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Transcript of Chapter 6 DC and AC Machines. Introduction An electrical machine is link between an electrical...
![Page 1: Chapter 6 DC and AC Machines. Introduction An electrical machine is link between an electrical system and a mechanical system. Conversion from mechanical.](https://reader036.fdocuments.net/reader036/viewer/2022062407/56649cbe5503460f94984026/html5/thumbnails/1.jpg)
Chapter 6DC and AC Machines
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Introduction
• An electrical machine is link between an electrical system and a mechanical system.
• Conversion from mechanical to electrical: generator
• Conversion from electrical to mechanical: motor
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Introduction
Machines are called
• AC machines (generators or motors) if the electrical system is AC.
• DC machines (generators or motors) if the electrical system is DC.
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DC machines can be divide by:
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DC Machines Construction
cutaway view of a dc machine
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DC Machines Construction
cutaway view of a DC machine
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DC Machines Construction
Rotor of a DC machine
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DC Machines Construction
Stator of a dc machine
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DC Machines Fundamentals• Stator: is the stationary part of the machine. The
stator carries a field winding that is used to produce the required magnetic field by DC excitation.
• Rotor (Armature): is the rotating part of the machine. The rotor carries a distributed winding, and is the winding where the e.m.f. is induced.
• Field winding: Is wound on the stator poles to produce magnetic field (flux) in the air gap.
• Armature winding: Is composed of coils placed in the armature slots.
• Commutator: Is composed of copper bars, insulated from each other. The armature winding is connected to the commutator.
• Brush: Is placed against the commutator surface. Brush is used to connect the armature winding to external circuit through commutator
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DC Machines Fundamentals
In DC machines, conversion of energy from electrical to mechanical form or vice versa results from the following two electromagnetic phenomenaGenerator action:An e.m.f. (voltage) is induced in a conductor if it moves through a magnetic field.Motor action: A force is induced in a conductor that has a current going through it and placed in a magnetic field•Any DC machine can act either as a generator or as a motor.
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DC Machines Equivalent Circuit
The equivalent/modelling circuit of DC machine has two components:
Armature circuit: • It can be represented by a voltage source and a
resistance connected in series (the armature resistance). The armature winding has a resistance, RA.
The field circuit: • It is represented by a winding that generates the
magnetic field and a resistance connected in series. The field winding has resistance RF.
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DC Motor
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Basic Operation of DC Motor
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Classification of DC Motor1. Separately Excited DC Motor • Field and armature windings are either connected
separate.
2. Shunt DC Motor• Field and armature windings are either connected in
parallel.
3. Series DC Motor• Field and armature windings are connected in series.
4. Compound DC Motor• Has both shunt and series field so it combines features
of series and shunt motors.
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Equivalent Circuit of a DC Motor
Armature circuit - voltage source, EA and a resistor, RA.
The field coils, which produce the magnetic flux are represented by inductor, LF and resistor, RF.
The separate resistor, Radj represents an external variable resistor used to control the amount of current in the field circuit. Basically it lumped together with Rf and called Rf
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Equivalent Circuit of DC Motor
F
TF R
VI
AAAT RIEV
F
FF R
VI
AAAT RIEV
AL II
1. Separately Excited DC Motor
2. Shunt DC Motor
FAL III
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)( SAAAT RRIEV
LSA III
3. Series DC Motor
)( SAAAT RRIEV
F
TF R
VI FLA III
4. Compound DC Motor
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Important terms in DC motor equivalent circuit
• VT – supply voltage• EA – internal generated voltage/back e.m.f.• RA – armature resistance• RF – field/shunt resistance• RS – series resistance• IL – load current• IF – field current• IA – armature current• IL – load current•
n – speed
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Speed of a DC Motor
• For shunt motor
• For series motor
1
2
1
2
12
2
1
1
2
1
2
,
A
A
A
A
E
E
n
n
thenIf
E
E
n
n
2
1
1
2
1
2
2
1
1
2
1
2
A
A
A
A
A
A
I
I
E
E
n
n
E
E
n
n
If Constant field excitation, means; if1 = if2 or constant flux; 1 = 2
Flux, ϕ produce proportional to the current produce
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Example 1
A 250 V, DC shunt motor takes a line current of 20 A. Resistance of shunt field winding is 200 Ω and resistance of the armature is 0.3 Ω. Find the armature current, IA and the back e.m.f., EA.
