CHAPTER 3
Basic AC
Motor.
School of Computer and Communication
Engineering, UniMAP
Prepared By:
Amir Razif b. Jamil Abdullah
EMT 113: V-2008
1
3.0 Application of Alternate Current
(AC) Motor.
3.1 Synchronous Machine.
3.1.1 Introduction to Synchronous Generator.
3.1.2 Synchronous Generator Construction.
3.1.3 Synchronous Machine Characteristics.
3.1.4 Induce Voltage in Armature Windings.
3.1.5 Synchronous Generator and Circuit Model.
3.1.6 Synchronous Motor and Circuit Model.
3.2 Induction Motors.
3.2.1 Construction of Induction Motor.
3.2.2 Basic Induction Motor Concept.
3.2.3 Equivalent Circuit of Three-Phase Induction Motor.
3.2.4 Performance Calculation Using Circuit Model.
3.2.5 Power and Torque in Induction Motors.
2
3.2.6 Induction Motor Torque-Speed Characteristics.
3.0 AC Machinery Fundamental.
 AC machines are the generators and motors.
 Two major classes of machines;
(i) Synchronous machines.
(ii) Induction machines.
3.1 Synchronous Machine.
3
3.1.1 Introduction to a Synchronous
Generators.
 Synchronous generators or alternators are synchronous
machines used to convert mechanical power to ac electric
power, depending on the mode of operation of the machine.
 A machine has two modes of operation;
(a) motor
An electric motor converts electrical energy into mechanical
energy.
(b) generator.
An electric generator is used to convert mechanical energy into
electrical energy. For this reason, electric machines are also called
electromechanical energy converters.
4
Cont’d…
 Below is the condition of motor and generator operations.
Figure 3.1. Machine Modes
of Operation:
(a) Motor, and
(b) Generation.
 The magnetic field in a machine forms the energy link between
the electrical and mechanical systems.
 Current flowing in coils of wire inside the machine produces the
magnetic field. The magnetic field performs two functions:
(a) Magnetic attraction and repulsion produces mechanical torque
(motor operation),
(b) The magnetic field by Faraday’s law induces voltages in the
coils of wire (generator operation).
5
Cont’d…
 Motor operation in Figure 3.1(a), the field-induced voltage, E,
permits the motor to draw power from the line to be converted
into mechanical power.
 While the magnetic field is inducing this voltage, it is also
developing the mechanical output torque, t. Note that the induced
voltage is in opposition to the current flow. It is often called the
counter emf .
 Generator operation in Figure 3.1(b), the field induced voltage E
is in the same direction as the current and is called the “generated
voltage.” In this case the machine torque opposed the input
mechanical torque that is trying to drive the generator, and it is
called counter torque.
6
3.1.2 Synchronous Generator
Construction.
 In a synchronous generator, a dc
current is applied to the rotor
winding, which produces a rotor
magnetic field.
 The rotor is turned by a prime
mover which producing a
rotating magnetic field within
the machine.
Figure 3.2: Cut-Way View of a
Synchronous Machine.
 The outside stationary part, called
the stator. The inner rotating part is called the rotor.
 The rotor is mounted on a stiff rod, called a shaft. The rotor is
solidly fastened to the shaft so that the rotor and the shaft turn at
the same speed.
7
Cont’d…
 The terms “rotor speed,” “shaft speed,” and “machine speed”
all mean the same thing.
 The rotor and the stator have three basic parts;
(i) Core.
(ii) Windings,
(iii) Insulation.
 The purpose of the rotor and stator cores is to conduct the
magnetic field through the coils of the windings.
(i) Cores.
 The cores are always made of iron steel, similar to the transformer
core.
8
Cont’d…
Windings.
 Two common term for windings; “field windings” and
“armature windings”.
 The “field windings” applies to the winding the produce the
main magnetic field to the machine.
 The “armature windings” applies to the windings where the
main voltage is induced.
 The “field windings” = “rotor windings” is on the rotor which
carries the magnetization current.
 The “armature windings” = “stator windings” is on the stator
which carries the load current.
Rotor.
 Rotor can either be salient or non-salient pole.
 Salient pole is a magnetic pole that protruding or stick out from
the surface of the rotor. It is normally used for the rotor with four
or more poles.
9
Cont’d…
 Non-salient pole is a magnetic pole constructed flush with the
surface of the rotor. It is normally used for the rotor with two- and
four-poles.
 There are two common approaches to supplying dc power to the
field circuit of the rotor.
(1) From the external dc source, by means of slipping and
brushes.
(2) From a special dc power source mounted directly on the shaft
of the synchronous generator.
10
Cont’d…
Figure 3.3: Four Pole Synchronous Machines:
(a) with Salient-Pole Rotor, (b) with Cylindrical Rotor Having three Coils per
Pole.
 Figure 3.3 shows the two types of rotors in a four-pole
synchronous machine.
 The rotor carries the field winding. The field current or the
excitation current is provided by an external dc source.
Synchronous machine rotors are simply rotating electromagnets
built to have as many poles as are produced by the stator windings.
11
3.1.3 Synchronous Machine
Characteristics.
 Synchronous machines are called “synchronous” because their
mechanical shaft speed is directly related to the power system’s
line frequency.
n P
fe  s
(rpm )
 Stator electric frequency;
120
ns 
where,
120 f e
P
or
s 
4 . f e
P
P is the number of magnetic poles.
nm is the mechanical speed of magnetic field in r/min.
fs is electrical frequency in Hz.
(rad / s )
 Typical machines have two-poles, four-poles, and six-poles.
 There is a set of relationships between angular quantities
(position and velocity) of the mechanical shaft and the electrical
system.
12
Cont’d…
 The electrical angle between two consecutive magnetic poles
(north pole and south pole) is 180°, but the mechanical angle is
based on the number of poles. The angle is found as follows;
 For example, a two-pole machine has a mechanical angle of 180°,
and a four-pole motor has a mechanical angle of 90°. Therefore,
the relation between the electrical angle and the mechanical angle is
written as:
 The angular velocity is the time-derivative of the angular
position,
13
Cont’d…
 The relation between the mechanical speed and the electrical
speed can be obtained as:
 If one machine has two-poles and another has eight-poles, the
eight-pole machine will run at precisely one-forth the speed of
the two-pole machine.
 The synchronous machines, may be operated either as motors
or as generators.
 Since electrical energy can be transported most economically by
three-phase transmission systems, nearly all synchronous
generators larger than 10 kVA, and most industrial motors are
designed for three-phase operation.
14
3.1.4 Induce Voltage in the
Armature Winding.
Distribution Factor
 Each phase winding consists of Nf-turns of conductor, which is
made up of coils and groups.
 There is a group of coils for each pole in the machine. Within
each group, coils are distributed over several slots in the stator.
An example is shown in Figure 3.4.
 A voltage is induced in each coil with the movement of the
rotating air-gap flux. The coil voltages are displaced from one
another in phase by the slot angle a. The slot angle is the distance
between adjacent slots, and is given (in electrical degrees) by
where, P is the number of poles,
Q is the number of slots,
m is the number of phases, and
q is the number of slots per pole per phase.
15
Cont’d…
 The resultant voltage for each group of coils is the phasor sum of
the coil voltages. It is defined as,
 The phasor addition of the voltages is shown in Figure 3.8. From
