Set5 Induction Motor Final

8/6/2019 Set5 Induction Motor Final

http://slidepdf.com/reader/full/set5-induction-motor-final 1/90

ELE 411: Induction Motors

Instructor: Dr. Abbas A. Fardoun

Department of Electrical Engineering

United Arab Emirates University



2

Home Work

7.2, 7.3, 7.4, 7.5, 7.7, 7.8, 7.14, 7-16 & (7-19parts a,b & c)



3

Chapter # 7 Induction Motors

7.1- Induction Motors Construction

7.2- Basic Induction Motor Concept- Development of Induced Torque

- Concept of rotor sip

- Electrical frequency of rotor

7.3- Equivalent Circuit of Induction Motor

7.4- Power & Torque of Induction Motor- Losses & Power flow diagram

- Power & torque

- Separating rotor losses & power converted in IM

7.5- Induction Motor Torque Speed Characteristics- Derivation of Induced Torque Equation

- Comments on Torque-Speed Characteristics

- Maximum Torque in IM

7.6- Variation in IM Torque-Speed Characteristics

7.11- Determining Circuit Model parameters



Objectives

• The Objective of this lecture to discuss Induction Machine

construction. Reasons & applications for different type of rotor construction

• Basic operation theory of induction machine; definition of slipas related to machine rotor & rotational field speeds.

4



Outcomes

• Understand different types of rotors of IM.

• Understand theory of operation.

• Understand what “slip” means.

5



6

• Induction motor is the most rugged and the most widely used motor in

industry.

•Like any AC machine, the induction machine has a stator and a rotor mounted

on bearings and separated from stator by an air gap.

•

Both stator and rotor windings carry alternating currents. The alternatingcurrents (ac) are supplied directly to the stator winding and to the rotor winding

by induction.

• Reasons for the popularity of induction motors include:

- Simplicity- Reliability and low cost

- Minimal service requirement

- Good efficiency

Introduction



7

Large three-phase induction motors (in tens or hundreds of

horsepower) are used in- pumps,

- fans,

- compressors,

- paper mills,

- textile mills, and so forth.

The induction motor is used in various sizes. Small single-phase

induction motors (in fractional horse-power) are used in manyhousehold appliances, such as

- blenders,

- lawn mowers,

- juice mixers,- washing machines,

- refrigerators.

Applications of Induction Motors



8

Applications of Induction Motor



9

Introduction

• It is named induction machine

because the rotor voltage (which

produces the rotor current and

rotor magnetic field) is induced

in the rotor winding rather thanbeing physically connected by

wires. It is similar to a

transformer

• Distinguishing feature: is that

NO DC field current is required

to run this machine as was the

case for synchronous generators.

1 2

Vdc+ -

Synchronous

Rotor

Synchronous GeneratorRotor or Field flux



10

• Two Types of Induction machines known as Wound rotor & Squirrel-cage machines

•Both Machines has a stator similar to that of Synchronous Generator.

•They differ in rotor construction (same stator):

- Wound-rotor

- Squirrel-cage rotor

7.1- Construction Of Induction Motors/Rotor



11

Induction Motor Cross Section

Air gap



12

Input voltages to the stator of induction motor;

Stator for aThree Phase Induction Machine

))(240sin()(

))(120sin()(

))(sin()(

V t V t v

V t V t v

V t V t v

mc

mbn

man

1 : n

Stator

v an

v bn

v cn

Stator winding

na

a

n

b

b

c

c

Three Phase Stator winding for Induction Motor (IM) isconstructed in the same as for synchronous Generators.



13

• Wound-rotor motor: Rotor has distributed windings are similar tostator windings.

•Rotor terminals are made accessible to external loads through

brushes bearing on slip rings.

•

Connected to external circuit such as to a resistance is very handyfor the purpose of speed control of the induction motor.

•Squirrel-cage motor: windings are made of conducting bars embedded

in the rotor and short-circuited at each end by conducting end rings.

•The rotor terminals are thus inaccessible in a squirrel-cage

construction.

•Squirrel cage induction machine is simpler, more reliable,

efficient, economical and more rugged than the wound-rotor

induction machine because there are no brushes, no slip rings.

7.1- Construction Of Induction Motors/Rotor



14

Wound Rotor of Three Phase Induction Machine

Slip ring

Brushea b cLoad

(a) Wound rotor with shorted rotor

terminals

Wound-rotor motor: Rotor has

distributed windings are similar tostator windings.

-Rotor terminals are made

accessible to external loads

through brushes bearing on sliprings.

- Rotor winding can be connected

to external resistive load via slip

rings for the purpose of speed

control of the induction motor.

ExternalLoadWound Rotor

Equivalent

Circuit



15

Squirrel Cage Rotor of a three Phase induction machine.

