9. Induction Motor

8/14/2019 9. Induction Motor

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CHAPTER 9

POLYPHASE INDUCTION MOTORS

9.1 Introduction

The direct-current (dc) and synchronous motors we have discussed thus farhave one thing in common: both are the doubly-fed type. These motors havedirect current in their field windings and alternating current (ac) in their armaturewindings.

We now consider a motor in which the rotor receives its power not byconduction but by induction and is therefore called an induction motor.

A winding t!t r"c"i#"$ it$ %ow"r "&c'u$i#"'( )( induction con$titut"$ !tr!n$*orm"r. T"r"*or"+ !n induction motor i$ ! tr!n$*orm"r wit ! rot!ting

$"cond!r( winding.From the above discussion, the following must be evident:

. !n induction motor is a singly-fed motor. Therefore, it does not re"uire a

commutator, slip-rings, or brushes. #n fact, there are no moving contactsbetween the stator and the rotor. This results in a motor that is rugged, reliable,and almost maintenance free.

$. The absence of brushes eliminates the electrical loss due to the brushvoltage drop and the mechanical loss due to friction between the brushes andcommutator or the slip-rings. Thus, an induction motor has a relatively highefficiency.

%. !n induction motor carries alternating current in both the stator and therotor windings.

&. !n induction motor is a rotating transformer in which the secondary

winding receives energy by induction while it rotates.

There are two basic types of induction motors:$ing'",%!$" induction motor$ !nd %o'(%!$" induction motor$. 'ingle-phaseinduction motors are favored for domestic application. ! large number of thesemotors are built in the fractional-horsepower range.

n the other hand, polyphase induction motors cover the entire spectrumof horsepower ratings and are preferably installed at locations where a polyphase

power source is easily accessible.

wing to the widespread generation and transmission of three-phasepower, most polyphase induction motors are of the three-phase type. #n this


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chapter, we confine our discussion eclusively to three-phase induction motors.The theoretical development, however, can be easily etended to an n-phase

induction motor where n$.

9. - Con$truction

The essential components of an induction motor are a $t!torand a rotor.

St!tor

The stationary member of an induction motor is called the stator and isformed by stac*ing thin-slotted, highly permeable steel laminations inside a steelor cast-iron frame. The frame provides mechanical support to the motor.!lthough the frame is made of a magnetic material, it is not designed to carry

magnetic flu.#dentical coils are wound (or placed) into the slots and then connected to

form a balanced three-phase winding.

Rotor

The rotor is also composed of this-slotted, highly permeable steellaminations that are pressed together onto a shaft. There are two types of rotors: a

$uirr"',c!g" rotorand a wound rotor.The $uirr"',c!g" rotor is commonly used when the load re"uires littlestarting tor"ue. For small motors, such a winding is molded by forcing a moltenconducting material ("uite often, aluminum) into the slots in a die-casting process.

+ircular rings called the "nd,ring$are also formed on both sides of thestac*. These end-rings short-circuit the bars on both ends of the rotor, aseplained in +hapter , where we referred to the s"uirrel-cage winding as adamper winding. For large motors, the s"uirrel-cage winding is formed byinserting heavy conducting bars (usually of copper, aluminum, or their alloys) intothe slots and then welding or bolting them to the end-rings.

ach pair of poles has as many rotor phases as there are bars because eachbar behaves independently of the other. #t is a common practice to $/"wthe rotorlaminations to reduce cogging and electrical noise in the motor.

#t becomes necessary to use a wound rotorwhen the load re"uires a highstarting tor"ue. ! wound rotor must have as many poles and phases as the stator.#n fact, the placement of coils in a wound rotor is no different from that in thestator.

The three-phase windings on the rotor are internally connected to form anint"rn!' n"utr!'connection. The other three ends are connected to the slip-rings,

as eplained in +hapter . With the brushes riding on the slip-rings, we can addeternal resistances in the rotor circuit. #n this way the total resistance in the rotor

$


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circuit can be controlled. y controlling the resistance in the rotor circuit we are,in fact, controlling the tor"ue developed by the motor. We will show that thespeed at which an induction motor develops the maimum tor"ue (called the)r"!/down $%""d) depends upon the rotor resistance. !s the rotor resistance

increases, the brea*down speed decreases. Therefore, it is possible to obtainmaimum tor"ue at starting (/ero speed) by inserting 0ust the right amount ofresistance in the rotor circuit. 1owever, a wound-rotor induction motor is moreepensive and less efficient than a s"uirrel-cage induction motor of the samerating. For these reasons, a wound-rotor induction motor is used only when as"uirrel-cage induction motor cannot deliver the high starting tor"ue demanded

by the load.

9. 0 Princi%'" o* O%"r!tion

When the stator winding of a three-phase induction motor is connected to athree-phase power source, it produces a magnetic field that

(a) is constant in magnitude and(b) revolves around the periphery of the rotor at the synchronous speed.The details of how the revolving field is produced and the tor"ue isdeveloped are given in +hapter %.! brief review is presented here.#ffis the fre"uency of the current in the stator winding andPis the number

of poles, the synchronous speed of the revolving field is

N f

Ps =

$2(3.a)

in revolutions per minute (rpm), or

s

f

P=

&(3.b)

in radians per second.

The revolving field induces electromotive force (emf) in the rotor winding.'ince the rotor winding forms a closed loop, the induced emf in each coil givesrise to an induced current in that coil. When a current-carrying coil is immersed ina magnetic field, it eperiences a force (or tor"ue) that tends to rotate it.

The tor"ue thus developed is called the $t!rting toru". #f the load tor"ueis less than the starting tor"ue, the rotor starts rotating. The force developed andthereby the rotation of the rotor are in the same direction as the revolving field.

Ti$ i$ in !ccord!nc" wit !r!d!(2$ '!w o* induction.4nder no load, the rotor soon achieves a speed nearly e"ual to the

synchronous speed. 1owever, the rotor can never rotate at the synchronous speed

because the rotor coils would appear stationary with respect to the revolving fieldand there would be no induced emf in them.

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#n the absence of an induced emf in the rotor coils, there would be nocurrent in the rotor conductors and conse"uently no force would be eperienced

by them. #n the absence of a force, the rotor would tend to slow down. !s soonas the rotor slows down, the induction process ta*es over again.

#n summary, the rotor receives its power by induction only when there is arelative motion between the rotor speed and the revolving field. 'ince the rotorrotates at a speed lower than the synchronous speed of the revolving field, aninduction motor is also called an !$(ncronou$ motor.

5etNm (or m) be the rotor speed at a certain load. With respect to the

motor, the revolving field is moving ahead at a relative speed of

N N Nr s m= (3.$a)or

r s m= (3.$b)

The relative speed is also called the $'i% $%""d. This is the speed withwhich the rotor is slipping behind a point on a fictitious revolving pole in order to

produce tor"ue. 1owever, it is a common practice to epress slip speed in termsof the slip (s), which is a ratio of the slip speed to the synchronous speed.

