evaluation of induction machines

7/29/2019 Evaluation of Induction Machines

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Massachusetts Institute of TechnologyDepartment of Electrical Engineering and Computer Science

6.685 Electric Machinery

Class Notes 8: Analytic Design Evaluation of Induction Machines

January 12, 2006

c2005 James L. Kirtley Jr.

1 Introduction

Induction machines are perhaps the most widely used of all electric motors. They are generallysimple to build and rugged, offer reasonable asynchronous performance: a manageable torque-speedcurve, stable operation under load, and generally satisfactory efficiency. Because they are so widelyused, they are worth understanding.

In addition to their current economic importance, induction motors and generators may findapplication in some new applications with designs that are not similar to motors currently incommerce. An example is very high speed motors for gas compressors, perhaps with squirrel cagerotors, perhaps with solid iron (or perhaps with both).

Because it is possible that future, high performance induction machines will be required tohave characteristics different from those of existing machines, it is necessary to understand themfrom first principles, and that is the objective of this document. It starts with a circuit theoreticalview of the induction machine. This analysis is strictly appropriate only for wound-rotor machines,but leads to an understanding of more complex machines. This model will be used to explain thebasic operation of induction machines. Then we will derive a model for squirrel-cage machines.Finally, we will show how models for solid rotor and mixed solid rotor/squirrel cage machines canbe constructed.

The view that we will take in this document is relentlessly classical. All of the elements thatwe will use are calculated from first principles, and we do not resort to numerical analysis orempirical methods unless we have no choice. While this may seem to be seriously limiting, it serves

our basic objective here, which is to achieve an understanding of how these machines work. It isour feeling that once that understanding exists, it will be possible to employ more sophisticatedmethods of analysis to get more accurate results for those elements of the machines which do notlend themselves to simple analysis.

An elementary picture of the induction machine is shown in Figure 1. The rotor and stator arecoaxial. The stator has a polyphase winding in slots. The rotor has either a winding or a cage, alsoin slots. This picture will be modified slightly when we get to talking of solid rotor machines,anon. Generally, this analysis is carried out assuming three phases. As with many systems, thisgeneralizes to different numbers of phases with little difficulty.

2 Induction Motor Transformer Model

The induction machine has two electrically active elements: a rotor and a stator. In normaloperation, the stator is excited by alternating voltage. (We consider here only polyphase machines).The stator excitation creates a magnetic field in the form of a rotating, or traveling wave, whichinduces currents in the circuits of the rotor. Those currents, in turn, interact with the traveling

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Stator Core Stator Windingin Slots

Rotor Windinor Cage inSlots

Rotor

AirGap

Figure 1: Axial View of an Induction Machine

wave to produce torque. To start the analysis of this machine, assume that both the rotor and thestator can be described by balanced, three phase windings. The two sets are, of course, coupledby mutual inductances which are dependent on rotor position. Stator fluxes are (a,b,c) androtor fluxes are (A,B ,C). The flux vs. current relationship is given by:

a ia b ib LS MSR c ic

=

(1)

A

iA

MT

SR

LR B iB C iC

where the component matrices are: La Lab Lab

L = Lab La Lab (2)SLab Lab La

LA LAB LAB

L = LAB LA LAB (3)RLAB LAB LA

The mutual inductance part of (1) is a circulant matrix: 2Mcos(p) Mcos(p+ 3 ) Mcos(p 23 )

M = Mcos(p 23 ) Mcos(p) Mcos(p+ 23 ) (4)SR2Mcos(p+ 23 ) Mcos(p 3 ) Mcos(p)

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To carry the analysis further, it is necessary to make some assumptions regarding operation.To start, assume balanced currents in both the stator and rotor:

ia = IScos(t)ib = IScos(t 23 ) (5)ic = IScos(t+ 23 )

iA = IR cos(Rt+ R)iB = IR cos(Rt+ R 23 ) (6)iC= IR cos(Rt+ R + 23 )

The rotor position can be described by= mt+ 0 (7)

Under these assumptions, we may calculate the form of stator fluxes. As it turns out, we needonly write out the expressions for a and A to see what is going on:

a = (La

Lab)Is cos(t) + MIR(cos(Rt+ R)cosp(m + 0) (8)

2 2 2 2+ cos(Rt+ R + )cos(p(mt+ 0) ) + cos(Rt+ R )cos(p(mt+ 0) + )3 3 3

which, after reducing some of the trig expressions, becomes:

3a = (La Lab)Is cos(t) + M IR cos((pm + R)t+ R +p0) (9)

2

Doing the same thing for the rotor phase A yields:

2 2A = M Is(cosp(mt+ 0)cos(t)) + cos(p(mt+ 0) )cos(t ) (10)

