3-epe452_induction motors-3ph.pdf

8/18/2019 3-EPE452_Induction Motors-3Ph.pdf

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1

ThreeThree--Phase Induction MotorPhase Induction Motor(Asynchronous )(Asynchronous )


2/87

2

Contents

Introduction

Rotating Magnetic field

Construction and Principle of Operation

Equivalent Circuit

Performance Characteristics

Starting Methods

Speed Control


3/87

3

Laws of Electromagnetism

Faraday’s Law

Lenz’s Law

Fleming’s Right Hand rule

Fleming’s Left Hand rule

Interaction of two magnetic fields


4/87

4

Faraday’s Law ofElectromagnetic Induction

When the magnetic fluxthrough a circuit ischanging an induced

EMF is setup in thatcircuit and its magnitudeis proportional to the

rate of change of flux”


5/87

5

Lenz’s Law

“

The direction of an

induced EMF is suchthat its effect tendsto oppose the changeproducing it”

http://h/123/Lenz.avi


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6

Fleming’s Right Hand rule

Used to measure the direction of

induced current in a conductor when

cut by a magnetic field.

Fleming’s Left Hand rule

Used to measure the direction of

motion of a current carryingconductor when placed in

magnetic field.


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7

Interaction of two magnetic fields

Electromagnetics

Stator

Rotor

http://c/Documents%20and%20Settings/user/Local%20Settings/Temporary%20Internet%20Files/Content.IE5/4LU4EGW8/DC_motor.avi


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About 65% of the electric energy in the United States isconsumed by electric motors.

In the industrial sector alone, about 75% is consumed by

motors and over 90% of them are induction machines.

Simple construction, Robust, Cheap

Introduction


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MOTOR

9


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10

Production of three phase

rotating magnetic field (RMF)

RMF may be set up in two-phase or three-phasemachines.

The number of pole pairs must be the same as the

number of phases in the applied voltage.

The poles are displaced from each other by an

angle equal to the phase angle between theindividual phases of the applied voltage.


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11

+ When Current positive and going into

When Current negative and coming from


12/87

12

Rotating Magnetic Field

p f n s

s120

When three-phase balanced currents

are applied to a three-phase winding,

(aa', bb', cc', displaced from each other

by 120 electrical degrees in space), a

rotating magnetic flux is produced.

The speed at which the magnetic flux rotates is called the

synchronous speed ns

,

Where f s is the supply frequency and p is the total numberof poles.


13/87

13

Rotating Magnetic Field (Cont.)

Currents in different phases of AC Machine

1 Cycle

Amp

timet0 t1 t2 t3 t4

t01 t12


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14

MMF due to ac current in phase “a”

Axis of phase a

a’a’

-90 -40 10 60 110 160 210 260-1

-0.8

-0.6

-0.4

-0.2

0

0.

2

0.4

0.6

0.8

1

Fa

Space angle (

) in degrees

t0

t01

t12

t2

at1

a

a

’

Fa

+

.

Axis of

phase a

Pulsating mmf


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15

a

a’

c’ b’

b c

Fc

F b

Fa F

t = t0

a

a’

c’ b’

b c

Fc

F

F b

t = t1

F

a

a’

c’ b’

b c

F b

Fa

Fc

t = t2

a

a’

c’ b’

b c

Fc F bF t = t3

-93 10 113 216-1.5

-1

-0.5

0

0.5

1

1.5

Space angle () in degrees

FFa Fc

F b

t = t0

MMF due to three-phase currents in three-

phase winding

MMF’s at various instant ( Rotating mmf )


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16

b a c

Time

t 1 t 3t 2



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17

b a c

Time

t 1 t 3t 2

ia

ic

ib

b

c

a a

c

b



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18

b a c

Tim

t 1 t 3t 2

ia

ic

ib

b

c

a

max2

3

max

2

3

max23

max2

3

max2

3 max2

3

max2

3

max2

3

max

2

3

At t1

At t 2

At t 3



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19


www.ewh.ieee.org


20/8720

Principle of Operation

If the stator windings are connected

to a three-phase supply; a rotating field

will be produced in the air-gap. This field

rotates at synchronous speed ns. This

rotating field induces voltages in the

rotor windings. Since the rotor circuit is

closed, the induced voltages in the rotor

windings produce rotor currents thatinteract with the air gap field to produce

torque. The rotor will eventually reach a

steady-state speed nm that is less than

the synchronous speed ns.

