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Chapter Four
Three Phase Induction Machine
4
...
They are simple, rugged, low-priced, and
easy to maintain
...
The speed is frequency-dependent and, consequently,
these motors are not easily adapted to speed control
...

In this chapter we cover the basic principles of the 3-phase
induction motor and develop the fundamental equations describing
its behavior
...

Squirrel-cage, wound-rotor ranging from a few horsepower to
several thousand horsepower permit the reader to see that they all
operate on the same basic principles
...
2 Principal components
A 3-phase induction motor (Fig
...
1) has two main parts: a
stationary stator and a revolving rotor
...
4 mm to 4 mm,
depending on the power of the motor
...
4
...

The stator (Fig
...
2) consists of a steel frame that supports a
hollow, cylindrical core made up of stacked laminations
...

The rotor is also composed of punched laminations
...
We use two types of rotor windings: (1)
conventional 3-phase windings made of insulated wire and (2)

218 Chapter Four
squirrel-cage windings
...


Fig
...
2 Exploded view of the cage motor of Fig
...
1, showing the
stator, rotor, end-bells, cooling fan, ball bearings, and terminal box
...

A squirrel-cage rotor is composed of bare copper bars, slightly
longer than the rotor, which are pushed into the slots
...
The entire construction (bars and
end-rings) resembles a squirrel cage, from which the name is
derived
...


Three-Phase Induction Machine

219

A wound rotor has a 3-phase winding, similar to the one on the
stator
...
The terminals are connected to
three slip rings, which turn with the rotor
...
The external resistors are
mainly used during the start up period; under normal running
conditions, the three brushes are short-circuited
...
3 Principle of operation
The operation of a 3-phase induction motor is based upon the
application of Faraday Law and the Lorentz force on a conductor
...

Consider a series of conductors of length l, whose extremities
are short-circuited by two bars A and B (Fig
...
3 a)
...
The following sequence of events then takes place:
1
...


220 Chapter Four
2
...

3
...

4
...

If the conducting ladder is free to move, it will accelerate
toward the right
...

Consequently, the force acting on the conductors wilt also
decreases
...

In an induction motor the ladder is closed upon itself to form a
squirrel-cage (Fig
...
3b) and the moving magnet is replaced by a
rotating field
...


221

Three-Phase Induction Machine

(a)

(b)
Fig
...
3 Moving magnet cutting across a conducting ladder
...
4 The Rotating Field and Induced Voltages
Consider a simple stator having 6 salient poles, each of which
carries a coil having 5 turns (Fig
...
4)
...
This creates three
identical sets of windings AN, BN, CN, that are mechanically
spaced at 120 degrees to each other
...


222 Chapter Four

Fig
...
4 Elementary stator having terminals A, B, C connected to
a 3-phase source (not shown)
...

The three sets of windings are connected in wye, thus forming a
common neutral N
...
In other
words, as regards terminals A, B, C, the windings constitute a
balanced 3-phase system
...
e
...
The result is illustrated in
Fig
...
5
...
Expressions for the induced voltages can be
obtained by using Faraday laws of induction
...
4
...

The flux density distribution in the air gap can be expressed as:

B(θ ) = Bmax cos θ

(4
...
2)

Where,
l is the axial length of the stator
...

Let us consider that the phase coils are full-pitch coils of N
turns (the coil sides of each phase are 180 electrical degrees apart
as shown in Fig
...
5)
...

The

flux

( = N φ P at
linkage

linkage

for

coil

aa'

will

be

maximum

ωt = 0o ) (Fig
...
5a) and zero at ωt = 90o
...
Hence;

224 Chapter Four

λa (ωt ) = Nφ p cos ωt

(4
...
4)

The voltages induced in the other phase coils are also
sinusoidal, but phase-shifted from each other by 120 electrical
degrees
...
5)

ec = E max sin (ωt + 120)
...
6)

From Equation (4
...
44 fNφ p
2

(4
...
Equation (4
...
However,

φP

in Equation (4
...