the geometry, the distribution factor can be obtained as;
 Where Nf is the total number of series turns for a phase winding.
16
Example 3.1:
Synchronous Motor.
A three-phase, wyeconnected, 60 Hz
synchronous machine has
six poles and 36 slots in its
stator. The flux per pole
(fmax) is 0.024 Wb and
each coil in the stator
winding has 2 turns. Each
coil has a span of 5 slots.
Find:
(a)the coil pitch r and the
pitch factor kp, and the
induced coil voltage Ec and
(b) the induced phase
voltage Ef on a line-to-line
basis.
Solution:
17
3.1.5 Synchronous Generator and
its Circuit Model.
 A synchronous generator functions on the basis of Faraday’s Law.
If the flux linking the coil changes in time, a voltage is
induced in a coil. The voltage is induced in a conductor if it cuts
magnetic flux lines.
 Figure 3.4 is a three-phase machine.
Figure 3.4: A Three-Phase Machine
with a Graph of its Associated Flux
Density Distribution Produced by the
Rotor Excitation and a Graph of the
Three-Phase Voltages Generated.
 The flux density in the air gap is uniform implies that sinusoidally
varying voltages will be induced in the three coils aa’, bb’, and cc’
18
if the rotor carrying dc, rotates at a constant speed, ns.
Cont’d…
 The voltage induced in phase a is given by;
Where,
j is the flux per pole,
e is the electrical angular frequency, and
Nf is the number of turns in phase a or the coil aa’.
 The induced voltages for the other two phases are equal in
magnitude and displaced by ±120°.
 The phase voltages may be written as;
and fe is the frequency of the induced voltage.
19
Cont’d…
 E be defined as the open-circuit, induced phase voltage for a
certain field current If.
 Then E is the internal voltage of the generator. It is assumed that
If is such that the machine is operating under unsaturated
conditions.
 The combination of the leakage reactance, the armature
magnetizing reactance and the winding resistance forms the
“synchronous impedance”, Zs.
 The summation of the two reactance is called the “synchronous
reactance”, Xs, but is often designated as Xd for the direct-axis
reactance (from the more detailed models of machines).
 The winding resistance is also called the armature resistance, Ra.
 The equivalent circuit of a synchronous generator and the
combined phasor/vector diagram with a lagging power-factor
current flow are shown in Figure 3.5.
20
Cont’d…
 If the generator operates at a terminal voltage VT while supplying a
load corresponding to an armature current Ia, then;
 In an actual synchronous machine, the reactance is much greater
than the armature resistance, in which case;
 Among the steady-state characteristics of a synchronous generator,
its voltage regulation and power-angle characteristics are the most
important ones.
 The voltage regulation of a synchronous generator is defined at a
given load as;
21
Cont’d…
 where VT is the terminal voltage for a given loading condition,
and E is the no-load terminal voltage.
 The voltage regulation can be illustrated by the following
example.
Figure 3.5: Synchronous Generator Model and Phasor-Vector Diagram.
22
Example 3.2: Synchronous Generator.
A three-phase, wye-connected 2500 kVA and 6.6 kV generator
operates at full-load. The per-phase armature resistance Ra and the
synchronous reactance, Xd, are (0.07W+j10.4W).
Calculate the percent voltage regulation at
(a) 0.8 power-factor lagging, and
(b) 0.8 power-factor leading.
Solution.
23
Cont’d…
Distribution Factor
 Each phase winding consists of Nf- turns of conductor, which is
made up of coils and groups.
 There is a group of coils for each pole in the machine. Within each
group, coils are distributed over several slots in the stator. An
example is shown in Figure 3.4.
 A voltage is induced in each coil with the movement of the rotating
air-gap flux. The coil voltages are displaced from one another in
phase by the slot angle a. The slot angle is the distance between
adjacent slots, and is given (in electrical degrees) by
where,
P is the number of poles,
Q is the number of slots,
m is the number of phases, and
q is the number of slots per pole per phase.
24
Cont’d…
 The resultant voltage for each group of coils is the phasor sum of
the coil voltages. It is defined as,
 The phasor addition of the voltages is shown in Figure 3.8. From
the geometry, the distribution factor can be obtained as;
 Where Nf is the total number of series turns for a phase winding.
25
3.1.6 Synchronous Motor and
Circuit Model.
Figure 3.6: Synchronous Motor Circuit Model (lossless) and Phasor-Vector Diagram.
 The rotor (or field) winding 17 be fed by a dc source that produces
the rotor magnetic field with definite polarities.
 The rotor will tend to align with the stator field and will tend to
rotate with the rotating magnetic field. To recapitulate, it can be
said that the stator rotating magnetic field has a tendency to
“drag” the rotor along, as if the north pole on the stator “locks
in” with a south pole of the rotor.
 The armature resistance (Ra) may be neglected as compared to
26
the synchronous reactance.
Cont’d…
 The steady-state per-phase equivalent circuit of a synchronous
machine simplifies to the one shown in Figure 3.6.
 In the figure, the armature current Ia goes into the machine for
motor operation, and out of it for generator operation where VT
is the terminal voltage for a given loading condition, and E is the
no-load terminal voltage.
27
3.2 Induction Motor.
 In the induction machine the rotor voltage which produces the
rotor current and the rotor magnetic field is induced in the rotor
windings rather than physically connected by wires, Figure 3.7.
 No dc field current is required to run the machine.
 Induction machine are cheap to manufacture, rugged and reliable
and find their way in most possible applications.
 Variable speed drives require inexpensive power electronics and
computer hardware, and allowed induction machines to become
more versatile.
 In particular, vector or field
oriented control allows
induction motors to replace
DC motors in many applications
Figure 3.7: Induction Motor.
28
3.1.1 Construction of Induction
Motor.
 Induction motor has the same physical stator as a synchronous
machine, with a different rotor construction.
 There are two different types of induction motor rotors ;
(a) Cage rotor.
(b) Wound rotor.
29
Cont’d…
(a) Cage rotor.
 Cage induction Motor rotor consists of a series of conducting
bars laid into slot carved in the face of rotor and shorted at either
end by large shorting ring; Figure 3.8.
Figure 3.8: Rotor of the Squirrel
Cage Induction Motor.
Figure 3.9: Rotor Components of
the Squirrel Cage Induction
Motor.
30
Cont’d…
(b) Wound rotor.
 A wound rotor has a complete set of three-phase winding that
are mirror images of the winding on the stator.
 The three phases of the rotor windings are usually Y-connected,
the end of the three rotor wires are tied to slip ring on the rotor
shaft.
 Rotor windings are shorted through brushes riding on the slip
rings. Wound-rotor induction motors are more expansive than the
cage induction motors, they required much more maintenance
because the wear associated with their brushes and slip rings.
Figure 3.10: Wound-Rotor Induction Motor.
31
3.1.2 Basic Induction Motor
Concept.
(a) The Development of Induce Torque in an Induction Motor.
 The three-phase of voltages has been applied to the stator, and
three-phase set of stator current is flowing . These currents
produce a magnetic field Bs, rotating counterclockwise direction.
 The speed of the magnetic field’s rotation is given by;
where,
- fe is the system frequency in hertz.
- P is the number of poles in the machine.
nsync
120 f e