That is,

s N

N

r

s

r

s

= =

or

s N N

N

s m

s

s m

s

=

=

(3.%)

!lthough the above e"uation yields the slip on a per-unit basis, it iscustomary to epress it as a percentage of synchronous speed (%"rc"nt $'i%).

#n terms of the synchronous speed and the per-unit slip, we can epress therotor speed as

N s Nm s= ( ) (3.&a)or

m s

s= ( ) (3.&b)

When the rotor is stationary, the per-unit slip is and the rotor appearseactly li*e a short-circuited secondary winding of a transformer.

The fre"uency of the induced emf in the rotor winding is the same as thatof the revolving field. 1owever, when the rotor rotates, it is the relative speed of

the rotorNr(or r) that is responsible for the induced emf in its windings. Thus,

the fre"uency of the induced emf in the rotor is

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f P N

rr=

$2

=

=

P N N PN N N

N

s m s s m

s

( )

$2 $2

= s f (3.6)

The above e"uation highlights the fact that the rotor fre"uency dependsupon the slip of the motor. !t standstill, the slip is and the rotor fre"uency is thesame as that of the revolving field.

1owever, the rotor fre"uency decreases with the decrease in the slip. !sthe slip approaches /ero, so does the rotor fre"uency. !n induction motor usuallyoperates at low slip. 1ence the fre"uency of the induced emf in the rotor is low.

For this reason, the core loss in the rotor magnetic circuit is most oftenignored.

9. 3 D"#"'o%m"nt o* !n Eui#!'"nt Circuit

When a balanced three-phase induction motor is ecited by a balancedthree-phase source, the currents in the phase windings must be e"ual inmagnitude and $2oelectrical apart in phase. The same must be true for thecurrents in the rotor windings as the energy is transferred across the air-gap fromthe stator to the rotor by induction.

The fre"uency of the induced emf in the rotor is proportional to its slip 7".(3.6)8. 'ince the stator and the rotor windings are coupled inductively, aninduction motor resembles a three-phase transformer with a rotating secondarywinding.

The similarity becomes even more stri*ing when the rotor is at rest()'oc/"d,rotor condition,s9 ).

Thus, a three-phase induction motor can be represented on a per-phasebasis by an e"uivalent circuit at any slips as depicted in Figure 3..

#n this figure,

:V 9 applied voltage on a per-phase basis

R 9 per-phase stator winding resistanceL 9 per-phase stator winding lea*age inductance

; 9 $fL9 per-phase stator winding lea*age reactance

Rr 9 per-phase rotor winding resistanceLb 9 per-phase rotor winding lea*age inductance

;b 9 $fLb9 per-phase rotor winding lea*age reactance under bloc*ed-

rotor condition (s 9 )

;r 9 $sfLb9sXb9per-phase rotor winding lea*age reactance at slips.

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;m9 per-phase magneti/ation reactanceRc 9 per-phase e"uivalent core-loss resistanceN 9 actual turns per phase of the stator windingN$ 9 actual turns per phase of the rotor winding

*9 winding factor for the stator winding*$9 winding factor for the rotor winding

m9 amplitude of the per-phase flu:E

9 &.&&fN*m9 per-phase induced emf in the stator winding

igur" 9.1


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ased upon the above e"uation, we can develop another circuit of aninduction motor as given in Figure 3.$.

#n this circuit, the hypothetical resistanceRr=sin the rotor circuit is calledthe "**"cti#" r"$i$t!nc". The effective resistance is the same as the actual rotor

resistance when the rotor is at rest (standstill or bloc*ed-rotor condition). n theother hand, when the slip approaches /ero under no-load condition, the effective

resistance is very high (Rr=s).

y defining the ratio of transformation, the a-ratio, as

a N k

N k=

$ $

(3.?)

we can represent the induction motor by its per-phase e"uivalent circuit asreferred to the stator. 'uch an e"uivalent circuit is shown in Figure 3.%, where

R a Rr$$= (3.a)

X a Xb$$= (3.b)

and

::

I I

a

r$ = (3.c)

igur" 9.-@odified e"uivalent circuit of a balanced three-phase motor on a per-phase basis.

For this e"uivalent circuit

: : :E I R

sjI X

$ $

$ $= +

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: : :I I Ic m = +

where

::

IE

Rc

c

= and:

:I

E

J Xm

m

=

The per-phase stator winding current and the applied voltage are

: : :I I I

$= +

and: : :

( )V E I R j X

= + +

The e"uivalent circuit of the rotor in Figure 3.% is in terms of thehypothetical resistance A$=s. #n this circuit, I R s$

$

$= represents the per-phase

power delivered to the rotor. 1owever, the %"r,%!$" co%%"r 'o$$ in the rotor

must be I R$

$

$. Thus, the per-phase power developed by the motor is

I R

sI R I R

s

s$

$ $$

$

$ $

$

$

=

orR

sR R

s

s

$

$ $

= +

(3.3)

igur" 9. 0P"r,%!$" "ui#!'"nt circuitof a balanced three-phase inductionmotor as referred to the stator side.


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The above e"uation establishes the fact that the hypothetical resistanceR$=scan be divided into two components: the actual resistance of the rotor A$andan additional resistanceR$7( -s)=s8.

The additional resistance is called the 'o!d r"$i$t!nc" or the d(n!mic

r"$i$t!nc". The load resistance depends upon the speed of the motor and is saidto represent the load on the motor because the mechanical power developed bythe motor is proportional to it.

In ot"r word$+ t" 'o!d r"$i$t!nc" i$ t" "'"ctric!' "ui#!'"nt o* !

m"c!nic!' 'o!d on t" motor.!n e"uivalent circuit of an induction motor in terms of the load resistance

is given in Figure 3.&.This circuit is proclaimed as the "&!ct "ui#!'"nt circuit o* )!'!nc"d

tr"",%!$" induction motor on ! %"r,%!$" )!$i$.

igur" 9.3The e"uivalent circuit of Figure 3.% modified to show the rotor andthe load resistance.

Pow"r R"'!tion$

'ince the load resistance varies with the slip and the slip ad0usts itself tothe mechanical load on the motor, the power delivered to the load resistance ise"uivalent to the power developed by the motor. Thus, the performance of themotor at any slip can be determined from its e"uivalent circuit, as given in Figure3.&.

For a balanced three-phase induction motor

P V Iin

= %

cos (3.2)

where is the phase difference between the applied voltage:V and the stator

winding current:I . 'ince the power input is electrical in nature, we must

account for the electrical losses first. The immediate electrical loss that must beta*en into consideration is the stator copper loss.

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The tot!' $t!tor co%%"r 'o$$is

P I Rscl

= %

$

(3.)