Squirrel cage rotor Equivalent circuit

Squirrel-cage motor: windings

are made of conducting barsembedded in the rotor andshort-circuited at each end byconducting end rings.

-The rotor terminals are thusinaccessible in a squirrel-cageconstruction.

-Squirrel cage induction

machine is simpler, morereliable, efficient, economicaland more rugged than thewound-rotor inductionmachine because there areno brushes, no slip rings.



16

Three Phase Induction Machine Construction/Summary

1 : n

Stator

v a

n

v b

nv c

n3-Phase Stator windings

na

a

Slip ring

Brushesa b cLoad

A wound rotor has a set

of 3-phase windingsthat are mirror image ofstator windings, withterminal short circuitedor connected to sliprings to have access to

speed control resistor

In squirrel-cage rotor,the windings are simplyconducting barsembedded in the rotorand short-circuited at

each end by conductingend rings.

Wound Rotor

EquivalentCircuit

Squirrel cageRotor Equivalent

Circuit



17

• We know that if two magnetic fields are present in a machine,

then the torque will be created which will tend to line up thetwo magnetic fields.

• If one magnetic field is produced by the stator of an ACmachine and other one is produced by the rotor of themachine, then a torque will be induced in the rotor which willcause the rotor to turn and align itself with the statormagnetic field.

• If there were some way to make the stator magnetic fieldrotate , then the induced torque in the rotor would cause it toconstantly “chase” the stator magnetic around in a circle. Thisin a nutshell, is the basic principle of all ac motor operation.

• Now the question is how can the stator magnetic field bemade to rotate.

7.2 Basic Induction Motor Concepts4.2: The Rotating Magnetic Field/Review

S Rind B BK



18

7.2 Induced Torque in an AC Machine

In an AC machine under normal operatingcondition, there are two magnetic fieldspresent- a magnetic field from the rotorcircuit and another magnetic field from thestator circuit. The interaction of these twomagnetic fields produces the torque in themachine.

S Rind B BK



19

Development of Induced Torque

Three phase set of voltages

are applied to stator winding,three phase currents flow inthe stator winding.

•A rotating magnetic field BS isproduced.

•The speed of magnetic field

• f e is frequency in hertz

• P is # of poles

p

f n e

synch

120

))(240sin()(

))(120sin()(

))(sin()(

V t V t v

V t V t v

V t V t v

mc

mbn

man



20

• when rotating magnetic field BS passes over rotorbars, it induces voltage in them

V: velocity of rotor bars relative to magnetic field

Bs: Stator magnetic flux density

I: length of conductor in the magnetic field

• Induced voltage cause current to flow in rotorwindings, hence rotor field Br is produced.

Then

l BV e Sind ).(

S Rind B BK

Development of Induced Torque



21

Approximate explanation of induced voltage

•AC Voltage is applied to

stationary winding

•Current flows through winding

•Field is produced bystationary winding Bs.

•Flux flows through rotor(assume rotating field).

•Voltage is producede=(vXB).l.

•Current flows in rotorwinding.

•Rotor field (Br) is produced towinding carrying current

Rotor

S Rind B BK



22

Induction Motors/review

Operation Principle

• The three-phase stator is supplied by balancedthree-phase voltage that drives an ac magnetizingcurrent through each phase winding

• The magnetizing current in each phase generates

a pulsating ac flux.• The total flux in the machine is the sum of the

three fluxes

• The summation of the three ac fluxes results in arotating flux, which turns with constant speed and

has constant amplitude



23

Induction Motors/review

Operation Principle• The rotating flux induces avoltage in the short-circuitedbars of the rotor. This voltagedrives current through the bars.

• The induced voltage isproportional with the differenceof motor and synchronousspeed. Consequently the motorspeed is less than the

synchronous speed• The interaction of the rotating

flux and the rotor currentgenerates a force that drivesthe motor.

S Rind B BK

l BV e Sind ).(



24

• The motor will start rotating, there is a limit to the

speed of rotor.

• If rotor speed becomes equal to synchronous speedthen rotor bars will be stationary relative to themagnetic field, and there would be no induced

voltage.• If then current in rotor bars is zero

No rotor current

No torque

• How about when the rotor is not rotating (0 speed)?

Voltage is induced at the same frequency as thatof the stator.

What is output power in this case?

0ind e

Concept of Rotor Slip



25


Induced Voltage depends on the

speed difference betweenSynchronous speed of rotatingfield & rotor mechanical speed.

Rotor Slip (difference betweenrotor & synchronous speeds)

nslip : Slip Speed

nsync: Synchronous Speednm : Mechanical Shaft Speed

%%

synch

msynch

synch

slip

n

nn

n

nSSlip

msyncslip nnn

l BV e Sind ).(



26


Rotating Transformer

- Induction motor works by induced voltage and current in rotor,therefore some times called as rotating transformer.