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SolutionGiven parameters:• Terminal voltage, VT = 250 V• Field resistance, RF = 200 Ω• Armature resistance, RA = 0.3 Ω• Line current, IL = 20 A
Figure 1
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Solution (cont..)
the field current,
the armature current,
VT = EA + IARA
the back e.m.f.,
EA = VT – IARA = 250 V – (18.75)(0.3) = 244.375 V
A25.1200
V250
F
TF
FAL
R
VI
III
18.75A
A25.1A20
FLA III
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Example 2
A 50hp, 250 V, 1200 rpm DC shunt motor with compensating windings has an armature resistance (including the brushes, compensating windings, and interpoles) of 0.06 Ω. Its field circuit has a total resistance Radj + RF of 50 Ω, which produces a no-load speed of 1200 rpm. There are 1200 turns per pole on the shunt field winding.
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Example 2 (cont..)
a) Find the speed of this motor when its input current is 100 A.
b) Find the speed of this motor when its input current is 200 A.
c) Find the speed of this motor when its input current is 300 A.
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SolutionGiven quantities:• Terminal voltage, VT = 250 V• Field resistance, RF = 50 Ω• Armature resistance, RA = 0.06 Ω• Initial speed, n1 = 1200 r/min
Figure 2
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Solution (cont..) (a) When the input current is 100A, the armature
current in the motor is
Therefore, EA at the load will be A95A5A100
50
V250A100
F
TLFLA R
VIIII
V3.244
V7.5V250
)06.0)(A95(V250
AATA RIVE
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Solution (cont..)
• The resulting speed of this motor is
min/r1173
min/r1200250
3.244
11
22
1
2
1
2
V
V
nE
En
E
E
n
n
A
A
A
A
![Page 28: Chapter 6 DC and AC Machines. Introduction An electrical machine is link between an electrical system and a mechanical system. Conversion from mechanical.](https://reader036.fdocuments.net/reader036/viewer/2022062407/56649cbe5503460f94984026/html5/thumbnails/28.jpg)
Solution (cont..)(b) When the input current is 200A, the armature
current in the motor is
Therefore, EA at the load will be A195A5A200
50
V250A200
F
TLFLA R
VIIII
V3.238
V7.11V250
)06.0)(195(V250
A
RIVE AATA
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Solution (cont..)
• The resulting speed of this motor is
min/r1144
min/r1200250
3.238
11
22
1
2
1
2
V
V
nE
En
E
E
n
n
A
A
A
A
![Page 30: Chapter 6 DC and AC Machines. Introduction An electrical machine is link between an electrical system and a mechanical system. Conversion from mechanical.](https://reader036.fdocuments.net/reader036/viewer/2022062407/56649cbe5503460f94984026/html5/thumbnails/30.jpg)
Solution (cont..)(c) When the input current is 300A, the armature
current in the motor is
Therefore, EA at the load will be A295A5A300
50
V250A300
F
TLFLA R
VIIII
V3.232
V7.17V250
)06.0)(295(V250
A
RIVE AATA
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Solution (cont..)
• The resulting speed of this motor is
min/r1115
min/r1200V250
V3.232
11
22
1
2
1
2
nE
En
E
E
n
n
A
A
A
A
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Example 3
The motor in Example 2 is now connected in separately excited circuit as shown in Figure 3. The motor is initially running at speed, n = 1103 r/min with VA = 250 V and IA = 120 A, while supplying a constant-torque load. If VA is reduced to 200 V, determine
i). the internal generated voltage, EA
ii). the final speed of this motor, n2
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Example 3 (cont..)
Figure 3
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SolutionGiven quantities
• Initial line current, IL = IA = 120 A
• Initial armature voltage, VA = 250 V
• Armature resistance, RA = 0.06 Ω
• Initial speed, n1 = 1103 r/min
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Solution (cont..)
i) The internal generated voltage
EA = VT - IARA
= 250 V – (120 A)(0.06 Ω)
= 250 V – 7.2 V
= 242.8 V
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Solution (cont..)
ii) Use KVL to find EA2
EA2 = VT - IA2RA
Since the torque and the flux is constant, IA is constant. This yields a voltage of:
EA2 = 200 V – (120 A)(0.06 Ω) = 200 V – 7.2 V = 192.8 V
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Solution (cont..)
• The final speed of this motor
min/r876
min/r1103V8.242
V8.192
11
22
1
2
1
2
nE
En
E
E
n
n
A
A
A
A
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Example 4A DC series motor is running with a speed of 800 r/min while taking a current of 20 A from the supply. If the load is changed such that the current drawn by the motor is increased to 55 A, calculate the speed of the motor on new load. The armature and series field winding resistances are 0.2 Ω and 0.3 Ω respectively. Assume the flux produced is proportional to the current. Assume supply voltage as 200 V.