#f the core loss is modeled by an e"uivalent core-loss resistance, as shownin the figure, we must also ta*e into account the total core loss (magnetic loss) as

P I Rm c c= %$

(3.$)

The net power that is crossing the air-gap and is transported to the rotor byelectromagnetic induction is called the !ir,g!% %ow"r. #n this case, the air-gap

power is

P P P Pag in scl m= (3.%a)

The air-gap power must also e"ual the power delivered to the hypotheticalresistanceR$=s. That is,

P I R

sag

=% $

$

$ (3.%b)

T" "'"ctric!' %ow"r 'o$$ in t" rotor circuit i$

P I R sPrcl ag = =% $$

$ (3.&)

1ence, the power developed by the motor is

P P Pd ag rcl =

=

= =%

$$

$I s R

ss P SPag ag

( )( ) (3.6)

where

S s NN

m

s

m

s= = =

i$ t" %"r,unit 4norm!'i5"d6 $%""d o* t" motor.

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igur" 9.7)

y subtracting the rotational loss from the power developed, we obtain thepower output of the motor as

P P Pd r2

= (3.?)

'ince the core loss has already been accounted for, the rotational loss

includes the friction and windage lossPfand the stray-load lossPst.

The corresponding power-flow diagram is given in Figure 3.6a. When the

core loss Pm is also considered a part of the rotational loss, the core-lossresistanceRcin Figure 3.& must be omitted, the d. (3.%!) becomesPag9Pin-


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Pscl. The power-flow diagram when the core loss is a part of the rotational loss isgiven in Figure 3.6).

S%""d,Toru" C!r!ct"ri$tic

"uation (3.>) reveals that the tor"ue developed by an induction motor isdirectly proportional to the s"uare of the current in the rotor circuit and thee"uivalent hypothetical resistance of the rotor.

1owever, the two "uantities, the rotor current and the hypothetical rotorresistance, are inversely related to each other. For instance, if the rotor resistanceis increased, we epect the tor"ue developed by the motor to increase linearly.

ut any increase in the rotor resistance is accompanied by a decrease in therotor current for the same induced emf in the rotor. ! decrease in the rotorcurrent causes a reduction in the tor"ue developed. Whether the overall tor"ue

developed increases or decreases depends upon which parameter plays adominant role.

5et us eamine the entire speed-tor"ue characteristic of the motor. !tstand-still, the rotor slip is unity and the effective rotor resistance is R$. Themagnitude of the rotor current, from Figure 3.%, is

I E

R X$

$

$

$

$=

+(3.)

Bote that the rotor winding resistance R$is usually very small comparedwith its lea*age reactance ;$. That is,R$CC ;$.

The $t!rting toru"developed by the motor is

T I R

ds

s

= % $$ $

(3.3)

!s the rotor starts rotating, an increase in its speed is accompanied by adecrease in its slip. !ssdecreases,R$=sincreases. !s long asR$=sis smaller than;$, the reduction in the rotor current is minimal. Thus, in this speed range, therotor current may be approimated as

I E

X$

$

(3.$2)

'ince the rotor current is almost constant, the tor"ue developed by themotor increases with the increase in the effective resistance R$=s. Thus, the tor"uedeveloped by the motor *eeps increasing with the decrease in the slip as long as

the rotor resistance has little influence on the rotor current.

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When the slip falls below a certain value called the )r"!/down $'i%sb, thehypothetical resistance becomes the domin!ting *!ctor.

#n this range,R$=sDD ;$ , and the rotor current can be approimated as

I sE

R$

$= (3.$)

The tor"ue developed by the motor is now proportional to the slips. !s theslip decreases, so does the tor"ue developed. !t no load, the slip is almost /ero,the hypothetical rotor resistance is nearly infinite, the rotor current isapproimately /ero, and the tor"ue developed is virtually /ero. With thisunderstanding, we are able to s*etch the speed-tor"ue curve of an inductionmotor. 'uch a curve is depicted in Figure 3.>.

igur" 9.8Typical speed-tor"ue characteristic of a three-phase induction motor.

APPLICATION

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A 8,%o'"+ -0,:+ 8,H5+ Y,conn"ct"d+ tr"",%!$" induction motor !$ t"

*o''owing %!r!m"t"r$ on ! %"r,%!$" )!$i$; R1< .7 + R-< .-7 + =1< .>7

+ =-< .7 + =m< 1 + !nd Rc< 7 . T" *riction !nd wind!g" 'o$$ i$

17 ?. D"t"rmin" t" "**ici"nc( o* t" motor !t it$ r!t"d $'i% o* -.7@.

S O L U T I O N

The synchronous speed of the motor is

N rpms =

=( )$2 >2

>$22 or s rad s= $6>>. =

The per-phase applied voltage is

V V$%2

%%$ ?3= = .

The effective rotor impedance as referred to the stator is

.

.. .Z

R

sj X j j

$

$

$

2$6

22$626 2 26= + = + = +

The stator winding impedance is

. .Z R j X j

26 2?6= + = +

'inceRc, 0;m, and Z$ are in parallel, we can compute the e"uivalent impedanceZe as

622

22

2 26 .Z j je

= + ++

= 22$ 2 26. .j S

. .Z je = +3 >3 &?

1ence, the tot!' in%ut im%"d!nc" i$

. .Z Z Z jin e

= + = +

23 $>?

T" $t!tor curr"nt::

:

. .I

V

ZA

in

2$ ,%$ $ 23= =

T" %ow"r *!ctor: pf lagging= =cos( . ) .$ 23 23?,2

Pow"r in%ut: P V I in = =% &33,6& cos .

&


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St!tor co%%"r 'o$$: P I R scl= =% $&>33$

.

: : : :. .E V I Z V

2$& ?>% %?= =

Cor",'o$$ curr"nt: ::

. .I ER

Acc

= = 22 $6 %?

M!gn"ti5!tion curr"nt::

:

. .I E

j XAm

m

= = 2$&, 3%?

E&cit!tion curr"nt;: : :

. .I I I Ac m = + = $?% ,$ %,2

H"nc"+ t" rotor curr"nt:: : :

. .I I I A$

2$ &> >63= =

Cor" 'o$$; P I R m c c= =% 3%?6$

.

Air,g!% %ow"r; P P P P ag in scl m= = &>6?,.

Rotor co%%"r 'o$$; P I R rcl= =% >&>$$

$.

Pow"r d"#"'o%"d: P P P d ag rcl = = &6&%&.

Pow"r out%ut; P P d2 62 &%3%&= = .

E**ici"nc(; = =P

P!r

in

22 ,?3 ,? 3E. .

9. 7 An A%%ro&im!t" Eui#!'"nt Circuit

! well-designed three-phase induction motor usually meets most of thefollowing guidelines:. The stator winding resistance is *ept small in order to reduce the stator

copper loss.$. The stator winding lea*age reactance is minimi/ed by reducing the mean-

turn length of each coil.

6


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igur" 9.>!n approimate e"uivalent circuit on a per-phase basis of a balancedthree-phase induction motor.

%. Thin laminations of low-loss steel are used to cut down the core loss. Thus,the e"uivalent core-loss resistance is usually high.&. The permeability of steel selected for laminations is high, and the operating

flu density in the motor is *ept below the *nee of the magneti/ation curve.Thus, the magneti/ation reactance is usually high.

