ECE 320 Energy Conversion and Power Electronics

Dr. Tim Hogan

Chapter 6: Induction Machines (Textbook Sections 6.1-6.5)

Chapter Objectives

The popularity of induction machines has helped to label them as the ‘workhorse of industry’. They are relatively easy to fabricate, rugged and reliable, and find their way into most applications. For variable speed applications, inexpensive power electronics can be used along with computer hardware and this has allowed induction machines to become more versatile. In particular, vector or field-oriented control allows induction motors to replace DC motors in many applications.

6.1 Description The stator of an induction machine is a typical three-phase one, as described in the previous

chapter. The rotor can be one of two major types – either (a) it is wound in a fashion similar to that of the stator with the terminals connected to slip rings on the shaft, as shown in Figure 1, or (b) it is made with shorted bars.

Shaft

SlipRings

Rotor

Figure 1. Wound rotor, slip rings, and connections. Figure 2 shows the rotor of such a machine, while the images in Figure 3 show the shorted bars

and the laminations. The bars in Figure 3 are formed by casting aluminum in the openings of the rotor laminations. In

this case the iron laminations were chemically removed.

1- 1

Figure 2. (a) Cutaway view of a three-phase induction motor with a wound rotor and slip rings connected to the three-phase rotor winding shown in Figure 6.1 in your textbook [1]. (b) Cutaway view of a three-phase squirrel-cage motor as shown in Figure 6.3 in your textbook [1].

(a) (b)

Figure 3. (a) The rotor of a small squirrel-cage motor. (b) The squirrel-cage structure after the rotor laminations have been chemically etched away as shown in Figure 6.3 in your textbook [1].

6.2 Concept of Operation As these rotor windings or bars rotate within the magnetic field created by the stator magnetizing

currents, voltages are induced in them. If the rotor were to stand still, then the induced voltages would be very similar to those induced in the stator windings. In the case of a squirrel cage rotor, the voltage induced in a bar will be slightly out of phase with the voltage in the next bar, since the flux linkages will change in it after a short delay. This is depicted in Figure 4.

If the rotor is moving at synchronous speed, together with the field, no voltage will be induced in the bars or the windings.

1- 2

Bg12

34

567

13

19

-1

-0.5

0

0.5

1

0 50 100 150 200 250 300 350

e(t)

ωt

bar 3bar 1

bar 2

bar 7

(a) (b)

Figure 4. (a) Rotor bars in the stator field and (b) voltages in the rotor bars.

Generally when the synchronous speed is ss fπω 2= , and the rotor speed ω 0, the frequency of the

induced voltages will be fr, where osrf ωωπ −=2 . Maxwell’s equation becomes here:

gBv ×=E (6.1)

where E is the electric field and v is the relative velocity of the rotor with respect to the field:

( )rv os ωω −= (6.2) Since a voltage is induced in the bars, and these are short-circuited, currents will flow in them. The current density will be: ( )θJ

( ) Eρ

θ 1=J (6.3)

where ρ is the resistivity of the bars.

These currents are out of phase in different bars, just like the induced voltages. To simplify the analysis we can consider the rotor as one winding carrying currents sinusoidally distributed in space. This will be clearly the case for a wound rotor. It will also be the case for uniformly distributed rotor bars, but now each bar, located at an angle θ will carry different current, as shown in Figure 5(a).

1- 3

Bg

Bg

(a) (b)

Figure 5. (a) Currents in rotor bars and (b) equivalent current sheet in the rotor.

( ) θωρ gosω BJ ⋅−=1 ( ) (6.4)

( ) ( ) θωρ

θ sinˆ1gos BωJ −= ( ) (6.5)

The bars can also be replaced with a conductive cylinder as shown in Figure 5(b) with a

distributed current. Slip, s, is defined as the ratio:

s

ossω

ωω −= (6.6)

Thus at starting, the speed is zero and s = 1, and at synchronous speed, os ωω = and s = 0. Above synchronous speed s < 0, and when the rotor rotates in a direction opposite of the magnetic field, then s > 1. Example 6.2.1

The rotor of a two-pole 3-phase induction machine rotates at 3300 (rpm), while the stator is fed by a three-phase system of voltages at 60 (Hz). What are the possible frequencies of the rotor voltages?

