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Chapter 3:Chapter 3:INDUCTION MOTORINDUCTION MOTOR
BEE2133 BEE2133 ELECTRICAL MACHINE & ELECTRICAL MACHINE &
POWER SYSTEMPOWER SYSTEM
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Learning Outcomes
• At the end of the lecture, student should be able to:– Understand the principle and principle and
the nature of single phase and the nature of single phase and 3 phase induction machines.3 phase induction machines.
– Perform an analysis on induction machines which is the most rugged and the most most widely used machine widely used machine in industry.in industry.
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CHAPTER OUTLINECHAPTER OUTLINE3.1 Introduction3.2 Overview of single phase IM3.3 Overview of Three-Phase IM3.4 Construction3.5 Principle of Operation3.6 Equivalent Circuit
• Armature reaction
• Power Flow, Losses and Efficiency
• Torque-Speed Characteristics3.7 Speed Control
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3.1 INTRODUCTION• A induction machine can be used
as either a induction generator or a induction motor.
• IM transform electrical energy into mechanical energy
• IM is a type of asynchronous AC motor where power is supplied to the rotating device by means of electromagnetic induction
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3.1 INTRODUCTION• popularly used in the industry and are used
worldwide in many residential, commercial, industrial, and utility applications.
• Main features: cheap and low maintenance
(absence of brushes)
Main disadvantages: speed control
is not easy
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• Construction : similar to 3 induction motor• A single-phase motor is a rotating
machine that has both main and auxiliary windings and a squirrel-cage rotor.
• Supplying of both main and auxiliary windings enables the single-phase machine to be driven as a two-phase machine.
3.2 OVERVIEW OF SINGLE PHASE IM
3.2 OVERVIEW OF SINGLE PHASE IM
• Home air conditioners• Kitchen fans• Washing machines• Industrial machines• Compressors• Refrigerators
3.2 OVERVIEW OF SINGLE PHASE IM
• Types of 1 induction Motor– Split Phase Motor– Capacitor Start Motors– Capacitor Start, Capacitor Run– Shaded Pole Induction Motor– Universal Motor (ac series
motors)
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3.3 OVERVIEW OF 3 PHASE IM
• Simple and rugged construction • Low cost and minimum maintenance • High reliability and sufficiently
high efficiency • The speed is frequency dependent
not easily to control the speed
thanks to power e
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3.3 OVERVIEW OF 3 PHASE IM• can be part of a pump or fan, or connected to
some other form of mechanical equipment such as a winder, conveyor, or mixer.
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3.4 CONSTRUCTION• Basic parts of an AC motor : rotor, stator, enclosure• The stator and the rotor are electrical circuits that
perform as electromagnets.
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• The stator - stationary stationary part of the motor.part of the motor.• Stator laminations are stacked togetherstacked together forming a
hollow cylinderhollow cylinder. • Coils of insulated wire are inserted into slots of the Coils of insulated wire are inserted into slots of the
stator core.stator core.• Each grouping of coilsEach grouping of coils, together with the steel core it
surrounds, form an electromagnet.
3.4 CONSTRUCTION (stator)
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• The rotor is the rotating part of the motor• It can be found in two types:
– Squirrel cage (most common)– Wound rotor
3.4 CONSTRUCTION (rotor)
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/rotor winding/rotor winding
Short circuits allShort circuits allrotor bars.rotor bars.
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Squirrel cage type:Squirrel cage type:Rotor winding is composed of copper bars
embedded in the rotor slots and shorted at both end by end rings
Simple, low cost, robust, low maintenance
3.4 CONSTRUCTION (rotor)
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Wound rotor type:Wound rotor type:Rotor winding is wound by wires. The winding
terminals can be connected to external circuits through slip rings and brushes.(similar with DC motor, with the coils connected together that make contact with brushes)
Easy to control speed, more expensive.