Equation (4
...
The N is the
total number of series turns per phase with the turns forming a
concentrated full-pitch winding
...
For such a
distributed winding, the EMF induced in various coils placed in
different slots are not in time phase, and therefore the phasor sum
of the EMF is less than their numerical sum when they are
connected in series for the phase winding
...
For most
three-phase machine windings KW

is about 0
...
95
...
44 fN phφ p KW

(4
...

4
...
The rotor, if free to do so, will then start
rotating
...
The rotor will eventually
reach a steady-state speed n that is less than the synchronous speed
ns at which the stator rotating field rotates in the air gap
...

In a P-pole machine, one cycle of variation of the current will
make the mmf wave rotate by 2/P revolutions
...
9)

p

The difference between the rotor speed n and the synchronous
speed ns of the rotating field is called the slip s and is defined as

s=

ns − n

(4
...
The frequency f 2 of the induced voltage and
current in the rotor circuit will correspond to this slip rpm, because
this is the relative speed between the rotating field and the rotor
winding
...
9):

f2 =

p
120

(ns − n ) =

p
120

sns = sf1

(4
...

The voltage induced in the rotor circuit at slip s is:

E2 s = 4
...
44 sf1 N 2φ p KW 2 = sE2 (4
...

The induced currents in the three-phase rotor windings also
produce a rotating field
...
13)

Because the rotor itself is rotating at n rpm, the induced rotor
field rotates in the air gap at speed n + n2 = (1 - s)ns + sns = ns
rpm
...
The stator
magnetic field and the rotor magnetic field are therefore stationary
with respect to each other
...
As the magnetic
fields tend to align, the stator magnetic field can be visualized as
dragging the rotor magnetic field
...
1 A 3-phase, 460 V, 100 hp, 60 Hz, four-pole
induction machine delivers rated output power at a slip of 0
...

Determine the:
(a) Synchronous speed and motor speed
...

(c) Frequency of the rotor circuit
...

(e) Speed of the rotor field relative to the (i) rotor structure
...
(iii) Stator rotating field
...
5
...
05) * 1800 = 1710 rpm
(b) 1800 rpm (same as synchronous speed)
(c) f2 = sf1 = 0
...

(d) slip rpm = s ns = 0
...
Now,

E2 s = sE2 = s

N2
460
E1 = 0
...
5 *
= 6
...
6 Equivalent Circuit of the Induction Motor
In this section we develop the equivalent circuit from basic
principles
...

Finally, we develop the equivalent circuit of an asynchronous
generator and determine its properties under load
...
Thus, the motor has 3
identical primary windings and 3 identical secondary windings one
set for each phase
...

When the motor is at standstill, it acts exactly like a
conventional transformer, and so its equivalent circuit (Fig
...
6) is
the same as that of a transformer, previously developed
...
4
...


230 Chapter Four
In the case of a conventional 3-phase transformer, we would be
justified in removing the magnetizing branch composed of

jX m and Rm because the exciting current I o is negligible
compared to the load current I P
...
Consequently, we cannot eliminate the magnetizing
branch
...
4
...

R1 = per-phase stator winding resistance
...

E1 = per-phase induced voltage in the stator winding
Lm = per-phase stator magnetizing inductance
Rc = per-phase stator core loss resistance
...
e
...
The difference

Three-Phase Induction Machine

231

lies only in the magnitude of the parameters
...
In induction machines it is as high
a: 30 to 50 percent of the rated current, depending on the motor
size, whereas it is only 1 to 5 percent in transformers
...


232 Chapter Four

Fig
...
7 Development of the induction machine equivalent circuit
Note that the rotor circuit frequency and current are: f 2 and I 2
...
14)

The power involved in the rotor circuit is:
2
P2 = I 2 R 2

(4
...
Equation
(4
...
16)

Equation (4
...
4
...