At 3300 (rpm):

6.3456023300 ==πωo (rad/s) while 377=sω (rad/s)

These two speeds can be in the opposite or the same direction such that:

6.345377 ±=−= osr ωωω (rad/s) = 722.58 (rad/s) or 31.43 (rad/s) or f = 115 (Hz) or f = 5 (Hz)r r

1- 4

To better understand the currents induced in the squirrel cage as caused by the magnetic flux density from the stator coils consider the simplified view of the squirrel cage in Figure 6. Adjacent bars of the squirrel cage form a coil since the ends of the cage are electrically shorted and the largest coils are formed by pairs of opposite side bars of the cage. The bars of the squirrel cage work collectively to respond to any changing magnetic field. If the squirrel cage is rotating at the same angular velocity as the stator magnetic flux density, then the loops of the squirrel cage experience a constant magnetic field, and no current flows through the bars of the cage. Under this condition, no magnetomotive force is generated by the squirrel cage and no torque on the rotor exists. The rotor then begins to slow down relative to the stator magnetic field. As is does so, the magnetic flux density for a given loop of the squirrel cage changes (some loops experiencing an increase in magnetic flux density, and some loops experiencing a decrease in magnetic flux density). The response to this change in magnetic field is a current that flows in the bars of the squirrel cage so as to oppose this change in magnetic field. Since the stator magnetic field is sinusoidally distributed in space around the rotor, and the most rapid change in this field is determined by its spatial derivative, then as the squirrel cage slips with respect to ωs, the field from the rotor caused by this slip is spatially oriented 90º with respect to the stator magnetic field (derivative of the cosine field distribution).

For example, if the stator magnetic flux density is oriented from left to right as was shown in Figure 5 and is rotating clockwise, then as a motor with s near zero (near synchronous speed s > 0) the rotor is also rotating clockwise, but not quite as fast as the stator magnetic flux density. As this slip occurs, the largest change in the field occurs from the top to the bottom of the squirrel cage or at the zero magnetic field points of the stator field where the spatial slope of the stator field is largest. As the slip occurs, the field through the squirrel cage, BBg in , begins to rotate in a clockwise fashion relative to the rotor. The largest change in field occurs in a direction from top to bottom of the rotor and the coils of the squirrel cage collectively generate a counter magnetomotive force so as to oppose the change in magnetic field. Thus current flows in the bars of the squirrel cage to generate a counter field directed from bottom to top, or into the page on the right side, and out of the page on the left as indicated in .

Figure 5

Figure 5

B

ω − ωs o

ωs

ωo Figure 6. Squirrel cage isometric view.

1- 5

6.3 Torque Development To calculate the torque on the rotor we use the following two equations:

BliF = and rFT ⋅= (6.7)

since the flux density is perpendicular to the current producing it. For the length of the conductor, l, we will use the depth of the rotor. The thickness, de, of the equivalent conducting sheet shown in Figure 5 is set equal to the total area of the bars of the squirrel cage.

rdnd

rddnA

e

e

8

24

2barsrotor

2barsrotor

⋅=

== ππ

(6.8)

where d is the diameter of a single bar of the squirrel cage.

