6. induction motors - university of florida · an induction motor has the same physical stator as a...

Induction MotorsInduction Motors

Revised October 6, 2008

EEL 3211 (© 2008, H. Zmuda) 6. Induction Motors 1

Induction Motors:

We just learned how damper or amortisseur windings on a synchronous motor could develop a starting torque without the necessity of supplying an external current to the field windingnecessity of supplying an external current to the field winding.

This idea works so well that a motor can be made without the synchronous motors main field windings at all.

A machine with only amortisseur windings is called an induction ac e w o y a o sseu w d gs s ca ed a inductionmachine.

These machines are called induction machines because the rotorThese machines are called induction machines because the rotor voltage, and hence the rotor current and rotor magnetic field is induced in the rotor windings instead of being physicaly connected


by wires.

Induction Motors:

The distinguishing feature of an induction machine is that no DC field current is required to run the machine.

Although an induction machine can be used as either a motor or a generator, it has many problems when used as a generator and so these are rare.

For this reason, when studying induction machines are really o s easo , w e s udy g duc o ac es a e ea ystudying induction motors.

An induction motor is called a singly excited machine (as opposedAn induction motor is called a singly excited machine (as opposed to a doubly excited machine) because power is applied only to the stator.


Induction Motors:

An induction motor has the same physical stator as a synchronous machine but with a different rotor construction.

There are two different types of induction motor rotors.

1. A squirrel cage induction motor rotor consists of conducting bars laid into slots carved in the face of the rotor and shorted at either end by end rings or shorting rings (see picture next page).e d by e d gs o s o g gs (see p c u e e page).

2. A wound rotor that has a complete set of three phase windings that are mirror images of the windings on the statorthat are mirror images of the windings on the stator.


Al or Cu bars

End Rings

Squirrel Cage Rotor


LaminatedStator Pole

End Rings

Shaft

Squirrel Cage Rotor

Embedded Al or Cu barsLaminated Fe


X

X

X

AppliedAppliedAC Voltage


Induction Motors:

A wound rotor that has a complete set of three phase windings that are mirror images of the windings on the stator. The three phases of the rotor windings are usually Y Connected and the ends ofof the rotor windings are usually Y-Connected, and the ends of the three-phase wires are tied to slip rings on the motor’s shaft.

The rotor windings are accessed through brushed riding on the slip rings.

Such a rotor arrangement is more expensive, requires maintenance, and hence is rarely used.

We will not consider wound rotors any further in this course.


Induction Motors – Basic Operation:

Operation is basically the same as that of synchronous motors with amortisseur (damper) windings.

There are a few specific terms that go along with induction motors.

As before, a three-phase set of voltages is applied to the stator, and a three-phase set of currents thus flows. These currents produce a stator magnetic field BS rotating in a counterclockwise direction.s a o ag e c e d S o a g a cou e c oc w se d ec o .

The speed of the magnetic field rotation as previously found is:

120 , , rev/minelectricalm m

fn P poles nP


P

Induction Motors – Induced Torque:

As the rotating magnetic field passes over the rotor bars it induces a voltage in them equal to:

B

ind rele v B

SB

Rotor (Stator not shown)



The motion of the rotor relative to the stator that produces the induced voltage.

v

SB

B

relv Current out of page

ind rele v B


v


relv Current into page


But since the rotor is inductive, current lags the induced voltage

MaximumI

SB

Maximuminduced currentRI




This rotor current produces its own magnetic field BR.

I

SB RI

RB

Rotor Only


(Stator not shown)


The resulting torque is counterclockwise: ind R SkB B

SB

RB

Rotor Only


(Stator not shown)


Note how the maximum motor speed is the synchronous speed.

If the motor however was tuning at synchronous speed then theIf the motor however was tuning at synchronous speed then the relative velocity between the rotor and stator would be zero. With zero relative velocity,

,ind rel ind R Se v B kB B

and the rotor would slow down due to friction But then the relative

0 0 0 0 0rel ind R R indv e I B

and the rotor would slow down due to friction. But then the relative velocity would no longer be zero and the rotor would once again accelerate.


Induction Motors – Rotor Slip

Consequently the motor will speed up to near synchronous speed but never actually reach it.

Note also that both the rotor and stator magnetic fields rotate together at synchronous speed nsync, while the rotor itself always turns at a slower speed.

