THREE-PHASE INDUCTION MOTORS
1.0
Introduction
.Unlike other types of motors an induction motor is singly excited. The stator winding is excited by the
supply voltage while the rotor is excited by electromagnetic induction. Three-phase induction motors are
capable of developing toque at any speed except at synchronous speed. They capable develop starting toque
and are thus self starting. The 3-phase induction motor is the most common type of AC motor used in
industry because:
❖ It is very rugged
❖ It is relatively inexpensive to build
❖ It requires little maintenance
Fig 1
2.0
Construction
Three-phase motors are classified into two types depending on the type of rotor winding.
❖ Squirrel cage induction motor
❖ Wound (Slip ring) induction motor
2.1 Stator
(a)
Laminated Stator Core
The cores has slots which carry the stator winding
(b)
Frame, which also serves as the York
(c)
Distributed 3-phase stator winding
Fig 2 Group of Red Phase
For a 2-pole motor, the number of coil groups per phase is 2.
The total number of groups is
Where:
G = p = 2 x 3 = 6
p = Number of poles
 = Number of phases
If the stator has 60 coils, the number of coils per phase group is 10
In the two-pole machine let the Red Phase groups be placed at 0 o and 180o. Then:
The Yellow Phase groups will be at 60o and 240o
The blue Phase groups will be at 120o and 300o
u1
w2
v1
u2
Fig 2 Group of Red Phase
w1
v2
2.2
Rotor
(a) Laminated rotor core with slots cut on the outer circumference. It carries the rotor winding
(b) Rotor windings There are two types of windings
2.2.1
Squirrel Cage Rotor
The winding is made up of copper or aluminium bars inserted in the rotor core slots. The bars are shorted at
each end by a conducting end ring as shown in Fig. 2
Conductors
Conducting
end rings
Fig 3: Squirrel Cage Rotor
Winding
2.2.2
Wound Rotor
The winding is a distributed 3=phase winding resembling that of the stator winding. Three ends of the
winding are connected to external circuit through slip rings and carbon brushgear.
Starting Resistance
End Shield
Terminal block
Carbon brush
Shaft
Slip ring
Frame
Fig 3: Slip Rind Induction Motor
3.0
Production of rotating Magnetic field
The balanced three phase stator currents produce a rotating resultant airgap magnetic the resultant flux has
constant amplitude and rotates at synchronous speed. The three-phase currents are defined by the following
equations:
Red phase
iR = Im sin t
[1]
Yellow Phase
iR = Im sin t + 240o
[2]
Blue Phase
iR = Im
[3]
(
)
sin (t + 120 )
o
The magnetomotive force mmf is in phase with the current
Red phase
mmf R = N Im sin t = m sin t
Yellow Phase
mmfY = N Im sin t + 240o = m sin t + 240o
[5]
Blue Phase
mmf B = N Im
[6]
[4]
(
)
sin (t + 120 ) = 
o
m
(
)
sin (t + 120 )
o
m = Maximum mmf per phase
Where:
At t = 90 , the resultant mmf is
o
R = m +
m

3
sin 30o + m sin 30o = m
2
2
2
Fig 4(a)
AT t = 180 , the resultant mmf is
o
R = m +
Fig 4 (b)
3m
3m
3
cos 30o +
cos 30o = m
2
2
2
AT t = 270 , the resultant mmf is
o
R = −m −
m