3.4 CONSTRUCTION (rotor)
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• The enclosure consists of a frame (or yoke) and two end brackets (or bearing housings). The stator is mounted inside the frame. The rotor fits inside the The rotor fits inside the stator with a slight stator with a slight air gapair gap separating it from the separating it from the stator (stator (NONO direct physical connection) direct physical connection)
Stator
Rotor
Air gap
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3.4 CONSTRUCTION (enclosure)
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• The enclosure The enclosure protectsprotects the electrical and operating the electrical and operating parts of the motor parts of the motor from harmful effects of the from harmful effects of the environmentenvironment in which the motor operates. in which the motor operates.
• Bearings, mounted on the shaft, support the rotor Bearings, mounted on the shaft, support the rotor and allow it to turn. A fan, also mounted on the shaft, and allow it to turn. A fan, also mounted on the shaft, is used on the motor shown below for cooling.is used on the motor shown below for cooling.
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3.4 CONSTRUCTION (enclosure)
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Nameplate
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3.5 PRINCIPLE OF OPERATION
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3.5 PRINCIPLE OF OPERATION
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Rotating Magnetic Field• When a 3 phase stator winding is connectedconnected to a 3 phase
voltage supply, 3 phase current will flow in the windingsflow in the windings, which also will inducedinduced 3 phase flux in the stator.
• These flux will rotate at a speed called a Synchronous Synchronous Speed, nSpeed, nss. The flux is called as Rotating magnetic Field
• Synchronous speed: speed of rotating flux
• Where; p = is the number of poles, and f = the frequency of supply
p
fns
120
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Slip and Rotor Speed1.1. Slip Slip ss– The rotor speed of an Induction machine is different from
the speed of Rotating magnetic field. The % difference of the speed is called slip.
– Where; ns = synchronous speed (rpm)
nr = mechanical speed of rotor (rpm)
– under normal operating conditions, s= 0.01 ~ 0.05, which is very small and the actual speed is very close to synchronous speed.
– Note that : s is not negligibles is not negligible
)1( snnORn
nns sr
s
rs
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Slip and Rotor Speed• Rotor SpeedRotor Speed
– When the rotor move at rotor speed, nr (rps), the stator flux will circulate the rotor conductor at a speed of (ns-nr) per second. Hence, the frequency of the rotor is written as:
• Where; s = slip
f = supply frequency
sf
pnnf rsr
)(
fsfiii
iipnn
f
nnRotorAt
ipn
f
nstatorAt
Note
r
rsr
pf
rs
s
pf
s
.:)()(
).....(120
)(
:
).....(120
:
:
120
120
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Principle of Operation
• Torque producing mechanismTorque producing mechanismWhen a 3 phase stator winding is connectedconnected to a
3 phase voltage supply, 3 phase current will flow flow in the windingsin the windings, hence the stator is energized.
A rotating flux Φ is produced in the air gap. The flux Φ induces a voltage Ea in the rotor winding (like a transformer).
The induced voltage produces rotor current, if rotor circuit is closed.
The rotor current interacts with the flux Φ, producing torque. The rotor rotates in the direction of the rotating flux.
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Direction of Rotor Rotates
• Q: How to change the direction of
• rotation?• • A: Change the phase
sequence of the• power supply.
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• Conventional equivalent circuit Note:
● Never use three-phase equivalent circuit. Always use per- phase equivalent circuit.
● The equivalent circuit always always bases on the Y connection bases on the Y connection regardless of the actual regardless of the actual connection of the motorconnection of the motor.
● Induction machine equivalent circuit is very similar to the single-phase equivalent circuit of transformer. It is composed of stator circuit and rotor circuit
Equivalent Circuit of Induction Machines
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• Step1 Rotor winding is openStep1 Rotor winding is open (The rotor will not rotate)
• Note: – the frequency of E2 is the same as that of E1 since the rotor
is at standstill. At standstill s=1.