!n induction motor conforming to the above stipulations can berepresented by an approimate e"uivalent circuit, as shown in Figure 3.?. #n thiscase, we have placed the parallel branch (the ecitation circuit) across the powersource. We admit that the analysis of an induction motor using the approimate

e"uivalent circuit is somewhat inaccurate, but the inaccuracy is negligible for awell-designed motor. n the other hand, the approimate e"uivalent circuit notonly simplifies the analysis but also aids in comprehending various characteristicsof the motor.

For instance, we use the approimate e"uivalent circuit to determine thespeed at which (a) the tor"ue developed is maimum, (b) the power developed ismaimum, and (c) the motor efficiency is maimum.


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above e"uation and setting the derivative e"ual to /ero. !fter differentiating andcanceling most of the terms, we obtain

R X R

s

se e

p

p

$ $ $

$

+ =

( )

or (3.$&)

Z R

ss

e

p

p= $ ( )

whereZeis the magnitude of the e"uivalent impedance of the stator and the rotorwindings at rest. That is

Z R j Xe e e

= + (3.$6)

Eu!tion 49.-36 $t!t"$ t!t t" %ow"r d"#"'o%"d )( ! tr"",%!$"induction motor i$ m!&imum w"n t" "ui#!'"nt 'o!d 4d(n!mic6 r"$i$t!nc"

i$ "u!' to t" m!gnitud" o* t" $t!nd$ti'' im%"d!nc" o* t" motor.This, of course, is the well-*nown result we obtained from the m!&imum

%ow"r tr!n$*"r t"or"mduring the study of electrical circuit theory.From ". (3.$&) we obtain the slip at which the induction motor develops

maimum power as

s R

R Zp e= +

$

$ (3.$>)

'ubstituting for the slip in ". (3.$%), we obtain an epression for themaimum power developed by a three-phase induction motor as

P V

R Zdm

e e

=+

%

$

$

(3.$?)

The net power output, however, is less than the power developed by an

amount e"ual to the rotational loss of the motor.

9.> M!&imum Toru" Crit"rion

The tor"ue developed by a three-phase induction motor, from ". (3.$%), is

T

V R

s

R X R s

s

R R s

s

d

e e

e

s

=

+ +

+

%

$

$

$

$ $ $

$

$( ) ( ) (3.$)

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where Re9RR$ and Xe9XX$.

igur" 9.ffect of rotor resistance on the brea*down slip.

Gifferentiating the above e"uation with respect to s and setting it e"ual to/ero, we obtain an epression for the brea*down slip sb at which the motordevelops the m!&imum 4)r"!/down6 toru"as

s R

R X Xb =

+ +

$

$

$

$( )(3.$3)

Bote that the brea*down slip is directly proportional to the rotor resistance.

'ince the rotor resistance can be easily ad0usted in a wound-rotor inductionmotor by means of an eternal resistor, we can obtain the maimum tor"ue at anydesired speed, including the /ero speed (starting). 'ubstituting the above

epression for the brea*down slip in ". (3.$), we obtain an epression for themaimum tor"ue developed by the motor as


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T V

R R X Xdm

s

=+ + +

%

$

$

$

$

$ ( )(3.%2)

Bote that the maimum tor"ue developed by the motor is independent ofthe rotor resistance. #n other words, the motor develops the same maimumtor"ue regardless of its rotor resistance.

The rotor resistance affects only the brea*down slip (or brea*down speed)at which the tor"ue is maimum, as illustrated in Figure 3..

9. M!&imum "**ici"nc( Crit"rion

When the core loss is considered a part of the rotational loss the powerinput to the motor using the approimate e"uivalent circuit, is

P V Iin = % $ cos (3.%>)

where is the power-factor angle between the applied voltage:V and the rotor

current:I$ .

The power output is

P V I I R R Pr2 $ $$

$% %= + cos ( ) (3.%?)

The motor efficiency is

=

+ % %%

$ $

$

$

$

V I I R R P

V I

rcos ( )

cos(3.%)

Gifferentiating with respect toI$and setting the derivative e"ual to /ero,

we obtain

%$

$

$I R R P

r( )+ = (3.%3)

as the criterion for the maimum efficiency of an induction motor.It $im%'( $t!t"$ t!t t" "**ici"nc( o* !n induction motor i$ m!&imum

w"n t" $um o* t" $t!tor !nd t" rotor co%%"r 'o$$"$ i$ "u!' to t"rot!tion!' 'o$$.

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9. 9 Som" Im%ort!nt Conc'u$ion$


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o%"r!t" !t 'ow $'i%.(b) T" rotor curr"nt #!ri"$ in#"r$"'( wit t" rotor r"$i$t!nc".

T" Motor E**ici"nc(

For an ideal motor we can assume that (a) the stator copper loss isnegligible and (b) the rotational loss is /ero.

#n this case, the air-gap power is e"ual to the power input. That is, Pin"Pag. 1owever, the power developed isPd"( -s)Pag" Spag, where Sis the per-unit speed.

'ince the rotational loss is /ero, the power output is e"ual to the powerdeveloped. 1ence, the motor efficiency under the ideal conditions is

= = s S (3.&$)

The above e"uation places a maimum limit on the efficiency of a three-phase induction motor. This e"uation highlights the fact that if a motor isoperating at >2E of its synchronous speed, the maimum efficiency under theideal conditions (theoretically possible) is >2E. Thus, the higher the speed ofoperation, the higher the efficiency. For eample, a motor operating at 6E slipcan theoretically have an efficiency of 36E.

APPLICATION , E = A M P L E

! $%2-H, >2-1/, &-pole, -connected, three-phase induction motor operates at a

full-load speed of ?2 rpm. The power developed at this speed is $ hp and the

rotor current is &.6 !. #f the supply voltage fluctuates 2E, determine

(!) the tor"ue range and()) the current range.

S O L U T I O N

N rpms =

=$2 >2

& ,22

s=

=,22 ?2

,22226.

m rad s=

=

$ ?2

>2?3 2?. =

The tor"ue developed at the rated voltage of $%2 H is

T N md = = $ ?&>

?3 2?,%%

..

$


22/43

When the supply voltage is down by 2E, the tor"ue developed by the motor is

T N mdL

=

= ,%%23 $%2

$%2

>?6

$

..

.

The corresponding rotor current is

I AL$

&623 $%2

$%2& 26=

=.

..

'imilarly, when the supply voltage is up by 2E, the tor"ue developed by themotor is

[ ]T N md# = = ,%% 22,$. . .

and the rotor current is

I A#$ &6 & 36= =. . .

1ence, the tor"ue varies from >.?6 Bm to 2.2 Bm, and the rotor current

fluctuates between &.26 ! and &.36 !.

9.1 Eui#!'"nt Circuit P!r!m"t"r$

The e"uivalent circuit parameters and the performance of a three-phaseinduction motor can be determined by performing four tests.

These are thr followings:(a) the stator resistance test,(b) the bloc*ed-rotor test,(c) the no-load test, and(d) the load test.