3 32 2

+ cos(p(mt+ 0) + ) cos(t+ ) + (LA LAB)IR cos(Rt+ R)3 3

This last expression is, after manipulating:

3A = M Is cos((pm)tp0) + (LA LAB)IR cos(Rt+ R) (11)

2

These two expressions, 9 and 11 give expressions for fluxes in the armature and rotor windingsin terms of currents in the same two windings, assuming that both current distributions are sinusoidal in time and space and represent balanced distributions. The next step is to make anotherassumption, that the stator and rotor frequencies match through rotor rotation. That is:

pm = R (12)It is important to keep straight the different frequencies here:

is stator electrical frequencyR is rotor electrical frequencym is mechanical rotation speed

3

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so that pm is electricalrotation speed.To refer rotor quantities to the stator frame (i.e. non- rotating), and to work in complex

amplitudes, the following definitions are made:

jt)a = Re(ae (13)

A = Re(AejRt) (14)

jt)ia = Re(I e (15)a

iA = Re(IAejRt) (16)

With these definitions, the complex amplitudes embodied in 56 and 64 become:

= LSIa +3

M IAej(R+p0) (17)a 2

A =3

M I ejp0 + LRIAejR (18)2 aThere are two phase angles embedded in these expressions: 0 which describes the rotor physical

phase angle with respect to stator current and R which describes phase angle of rotor currentswith respect to stator currents. We hereby invent two new rotor variables:

AR = Aejp) (19)

= IAej(p0+R) (20)IAR

These are rotor flux and current referred to armature phase angle. Note that AR and IARhave the same phase relationship to each other as do A and IA. Using 19 and 20 in 17 and 18,the basic flux/current relationship for the induction machine becomes:

3a LS 2M Ia= 3 (21)AR2

M LR IAR

This is an equivalent single- phase statement, describing the flux/current relationship in phasea, assuming balanced operation. The same expression will describe phases b and c.

Voltage at the terminals of the stator and rotor (possibly equivalent) windings is, then:

V = j + RaIa (22)a a= jRAR + RAIAR (23)VAR

or:3

V = jLSI +j M IAR + RaI (24)a a 2 a3= jR M Ia +jRLRIAR + RAIAR (25)VAR 2

To carry this further, it is necessary to go a little deeper into the machines parameters. Notethat LS and LR are synchronous inductances for the stator and rotor. These may be separatedinto space fundamental and leakage components as follows:

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l

3 4 0RlN2k2LS= La Lab = S S + LSl (26)

p22 g3 4 0RlN2k2

LR = LA LAB =p2

R R + LRl (27)2 g

Where the normal set of machine parameters holds:

R is rotor radiusis active length

g is the effective air- gapp is the number of pole- pairsN represents number of turnsk represents the winding factorS as a subscript refers to the statorR as a subscript refers to the rotorLl is leakage inductance

The two leakage terms LSl and LRl contain higher order harmonic stator and rotor inductances,slot inducances, end- winding inductances and, if necessary, a provision for rotor skew. Essentially,they are used to represent all flux in the rotor and stator that is not mutually coupled.

In the same terms, the stator- to- rotor mutual inductance, which is taken to comprise onlyaspace fundamental term, is:

4 0RlNSNRkSkRM= (28)

p2 gNote that there are, of course, space harmonic mutual flux linkages. If they were to be included,

they would hair up the analysis substantially. We ignore them here and note that they do have an

effect on machine behavior, but that effect is second- order.Air- gap permeance is defined as:

4 0Rlag = (29)

p2 gso that the inductances are:

3SN2LS= agk2 S+ LSl (30)2

3RN2LR = agk2 R + LRl (31)2

M= agNSNRkSkR (32)Here we define slip sby:

R = s (33)5


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so that

pms= 1 (34)

Then the voltage balance equations become:

3

S

N2V = j agk2S

+ LSl I +j3agNSNRkSkRIAR + RaI (35)a a2

a

2

3 3RN

2VAR = js agNSNRkSkRIa +js agk2 R + LRl IAR + RAIAR (36)2 2At this point, we are ready to define rotor current referred to the stator. This is done by

assuming an effective turns ratio which, in turn, defines an equivalent stator current to producethe same fundamental MMF as a given rotor current:

NRkRI2 = (37)IARNSkS

Now, if we assume that the rotor of the machine is shorted so that VAR = 0 and do somemanipulation we obtain:

V = j(XM+ X1)I +jXMI2 + RaI (38)a a aR2

0 = jXMI +j(XM+ X2)I2 + I2 (39)a swhere the following definitions have been made:

3XM= agN

2Sk

2 (40)2 S

X1 = LSl (41)

NSkS2

X2 = LRl (42)NRkR

NSkS2

R2 = RA (43)NRkR

These expressions describe a simple equivalent circuit for the induction motor shown in Figure 2.We will amplify on this equivalent circuit anon.