For a small angle dθ centered at a given angle θ, we find the contribution to the total force and torque as:

( )( )

( )osge

ge

g

BldrdTT

rdFdT

BrdJddF

lBJdAdF

ωωρ

π

θ

πθ

θ−⎟

⎟⎠

⎞⎜⎜⎝

⎛==

=

=

⋅⋅=

∫=

=

222

0ˆ2

(6.9)

where and ( )θsinˆgg BB = ( ) ( ) θωρ

θ sinˆ1gos BωJ −= ( ) from equation (6.5) with ρ equal to the

resistivity of the bars of the squirrel cage. Using the relationship between flux density (or flux linkages), Λs, and the rotor voltage, Es, the

torque can be expressed as:

( osss

elN

dT ωωρπ

−Λ⎟⎟⎠

⎞⎜⎜⎝

⎛= 2

28 ) where

s

ss

Eω

=Λ (6.10)

where the stator voltage is related to the flux density as gsss lrBNe2πω= . Focusing on the variables

of this equation we see the torque is proportional to the frequency of the rotor currents, ( )osω ω− and to the square of the flux density. This is so since the torque comes from the interaction of the flux density, BBg, and the rotor currents, but the rotor currents are induced due to the flux, BgB

), and the

relative speed ( osω ω− . Equation (6.10) gives torque as a function of more accessible quantities of stator induced voltage, Es, and frequency ωs. This comes as a result of the simple and direct relationship between stator induced voltage, flux (or flux linkages), and frequency.

1- 6

6.4 Operation of the Induction Machine near Synchronous Speed We already determined that the voltages induced in the rotor bars are of slip frequency,

fr = (ωs - ωo)/2π. At rotor speeds near synchronous, fr is small. The rotor bars in a squirrel cage machine possess resistance and leakage inductance, but at very low frequencies (near synchronous speed) we can neglect this leakage inductance. The rotor currents are therefore limited near synchronous speed by the rotor resistance only.

The induced rotor-bar voltages and currents form space vectors. These are perpendicular to the stator magnetizing current and in phase with the space vectors of the voltages induced in the stator as shown in Figure 7.

Bg

ir

is,m

Figure 7. Stator magnetizing current, airgap flux, and rotor currents.

These rotor currents, ir, produce additional airgap flux, which is 90º out of phase of the

magnetizing flux. The stator voltage, es, is applied externally and is proportional to and 90º out of phase with the airgap flux. Thus additional currents, isr, will flow in the stator windings in order to cancel the flux due to the rotor currents. These additional currents are shown in Figure 8(a), and the resulting stator current and space vectors are depicted in Figure 8(b).

Bg

ir

is,m

is,r

Bg ir

is

i s

(a) (b)

Figure 8. Rotor and stator currents in an induction motor (a) Rotor and stator current components (b) total stator current and space vector.

1- 7

is,mBg

ir

i sis,r

es

Figure 9. Space vectors of the stator and rotor current and induced voltages.

A few items to note from the above analysis: • isr is 90º ahead of the stator magnetizing current, is,m. This means that it corresponds to

currents in the windings i1r, i2r, i3r, leading by 90º the magnetizing currents i1m, i2m, i3m. • The amplitude of the magnetizing component of the stator current is proportional to the stator

frequency, fs, and induced voltage. On the other hand, the amplitude of this component of the stator currents, isr, is proportional to the current in the rotor, ir, which is proportional to the flux and slip speed, ωr = ωs – ωo, or proportional to the developed torque.

• The stator current of one phase, is1, can be split into two components. One in phase with the voltage, isr1, and one 90º behind it, ism1. The first reflects the rotor current, while the second depends on the voltage and frequency. In an equivalent circuit, this means that isr1 will flow through a resistor, and ism1 will flow through an inductor.

• Since isr1 is equal to the rotor current (through a factor), it will be inversely proportional to ωs – ωr, or better stated as proportional to ωs/(ωs – ωr). The equivalent circuit for the stator shown in Figure 10 reflects these considerations.

es

+

Xm RR

is,m

is,r

ωsω − ωs o

is,1

Figure 10. Equivalent circuit of one stator phase.

If the inductor motor is supplied with a three-phase, balanced sinusoidal voltage, then it is

expected that the rotor will develop a torque according to equation (6.10). The relationship between speed, ωo, and torque near synchronous speed is shown in Figure 11. This curve is accurate as long as the speed does not vary more than ±5% around the rated synchronous speed, ωs.