The difference between the rotor

speed and the synchronous speed is

called the slip, s ,

www.ewh.ieee.org

s

s

n

nnS

s

sS

60

n2 n

in rpmω in rad/s


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21

Analytical expression for the rotating mmf

An analytical expression may be obtained for the resultant mmf wave at

any point in the air gap, defined by an angle .

)120cos()120cos(cos

)()()()(

cba

cba

Ni Ni Ni

F F F F

The currents ia

, i b

and ic are function of time thus

)120cos()120cos(

)120cos()120cos(coscos)(

t NI

t NI t NI F

m

mm

)cos(2

3

)240cos(

2

1)cos(

2

1

)240cos(2

1)cos(

2

1

)cos(

2

1)cos(

2

1)(

t NI

t NI t NI

t NI t NI

t NI t NI F

m

mm

mm

mm

0 /


22/87

22


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23

Construction

1- STATOR

A three-phase windings is put in slots cut on the inner

surface of the stationary part. The ends of these

windings can be connected in star or delta to form athree phase connection. These windings are fed from a

three-phase ac supply.


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24

Construction (Cont.)

2- Rotorit can be either:

a-

Squirrel-cage (brushless)

The squirrel-cage windingconsists of bars embedded inthe rotor slots and shorted at both ends by end rings.

The squirrel-cage rotor is the

most common type because itis more rugged, moreeconomical, and simpler.


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/rotor winding/rotor winding

Short circuits allShort circuits all

rotor bars.rotor bars.

25

Squirrel Cage Induction Motor (SCIM)


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Construction (Cont.)

b-

Slip ring (wound-rotor)The wound-rotor winding has the same

form as the stator winding. The windings

are connected in star. The terminals ofthe rotor windings are connected to three

slip rings. Using stationary brushes

pressing against the slip rings, the rotorterminals can be connected to an

external circuit.

26


27/87

Construction-Wound Rotor Induction Motor (WRIM)

Cutaway in a

typical wound-

rotor IM.

Notice the

brushes and the

slip rings

Brushes

Slip rings

27


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APPLICATION OF SLIP RING

MOTOR

compressor Central air conditoining

29

Wound rotr


29/87

ADVANTAGES OF SLIP RING AND

SQUIRREL CAGE MOTOR

SQUIRREL CAGE SLIP RING

cheaper and more robustslightly

the starting torque is muchhigher and the starting current

much lower

higher efficiency and power

factor

the speed can be varied by means

of external rotor resistors

explosion proof, since the

absence of slip-rings and brusheseliminates risk of sparking.


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31

Principle of Operation

If the stator windings are connected to athree-phase supply; a rotating field will be produced in the air-gap. This field

rotates at synchronous speed ns. Thisrotating field induces voltages in therotor windings. Since the rotor circuit isclosed, the induced voltages in the rotor

windings produce rotor currents thatinteract with the air gap field to producetorque. The rotor will eventually reach a

steady-state speed nm that is less than thesynchronous speed ns.