- Primary (Stator) induces voltage in Secondary (Rotor)

- Secondary frequency:

If rotor is locked (S=1), then secondary frequency is sameas primary. However output power is zero because

P= torque X speed = If rotor turns at synchronous speed the Rotor frequency

will be zero since no voltage is induced.



27


Or in terms of angular velocity

If => S = 0 (rotor @ sync speed)

=> rotor freq=stator freq

=> S = 1 (locked/stalled/stationary rotor)Normal operation S is between 0 & 1

synchm nSn )1(

syncm

0m

%100

synch

msynchS

sync

msync

n

nn

s

synchm S )1(

πf ω p

f n esynch 2 &

120



28

Electrical frequency on the Rotor

0& 0

s f

nn

r

syncm

synch

msynch

n

nnS

1s& f f 0n

er

m

Slip Definition

When Rotor is not moving, the rotor seessynchronous speed.It means induced voltage has the samefrequency as that of the synchronous speed

(electrical frequency fe).

When rotor is moving at the same speed assynchronous speed. It means relative speedbetween rotating field & rotor is zero.

In between Zero & Synchronous speeds,induced voltage frequency is proportional torelative speed of rotor w.r.t. rotating field orthe difference between synchronous speed &

mechanical speeds which is slip speed. er

Sync

mSync

sf f

n

nns



29

Concept of Rotor Slip /Review

• nm= speed of rotor shaft/motor.

• ns= synchronous speed

• Rotor sees a relative speed of amagnetic field equals to nr

• Voltage is induced in rotor atrelative magnetic field speed=nr

• The induced voltage in the rotor

has an angular speed r

• Frequency of rotor inducedvoltage=f r sr sf f

msyncslip nnn

msyncelecr

_

sync

msync

n

nns

Rotor induced voltage & rotor electrical



30

Rotor voltage frequency, fR

Rotor speed (rpm)Slip

ERof stator


0 no load 1 0

0 no load1 0

Rotor voltage, ER

Rotor induced voltage & rotor electricalfrequency as function of slip/speed

•Voltage induced at locked rotor condition E R0 =a eff *V stator

-Use locked rotor condition as reference point - reference freq. & voltage at different speeds to that of locked rotor

•Voltage induced at any Slip E R =SE Ro

• Rotor current/Voltage Frequency f r =Sf e



31

Class Activity



32

Agenda



33

Agenda

• Review of Slip concept• Derivation of IM Equivalent Circuit Parameters

measurement• Example problem• Power & Torque of Induction Motor

- Losses & Power flow diagram- Power & torque- Separating rotor losses & power converted in IM



34

Chapter # 7 Induction Motors

Introduction/Review

7.1- Construction Of Induction Motors7.2- Basic Induction Motor Concept

- Development of Induced Torque

- Concept of rotor sip

- Electrical frequency of rotor

7.3- Equivalent Circuit of Induction Motor7.4- Power & Torque of Induction Motor

- Losses & Power flow diagram

- Power & torque

- Separating rotor losses & power converted in IM

7.5- Induction Motor Torque Speed Characteristics- Derivation of Induced Torque Equation

- Comments on Torque-Speed Characteristics

- Maximum Torque in IM

7.6- Variation in IM Torque-Speed Characteristics

7.11- Determining Circuit Model parameters

C t f R t Sli /R i



35

Concept of Rotor Slip /Review

• nm= mechanical speed of rotor

shaft/motor.

• ns= synchronous speed

• Rotor sees a relative speed of amagnetic field equals to nslip

• Voltage is induced in rotor at relativemagnetic field speed=n

slip

• Frequency of rotor induced voltage=f rsr sf f

msyncslip nnn

sync

msync

n

nns

synchm nSn )1(



36

7.3 Equivalent Circuit of an Induction Machine

• An induction motor is called a singly excitedmachine (as opposed to a doubly excitedsynchronous machine), since power is suppliedonly to the stator circuit. Because induction

motor does not have an independent field circuitits model will not contain an internal voltagesource such as internal generated voltage EA in asynchronous machine.

• We will begin with the transformer model.



37

Transformer Model of an Induction Machine

jXR

RR

IR

ERE1 jXM

I2 jX1R1I1

RC

+

-

Stator Equivalent Circuit:

Similar to a transformer, stator winding has a self inductance & copper loss.Can be represented as Rs & Ls, Xs=?

aeff:1



38

IM Equivalent Circuit/Magnetizing Impedance

• Due to the air gap in an induction

machine, the reluctance of the fluxpath is increased greatly. The higherreluctance caused by the air gapmeans that a higher magnetizingcurrent is required to obtain a givenflux. Ni=fR

• Due to higher reluctance (air gap),magnetizing reactance will be muchsmaller than that of a transformer.

• Rfe, similar to a transformerrepresents core losses (Hysterises & eddy current losses).

