Massachusetts Institute of Technology

Department of Electrical Engineering and Computer Science 6.685 Electric Machinery Class Notes 8: Analytic Design Evaluation of Induction Machines c 2005 James L. Kirtley Jr.

January 12, 2006

Introduction

Induction machines are perhaps the most widely used of all electric motors. They are generally simple to build and rugged, oer reasonable asynchronous performance: a manageable torque-speed curve, stable operation under load, and generally satisfactory eciency. Because they are so widely used, they are worth understanding. In addition to their current economic importance, induction motors and generators may nd application in some new applications with designs that are not similar to motors currently in commerce. An example is very high speed motors for gas compressors, perhaps with squirrel cage rotors, perhaps with solid iron (or perhaps with both). Because it is possible that future, high performance induction machines will be required to have characteristics dierent from those of existing machines, it is necessary to understand them from rst principles, and that is the objective of this document. It starts with a circuit theoretical view 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 the basic 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 can be constructed. The view that we will take in this document is relentlessly classical. All of the elements that we will use are calculated from rst principles, and we do not resort to numerical analysis or empirical 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 is our feeling that once that understanding exists, it will be possible to employ more sophisticated methods of analysis to get more accurate results for those elements of the machines which do not lend themselves to simple analysis. An elementary picture of the induction machine is shown in Figure 1. The rotor and stator are coaxial. The stator has a polyphase winding in slots. The rotor has either a winding or a cage, also in slots. This picture will be modied slightly when we get to talking of solid rotor machines, anon. Generally, this analysis is carried out assuming three phases. As with many systems, this generalizes to dierent numbers of phases with little diculty.

Induction Motor Transformer Model

The induction machine has two electrically active elements: a rotor and a stator. In normal operation, the stator is excited by alternating voltage. (We consider here only polyphase machines). The stator excitation creates a magnetic eld in the form of a rotating, or traveling wave, which induces currents in the circuits of the rotor. Those currents, in turn, interact with the traveling 1

Stator Core

Stator Winding in Slots Rotor Winding or Cage in Slots

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 the stator can be described by balanced, three phase windings. The two sets are, of course, coupled by mutual inductances which are dependent on rotor position. Stator uxes are (a , b , c ) and rotor uxes are (A , B , C ). The ux vs. current relationship is given by:

a b c A B C

L S MT SR

M SR

LR

ia ib ic iA iB iC

(1)

where the component matrices are: La Lab Lab LS = Lab La Lab Lab Lab La

(2)

The mutual inductance part of (1) is a circulant matrix: M cos(p ) = M cos(p M cos(p +

LA LAB LAB LAB LR = LAB LA LAB LAB LA M cos(p + 2 ) M cos(p ) 3 2 3 ) M cos(p 2

2 3 )

(3)

M SR

M cos(p M cos(p + 2 3 ) M cos(p )

2 3 ) 2 3 )

(4)

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 = IS cos(t)
ib = IS cos(t ic = IS cos(t +
2 3 ) 2 3 )

(5)

iA = IR cos(R t + R )
iB = IR cos(R t + R iC = IR cos(R t + R + The rotor position can be described by = m t + 0

2 3 ) 2 3 )

(6)

(7)

Under these assumptions, we may calculate the form of stator uxes. As it turns out, we need only write out the expressions for a and A to see what is going on: a = (La Lab )Is cos(t) + M IR (cos(R t + R ) cos p(m + 0 ) (8) 2 2 2 2 + cos(R t + R + ) cos(p(m t + 0 ) ) + cos(R t + R ) cos(p(m t + 0 ) + ) 3 3 3
3 which, after reducing some of the trig expressions, becomes:
3 a = (La Lab )Is cos(t) + M IR cos((pm + R )t + R + p0 ) 2
Doing the same thing for the rotor phase A yields:
A = M Is (cos p(m t + 0 ) cos(t)) + cos(p(m t + 0 ) + cos(p(m t + 0 ) + 2 2 ) cos(t ) 3 3 (9)

(10)

2 2 ) cos(t + ) + (LA LAB )IR cos(R t + R ) 3 3

This last expression is, after manipulating:

3
A = M Is cos(( pm )t p0 ) + (LA LAB )IR cos(R t + R ) 2 (11)

These two expressions, 9 and 11 give expressions for uxes in the armature and rotor windings in 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 another assumption, that the stator and rotor frequencies match through rotor rotation. That is: pm = R It is important to keep straight the dierent frequencies here: R m is stator electrical frequency is rotor electrical frequency is mechanical rotation speed (12)

so that pm is electrical rotation speed. To refer rotor quantities to the stator frame (i.e. non- rotating), and to work in complex amplitudes, the following denitions are made: a = Re(a ejt ) A = Re(A ejR t ) ia = Re(I a ejt ) iA = Re(I A ejR t ) With these denitions, the complex amplitudes embodied in 56 and 64 become: 3 a = LS I a + M I A ej (R +p0 ) 2 (17) (13) (14) (15) (16)

3 A = M I a ejp0 + LR I A ejR (18) 2 There 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 currents with respect to stator currents. We hereby invent two new rotor variables: AR = A ejp) I AR = I A ej (p0 +R ) (19) (20)

These are rotor ux and current referred to armature phase angle. Note that AR and I AR have the same phase relationship to each other as do A and I A . Using 19 and 20 in 17 and 18, the basic ux/current relationship for the induction machine becomes: a AR = LS 3 2M
3 2M LR

Ia I AR

(21)

This is an equivalent single- phase statement, describing the ux/current relationship in phase a, 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 a = j a + Ra I a V AR = jR AR + RA I AR or: 3 V a = jLS I a + j M I AR + Ra I a 2 (22) (23) (24)

3 (25) V AR = jR M I a + jR LR I AR + RA I AR 2 To carry this further, it is necessary to go a little deeper into the machines parameters. Note that LS and LR are synchronous inductances for the stator and rotor. These may be separated into space fundamental and leakage components as follows: 4

LS = La Lab = LR = LA LAB =

2 k2 3 4 0 RlNS S + LSl 2 2 p g 2 k2 3 4 0 RlNR R + LRl 2 p2 g

(26) (27)

Where the normal set of machine parameters holds: R l g p N k S R Ll is rotor radius is active length
is the eective air- gap
is the number of pole- pairs
represents number of turns
represents the winding factor
as a subscript refers to the stator
as a subscript refers to the rotor
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 ux 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 only a space fundamental term, is: M= 4 0 RlNS NR kS kR p2 g (28)

Note that there are, of course, space harmonic mutual ux 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 eect on machine behavior, but that eect is second- order. Air- gap permeance is dened as: ag = so that the inductances are: 3 2 2 LS = ag kS NS + LSl 2 3 2 2 LR = ag kR + LRl NR 2 M = ag NS NR kS kR Here we dene slip s by: R = s 5 (33) (30) (31) (32) 4 0 Rl p2 g (29)

so that
s=1 Then the voltage balance equations become: V a = j 3 3 2 2 NS + LSl I a + j ag NS NR kS kR I AR + Ra I a ag kS 2 2 (35) (36) pm (34)

3 3 2 2 ag kR V AR = js ag NS NR kS kR I a + js NR + LRl I AR + RA I AR 2 2

At this point, we are ready to dene rotor current referred to the stator. This is done by assuming an eective turns ratio which, in turn, denes an equivalent stator current to produce the same fundamental MMF as a given rotor current: I2 = NR kR I NS kS AR (37)

Now, if we assume that the rotor of the machine is shorted so that V AR = 0 and do some manipulation we obtain: V a = j (XM + X1 )I a + jXM I 2 + Ra I a 0 = jXM I a + j (XM + X2 )I 2 + where the following denitions have been made: XM = 3 2 2 kS ag NS 2 (40) (41)
2

(38) (39)

R2 I s 2

X1 = LSl X2 = LRl R2 = RA NS kS NR kR NS kS NR kR
2

(42) (43)