3
sin 30o − m sin 30o = − m
2
2
2
Fig 4 (c)
AT t = 360 , the resultant mmf is
o
R = −
3m
3m
3
cos 30o −
cos 30o = − m
2
2
2
Fig 4 (d)
We have demonstrated that the mmf at any point in time has constant amplitude of
3
m . The stator mmf
2
rotates at constant amplitude and constant speed. The speed of rotation of the stator field is called
synchronous speed and is given by:
ns =
Where-:
4.0
120 f
p
[7]
ns = synchronous speed
f = supply frequency in Hz
P = number of poles
Principles of Operation
The principle of operation of a three-phase induction motor can be demonstrated by using step ladders. Let
us move the magnet in the direction shown. The magnetic flux is entering the paper and the flux cuts the
conductor as the pole moves. We can establish the direction of the induced emf across the conductor by
using the cross product. The emf across the conductor has the direction shown.
-
NORTH
POLE
v
I
+
(a)
v
e = lv  B
e
B
(b)
Fig 5 (a) Step Ladder and (b) Direction of induced emf
Since there is a conducting path the emf will push current through the steps in the direction shown in the
step ladder. The magnetic field created by the current will interact with the main field to produce a force on
the conductor. The direction of the force is determined by the cross product rule.
F
I
F = li  B
B
(a)
(b)
Fig 5(c) and (d): Production of Force on the step Ladders
Thus we note that if the ladder was free to move it would move in the direction of the moving field. The
ladder represents the squirrel cage winding of a three-phase induction motor. When the stator rotating
magnetic field sweeps across the conductors emf is induced in the conductors. Since the conductors are
shorted by the end rings, current flows. The current establishes its own field around the conductor. Like in
the step ladder, force acts on each conductor, thus producing electromagnetic torque. The rotor rotates in
the direction of the rotating magnetic field. Therefore we have demonstrated the principle of operation of
the three-phase induction motor.
4.1
Transformer action
A 3-phsae induction motor behaves like a 3-phase transformer when the rotor winding is open circuited.
The stator winding is the primary and the rotor is the secondary winding. When the stator winding is
energized, a no-load current flows in the stator winding creating a magnetic field, which rotates at constant
amplitude around the airgap. Emf of self-induction is induced in the stator winding. The stator field links
with the rotor winding inducing an emf of mutual induction E2
E1 = 4.44 fN1 and E2 = 4.44 fN 2
[8]
The stator current at this moment is responsible for creating the magnetic field and supplying the core
losses.
s
F
Fig 6 (a): Transformer action
Fig 6 (b): Induction Current
Fig 6 (c): Production of Starting
Torque
Let us short circuit the rotor winding while the stator is still energized. Current will flow in the rotor
winding. The rotor current creates its on field. This field reacts with the rotating stator filed to produce
torque acting on each rotor conductor. Since the rotor is free to rotate, the net torque on the rotor makes the
rotor to start rotating in the direction of the rotating magnetic field.
4.2
Slip Speed
The relative speed between the stator field and the rotor conductors deceases as the rotor accelerates If the
rotor speed were to reach synchronous speed, the relative speed would be zero and hence torque would be
zero. In practice this condition does not happen. The motor must develop torque to balance a small no-load
mechanical torque due to windage and bearing friction. Thus the rotor will settle for speed, which is close
but less that the synchronous speed. The relative speed between the rotating stator air gap flux and the rotor
conductors is called slip speed
Slip speed = synchronous speed – actual rotor speed = ns − n
Usually the slip speed is expressed as percentage of synchronous speed and is given the name slip (s)
s=
ns − n s − 
=
ns
s
[9]
The frequency of the rotor emf and current depends on the relative speed between the stator mmf and the
rotor conductors. At stand still the rotor frequency equals the supply frequency. If n is the speed of the
rotor conductors in r/min, then the speed at which the rotor conductors are being cut is:
Frequency of the rotor emf is:
But
fr = p
ns − n
120
fr = p
sns
pn
= s s = sf
120
120
ns − n = sns
[9]
The rotor emf at stand still
At any slip, the rotor emf is given by:
E2 = 4.44 fN 2
Er = 4.44 f r N2 = 4.44sfN2 = sE2
[10]
Thus the magnitude of the rotor emf is directly proportional slip and therefore depends on the relative speed
between the stator mmf and the rotor conductors.
Example 1
A six pole, 50 Hz three-phase induction motor operates at 2% slip.
(a)
(b)
(c)
Express the rotor emf and frequency in terms of slip
At what speed to the stator field rotor and rotor field rotate?
What is the frequency of the rotor currents?
Solution
ns − n
sn
sp 120 f
p= s p=