P
32
Cont’d…
 The rotating magnetic field Bs passes over the rotor bars and
induce a voltage in them.
 The voltage induced in a given rotor bar is given by the
equation;
eind  (v  B)  I
where,
v = velocity of the bar relative to the magnetic field.
B = magnetic flux density vector.
I = length of conductor in the magnetic field.
 The relative motion of the rotor compared to the stator magnetic
field that produces induce voltage in a rotor bar.
 The induce torque in the machine is given by;
t ind  kBR  BS
 The rotor induced torque in counterclockwise, the rotor
accelerates in that direction.
33
Cont’d…
Speed.
 There is a finite upper limit to the motor speed.
 If the induction motor rotor were tuning at synchronous speed,
then the rotor bar would be stationary relative to the magnetic
field (no induce voltage).
 If eind is zero, there would be;
- no rotor current,
- no rotor magnetic field,
- the induce torque is zero,
- and the rotor would slow down as a result of friction lost.
34
Cont’d…
(b) The Concept of Rotor Slip.
 The voltage induced in a rotor bar of an induction motor
depends on the speed of the rotor relative to the magnetic
field.
 Slip speed is defined as the difference between synchronous
speed and rotor speed;
n slip  n sync  nm
Where,
nslip = slip speed of the machine.
nsync = speed of the magnetic fields.
nm = mechanical shaft speed of motor.
 Another term used to describe the relative motion is slip;
s
s
 Slip in angular velocity, ;
s
n slip
n sync
 100%
n slip  n m
n sync
 sync   m
n sync
 100%
 100%
35
Cont’d…
(c) The Electrical Frequency on the Rotor.
 The induction motor works by inducing voltages and current in
the rotor of the machine.
 If the rotor of a motor is locked so that it cannot move, the rotor
will have the same frequency as the stator.
 If the rotor turns at synchronous speed, the frequency on the rotor
will be zero.
 For nm= 0 r/min & the rotor frequency fr = fe  slip, s = 1
nm=nsync & the rotor frequency fr = 0  slip, s = 0
- For any speed in between, the rotor frequency is directly
proportional to the difference between the speed of the magnetic
field nsync and the speed of the rotor nm.
 Rotor frequency can be express as;
f r  sf e
P
fr 
(n sync  nm )
120
36
Example 3.3: Induction Motor.
A 208-V, 10-hp, four-pole, 60-Hz, Y- connected induction motor has a
full-load slip of 5 percent.
(a) What is the synchronous speed of this motor?
(b) What is the rotor speed of this motor at the rated load?
(c) What is the rotor frequency of this motor at the rated load?
(d) What is the shaft torque of this motor at the rated load?
Solution:
(a)The synchronous speed of this motor is,
120
fe
P
120(60 Hz )