This called rotor slip.s ca ed oto slip.

In a synchronous motor, the voltage induced in a rotor bar depends on the speed of the rotor relative to the magnetic fieldson the speed of the rotor relative to the magnetic fields.

The speed of an induction motor depends on the rotor voltage and


current.


The difference between the synchronous speed (the speed of the magnetic fields) in an induction motor and the rotor speed is known as the slip speed nknown as the slip speed nslip.

slip sync mn n n

The notion of slip s is the slip speed on a per-unit basis:

slip sync m

1 1slip sync m mm sync

sync sync sync

n n n ns s n s nn n n

If the motor is tuning at synchronous speed then s = 0. If the motor is stationary then s = 1.

y y y



Miscellaneous relationships:

sync m sync mn ns

n

sync syncn

1 1m sync m syncn s n s


Electrical Frequency of the Rotor

The induction motor operates by inducing voltages and currents in the rotor and as such can be viewed as a rotating transformer.

The stator serves as the primary,

The rotor serves as the secondary,

BUT the secondary frequency is not necessarily the same as the U e seco da y eque cy s o ecessa y e sa e as eprimary frequency (unlike a transformer).



If the rotor is locked and cannot move, the frequency of the rotor current will be the same as that in the primary.

If the rotor turns at synchronous speed, the frequency of the rotor current will be zero.

What is the rotor frequency for any other speed of rotor rotation?



sync m sync mn n f fs

n f

sync syncn f

0 , 1m r en f f s

0, 0m sync rn n f s

r eff s



120 , sync mesync r e e

sync

n nfn f sf fP n

1 02

sync

sync mr e

n nf ff

1 02 e

P

fP

Notation: f denotes the electrical frequency of the rotor while f is

1 02r sync m

Pf n n

Notation: fr denotes the electrical frequency of the rotor, while fm is its mechanical frequency. This is why it’s smart to use nm (in rpm) for mechanical speeds and fr (in hertz for electrical


frequency).

Circuit Model (per Phase) for the Induction Motor

Because of the transformer-like operation, the model for the induction motor will be essentially that of the circuit model for a transformertransformer.

Because the induction motor is a single excited machine (no power applied to a field circuit), the model will not have an internally generated voltage EA as we had in the synchronous (doubly excited) machine.e c ed) ac e.

To develop the model for the induction motor, let’s start with the transformer model and go from theretransformer model and go from there.



Recall the transformer model from Note Set 3:

PjXPR SjXSR

PN SNCRMjX P SCRMjX

IDEAL



jXI Stator Leakage

1jX1R RjX1I RI

I

2I

a

RRCRMjXV

EE

MI effa

RCRMjXPV

RE

1E

Magnetizing Reactance (much smaller than that of a good transformer because...)

IDEAL



Magnetizing Reactance:

Transformer

Induction Motor

Y



Note how the slope of the motor’s flux-mmf curve is much shallower that that of a good transformer.

This is because of the presence of an air gap in the motor, not present in a transformer, which greatly increases the reluctance of the flux path and thereby reducing the coupling between primary and secondary.

The higher reluctance means that a higher magnetizing current is needed to obtain a given flux level.

This results in a much smaller value of XM than in an ordinary transformer.



The major difference between the model of the induction motor and that of a transformer is due the difference in frequency between th i d th dthe primary and the secondary.

How do we model this?

In general for an induction motor, a voltage is applied to the stator windings inducing a voltage in the rotor windings.windings inducing a voltage in the rotor windings.

The greater the relative motion between the rotor and stator magnetic field the greater the resulting rotor voltage and rotormagnetic field, the greater the resulting rotor voltage and rotor frequency, since

,i d lPe v B f n n


,120ind rel r sync me v B f n n

Rotor Model

The largest relative motion occurs when the rotor is locked or stationary. The largest voltage and rotor frequency occur under thi ditithis condition.

The smallest voltage (zero) and frequency (zero) occur when the rotor turns at the same speed as the stator magnetic field yielding no relative motion.

The magnitude of the induced voltage and rotor frequency at any speed between these extremes is directly proportional to the rotor slip:slip: ,R LR r eE sE f s sf


Locked Rotor

Rotor Model

The rotor resistance is essentially constant, independent of slip. The rotor reactance is not.

