induction motor 1

AC Machine Stator

‘a’ phase axis

‘b’ phase axis

‘c’ phase axis

1200

1200

1200

induction motor 21 Cycle

Amp

timet0t1 t2 t3 t4

t01 t12

Currents in different phases of AC Machine

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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 (theta) in degrees

t0

t01

t12

t2

a

MMF Due to ‘a’ phase current

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a Fc

-93 10 113 216-1.5

-1

-0.5

0

0.5

1

1.5

a’

c’ b’

b c

a

a’

c’ b’

b c

a

a’

c’ b’

b c

a

a’

c’ b’

b c

Fb

Fa F

FbFc

F

Fa

F

Fb

Fc Fc Fb

F

Space angle () in degrees

FFa Fc

Fb

t = t0= t4

t = t1t = t2 t = t3

t = t0= t4

RMF(Rotating Magnetic Field)

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Video of the unfolded rotating magnetic field

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RMF(Rotating Magnetic Field)-Analogy with DC machines

The salient field structure in DC machines is mimicked along with speed in an AC machines by a multiphase (2 or more) winding. The number of poles are

determined by winding distribution and is independent of the number of phases.

The rotational speed is determined by the supply frequency and the number of poles, such that an observer in air-gap counts same

number of poles per second, meaning the more the number of poles the slower the machine will run and vice-versa.

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Induction Motor•Most popular motor today in the low and medium horsepower range

•Very robust in construction

•Speed easily controllable using V/f or Field Oriented Controllers

•Have replaced DC Motors in areas where traditional DC Motors cannot be used such as mining or explosive environments

•Of two types depending on motor construction: Squirrel Cageor Slip Ring

•Only Disadvantage: Most of them run with a lagging power factor

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Squirrel Cage Rotor

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Slip Ring Rotor

•The rotor contains windings similar to stator.

•The connections from rotor are brought out using slip rings thatare rotating with the rotor and carbon brushes that are static.

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Torque Production in an Induction Motor

•In a conventional DC machine field is stationary and the current carrying conductors rotate.

•We can obtain similar results if we make field structure rotating and current carrying conductor stationary.

•In an induction motor the conventional 3-phase winding sets up the rotating magnetic field(RMF) and the rotor carries the current carrying conductors.

•An EMF and hence current is induced in the rotor due to the speed difference between the RMF and the rotor, similar to that in a DC motor.

•This current produces a torque such that the speed difference between the RMF and rotor is reduced.

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Slip in Induction Motor

•However, this speed difference cannot become zero because that would stop generation of the torque producing current itself.

•The parameter slip ‘s’ is a measure of this relative speed difference

s

s

s

s

n

nns

where ns,s,f1 are the speeds of the RMF in RPM ,rad./sec and supply frequency respectivelyn, are the speeds of the motor in RPM and rad./sec respectively

•The angular slip frequency and the slip frequency at which voltage is induced in the rotor is given by

11

22122 ,, E

N

NsEsffs s turnsRotorNturnsStatorN 21

polesofpp

fns #;

120 1

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Induction Motor Example

A 100 hp, 460V, 8 pole, 60 Hz, star connected3 phase induction motor runs at 891 rpm under full load. Determine the synchronous speed in rpm, slip, slipfrequency (frequency of the rotor circuit),slip rpm at full load. What is the speed of the rotor field relative to (i) rotor structure, (ii)stator structure, (iii) stator rotating field?

Voltage induced in rotor under full load? N2/N1=0.5

Solution on Greenboard

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Induction motor Equivalent Circuit

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Relation between air-gap, gross mechanical power and rotor copper loss

ssPPP mechag :1:1:: 2

Internal efficiency = sP

P

ag

mech 1

Implies lower the slip higher is the induction motor efficiency

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Example problem related to theformula shown in the previous

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Approximate Equivalent Circuit

•Assumes negligible magnetizing current

• Note Rc has been removed.The sum of core losses and the windage and friction loses are treated as constant. This is because as speed increases rotor core lossdecreases (lower f2) but windage and frictionloses increase.With decrease of speed the converse is true. Thus the sum is constant at any speed and is termed as rotational loss.

j

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IEEE Equivalent Circuit

•Assumes 30-50% magnetizing current and drop across R1+jX1 notnegligible

• As before, the sum of core losses and the windage and friction loses are treated as constant.

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Thevnin’s equivalent of theIEEE Equivalent Circuit

• This is done by applying Thevenin’s theorem and treating the rotor side as load

111

21 ,,,

XX

XKXXRKRVKV

m

mthththththth

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Determining equivalent circuit parameters

j

Uses no-load test and blocked rotor tests to determine them

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Example problem related to no-load and blocked rotor test

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Performance Characteristics(1)

syn

ag

mech

mechmech

pf

synsynmech

agmechmechmech

PPT

s

sPss

RITP

14

222

;)1(

);1()1(

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Performance Characteristics(2)

s

R

XXsRR

VT

s

RI

PPT

thth

th

synmech

synsyn

ag

mech

mechmech

'2

2'2

2'2

2

'22'

2

)()/(

1

1

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Performance Characteristics(3)

Case 1: )( zerotoclosesmalls

'2

'2

'2

XXs

Rand

Rs

RThen

th

th

sT

sR

VT

mech

th

synmech

)(

1'2

2

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Performance Characteristics(4)

Case 2: )(arg onetocloseels

'2

'2 XX

s

RRThen thth

sT

s

R

XX

VT

mech

th

th

synmech

1

)(

1 '2

'2

2

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Performance Characteristics(5)Combining case 1 and 2 the approximate torque speed characteristics would look approximately like:

Tmech

Speed (n)

Tmax

nm ns

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Performance Characteristics(6)How to obtain Tmax? By differentiating the following equation with respect to s and equating it to zero.

s

R

XXsRR

VT

thth

th

synmech

'2

2'2

2'2

2

)()/(

1

One can obtain the following:

Slip at maximum torque =])([ 2'

22

'2

max

XXR

Rs

thth

T

)()(

1

2)(2

1

)(2

1

1'2

2

1'2

2

2'2

2

2

max

RsmallLL

Vp

XX

V

XXRR

VT

th

th

th

th

syn

ththth

th

syn

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Performance Characteristics(7)(Speed Control)

Speed control by varying rotor resistance (varyTmax by varying sTmax)

(inefficient)

Speed control by varying supply voltage and frequency

(Vth/1)(efficient)

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Performance Characteristics(8)

s

R

XXsR

VT

th

th

synmech

'2

2'2

2'2

2

)()/(

1

Also using

)( '2

'2

max XX

Rs

thT

and

for small R1 one can write the following:

ss

ss

T

T

T

T

mech max

max

2

22max

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Example problem based on the formula on previous to express maximum torque and starting torque in terms of rated torque

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Performance Characteristics(9)

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Example problem related to efficiency calculation of induction motor based on equivalent circuit parameters

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Related to the problem in the previous slide

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Different modes of IM operation

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Different modes of IM operation

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Example problem on variable frequency supply using a slip-ringinduction motor

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Speed control of SRIM with ext. resistors

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Applications of SRIM

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Wind Power applications of SRIM

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