= sf
120
120
120
p
Rotor frequency
fr = 1
Rotor emf
Er = 4.44 f r N2 = 4.44  sfN2 = s 4.44 fN 2 = sE2
Speed of magnetic field
ns =
Speed of the rotor
120 f 120  50
=
= 1000 r/min
p
6
n = ns (1 − s ) = 1000 (1 − 0.02 ) = 980 r/min
Speed of the rotor field equals speed of the rotor
Frequency of the rotor currents
f r = sf = 0.02  50 = 1 Hz
5.0
Equivalent Circuit
The equivalent circuit of a three-phase induction motor is identical to that of the transformer. Since the
motor is dynamic, the equivalent circuit must include a dynamic load. First let us consider the secondary
winding equivalent circuit shown in Fig 7 (a).
The rotor current is given by:
Where:
The rotor power is
I2 =
sE2
R2 + jsX 2
[11]
R2 = Rotor winding resistance at standstill
X 2 = Rotor leakage reactance at standstill
P = 3I 2 2 R2 which is the power dissipated in the rotor resistance (Pcu2). We notice that
the mechanical power is not accounted for. If we divide top and bottom of equation 11 by s we don’t
change the value of the current but the current is now defined by equation 12
I2 =
E2
R2
+ jX 2
s
[12]
The equivalent circuit corresponding to equation 12 is shown in Fig 7 (b). The rotor power is given by-
P = 3I 2 2
R2
s
Please note that the power is greater than that given by equation 12. This is the input power to the rotor. It
is the power transferred from the stator to the rotor (electromagnetic power developed by the motor). We
R2
is the equivalent dynamic resistance representing the
s
will simply refer to it as air gap power Pg and
electromagnetic power developed by the motor.
Pg = 3I 2 2
Where-
R2
= T s
s
[13]
T = electromagnetic torque developed
jsX 2
I2
jX 2
I2
sE2
R2
R2
s
E2
(a)
(b)
Fig 7: Equivalent Rotor Circuit
At this moment we are still unable to account for the mechanical power developed by the motor. Let us do
the following operation;
R2 R2
=
− R2 + R2
s
s
[14]
This does not alter the value of the equivalent air gap power dynamic resistance. But we can sketch a
circuit representing this condition as shown in Fig 8. The power in the rotor is given by;
Pg = Pm + Pcu 2 = 3I 2 2
Pg = Pg (1 − s ) + sPg
The equivalent dynamic resistance
R2
(1 − s ) + 3I 22 R2
S
[15]
R2
(1 − s ) represents the mechanical power (Pm) developed by the
s
motor
Pm = 3I 2 2
Where:
R2
(1 − s ) = Pg (1 − s ) = T s (1 − s ) = T  = Tm
S
Tm = Mechanical torque
I2
jX 2
R2
E2
Fig 9: Equivalent Rotor Circuit
R2
(1 − s )
s
[16]
The output shaft power is the mechanical power minus the windage and friction (rotational) losses.
Po = Pm − Prot = Tm + Trot = TL
Where:
[17]
TL = shaft load torque
Example 2
A 4-pole induction motor has open-circuit rotor voltage of 200 v, 50 Hz. The rotor resistance is 0.1 Ω and
the rotor leakage reactance is 0.3 Ω. The rotor winding is star connected. If the motor operates at a speed of
1455 r/min, calculate:
(a)
(b)
(c)
The rotor current
Electromagnetic power and torque
The output power if the rotational losses amount to 54.3 W
Solution
I2 =
(a)
Rotor current
(b)
Electromagnetic power;
Electromagnetic torque
(c)
E2
200 / 3
=
= 34.5 − 5.14o A
R2
0.1
+ j 0.3
+ jX 2
0.03
s
R2
0.1
= 3  34.52 
= 1190 W
s
0.03
Pg 1190  60
T=
=
= 75.8 Nm
s
2 1500
Pg = 3I 2 2
Po = Pm − Prot
Output shaft power
Po = Pg (1 − s ) + Prot = 1190 (1 − 0.03) − 54.3 = 1100 W
5.1
Equivalent circuit referred to the stator winding
Please note that the stator equivalent circuit is exactly the same as that of the transformer primary winding
as shown in Fig 10.
I1
R1
jX 1
Im
− jI
I
V1
c
Rc
jX 
E1
Fig 9: Stator Equivalent Circuit
Just as we did with the transformer we can simplify the motor equivalent circuit by referring rotor
quantities to the stator as follows:
R2
R
= n2 2
s
s
X 2
X
= n2 2
s
s
E2 = nE2 = E1
[16]
[17]
[18]
I
I 2 = 2
n
[19]
Since the emf E2 = E1 , we can join their terminals to obtain the exact equivalent circuit shown in Fig 10.
The circuit can be simplified further by moving the magnetizing branch and connecting it across the supply
to obtain the approximate equivalent circuit of Fig 11
jX 2
jX 1
R1
I1
I 2
Im
− jI
Ic
Rc
V1
R2
s
jX 
Fig 10: Exact Equivalent Circuit referred to the Stator
R1
I1
(
j X 1 + X 2
)
I 2
Im
− jI
Ic
V1
Rc
R2
s
jX 
Fig 12: Approximate Equivalent Circuit referred to the stator
Now we can use the approximate equivalent circuit to analyze the performance of the motor.
The rotor current referred to the stator is given by;
V1
I 2 =
R1 +
(
R2
+ j X 1 + X 2
s
)
I o = I c + I =
The no load current
V1
V
+ 1
Rc jX 
I1 = I 2 + I o
The input current
The air gap power
R
Pg = 3I 22 2
s
= 3I 22 R2
The rotor copper losses
Pcu 2
The mechanical power
R
Pm = 3I 22 2 (1 − s )
s
Torque – Speed Characteristic
6.0
The important characteristic of the motor is that relating speed to torque because the motor looks at the load
as a torque applied to its shaft. Electromagnetic torque is;