 1800r / min
4 Poles
n sync 
n sync
(b)The rotor speed of the motor is given by,
nm  (1  s)nsync
nm  (1  0.95)(1800r / min)  1710r / min
37
Cont’d…
(c) The rotor frequency of this motor is given by ,
f r  sf e  (0.05)(60Hz )  3Hz
or
P
(n sync  nm )
120
4
fr 
(1800r / min  1710r / min)  3Hz
120
fr 
(d) The shaft load torque is given by,
 .
t load 
Pout
t load 
(10hp )(746W / hp )
 41.7 N .m
(1710r / min)( 2rad / r )(1 min/ 60 s )
m
38
3.1.3 The Equivalent Circuit of an
Induction Motor.
 Induction motor operates on the induction of voltage and
current in its rotor circuit from the stator circuit (transformer
action).
 An induction motor is called a singly excited machine, since
power is supply to only the stator circuit.
 Figure 3.11 is the equivalent circuit of an induction motor.
Figure 3.11:
The Two Forms of Exact Equivalent
Circuit of an Induction Motor.
39
Cont’d…
 R1
40
Cont’d…
 R1
41
Cont’d…
(a) The Transformer Model of an Induction Motor.
 R1 is the stator resistance and X1 is the stator leakage
reactance.
 The flux in the machine is related to the integral of the applied
voltage E1.
 The curve of magnetomotive force versus flux (magnetization
curve) for this machine is compared to a similar curve for a power
transformer.
42
Cont’d…
The Final Equivalent Circuit.
 The core loss resulting from stator leakage flux is not negligible, as it is in the
transformer.
Figure 3.12: (a) An Equivalent Circuit of an Induction Motor with Rc Neglected.
(b) The Power Flow and Associated Losses of
43
Example 3.4: Induction Motor Equivalent Circuit.
An induction motor draws 25 A from a 460 V, three-phase line at a
power of 0.85 lagging. The stator copper lost is 1000W and the
rotor copper loss is 500W. The rotational losses include 250W of
friction and wind-age, 800W of core and 200W of stray load
losses. Calculate,
(a)The air-gap power,
(b)The developed mechanical power, Pd,
(c)The output horse power,
(d)The efficiency.
Solution:
(a)The air-gap power,
Pg  Pin  SCL
Pg  3 (460V )( 25 A)(0.85 pf )  1000W
15.93KW
(b)The developed mechanical power, Pd,
Pd  Pg  RCL
Pd  15.93KW  500W
 15.43KW
44
Cont’d…
(c) The output horse power,
Pout  Pd  Prot  Pd  ( Pc  Pjw  Psll )
Pout  15.43KW  (800W  250W  200W )
14.18KW
Pout 
14.18KW
 19.0hp
746hp / KW
(d) The efficiency,
Pout 14.18KW