RjX sRI

RRR RLE sE ~


Rotor Model

For a rotor inductance LR, 2R R R r RX s L f s L

but r ef s sf

thus 2X s s L sf L

2

2R R R e R

e R LR

X s s L sf L

s f L sX

Locked Rotor Reactance


Rotor Model

LR RjsX jX sRI

R

RRR RLE sE

~


Rotor Model

Note that the rotor current is:

E sE E

R LR LRR

RR R R LRLR

E sE EI s RR jX s R jsX jX

LR

RR eq LR

sRZ s jX

It is thus possible to model all of the rotor effects due to varying rotor speed as being caused by a varying resistance supplied by a

,R eq LRjs

rotor speed as being caused by a varying resistance supplied by a constant voltage source ELR.


Rotor Model

jXFinal Rotor Model: LRjX

RI

RRLRE ~ sLRE

Note that at very low slip the resistive term will dominate and the rotor current will vary linearly with slip At high slip therotor current will vary linearly with slip. At high slip, the inductor dominates, and the rotor current approaches a steady-state current.


Final Equivalent Circuit for Induction Motor

All rotor quantities referred back to the stator side.

1jX1R2

2eff LRja X jX1I 2I

2 RR R MI

22

Reffa R

s

CRMjXV

1E


Equivalent Circuit for Induction Motor

The rotor resistance RR and locked rotor reactance XLR are verydifficult to determine for induction motors as is the effective turns

tiratio aeff.

Fortunately, it is possible to take measurements that give R2 and X2even though RR, XLR, and aeff remain unknown.


Power and Torque in an Induction Motor

1jX1R 2jX1I

2R sCRMjXV 1EeqZ

1 1 2 21

eq C M

VZ R jX R jX R s jX

I


1I

Power and Torque in an Induction Motor

1 1 2 2

1eq C M

VZ R jX R jX R s jX

I

1 11

1 1R jX

R jX R s jX

2 2

1 11

1 1

C MR jX R s jX

R jX

2 21

1 1R s jX

R jX

1 11

1

C MR jX

R jX

EEL 3211 (© 2008, H. Zmuda) 6. Induction Motors 392 2

1C MG jB

R s jX

Power and Torque in an Induction Motor – Losses:

3in T L

St t C L

P V IP

StatorCopperLoss

CoreLosses

P

P

converted induced mechanical

RotorCopperLoss

AirGap PP

P

RotorCopperLoss

Friction and WindageLossP

P

.Stray and Mis

out load mechanic

c Losse

l

s

a

P

P



1jX1R 2jX1I

R sRjXV

EZ

Stator Copper Losses

2R sCRMjXV

1E

eqZ

Core Losses

1VZ R jX

1 1

1

2 2

1eq

C M

Z R jXI G jB

R s jX


2 2j


Stator Copper Losses: 21 13SCLP I R

Core Losses: 213Core CP E R

Power in Air Gap: AG in SCL CoreP P P P

Looking at the model, the only place for the air gap power to be consumed is in the resistor R2/s, thus

2 223AG

RP Is


s


The actual resistive losses in the rotor are:23RCL R RP I R

But the power is unchanged when transferred across an ideal transformer. Thus this rotor copper loss can also be expressed as:

RCL R R

22 23RCLP I R

The electrical power converted to mechanical power is thus,

.conv AG RCLP P P


Power and Torque in an Induction Motor:

Converted Power:

P P P.

2 223 3

conv AG RCLP P PRI I R

2 2 2

2

3 3

13 1

I I Rs

I R

2 2

2

3 1

1

I Rs

s

22 2

13 sI Rs



Note:

2 22 13 3RP I I R 2 2 23 3AGP I I Rs s

22 23RCLP I R

1 ,AG RCL RCL AGP sPP P

Therefore the lower the slip, the lower the rotor losses.

AG RCL RCL AGs



Note also that if the rotor is not turning, s = 1, and the air gap power is entirely consumed by the rotor. Clearly if the rotor is not t i th h i l t t i Alturning, the mechanical output power is zero. Also,

.conv AG RCLP P P

1AG AG

AG

P sPP s

If the friction, windage, and miscellaneous losses are known then,

1AGP s

. & .out conv F W miscP P P P


Power and Torque in an Induction Motor:P

Induced Torque: .convind

m

P

Recall: 1m syncs

and . 1conv AGP P s

1AGconv AGP sP P

.