1- 8

ωsωo

T

Figure 11. Torque-speed characteristics near synchronous speed.

As the speed exceeds synchronous, the torque produced by the machine is in the opposite direction to the speed (i.e. the machine operates as a generator), developing a torque opposite to the rotation (or counter torque) and transferring power from the shaft to the electrical system.

We already know the relationship of the magnetizing current, Ism, to the induced voltage, Esm, through our analysis of the three-phase windings. Now we relate the currents, ir and isr to the same induced voltage.

The current density, , on the rotor conducting sheet is related to the air gap flux density as: J

( ) gos BωJ ⋅−= ωρ1 (6.11)

This current density corresponds to a space vector ir that is opposite to the isr in the stator. This current space vector will correspond to the same current density:

rd

NiJ ssr1

= (6.12)

while the stator voltage es is also related to the flux density BBg. The amplitude of es is:

gsss lrBNe2πω= (6.13)

Substitution of the relating phasors instead of space vectors in equation (6.11) we obtain:

sros

sRs IRE

ωωω−

= (6.14)

The torque is then three times (three phases) the power Es·Isr over the stator speed or

s

gr

R

s

s

os

Rs

s PRR

ETω

ωω

ωωω

331322

=Λ

=−

= (6.15)

1- 9

where Λ = (Es/ωs). Here Pg is the power transferred to the resistance os

sRR

ωωω−

, through the

airgap. Of this power a portion is converted to mechanical power represented by losses on the

resistance os

oRR

ωωω−

, and the remaining is losses in the rotor resistance, represented by the losses

on resistance RR. In Figure 12 this split in the equivalent circuit is shown; note the resistance

os

oRR

ωωω−

can be negative, indicating that mechanical power is absorbed in the induction machine.

es

+

Xm

RR

is,m

is,r

ωoω − ωs o

is,1

RR

RRωs

ω − ωs o

Figure 12. Equivalent circuit of one stator phase separating the loss and torque rotor

components. 6.4.1 Example

A 2-pole three-phase induction motor is connected in a Y configuration and is fed from a 60 (Hz), 208 (V) (l-l) system. Its equivalent one-phase rotor resistance is RR = 0.1125 (Ω). At what speed and slip is the developed torque 28 (N·m)?

(rad/s) 6.366

0275.0377

364.10(rad/s) 364.10

1125.01

377120328

)V( 120 with 13

s

r

2

2

=−=

===

=

⎟⎠⎞

⎜⎝⎛=

=⎟⎟⎠

⎞⎜⎜⎝

⎛=

rso

r

r

srRs

s

s

VR

VT

ωωωωω

ω

ω

ωω

1- 10

6.5 Leakage Inductances and Their Effects In the previous discussion we assumed that all the flux crosses the airgap and links both the stator

and the rotor windings. In addition to this flux there are flux components which link only the stator or the rotor windings and are proportional to the currents there, producing voltages in these windings 90º ahead of the stator and rotor currents and proportional to the amplitude of these currents and their frequency.

This is a simple model for the stator windings, since the equivalent circuit we are using is for the stator, and we can model the effects of this flux with an inductor added to the circuit. The rotor leakage flux can be modeled in the rotor circuit with an inductance L1s, as well, but corresponding to

frequency of πωω

2os

rf−

= , the frequency of the rotor currents. Its effects on the stator can be

modeled with an inductance L1r at frequency fs, as shown in the complete 1-phase equivalent circuit in Figure 13.

Es

+

Xm

RR

Is,mIs,r

ωoω − ωs o

Is,1 RR

RRωs

ω − ωs o

Xls XlrRs

Vs

+

Figure 13. Complete equivalent circuit of one stator phase including leakage flux contributions. Here is the phasor of the voltage induced into the rotor windings from the airgap flux, while

is the phasor of the applied 1-phase stator voltage. The torque equation (sE

sV 6.15) still holds here, but give us slightly different results. We can develop torque-speed curves, by selecting speeds, solving the equivalent circuit, calculating power Pg, and using equation (6.15) for the torque. Figure 14 shows the stator current per phase, torque for the three phase induction machine, and the power factor as a function of ωo for ωo given as a percentage of ωs and ranging from negative values to greater than ωs.