2 R2
3
I
2
Pg
s =
T=
= Pg =
s
s
3V12
R2
s
2



R
s  R1 + 2  + X 1 + X 2

s 

(
)
2




Starting torque occurs at slip s = 1
Ts =
Pg
s
= Pg =
3I 22 R2
s
=

3V12 R2
) (
(
)
2
2
s  R1 + R2 + X 1 + X 2 


It can be shown that the maximum torque is given by:
Tmax =
3V12
(

2s  R1 + R12 + X 1 + X 2

) 
2
Where corresponding slip at maximum torques is
sm =
(
R2
R12 + X 1 + X 2
)
2
Example 3
The motor in example 2 has a stator resistance of 0.4 Ω and stator leakage reactance of
1.2 Ω. The rotor resistance is 0.1 Ω and rotor leakage reactance is 0.3 Ω. The effective stator to rotor
winding turns ration is 2. The stator winding is connected in star and fed from a 400V, 50 Hz supply. The
motor operates at a rated slip of 0.03 and the no-load current is 2.2975.97 A. If the rotational loss
torque is 3.29 Nm, determine:
o
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
The rotor current referred to the stator
The electromagnetic torque
The mechanical power
The load torque and shaft output power
The core and total copper losses
The efficiency
The input stator current
The core loss resistance and magnetizing reactance
Solution
(a)
The rotor current referred to the stator
I 2 =
T=
240
= 17.21 − 9.91o
0.4 + j (1.2 + 1.2 )
0.4 +
0.03
3  2402  13.33
= 75.48 Nm
(b)
Electromagnetic torque
(c)
Mechanical power
Pm = T s (1 − s ) = 75.48  50 (1 − 0.03) = 11501 W
(d)
Load torque
TL = T − Trot = 75.48 − 3.29 = 72.19
(e)
Shaft output power
Po = To = 72.19  50 (1 − 0.03 ) = 11 kW
Core losses
Pc = 3V1I c = 3  240  2.29cos 75.97o = 400 W
Total copper losses
Pcu = 3I 22 R1 + R2 = 3 17.212  0.8 = 711 W
(
)
Prot = Trot = 3.29  50  0.97 = 501 W
Po
11000
=
100 =
100 = 90.8 %
Po +  Losses
11000 + 400 + 711 + 501
(f)
Efficiency
(g)
Input current:
(h)
2
50 ( 0.4 + 13.33) + 2.4 2 