Although the magnitude and phase angle of I 2 are the same in
Equations (4
...
16), there is a significant difference
between these two equations and the circuits (Figs
...
7b and 4
...
The current I 2 in Equation (4
...
16) is at line frequency

f1
...
14) the rotor leakage reactance sX 2 varies with
speed but resistance R2 remains fixed, whereas in Equation (4
...
The per phase power associated with the
equivalent circuit of Fig
...
7c is:
2
P = I2

R2 P2
=
s
s

(4
...
01 to 0
...
4
...
Note that the equivalent circuit of
Fig
...
7c is at the stator frequency, and therefore this is the rotor

234 Chapter Four
equivalent circuit as seen from the stator
...
17) therefore represents the power that crosses the air gap and
thus includes the rotor copper loss as well as the mechanical power
developed
...
17) can be rewritten as:

R

2
2 R
P = Pag = I 2  R2 + 2 (1 − s ) = I 2 2
s
s



(4
...
4
...
The
speed

dependent

resistance

R2 (1 − s ) / s

represents

the

mechanical power developed by the induction machine
...
19)
(4
...
21)

and, P2 = I 2 R2 = sPag

(4
...
23)

Equation (4
...
e
...


Three-Phase Induction Machine

235

Therefore, for efficient operation of the induction machine, it
should operate at a low slip so that more of the air gap power is
converted into mechanical power
...
The remainder
of the mechanical power will be available as output shaft power
...
7 IEEE-Recommended Equivalent Circuit
In the induction machine, because of its air gap, the exciting
current 10 is high-of the order of 30 to 50 percent of the full-load
current
...
The IEEE
recommends that, in such a situation, the magnetizing reactance

X m not be moved to the machine terminals but be retained at its
appropriate place, as shown in Fig
...
8
...
This equivalent circuit (Fig
...
8) is to be
preferred for situations in which the induced voltage E1 differs
appreciably from the terminal voltageV1
...
4
...

4
...
4
...
24)

If R12 << ( X 1 + X m )2 as is usually case,
Vth ≈

Xm
V1 = K thV1
X1 + X m

(4
...
26)

Three-Phase Induction Machine

237

If R1 << ( X 1 + X m ) , then,
2

2

2

 Xm 
2

Rth ≅ 
 X + X  R1 = K th R1
 1
m

(4
...
28)

Fig
...
9 Thevenin equivalent circuit
...
9 Tests To Determine The Equivalent Circuit
The approximate values of R1 , R2 , X m , Rm and X in the
equivalent circuit can be found by means of the following tests:
No-load test When an induction motor runs at no load, the slip
is exceedingly small
...
4
...
Thus, at no-load the circuit consists essentially

238 Chapter Four
of the magnetizing branch X m , Rm
...


Measure the stator resistance RLL between any two terminals
...


(4
...
4
...
Measure the no load current I NL and the total
3-phase active power PNL
...
4
...

IEEE-recommended equivalent circuit
...
Therefore, in the equivalent circuit of Fig
...
9,

239

Three-Phase Induction Machine

the magnetizing reactance X m is shunted by a very high resistive
branch representing the rotor circuit
...
Therefore the total
reactance X NL , measured at no load at the stator terminals, is
essentially X 1 + X m
...
4
...


(a) No-load equivalent circuit

(b) Blocked-rotor equivalent circuit
...

Fig
...
11
The primary phase voltage can be obtained from the following
equation:

240 Chapter Four

V1 =

VLL
V / Phase
3

(4
...
31)

The no-load resistance is:

RNL =

PNL
3I12

(4
...
33)

In the IEEE recommended equivalent circuit we assume that

X 1 + X m = X NL

(4
...
Furthermore, the slip s is equal to one
...
Because I p is much greater than the
exciting current I o , we can neglect the magnetizing branch
...
4
...
Their values can be
determined by measuring the voltage, current, and power under
locked-rotor conditions, as follows:
a
...
Sometimes it is recommended to use lower
frequency than the rated to avoid the errors due to skin effect in
the rotor circuit
...
Take readings of VLL

BL

(line-to-line), I1 BL , and the total

3-phase power PBL (Fig
...
12)
...
In the equivalent
circuit of Fig
...
9, the magnetizing reactance X m is shunted by the

′
′
′
′
low-impedance branch R2 + jX 2
...
4
...