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

I a Ra

X1

X2 I2 < > R2 Xm < > < s

Figure 2: Equivalent Circuit

Operation: Energy Balance

Now we are ready to see how the induction machine actually works. Assume for the moment that Figure 2 represents one phase of a polyphase system and that the machine is operated under balanced conditions and that speed is constant or varying only slowly. Balanced conditions means that each phase has the same terminal voltage magnitude and that the phase dierence between phases is a uniform. Under those conditions, we may analyze each phase separately (as if it were a single phase system). Assume an RMS voltage magnitude of Vt across each phase. The gap impedance, or the impedance looking to the right from the right-most terminal of X1 is: R2 Zg = jXm ||(jX2 + ) (44) s A total, or terminal impedance is then Zt = jX1 + Ra + Zg and terminal current is It = Rotor current is found by using a current divider: I2 = It jXm 2 jX2 + R s + jXm R2 s (47) Vt Zt (45) (46)

Air-gap power is then calculated (assuming a three-phase machine): Pag = 3|I2 |2 (48)

This is real (time-average) power crossing the air-gap of the machine. Positive slip implies rotor speed less than synchronous and positive air-gap power (motor operation). Negative slip means rotor speed is higher than synchronous, negative air-gap power (from the rotor to the stator) and generator operation. Now, note that this equivalent circuit represents a real physical structure, so it should be possible to calculate power dissipated in the physical rotor resistance, and that is: Ps = Pag s 7 (49)

(Note that, since both Pag and s will always have the same sign, dissipated power is positive.) The rest of this discussion is framed in terms of motor operation, but the conversion to generator operation is simple. The dierence between power crossing the air-gap and power dissipated in the rotor resistance must be converted from mechanical form: Pm = Pag Ps and electrical inputpower is: Pin = Pag + Pa where armature dissipation is: Pa = 3|It |2 Ra Output (mechanical) power is Pout = Pag Pw Where Pw describes friction, windage and certain stray losses which we will discuss later. And, nally, eciency and power factor are: = Pout Pin Pin 3Vt It (53) (52) (51) (50)

(54) (55)

cos =

3.1

Example of Operation

The following MATLAB script generates a torque-speed and power-speed curve for the simple induction motor model described above. Note that, while the analysis does not require that any of the parameters, such as rotor resistance, be independent of rotor speed, this simple script does assume that all parameters are constant.

3.2

Example

That MATLAB script has been run for a standard motor with parameters given in Table 1. Torque vs. speed and power vs. speed are plotted for this motor in Figure 3. These curves were generated by the MATLAB script shown above.

Squirrel Cage Machine Model

Now we derive a circuit model for the squirrel-cage motor using eld analytical techniques. The model consists of two major parts. The rst of these is a description of stator ux in terms of stator and rotor currents. The second is a description of rotor current in terms of air- gap ux. The result of all of this is a set of expressions for the elements of the circuit model for the induction machine. To start, assume that the rotor is symmetrical enough to carry a surface current, the fundamental of which is: K r = z Re K r ej (stp ) = z Re K r ej (tp) 8 (56)

% ------------------------------------------------------
% Torque-Speed Curve for an Induction Motor
% Assumes the classical model
% This is a single-circuit model
% Required parameters are R1, X1, X2, R2, Xm, Vt, Ns
% Assumed is a three-phase motor
% This thing does a motoring, full speed range curve
% Copyright 1994 James L. Kirtley Jr.
% -------------------------------------------------------
s = .002:.002:1; % vector of slip N = Ns .* (1 - s); % Speed, in RPM oms = 2*pi*Ns/60; % Synchronous speed Rr = R2 ./ s; % Rotor resistance Zr = j*X2 + Rr; % Total rotor impedance Za = par(j*Xm, Zr); % Air-gap impedance Zt = R1 + j*X1 +Za; % Terminal impedance Ia = Vt ./ Zt; % Terminal Current I2 = Ia .* cdiv (Zr, j*Xm); % Rotor Current Pag = 3 .* abs(I2) .^2 .* Rr; % Air-Gap Power Pm = Pag .* (1 - s); % Converted Power Trq = Pag ./ oms; % Developed Torque subplot(2,1,1) plot(N, Trq) title(Induction Motor); ylabel(N-m); subplot(2,1,2) plot(N, Pm); ylabel(Watts); xlabel(RPM);

Table 1: Example, Standard Motor

Rating Voltage Stator Resistance R1 Rotor Resistance R2 Stator Reactance X1 Rotor Reactance X2 Magnetizing Reactance Xm Synchronous Speed Ns 300 440 254 .73 .64 .06 .06 2.5 1200 kw VRMS, l-l VRMS, l-n RPM

Induction Motor
300
250
200
Nm 150
100
50
0
0 200 400 600 800 1000 1200

3 2.5
2

x 10

Watts

1.5
1
0.5
0
0 200 400 600 RPM 800 1000 1200

Figure 3: Torque and Power vs. Speed for Example Motor Note that in 56 we have made use of the simple transformation between rotor and stator coordinates: = m t (57) and that pm = r = (1 s) Here, we have used the following symbols: Kr s r m is is is is is complex amplitude of rotor surface current per- unit slip stator electrical frequency rotor electrical frequency rotational speed (58)

The rotor current will produce an air- gap ux density of the form: Br = Re B r ej (tp) where B r = j0 10 R K pg r (59)

(60)

Note that this describes only radial magnetic ux density produced by the space fundamental of rotor current. Flux linked by the armature winding due to this ux density is:
0

AR = lNS kS This yields a complex amplitude for AR :

Br ()Rd

(61)

AR = Re AR ejt where AR = 2l0 R2 NS kS Kr p2 g

(62)

(63)

Adding this to ux produced by the stator currents, we have an expression for total stator ux: a =
2 Rlk 2 3 4 0 NS 2l0 R2 NS kS S + + L Kr I Sl a 2 p2 g p2 g

(64)

Expression 64 motivates a deniton of an equivalent rotor current I2 in terms of the space fundamental of rotor surface current density: I2 = R K 3 NS kS z (65)

Then we have the simple expression for stator ux: a = (Lad + LSl )I a + Lad I 2 where Lad is the fundamental space harmonic component of stator inductance: Lad =
2 k 2 Rl 3 4 0 NS S 2 p2 g

(66)

(67)

4.1

Eective Air-Gap: Carters Coecient

In induction motors, where the air-gap is usually quite small, it is necessary to correct the air-gap permeance for the eect of slot openings. These make the permeance of the air-gap slightly smaller than calculated from the physical gap, eectively making the gap a bit bigger. The ratio of eective to physical gap is: t+s (68) ge = g t + s gf () where f () = f s 2g = tan() log sec (69)

11

4.2

Squirrel Cage Currents

The second part of this derivation is the equivalent of nding a relationship between rotor ux and I2 . However, since this machine has no discrete windings, we must focus on the individual rotor bars. Assume that there are NR slots in the rotor. Each of these slots is carrying some current. If the machine is symmetrical and operating with balanced currents, we may write an expression for current in the kth slot as: ik = Re I k ejst where I k = Ie
p j 2 N R

(70)

(71)

and I is the complex amplitude of current in slot number zero. Expression 71 shows a uniform progression of rotor current phase about the rotor. All rotor slots carry the same current, but that current is phase retarded (delayed) from slot to slot because of relative rotation of the current wave at slip frequency. The rotor current density can then be expressed as a sum of impulses: Kz = Re

NR 1 k =0
p 1 j (r tk 2 2k ) NR Ie ) ( R NR

(72)

The unit impulse function () is our way of approximating the rotor current as a series of impulsive currents around the rotor. This rotor surface current may be expressed as a fourier series of traveling waves:

Kz = Re
n=

K n ej (r tnp )

(73)

Note that in 73, we are allowing for negative values of the space harmonic index n to allow for reverse- rotating waves. This is really part of an expansion in both time and space, although we are considering only the time fundamental part. We may recover the nth space harmonic component of 73 by employing the following formula: K n =< 1
2 0

Kr (, t)ej (r tnp) d >

(74)

Here the brackets <> denote time average and are here beause of the two- dimensional nature of the expansion. To carry out 74 on 72, rst expand 72 into its complex conjugate parts: Kr = 1 2
NR 1 k =0
p p I j (r tk 2 I j (r tk 2 2k ) ) NR NR ) e + e ( R R NR

(75)

If 75 is used in 74, the second half of 75 results in a sum of terms which time average to zero. The rst half of the expression results in:

12

Kn =

I 2R

2 NR 1 0 k =0

pk j 2N jnp R

2k )d NR

(76)