I1 = I 2 + I o = 17.21 − 9.91 + 3.29 − 75.97o
Core loss resistance
Magnetizing reactance
I1 = 17.751 − j 6.154 = 18.88 −1912o
V
240
Rc = 1 =
= 300 Ω
I c 3.29  cos 75.97
V
240
X = 1 =
= 75.2 Ω
I 3.29  sin 75.97
Let us develop the torque equation using the rotor circuit.
3E2
T
R2
s
 R2 2

2
 + X2 
 s 

s 
2
 R2 
2
2
 >> X 2 thus X 2 can be neglected
 s 
For small valus of slip, 
R2
2
s = 3E2 s
T=
 R2 2  s R2
 s   
 s  
3E2 2
T = ks
Electromagnetic torque is directly proportional to slip
2
2
R 
R 
2
For very large values of slip,  2  << X 2 thus  2  can be neglected
 s 
 s 
R2
3E2 2 R2 1
s
T=
=

s  X 2 2  s X 2 2 s
3E2 2
Electromanetic torque is inversely proportional to slip s
T=
k
s
The maximum torque: It can be shown that muxumum torque occurs when
Tmax
R2
= X2
s
3E2 2
=
=k
2s X 2
Maximum torque is a constant for a particuler motor. The slip at which maximum toque occurs increases
with increasing rotor resistance
At starting s =1:
Ts =
3E2 2 R2
3E2 2 R2
=
= kR2
s  R2 2 + X 2 2  s X 2 2
Starting torque is directly proportional to rotor resistance. We can start the motor at maximum torque if we
make rotor resistance equal to rotor reactance
Maximum starting torque
Ts max = Tmax = Tmax
3E2 2 X 2 3E2 2
=
=
s X 2 2 s X 2
The torque speed characteristic is shown in Fig 13
TORQUE - SPEED CHARACTERISTIC
3
2.5
TORQUE (P.U.)
2
1.5
Toqrque
1
0.5
0
0
0.2
0.4
0.6
0.8
1
1.2
SPEED (P.U.)
Example 4
The speed of the motor in Example 3 dives a constant torque load. If the Mechanical torque is 75.48 Nm,
(a)
Plot the torque – speed characteristic of the motor and the load
(i)
(ii)
What is the maximum torque?
What is the slip at maximum torque?
(b)
Determine the voltage range and corresponding speed control range when speed is controlled by
varying the supply voltage
(c)
Determine the rotor resistance range and corresponding speed control range when speed is
controlled by varying the rotor resistance
Solution
(a)
plot the toque speed characteristic
From graph (i) maximum torque Tmax
= 194 Nm
(b) The minimum supply voltage is limited by
(ii) slip
sm = 0.164
Tmax 2 = 75.48 Nm
2
Voltage
V2 = V1
Tmax 2
Tmax1
V 
Tmax 2 = Tmax1  2 
 V1 
75.48
= 240
= 149.7
194
nmin = ns (1 − sm ) = 1500 (1 − 0.164 ) = 1254 r/min
Corresponding minimum speed
The speed control range is 1254 r/min to 1455 r/min, corresponding to voltage range of 149.7 V to 240 V
(c)
Speed is limited by the rotor resistance at s=1