1- 11

-300

-200

-100

0

100

200

0

0.2

0.4

0.6

0.8

1

-50 0 50 100 150

I s1 (A

), an

d T

(N·m

)

pf

ωo (percent of synchronous speed)

Is1

(A)

T (N·m)

pf

Figure 14. Stator current (single phase), torque and power factor of an induction machine vs. speed.

6.6 Operating Characteristics Figure 14 shows the developed torque, current, and power factor of an induction machine over a

speed range from below zero (slip > 1 or braking region) to above synchronous (slip < 0 or generator region). There are three regions of interest:

1. For speeds in the range 0 ≤ ωo ≤ ωs the torque is of the same sign as the speed, and the machine operates as a motor. There are a few interesting points on this curve and on the corresponding current and power factor curves.

2. For speeds in the range ωo ≤ 0, torque and speed have opposite signs, and the machine is in breaking mode. Notice the current is very high, resulting in high winding losses.

3. For speeds in the range ωo ≥ ωs the speed and torque are of opposite signs and the machine is in the generating mode.

These regions are identified on an extended plot of torque in Figure 15. If we consider the motor

operation region 0 ≤ ωo ≤ ωs the operating point is often designed to be near, or at, the point where the power factor is maximized. It is for this point that the motor characteristics are given on the nameplate, rated speed, current, power factor, and torque. When designing an application it is this point that we have to consider primarily. Will the torque suffice? Will the efficiency and power factor be acceptable?

Starting the motor is also of interest (where slip s = 1) where the torque is not necessarily high, but the current often is. When selecting a motor for an application, we have to make sure this starting torque is adequate to overcome the load torque which may also include a static component. In

1- 12

addition, the starting current is often 3-5 times the rated current of the machine. If the developed torque at starting is not adequately higher than the load starting torque, their difference, called the accelerating torque, will be small and it may take too long to reach the operating point. This means that the current will remain high for a long time, and fuses or circuit breakers could have their limits exceeded.

0

-100 -50 0 50 100 150 200 250

Torq

ue

ωo (percent of synchronous speed)

Mot

orG

ener

ator

Breakingregion

Motorregion

Generator region

Figure 15. Stator torque vs. speed identifying three regions of operation. A third point of interest is the maximum torque, Tmax, corresponding to speed ωTmax. We can find

it by analytically calculating torque as a function of slip, and equating the derivative to zero. This point is interesting, since speeds higher than this generally correspond to stable operating conditions, while lower speeds generally correspond to unstable operating conditions. To study this point of operation we use the Thevenin equivalent circuit for the left side of the circuit as seen looking into terminal A-B in Figure 17.

Es

+

Xm

RR

Is,mIs,r

ωoω − ωs o

Is,1 RR

RRωs

ω − ωs o

Xls XlrRs

Vs

+A

B

Thevenin

Figure 16. Equivalent circuit for induction machine indicating terminals for Thevenin equivalent circuit.

1- 13

Es

+

RR

Is,r

ωoω − ωs o

RR

RRωs

ω − ωs o

XTH XlrRTH

VTH

+A

B

Figure 17. Equivalent circuit for induction machine with the stator circuit replaced with a Thevenin equivalent circuit.