The impulse function turns the integral into an evaluation of the rest of the integrand at the impulse. What remains is the sum: Kn = I 2R
NR 1 k =0

kp j (n1) 2N R

(77)

The sum in 77 is easily evaluated. It is:

NR 1 k =0

2kp(n1) NR

P = integer NR if (n 1) N R 0 otherwise

(78)

The integer in 78 may be positive, negative or zero. As it turns out, only the rst three of these (zero, plus and minus one) are important, because these produce the largest magnetic elds and therefore uxes. These are: (n 1) p NR
p = 1 or n = NR p

=0 =1

or n = 1 or n =
NR + p p

(79)

Note that 79 appears to produce space harmonic orders that may be of non- integer order. This is not really true: is is necessary that np be an integer, and 79 will always satisfy that condition. So, the harmonic orders of interest to us are one and n+ = n NR +1 p NR = 1 p (80) (81)

Each of the space harmonics of the squirrel- cage current will produce radial ux density. A surface current of the form: Kn = Re produces radial magnetic ux density:
Brn = Re B rn ej (r tnp ) where 0 NR I (84) 2npg In turn, each of the components of radial ux density will produce a component of induced voltage. To calculate that, we must invoke Faradays law: B rn = j 13

NR I j (r tnp ) e 2R

(82)

(83)

B t
The radial component of 85, assuming that the elds do not vary with z , is:
E = Br 1 Ez = R t Or, assuming an electric eld component of the form: Ezn = Re E n ej (r tnp)

(85)

(86)

(87)

Using 84 and 87 in 86, we obtain an expression for electric eld induced by components of airgap ux: r R En = (88) B np n E n = j 0 NR r R I 2g(np)2 (89)

Now, the total voltage induced in a slot pushes current through the conductors in that slot. We may express this by: E 1 + E n + E n+ = Z slot I (90)

Now: in 90, there are three components of air- gap eld. E1 is the space fundamental eld, produced by the space fundamental of rotor current as well as by the space fundamental of stator current. The other two components on the left of 90 are produced only by rotor currents and actually represent additional reactive impedance to the rotor. This is often called zigzag leakage inductance. The parameter Zslot represents impedance of the slot itself: resistance and reactance associated with cross- slot magnetic elds. Then 90 can be re-written as: E 1 = Z slot I + j 0 NR r R 2g 1 1 + I 2 (n+ p) (n p)2 (91)

To nish this model, it is necessary to translate 91 back to the stator. See that 65 and 77 make the link between I and I 2 : I2 = NR I 6NS kS 1 1 + 2 (n+ p) (n p)2 (92)

Then the electric eld at the surface of the rotor is: E1 = 6NS kS 3 0 NS kS R Z slot + jr NR g I2 (93)

This must be translated into an equivalent stator voltage. To do so, we use 88 to translate 93 into a statement of radial magnetic eld, then nd the ux liked and hence stator voltage from that. Magnetic ux density is:

14

Br = =

pE 1 r R 6NS kS p NR R

Rslot 3 0 NS kS p + jLslot + j r g

1 1 + 2 (n+ p) (n p)2

I2

(94)

where the slot impedance has been expressed by its real and imaginary parts: Z slot = Rslot + jr Lslot Flux linking the armature winding is:
0

(95)

ag = NS kS lR Which becomes:

2 p

Re B r ej (tp) d

(96)

ag = Re ag ejt where: ag = j Then air- gap voltage is: V ag = j ag = = I 2 2NS kS lR Br p 2 k2 12lNS R2 S jLslot + NR s 2NS kS lR B r p

(97)

(98)

+ j

2 k2 6 0 RlNS S g

1 1 + 2 (n p)2 (n+ p)

(99)

Expression 99 describes the relationship between the space fundamental air- gap voltage V ag and rotor current I 2 . This expression ts the equivalent circuit of Figure 4 if the denitions made below hold: X2 I2

< > R2 < > s <

Figure 4: Rotor Equivalent Circuit

X2 R2

2 k2 2 k2 12lNS 6 0 RlNS S S = Lslot + NR g 2 k2 12lNS S = Rslot NR

1 1 + 2 (NR + p) (NR p)2

(100) (101)

15

The rst term in 100 expresses slot leakage inductance for the rotor. Similarly, 101 expresses rotor resistance in terms of slot resistance. Note that Lslot and Rslot are both expressed per unit length. The second term in 100 expresses the zigzag leakage inductance resulting from harmonics on the order of rotor slot pitch. Next, see that armature ux is just equal to air- gap ux plus armature leakage inductance. That is, 66 could be written as: a = ag + Lal I a (102)

4.3

Stator Leakage

There are a number of components of stator leakage Lal , each representing ux paths that do not directly involve the rotor. Each of the components adds to the leakage inductance. The most prominent components of stator leakage are referred to as slot, belt, zigzag, end winding, and .skew Each of these will be discussed in the following paragraphs. 4.3.1 Belt Leakage

Belt and zigzag leakage components are due to air- gap space harmonics. As it turns out, these are relatively complicated to estimate, but we may get some notion from our rst- order view of the machine. The trouble with estimating these leakage components is that they are not really independent of the rotor, even though we call them leakage. Belt harmonics are of order n = 5 and n = 7. If there were no rotor coupling, the belt harmonic leakage terms would be: Xag5 = Xag7 =
2 k 2 Rl 3 4 0 NS 5 2 52 p2 g

(103)

2 k 2 Rl 3 4 0 NS 7 (104) 2 72 p2 g The belt harmonics link to the rotor, however, and actually appear to be in parallel with components of rotor impedance appropriate to 5p and 7p pole- pair machines. At these harmonic orders we can usually ignore rotor resistance so that rotor impedance is purely inductive. Those components are: 2 k2 2 k2 6 0 RlNS 12lNS 5 5 Lslot + NR g 2 k2 2 k2 12lNS 6 0 RlNS 7 7 Lslot + NR g

X2,5 = X2,7 =

1 1 + 2 (NR + 5p) (NR 5p)2 1 1 + 2 (NR + 7p) (NR 7p)2

(105) (106)

In the simple model of the squirrel cage machine, because the rotor resistances are relatively small and slip high, the eect of rotor resistance is usually ignored. Then the fth and seventh harmonic components of belt leakage are: X5 = Xag5 X2,5 X7 = Xag7 X2,7 16 (107) (108)

4.3.2

Zigzag Leakage

Stator zigzag leakage is from those harmonics of the orders pns = Nslots p where Nslots . Xz =
2 Rl 3 4 0 NS 2 g

Note that these harmonic orders do not tend to be shorted out by the rotor cage and so no direct interaction with the cage is ordinarily accounted for. 4.3.3 Skew Leakage

kns + kns + 2 (Nslots + p) (Nslots p)2

(109)

In order to reduce saliency eects that occur because the rotor teeth will tend to try to align with the stator teeth, induction motor designers always use a dierent number of slots in the rotor and stator. There still may be some tendency to align, and this produces cogging torques which in turn produce vibration and noise and, in severe cases, can retard or even prevent starting. To reduce this tendency to cog, rotors are often built with a little skew, or twist of the slots from one end to the other. Thus, when one tooth is aligned at one end of the machine, it is un-aligned at the other end. A side eect of this is to reduce the stator and rotor coupling by just a little, and this produces leakage reactance. This is fairly easy to estimate. Consider, for example, a space-fundamental ux density Br = B1 cos p , linking a (possibly) skewed full-pitch current path: =
l 2 l 2 x +p 2p l 2 +p p x l

B1 cos pRddx

Here, the skew in the rotor is electrical radians from one end of the machine to the other. Evaluation of this yields: 2B1 Rl sin 2 = p 2 Now, the dierence between what would have been linked by a non-skewed rotor and what is linked by the skewed rotor is the skew leakage ux, now expressible as: Xk = Xag 1 4.3.4 Stator Slot Leakage

sin 2 2 2

Currents in the stator slots produce uxes that link the stator conductors but not the rotor. To estimate these uxes, refer to the slot geometry shown in Figure 4.3.4. This shows a possibly unrealistic straight-sided stator slot. Typical in induction machines is for such slots to be trapezoidal in shape. A more careful eld analysis than we will do here shows that this analysis will be no more than a few percent in error if the slot width used in the calculation is the slot top (the end of the slot closest to the air-gap). There are ve important dimensions here: the slot height h, width w and the slot depression height d and width u, and (not shown) length . To estimate slot leakage inductance we assume some current in the slot, calculate the magnetic energy that results and then use the expression:

17

1 wm = L I 2 2 If there are N conductors in the slot, each carrying current I , the current density in the slot is: J= NI hw

Using Amperes Law around a loop (shown dotted in the gure), magnetic eld in the x direction at height y from the bottom of the slot is: Hx = In the slot depression that eld is: NI y w h

NI u Magnetic energy stored in the slot and slot depression are then conveniently calculated as: Hxd = 1 wm = L I 2 = w 2 Noting the slot permeance as:
h 0

1 1h d 1 2 2 + dy + ud Hxd = 0 0 H x N 2I 2 2 2 3w u P = 0

1h d + 3w u We have the total inductance of the slot to be: L = P N 2 For the purpose of this estimate we will assume an ordinary winding consisting of coils of Nc turns each. For such a winding if there are m slots per pole per phase and p pole pairs and if the winding is short-pitched by Nsp slots, there will be 2p(m Nsp ) slots per phase with two coils from the same phase and 2pNsp slots per phase sharing another phase. (We assume here a three phase machine). Then the self slot leakage inductance must be:
2 2 Ls = P 4Nc 2p(m Nsp ) + Nc 2pNsp

Since there are a total of pNsp mutual slots between each pair of phases, and the sense of the windings is opposite, the mutual component of slot leakage is:
2 Lm = P pNsp Nc

Total slot leakage is then:

2 L = Ls Lm = P pNc (8m 5Nsp )

Expressed in terms of the total number of stator turns, Na = 2pmNc , L = P

2 Na p

2 5 Nsp m 4 m2

18

0 1 1111 0000 00000 011111 1 000 111 000 111 0 1 0000 1111 0 1 00000 0 1 011111 1 000 111 000 111 0 1 0 1 0 0J 1 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 000000 111111 0 1 0 1 000000 111111

y x h

Figure 5: Stator Slot Geometry for Leakage Calculation 4.3.5 End Winding Leakage

The nal component of leakage reactance is due to the end windings. This is perhaps the most dicult of the machine parameters to estimate, being essentially three-dimensional in nature. There are a number of ways of estimating this parameter, but for our purposes we will use a simplied parameter from Alger[1]: Xe =
2 14 q 0 RNa (p 0.3) 4 2 2 p2

As with all such formulae, extreme care is required here, since we can give little guidance as to when this expression is correct or even close. And we will admit that a more complete treatment of this element of machine parameter construction would be an improvement.

4.4

Stator Winding Resistance

Estimating stator winding resistance is fairly straightforward once end winding geometry is known. Total length of the armature winding is, per phase: w = Na 2 ( + e ) Estimating e , the length of one end winding, requires knowing how the winding is laid out and is beyond our scope here. (But once you see it you will know that length.) The area of the winding may be estimated by knowing wire diameter and how many strands are in parallel: Aw = 2 d Ninh 4 w 2Nc Aw Aslot 19

The area of the winding is related to slot area by a winding factor: a =

Winding resistance, per phase, is simply Ra = w Aw

where is wire conductivity. Note that conductivity of the materials used in induction machines is a function of temperature and so will be winding resistance (and rotor resistance for that matter). The Fitzgerald, Kingsley and Umans textbook[2] gives the following correction for resistance of copper: RT = R t T0 + T T0 + t

where RT and Rt are resistances at temperatures T and t. T0 = 234.5 for copper with basic conductivity of IACS (5.8 107 S/m)[3]. For aluminum with conductivity of 63% of IACS, T0 212.9 Temperatures are given in Celcius.

4.5

Harmonic Order Rotor Resistance and Stray Load Losses

It is important to recognize that the machine rotor sees each of the stator harmonics in essentially the same way, and it is quite straightforward to estimate rotor parameters for the harmonic orders, as we have done just above. Now, particularly for the belt harmonic orders, there are rotor currents owing in response to stator mmfs at fth and seventh space harmonic order. The resistances attributable to these harmonic orders are: R2,5 = R2,7 =
2 k2 12lNs 5 Rslot,5 NR 2 k2 12lNs 7 Rslot,7 NR

(110) (111)

The higher-order slot harmonics will have relative frequencies (slips) that are: sn = 1 (1 s)n n = 6k + 1 n = 6k 1 k an integer (112)

The induction motor electromagnetic interaction can now be described by an augmented magnetic circuit as shown in Figure 20. Note that the terminal ux of the machine is the sum of all of the harmonic uxes, and each space harmonic is excited by the same current so the individual harmonic components are in series. Each of the space harmonics will have an electromagnetic interaction similar to the fundamental: power transferred across the air-gap is:
2 Pem,n = 3I2 ,n

R2,n sn

Of course dissipation in each circuit is:

2 Pd,n = 3I2 ,n R2,n

20

leaving

R2,n (1 sn ) sn Note that this equivalent circuit has provision for two sets of circuits which look like cages. In fact one of these sets is for the solid rotor body if that exists. We will discuss that anon. There is also a provision (rc ) for loss in the stator core iron. Power deposited in the rotor harmonic resistance elements is characterized as stray load loss because it is not easily computed from the simple machine equivalent circuit.
2 Pm,n = 3I2 ,n

4.6

Slot Models

Some of the more interesting things that can be done with induction motors have to do with the shaping of rotor slots to achieve particular frequency-dependent eects. We will consider here three cases, but there are many other possibilities. First, suppose the rotor slots are representable as being rectangular, as shown in Figure 6, and assume that the slot dimensions are such that diusion eects are not important so that current in the slot conductor is approximately uniform. In that case, the slot resistance and inductance per unit length are: Rslot = 1 ws hs hs Lslot = 0 3ws (113) (114)

The slot resistance is obvious, the slot inductance may be estimated by recognizing that if the current in the slot is uniform, magnetic eld crossing the slot must be: Hy = I x ws hs
2

Then energy stored in the eld in the slot is simply: 1 L I 2 = ws 2 slot

hs 0

0 2

Ix ws hs

dx =

1 0 hs 2 I 6 ws

4.7

Deep Slots

Now, suppose the slot is not small enough that diusion eects can be ignored. The slot becomes deep to the extent that its depth is less than (or even comparable to) the skin depthfor conduction at slip frequency. Conduction in this case may be represented by using the Diusion Equation: 2 H = 0 H t

In the steady state, and assuming that only cross-slot ux (in the y direction) is important, and the only variation that is important is in the radial (x) direction: 2 Hy = js 0 Hy x2 21

wd

hd hs y

ws
Figure 6: Single Slot This is solved by solutions of the form: Hy = H e(1+j ) where the skin depth is = 2 s 0
x

Since Hy must vanish at the bottom of the slot, it must take the form: Hy = Htop Since current is the curl of magnetic eld, Jz = Ez =
s Hy 1 + j cosh(1 + j ) h = Htop s x sinh(1 + j ) h

sinh(1 + j ) x s sinh(1 + j ) h

Then slot impedance, per unit length, is: 1 1+j hs Zslot = coth(1 + j ) ws Of course the impedance (purely reactive) due to the slot depression must be added to this. It is possible to extract the real and imaginary parts of this impedance (the process is algebraically a bit messy) to yield: Rslot =
hs s 1 sinh 2 h + sin 2 hs s ws cosh 2 h cos 2

Lslot = 0

hs s 1 1 sinh 2 h hd sin 2 + hs s wd s ws cosh 2 h cos 2

22

4.8

Arbitrary Slot Shape Model

It is possible to obtain a better model of the behavior of rotor conductor slots by using simple numerical methods. In many cases rotor slots are shaped with the following objectives in mind: 1. A substantial part of the periphery of the rotor should be devoted to active conductor, for good running performance. 2. The magnetic iron of the rotor must occupy a certain fraction of the periphery, to avoid saturation. 3. For good starting performance, some means of forcing current to ow only in the top part of the rotor bar should be devised. Generally the rotor teeth, which make up part of the machines magnetic circuit, are of roughly constant width to avoid ux concentration. The rotor conductor bars are therefore tapered, with their narrow ends towards the center of the rotor. To provide for current concentration on starting they often have a starting bar at the outer periphery of the rotor with a much narrower region which has high inductance just below. The bulk of the rotor bar occupies the tapered region allowed between the teeth. This geometry is quite a bit more complicated than that described in the previous section. Note that, if we can describe the slot impedance per unit length as a function of frequency: Zs ( ) = Rs ( ) + jXs ( ), we can carry out the analysis of the machine as described previously. Thus our analysis is directed toward frequency response modeling of the rotor slot. Focusing then on a single slot, use the notation as described in Figure 7.