Magnetizing Impedance: Flux vs. Primary/stator



39

• Why slope of transformer is steeper?

Transformer

Induction Motor

φ

mmf

Magnetizing Impedance: Flux vs. Primary/statorcurrent for induction machine & transformer

Rotor induced voltage & rotor electrical



40

Rotor voltage frequency, fR


ERof stator


0 no load 1 0

0 no load1 0

Rotor voltage, ER

Rotor induced voltage & rotor electricalfrequency as function of slip/speed

•

Voltage induced at locked rotor condition E R0 =a eff *V stator -Use locked rotor condition as reference point - reference freq. & voltage at different speeds to that of locked rotor

•Voltage induced at any Slip E R =SE Ro

• Rotor current/Voltage Frequency f r =Sf e



41

Rotor Equivalent Circuit (Resistance & reactance)

•Rotor resistance is fixed RR,

It represents rotor copperloss.

•Rotor Reactance is defined

as, (rotor reactance changeswith motor speed)

•Let XR0 be the rotorreactance at blocked rotorcondition (S=1, my referencepoint)

2

)(since 2

Re R

er

Rr R

L f s X

sf f L f X

Ro R

Re R

Re Ro

sX X

L f X

L f X

s2

2

jXR

RR

IR

ER

+

-



42

0

0

R R

R R

jX S

R E I

Rotor Equivalent Circuit

Original Rotor Equivalent circuit

Divide by slip S

0

0

R R

R R

R R

R R

jSX R

SE

I

jX R

E I

RR

jXR=jSXRo

ER=SERo

RR /s

jXRo

ERo

Rotor Circuit Model with all

the frequency effectsconcentrated in Resistor RR

o

jXR

RR

IR

ER

+

-



43

Rotor Equivalent Circuit

=>It is possible to treat all of

the rotor effects due tovarying rotor speed as beingcaused by varyingimpedance supplied withpower from a constant

voltage source ER0.

-very low Slip RR /S >> XR0, sorotor resistance dominates.

at high Slip XR0 >> RR /S

RR /s

jXRo

ERo

0, R R

eq R jX S

R Z



44

Final Rotor Equivalent Circuit

• To solve circuit, it is

required to refer rotorparameters to stator side.

• Speed effects are

concentrated in resistance.

• Similar to transformersecondary/rotor can be

referred to primary/statorby turn ratio.

• Rotor Impedance is

represented as 0, R Req R jX S R

Z

RR /s

jXRo

ERo



45

Equivalent Circuit

jXR

RR

IR

ERE1 jXM

I2 jX1R1

I1

RC

+

-

E1 jXM

I2 jX1R1

I1

RC RR /s

jXRo

ERo

IR

+

-

Original

With Rotor circuit manipulation

aeff

:1

aeff :1



46

Let, a eff = turn ratio

-Similar to transformer

Equivalent circuit of IM

eff

R

Reff R

a I I

E a E E

2

0

'

1

0

2

2

2

2

Reff

Reff

X a X

Ra R

I2

XM

IM

jX1

R1I1

RCR2 /s

jX2

f V E1

jXR

RR

IR

ERE1 jXM

I2 jX1R1

I1

RC

+

-

Class Activity



47

Class Activity

Example 7-5: A 460V wound rotor,

60 Hz, 4 pole, 25hp, 3 phase Yconnected IM has the followingimpedances referred to stator:

R1=0.641 R2= 0.332 X1=1.106X2=0.464 & Xm=26.3 ( all in ohms).

Neglect RC.a)What is the motor speed

b)What is the stator current at a slipof 2.2%?

c)What is the rotor current?

d)what is the airgap power?

I2

XM

IM

jX1

R1I1

RC

R2 /s

jX2

f V E1

Agenda



48

Agenda

• Review of Equivalent Circuit• Power & Torque of Induction Motor

- Losses & Power flow diagram- Power & torque- Separating rotor losses & power converted in IM

• Derivation of Torque-Speed Curve- Salient features of torque-speed curve- Variation of curve as function of rotor resistance



49

7.4 Power and Torque in Induction Machine

• Induction motor same like transformer.

• Input is 3-φ voltage & currents. • Output of transformer is electric power at

secondary winding.

• The secondary (rotor) of an induction machine is

shorted, therefore no electrical output forminduction machine instead mechanical output.