1AGconv AG

indm sync syncs

PAGind

sync

P



Also notice that we can separate the rotor copper losses and the power converted in the equivalent circuit.

. 1conv AG RCL AGP P P P s

The power crossing the air gap is partly consumed in the rotor copper losses and the rest is converted to mechanical power to drive the shaft. Recall that the air gap power was found asdrive the shaft. Recall that the air gap power was found as

2 223AG

RP Is

while the rotor loss was found to bes

22 2RCLP I R



Thus,

AG RCLP P P .

2 222 2 2

conv AG RCLP P PRI I R

22 2

1 1

s

I R

2 2

2 1 11

ssI R R R R

2 . . 2 2, 1conv convI R R R Rs s



Also notice that we can separate the rotor copper losses and the power converted in the equivalent circuit.

1jX1R 2jX 2R Rotor Copper Loss

1 s CRMjX 2

1 sRs

ConvertedConverted


Torque-Speed Characteristics of an Induction Motor:

How does the torque of an induction motor change as the load changes?

How much torque can an induction motor supply at starting conditions?

Hw much does the speed of an induction motor shaft drop as the load increases?load increases?

All these questions can be answered with our circuit model.

Recall two important results: .conv AGind

m sync

P P


m sync


Since sync is constant, if we know the air gap power, we know the induced torque.

PAGind

sync

P

The air gap power is the power crossing the gap from the stator to the rotor and is equal to the power absorbed in the resistance R2/s. This power be found via simple circuit analysis.This power be found via simple circuit analysis.

Consider just one phase for the analysis.



Compute the Thevenin Equivalent Circuit at the place shown below.

1jX1R 2jX1I

2I

RRjXV

E Z

2R sCRMjXV

1E

THZ

22 2AGP I R s



By voltage divider:

Z

1

MTH

M

ZV VZ Z

1 1

C M

C M

R jXVR jX R jX

1 1 C M

C MjR XVR R X X j R X X R R X

1 1 1 1C M M C C MR R X X j R X X R R X



For small core losses CR

1 1

C MTH

M M

jR XV VX X R XR R j X X

1 1

1 1M M

C MC C

R R j X XR R

jX

1 1

M

M

jXVR j X X

also, is often the case that 1 1 MR X X



To a very good approximation,

1

MTH

M

XV VX X

Thevenin Impedance:

1 1TH C MZ R jX R jX



1 1C M

C M

jR XR jXR jX

Z

1 1

C MTH

C M

C M

jZ jR XR jX

R jX

1 1

C M

C M

R jXjR X R jX

R R X X j R X R X R X

1 1 1 1

1 1

C M C M C M

C M

R R X X j R X R X R X

jR X R jX

1 1

1 1M M

C MX X R XR R j X X

R R


C CR R


Using the same approximation as before,

CR

1 1

1 1

MTH

M

jX R jXZ

R j X X

and

1 1 MR X X

1 1MTH

X R jXZ

1TH

M

ZX X



Thus,1 1M M

TH TH THX R X XZ R jX j 1 1

TH TH THM M

Z R jX jX X X X

2Z

2jX2I

THZ

2R s~MTH

XV V 2R s

1TH

MX X



Clearly,2

2

TH THV VI RZ Z

22

2

2

THTH TH

RZ Z R jX jXs

V

2 22 2 2

22

THVI I

RR X X

22

2

TH THR X Xs

R

2 223AG

RP Is



Clearly,

2 23 RP I 2

2

3

3

AGP Is

V R

2

222

2

3 TH

TH TH

V RsRR X X

2TH TH

AG

R X Xs

P

Gind

sync



Plot ind vs. nm. Use:

2

22

3 THind

V RsR

222

syncTH TH

sRR X Xs

120r sync mPf n n

1 1m sync m syncn s n s



Plot ind vs. nm. Use:

23 V

2

222

3 THind

sync

V RsRR X X

2

2y

TH THR X Xs

n n

sync m

sync

n ns

n

120 1202 60

syncsync sync sync syncn f P n

P P



Plot ind vs. nm.

2

2

22

22

3 THind

sync

V RsRR X X

2

2

23

yTH THR X X

s

V R

22

22 2

3

60

TH

sync msync sync

TH TH

V Rn nn P nR R X X

2 2 60 yTH TH sync

sync m

R R X X nn n


Observations:

0m sync indn n


Observations:

Nearl linearNearly linearbetween no-loadand full-load

This is because in this range the rotor resistance is much larger than th t t th tthe rotor reactance, so the rotor current, the rotor magnetic field, and the induced torque increase linearly with increasing slip s


with increasing slip s.