Ez [n]

w[n]

x = n x

Ez [n1]

x y z

Figure 7: Slot Geometry Notation The impedance per unit length is the ratio between slot current and axial electric eld: Zs = Ez I

For the purpose of this analysis we will use the symbol x as the radial distance from the bottom of the slot. Assume the slot can be divided radially into a number of regions or slices, each with 23

radial height x. We further assume that currents are axially (z ) directed and that magnetic eld crosses the slot in the y direction. Under these assumptions the electric eld at the top of one of the slices is related to the electric eld at the bottom of the slice by magnetic eld crossing through the slice. Using the trapezoidal rule for integration: E z (x) E z (x x) = j0 The magnetic eld is simply: H y (x) = 1 w(x)
x 0

x H y (x) + H y (x x) 2

w(x)Ez (x)dx =

1 wn

nI n
i=1

where In is the total current owing in one slice. Note that this can be reformulated into a ladder network by again using the trapezoidal rule for integration: current owing in slice number n would be: x (wn E n + wn1 E n1 ) In = 2 Now the slot may be described as is shown in the ladder network of Figure 8. The incremental reactance of one slice is: Xn = 0 x and the resistance of a slice is: Rn =
L[n] 2

2 (wn + wn1 )

1 2 x wn + wn1
L[n1] 2 L[n1] 2

L[n] 2

R[n]

R[n1]

Figure 8: Slot Impedance Ladder Network The procedure is to start at the bottom of the slot, corresponding to the right-hand end of the ladder (the inductance at the bottom of the slot is innite so the rst slice has only the resistance), and building toward the top of the slot.

4.9

Multiple Cages

In some larger induction motors the rotor cage is built in such a way as to separate the functions of starting and running. The purpose of a deep slot is to improve starting performance of a motor. When the rotor is stationary, the frequency seen by rotor conductors is relatively high, and current crowding due to the skin eect makes rotor resistance appear to be high. As the rotor accelerates the frequency seen from the rotor drops, lessening the skin eect and making more use of the rotor conductor. This, then, gives the machine higher starting torque (requiring high resistance) without compromising running eciency. 24

This eect can be carried even further by making use of multiple cages , such as is shown in Figure 9. Here there are two conductors in a fairly complex slot. Estimating the impedance of this slot is done in stages to build up an equivalent circuit.
wd hd h2 hs w2 ws h1

w1

Figure 9: Double Slot Assume for the purposes of this derivation that each section of the multiple cage is small enough that currents can be considered to be uniform in each conductor. Then the bottom section may be represented as a resistance in series with an inductance: Ra = La = 1 w1 h1 0 h1 3 w1

The narrow slot section with no conductor between the top and bottom conductors will contribute an inductive impedance: hs Ls = 0 ws The top conductor will have a resistance: Rb = 1 w2 h2

Now, in the equivalent circuit, current owing in the lower conductor will produce a magnetic eld across this section, yielding a series inductance of Lb = 0 h2 w2

By analogy with the bottom conductor, current in the top conductor ows through only one third of the inductance of the top section, leading to the equivalent circuit of Figure 10, once the inductance of the slot depression is added on: hd Lt = 0 wd Now, this rotor bar circuit ts right into the framework of the induction motor equivalent circuit, shown for the double cage case in Figure 11, with R2a =
2 k2 12lNS S Ra NR

25

Lt

1 3 Lb

< > < > Rb <

2 3 Lb

Ls

La

< > < > Ra <

Figure 10: Equivalent Circuit: Double Bar

2 k2 12lNS S Rb NR 2 k2 2 12lNS S ( Lb + Ls + La ) = NR 3 2 k2 12lNS 1 S = (Lt + Lb ) NR 3

R2b = X2a X2a

I a Ra

X1

X2b I2 X2a < < > R2b < > R2a Xm < > s > s < <

Figure 11: Equivalent Circuit: Double Cage Rotor

4.10

Rotor End Ring Eects

It is necessary to correct for end ring resistance in the rotor. To do this, we note that the magnitude of surface current density in the rotor is related to the magnitude of individual bar current by: 2R Iz = Kz (115) NR Current in the end ring is: IR = Kz R p (116)

Then it is straightforward to calculate the ratio between power dissipated in the end rings to power dissipated in the conductor bars themselves, considering the ratio of current densities and volumes. Assuming that the bars and end rings have the same radial extent, the ratio of current densities is: JR NR wr = Jz 2p lr 26 (117)

where wr is the average width of a conductor bar and lr is the axial end ring length. Now, the ratio of losses (and hence the ratio of resistances) is found by multiplying the square of current density ratio by the ratio of volumes. This is approximately: Rend = Rslot NR wr 2p lr
2

2R lr NR Rwr = NR l wr llr p2

(118)

4.11

Windage

Bearing friction, windage loss and fan input power are often regarded as elements of a black art. We approach them with some level of trepidation, for motor manufacturers seem to take a highly empirical view of these elements. What follows is an attempt to build reasonable but simple models for two eects: loss in the air gap due to windage and input power to the fan for cooling. Some caution is required here, for these elements of calculation have not been properly tested, although they seem to give reasonable numbers The rst element is gap windage loss. This is produced by shearing of the air in the relative rotation gap. It is likely to be a signigant element only in machines with very narrow air gaps or very high surface speeds. But these include, of course, the high performance machines with which we are most interested. We approach this with a simple couette ow model. Air-gap shear loss is approximately: Pw = 2R4 3 la f (119) where a is the density of the air-gap medium (possibly air) and f is the friction factor , estimated by: .0076 (120)
f= 1 4 Rn and the Reynolds NumberRn is Rg (121) Rn = air and air is the kinematic viscosity of the air-gap medium. The second element is fan input power. We base an estimate of this on two hypotheses. The rst of these is that the mass ow of air circulated by the fan can be calculated by the loss in the motor and an average temperature rise in the cooling air. The second hypothesis is the the pressure rise of the fan is established by the centrifugal pressure rise associated with the surface speed at the outside of the rotor. Taking these one at a time: If there is to be a temperature rise T in the cooling air, then the mass ow volume is: m = and then volume ow is just v = Pressure rise is estimated by centrifugal force: P = air 27 r p fan
2

Pd Cp T m air

then power is given by: Pfan = P v For reference, the properties of air are: Density Kinematic Viscosity Heat Capacity air air Cp 1.18 1.56 105 1005.7 kg/m2 m2 /sec J/kg

4.12

Magnetic Circuit Loss and Excitation

There will be some loss in the stator magnetic circuit due to eddy current and hysteresis eects in the core iron. In addition, particularly if the rotor and stator teeth are saturated there will be MMF expended to push ux through those regions. These eects are very dicult to estimate from rst principles, so we resort to a simple model. Assume that the loss in saturated steel follows a law such as: Pd = PB e B
f

B BB

(122)

This is not too bad an estimate for the behavior of core iron. Typically, f is a bit less than two (between about 1.3 and 1.6) and b is a bit more than two (between about 2.1 and 2.4). Of course this model is good only for a fairly restricted range of ux density. Base dissipation is usually expressed in watts per kilogram, so we rst compute ux density and then mass of the two principal components of the stator iron, the teeth and the back iron. In a similar way we can model the exciting volt-amperes consumed by core iron by something like: Qc = V a1 B BB
v 1

+ V a2

B BB

v 2

(123)

This, too, is a form that appears to be valid for some steels. Quite obviously it may be necessary to develop dierent forms of curve ts for dierent materials. Flux density (RMS) in the air-gap is: Br = Then ux density in the stator teeth is: Bt = B r wt + w1 wt (125) pVa 2RlNa k1 s (124)

where wt is tooth width and w1 is slot top width. Flux in the back-iron of the core is Bc = B r where dc is the radial depth of the core. R pdc (126)

28

One way of handling this loss is to assume that the core handles ux corresponding to terminal voltage, add up the losses and then compute an equivalent resistance and reactance: rc = xc = 3|Va |2 Pcore 3|Va |2 Qcore

then put this equivalent resistance in parallel with the air-gap reactance element in the equivalent circuit.