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SolutionGiven quantities• Supply voltage, VT = 200 V• Armature resistance, RA = 0.2 Ω• Series resistance, RS = 0.3 Ω• Initial speed, n1 = 800 r/min• Initial armature current, Ia1 = IL1 = 20 A
Figure 4
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Solution (cont..) When the armature current increased, Ia2 = 55 A, the back emf
EA2 = V – Ia2 (RA + RS) = 200 – 55(0.2 + 0.3) = 225 V
min/r30050
20
240
225800
2
1
1
212
2
1
1
2
1
2
2
1
1
2
1
2
I
I
E
Enn
I
I
E
E
n
n
E
E
n
n
A
A
A
A
A
A
The speed of the motor on new load
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Solution (cont..)
For initial load, the armature current, Ia1 = 20 A and the speed n1 = 800 r/min
V = EA1 + Ia1 (RA + RS)
The back e.m.f. at initial speed
EA1 = V - Ia1 (RA + RS) = 200 – 20(0.2 + 0.3) = 190 V
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DC Generator
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Generating of an AC Voltage
• The voltage generated in any DC generator inherently alternating and only becomes DC after it has been rectified by the commutator
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Armature windings
• The armature windings are usually former-wound. This are first wound in the form of flat rectangular coils and are then puller.
• Various conductors of the coils are insulated each other. The conductors are placed in the armature slots which are lined with tough insulating material.
• This slot insulation is folded over above the armature conductors placed in the slot and is secured in place by special hard wooden or fiber wedges.
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Generated or back e.m.f. of DC Generator
• General form of generated e.m.f.,
Φ = flux/pole (Weber)Z = total number of armature conductors = number of slots x number of conductor/slotP = number of polesA = number of parallel paths in armature[A = 2 (for wave winding), A = P (for lap winding)]N = armature rotation (rpm)E = e.m.f. induced in any parallel path in armature
A
PZNE
60
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Classification of DC Generator1. Separately Excited DC Generator• Field and armature windings are either connected
separate.
2. Shunt DC Generator• Field and armature windings are either connected in
parallel.
3. Series DC Generator• Field and armature windings are connected in series.
4. Compound DC Generator• Has both shunt and series field so it combines features of series and shunt motors.
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Equivalent circuit of DC generator
FAL III
AL II
Separately excited DC generator
F
FF R
VI
AAAT RIEV
Shunt DC generator
F
TF R
VI
AAAT RIEV
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FAL III
ASL III
Series DC generator
)( SAAAT RRIEV
Shunt DC generator
F
TF R
VI
AAAT RIEV
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Example • A DC shunt generator has shunt field winding
resistance of 100Ω. It is supplying a load of 5kW at a voltage of 250V. If its armature resistance is 0.02Ω, calculate the induced e.m.f. of the generator.
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Solution
Given quantities• Terminal voltage, VT = 250V
• Field resistance, RF = 100Ω
• Armature resistance, RA = 0.22Ω
• Power at the load, P = 5kW
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Solution (cont..)
A5.2100
V250
F
TF
FLA
R
VI
III
The field current,
A20V250
W5000
TL V
PIThe load current,
The armature current, IA = IL + IF = 20A + 2.5A = 22.5A
The induced e.m.f.,
EA = VT + IA RA = 250V + (22.5)(0.22) = 254.95V
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Power flow and losses in DC machines
DC generators take in mechanical power and produce electric power while DC motors take in electric power and produce mechanical power
Efficiency
%xP
P
in
out 100
%xP
PP
in
lossout 100
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The losses that occur in DC machine can be divided into 5 categories
1.Copper losses (I2R)2.Brush losses3.Core losses4.Mechanical losses5.Stray load losses
aaa RIP 2
fff RIP 2 Ia = armature current
If = field current
Ra = armature resistance
Rf = field resistance
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Core losses – Hysteresis losses and Eddy current losses
Mechanical losses – The losses that associated with mechanical effects.
Two basic types of mechanical losses: Friction & Windage.
Friction losses caused by the friction of the bearings in the machine.
Windage are caused by the friction between the moving parts of the machine and the air inside the motor casing’s
Stray losses (Miscellaneous losses) – Cannot placed in one of the previous categories.
Power Losses
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The Power Flow Diagram
For generator
Pout = VTIL
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The Power Flow Diagram
For motor
appoutP
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Example
A short-shunt compound generator delivers 50A at 500V to a resistive load. The armature, series field and shunt field resistance are 0.16, 0.08 and 200, respectively.
Calculate the armature current if the rotational losses are 520W, determine the efficiency of the generator
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Solution
W520Pu W25000A50Vx500Pout
A52200
500If . A552A50A52III Lfa ..
Armature Copper Loss
W441160552RIP 2a
2aca ).().()(
Series Field Copper Loss
W5220080552RIP 22f
2a2cf .).().()(
Shunt Field Copper Loss
W125020052RIP 21f
2f1cf )().()(
Friction + Stray + windage + etc:
W520Pu
Total Losses = W5243152012505220441 .).(
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Efficiency, η =losses Totall
Pout
Pout
Pin
Pout
%.@..