Mechanical Electrical

I d i M b l fl di



50

Induction Motor energy balance flow diagram

Input power

Psup

Stator Copper loss

3 Ista2 Rsta

Rotor Copper loss

3 Irot2 Rrot

Stator Iron loss

3 Vsta2 / Rc

Windage, stray

and friction losses

Output power

Pout

Air gap

power Pag

Developed power

Pdv = 3 Irot2 Rrot (1-s)/s

Air gap P_converted

ind.m

Pout=load.m

..3 f P I V P t t in

Equivalent Circuit



51

Equivalent Circuit

I2

E1

iXM

IM

jX1

R1I1

RCR2 /s

jX2

f V

+

-

Stator Copper Losses:

Core Losses:

Air gap power:

In the equivalent circuit itcan be seen that the onlyelement where the air gappower can be consumed isR2/s

1

2

13 R I PSCL

C CORE R E P/ 3

2

1

coreSCLin AG PPPP Air gap

RotorStator

S

R I P AG

22

23

Induction Motor: Power & Torque



52

Induction Motor: Power & Torque

•Air Gap Power:Rotor copper losses = Air gappower * Slip

•The lower the slip the lowerthe rotor copper losses.

•When the rotor is stand still(s=1) all the air gap power is

consumed in rotor.

•Assume friction, windage & other miscillenous losses arelumped as mechanical powr

loss, output power is:

S R I P AG 2223

AG RCL PSP

AGConv

AG AG

RCL AGConv

PSP

SPP

PPP

)1(

mechlossconvout PPP _

Induction Motor: Power & Torque in



53

Induction Motor: Power & Torque inInduction Motor

• Torque generated by

converted power is definedas induced torque Tind where:

• Tind

can be re-written asfunction of air gap power & synchronous speed.

• Interesting equation

because it provides Tind as afunction PAG & ωsynch,where ωsynch is constant.

• Output Torque:

synch

AGind

PT

synch

AG

ind S

PS

T )1(

)1(

m

convind

PT

m

out out

PT

Induction Motor Energy Balance Flow



54

Input power

Psup

Stator Copper loss

3 Ista2 Rsta

Rotor Copper loss

3 Irot2 Rrot

Stator Iron loss

3 Vsta2 / Rc

Windage, strayand friction losses

Output power

Pout

Air gap

power Pag

Developed power

Pdv = 3 Irot2

Rrot (1-s)/s

Air gap P_converted

ind.m

Pout=load.m

duct o oto e gy a a ce oDiagram/Summary

S R I P AG

2223

AG RCLPSP

AGConv PSP )1(

Pin=3VphIph*p.f. =sqrt(3)VLIL*p.f

2

2

23 R I P RCL


synchind AG T P

Equivalent Circuit



55

Equivalent Circuit

I2

E1

iXM

IM

jX1

R1I1

RCR2 /s

jX2

f V

+

-

Stator Copper Losses:

Core Losses:

Air gap power:

In the equivalent circuit itcan be seen that the onlyelement where the air gappower can be consumed isR2/s

1

2

13 R I PSCL

C C

CORE G E R

E

P

2

1

2

1

3

3

coreSCLin AG PPPP Air gap

RotorStator

S

R I P AG

22

23

E i l t Ci it ith R t L S t d



56

I2

E1

jXM

jX1

R1I1

RCR2(1-s)/s

jX2

R2

Coreloss

Equivalent Circuit with Rotor Losses Separated

S

S R I P

CONV

13 2

2

2

•Rotor Losses= 3*R2I22 or R2 /s=R2 + R2(1-s)/s

•P_conv= Power converted mechanical power

Converted power

Airgap power

2

2

222

2 33 R I S

R I Por PPP conv RCL AGconv

RCL AGconv PPP

Separating Rotor Copper Losses and Power



57

p g ppConverted in induction Machine

2

2

23 R I P RCL

S

S

R R

S R I P

R I S

R I P

PPP

conv

conv

conv

RCL AGconv

1

11

3

33

2

222

222

222

Actual resistive losses inrotor

When referred to stator

Electrical Power convertedto mechanical is expressedas the difference betweenair gap & rotor losses& can

be expressed as twoseparate terms.

A “power conversionresistance” can be

expressed as:

R R RCL R I P2

3

Per phase IM equivalent circuit with rotor



58

I2

E1

jXM

jX1

R1I1

RCR2(1-s)/s

jX2

R2

Core

loss

p qlosses separated from converted power

S

S R I PCONV

13 2

2

2

•Rotor Losses= 3*R2I22

•P_conv= Power converted mechanical power

Converted power

Equivalent Circuit



59

Rotor Circuit

Induction machine equivalent circuit

q

P Effi i d T /R i



60

Power, Efficiency and Torque/Review

The input power Pin supplies the stator resistance & core losses ccoreSCL R E P R I P / 3&32

11

2

1 The air-gap power,

The rotor copper loss,

s

P

s

R I P RCL

ag 2223

2222 3 R I P

The mechanical or converted or developed power output (neglecting mechanical losses),

2222

13 R

s

s I PPP agmech

Power

Efficiency

The input power to the motor is given by, 111 cos3 f I V PinOutput power, lossesmechmechout PPP _

Efficiency,

in

out

P

P

St d d /L



Standards/Losses

• The rating of Electrical machines is determined by mechanical & thermal consideration.