T" St!tor,R"$i$t!nc" T"$t

$$


23/43

This test is performed to determine the resistance of each phase winding ofthe stator. 5etRbe the dc value of the resistance between any two terminals ofthe motorI then the per-phase resistance is

A9 2.6 A for J-connection

A9 .6 A for -connection

The measured value of the resistance may be multiplied by a factor rangingfrom .26 to .$6 in order to convert it from its dc value to its ac value. This isdone to account for the $/in "**"ct. The multiplying factor may be debatable at

power fre"uencies of 62 or >2 1/, but it does become significant for a motoroperating at a fre"uency of &22 1/.

T" 'oc/"d,Rotor T"$t

This test, also called the loc*ed-rotor test, is very similar to the short-circuit test of a transformer. #n this case, the rotor is held stationary by applyingeternal tor"ue to the shaft. The stator field winding is connected to a variablethree-phase supply. The voltage is carefully increased from /ero to a level atwhich the motor draws the rated current. !t this time, the readings of the linecurrent, the applied line voltage, and the power input are ta*en by using the two-wattmeter method, as illustrated in Figure 3.3.

'ince the rotor-circuit impedance is relatively small under bloc*ed-rotorcondition (s9 ), the applied voltage is considerably lower than the rated voltageof the motor. Thus, the ecitation current is "uite small and can be neglected.

4nder this assumption, the approimate e"uivalent circuit of the motor isgiven in Figure 3.2 on a per-phase basis.

The total series impedance is

( )Z R R j X X R j Xe e e= + + + = + $ $ (3.&%)

5et Vbr,Ibr, andPbrbe the applied voltage, the rated current, and the powerinput on a per-phase basis under the bloc*ed-rotor conditionI then

R P

Ie

br

br

=$ (3.&&)

$%


24/43

igur" 9.9Typical connections to perform test on a three-phase induction motor.

igur" 9.1!n approimate per-phase e"uivalent circuit of an induction motorunder bloc*ed-rotor condition.

'ince R is already *nown from the stator-resistance test, the e"uivalentrotor resistance is

R R Re$

= (3.&6)

1owever,

Z V

Ie

br

br

= (3.&>)

Therefore,

$&


25/43

X Z Re e e= $ $ (3.&?)

#t is rather difficult to isolate the lea*age reactances ; and ;$. For all

practical purposes, these reactances are usually assumed to be e"ual. That is

X X Xe $ 26= = . (3.&)

T" No,Lo!d T"$t

#n this case the rated voltage is impressed upon the stator windings and themotor operates freely without any load. This test, therefore, is similar to the open-circuit test on the transformer ecept that friction and windage loss is associatedwith an induction motor. 'ince the slip is nearly /ero, the impedance of the rotor

circuit is almost infinite. The per-phase approimate e"uivalent circuit of themotor with the rotor circuit open is shown in Figure 3..

5et !c, I!c, and V!cbe the power input, the input current, and the ratedapplied voltage on a per-phase basis under no-load condition. #n order torepresent the core loss by an e"uivalent resistance Rc, we must subtract thefriction and windage loss from the power input.

The friction and windage loss can be measured by coupling the motorunder test to another motor with a calibrated output and running it at the no-load

speed of the induction motor. 5etPfbe the friction and windage loss on a per-

phase basis. Then, the power loss inRcis

P P!c !c f = (3.&3)

1ence, t" cor",'o$$ r"$i$t!nc"is

R V

Pc !c

!c=

$

(3.62)

The power factor under no load is

cos!c

!c

!c !c

V I= (3.6)

The magneti/ation reactance is

$6


26/43

X V

Im

!c

!c !c

=sin (3.6$a)

The magneti/ation reactance can also be computed as follows:

S V I!c !c !c=

$ S !c !c !c

= $ $

and

X V

$m

!c

!c

=$

(3.6$b)

When using the two,w!ttm"t"r m"todto measure the power under no

load the reading on one wattmeter may actually be negative because the powerfactor of the motor under no load may be less than 2.6.#f this is the case, the total power input is simply the difference of the two

wattmeter readings.

T" Lo!d T"$t

To eperimentally determine the speed-tor"ue characteristics and theefficient of an induction motor, couple the motor to a dynamometer and connectthe times phase stator windings to a balanced three-phase power source. #f need

be the direction of rotation may be reversed by interchanging any two supplyterminals.

'tarting from the no-load condition, the load is slowly increased and thecorresponding readings for the motor speed, the shaft tor"ue, the power input theapplied voltage, and the line current are recorded.

From these data, the motor performance as a function of motor speed (orslip) can be computed. 4sing analog-to-digital converters, the data can be storedon a magnetic dis* for further manipulations.

APPLICATION

The test data on a $2-H, >2-1/, &-pole, J-connected, three-phaseinduction motor rated at ?2 rpm are as follows:

The stator resistance (dc) between any two terminals 9 $.&

Bo-5oad Test loc*ed-Aotor Test

$ ! $.?? !

5ine voltage $2 H $? H

$>


27/43

Friction and windage loss 9 W+ompute the e"uivalent circuit parameters of the motor.

S O L U T I O N

For J-connected motor, the per-phase resistance of the stator winding is

R

$ &

$$= =

..

K

Ze = =

66,,

$ ??

6>$?.

.

.

Xe = =6>$? $ 6, 6$ $. .

Thus,X X

$$6= = .

9. 11 St!rting o* Induction Motor$

!t the time of starting, the rotor speed is /ero and the per-unit slip is unity.Therefore, the starting current, from the approimate e"uivalent circuit given inFigure 3.?, is

:I

V

R j Xs

e e

$

=+

(3.6%)

where Re9R R$ and ;e9 ; ;$. The corresponding value of the startingtor"ue is

[ ]T

V R

R Xds

s e e

=+

%

$

$

$ $(3.6&)

'ince the effective rotor resistance,R$, is very small at the time of startingcompared with its value at rated slip,R$=s, the starting current may be as much as&22E to 22E of the full-load current. n the other hand, the starting tor"ue mayonly be $22E to %62E of the full-load tor"ue. 'uch a high starting current isusually unacceptable because it results in an ecessive line voltage drop which in

turn, may affect the operation of other machines operating on the same powersource.

$?


28/43

'ince the starting current is directly proportional to the applied voltage, ".(3.6%) suggests that the starting current can be reduced by impressing a lowvoltage across motor terminals at the time of starting. 1owever, it is evident from". (3.6&) that a decrease in the applied voltage results in a decline in the starting

tor"ue.T"r"*or"+ w" c!n "m%'o( t" 'ow,#o't!g" $t!rting on'( *or to$"!%%'ic!tion$ t!t do not r"uir" ig $t!rting toru"$. or in$t!nc"+ ! *!n'o!d r"uir"$ !'mo$t no $t!rting toru" "&c"%t *or t" 'o$$ du" to *riction.

T" induction motor dri#ing ! *!n 'o!d c!n )" $t!rt"d u$ing 'ow,#o't!g"$t!rting.