Observations:

Maximum possible orbreakdown torque.

(We’ll compute thismomentarily.)


Observations:

The starting torque is greater than the full-load torque.


Observations:

Since2 2

ind THV V

The torque for a given slip varies as the square of the applied voltage.

ind TH

This is an important observation for one form of motor speed control to be discussed later.


Observations:

If the rotor is driven faster then synchronous speed, the direction of the induced torque reverses and the machine becomes a generator.

If the motor is turning backwards relative to the direction of the stator magnetic fields, the motor will stop very rapidly and try to t i th th di titurn in the other direction.

Recall that reversing the direction of magnetic field rotation is simply a matter of switching any two stator phases.

This method is used to rapidly stop an induction motor and is calledThis method is used to rapidly stop an induction motor and is called plugging.


Observations: Overdriven Motor


Observations: A plot of Power and Torque

.conv m indP


Maximum, Pullout, or Breakdown Torque

SinceAG

indsync

P

Maximum torque occurs when the air gap power is maximum.

sync

The air gap power is the power consumed by 2R s

When is this power maximum?



We can’t change VTH, ZTH, or X2, or R2.

sourceZ

2jX2I

Z

R

THZ

X R jX2R s~

MTH

XV VX X

1 1

1

MTH

M

X R jXZ

X X

1 MX X



We can’t change VTH, ZTH, or X2, or R2.

I2IsourceZ

2R s~THV

22 2AGP I R s



22 2

2

AGP I R s

V

2

222

TH

TH

V R sZ R s

2

22 2THV R s

R R s X X

2 2TH THR R s X X



2 2 2R R s X X R R s R sP

2 2 2 2222 2

22 2

20TH TH THAG

TH

TH TH

R R s X X R R s R sP VR s R R s X X

22 2 2 2 22 2 2 2 2

22 2 22 2

2 2 2 0

0TH TH TH TH

TH TH

R R s R R s X X R R s R s

R R s X X

2 2

22 2 22 2

TH TH

TH THR s R X X

R s Z

2 sourceR s Z



2 222 2max source TH THR s Z R X X

The slip at pullout torque is:

2max 22

RsR X X

2TH THR X X

23 THV max 2 max2 2

2 max 2

3 THAG

TH TH

VP R sR R s X X

maxAGpullout

sync

P


sync


2

max

2

2 max2 22 max 2

3 THAG

TH TH

VP R sR R s X X

2 max 2

222 2max

TH TH

source TH THR s Z R X X

max

2

2 2

3 TH sourceAG

V ZP

R Z X X

2

23TH source TH

TH source

R Z X X

V Z

222

2 2TH TH source TH sourceR X X Z R Z



23 TH sourceV ZP

max

2

2222 2

AGTH TH source TH source

PR X X Z R Z

2

23sourceZ

TH sourceV Z 2

23

2 2source TH source

AG

Z R ZP V

max 3

2AG TH

pulloutsync sync source TH

P VZ R



Since

2max

Rs

th li t hi h i t i li l ti l t

maxsourceZ

the slip at which maximum torque occurs is linearly proportional to the rotor resistance though the value of the maximum torque is independent of the value of the rotor resistance.

max

23AG THP V

max

2TH

pulloutsync sync source THZ R



Recall the plot on Slide 65:


Torque-Speed Characteristics

Also recall the torque speed characteristics derived and examine its dependence on R2:

23 V

22

2

3 THind

sync msync sync

Vn nn P nR X X

R

R

2 2 60

y syncTH TH sync

sync m

R X X nnR

n



If we could somehow vary R2 we would find this kind of behavior (verify this for yourself by plotting the equation on the last slide):



If a rotor is designed with high resistance, the starting torque is quite high, but the slip is also quite high under normal operating

di iconditions.

Recall from Slide 47: . 1conv AGP P s

The higher the slip, the smaller the fraction of air-gap power converted to mechanical power thus the smaller the efficiency

converted to mechanical power, thus the smaller the efficiency.

Hence a motor with high starting torque has poor efficiency at l ti dnormal operating speeds.