• The maximum current in an IM winding is determined by themaximum operating temperature which the insulation canwithstand without damage or excessive life degradation.

• NEMA Standard MG1-1993

Allowable Temperature rise per insulation classes, OC

Motor Type Class B Class F Class H

Fan Cooled 80 105 125

Non-Ventilated 85 110 135

http://www.nema.org/



Insulation vs. Mean Life of AC Machine

H

o u r s

i n

T h

o u s a n d s

Winding Temperature

Class Activity



63

Class Activity

• A 480V, 60 Hz, 50hp, 3 phase

IM is drawing 60A at 0.85 pf lagging.

• Stator copper losses=2KW,• rotor copper losses=700W,• Mechanical losses=600W, core

(iron) losses=1800 W,determine:

a)Air gap power

b)Converted powerc)Output Powerd)Motor efficiencye)slip

Input power

Psup

Stator Copper loss

3 Ista

2 Rsta

Rotor Copper loss

3 Irot2 Rrot

Stator Iron loss

3 Vsta2 / Rc

Windage, stray

and friction losses

Output power

Pout

Air gap

power Pag

Developed power


Air gap P_converted

ind.m

Pout=load.m

S

R I P AG

22

23

AG RCLPSP

AGConv PSP )1(

Pin=3VphIph*p.f.

=sqrt(3)VLIL*p.f

2

2

2

3 R I P RCL


synchind AG T P

Class Activity



64

Class Activity

• A 480V, 60 Hz, 50hp, 3 phase IM is drawing 60A at 0.85 pf lagging.

• Stator copper losses=2KW, rotor copper losses=700W, Mechanicallosses=600W, core (iron) losses=1800 W, determine:a)Air gap powerb)Converted powerc)Output Powerd)Motor efficiency

e)slip

Input power

Psup

Stator Copper loss

3 Ista2Rsta

Rotor Copper loss

3 Irot2 Rrot

Stator Iron loss

3 Vsta2 / Rc

Windage, stray

and friction losses

Output powerPout

Air gap

power Pag

Developed power


Air gap P_converted

ind.m

Pout=load.m

S

R I P AG22

23

AG RCL PSP

AGConv PSP )1(

Pin=3VphIph*p.f.

=sqrt(3)VLIL*p.f

2

2

23 R I P RCL


synchind AG T P

2KW

700W

600W1800 W

kW

f P I V Pt t in

4.42

..3



Outcomes

• Calculate power and torque in the induction motor as

function of motor operating point.• Calculate Converted power from equivalent circuit.

• Derive the torque-speed characteristics of the inductionmachine.

• Understand and predict the effect of various parameters

(rotor resistance, slip & input voltage) variation on the torquespeed characteristics.

65

7.5 Derivation of Induction Motor Induced



66

Torque-speed Characteristics• Requirement: find induced torque

equation as function of known quantities

(operating point motor speed, inputvoltage & IM impedances.

• The goal is to express the induced torqueas function of circuit parameters (R’s & x’s), operating point (slip) & input

voltage.

• Airgap power is expressed as function of I2. Solving for I2, one can solve for theinduced torque.

• We will use the equivalent circuit of aninduction machine and power flowdiagram to drive a general expression forinduced torque as a function of speed.

syncind AG

S

R I P 22

23

m

conv

synch

AGind

PPT

Derivation of Induction Motor Induced Torque-d Ch t i ti



67

speed Characteristics

• To solve for I2 apply

Thevenin equivalent on IMper phase equivalentcircuit with Rc neglected atnodes a & b

• Thevenin voltage:

For XM+X

1>>R

1

R1

VTHVf

jX1

jXM

f V

I2

E2 jXM

jX1R1

R2 /s

jX2

M

M TH

jX jX R

jX V V

11

.

21

2

1

.

M

M

TH

X X R

X V V

a

b

M

M TH

X X

X V V

1

.

Derivation of Induction Motor Induced Torque-d Ch i i



68

•Thevenin Impedance:

Short circuit sources & solvefor equivalent impedance

XM >> X1

& X1+ XM>> R1

R1

ZTH

jX1

jXM


TH TH

M

M

M

M TH jX R

X X j R

jX R jX

Z Z

Z Z Z

11

11

1

1.

2

1

1

M

M TH

X X

X

R R

1 X X TH

a

b

11

1 )( jX X X

X R jX R Z

M

M TH TH TH

Thevenin Impedance:

Derivation of Induction Motor Induced Torque-d Ch i i



69


• Use Thevenin equivalent

circuit to solve for I2 & substitute in torque Eq.