Solid Iron Rotor Bodies

Solid steel rotor electric machines (SSRM) can be made to operate with very high surface speeds and are thus suitable for use in high RPM situations. They resemble, in form and function, hysteresis machines. However, asynchronous operation will produce higher power output because it takes advantage of higher ux density. We consider here the interactions to be expected from solid iron rotor bodies. The equivalent circuits can be placed in parallel (harmonic-by-harmonic) with the equivalent circuits for the squirrel cage, if there is also a cage in the machine. To estimate the rotor parameters R2s and X2s , we assume that important eld quantities in the machine are sinusoidally distributed in time and space, so that radial ux density is: Br = Re B r ej (tp) and, similarly, axially directed rotor surface current is: Kz = Re K z ej (tp) Now, since by Faradays law: E = we have, in this machine geometry: B t (129) (128) (127)

1 Br Ez = R t

(130)

The transformation between rotor and stator coordinates is: = m t where m is rotor speed. Then: pm = r = (1 s) and Now, axial electric eld is, in the frame of the rotor, just: Ez = Re E z ej (tp) = Re E z ej (r 29
tp )

(131) (132)

(133) (134)

and Ez =

r R Br p

(135)

Of course electric eld in the rotor frame is related to rotor surface current by: Ez = Z sK z (136)

Now these quantities can be related to the stator by noting that air-gap voltage is related to radial ux density by: p (137) Br = V 2lNa k1 R ag The stator-equivalent rotor current is: I2 = R K 3 Na ka z (138)

Then we can nd stator referred, rotor equivalent impedance to be: Z2 = V ag 3 4 l 2 2 Ez N k = I2 2 R a a r K z (139)

Now, if rotor surface impedance can be expressed as: Z s = Rs + jr Ls then Z2 = where R2 = X2 = 34 l 2 2 N k Rs 2R a 1 34 l 2 2 N k Xs 2R a 1 (142) (143) R2 + jX2 s (140)

(141)

Now, to nd the rotor surface impedance, we make use of a nonlinear eddy-current model proposed by Agarwal. First we dene an equivalent penetration depth (similar to a skin depth): = 2Hm r B0 (144)

where is rotor surface material volume conductivity, B0 , saturation ux density is taken to be 75 % of actual saturation ux density and Hm = |K z | = 3 Na ka |I 2 | R (145)

Then rotor surface resistivity and surface reactance are: Rs = Xs 16 1 3 = .5Rs 30 (146) (147)

Note that the rotor elements X2 and R2 depend on rotor current I2 , so the problem is nonlinear. We nd, however, that a simple iterative solution can be used. First we make a guess for R2 and nd currents. Then we use those currents to calculate R2 and solve again for current. This procedure is repeated until convergence, and the problem seems to converge within just a few steps. Aside from the necessity to iterate to nd rotor elements, standard network techniques can be used to nd currents, power input to the motor and power output from the motor, torque, etc.

5.1

Solution

Not all of the equivalent circuit elements are known as we start the solution. To start, we assume a value for R2 , possibly some fraction of Xm , but the value chosen doesn not seem to matter much. The rotor reactance X2 is just a fraction of R2 . Then, we proceed to compute an air-gap impedance, just the impedance looking into the parallel combination of magnetizing and rotor branches: R2 (148) Zg = jXm ||(jX2 + ) s (Note that, for a generator, slip s is negative). A total impedance is then Zt = jX1 + R1 + Zg (149) and terminal current is It = Rotor current is just: I2 = It Vt Zt (150)

jXm 2 jX2 + R s

(151)

Now it is necessary to iteratively correct rotor impedance. This is done by estimating ux density at the surface of the rotor using (145), then getting a rotor surface impedance using (146) and using that and (143 to estimate a new value for R2 . Then we start again with (148). The process drops through this point when the new and old estimates for R2 agree to some criterion.

5.2

Harmonic Losses in Solid Steel

If the rotor of the machine is constructed of solid steel, there will be eddy currents induced on the rotor surface by the higher-order space harmonics of stator current. These will produce magnetic elds and losses. This calculation assumes the rotor surface is linear and smooth and can be characterized by a conductivity and relative permeability. In this discussion we include two space harmonics (positive and negative going). In practice it may be necessary to carry four (or even more) harmonics, including both belt and zigzag order harmonics. Terminal current produces magnetic eld in the air-gap for each of the space harmonic orders, and each of these magnetic elds induces rotor currents of the same harmonic order. The magnetizing reactances for the two harmonic orders, really the two components of the zigzag leakage, are: Xzp = Xm
2 kp 2 k2 Np 1

(152)

31

Xzn = Xm

2 kn 2 2 Nn k1

(153)

where Np and Nn are the positive and negative going harmonic orders: For belt harmonics these orders are 7 and 5. For zigzag they are: Np = Nn = Ns + p p Ns p p (154) (155)

Now, there will be a current on the surface of the rotor at each harmonic order, and following 65, the equivalent rotor element current is: I 2p = I 2n = R K 3 Na kp p R K 3 Na kn n (156) (157)

These currents ow in response to the magnetic eld in the air-gap which in turn produces an axial electric eld. Viewed from the rotor this electric eld is: E p = sp RB p E n = sn RB n where the slip for each of the harmonic orders is: sp = 1 Np (1 s) sn = 1 + Np (1 s) and then the surface currents that ow in the surface of the rotor are: Kp = Kn = Ep Zsp En Zsn (162) (163) (160) (161) (158) (159)

where Zsp and Zsn are the surface impedances at positive and negative harmonic slip frequencies, respectively. Assuming a linear surface, these are, approximately: Zs = 1+j (164)

where is material restivity and the skin depth is = 2 s (165)

32

and s is the frequency of the given harmonic from the rotor surface. We can postulate that the appropriate value of to use is the same as that estimated in the nonlinear calculation of the space fundamental, but this requires empirical conrmation. The voltage induced in the stator by each of these space harmonic magnetic uxes is: Vp = Vn = 2Na kp lR Bp Np p 2Na kn lR Bn Nn p (166) (167)

Then the equivalent circuit impedance of the rotor is just: Z2p = Z2n =
2 k2 l 3 4 Na Vp p Zsp = Ip 2 Np R s p 2 k2 l Z Vn 3 4 Na sn n = In 2 Nn R s n

(168)

(169)

The equivalent rotor circuit elements are now: R2p = R2n =

2 k2 l 1 3 4 Na p 2 Np R p

(170) (171) (172) (173)

2 k2 l 1 3 4 Na n 2 Nn R n 1 X2p = R2p 2 1 X2n = R2n 2

5.3

Stray Losses

So far in this document, we have outlined the major elements of torque production and consequently of machine performance. We have also discussed, in some cases, briey, the major sources of loss in induction machines. Using what has been outlined in this document will give a reasonable impression of how an induction machine works. We have also discussed some of the stray load losses: those which can be (relatively) easily accounted for in an equivalent circuit description of the machine. But there are other losses which will occur and which are harder to estimate. We do not claim to do a particularly accurate job of estimating these losses, and fortunately they do not normally turn out to be very large. To be accounted for here are: 1. No-load losses in rotor teeth because of stator slot opening modulation of fundamental ux density, 2. Load losses in the rotor teeth because of stator zigzag mmf, and 3. No-load losses in the solid rotor body (if it exists) due to stator slot opening modulation of fundamental ux density. 33

Note that these losses have a somewhat dierent character from the other miscellaneous losses we compute. They show up as drag on the rotor, so we subtract their power from the mechanical output of the machine. The rst and third of these are, of course, very closely related so we take them rst. The stator slot openings modulate the space fundamental magnetic ux density. We may estimate a slot opening angle (relative to the slot pitch): D = wd Ns 2wd Ns = 2r r 2 D sin 2

Then the amplitude of the magnetic eld disturbance is: BH = B r 1

In fact, this ux disturbance is really in the form of two traveling waves, one going forward and one backward with respect to the stator at a velocity of /Ns . Since operating slip is relatively small, the two variations will have just about the same frequency as viewed from the rotor, so it seems reasonable to lump them together. The frequency is: H = Ns p