Equivalent Circuit of Induction Machines
f f
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Equivalent Circuit of Induction Machines
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Equivalent Circuit of Induction Machines• Step2 Rotor winding is shortedStep2 Rotor winding is shorted
(Under normal operating conditions, the rotor winding is shorted. The slip is s)
• Note:
– the frequency of E2 is fr=sf because rotor is rotatingrotor is rotating.
f fr
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• Step3 EliminateStep3 Eliminate ff22
Keep the rotor current same:
Equivalent Circuit of Induction Machines
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• Step 4 Referred to the stator sideStep 4 Referred to the stator side
• Note:– X’2 and R’2 will be given or measured. In practice, we
do not have to calculate them from above equations.– Always refer the rotor side parameters to stator side.– Rc represents core loss, which is the core loss of stator side.
Equivalent Circuit of Induction Machines
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• IEEE recommended equivalent circuitIEEE recommended equivalent circuit
• Note:– Rc is omitted. The core loss is lumped with
the rotational loss.
Equivalent Circuit of Induction Machines
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• IEEE recommended equivalent circuitIEEE recommended equivalent circuit
Note: can be separated into 2 PARTSNote: can be separated into 2 PARTS
• Purpose : Purpose : – to obtain the developed mechanicalto obtain the developed mechanical
Equivalent Circuit of Induction Machines
I1 1R1X
mX
'2X
'2R
s
sR
1'21V
s
R2
s
sRR
s
R )1(22
2
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Analysis of Induction Machines• For simplicity, let
assume
IIss=I=I1 1 , I, IRR=I=I22
(s=stator, R=rotor)
RmsTotal
sss
cmm
cmcm
RR
R
ZZZZ
jXRZ
neglectedRjXZ
neglectedRjXRZ
jXs
RZ
//
;
;
;//
;''
ZZRRZZmm
ZZss
VVs1s1
IIs1s1 IIm1m1 IIR1R1
T
s
s Z
VI
1
1
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Analysis of Induction Machines
m
RMm
R
RMR
sT
mRRM
Z
VI
Z
VIHence
VZ
ZZV
RulesDividingVoltage
11
11
11
,
//
,
ZZRRZZmm
ZZss
VVs1s1
IIs1s1 IIm1m1 IIR1R1
11
11
,
sRm
mR
sRm
Rm
IZZ
ZI
IZZ
ZI
RulesDividingCurrent
OR
Note : 1hp =746WattNote : 1hp =746Watt
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EXAMPLE 1A 4 poles, 3 Induction Motor operates from a supply which frequency is 50Hz. Calculate:
a. The speed at which the magnetic field is rotating
b. The speed of the rotor when slip is 0.04
c. The frequency of the rotor when slip is 3%.
d. The frequency of the rotor at standstill
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EXAMPLE 2A 500hp, 3 6 poles, 50Hz Induction Motor has a speed of 950rpm on full load. Calculate the slip.
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EXAMPLE 3If the emf in the rotor of an 8 poles Induction Motor has a frequency of 1.5Hz and the supply frequency is 50Hz. Calculate the slip and the speed of the motor.
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EXAMPLE 4A 440V, 50Hz, 6 poles, Y connected induction motor is rated at 135hp. The equivalent circuit parameters are :
Rs=0.084 RR’=0.066
Xs=0.2 XR’=0.165
s = 5% Xm=6.9
Determine the stator current, magnetism current and rotor current.