The difference between the rotor speedand the synchronous speed is called the

slip, S , and is defined as

s

m s

n

n nS


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Definitions

nm = the rotor speed (the motor speed) w.r.t. stator

ns

= the speed of stator field w.r.t. stator or the synch. speed

nr = the speed of rotor field w.r.t rotor

S = the slip

f s = the frequency of the induced voltage in the stator (stator

or supply frequency)

f r = the rotor circuit frequency or the slip frequency

S mS r snnnn Slip rpm

S S mS r r f S nS

pnn

pn

p f )(

120)(

120)(

120


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33

Induced EMF

w p phrms K N f E 44.4

The instantaneous value of the induced voltage

in N turns coil is given by:

)90sin(2)cos(

)sin(

t f N t N e

t Let

dt

d

N e

mm

m

The r.m.s. value of the induced voltage per phase is

where

N ph is the number of turns in series per phase

f is the frequency

p

is the flux per pole

K w is the winding factor

phase voltage is 1/√3

of the normal voltage

phase voltage is equal

to the line voltage.


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341

2

1

2

N

N

E

E

I 1

X 1 R1

N 1

I

E 1V

I c

X c Rc

I r I 2

2 R2

2 E 2

1

2

2

12

I

I

N

N

I

I

•

At standstill

(nm= 0 , S = 1)The equivalent circuit of an induction motor at standstill is the

same as that of a transformer with secondary short circuited.

Equivalent Circuit Per Phase


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35

•

At standstill

(nm= 0 , S = 1)

The equivalent circuit of an induction motor at standstill is the

same as that of a transformer with secondary short circuited.

V 1

I 2’

R1 R2’ X 1 X 2

’

X c Rc

I c I 1

I 2’

E 1

= E 2’ All values are

per phase

Where

E2

= per-phase induced voltage in the rotor at standstill

X2 = per-phase rotor leakage reactance at standstill

Equivalent Circuit Per Phase


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36

Equivalent Circuit Per Phase (cont.)

At any slip S

When the rotor rotates with speed nm

the rotor circuit

frequency will be:

Therefore induced voltage in the rotor at any slip S will

be

E2S

= S E2

, similarly X2S

= SX2

and the rotor equivalent circuit per-phase will be:

S r f S f

sX2 X2

X2

R2

(1-S)/S

I2

sE2 R2

I2

E2 R2

/S

I2

E2

R2

2 2

2

2 2

2 2

jX ) s / R( E

jSX RSE I

Where


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37

I 2 I 1

X 2 X 1 R1 R2 /s N 1 : N 2

E 2

I

E 1

I c

X c RcV

Combined Equivalent Circuit


37/87

38

I 2’

I 1

2’1 R1 R2’ /s

E 1 E 2’

I c

c RcV

2

2

12

' 2

N

N R R

2

2

12

' 2

N

N X X

1

22

' 2

N

N I I

Combined Equivalent Circuit

2’

1 R1 R2’ /s


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39

I 2’

I 1

2’

1 R1 R2’

E 1 E 2’

I c

c RcV )s1(s

R' 2

)s1(s

R Rs

R'

2' 2

'

2

I 2

’

I 1

E 1 E 2’

I c

c RcV

Equivalent Circuit of IM


39/87

40

Equivalent Circuit of IM

I o

V 1

I 1

E 1

'

2 R

S

RS '

2)1(

Stator CircuitStator Circuit Rotor Circuit Rotor Circuit

Airgap & Airgap &

Magnetic Circuit Magnetic Circuit

1 X

'

2 X

m

X

I’2

c R

Load + Load +

Rotational losses Rotational losses

1 R

2’

1R1 R2’


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41

Equivalent to transformer’s

load

Equivalent to transformer’ssecondary windings

Equivalent to transformer’sprimary windings

I 2’

I 1

21 R1 R2

E 1 E 2’

I c

c RcV )s1(s

R' 2

2’

1R1 R2’


41/87

42

I 2’

I 12

’ R2’

E 1 E 2’

1 R1

c

I c

RcV )s1(s

R' 2

I 2’

I 1

21 R1 R2

E 1 E 2’

I c

c RcV )s1(s

R' 2


42/87

43

I 2’

I 1

X eq Req

X c

I c

RcV )s1(s R' 2

' 21eq

'