 83.3%
Pin 16.93KW
 .
45
Example 3.5: Induction Motor Equivalent Circuit.
If the frequency of the source in the above example is 60Hz, and the
machine has four poles, fine
(a)The slip,
(b)The operating speed,
(c)The developed torque,
(d)The output torque.
Solution:
(a)The slip,
RCL
500W
s

 0.0314
Pg
15.93KW
(b)The operating speed,
4f e 4 (60 Hz )

 188.5rad / s
P
4
120 f e 120(60 Hz )
ns 

 1800rpm
P
4
   s (1  s )  188.5rad / s (1  0.0314)  182.6rad / s
s 
n  n s (1  s )  1800(1  0.0314)  1744rpm
46
Cont’d…
(c) The developed torque,
Pd
15.43KW
td 

 84.5N .m

182.6
(d) The efficiency,
Pout
14.18KW
td 

 77.7 N .m

182.6
note: that the torque is 8% less than the developed torque.
 .
47
3.1.4 Performance Calculation
Using Circuit Model.
 The performance of the motor at any speed may be calculated on
the basis of relationships derived in previous session. Below is the
steps;
(1) Calculate the synchronous speed.
s 
4f1 120 f
or
P
P
(2) Calculate the slip for the desired speed.
s
 s   ns  n

s
ns
(3) Calculate the rotor impedance.
R' 2
z2 
 jX ' 2
s
48
Cont’d…
Figure 3.13: The Field Impedance Concept in the Induction Machine.
(a) the Field Impedance, Zf. (b) The Model Employing ,Zf.
49
Cont’d…
(4) Calculate the field impedance, Zf.
This is the effect of the resultant magnetic field on the impedance of
one phase of the stator. It is the parallel combination of the rotor
impedance, z2, and the core’s magnetizing reactance, jXm, referring to
Figure 4.5(a).
 R'