From the blocked-rotor test, the blocked-rotor resistance is:

RBL =

PBL
3I12

(4
...
37)

I1 BL

The blocked-rotor reactance at frequency of blocked rotor test is:

X BL

fBL

=

(Z

2
BL fBL

2
− RBL

)

(4
...
39)
Frequency at blocked rotor test

′
X BL ≅ X 1 + X 2

(4
...

From no load test we know that X 1 + X m = X NL and X 1 are
known then the magnetizing reactance is :

X m = X NL − X 1

(4
...
So, an

′
accurate determination of R2 is recommended by the IEEE as
follows:
The blocked resistance RBL is the sum of R1 and an equivalent

′
′
resistance, say R, which is the resistance of R2 + jX 2 in parallel
with X m as shown in Fig
...
11c; therefore,

Three-Phase Induction Machine
2
Xm
R= 2
R′
2 2
′
′
R2 + ( X 2 + X m )

243
(4
...
43)

2

 Xm 
or R ≅ 
 X ′ + X  R2
 ′
 2
m 

(4
...
So, we can use this value of R to determine

′
R2 from equation (4
...


Fig
...
12 A locked-rotor test permits the calculation of the total
leakage reactance x and the total resistance (R1 + R2 )
...


244 Chapter Four
Example 4
...
5 Ω
The locked-rotor test, conducted at reduced voltage, gave the
following results:
Locked-rotor voltage (line-to-line): 163 V
Locked-rotor power: 7200 W
Locked-rotor current: 60 A
Determine the equivalent circuit of the motor
...
5 / 2 = 0
...
143 Ω
I1 14

The no-load resistance is:

RNL =

PNL 1470
=
= 2
...
1432 − 2
...
97 Ω

In the IEEE recommended equivalent circuit we assume that

X 1 + X m = X NL = 17
...
6667 Ω
2
3 * 60 2
3 I1
BL

Note that in this example it does not mentioned the frequency of
the blocked rotor test so we can use the rated frequency as the
frequency of the blocked rotor test
...
5685 Ω
60

The blocked-rotor reactance is:

246 Chapter Four

X BL =

(Z

2
BL

)

2
− RBL = 1
...
6667 2 = 1
...
42 Ω
′
Assume, X 1 = X 2 (at rated frequency)
′
then X 1 = X 2 = 0
...
71 Ω, then the magnetizing reactance is :

X m = X NL − X 1 = 17
...
71 = 17
...
6667 − 0
...
4167 Ω
...
43):
2

2

 X′ + Xm 
0
...
26 
′ 
 R=
R2 =  2

 * 0
...
4517 Ω
Xm 
 17
...
3 The following test results are obtained from a threephase 60 hp, 2200 V, six-pole, 60 Hz squirrel-cage induction
motor
...
5 A, Input power = 1600 W
(2) Blocked-rotor test:
Frequency = 15 Hz, Line voltage = 270 V
Line current = 25 A, Input power = 9000 W

Three-Phase Induction Machine

247

(3) Average DC resistance per stator phase: R1 = 2
...

(b) Determine the parameters of the IEEE-recommended
equivalent circuit of Fig
...
9
...
4
...

Solution:
(a) From the no-load test, the no-load power is:

PNL = 1600 W
The no-load rotational loss is:

Prot = PNL − 3I12 R1
Prot = 1600 − 3 * 4
...
8 = 1429
...
For the no-load

′
condition, R2 / s is very high
...
4
...
The reactance of this
parallel combination is almost the same as X m
...
The equivalent circuit at no load is shown in
Fig
...
11a
...