We find

( ) ( ) ( )( )mlss

msmlsmlssTHTH XXjR

XjRXXjXjXRjXR++++−

=+=+ (6.16)

and

( )mlss

msTH XXjR

jXVV++

= ˆˆ (6.17)

Then from the circuit in Figure 17, we can find as: rsI ,ˆ

( )lrTH

RTH

THrs

XXjs

RR

VI++⎟

⎠⎞

⎜⎝⎛ +

=ˆˆ , (6.18)

where s is in the per unit system, that is ( ) soss ωωω /−= . From equation (6.15)

( )22

22,

333

lrTHR

TH

RTH

s

Rrs

sg

s XXs

RR

sRV

sRIPT

++⎟⎠⎞

⎜⎝⎛ +

⎟⎠⎞

⎜⎝⎛

=⎟⎠⎞

⎜⎝⎛==

ωωω (6.19)

This torque reaches a maximum as a slip, smaxT , that can be found by taking the derivative of

equation (6.19) and setting it equal to zero. When this is done we find:

( ) ( )

( ) ( )22max

22

maxor

lrTHTH

RT

lrTHTHT

R

XXR

Rs

XXRs

R

++=

++=

(6.20)

1- 14

This gives the maximum torque of:

( )22

2max 2

3

lrTHTHTH

TH

s XXRR

VT+++

=ω

(6.21)

If the stator resistance is negligible, then RTH ≈ 0 and

ss

ssTT

T

T

max

max

max2

+≈ (6.22)

If we neglect both the stator resistance and the magnetizing inductance, we can find simple equations for Tmax and ωTmax. To do so, we must assume operation near synchronous speed, where the value of

os

sRR

ωωω−

is much larger than ωsLlr.

lslr

RsT LL

R+

−≈ ωω max (6.23)

( ) ⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛=−⎟⎟

⎠

⎞⎜⎜⎝

⎛≈

lrlss

sTs

Rs

sLL

VR

VT 1231

23

2

max

2

max ωωω

ω (6.24)

The slip frequency at this torque, maxTsr ωωω −= , for a constant flux s

ss

Eω

=Λ is independent

of frequency, and is proportional to the resistance RR. We already know that RR is proportional to the rotor resistance, so if the rotor resistance is increased, the torque-speed characteristic is shifted to the left, as shown in Figure 18. If we have convenient ways to increase the rotor resistance, we can increase the starting torque, while decreasing the starting current. Increasing the rotor resistance can be easily accomplished in a wound-rotor induction machine. For the squirrel cage motor, more complex structures such as double or deep rotor bars can be used to increase the rotor resistance.

In the formula developed we notice the maximum torque is a function of the flux. This means that we can change the frequency of the stator voltage, but as long as the voltage amplitude changes so that the flux stays the same, the maximum torque will also stay the same as shown in Figure 19. This is called Constant Volts per Hertz Operation and is a first approach to controlling the speed of the motor through its supply.

Near synchronous speed the effect of the rotor leakage inductance can be neglected, and led to the torque-speed equation (6.15) repeated below:

s

gr

R

s

s

os

Rs

s PRR

ETω

ωω

ωωω

331322

=Λ

=−

=

Figure 20 shows both the exact and the approximate torque-speed characteristics. It is important to notice that the torque calculated from the approximate equation is grossly incorrect away from synchronous speed.

1- 15

0 20 40 60 80 100

I s

ωo (percent of synchronous speed)

RR = r

1RR = 2·r

1

0 20 40 60 80 100

Torq

ueω

o (percent of synchronous speed)

RR = r

1RR = 2·r

1

Figure 18. Effect of changing the rotor resistance on the torque-speed and

current-speed characteristics of an induction motor.

-50 0 50 100 150

I s

ωo (rad/s)

60 (Hz)

45 (Hz)20 (Hz)

-50 0 50 100 150

Torq

ue

ωo (rad/s)

60 (Hz)

45 (Hz)

20 (Hz)

Figure 19. Effect on the torque-speed characteristic of changing frequency

while keeping the flux constant.

1- 16

40 50 60 70 80 90 100 110

Torq

ueω

o (percent of synchronous speed)

LinearApprox.

Figure 20. Exact and approximate torque-speed characteristics.

6.7 Starting Characteristics of Induction Motors From the above analysis we see that a particular challenge with induction motors is the high

current and low torque during starting. A simple way to decrease the starting current is to decrease the stator terminal voltage during

startup. From equation (6.15), we see the torque is proportional to the stator voltage squared, while the current is directly proportional to the stator voltage. If a transformer is used to decrease the stator voltage, then both the developed torque and the line current will decease by the square of the turns ratio of the transformer.

