0 Prated
Power kW
GSR f rated Df
Rearranging the speed regulation equation gives: GD
Prated DP
Where: GD = governor droop
GSR = governor speed regulation 9
Df = change in f due to change in load
DP = change in load
Example 22-1: Parallel generators A and B share a total load of 300
kW at 60 Hz.
Machine Voltage Power GSR Freq Load
A 460 V 500 kW 2.0% 60 Hz 100 kW
B 460 V 500 kW 2.0% 60 Hz 200 kW
Lesson 22_et332b.pptx
If generator A trips off line determine a.) frequency of
generator A b.) frequency of generator B c.) frequency of the system
10
EXAMPLE 22-1 SOLUTION (1)
Show system operation using power/frequency plots
f nl f rated GSR f rated
fnl=61.2 Hz
f nl 60 Hz 0.02 60 Hz
f nl 61.2 Hz
Lesson 22_et332b.pptx
frated=60.0 Hz
0 kW 500 kW
f nl
Frequency (Hz)
f rated
Dfb
f new
A B
DPa DPb
Power kW
11
After A trips off-line PA =0 and PB=300 kW
DPA=DPB=100 kW
EXAMPLE 22-1 SOLUTION (2)
a.) For machine A
GSR f rated Df a
Prated DPa
Lesson 22_et332b.pptx
DPa = 100 kW A tripped off. Now delivers 0 kW
0.02 60 Hz Df a
500 kW 100
Df a 0.24 Hz
f a 60 0.24 60.24 Hz Answer
f b 60 0.24 59 .76 Hz
Lesson 22_et332b.pptx
Answer
fs = 59.76 Hz Answer
13
Example 22-2: 500 kW 60 Hz 2300 V alternator is
paralleled with a 60 Hz 300 kW machine. Both have
governor speed regulation values of 2.43%. Each
machine carries 200 kW at a frequency of 60.5 Hz. If total
load increases to 500 kW determine:
Lesson 22_et332b.pptx
a.) system frequency
b.) load carries by each machine
14
EXAMPLE 22-2 SOLUTION (1)
Machines have unequal governor droops.
61.458 61.458
A B
Lesson 22_et332b.pptx
B 60.0 60.0
500 kW 300 kW
A
60.5 Hz Define similar triangles
Df
f sys
DPb A
DPa B
0 PA = 200 kW PAnew= DPA+PA
PB = 200 kW PBnew=DPB+PB
Power kW
Lesson 22_et332b.pptx
16
EXAMPLE 22-2 SOLUTION (3)
Lesson 22_et332b.pptx
17
EXAMPLE 22-2 SOLUTION (4)
Lesson 22_et332b.pptx
18
EXAMPLE 22-2 SOLUTION (5)
Lesson 22_et332b.pptx
19
Lesson 22_et332b.pptx
END LESSON 22: ACTIVE POWER
DIVISION BETWEEN ALTERNATORS
20 ET 332b
Ac Motors, Generators and Power Systems