22

2

2

2

22

2 ||

X X S

R R

V I or

jX jX S

R R

V I

TH TH

TH

TH TH

TH

VTH

E2

jXTH

RTH

jX2

x

x

R2/sI2

2

2

2

2

22

222

3 / 3

X X S

R R

S

RV

s R I

TH TH SYNC

TH

sync

ind

a

b

Induced torque can be expressed as:



71

v_th = v_phase * ( xm / sqrt(r1^2 + (x1 + xm)^2) );

z1=r1 + j*x1;z2=j*xm;

z_th = ((j*xm) * (r1 + j*x1)) / (r1 + j*(x1 + xm));

z=z1*z2/(z1+z2);

r_th = real(z_th);

x_th = imag(z_th);

s = (0:0.02:1); % Slip

s(1) = 0.001;

nm = (1 - s) * n_sync; % Mechanical speed

% Calculate torque for original rotor resistance

for ii = 1:51

t_ind1(ii) = (3 * v_th^2 * r2 / s(ii)) /(w_sync *

((r_th + r2/s(ii))^2 + (x_th + x2)^2) );

end

S C



Plot of Torque Speed Characteristics

72

0 200 400 600 800 1000 1200 1400 1600 1800

50

100

150

200

250

Motor Speed [rpm]

I n d u c e d T o r q u e [ N m ]

Induction motor torque-speed characteristic

Original R2

IM Torque speed Characteristics



73

q p

1. T at synchronous speed is zero.

2. T vs ω is linear between no load and full load.

3. Maximum torque is 2 to 6 times rated full load torquedepending on IM class.

4. Starting torque (s=1) is larger than its full load torque.

5. For a given slip

6. If rotor runs at a speed faster than synchronous speed then Tind is negative, machine acts as a generator.

2V

2

2

2

2

22 3

X X S

R R

S

RV

TH TH SYNC

TH

ind

Rated torque

0 rpm motor speed ns (rpm)

Comments on Torque speed Characteristics



R2 increases

q p

2

2

2

2

22 3

X X S

R

R

S

RV

TH TH SYNC

TH

ind

Comments on Torque speed Characteristics asFunction of Input Voltage



75

Function of Input Voltage

Maximum (Pullout) Torque in IM



76

( ) q

•Max Torque can be

determined as :

•The induced torque is

proportional to PAG; Inducedtorque is max when PAG ismaximum.

• By power transfer theorem is

achieved when sourceimpedance is equal to loadimpedance.

VTH

Zsource

jXTH

RTH

jX2

I2

R2/s

s R Z

jX jX R Z

Z Z

load

TH TH source

load source

/ 2

2

0ds

d

syncind AG

S

R I P 22

23

2

2

2

2

22 3

X X S

R R

S

RV

TH TH SYNC

TH

ind

Maximum Pull out Torque in Induction Machine



77

•Pmax is defined when

source impedance = R2 /s

•Smax is expressed asfunction of equivalentcircuit impedances

•Substitute for Smax intorque equation: S=Smax

•Tmax is expressed as:

Tmax proportional to Vth2.

Tmax is independent of rotor resistance R2. as R2 increseas Tstart increases.

22

22 X X R

S

RTH TH

2

2

2

2max

X X R

RS

TH TH

2

2

2

2

max

2

3

X X R R

V T

TH TH TH SYNC

TH

2

2

2

2

22 3

X X S

R R

S

RV

TH TH SYNC

TH

ind

Classes of Induction MachinesNEMA Standards



78

NEMA Standards

Class A:-High inrush current (500-800% of rated)

-Low starting torque-7.5-200 hP-Applications: Blowers, pumps-Low slip at full load

Class B:-Same starting torque as class A

-Less starting current (special rotor-deep bar)-Same applications as class A

Class C:-Most expensive (Double cage rotor)-High starting torque

-Special applications (Compressors & conveyers)

Class D:-Same as class A but with higher rotor Resistance-High starting torque-Higher slip at full load-Max torque can occur at starting

-Applications: high inertia loads (flywheels)



Review of Torque-speed characteristics



80

• Equivalent Circuit

• Torque-speed equation

1. T at synchronousspeed is zero

2. For a given slip torqproportional V2.

3. Torque speedrelationship is linearnear rated load

11

1 )( jX X X

X R jX RZ M

M TH TH TH

2

2

2

2

22

222

3 / 3

X X

S

R R

S

RV

s R I

TH TH SYNC

TH

sync

ind

VTH

Zsource

jXTH

RTH

jX2

I2

R2/s

M

M TH

X X

X V V

1

.

Review of Torque-speed characteristics



81

• Equivalent Circuit

• Torque-speed equation

1. T at synchronous speedis zero

2. For a given slip torqproportional V2.

3. Torque speedrelationship is linear nearrated load

• Max slip:• Max Torque:

- Proportional to V2

- Independent of load

impedance

111 )( jX X X

X

R jX R Z M

M

TH TH TH

2

2

22

22

222

3 / 3

X X S

R R

S

RV

s R I

TH TH SYNC

TH

syncind

VTH

Zsource

jXTH

RTH

jX2

I2

R2/s

M

M TH

X X

X V V

1

.