The starting current can also be decreased by increasing the rotorresistance. !s mentioned earlier, an increase in the rotor resistance also results inan increase in the starting tor"ue which, of course, is desired for those loadsre"uiring high starting tor"ues.

1owever, a high rotor resistance (a) reduce the tor"ue developed at fullload, (b) produces high rotor copper loss, and (c) causes a reduction in motorefficiency at full load.

These drawbac*s, however, do not represent a problem for wound-rotormotors. For these motors, we can easily incorporate high eternal resistance inseries with the rotor windings at the time of starting and remove it when themotor operates at full load.

igur" 9.1-(!) Geep-bar and ()) double-cage rotors.

$


29/43

For rotors using s"uirrel-cage winding, the change in resistance from a highvalue at starting to a low value at full load is accomplished by using "uite a fewdifferent designs, as shown in Figure 3.$.

In "!c d"$ign+ t" und"r'(ing %rinci%'" i$ to !ci"#" ! ig rotor

r"$i$t!nc" !t $t!rting !nd ! 'ow rotor r"$i$t!nc" !t t" r!t" $%""d.!t starting, the fre"uency of the rotor is the same as the fre"uency of theapplied source. !t full load, however, the rotor fre"uency is very low (usuallyless than 2 1/). Thus, the s*in effect is more pronounced at starting than at fullload. 1ence, the rotor resistance is higher at starting than at full load owing to thes*in effect alone. !lso, as the currents are induced in the rotor bars, they producea secondary magnetic field.


30/43

igur" 9.13 'peed-tor"ue characteristics for single-cage and double-cage rotors.

igur" 9.17 (!)4ns*ewed and ()) s*ewed rotor bars.

!nother techni"ue that is commonly used to increase the rotor resistanceand lessen the effects of harmonics in an induction motor is called $/"wing. #n

%2


31/43

this case, the rotor bars are s*ewed with respect to the rotor shaft, as illustrated inFigure 3.6. '*ew is usually given in terms of bars. The minimum s*ew must beone bar to avoid cogging. '*ews of more than one bar are commonly used.

9. 1- Rotor Im%"d!nc" Tr!n$*orm!tion

Thus far we have tacitly assumed for both the s"uirrel-cage and woundrotors that the rotor circuit impedance can be transformed to the stator side interms of an a-ratio. The a-ratio was defined on a per-phase basis as the ratio ofthe effective turns in the stator winding to the effective turns in the rotor winding.That is

a k N

k N

E

Eb= =

$ $

(3.6>)

For a wound rotor having the same number of poles and phases as that of

the stator winding, the total turns per phaseN$and the winding factor k$can be

calculated the same way as that for the stator winding. 1owever, the problem issomewhat more perpleing for the s"uirrel-cage (die-cast) rotor.

5et us suppose that there arePpoles in the stator and $bars on the rotor.5et us assume that one of the bars is under the middle of the north pole of thestator at any given time. !nother bar also eists which is in the middle of the

ad0acent south pole.The induced emf in both bars is maimum but of opposite polarity. Thus,

these two bars carry the maimum current and can be visuali/ed as if they form asingle turn. 1ence, the total number of turns on the rotor is $=$. The emfs arealso induced in other bars. #f the flu is distributed sinusoidally, the induced emfsand thereby the induced currents also follow the same pattern.

1owever, the root-mean-s"uare (rms) value of the induced emf in each turnis the same. 'ince each turn is offset by one slot on the rotor, the induced emf ineach bar is offset by that angle. Therefore, we can assume that each turn ise"uivalent to a phase group, and there are $=$ phase groups in all. 'ince there are

$bars andPpoles, the number of bars per pole is $=P. !s each bar identifies adifferent phase group, the number of bars per pole is then e"uivalent to thenumber of phases m$on the rotor. That is

m $

P$ = (3.6?)

This reali/ation highlights the fact that the number of bars per pole perphase is

%


32/43

. 'tated differently, the number of turns per pole per phase is L. Bow we candetermine the total number of turns per phase by multiplying the number of turns

per pole per phase by the number of poles. That is,

N P

$$= (3.6)

'ince the two bars that are displaced by 2oelectrical form a turn, thewinding factor is unity because (a) the pitch factor is unity as each turn is a full-

pitch turn and (b) the distribution factor is unity as there is only one turn in eachphase group.

'ince we are trying to transform the rotor circuit elements to the statorside, let mbe the number of phases on the stator side,E$the induced emf, andI$the e"uivalent rotor current.

For the e"uivalent representation to be valid, the apparent powerassociated with the rotor circuit on the rotor side must be the same for thee"uivalent rotor circuit as referred to the stator side. Thus,

m E I m E I bar bar $ $ $= (3.63)

whereEbaris the induced emf in the rotor bar andIbaris the induced current.'ince the induced emf on the stator side is E, E$ must be e"ual to E.

1ence,

I m k N

m k NI

bar$

$ $ $

=

(3.>2)

!lso the rotor copper loss prior to and after the transformation must bee"ual. That is,

m I R m I Rbar bar $

$

$ $

$= (3.>)

R m

m

k N

k NRbar$

$

$ $

$

=

(3.>$)

whereRbaris the resistance of a rotor bar.Finally, the magnetic energy stored in the rotor lea*age inductance before

and after the transformation must also be the same. Therefore,

$ $

$ $ $

$ $

$

$m I X

fm I

X

fbar

bar

=

%$


33/43

X m

m

k N

k NX

bar$

$

$ $

$

=

(3.>%)

where ;baris the lea*age reactance of each rotor bar.

"uations (3.>$) and (3.>%) outline how the actual rotor parameters for as"uirrel-cage rotor can be transformed into e"uivalent rotor parameters on thestator side.R$and ;$are the rotor resistance and lea*age reactance that we haveused in the e"uivalent circuit of an induction motor.

From the above e"uations, it is evident that the a-ratio is

a mm

k Nk N=

$

$ $

(3.>&)

Bote that for a wound rotor m 9 m$.

9. 10 S%""d Contro' o* Induction Motor$

#t was pointed out in the preceding sections that the speed of an inductionmotor for a stable operation must be higher than the speed at which it develops

maimum tor"ue. #n other words, the slip at full load must be less than thebrea*down slip. For an induction motor having a low rotor resistance, thebrea*down slip is usually less than 2E. For such a motor, the speed regulationmay be within 6E. For all practical purposes, we can refer to a low-resistanceinduction motor as a constant-speed motor. Therefore, we must devise somemethods in order to vary its operating speed.

We already *now that the synchronous speed is directly proportional to thefre"uency of the applied power source and inversely proportional to the numberof polesI the motor speed at any slip is

N fP

sm =

$2 ( ) (3.>6)

From this e"uation it is evident that the operating speed of an inductionmotor can be controlled by changing the fre"uency of the applied voltage sourceand=or the number of poles.