21eq

X X X

R R R


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44


1-The Complete Equivalent Circuit per phase

I o

V 1

R1 I 1

E 1

'

2 R

S

RS '

2)1(

Stator CircuitStator Circuit Rotor Circuit Rotor Circuit Airgap & Airgap &

Magnetic Circuit Magnetic Circuit

1 X '

2 X

m X

I’2

c R

Load + Load +

Rotation losses Rotation losses

R 2‘

/S

I 1

V 1 X m=2 f 1 Lm

R1 X 1=2 f 1 L1

I

X2’

=2 f 1 L2

I2’

2-IEEE-Recommended Equivalent Circuit


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45


3-Thevenin Equivalent Circuit

In order to simplify the computations, IEEE recommended equivalent circuitcan be replaced by the following Thevenin’s equivalent circuit.

If

12

1

2

1)(

V X X R

X V

m

m

th

2 m1

21 ) X X ( R

1

2

th1

2

m1

m

th

R K R ) X X

X ( R

1 th1

m1

m th V K V

X X

X V

th th

m11

11 m th jX R

) X X ( j R

) jX R( jX Z

1 th X X and

Xth

Vth

Rth

Air gap

X2’

R’2

/S

I2’

DETERMINATION OF THE EQUIVALENT


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46

1. No-Load Test ( S = 0 )

Measured values V oL = V 1L , I L , and Pot = W 1 ± W 2

Calculate the per phase values V 0 , I

and P0= P ot

/3

)sin()cos()cos(0

omoc

o

o I I I I

I V

P

m

m

c

c I

V X

I

V R 00

V 0

I m

X m Rc

I c

I

V

W1A

W2

I.M.

No Load


CIRCUIT PARAMETERS (cont.)



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47

2. Blocked rotor Test (S =1)

Measured values V bL < V 1L , I bL , and Pbt = W 1

± W 2

Calculate the per phase values V b , I b and Pb = Pot /3

V

W1A

W2

I.M.

R1

+R ’2 X 1

+X’2

Z bV b

I b

Blocked

2&

,,

2112

22

21221

b

b

bbb

b

b

b

b

b

b

X X X R R R

R Z X X X I

V Z

I

P R R R


CIRCUIT PARAMETERS (cont.)


47/87

48

Power Flow In Induction Motors

= 3V 1 I 1cos

dev


48/87

49

Power Flow In Induction Motors

S

S R I P

R I P

S

R I P

dev

RCL

AG

)1()(3

)(3

)(3

22

2

2

2

2

22

2

)1(::1:: S S PPPdev RCLGA

P AG

I mV 1

I 2’

R1 R2’ X 1 X 2

’

X m Rc

I c I I 1

I 2’

R2’(1-S)

S

Power Flow In Induction Motor


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50

nput Power (Pin )

tator Losses:

Copper losses (P cu 1 )

Core losses (Piron

)

irgap Power (P g )

eveloped Power (P d ) otor Copper Losses (P cu 2 )

otational Losses (P rotational ) Output Power (P out )




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51

11in cos I V 3P

m

2

iron

1211cu

R

V 3P

R I 3P

sd ' 22'

2g T

s

R) I (3P

gcu Ps R I P '22'22 3

Protational T Pout

d gd T sPss R I P )1()1(3

'

22'2

)1(::1:: 2 ssPPP d cug

I 2’

I 1

X 2’

X 1 R1 R2’

E 1 E 2’

I c

X c RcV )s1(s

R' 2



51/87

Power relations

3 cos 3 cosin L L ph ph

P V I V I

21 13SCLP I R

( ) AG in SCL coreP P P P

2

2 23 RCLP I R

conv AG RCLP P P

( )out conv f w strayP P P P

52


52/87

Example

53


53/87

54

Solution

100 hp, 460 V, 8 pole, 60 Hz, star connected, 3-phase IMV ph

= 460/√3 = 265.6 V

a) Machine parameters

No Load Test :

Vnl

= 460/√3 = 265.6 V, Inl

= 40 A , Pnl

= 4200/3 = 1400 Wcos (nI

) = Pnl

/ (VnI

* Inl

) = 1400/ (265.6*40) = 0.132

nI

= 82.4°

Ic

= Inl

cos (nI

)= 40* cos(82.4°) = 5.27 A , R c

= Vnl

/ Ic

= 50.38 .