R2 = R12 + ( X 1 + X 2 ) = 0.42 + 2.4 2 = 2.4331 Ω
2
We must plot the characteristic
From graph slip at minimum speed is 0.18
Speed
n = 1500 (1 − 0.18 ) = 1230 r/min
The speed control range is 1230 r/min to 1455 r/min as the rotor resistance varies from 0.4 Ω to
2.4331Ω
Effect of varying rotor resistance
1
2
3
For given load the operating point will shift. (Slip increases while speed decreases and rotor
current increases
Maximum toque is constant but slip at maximum torques increases
The starting torque increases
TORQUE - SPEED CHARACTERISTCIC
210
200
190
180
170
160
150
140
TORQUE (Nm)
130
Series1
Series2
Series3
Series4
Series5
Series6
120
110
100
90
80
70
60
50
40
30
20
10
0
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
SLIP
Example 5
The rotor resistance of the motor in example 3 is increased by 100 % by adjusting the external rotor
resistance.
(a)
(b)
Calculate the stating current and torque
If the load torque is constant at rated value, calculate the slip and new operating speed
Solution
(a)
Staring current
I 2 =
I 2 =
V1
(
R1 + 2 R2 + j X 1 + X 2
)
=
400
0.4 + 0.8 + j1.2
400
= 235.7 − 45o
0.4 + 0.8 + j1.2
I1 = I 2 + I oc = 166.7 − j166.7 + 0.555 − j 2.222
I1 = 167.3 − j168.9 = 237.7 − 45.27o A
Magnitude of stator line current
I L = 3I1 = 3  237.7 = 412 A
0.9
1
7.0
Motor Speed Control Methods
The speed of a 3-phase induction motor can by varied by the following methods:
1
Varying the supply Voltage
The motor torque varies with the square of the voltage, reducing the supply voltage deceases toque
as shown in Fig 14. Speed at which the motor operates will decrease and the actual operating
point will depend upon the load characteristic
250
200
150
Series1
Series2
Series3
Series4
Series5
100
50
0
0
2
0.2
0.4
0.6
0.8
1
1.2
Varying the supply frequency
Speed can also be controlled by varying the supply frequency f which varies the synchronous
speed and hence motor speed
6000
5000
4000
Series1
Series2
Series3
Series4
3000
2000
1000
0
0
3
200
400
600
800
1000
1200
1400
1600
Pole changing
Pole changing varies the synchronous speed. 2 or 4 discrete speeds can be obtained by creating
consequential poles
12000
10000
8000
n=1500
n=750
n=375
n=187.5
6000
4000
2000
0
0
4
200
400
600
800
1000
1200
1400
1600
Varying voltage and frequency V/f = constant control
This method combines the supply frequency and voltage control methods. The ratio v/f is kept
constant to keep the air gap flux constant
E = 2.22 fZ 
=
E
1
E
=