(c) Blocked-rotor equivalent circuit for improved value for R2
...
4
...
2 V / Phase
3

The no-load impedance is:

Z NL =

V1 1270
...
27 Ω
I1
4
...
34 Ω
3I12 3 * 4
...
27 2 − 26
...
0 Ω
...
Because
′
′
X m >> R2 + jX 2 , the impedance X m can be neglected and the
equivalent circuit for the blocked-rotor test reduces to the form
shown in Fig
...
11b
...
8 Ω
3I12 3 * 252

′
Therefore, R2 = RBL − R1 = 4
...
8 = 2Ω , The blocked-rotor
impedance at 15 Hz is: Z BL =

V1
270
=
= 6
...
24

2

)

− 4
...
98 Ω

Its value at 60 Hz is X BL = 3
...
92 Ω
15

′
X BL ≅ X 1 + X 2

′
Hence, X 1 = X 2 =

15
...
96 Ω (at 60 Hz)
2

The magnetizing reactance is therefore:

250 Chapter Four

X m = 281 − 7
...
04 Ω
Now R = RBL − R1 = 4
...
8 = 2 Ω
...
96 + 273
...
12 Ω
273
...
31)

Vth ≅

273
...
97 V1
7
...
04

From Equation (4
...
97 2 R1 = 0
...
8 = 2
...
35)

X th ≅ X 1 = 7
...

4
...
The important performance characteristics in the steady
state are the efficiency, power factor, current, starting torque,
maximum (or pull-out) torque, and so forth
...
45)

Three-Phase Induction Machine

Where ω mech =

2πn
60

(4
...
47)
(4
...
49)

From Equations (4
...
47), and (4
...
50)
(4
...
52)

′
R2
s

(4
...
4
...
53)

Tmech

2
Vth
R′
=
*
* 2
ω syn (Rth + R2 / s )2 + ( X th + X 2 )2 s
′
′

1

(4
...
4
...
54), Vth , Rth , and X th
should be replaced by V1 , R1 , and X 1 , respectively
...
4
...
4
...
16
...
54) should be multiplied
by three to obtain the total torque developed by the machine
...
4
...
At low values of
slip,

Rth +

′
R2
R′
′
>> X th + X 2 and 2 >> Rth
s
s

Thus Tmech

2
Vth
* *s
≅
′
ω syn R2

1

(4
...
56)

The linear torque-speed relationship is evident in Fig
...
14 near
the synchrouns speed
...
4
...
56) Vth should be
replaced by V1
...
57)
(4
...
4
...


Fig
...
14 Torque-speed profile at different voltages
...
49) also indicates that at a particular speed (i
...
, a
fixed value of s ) the torque varies as the square of the supply
voltage Vth (hence V1 Fig
...
14) shows the T-n profile at various
supply voltages
...
Differentiating Equation (4
...
59)

254 Chapter Four
This expression can also be derived from the fact that the
condition for maximum torque corresponds to the condition for
maximum air gar power (Equation (4
...
This occurs, by the
familiar impedance-matching principle in circuit theory, when the

′
impedance of R2 / s equals in magnitude the impedance between
it and the supply voltage V1 (Fig
...
19) as shown in Equation (4
...

The slip STmax at maximum torque Tmax is:

STmax =

′
R2

2
Rth

(4
...
49) and
(4
...
61)

Equation (4
...
However, from Equation (4
...
The torque speed
characteristics for various values of R2 are shown in Fig
...
15
...
As the motor speeds up,

Three-Phase Induction Machine

255

the external resistance is gradually decreased and finally taken out
completely
...