'peed can also be controlled by either changing the applied voltage, thearmature resistance, or introducing an eternal emf in the rotor circuit. 'ome ofthese methods are discussed below.

r"u"nc( Contro'

%%


34/43

The operating speed of an induction motor can be increased or decreasedby increasing or decreasing the fre"uency of the applied voltage source. Thismethod enables us to obtain a wide variation in the operating speed of an

induction motor. The only re"uirement is that we must have a variable-fre"uencysupply.

To maintain constant flu density and thereby the maimum tor"uedeveloped, the applied voltage must be varied in direct proportion to thefre"uency. This is due to the fact that the induced emf in the stator winding isdirectly proportional to the fre"uency. The speed-tor"ue characteristics of aninduction motor at four different fre"uencies are given in Figure 3.?.

!lso shown in the figure is a typical load curve. !t each fre"uency themotor operates at a speed at which the load line intersects the speed-tor"ue

characteristic for that fre"uency.

igur" 9.1>'peed-tor"ue characteristics for various fre"uencies and ad0usted

supply voltages

C!nging St!tor Po'"$

This method is "uite suitable for an induction motor with a s"uirrel-cagerotor. #n this case, the stator can be wound with two or more entirely independentwindings. ach winding corresponds to a different number of poles and thereforedifferent synchronous speed. !t any time, only one winding is in operation.

!ll other windings are disconnected. For eample, an induction motorwound for & and > poles at a fre"uency of >2 1/ can operate either at a

%&


35/43

synchronous speed of 22 rpm (&-pole operation) or at $22 rpm (>-poleoperation).

This method of speed control, although somewhat limited, is very simple,provides good speed regulation, and ensures high efficiency at either speedsetting. 4se has been made of this method in the design of traction motors,elevator motors, and small motor driving machine tools.

!n induction motor is usually wound such that the current in each phasewinding produces alternate poles. Thus, the four coils for each phase of a &-poleoperation motor produce two north poles and two south poles, with a south polelocated between the two north poles and vice versa. 1owever, if the phase coilsare reconnected to produce either four north poles or four south poles, thewinding is said to constitute a conse"uent-pole winding. etween any two li*e

poles an unli*e pole is induced by the continuity of the magnetic field lines. Thus,a &-pole motor when reconnected as a conse"uent-pole motor behaves li*e an -

pole motor. Therefore, by simply reconnecting the phase windings on aninduction motor, a speed in the ratio of $: can be accomplished by a singlewinding.

Rotor R"$i$t!nc" Contro'

We have already discussed the effect of changes in the rotor resistance onthe speed-tor"ue characteristic of an induction motor. This method of speedcontrol is suitable only for wound-rotor induction motors. The operating speed ofthe motor can be decreased by adding eternal resistance in the rotor circuit.1owever, an increase in the rotor resistance causes

(a) an increase in the rotor copper loss,(b) an increase in the operating temperature of the motor, and(c) a reduction in the motor efficiency. ecause of these drawbac*s, this

method of speed control can be used only for short periods.

St!tor :o't!g" Contro'

'ince the tor"ue developed by the motor is proportional to the s"uare ofthe applied voltage, the (reduction=augmentation) in operating speed of aninduction motor can be achieved by (reducing=augmenting) the applied voltage.

The speed-tor"ue characteristics for two values of the applied voltage aredepicted in Figure 3.. The method is very convenient to use but is very limitedin its scope because to achieve an appreciable change in speed a relatively largechange in the applied voltage is re"uired.

%6


36/43

igur" 9.1'peed-tor"ue characteristics as a function of supply voltage.

InB"cting !n EM in t" Rotor Circuit

The speed of a wound-rotor induction motor can also be changed byin0ecting an emf in the rotor circuit, as shown in Figure 3.3. For properoperation, the fre"uency of the in0ected emf must be e"ual to the rotor fre"uency.

1owever, there is no restriction on the phase of the in0ected emf. #f thein0ected emf is in phase with the induced emf in the rotor, the rotor currentincreases. #n this case, the rotor circuit manifests itself as if it has a lowresistance.

n the other hand, if the in0ected emf is in phase opposition to the inducedemf in the rotor circuit, the rotor current decreases.

The decrease in the rotor current is analogous to the increase in the rotorresistance. Thus, changing the phase of the in0ected voltage is e"uivalent tochanging the rotor resistance.

The change in the rotor resistance is accompanied by the change in theoperating speed of the motor.

Further control in the speed can also be achieved by varying the magnitudeof the in0ected emf.

%>


37/43

igur" 9.19!n e"uivalent circuit of an induction motor with an eternal sourcein the rotor circuit.

9.13 T(%"$ o* Induction Motor$

The Bational lectrical @anufacturers !ssociation (B@!) hascategori/ed s"uirrel-cage induction motors into si different types by assigningthem the letter designations

C'!$$ A Motor$

A c'!$$ A motor is considered a $t!nd!rd motor and is suitable forconstant-speed applications. The motor can be started by applying the ratedvoltage. #t develops a starting tor"ue of $6E to ?6E of full-load tor"ue. Thestarting current at the rated voltage is 6 to ? times the rated current. The full-load

slip is usually less than 6E because the rotor resistance is relatively low. Thespeed regulation is $E to &E.

The rotor bars are placed close to the surface of the rotor laminations inorder to reduce the lea*age reactance. These motors drive low-inertia loads and

possess high accelerations. T"( !r" "m%'o("d in $uc !%%'ic!tion$ !$ *!n$+)'ow"r$+ c"ntri*ug!' t(%" %um%$+ !nd m!cin" too'$.

C'!$$ Motor$

%?


38/43

! class motor is considered a general-purpose motor and can be startedby applying the r!t"d #o't!g". The rotor resistance for a class motor issomewhat higher than for a class ! motor.

T" rotor conductor$ !r" %'!c"d d""%"r in t" $'ot$ t!n *or t" c'!$$

A motor. Therefore, the rotor reactance of a class motor is higher than that of aclass ! motor.The increase in the rotor reactance reduces the starting tor"ue, whereas an

increase in the rotor resistance increases the starting tor"ue. Thus, the startingtor"ue range for a class motor is almost the same as that of the class ! motor.

wing to the increase in reactance, the starting current is about &.6 to 6.6times the full-load current. The low starting current and almost the samestarting tor"ue ma*e class motors appropriate for class ! applications aswell. T"r"*or"+ c'!$$ motor$ c!n )" $u)$titut"d in !'' !%%'ic!tion$u$ing c'!$$ A motor$. T" $%""d r"gu'!tion *or c'!$$ motor$ i$ 0@ to

7@.

C'!$$ C Motor$

A c'!$$ C motor u$u!''( !$ ! dou)'",c!g" rotor !nd i$ d"$ign"d *or*u'',#o't!g" $t!rting.

The high-resistance rotor limits the starting current to %.6 to 6 times thefull-load current. The starting tor"ue is $22E to $?6E of the full-load tor"ue. Thespeed regulation is &E to 6E.