Now, for laminated rotors this magnetic eld modulation will aect the tips of rotor teeth. We assume (perhaps arbitrarily) that the loss due to this magnetic eld modulation can be estimated from ordinary steel data (as we estimated core loss above) and that only the rotor teeth, not any of the rotor body, are aected. The method to be used is straightforward and follows almost exactly what was done for core loss, with modication only of the frequency and eld amplitude. For solid steel rotors the story is only a little dierent. The magnetic eld will produce an axial electric eld: E z = R BH p and that, in turn, will drive a surface current Kz = Ez Zs

Now, what is important is the magnitude of the surface current, and since |Z s | = 1 + .52 Rs 1.118Rs , we can simply use rotor resistance. The nonlinear surface penetration depth is: = 2B0 H |K z |

A brief iterative substitution, re-calculating and then |K z | quickly yields consistent values for and Rs . Then the full-voltage dissipation is: Prs = 2Rl and an equivalent resistance is: Rrs = |K z |2

3|Va |2 Prs

34

Finally, the zigzag order current harmonics in the stator will produce magnetic elds in the air gap which will drive magnetic losses in the teeth of the rotor. Note that this is a bit dierent from the modulation of the space fundamental produced by the stator slot openings (although the harmonic order will be the same, the spatial orientation will be dierent and will vary with load current). The magnetic ux in the air-gap is most easily related to the equivalent circuit voltage on the nth harmonic: Bn = npvn 2lRNa kn

This magnetic eld variation will be substantial only for the zigzag order harmonics: the belt harmonics will be essentially shorted out by the rotor cage and those losses calculated within the equivalent circuit. The frequency seen by the rotor is that of the space harmonics, already calculated, and the loss can be estimated in the same way as core loss, although as we have pointed out it appears as a drag on the rotor.

6
6.1

Induction Motor Speed Control

Introduction

The inherent attributes of induction machines make them very attractive for drive applications. They are rugged, economical to build and have no sliding contacts to wear. The diculty with using induction machines in servomechanisms and variable speed drives is that they are hard to control, since their torque-speed relationship is complex and nonlinear. With, however, modern power electronics to serve as frequency changers and digital electronics to do the required arithmetic, induction machines are seeing increasing use in drive applications. In this chapter we develop models for control of induction motors. The derivation is quite brief for it relies on what we have already done for synchronous machines. In this chapter, however, we will stay in ordinary variables, skipping the per-unit normalization.

6.2

Volts/Hz Control

Remembering that induction machines generally tend to operate at relatively low per unitslip, we might conclude that one way of building an adjustable speed drive would be to supply an induction motor with adjustable stator frequency. And this is, indeed, possible. One thing to remember is that ux is inversely proportional to frequency, so that to maintain constant ux one must make stator voltage proportional to frequency (hence the name constant volts/Hz). However, voltage supplies are always limited, so that at some frequency it is necessary to switch to constant voltage control. The analogy to DC machines is fairly direct here: below some base speed, the machine is controlled in constant ux (volts/Hz) mode, while above the base speed, ux is inversely proportional to speed. It is easy to see that the maximum torque is then inversely to the square of ux, or therefore to the square of frequency. To get a rst-order picture of how an induction machine works at adjustable speed, start with the simplied equivalent network that describes the machine, as shown in Figure 12 Earlier in this chapter, it was shown that torque can be calculated by nding the power dissipated in the virtual resistance R2 /s and dividing by electrical speed. For a three phase machine, and assuming we are dealing with RMS magnitudes: 35

I a Ra

X1

X2 I2 < > R2 Xm < > < s

Figure 12: Equivalent Circuit

p R2 Te = 3 |I2 |2 s where is the electrical frequency and p is the number of pole pairs. It is straightforward to nd I2 using network techniques. As an example, Figure 13 shows a series of torque/speed curves for an induction machine operated with a wide range of input frequencies, both below and above its base frequency. The parameters of this machine are: Number of Phases Number of Pole Pairs RMS Terminal Voltage (line-line) Frequency (Hz) Stator Resistance R1 Rotor Resistance R2 Stator Leakage X1 Rotor Leakage X2 Magnetizing Reactance Xm 3 3 230 60 .06 .055 .34 .33 10.6

Strategy for operating the machine is to make terminal voltage magnitude proportional to frequency for input frequencies less than the Base Frequency, in this case 60 Hz, and to hold voltage constant for frequencies above the Base Frequency. For high frequencies the torque production falls fairly rapidly with frequency (as it turns out, it is roughly proportional to the inverse of the square of frequency). It also falls with very low frequency because of the eects of terminal resistance. We will look at this next.

6.3

Idealized Model: No Stator Resistance

Ignore, for the moment, R1 . An equivalent circuit is shown in Figure 14. It is fairly easy to show that, from the rotor, the combination of source, armature leakage and magnetizing branch can be replaced by its equivalent circuit, as shown in in Figure 15. In the circuit of Figure 15, the parameters are: V = V X Xm Xm + X1 = Xm ||X1 36

Induction Motor Torque

250

200

Nm

150

100

50

0 0

50

100

150 Speed, RPM

200

250

Figure 13: Induction Machine Torque-Speed Curves

X2 I2 < + > R2 Xm < > s V <

I a X1

Figure 14: Idealized Circuit: Ignore Armature Resistance If the machine is operated at variable frequency , but the reactance is established at frequency B , current is: I= and then torque is Te = 3|I2 |2
2 |V |2 R 3p R2 s = 2 2 s (X1 + X2 )2 + ( R s )

V +X ) + j (X1 2 B

R2 s

Now, if we note that what counts is the absolute slip of the rotor, we might dene a slip with respect to base frequency: r r B B s= = = sB B

37

X1 I a

X2 I2

< > R2 < > < s

Figure 15: Idealized Equivalent Then, if we assume that voltage is applied proportional to frequency: V = V0 and with a little manipulation, we get:
2 |V0 |2 R 3p sB Te = + X )2 + ( R2 )2 B (X1 2 s B

This would imply that torque is, if voltage is proportional to frequency, meaning constant applied ux, dependent only on absolute slip. The torque-speed curve is a constant, dependent only on the dierence between synchronous and actual rotor speed. This is ne, but eventually, the notion of volts per Hz runs out because at some number of Hz, there are no more volts to be had. This is generally taken to be the base speed for the drive. Above that speed, voltage is held constant, and torque is given by: Te =
2 |V |2 R 3p sB + X )2 + ( R2 )2 B (X1 2 sB

The peak of this torque has a square-inverse dependence on frequency, as can be seen from Figure 16.

6.4

Peak Torque Capability

Assuming we have a smart controller, we are interested in the actual capability of the machine. At some voltage and frequency, torque is given by: Te = 3|I2 |2
p 2 |V |2 R 3 R2 s = + X )( ))2 + (R + s ((X1 2 B 1

R2 2 s )

Now, we are interested in nding the peak value of that, which is given by the value of R2 s which maximizes power transfer to the virtual resistance. This is given by the matching condition: R2 = s
2 + ((X + X )( R1 2 1

2 )) B

38

Induction Motor Torque

250

200

Nm

150

100

50

0 0

500

1500 1000 Speed, RPM

2000

Figure 16: Idealized Torque-Speed Curves: Zero Stator Resistance Then maximum (breakdown) torque is given by: Tmax =
3p 2 |V | 2 + ((X + X )( ))2 R1 2 B 1 2 + ((X + X )( ))2 )2 R1 2 B 1

+ X )( ))2 + (R + ((X1 2 B 1

This is plotted in Figure 17. Just as a check, this was calculated assuming R1 = 0, and the results are plotted in gure 18. This plot shows, as one would expect, a constant torque limit region to zero speed.