2.22 fZ 2.22Z f
k
V
f
1400
1200
1000
800
f=50
f=40
F=30
f=25
Tm
600
400
200
0
0
400
600
800
1000
1200
1400
1600
Varying rotor circuit resistance
This method is used in Slip ring type induction motors. Speed is controlled by adjusting the
external rotor resistance.
TORQUE - SPEED CHARACTERISTCIC
210
200
190
180
170
160
150
140
130
TORQUE (Nm)
5
200
Series1
Series2
Series3
Series4
Series5
Series6
120
110
100
90
80
70
60
50
40
30
20
10
0
0
0.1
0.2
0.3
0.4
0.5
SLIP
0.6
0.7
0.8
0.9
1
8.0
Motor Starting Methods
1
2
3
4
5
6
Direct on line Starting: no current limit
Star delta Starting
Autotransformer Starting
Stator Resistance Starting
Rotor Resistance Starting
Power Electronic Starting
8.1
Direct on line Starting
I 2 =
Starting Current I
VL
((R + R  + ) + ( X + X  )
2
1
2
1
M
O
L
3
Phase
3
M
Mcb
Stop
Start
M
M
OL
8.2
Fig 10: Direct On
Lone Starter
Autotransformer Starting
The starting Current
2
2
I 2 =
nVL
((R + R  + ) + ( X + X  )
2
1
2
1
2
2
Fig 11: Autotransformer Starter
8.3
Star – Delta Starting
I 2 =
(
VL
2
) (
2
( R1 + R2 + + X 1 + X 2
)
2
R
Y
B
Mcb
OL
K2
K1
K3
Fig 12: Star – delta starter Power Circuit
L1
OL
Stop
KT
1
Start
K
N
K1
KT1
K3
K2
K2
1
Fig 13: Star – delta Starter Control Circuit
K3
8.4
Rotor Resistance Starting
I 2 =
(
VL
( R1 + R2 + Rext
) (
2
+ X 1 + X 2
)
2
Fig 14: Rotor Resistance Starter
8.5
Power Electronic Starting
Employs power electronic techniques to start the motor. Power Electronic Starters are known as Soft
Starters because they are capable of starting the motor smoothly.
9.0
Determining Equivalent Circuit Parameters
9.1
No-load Test
The motor is run without load coupled to the shaft. The no-load power is the sum of the core losses, the
rotational losses and the stator no-load copper losses.
Poc = 3I oc 2 R1 + Pc + Prot
If we assume that the core losses are equal to rotational losses then the core losses are given by:
Pc 
The core loss resistance is;
Rc =
Poc − 3I oc 2 R1
2
Poc − 3I oc 2 R1
2Voc
I = I oc sin oc
Magnetizing reactance
X 
Voc
I oc sin  oc
9.2
Blocked Rotor Test
During the blocked rotor test the rotor is locked so that it does not rotate. The power measured at full load
current is approximately equal to the total full load copper losses.
Equivalent winding resistance referred to the stator:
Re1 =
Psc
I sc 2
R2 = Re1 − R1
Vsc
I sc
The equivalent impedance referred to the stator:
Z e1 =
The equivalent reactance referred to the stator
X e1 = Z e12 + X e12
X
X 2 = e1
2
9.0
Questions
1
A 16 pole three-phase induction motor operates from 50 Hz supply
(a)
What is the slip if the rotor speed is 369r / min ?
(b)
If the rotor emf at standstill is 100 V/phase, what is the magnitude of the rotor emf/phase
and its frequency when the rotor speed is 369r / min ?
Ans:
(a)
s = 0.016
(b)
Er = 1.6V
2
Show that the three phase balanced current flowing through the stator winding of a three-phase
induction motor create a rotating magnetic field of constant amplitude
3
Explain with the aid of suitable diagram how starting torque is produced in a three-phase induction
motor
4
Explain why the three-phase induction motor will not run at synchronous speed.
5
A 440 V, 50 Hz 8-pole delta connected three-phase squirrel cage induction motor has the
following equivalent circuit parameters per phase:
R1 = R2 = 0.10
X 1 = X 2 = 0.50
Rc = 100
X  = 25
The blocked rotor test is conducted with a line current of 156 A and the no-load test is conducted
by applying rated voltage across the stator winding and at the same time driving the rotor in the
direction of the rotating field at synchronous speed ( s = 0 ) . Determine;
(a)
(b)
The line voltage and power factor on blocked rotor test
The line current and power factor on no-load test
Ans:
(a)
Vsc = 159.1V pf = 0.1961 lag
(b)
I sc = 31.4 A pf = 0.2425 lag
6
If the machine in Q5, operates at a slip of 2 %. calculate:
(a)
The magnitude of input stator phase and line currents and power factor
(b)
The rotor current referred to stator winding
(c)
The electromagnetic torque
(d)
The efficiency if mechanical losses amount to 1 kW
Ans:
(a)
(c)
7
I P = 94.26 A
T = 1396 Nm
I L = 163.3 A
(d)
 = 93.6%
(b)
84.66 − 11.94o A
A three-phase 400 V, 50 Hz star connected 4-pole induction motor runs at a speed of 1440 r/min.
when operating at its rated load. The equivalent circuit has the following parameters per phase:
R1 = 0.20
R2 = 0.40
X 1 = X 2 = 2.0