Fig
...
15 Torque speed characteristics for varying R2
...
60) and (4
...
62)

(4
...
63) indicates that the maximum torque developed
by an induction machine is inversely proportional to the sum of the
leakage reactances
...
54), the ratio of the maximum developed
torque to the torque developed at any speed is:

′
′
(Rth + R2 / s )2 + ( X th + X 2 )2 * s
Tmax
=
2
T
′
′ 2
Rth + R2 / sTmax + ( X th + X 2 ) sTmax

(

)

(4
...
65)

From Equations (4
...
65)

(

′
′
Tmax (R2 / s ) + R2 / sTmax
=
2
T
′
2 R2 / sTmax
2

(

2
sTmax + s 2
Tmax
=
T
2 * sTmax * s

)

)

2

*

s
sTmax

(4
...
67)

Equation (4
...


Three-Phase Induction Machine

257

4
...
These losses are illustrated in the power flow diagram
of Fig
...
16
...
68)

The power loss in the stator winding is:

P = 3I12 R1
1

(4
...

Power is also lost as hysteresis and eddy current loss in the
magnetic material of the stator core
...
Part of it is lost
in the resistance of the rotor circuit
...
70)

Where R2 is the ac resistance of the rotor winding
...

Power is also lost in the rotor core
...


258 Chapter Four

Fig
...
16 Power flow in an induction motor
...
Part of
this is lost as windage and friction losses, which are dependent on
speed
...

The efficiency of the induction motor is: η =

Pout
Pin

(4
...
If all losses are
neglected except those in the resistance of the rotor circuit,

Pag = Pin

(4
...
73)

Pout = Pmecj = Pag (1 − s )

(4
...
75)

Three-Phase Induction Machine

Sometimes

259

ηideal is also called the internal efficiency as it

represents the ratio of the power output to the air gap power
...
This is why the
slip is very low for normal operation of the induction machine
...
75)
...

4
...
The power flow in the machine will
depend on the mode of operation
...
11 for various power relationships hold good for all
modes of operation
...
For example, in the generating mode, the slip is
negative
...
185) the air gap power Pag
is negative (note that the copper loss P2 in the rotor circuit is
always positive)
...


260 Chapter Four
The power flow diagram in the three modes of operation is
shown in Fig
...
17
...

In the motoring mode, slip s is positive
...
18) and the developed mechanical power Pmech
Equation (4
...
4
...

In the generating mode s is negative and therefore both Pag , and

Pmech are negative, as shown in Fig
...
17b
...
4
...

In the plugging mode, s is greater than one and therefore Pag is
positive but Pmech is negative as shown in Fig
...
17c
...
Power therefore
flows from both sides, and as a result the loss in the rotor circuit,

P2 , is enormously increased
...
4
...


Three-Phase Induction Machine

261

Fig
...
17 Power flow for various modes of operation of an
induction machine
...
(b) Generating
mode, s < 0
...


262 Chapter Four
Example 4
...
25 Ω, R2 = 0
...
5 Ω, X m = 30 Ω
The rotational losses are 1700 watts
...

(ii) Starting torque
...

(ii) Full-load current
...

(iv) Full-load power factor
...

(iv) Internal efficiency and motor efficiency at full load
...

(ii) Maximum torque developed
...
11 N
...
3
*100 = 87
...
4

ηint ernal = (1 − s ) *100 = (1 − 0
...
7%
(c) (i) From Equation (4
...
61)

Note that for parts (a) and (b) it is not necessary to use Thevenin
equivalent circuit
...
4
...
5 A three-phase, 460 V, 60 Hz, six-pole wound-rotor
induction motor drives a constant load of 100 N - m at a speed of
1140 rpm when the rotor terminals are short-circuited
...
Determine the value of the
resistance if the rotor winding resistance per phase is 0
...

Neglect rotational losses
...