Alternatively a rotor is designed with low resistance has a low starting torque and a high starting current but its efficiency at

l i d i i hi hnormal operating speed is quite high.

It would be very desirable to have a motor with a high R2 on starting y g 2 gbut a low R2 at normal speed.

Large bars (small R ) near surface

Small bars (large R ) near surfaceR2) near surface

(small X2) Pullout torque near synchronous

R2) near surface (X2 still small ) Pullout torque near synchronous synchronous

speed. Good efficiency.

yspeed. Poorer efficiency.



The desirable feature of having a motor with a high R2 on starting but a low R2 at normal speed can actually be accomplished by taking d f h l k i h d i f i d iadvantage of the leakage reactance in the design of an induction

motor.

Keep in mind:

Mechanical Speeds: 1n s n Mechanical Speeds:

El t i l F i

1Rotor Statorn s n

f sfElectrical Frequencies: r syncf sf



An ingenious and simple way of obtaining a rotor resistance which will automatically vary with speed makes use of the fact that at

d ill h f l h f hstandstill the rotor frequency equals the stator frequency; as the motor accelerates, the rotor frequency decreases to a very low value, perhaps 2 or 3 Hz at full load in a 60-Hz motor. p p

Squirrel-cage rotors can be designed so that their effective resistance at 60 Hz is several times their resistance at 2 or 3 Hz The variousat 60 Hz is several times their resistance at 2 or 3 Hz. The various schemes all make use of the inductive effect of the slot-leakage flux on the current distribution in the rotor bars.

One way that this happens is by the skin effect.


Skin Effect: Gives depth of penetration into a (real) conductor as a function of frequency asfunction of frequency as

2 1f

z

oI z I e

oI value at surface

1

oI I e

Perfect Conductor

Real Conductor

z

Current flows entirely on the conductor surface.

Current penetratesinto the conductor.


Skin Effect:

Current distribution due to skin effect. The higher the frequency the more the current flows on the conductor surface giving a smaller effective area and hence a larger R


smaller effective area and hence a larger R2.


Leakage Resistance Skin Depth:

2 12 1f

Resistance:

R A



Leakage Resistance

d

d

w w

2 1f


f


Leakage Resistance

dA wd 2 2A A w d

dA wd 2 2effA A w d

w w

2 1f


f


Leakage Resistance7 85.813 10 1.72 10C C 5.813 10 1.72 10Cu Cu

22 4R A d 2, 2 4effR A w dA

22 4 1 12 4

effAC

DC eff

AR A wd wdR A w d

w dA f f

A f f


Torque-Speed Characteristics - Leakage Resistance



Leakage Reactance: Recall that the reactance X2 in an induction motor equivalent circuit is the rotor’s leakage reactance referred b k h i (S Slid 32 36)back to the primary. (See Slide 32, 36)

2X a X

The leakage reactance is due to the rotor flux lines that do not couple

2 eff LRX a X

The leakage reactance is due to the rotor flux lines that do not couple with the stator windings.

Th f th t b ( t f t b ) i f th t tThe farther away a rotor bar (or part of a rotor bar) is from the stator, the greater its leakage reactance.



Leakage Reactance: If the squirrel cage bars are placed close to the surface of the rotor, the leakage reactance will be small (X2 will be

ll) If h b l d d i h f f hsmall). If the rotor bars are placed deep in the surface of the rotor, the leakage will be greater and X2 will be larger.


X

Torque-Speed CharacteristicsX

Consider a squirrel-cage rotor havingdeep, narrow bars:

Low Leakage(X2 small)(X2 small)

X

Large LeakageLeakage inductance for the bottom portion of the baris greater than that of the top portion becauseth b tt i h l fl li k


(X2 large)the bottom experiences has less flux linkage.


Because of the variation in the reactance, under ac conditions the current in the low-reactance upper layers will be greater than that i h hi h l lin the high-reactance lower layers.

As a result, the current will be forced toward the top of the slot, and pthe phase of current in the upper layers will lead that of the current in the lower ones.

This non-uniform current distribution results in an increase in the effective bar resistance.



Low X2 More currenthere

Less current

Rotor Bar

High X2

Current distribution

Current distribution for f > 0. Note how the current is forced

here

is uniform for f ~ 0(Low R2)

to the top reducing the effective area and thus increasing R2 for low speeds (large fr).