Rc = 200
X  = 40
Determine for rated load:
(a)
The value of line current and power factor
(b)
The electromagnetic power and torque
(c)
The mechanical power
(d)
The output power
(e)
The efficiency
NB; rotational losses amount to 1700 kW
Ans:
(a)
I L = 23.83 A
T = 84.2 Nm
 = 75.8%
8
pf = 0.9138 lag.
(c)
(b)
Pm = 12.7kW (d)
Pg = 13.23kW
Po = 11kW
(e)
8when operating at its rated load. The equivalent circuit has the following parameters per phase:
R1 = 0.20
R2 = 0.20
X 1 = X 2 = 1.0
Rc = 200
X  = 40
Determine for rated load:
(a)
The value of line current and power factor
(b)
The electromagnetic power and torque
(c)
The mechanical power
(d)
The output power
(e)
The efficiency
NB; rotational losses amount to 2541 kW
Ans:
9
(a)
I L = 45.54 A
(e)
T = 162.7 Nm (c)
 = 78.1%
pf = 0.8918 lag.
(b)
Pm = 24.541kW
Pg = 25564kW
(d)
Po = 22kW
In a certain 3-phase induction motor, the leakage reactance is 4 times the resistance for both stator
and rotor windings. The stator impedance is identical to the referred rotor impedance. The slip at
full load is 3 %. It desired to limit the starting current to three times the full load current. By how
much:
(a)
Would the stator resistance be increased?
(b)
Would the rotor resistance referred to the stator be increased?
Ans:
(a)
R1ext = 6.607 R1
(b)
R2ext = 6.607 R2
10
A 3-phase, 4-pole, 3300V, 50 Hz star connected induction motor has identical stator and referred
rotor impedance of value 3 + j 9 per phase. The stator to rotor turns ratio is 3 and the rotor is
star connected and brought out to slip rings. Calculate:
(a)
(b)
(c)
(d)
(e)
The full load developed torque at rated slip 0f 5 %
The maximum torque at normal voltage and frequency
The supply voltage which can be with stood without the motor stalling
The maximum torque if the supply voltage and frequency both fall to half normal value
The increase in rotor circuit resistance which at normal voltage and frequency will
permit maximum torque to be developed at starting
T = 969 Nm (b)
(d) Tmax = 1388 Nm
Ans:
11
12
Speed
torque
(a)
(c)
Tmax = 1631Nm
(e)
R2ext = 1.028
VL = 2543V
A 3-phase 6-pole 50 Hz induction motor has a peak torque of 6 Nm and a starting torque of 3 Nm
when operating at full voltage. Maximum torque occurs at a slip of 25 %. The current is 2 A when
started at 1/3 of normal voltage.
(a)
(b)
(c)
(d)
What is the mechanical power at peak torque when operating at normal voltage?
What maximum torque would the machine produce at 1/3 of normal voltage?
What starting current would the machine take when supplied at normal voltage?
What extra rotor circuit resistance as a percentage would be required to give maximum
torque at starting and what would then be the current in terms of that at peak torque
without external resistance?
Ans:
(a)
Pm = 707W
(d)
%R = 300%
(b)
Tmax = 0.667 Nm
Is = 6 A
(c)
Is = 6 A
An induction motor has the following speed/torque characteristic:
1470
3
1440
6
1410
9
1300
13
1100
16
900
13
750
11
350
7
0
5
It is drives a load requiring a torque including losses of 4 Nm at starting and which increases
linearly with speed to be 8 Nm at 1500 r/min.
(a)
(c)
Determine the range of speed control obtainable without stalling by providing supply
voltage reduction
If the rotor was replaced with one having the same leakage reactance but double
resistance, what would then be the possible range of speed variation with voltage control
For part (a) and (b) give the range of voltage variation required
Ans:
(a)
(b)
(c)
13
Effect of varying rotor resistance on the following:
(a)
(b)
(c)
14
(b)
1420r / mn. − 1100r / min.
1344r / min − 700r / min
V1 − 0.6055V1
V1 − 0.6583V1
Motor steady state operating point
Maximum torque and current
Starting torque and current
Explain the effect of varying the supply voltage on the following
(a)
(b)
(c)
(d)
15
Explain the effect of varying the supply frequency on the following
(a)
(b)
(c)
(d)
16
Motor steady state operating point
Maximum torque and current
Starting torque and current
Motor steady state operating point
Motor steady state operating point
Maximum torque and current
Starting torque and current
Motor steady state operating point
Explain the effect of varying the supply voltage and frequency (V/f = constant) on the following
(e)
(f)
(g)
(h)
Motor steady state operating point
Maximum torque and current
Starting torque and current
Motor steady state operating point