1391911305243125000
25000
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AC Machine Fundamentals & Induction Machines
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The induction machine is the most rugged and the most widely used machine in industry.Like dc machine, the induction machine has a stator and a rotor mounted on bearings and separated from the stator by an air gap.However, in the induction machine both stator winding and rotor winding carry alternating currents.The induction machine can operate both as a motor and as generatorAs motors, they have many advantages. They are rugged, relatively inexpensive and require very little maintenance. They range in size from a few watts to about 10,000 hp. The speed of an induction motor is nearly but not quite constant, dropping only a few percent in going from no load to full load.
INDUCTION MACHINE
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The main disadvantages of induction motors are
a. The speed is not easily controlled.b. The starting current may be five to eight times
full-load current.c. The power factor is low and lagging when the
machine is lightly loaded
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INDUCTION MOTOR CONSTRUCTION
Two different types of induction motor which can be placed in stator
a) squirrel cage rotorb) wound rotor
Squirrel Cage rotor Wound rotor
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Squirrel cage rotor – consists of conducting bars embedded in slots in the rotor magnetic core and these bars are short circuited at each end by conducting end rings. The rotor bars and the rings are shaped like squirrel cage.
Wound rotor – carries three windings similar to the stator windings. The terminals of the rotor windings are connected to the insulated slip rings mounted on the rotor shaft. Carbon brushes bearing on these rings make the rotor terminals available to the user of the machine. For steady state operation, these terminals are short circuited.
Types of rotor
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Rotor bars (slightly skewed)
End ring
Squirrel Cage Rotor
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Wound Rotor• Most motors use the squirrel-cage rotor because of the
robust and maintenance-free construction.• However, large, older motors use a wound rotor with three
phase windings placed in the rotor slots.• The windings are connected in a three-wire wye. • The ends of the windings are connected to three slip rings. • Resistors or power supplies are connected to the slip rings
through brushes for reduction of starting current and speed control
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Induction Motor Components
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BASIC INDUCTION MOTOR CONCEPT
A single/three phase set of voltages has been applied to the stator, and single/three phase set of stator currents is flowing. These produce a magnetic field Bs, which is rotating in a counterclockwise direction .
The speed of the magnetic field’s rotation is P
fn e
sync
120
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THE CONCEPT OF ROTOR SLIP
The voltage induced in a rotor depends on the speed of the rotor relative to the magnetic field.
Slip speed is defined as the difference between synchronous speed and rotor speed
msyncslip n - n n wherenslip = slip speed of the machinensync = speed of the magnetic fieldsnm = mechanical shaft speed of motor
Slip is the relative speed expressed on a per unit or a percentage basis
100% xn
n s
sync
slip100% x
n
n - n s
sync
msync
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In term angular velocity (radians per second, rps)
100% x -
ssync
msync
If the rotor turns at synchronous speed, s = 0 while if the rotor is stationary/standstill, s = 1
synxm s)n- (1 n
synxm s)- (1
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THE ELECTRICAL FREQUENCY CONCEPT
Like a transformer, the primary (stator) induces a voltage in the secondary (rotor) but unlike a transformer, the secondary frequency is not necessary the same as the primary frequency.
If the rotor of a motor is locked, then the rotor will have same frequency as the stator.
The rotor frequency can be expressed
er sf f )( msyncr n - n120
P f
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A 208V, 10hp, 4 pole, 60Hz, Y connected induction motor has full load slip of 5%.
Calculate,
a. synchronous speed, nsync (Ans:1800rpm)
b. rotor speed, nm (Ans: 1710rpm)
c. rotor frequency, fr at the rated load (Ans: 3 Hz)
d. Shaft torque at the rated load (Ans: 41.7Nm)
Example
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The derivation of the induction motor induced-torque equation
m
convind
P
sync
AGind
P
The induced torque in induction motor is
s
RIPAG
222
s
RIPAG
2223
Air gap power
Total Air gap power
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a) What is the motor’s slip? (Ans:1.67%)
b) What is the induced torque in the motor in Nm under these conditions? (48.6Nm)
c) What will the operating speed of the motor be if its torque is doubled? (2900 rpm)
d) How much power will be supplied by the motor when the torque is doubled? (29.5kW)
A two pole, 50hz induction motor supplies 15kW to a load at speed 2950 rpm.
Assignment 6.5
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Speed control of induction motors
i. Induction motor speed control by pole
ii. changing
iii. Speed control by changing the line frequency
iv. Speed control by changing the line voltage
v. Speed control by changing the rotor
vi. resistance