The result is that the current in the low-reactance upper layers will be greater than that in the high-reactance lower layers. As a result, the current will be forced toward the top of the slot and the phase ofcurrent will be forced toward the top of the slot, and the phase of current in the upper layers will lead that of the current in the lower ones.

This non-uniform current distribution results in an increase in the effective bar resistance and a smaller decrease in the effective leakage inductance of the bar.



Consequently a squirrel-cage rotor with deep bars can bereadily designed to have an effective resistance at stator frequency

(corresponding to rotor standstill conditions) several times greater(corresponding to rotor standstill conditions) several times greater than its dc resistance.

As the motor accelerates, the rotor frequency decreases and therefore the effective rotor resistance decreases, approaching its dc value at small slips.p



An alternative way of attaining similar results is the double-cage arrangement

shown on the next slideshown on the next slide.

The squirrel-cage winding consists of two layers of bars short-circuited by end rings. The upper bars are of smaller cross-sectional area than the lower bars and consequently have higher resistance.

As before, the inductance of the lower bars is greater than that of the upper ones At standstillupper ones. At standstill,


Torque-Speed Characteristics – Double Cage Rotor Design

An alternative way of attaining similar results is the double-cage design:

Top Bar (High R2)

Bottom Bar (Low R2)( 2)



At standstill, when rotor frequency equals stator frequency, there is relatively little current in the lower bars because of their high reactance; the effective resistance of the rotor at standstill is thenreactance; the effective resistance of the rotor at standstill is then approximately equal to that of the high-resistance upper layer.

At the low rotor frequencies corresponding to small slips, however, reactance effects become negligible, and the rotor resistance then approaches that of the two layers in parallel.pp y p



The upper bars are of smaller cross-sectional area than the lower bars and consequently have higher resistance.

The inductance of the lower bars is greater than that of the upper ones because of the flux crossing the slot between the two layers. g y

The difference in inductance can be made quite large byproperly proportioning the constriction in the slot between the twoproperly proportioning the constriction in the slot between the two

bars.


Summary: By use of double-cage and deep-bar rotors, squirrel-cage motors can be designed to have the good starting characteristics resulting from high rotor resistance and, at the same time, the good running characteristics resulting from low rotor resistance. Induction motorsresulting from low rotor resistance. Induction motors generally fall into four types:

Class A: Normal Starting Torque Normal StartingClass A: Normal Starting Torque, Normal Starting Current, Low Slip

Class B: Normal Starting Torque, Low Starting Current, Low Slip

Class C: High Starting Torque, Low Starting Current

Class D: High Starting Torque High SlipEEL 3211 (© 2008, H. Zmuda) 6. Induction Motors 109

Class D: High Starting Torque, High Slip

Starting Induction Motors and Speed Control

Clearly, by their very nature, inductions motors tend to be self starting. Electronics is used to limit the starting current.

The speed can be somewhat controlled with line voltage. Another method of speed control uses power electronic circuits to vary the p p yline current.

Read Section 7 7 and 7 8 for a discussion on these topicsRead Section 7.7 and 7.8 for a discussion on these topics.


Determination of Circuit Model Parameters

We still need to determine:

F l F Slid 50

1 2 1 2, , , , MR R X X and X

For a real motor. From Slide 50:

jXR jX R1jX1R 2jX2R Rotor Copper Loss

CRMjX 21 sR

s

s

Converted


Determination of Circuit Model Parameters

DC Stator Resistance1,DC RatedI I

R

A

R RR1

V

R1 R1

DCV

R1

Motor (No-Load)1 2DCVRI


Motor (No Load)2 DCI

Determination of Circuit Model Parameters No-Load Test –

Measures rotational losses and provides magnetization current.

1 Th i f l h l l d h f i i d1. The motor spins freely, so the only loads are the friction and windage losses.

2. All Pconv in this motor is consumed by mechanical losses.

3 The slip is very small3. The slip is very small.


Determination of Circuit Model Parameters No-Load test –Measures rotational losses and provides magnetization currentMeasures rotational losses and provides magnetization current.