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3

Ia Ra X1 X2 I2

Xm (74)n 0Here the brackets denote time average and are here beause of the two- dimensional nature

of the expansion. To carry out 74 on 72, first expand 72 into its complex conjugate parts:

NR1 j(rtk 2p j(rtk 2pI ) 2kKr =

1 Ie NR )

+ e NR

(

) (75)2 R R NRk=0If 75 is used in 74, the second half of 75 results in a sum of terms which time average to zero.

The first half of the expression results in:

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I 2 NR1 j 2pk 2kjnp(K = e NR e )d (76)n 2R 0 NRk=0The impulse function turns the integral into an evaluation of the rest of the integrand at the

impulse. What remains is the sum:

NR1 j(n1) 2kpIK = e NR (77)n 2Rk=0

The sum in 77 is easily evaluated. It is:

NR1 j 2kp(n1) NR if (n 1) P = integere NR = NR (78)

0 otherwisek=0

The integer in 78 may be positive, negative or zero. As it turns out, only the first three of these(zero, plus and minus one) are important, because these produce the largest magnetic fields andtherefore fluxes. These are:

(n 1) p = 1 or n= NRppNR

= 0 or n= 1= 1 or n= NR+p (79)p

Note that 79 appears to produce space harmonic orders that may be of non- integer order. Thisis not really true: is is necessary that npbe an integer, and 79 will always satisfy that condition.

So, the harmonic orders of interest to us are one and

NRn+ = + 1 (80)

pN

Rn = 1 (81) pEach of the space harmonics of the squirrel- cage current will produce radial flux density. A

surface current of the form:

Kn = Re NRIej(rtnp ) (82)2R

produces radial magnetic flux density:

Brn = Re Brnej(rtnp ) (83)where

0NRIB = j

2npg (84)rnIn turn, each of the components of radial flux density will produce a component of induced

voltage. To calculate that, we must invoke Faradays law:

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B(85) E=

tThe radial component of 85, assuming that the fields do not vary with z, is:

1 BrEz = (86)

R tOr, assuming an electric field component of the form:

Ezn = Re E ej(rtnp) (87)nUsing 84 and 87 in 86, we obtain an expression for electric field induced by components of air-

gap flux:rR

E = B (88)n nnp0NRrR

E = j I (89)n 2g(np)2Now, the total voltage induced in a slot pushes current through the conductors in that slot. Wemay express this by:

= ZslotI (90)E1 + En + En+Now: in 90, there are three components of airgap field. E1 is the space fundamental field,

produced by the space fundamental of rotor current as well as by the space fundamental of statorcurrent. The other two components on the left of 90 are produced only by rotor currents andactually represent additional reactive impedance to the rotor. This is often called zigzagleakageinductance. The parameter Zslot represents impedance of the slot itself: resistance and reactanceassociated with cross- slot magnetic fields. Then 90 can be re-written as:

0NRrR

1 1E1 = ZslotI+j 2g (n+p)2 + I (91)(np)2To finish this model, it is necessary to translate 91 back to the stator. See that 65 and 77 make

the link between I and I2:NR

I2 = I (92)6NSkSThen the electric field at the surface of the rotor is:

6NSkS 3 0NSkSR 1 1E1 = Zslot +jr g + (93)NR (n+p)2 (np)2 I2

This must be translated into an equivalent stator voltage. To do so, we use 88 to translate 93into a statement of radial magnetic field, then find the flux liked and hence stator voltage fromthat. Magnetic flux density is:

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pE1B =r rR 6NSkSp Rslot 3 0NSkSp 1 1

+ (94)=NRR r

+jLslot +j g (n+p)2 (np)2 I2

where the slot impedance has been expressed by its real and imaginary parts:

Zslot = Rslot +jrLslot (95)

Flux linking the armature winding is:

0ag = NSkSlR R e B ej(tp) d (96)r

2p

Which becomes:ag = Re ejt (97)ag

where:

2NSkSlR = j B (98)ag rpThen airgap voltage is:

2NSkSlRVag = jag = p Br

= I212lN2Sk2S

NR

jLslot +

R2s

+j6

0RlN2Sk2Sg

1

(n+p)2+

1

(np)2

(99)

Expression 99 describes the relationship between the space fundamental airgap voltage Vagand rotor current I2. This expression fits the equivalent circuit of Figure 4 if the definitions madebelow hold:

X2 I2 R2 s

evaluation of induction machines

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