Im

= Inl

sin (nI

) = 40* sin(82.4°) = 39.65 A , Xm

= Vnl

/ Im

= 6.7 . Blocked rotor test :V bl = l00/√3 = 57.74 V, I bl = 140 A , P bl

= 8000/3 = 2666.67 W

R 1

+R '2

= P bl

/ (l bl

)2

= 0.136

But R 1

= (0.152/2) * 1.11 = 0.084 , so R '2

= 0.052Z bl

= V bl

/ I bl

= 0.412 .X bl =√{(Z bl

2

-(R 1

+R '2

)2} = 0.39 X1

= X'2

.= 0.39/2 = 0.195


54/87

55

195.0084.0895.6732.1

7.6*)195.0732.1( j

j

j j Z

in

in Z

b) For nm

= 873 rpm

ns

= 120*f s

/ P = 120*60/8 = 900 rpm

S = (ns

- nm

)/ns

= 0.03

IEEE recommended equivalent circuit per phase

1.623 + j0.771 = 1.797 25.41I1 = V 1

/ Zin = 147.75 A

p.f. = cos(25.41)= 0.903 lagPin

= 3 V1

I1

cos() = 3 *265.6*147.75*cos (25.41) = 106308 WP

AG

= Pin

-3I1

2

R l

= 100806 W

PRCL

= s * PAG

= 3024 W

Pdev

= PAG

–PRCL

= 97782 W

Rotational losses = Pnl (total)

– 3 (Inl

)2

R 1

= 4200 –

3 (40)2

(0.084) = 3797 W

Pout

= Pdev

–Rotational losses = 97782-3797 = 93985 W

Efficiency = Pout

/ Pin

= 0.884


55/87

Problem

56


56/87

57

Performance Characterstics

2

2

2'

2

'

2

)( X X s

R R

V I

thth

th

22

2'

2

2'

2

)(

3

X X S

R R

V

S

RT

thth

th

s

Xth

Vth

Rth

Air gap

X2’

R’2

/S

I2’

Torque / speed Curve

Consider Thevenin’s equivalent circuit

The rotor current referred to stator side is:

And the developed torque T is :

ss

AG

s

AG

m

dev R I P

S

S PP

T

2

2

2)(3

)1(

)1(

Speed p.u.


57/87

58

Torque/Speed Curve (cont.)

Maximum torque

Starting torque

Starting Torque:

at starting S =1, so the starting torque is:

22

2

2

2'

2

)(3

X X R RV RT

thth

th

s

St

S Tmax

For maximum torque

which fromdS

dT ,0

2/12'

2

2

'2

max])([ X X R

RS

thth

T

2/12'

2

2

2

max])()[(2

3 X X R R

V T ththth

th

S

Stable

Unstable


58/87

Torque-speed characteristics

Typical torque-speed characteristics of induction motor 59


59/87

60

Torque/Speed Curve (cont.)

)(2

3

)( '2

2

max'

2

'

2

max X X

V

T and X X

R

S th

th

S th

T

)()/()/(

)/()/()(

)()/(

)()/(

max

2

max

'

2

2

max

'

2

2

max

'

2

2'

2

max

2'

2

2

max

'

2

2'

2

2'

2max

T T T

T

T thT

th

S

S

S RS R

S RS R

S

S

X X S R

X X S R

T

T

S S

S S

T

T

T

T

max

2max

2

max

2

If the stator resistance is small, so R th

may be neglected

and the ratio between the maximum torque and the torque developed at

any speed is given by:


60/87

Problem

61

Torque Characteristics


61/87

62


2

2'