SolutionGiven V=440V, p=6, f=50Hz, 135hp
]63.9685.32,76.1344.170,34.246.177[ AAA ooo
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Example 4 (Cont)s
RR ' 'RX
mX
sXsR
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Example 4 (Cont)
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Example 4 (Cont – 1st Method)
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Example 4 (Cont – 2nd Method)
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Power Flow Diagram
PPinin (Motor) (Motor)
PPinin (Stator) (Stator)
PPcore losscore loss
(P(Pcc))
PPair Gapair Gap
(P(Pagag))
PPdevelopeddeveloped
PPmechanicalmechanical
PPconvertedconverted
(P(Pmm))
PPout, out, PPoo
PPstator copper stator copper
loss, loss, (P(Pscuscu))PProtor copper rotor copper
lossloss (P (Prcurcu))PPwindage, friction, etcwindage, friction, etc
(P (P - -
Given)Given)
cos3 ss IV
s
RI RR
''3 2
2
3
c
RM
R
V ''3 2RR RI
s
sRI RR
1''3 2
Whp 7461
ss RI23
PPinin (Rotor) (Rotor)
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Power Flow Diagram• Ratio:Ratio:
Pag Prcu Pm
s
RI RR
''3 2 ''3 2
RR RI
s
sRI RR
1''3 2
s
11
1s
1
1 s s1
Ratio makes the analysis simpler to find the value of the particular power if we haveanother particular power. For example:
s
s
P
P
m
rcu
1
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Efficiency
WattxWhpxP
IVP
otherwise
PPP
PPP
givenarePif
P
P
out
ssin
mo
lossesino
losses
in
out
746746
cos3
,
,
%100
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Example 5 (Cont from Ex 4)• Calculate
i. Stator Copper Lossii. Air Gap Poweriii. Power converted from electrical to mechanical poweriv. Output powerv. Motor efficiency
• SolutionSolution
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Example 5 (Cont from Ex 4)
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Torque-Equation• Torque, can be derived from power equation in term of
mechanical power or electrical power.
n
PTHence
sradn
whereTPPower
2
60,
)/(60
2,,
r
oo
r
mm
n
PTTorqueOutput
n
PTTorqueMechanical
Thus
2
60,
2
60,
,
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Torque-Equation• Note that, Mechanical torque can written in terms of Note that, Mechanical torque can written in terms of
circuit parameters. This is determined by using circuit parameters. This is determined by using approximation methodapproximation method
...
...
...
)1('
'3
)1('
'3
2
2
r
RR
r
mm
mrmR
Rm
ssR
IPT
TPandss
RIP
22
2
)'()'(
'
2
)(3
RR
R
s
RMm sXR
sR
n
VT
Hence, Plot Tm vs s
Tm
ns
ssmaxmax is the slip for T is the slip for Tmaxmax to occur to occur
s=1
Tst
Tmax
smax
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Torque-Equation
22
2
)'()'(
'
602
)(3
1,
RsRs
R
s
sst XXRR
Rn
VT
sTorqueStarting
22
2
max
22max
)'()(
1
6022
)(3
)'()(
'
Rssss
s
Rs
R
XXRRn
VT
XR
Rs
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Example 6 (Cont from Ex 4)• Calculate
v.Mechanical torquevi.Output torquevii.Starting torqueviii.Maximum torque and maximum slip
solutionsolution
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Example 6 (Cont from Ex 4)
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Example 6 (Cont from Ex 4)
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Speed Control• There are 3 types of speed control of 3
phase induction machinesi.i. Varying rotor resistanceVarying rotor resistance
ii.ii. Varying supply voltageVarying supply voltage
iii.iii. Varying supply voltage and supply frequencyVarying supply voltage and supply frequency
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Varying rotor resistance• For wound rotor only• Speed is decreasing• Constant maximum
torque• The speed at which
max torque occurs changes
• Disadvantages: – large speed regulation
– Power loss in Rext – reduce the efficiency
T
ns~nNL
T
nr1nr2nr3 n
nr1< nr2< nr3R1R2R3
R1< R2< R3
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Varying supply voltage• Maximum torque changes• The speed which at max
torque occurs is constant (at max torque, XR=RR/s
• Relatively simple method – uses power electronics circuit for voltage controller
• Suitable for fan type load• Disadvantages :
– Large speed regulation since ~ ns
T
ns~nNL
T
nr1nr2nr3 n
nr1> nr2 > nr3
V1
V2
V3
V1> V2 > V3
V decreasing
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• The best method since supply voltage and supply frequency is varied to keep VV//ff constant
• Maintain speed regulation• uses power electronics
circuit for frequency and voltage controller
• Constant maximum torque
Varying supply voltage and supply frequency
T
nNL1
T
nr1nr2nr3 n
f decreasing
nNL2nNL3