A commonly used method is to change the connection configuration during starting of a motor. For example a motor designed to operate with the stator windings connected in Δ are changed to a Y connection during starting. The voltage ratio of these configurations is:

Δ= ,, 31

sYs VV (6.25)

then

Δ= ,, 31

sYs II (6.26)

Δ= ,, 31

sYs TT (6.27)

+

Vl-l

I line Y

Is,Y+

Vs,Y

I line Δ

Is,Δ +

Vs,Δ

+

Vl-l

Figure 21. Y – Delta starting of an induction motor.

1- 17

In the Δ connection, phline II 3= , leading to:

Δ= ,, 31

lineYline II (6.28)

Once the machine has approached the desired operating point, we can reconfigure the connection

to Δ, and provide better efficiency. This decrease in current is often adequate to allow a motor to start low load starting torque.

Using a variable frequency and voltage supply we can comfortably increase the starting torque, as shown in Figure 19, while decreasing the starting current.

6.8 Multiple Poles If we consider that an induction machine will operate close to synchronous speed 3000 (rpm) for

50 (Hz) and 3600 (rpm) for 60 (Hz) we may find that the speed of the machine is too high for an application. If we recall the pictures of the flux in AC machines we have shown before, we notice the flux has a relatively long path to travel in the stator. This makes the stator heavy and lossy.

In machines with more than one pair of poles, the sinusoidal distribution for the windings covers a smaller angle. For example, in a 4 pole machine each side of a sinusoidally distributed winding of one phase covers only 90º instead of 180º (as was the case for a 2 pole machine).

Figure 22 shows at one instant the equivalent windings resulting from rotor windings of a 2-pole, 4-pole, and 6-pole machines.

Figure 22. Multipole induction machines 2-pole, 4-pole, 6-pole rotor windings

(stator windings not shown). The effects of a large number of poles on the operation of the machine are not difficult to predict.

If the machine has p poles, or p/2 pole pairs, then in one period of the voltage, the flux will travel 2ωs/p (rad/s). This leads to the rotor speed corresponding to synchronous of ωsm:

ssm pωω 2

= (6.29)

We now introduce the actual mechanical speed of the rotor as ωm, while we keep the term ωo as

the rotor speed of a two pole motor. We generally measure ωm in (rad/s), while we measure ωo in (electrical rad/s). We retain the same definition for slip based on the electrical speed ωo.

1- 18

om pωω 2

= (6.30)

s

ms

s

os

p

sω

ωω

ωωω 2

−=

−= (6.31)

This shows that for a 4-pole machine, supplied by a 60 (Hz) source, and operating close to rated

conditions, the speed will be near 1800 (rpm), while for a 6-pole machine, the speed will be near 1200 (rpm). While increasing the number of poles results in a decrease of the synchronous and operating speeds of the machine, it also results in an increase of the developed torque of the machine by the same ratio. Hence, the corrected torque formula will be:

o

m

s

g PpPpTωω 2

32

3 == (6.32)

Similarly, the torque near the synchronous speed is:

s

g

R

rs

s

os

Rs

s PpR

pR

EpTω

ωω

ωωω 2

32

312

322

=Λ

=−

= (6.33)

while the previously developed formulas for maximum torque will become:

( )22

2max 2

12

3lrTHTHTH

TH

s XXRR

VpT+++

=ω

(6.34)

and

( )lrlss

sTs

Rs

sLL

VpR

VpT+⎟⎟

⎠

⎞⎜⎜⎝

⎛=−⎟⎟

⎠

⎞⎜⎜⎝

⎛≈

122

3122

32

max

2

max ωωω

ω (6.35)

Example 6.8.1

A 3-phase, 2-pole induction motor is rated at 190 (V), 60 (Hz), is connected in the Y configuration, and has RR = 6.6 (Ω), Rs = 3.1 (Ω), XM = 190 (Ω), Xlr = 10 (Ω), Xls = 3 (Ω). Calculate the motor starting torque, starting current and starting power factor under rated voltage. What will be the current and power factor if no load is connected to the shaft?