jX f  2  jX ' 2 
 s

Z f  R f  jX f 
R' 2
 j( X m  X '2 )
s
(5) Calculate the impedance looking in from the phase terminals
of the machine, as shown in Figure 4.5(b).
Z in  z1  Z f  (r1  R f )  j ( x1  X f )
(6) Calculate the phase current.
I1 
V1 0 o
Z in
50
pf  cos( f )
(7) Calculate the power factor.
(8) Calculate the input power.
Pin  3VLL I L cos(f )
(9) Calculate the stator copper loss.
(10) Calculate the air gap power.
SCL  3I 1 r1
2
Pg  3I1 R f
2
(11) Calculate the rotor copper loss.
Pin  3VLL I L cos(f )
(12) Calculate the developed mechanical power.
Pd  (1  s ) Pg
51
Cont’d…
(13) Calculate the developed torque.
Pg
td 
N .m
s
or 7.04
Pg
ns
ft.lb.
(14) Calculate the output power.
Pout  Pd  Prot
(15) Calculate the output torque.
t out 
Pout

N .m
Pout
or 7.04
ft.lb.
n
(16) Calculate the efficiency.

Pout
Pin
52
3.1.6 Induction Motor TorqueSpeed Characteristics.
 Apply the Thevenin’s theorem to the exact equivalent circuit model
of an induction motor.
 According to Thevenin’s theorem, the equivalent source voltage
VTh is the voltage that would appear across the terminals a and b
of Figure 4.6 with the rotor circuit open.
Vin ( jX m )
VTh 
r1  j ( x1  X m )
(3) Calculate the rotor impedance.
53
Cont’d…
 The Thevenin equivalent stator impedance RTh + j XTh is the
impedance between the terminals a and b of Figure 3.14, viewed
towards the source with the source voltage Vin short circuited.
 The impedance is calculated as:
Z Th  RTh  jX Th
 ( r1  jx1 ) //( jX m )
( r1  jx1 )( jX m )

( r1  jx1 )  ( jX m )
 The torque is;
3 I 2 2 R' 2
td  td 
s
s s
Pg
Figure 3.14: Application of Thevenin’s Theorem to the Circuit Model of an
Induction Motor.
54
Cont’d…
I2 
VTh
R' 

 RTh  2   j ( X Th  X ' 2 )
s 

 By combining equation 4.31 and 4.32, results in,
R' 2
V 2 Th
td 
R' 
s 
s
2
 RTh  2   ( X Th  X ' 2 )
s 

3
 Figure 3.15 is the general shape of the torque-speed (torque-slip)
curve. The region of (s>0) and the generator region (s<0) are
shown in the Figure.
55
Cont’d…
 The internal torque is
maximum when the power
delivered to R’2/s is at
maximum, Figure 3.14(b).
 The maximum torque and
power happen at the
matching of the source
impedance and the slip
value, smax,
R' 2
 Z Th  jX ' 2  R 2 Th  ( X Th  X ' 2 ) 2
s
 The slip at maximum
torque, smax,
s max 
R' 2
R
2
Th
 ( X Th  X ' 2 )
2
Figure 3.15: Torque of an Induction Machine
(a) Developed Torque as a Function of Slip.
(b) Developed Torque as a Function of Speed.
56
Cont’d…
t max 
3

V 2 Th
 s 2 RTh  R 2 Th  ( X Th  X ' 2 ) 2

 The maximum torque is independent of the rotor resistance
while the slip at which it occurs is proportional to the rotor
resistance.
57
Example 3.6: Torque Speed Characteristics on an
Induction Motors.
A 220 V line-to-line, three-phase, wye-connected, 10 hp, 60 Hz 6-pole
induction motor has the following constants in ohms per phase with
reference to the stator:
r1 = 0.294 W
R'2 = 0.144 W
x1 = 0.503 W
X'2 = 0.209 W
Xm = 13.25 W
(a) For a slip s = 0.03 determine the load current, I2, the developed
torque, td, and the developed power Pd.
(b) Find the maximum torque and the corresponding speed.
(c) Calculate the starting torque and the corresponding current I2.
Solution:
Using the circuit in Figure 3.14, V  122.3V
Th
Z Th  RTh  jX Th  0.273  j 0.490
(b) At the maximum torque point, the slip is calculated,
-Note how the resistance part dominates this impedance.
58
Cont’d…
59