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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