6.5

Field Oriented Control

One of the more useful impacts of modern power electronics and control technology has enabled us to turn induction machines into high performance servomotors. In this note we will develop a picture of how this is done. Quite obviously there are many details which we will not touch here. The objective is to emulate the performance of a DC machine, in which (as you will recall), torque is a simple function of applied current. For a machine with one eld winding, this is simply: T = GIf Ia This makes control of such a machine quite easy, for once the desired torque is known it is easy to translate that torque command into a current and the motor does the rest. Of course DC (commutator) machines are, at least in large sizes, expensive, not particularly ecient, have relatively high maintenance requirements because of the sliding brush/commutator interface, provide environmental problems because of sparking and carbon dust and are environmentally sensitive. The induction motor is simpler and more rugged. Until fairly recently the induction motor has not been widely used in servo applications because it was thought to be hard to control. As we will show, it does take a little eort and even some computation to do the controls right, but this is becoming increasingly aordable. 39

Breakdown Torque
300

250

200 NewtonMeters

150

100

50

0 0

20

40

60 80 Drive Frequency, Hz

100

120

Figure 17: Torque-Capability Curve For An Induction Motor

6.6

Elementary Model:

We return to the elementary model of the induction motor. In ordinary variables, referred to the stator, the machine is described by ux-current relationships (in the d-q reference frame): dS dR qS qR = = LS M LS M M LR M LR idS idR iqS iqR

Note the machine is symmetric (there is no saliency), and since we are referred to the stator, the stator and rotor self-inductances include leakage terms: LS = M + LS

LR = M + LR The voltage equations are: ddS qS + rS idS dt dqS + dS + rS iqS dt ddR s qR + rR idR dt dqR + s dR + rR iqR dt 40

vdS vqS

= =

0 = 0 =

Breakdown Torque
300

250

NewtonMeters

200

150

100

50 0

20

40

60 80 Drive Frequency, Hz

100

120

Figure 18: Idealized Torque Capability Curve: Zero Stator Resistance Note that both rotor and stator have speed voltage terms since they are both rotating with respect to the rotating coordinate system. The speed of the rotating coordinate system is w with respect to the stator. With respect to the rotor that speed is , where wm is the rotor mechanical speed. Note that this analysis does not require that the reference frame coordinate system speed w be constant. Torque is given by:
3
T e = p (dS iqS qS idS ) 2

6.7

Simulation Model

As a rst step in developing a simulation model, see that the inversion of the ux-current relationship is (we use the d- axis since the q- axis is identical): idS = LR M dS dR LS LR M 2 LS LR M 2 LS M dS dR LS LR M 2 LS LR M 2

idR =

Now, if we make the following denitions (the motivation for this should by now be obvious): Xd = 0 LS Xkd = 0 LR Xad = 0 M
Xd = 0 LS

M2 LR

41

the currents become: idS = 0 Xad 0 dS dR Xd Xkd Xd Xd 0 Xad 0 dS X X dR Xkd Xd kd d

idR =

The q- axis is the same. Torque may be, with these calculations for current, written as: 3 0 Xad 3 Te = p (dS iqS qS idS ) = p (dS qR qS dR ) 2 2 Xkd Xd Note that the usual problems with ordinary variables hold here: the foregoing expression was written assuming the variables are expressed as peak quantities. If RMS is used we must replace 3/2 by 3! With these, the simulation model is quite straightforward. The state equations are: ddS dt dqS dt ddR dt dqR dt dm dt = VdS + qS RS idS = VqS dS RS iqS = s qR RR idR = s dR RS iqR = 1 (Te + Tm ) J

where the rotor frequency (slip frequency) is: s = p m For simple simulations and constant excitaion frequency, the choice of coordinate systems is arbitrary, so we can choose something convenient. For example, we might choose to x the coordinate system to a synchronously rotating frame, so that stator frequency = 0 . In this case, we could pick the stator voltage to lie on one axis or another. A common choice is Vd = 0 and Vq = V .

6.8

Control Model

If we are going to turn the machine into a servomotor, we will want to be a bit more sophisticated about our coordinate system. In general, the principle of eld-oriented control is much like emulating the function of a DC (commutator) machine. We gure out where the ux is, then inject current to interact most directly with the ux. As a rst step, note that because the two stator ux linkages are the sum of air-gap and leakage ux, dS qS = agd + LS idS = agq + LS iqS

42

This means that we can re-write torque as: 3 T e = p (agd iqS agq idS ) 2 Next, note that the rotor ux is, similarly, related to air-gap ux: agd = dR LR idR agq = qR LR iqR

Torque now becomes: 3 3 T e = p (dR iqS qR idS ) pLR (idR iqS iqR idS ) 2 2 Now, since the rotor currents could be written as: idR = iqR = That second term can be written as: idR iqS iqR idS = So that torque is now: Te = 3 LR p 1 2 LR 3 M (dR iqS qR idS ) = p (dR iqS qR idS ) 2 LR 1 (dR iqS qR idS ) LR M dR idS LR LR M qR iqS LR LR

6.9

Field-Oriented Strategy:

What is done in eld-oriented control is to establish a rotor ux in a known position (usually this position is the d- axis of the transformation) and then put a current on the orthogonal axis (where it will be most eective in producing torque). That is, we will attempt to set dR = 0 qR = 0 Then torque is produced by applying quadrature-axis current: Te = 3 M p 0 iqS 2 LR

The process is almost that simple. There are a few details involved in guring out where the quadrature axis is and how hard to drive the direct axis (magnetizing) current.

43

Now, suppose we can succeed in putting ux on the right axis, so that qR = 0, then the two rotor voltage equations are: 0 = 0 = Now, since the rotor currents are: idR = iqR = dR M idS LR LR M qR iqS LR LR ddR s qR + rR IdR dt dqR + s dR + rR IqR dt

The voltage expressions become, accounting for the fact that there is no rotor quadrature axis ux: 0 = M ddR dR + rR idS LR dt LR M iqS 0 = s dR rR LR

Noting that the rotor time constant is TR = we nd: TR ddR + dR = M idS dt M iqS s = TR dR LR rR

The rst of these two expressions describes the behavior of the direct-axis ux: as one would think, it has a simple rst-order relationship with direct-axis stator current. The second expression, which describes slip as a function of quadrature axis current and direct axis ux, actually describes how fast to turn the rotating coordinate system to hold ux on the direct axis. Now, a real machine application involves phase currents ia , ib and ic , and these must be derived from the model currents idS and iqs . This is done with, of course, a mathematical operation which uses a transformation angle . And that angle is derived from the rotor mechanical speed and computed slip: = (pm + s ) dt

A generally good strategy to make this sort of system work is to measure the three phase currents and derive the direct- and quadrature-axis currents from them. A good estimate of direct-axis ux is made by running direct-axis ux through a rst-order lter. The tricky operation involves dividing quadrature axis current by direct axis ux to get slip, but this is now easily done numerically (as 44

dR

ND
N

M 1 + STa M TR

T
* ia

ia
o

* d

T
* iq

-1

* ib

Amp

o o

ib ic

Motor Load

* c

Figure 19: Field Oriented Controller are the trigonometric operations required for the rotating coordinate system transformation). An elmentary block diagram of a (possbly) plausible scheme for this is shown in Figure 19. In this picture we start with commanded values of direct- and quadrature- axis currents, corresponding to ux and torque, respectively. These are translated by a rotating coordinate transformation into commanded phase currents. That transformation (simply the inverse Parks transform) uses the angle q derived as part of the scheme. In some (cheap) implementations of this scheme the commanded currents are used rather than the measured currents to establish the ux and slip. We have shown the commanded currents i a , etc. as inputs to an Amplier. This might be implemented as a PWM current-source, for example, and a tight loop here results in a rather high performance servo system.

References
[1] P.L. Alger, Induction Machines, Gordon and Breach, 1969 [2] A.E. Fitzgerals, C. Kingsley Jr., S.D. Umans, Electric Machinery, Sixth Edition, McGraw Hill, 2003

45

[3] D. Fink, H. W. Beaty, Standard Handbook for Electrical Engineers, Thirteenth Edition,
McGraw-Hill, 1993

46

r1

x1 jxag
< >r < > c <

xc jx2s5
< r > 2s5 < > s5 <

jx2s
< > r2s < > s <

jx2c
< > r2c < > s <

jxa5

jx2c5
< r > 2c5 < > s5 <

jxa7

jx2s7
< > r2s7 < > s7 <

jx2c7
< > r2c7 < > s7 <

jxam

jx2sm
< > r2sm < > sm <

jx2cm
< > r2cm < > sm <

jxap

jx2sp
< > r2sp < > sp <

jx2cp
< > r2cp < > sp <

Figure 20: Extended Equivalent Circuit

47