21

'

2

eq X s

R

R

V I

I 2’

I 1 X eq Req

X c

I c

RcV )s1(s

R' 2

d d

PT )1()(

3 '22'2 s

s

R I

2

2'

2

1

'

2

2 )1(3

eq X s

R

Rs

s RV

Consider The approximate equivalent circuit. The rotor current referred

to stator side is



62/87

63

Torque

n s

1

0ns

Tst

T

max

smax

s

s

s

s

n

nns

2

2'

21

'

2

2

3

eqs

d d

X s

R Rs

RV PT




63/87

64

smax

T maxT

Torque

s

1

0

st

Small Slip

Large Slip

Maximum Torque


Maximum Torque


64/87

65

Maximum Torque

2

2'

21

'

2

23

eqs

d

d

X s

R

Rs

RV PT

2eq

21

'

2max X R

Rs

2

eq211s

2

max

X R R2V 3T

0s

T d

Set

Starting of IM


65/87

66

Starting of IM

•

Problems: –

High starting current

–

Low starting torque

Starting of Induction Motor


66/87

67

Starting of Induction Motor

2

eq

2'

21

' 2

X R R

V I

st

I 2’

I 1 X eq Req

X c

I c

RcV )s1(s

R' 2

Starting by Reducing Voltage


67/87

68

Sta t g by educ g o tage

Torque

n

n s

s max

T st 1

T max T st 2

V 2 < V 1

V 1

2eq2'

21

' 2

X R R

V I

st

2eq

21

' 2

max X R

Rs

2eq

211s

2

max

X R R2

V 3T

22212

23

eq

'

s

'

st

X R R

RV T

Starting by Reducing Voltage


68/87

69

g y g g

•

Starting current is reduced (good)

•

Starting torque is reduced (cannot start heavy

loads)

•

Maximum torque is reduced (Motor acceleration

is low)•

Speed at maximum torque is unchanged

Starting by Adding Rotor Resistance


69/87

70

g y g

Torque

n

n s

s max

T st1 T st3=T max

T st2

Radd3 Radd2

Radd1

Radd3 > Radd2 > Radd1

2eq2'

21

' 2

X R R

V I

st

2eq

21

' 2

max X R

Rs

2eq

211s

2

max

X R R2

V 3T

22212

2

3

eq

'

s

'

st

X R R RV T

Starting by Adding rotor resistance


70/87

71

g y g

•

Starting current is reduced (good)

•

Starting torque is increased (good)

•

Maximum torque is unchanged (Motoracceleration is high)

•

Speed at maximum torque is reduced

Methods Of Impro ing Starting Torq e


71/87

72

Methods Of Improving Starting Torque

In creasing the resistance of the rotor conductors

Using a combination of high and low resistance

Conductors

Using a wire wound rotor connected to variable

resistor

Torque/Speed Curve for varying R


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73

Torque/Speed Curve for varying R2

The maximum torque isindependent of the rotorresistance. However, the value of

the rotor resistance determinesthe slip at which the maximumtorque will occur. The torque-slipcharacteristics for various values

of are shown.

To get maximum torque atstarting::

)(

)(23

'

2

'

2

max

'

2

2

max

X X

RS

X X V T

th

T

th

th

S

)(..1 22max X X ReiS thT

T

NL~nsn

T

r1nr2nr3nn

r3n


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Maximum Torque

Effect of rotor resistance on torque-speed characteristic74


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75

Stator current and input power factor

From the IEEE recommended equivalent circuit, the input impedance Zinis:

The stator current I1

is :

And the input power factor is :

p.f. = cos ()

inin

m

m

in

Z Z

X X jS

R

jX S

R jX

jX R Z '

2

'

2

'2

'

2

11

in Z V I 11

R 2‘ /SV 1 jX m

R1 jX 1 jX2

’

I 1

I 1

T

p.f.