IA

AIWattmeter

Variable Voltage

Variable Frequency

V

ABI

Motor(N L d)

Wattmeter

Three-Phase SourceCI

(No-Load)

(spins freely)A (spins freely)Wattmeter


Determination of Circuit Model Parameters No-Load Test

1jX1R 2jX2R2 0I

1I

0.0011

s

I

CRMjX 21 sR

s

MI

21000R

2 2 2 21 1,s sR R R X

s s


s s


1jX1R1I

1 sR

CRMjX 2Rs

R

&F WR



1jX1R1I

2

1C

sR Rs

MjX



The input power measured by the meters must equal the losses in the motor.

The rotor copper losses can be neglected since I2 is extremely small since the slip is so small.p

The (known) stator copper losses are, 21 13SCLP I R

So the input power must equal,

P P P P P &in SCL core F W misc

Rotational Losses

P P P P P



1jX1R1I2

1M C

sX R R

1 s

2M C s

21

CsR R

s

MjX

Th t d d t t bli h ti fi ld i it lThe current needed to establish a magnetic field is quite large because of the large reluctance of its air gap, hence the reactance XM will be much smaller than the resistances in parallel with it.



I 1jX1R1,no loadI 2

1M C

sX R Rs

21

CsR R

MjXV

2C s

MjXV

The power factor will be small and lagging, and most of the voltage d ill b h i d Thdrop will be across the inductors. Thus,

1in M

VZ X X


11,

in Mno loadI

Determination of Circuit Model Parameters Locked-Rotor Test

Equivalent to a short circuit transformer test.

IA

AIWattmeter

Variable Voltage

Variable Frequency

V

ABI

Motor(Rotor Locked)Wattmeter

Three-Phase Source

A

CI

(Rotor Locked)

A Wattmeter



With the rotor blocked, the current is adjusted to be that at full-load (approximately). The voltage, current, and power to the motor are measuredmeasured.

Since the rotor is not moving, the slip s = 1. Hence R2/s = R2 and R2i i ll F l k d X i l i ll d his quite small. For a locked rotor, X2 is also quite small, and the model can be approximated as...



1jX1R 2jX

Rl tCRMjX 2

2R Rsneglect

these

2 2

2 2

M

C

X R jX

R R jX


2 2C j


1jX1R 2jX1,locked rotorI

R

22

R RsV



A major problem with this test is that although the stator frequency equals the line frequency, the rotor frequency is different. Under normal operating conditions the slip for most motors is on thenormal operating conditions, the slip for most motors is on the order of a few percent, so that the rotor frequency is in the range of 1 – 3 Hertz.

When the rotor is locked, then the rotor frequency equals the line frequency. q y

Particularly, difficulties are encountered with variable resistance rotorsrotors.



This is why a variable frequency source is used. A typical compromise is to use a test frequency that is 25% or less than the rated frequencyrated frequency.

At lower than normal operating frequency, the inductive reactances ll h l l l d b dj dare smaller, so the voltage level needs to be adjusted so as not to

exceed maximum current ratings.

IEEE standards carefully guideline the actual test procedure.

Nevertheless great care must be used when using this testNevertheless, great care must be used when using this test.



For a given test voltage and frequency, the power meter gives:

3 cosP V I

the locked-rotor power factor is

3 cosin T LP V I

cos3

in

T L

PPFV I

The magnitude of the impedance is

3 T LV I

1 3T

locked rotorL

V VZI I


1 L


Thus (recall Slide 32):

cos sinlocked rotor RL RL RL

RL RL

Z Z R jXZ j Z

RL RLj

R R R R R R1 2 2 1RL RLR R R R R R

1 2RLX X X (At the test frequency)



Thus (recall Slide 32):

1 2 1 2RL t t RL t tX X X L L L

1 2 1 2RL test RL testX X X L L L

X L L L X X

1 2 1 2RL rated RL ratedX L L L X X

1 2 ,ratedRL rated ratedRL RL

fX Xf

L LXX L L

1

1 2

2

R test

ratedRL

L test test ffX X

X L

X

L


1 2RLtestf


Unfortunately there is no way to isolate X1 and X2. In practice, the following rule is generally observed:

RotorType

X1 and X2 vs. XLR

X XX1 X2

Class A 0.5 XLR 0.5 XLR

Class B 0.4 XLR 0.6 XLR

Cl C 0 3 X 0 7 XClass C 0.3 XLR 0.7 XLR

Class D 0.5 XLR 0.5 XLR


Induction Motor ExamplesSee Slide Set 6aSee Slide Set 6a


6. induction motors - university of florida · an induction motor has the same physical stator as a...

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