C'!$$ C motor$ !r" u$"d in !%%'ic!tion$ t!t r"uit" ig $t!rting

toru"$+ $uc !$ com%r"$$ion %um%$+ cru$"r$+ )oring mi''$+ con#"(or"ui%m"nt+ t"&ti'" m!cin"r(+ !nd wood,wor/ing "ui%m"nt.

C'!$$ D Motor$

! class G motor is a high-resistance motor capable of developing a startingtor"ue of $62E to %22E of the rated tor"ue. The high rotor resistance is created

by using high-resistance alloys for the rotor bars and by reducing the cross-sectional area of the bar. Gepending upon the design, the starting current may be

% to times the rated current.The efficiency of a class G motor is lower than that of those discussed

above. The speed regulation may be as high as 2E.T"$" motor$ !r" u$"d in $uc !%%'ic!tion$ !$ )u''do5"r$+ $"!ring

m!cin"$+ %unc %r"$$"$+ $t!m%ing m!cin"$+ '!undr( "ui%m"nt+ !ndoi$t$.

C'!$$ E Motor$

+lass motors in general have low starting tor"ue and operate at low slipat rated load. The starting current is relatively low for motors below ?.6

%


39/43

horsepower. These motors may be started at rated voltage. 1owever, for motorsabove ?.6 horsepower the starting current may be high enough to re"uire a low-voltage starting circuit.

C'!$$ Motor$

! class F motor is usually a double-cage motor. #t is a low-tor"ue motorand re"uires the lowest amount of starting current of all motors. The startingtor"ue is usually .$6 times the rated tor"ue, whereas the starting current is $ to &times the rated current. The speed regulation is over 6E. These motors can bestarted by applying the rated voltage. They are designed to replace class motorsand are built in si/es above $6 horsepower.

S U M M A R Y

#n this chapter we shed some light on a three-phase induction motor, whichessentially consists of a stator and a rotor. The stator is wound using double-layerwinding 0ust li*e the stator of a synchronous machine. There are two types ofrotors: a s"uirrel-cage rotor and a wound rotor. ! wound rotor, althoughepensive, is wound for the same number of poles as the stator. #t provides meansto add eternal resistance in series with the rotor circuit.

! s"uirrel-cage rotor uses bars in the slots that are shorted at either end bythe end-rings. For a low-horse power motor, the bars and the end-rings areformed in a die-casting process.

When the three-phase stator winding is connected to a balanced three-phase source, it sets up a revolving field that rotates around the periphery of therotor at the synchronous speed given by the following e"uation:

N f

Ps =

$2

wherefis the fre"uency of the applied voltage source andP is the number ofpoles in the stator.

The uniform revolving field induces emf in the rotor conductors. 'ince therotor winding forms a closed circuit, the induced emf gives rise to a current therotor conductors.

The interaction of the current in the rotor conductors with the magneticfield in the motor creates a tor"ue in accordance with the 5orent/ force e"uation.

Therefore, the rotor starts rotating and attains a speed slightly less than thesynchronous speed. For this reason, an induction motor is also called anasynchronous motor.

The difference between the synchronous speed and the rotor speed is

called the slip speed. The per-unit slip is then defined as

%3


40/43

s N N

N

s m

s

=

y using the transformer analogy we developed an e"uivalent circuit of an

induction motor as referred to the stator side. The rotor circuit parameters weretransformed to the stator side by using the a-ratio given as

a m

m

k N

k N=

$

$ $

where m and m$are the number of phases, k and k$are the winding factors

andN andN$are the number of turns in each phase of the stator and the rotor

windings. For s"uirrel-cage rotor, k$9 , m$9 $%P, andN$9P=$, where $is the

number of bars in the rotor. For a wound-rotor induction motor m9 m$.We definedR$( -s)=sas the dynamic (or effective) resistance because the

power developed is proportional to it. Bote that R$is the rotor resistance. Fromthe per-phase e"uivalent circuit, we can compute the power input as

P V Iin

= %

cos

where VandIare the per-phase applied voltage and the input current. is the

power factor angle between the two.

The stator copper loss can be computed as

P I Rscl

= %

$

#fIcis the core-loss current throughRc, the core loss is

P I Rm c c= %$

The air-gap power is

P P P P I R

sag in scl m= =

% $$

$

The rotor copper loss is

P I Rrcl = % $$

$

1ence, the power developed is

&2


41/43

P P P s P I R s

sd ag scl ag = = =

( )

( )

% $

$

$

The tor"ue developed is

T P P I R

sd

d

m

ag

s s

= = =

%$

$

$

( )

4sing the approimate e"uivalent circuit, we found out that t" "**ici"nc(of an induction motor is maimum when

[ ]% $$

$I R R Pr+ =

The motor develops m!&imum toru"at a slip *nown as the brea*downslip such that

s R

R Xb

e

=+

$

$ $

where ;e9 ; ;$. The epression for the maimum tor"ue developed is

T V

R R X

dm

s e

=

+ +

%

$

$

$ $

When the stator winding impedance is negligible, the approimateepressions for the brea*down slip and the brea*down tor"ue are

s R

Xb

= $

$

T V

Xdm

s

=%

$

$

$

#f Tdis the tor"ue developed at a slips, then

T

T

ss

s s

d

dm

b

b

=+

$$ $

The power developed by an induction motor is maimum when

[ ]P V

R Zdm

e e

= +%

$

$

&


42/43

The motor circuit parameters can be determined by performing thebloc*ed-rotor test, the no-load test, and the stator-resistance test. From thebloc*ed-rotor test

R P

Ie

br

br

=$

where Pbr and Ibr are the per-phase power input and the current. The test isconducted when the rotor is held stationary and the motor draws the rated currentfrom a carefully applied low voltage, Vbr. The magnitude of the stator and therotor winding impedance is

Z

V

Iebr

br=Thus,

X Z Re e e= $ $

The individual values of t" '"!/!g" r"!ct!nc"are

X X Xe $

26= = .

and the rotor resistance is A$9 Ae- A.The no-load test is conducted at the rated voltage when the rotor is free to

rotate without load. #f !c, I!c and V!care the power input, the current, and theapplied voltage on a per-phase basis, then

P P!c !c f =

wherePfis the per-phase friction and windage loss.

The core-loss resistance is

R V

Pc

!c

!c

=$

The magneti/ation reactance is

X V

$m

!c

!c

=$

where

$ S !c !c !c= $ $

and S V I!c !c !c=

&$


43/43

We have also eamined the effect of changes in the rotor resistance on thespeed-tor"ue characteristic of an induction motor. !n increase in the rotorresistance increases the starting tor"ue, reduce the starting current, and enablesthe operation of the motor at a somewhat lower speed.

#n a wound motor, the rotor resistance is increased by adding eternalresistance to the rotor circuit via slip-rings. #n a s"uirrel-cage induction motor, thechange in the rotor resistance is reali/ed by using a multicage rotor.

We also eamined various schemes that enable us to control the speed ofan induction motor. 'ome of the methods we have discussed are fre"uencycontrol, changing stator poles, rotor resistance control, stator voltage control, andin0ecting an emf in the rotor circuit.

Documents

9. Induction Motor