22

2

2

max X X R

RS

TH TH

2

2

2

2

max

2

3

X X R R

V T

TH TH TH SYNC

TH

Class Activity



82

Class Activity

• Example 7-4: a 2 pole 50 Hz IM

supplies 15kW (converted power) to aload at speed of 2950 rpm.

a- What is the motor slip

b- What is the induced torque underconditions in a?

c- What will be the operating speed if the torque doubled (Input voltageis constant)? Hint: at low slip,torque is linearly proportional totorque.

d- What is the converted power at thenew operating point.

sync

msync

n

nns

synch

AG

M

conv

ind

PP



Class Activity



84

Example 7-5: A 460V wound rotor,60 Hz, 25hp, 3 phase Y connected IM

has the following impedancesreferred to stator:

R1=0.641 R2= 0.332 X1=1.106X2=0.464 & Xm=26.3 ( all in ohms).

a- What is the max torque of themotor? At what speed does itoccur?

b- What is the starting torque of the motor?

c- The rotor resistance is doubled,what is the new max torque? Atwhat speed does it occur? What isthe new starting torque of the

motor?

2

2

2

2

22 3

X X

S

R R

S

RV

TH TH SYNC

TH

ind

22

2

2max

X X R

RS

TH TH

2

2

2

2

max

2

3

X X R R

V T

TH TH TH SYNC

TH

11

1 )( jX X X

X R jX R Z

M

M TH TH TH

M

M TH

X X

X V V

1

.

I2

E1

jXM

jX1

R1I1

RC

R2(1-s)/s

jX2

R2

Core

loss

7 11 Determining Equivalent Circuit Parameters



85

7.11 Determining Equivalent Circuit Parameters

Equivalent Circuit Parameters for Induction Machine

• DC test

Calculate stator resistance

• No load Test

Approximate all rotational losses (core,windage, friction, misc)

• Locked rotor test

Approximate rotor & stator impedances.

DC Stator test



86

• Apply dc voltage across stator winding.

Calculate dc resistance of stator by applyingDC voltage across L-L winding

Measure dc Voltage & current

2Rstator= Vdc /Idc for Y connection

2Rstator /3=Vdc /Idc for delta connection

Locked Rotor Test



87

• Rotor is locked (S=1)

• Neglect (Xm // Rc) since Xm // Rc>> X2 +R2

• Measure, PLR, VLR, ILR

• Approximate, R2, X1 & X2

• Assume X1=X2 (depends on IM

class)

• R1 is already calculated from DCtest, R2=Req-R1

LRt LRt

LR

V I

P pf

__

1

3cos

1221

2121

'

'

)()(

0

R R R R R R

X X j R R jX R

jX R I

V Z

LR LR

LR

LR LR

LR

LR

LR

LR

I2

E1

iXM

IM

jX1

R1I1

RCR2 /s

jX2

f V

+

-

No load test



88

No load test• Similar to transformer no load test.

• Pnoload=Pcore+PSCL+Pmech_loss

• Pmech_Loss= Pwindage+ Pfriction + Pmisc

• Measure power (Poc), voltage (Voc) & current (Ioc).

• Prot=Pnoload-PSCL=Pmech_loss + Pcore

• where PSCL =R1I2oc jXM

jX1R1I1

RC

+

Voc

-

nL

M c

I

V X

X R

X R

1

1M

1M1

X

X

0)(s load noat

No load test



89

No load test• Similar to transformer no load test.

• Pnoload=Pcore+PSCL+Pmech_loss

• Pmech_Loss= Pwindage+ Pfriction + Pmisc

• Measure power (Poc), voltage (Voc) & current (Ioc).

• Asssume mechanical losses are

provided/approximated or combined in Rc• Pcore=Pnoload-PSCL-Pmech_loss

• where PSCL =RsI2oc

)(cos1

OC OC

OC

I V

P

sin

cos

OC

OC sOC

lm I

I RV

X X

Core

sococ

c P

R I V

R

2)cos(

jXM

jX1R1I1

RC

+

Voc

-

+

Vx

-

For XM>>X1 & XM >>R1nL

c M M

I

V

R X X X

1

M

1

X

1)(sloadnoat&

Conclusion



Conclusion

• Stator and rotor IM construction has been

discussed• Basic IM concept of operation has been detailed

• Construction of IM equivalent circuit has beenexplained

• Power & Torque of IM theory and flow has beenpresented

• IM Torque Speed Characteristics has been derived

• IM torque-speed curve as function of salientpoints and characteristics have been discussed

• Tests to determining IM equivalent circuitparameters have been studied.

Set5 Induction Motor Final

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Transcript of Set5 Induction Motor Final