01

s

T st

art

No-load Starting

Typical Induction MotorTypical Induction Motor

CharacteristicsCharacteristics

Maximum or

pull-out torque


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76

Classes of squirrel-cage motors

According to the National Electrical Manufacturing Association

(NEMA) criteria, squirrel-cage motors are classified into class A, B,

C or D. The torque-speed curves and the design characteristics for

these classes are :

Rated

Load Slip

Starting

Torque

Starting

Current

Class

< 5% Normal Normal A

< 5% NormalLow B

< 5%HighLowC

8-13 %Very HighLow D

A D

C

B

Speed control of induction motorsSpeed control of induction motors


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77


1- Stator voltage Control

Rotor

T

Vs2

i

a

i

L

v A

N

T

1

T

4

v B

N vC

N

T

3

T

6T

5

T 2

i

b

i

c

N B

C

A

Rotor

T

NL~nsn

T

r1nr2nr3nn

r3n>r2n>r1n

1V

2V

3V

3V>2V>1V

V decreasing



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78


2- Stator Frequency Controlnm=(1-S)ns , ns= 120 f s /P

Torque-Speed curves at different

stator frequencies with constantvoltage supply

Rotor

i L

N B

C

A

Rectifie

r

AC→

DC

Inverter

DC →

AC

f Control

filter

ns1

ns3

1

3

2

1 > f 2 > f 3

Speed

Torque

ns2

2

2

21

2

23

eq

'

s

'

d

X s

R Rs

RV T


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79


3. Rotor voltage Control (Slip-Energy Recovery)

For wound rotor induction motor only

Advantages: Advantages:• Very simple

• Very useful at starting:

(high starting torque & low starting current)

Drawbacks: Drawbacks:• Low efficiency:extra losses

• Slow control

• unbalance problems if the three

resistors are unequal.

a-

Classical technique:

Variation of rotor circuit resistance


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80


b-

An alternative methods of varying rotor-circuit resistance

Using three-phase diode bridge and a single variable resistor


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81


c-

Slip-Energy Recovery System

Diode Rectifier Controlled Inverter

Starting of Induction Motors


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82


1-

Direct-On-Line Starting

Induction motors when started by connecting them directly across

the

supply line take 5 to 8 times their full-load current. This initial excessive

current may cause large line voltage drop that affects the operation of otherelectrical equipment connected to the same lines. This method is

suitable

for small motors up to 10-hp rating.

2-

Auto-transformer

A three-phase step-down autotransformer may be employed as a reducedvoltage starter. As the motor approaches full speed, the autotransformer is

switched out of the circuit.



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83

R ext

decreases


3-

Star-delta methodThe normal connection of the stator windings is delta

while running. If these windings are connected in star

at starting, the phase voltage is reduced, resulting in

less current at starting. As the motor approaches thefull speed, the windings will be connected in delta.

4-

A solid-state controller

The controller can provide smooth starting.

5-

Using an external resistance in series with the rotor

For wound rotor induction motor only. The external

resistance can be chosen to get high torque and low

current at starting and can be decreased as motor

starts up.

SINGLE PHASE INDUCTION


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The single-phase induction machine is the most

frequently used motor for refrigerators, washing

machines, clocks, drills, compressors, pumps, andso forth.

The single-phase motor stator has a laminated iron

core with two windings arranged perpendicularly.

One is the main and

The other is the auxiliary winding or starting

winding

MOTOR

84

Nameplate


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85

Nameplate



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MOTOR

86

MOTOR


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MOTOR

This “single-phase”

motors are truly two-

phase machines.

The motor uses a

squirrel cage rotor,

which has a laminated

iron core with slots.

Aluminum bars are

molded on the slots and

short-circuited at bothends with a ring.

Stator with laminated

iron core Slots with winding

Bars

Ring to short

circuit the barsStarting winding

+

_

Main winding

Rotor withlaminated

iron core+

_

Single-phase induction motor.

87


MOTOR


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MOTOR

Squirrel cage rotor