Solution:

Three-Phase Induction Machine

267

From the equivalent circuits, it is obvious that if the value of

′
R2 / s remains the same, the rotor current I 2 and the stator current
I1 will remain the same, and the machine will develop the same
torque (Equation (4
...
Also, if the rotational losses are
neglected, the developed torque is the same as the load torque
...
6 The following test results are obtained from three
phase 100hp,460 V, eight pole star connected induction machine
No-load test : 460 V, 60 Hz, 40 A, 4
...
Blocked rotor test is
100V, 60Hz, 140A 8kW
...
152 Ω
(a) Determine the parameters of the equivalent circuit
...
Determine the input current, input power, air
gap power, rotor cupper loss, mechanical power developed,
output power and efficiency of the motor
...
64 Ω
40

PNL
4200
=
= 0
...
64 2 − 0
...
58 Ω
Then, X 1 + X m = 6
...
136 Ω
3 *140 2

0
...
076 Ω (from resistance between two stator
2

terminals)
...
412 Ω
140

X BL = 0
...
136 2 = 0
...
389 Ω

269

Three-Phase Induction Machine

′
Then, X 1 = X 2 =

0
...
1945 Ω
2

X m = 6
...
1945 = 6
...
136 − 0
...
06 Ω
2

0
...
3855 
′ 
Then, R2 = 
 * 0
...
0637 Ω
6
...
076 Ω

j0
...
195 Ω

j6
...
0637
s

120 f 120 * 60
=
= 900rpm
8
P

ns − n 900 − 873
=
= 0
...
0637
=
= 2
...
03
s
Input impedance Z1

Z1 = 0
...
195 +

I1 =

( j 6
...
123 + j 0
...
121∠27
...
123 + j (6
...
195)

V1
460 / 3
=
= 125
...
16o
Z1 2
...
16

270 Chapter Four
Input power:

Pin = 3 *

(

)

460
*125
...
16o = 88
...
22 2 * 0
...
575 kW
Air gap power

Pag = 88
...
575 = 85
...
03 * 85
...
556 kW
Mechanical power developed:

Pmech = (1 − s ) Pag = (1 − 0
...
192 = 82
...
076 = 3835
...
636 *103 − 3835
...
8 kW
Then the efficiency of the motor is:

η=

Pout
78
...
77 %
88
...
7 A three phase, 460 V 1450 rpm, 50 Hz, four pole
wound rotor induction motor has the following parameters per

′
′
phase ( R1 =0
...
18 Ω, X 1 = X 2 =0
...
The
rotational losses are 1500 W
...

Also find starting torque
...


(c)

Maximum torque and slip at which maximum torque will
be developed
...
6 V / phase
3

Z1 = 0
...
2 +
Then, I st =
(b) s =

j 40 * (0
...
2)
= 0
...
59o Ω
0
...
2

V1
265
...
91 ∠ − 46
...
55∠46
...
0333
1500

272 Chapter Four

′
R2
0
...
4 Ω
s 0
...
2 + j 0
...
4 + j 0
...
959 ∠10
...
4 + j 45
...
6
= 53
...
83o A
4
...
83o
o

Then the power factor is: cos 10
...
9822 lag
...
08 rad / sec
...
6 * ( j 40 )
= 264
...
285o V
(0
...
2)

Then,

Z th =

j 40 * (0
...
2 )
= 0
...
285o = 0
...
2 Ω
0
...
2

Then,

3 * (264
...
4
T=
= 228
...
08 * (0
...
4 ) + (0
...
2 )
2

Then, Pag = T * ω sys = 228
...
08 = 35921
...
0333 * 35921
...
7W
Then, Pout = Pm − Prot = 34723
...
7W

273

Three-Phase Induction Machine

Pin = 3 * 265
...
56 * 0
...
7
=
= 79
...
275)2

(

2 *188
...
198 + 0
...
2 + 0
...
18
=
= 0
...
198 + (0
...
2 )

[

(d) sTmax = 1 =

)]

2 1/ 2

2

= 862
...
198

′
′
R2 + Rext
2

]

2 1/ 2

+ (0
...
2 )

′
′
Then, R2 + Rext = 0
...
446323 − 0
...
26632 Ω
Example 4
...