1. At starting, s = 1

[ ] ( ) 06.7106.619031.3

3190

ˆ =+++

=jjj

Is /-54.5º (A)

7.6190106.6

190ˆˆ =++

=jj

jII sr /-52.6º (A)

( ) 36.222

3776.67.63

23

2===

pPT

s

gω

(N·m)

1- 19

2. Under no load, the speed is synchronous and s = 0

57.019031.3

3190

ˆ =++

=jj

Is /-89.1º (A)

57.0=sI (A)

016.0=pf lagging

Example 6.8.2

A 3-phase, 2-pole induction motor is rated at 190 (V), 60 (Hz), is connected in the Y configuration, and has RR = 6.6 (Ω), Rs = 3.1 (Ω), XM = 190 (Ω), Xlr = 10 (Ω), Xls = 3 (Ω). It is operating from a variable speed – variable frequency source at a speed of 1910 (rpm), under a constant (V/f ) policy and the developed torque is 0.8 (N·m). What is the voltage and frequency of the source? (Hint: First calculate the slip).

The ratio (Vs/ωs) stays at 377

1903

1

.

)(V 110or (V) 4.6437711066.220(Hz) 35

(rad/s) 66.22065.2001.2202

(rad/s) 65.206.6

1377110318.0

132

-

2

2

llss

rms

rr

rRs

s

Vf

p

RVpT

==⇒=

=+=+=

=⇒⎟⎠⎞

⎜⎝⎛⋅=

⎟⎟⎠

⎞⎜⎜⎝

⎛=

ωωω

ωω

ωω

Example 6.8.3

A 3-phase, 4-pole induction machine is rated 230 (V), 60 (Hz). It is connected in the Y configuration, and has RR = 0.191 (Ω), LM = 35 (mH), Lls = 1.2 (mH). It is operated as a generator connected to a variable frequency/variable voltage source. Its speed is 2036 (rpm), with a counter-torque of 59 (N·m). What is the efficiency of this generator? (Hint: Here power in is mechanical, power out is electrical; also first calculate the slip).

Although we do not know the voltage or the frequency, we know their ratio is (132.8/377).

(rad/s) 14.15191.01

3778.1322359

132

2

2

−=

⎟⎠⎞

⎜⎝⎛⋅=−

⎟⎟⎠

⎞⎜⎜⎝

⎛=

r

r

rRs

sR

VpT

ω

ω

ωω

1- 20

Now the synchronous speed can be found by adding slip and rotor speeds as:

(V) 14460

1325.65(Hz) 5.65

(rad/s) 3.41114.152602036·2

2

=⋅=⇒=

=−=+=

ss

rms

Vf

p πωωω

We have to calculate the impedances of the equivalent circuit for the frequency of 65.5 (Hz):

)( 617.0)( 49.0

)( 4.143.4111035 3

Ω=

Ω=Ω=⋅×= −

lr

ls

m

XXX

then

)( 38.52 Ω−=+

r

mrR

p

Rω

ωω

[ ] [ ] 30617.038.5191.04.1449.02.0

144ˆ =+−++

=jjj

Is /-148º (A)

2.27ˆ =rI /-166.9º (A) Notice that with generation operation RR < 0. We can calculate now the losses, etc.:

919.0

(kW) 98.10

(W) 5402.0303

(W) 423191.02.273

(kW) 941.1138.52.273

,,

2,

2,

2

==

=−−=

=⋅⋅=

=⋅⋅=

=⋅⋅=

m

out

lossstatorlossrotormout

lossstator

lossrotor

m

PP

PPPP

P

P

P

η

1 A. E. Fitzgerald, C. Kingsley, Jr., S. D. Umans, Electric Machinery, 6th edition, McGraw-Hill, New York, 2003.

1- 21