(a) Determine the starting torque as percent of full load torque
...

(c) Determine the maximum torque developed by the motor as
percent of full load torque
...

Therefore equivalent circuit parameters cannot be used directly for
computation
...
57)
2
2
I 2 R2 I 2 R2
α
, Thus,
T=
sω syn
s

From Equation(4
...
72)

Example 4
...
Determine (i) full load speed (ii) speed
at maximum torque
...
5
Tst
T
= 1
...
75
TFL
TFL
Tmax
2
From above and equation (4
...
75
2
Tmax 1 + sTmax

2

Then, 0
...
75 = 0
Then sTmax = 2
...
451416
Also from Equation 4
...
451416
2
Tmax
0
...
451416 * s FL
2
s FL − 4 * 0
...
451416 2 = 0
2
s FL − 1
...
203777 = 0

s FL = 1
...
120957

ns =

120 * 50
= 1500 rpm
4

Three-Phase Induction Machine

277

then (a) nFL = (1 − s FL ) * ns

nFL = (1 − 0
...
451416) *1500 = 823 rpm
Example 4
...


′
′
R1 = 0
...
1 Ω, X 1 = X 2 = 0
...
For 5% slip, determine (a) The
motor speed in rpm and radians per sec
...
(c)
The stator cu-loss
...
(e) The rotor cu-loss
...
(g) The developed torque and the shaft torque
...

Solution:

120 * 60
= 1800 rpm
4
1800
ωs =
* 2π = 188
...
12 Ω

j0
...
25 Ω

0
...
05

j10 Ω

Z1 = 0
...
25 + Re + X e
Z1 = 0
...
25 +
V1 =

I1 =

208

j10 * (2 + j 0
...
1314∠23
...
25

= 120
...
1
= 2
...
55o A
o
2
...
55
2

(c) P = 3 * 56
...
12 = 1143
...
1 * 56
...
55o = 18610
...
9485 W
1
(e) P2 = sPag = 0
...
9785 = 873
...
5511W
(g) T =

Pag
188
...
9485
= 92
...
m
188
...
5
Pshaft
Ps

=

279

16194
...
9127 Nm
188
...
02%

For operation at low slip, the motor torque can be considered
proportional to slip
...
11 A 30, 100 WA, 460 V, 60 Hz, eight-pole induction
machine has the following
equivalent circuit parameters:

′
′
R1 = 0
...
05 Ω, X 1 = X 2 = 0
...
5 Ω
(a)

Derive the Thevenin equivalent circuit for the

induction machine
...

(c) If the maximum torque is to occur at start, determine the
external resistance required in each rotor phase
...
2
...
5
* V1 =
* 265
...
7 V
X1 + X m
0
...
5

Vth =

Rth + jX th =

( j 6
...
2 + 0
...
06589 +
0
...
2 + j 6
...
06589 j0
...
1947 Ω

j0
...
05
s

257
...
7 2 * 0
...
25 (0
...
05) + (0
...
2 )

Tmax =

2

2

]

= 624
...
7 2

[

2 * 94
...
06589 + 0
...
1947 + 0
...
8 Nm
sTmax =
Speed

(

0
...
06589 + (0
...
2 )

2

2

in

)

rpm

for

which

= 0
...
1249 ) * 900 = 787
...
1249

* 0
...
4 Ω

Then Rext = (0
...
05) / 1
...
243 Ω

1-speed characteristic of an induction machine
We have seen that a 3-phase squirrel-cage induction motor can
also function as a generator or as a brake
...
4
...
This curve,
together with the adjoining power flow diagrams, illustrates the
overall properties of a 3-phase squirrel-cage induction machine
...
But to operate in the
generator mode, the shaft must turn faster than synchronous speed
...

Finally, in order to operate as a brake, the shaft must turn in the
opposite direction to the revolving

Fig
...
18 Complete torque speed curve of a 3-phase induction
machine



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