1
K.Rajkumar, Associate Professor/EEE, Dhanalakshmi Srinivasan Institute of Technology, Samayapuram.
Unit 3 - Induction Motor
PART - A
1. Why the rotor slots are slightly skewed in squirrel cage IM? (or)Why are the clots on the
cage rotor of induction motor usually skewed?
The rotor bars are not parallel to the shaft but slightly skewed for the following purposes,
i. It helps to make the motor run quietly by reducing magnetic hum and
ii. It avoids the tendency of magnetic locking between stator and rotor teeth.
iii. It gives more uniform torque in running.
2. A 50 Hz, 6 pole, 3–phase induction motor runs at 970 rpm. Find slip.
Rotor Speed, Nr=970 r.p.m
Synchronous Speed, mprp
fN s ..1000
6
50120120
%)(., 30301000
9701000or
N
NNsSlip
s
rs
3. Under what condition, the slip in an induction motor is
(a) Negative
(b) Greater than one.
The slip of the induction motor is negative when it is operated above the rated speed (as
induction generator).
The slip is greater than one when the motor is operated at brake mode.
4. What are the two fundamental characteristics of a rotating magnetic field?
i. The resultant flux is constant i.e, 1.5 times the maximum value of the flux due to any
phase.
ii. The resultant flux rotates around the stator at synchronous speed given by ,
mprp
fN s ..
120
5. Define slip of an induction motor.
The difference between the synchronous speed Ns and the actual speed Nr of the rotor is
known as slip.
100
s
rs
N
NNSlip%
6. What are the merits and demerits of double squirrel cage induction motor?
Merits (or) Advantages:
i. Higher starting torque
ii. Wide range of torque-speed characteristics can be obtained
Demerits (or) Disadvantages:
i. Costlier about 20-30% of single cage rotor
ii. Effective rotor resistance is high, so rotor heating at the starting is more.
iii. Due to higher effective leakage reactance, the powerfactor is reduced.
2
K.Rajkumar, Associate Professor/EEE, Dhanalakshmi Srinivasan Institute of Technology, Samayapuram.
7. How do change in supply voltage and frequency affect the performance of a 3 phase
induction motor?
The torque developed by the induction motor is proportional the voltage applied ( T α E2)
The speed of the motor is inversely proportional to the supply frequency . (Ns α 1/f).
8. The starting torque of a squirrel cage induction motor cannot be altered when the applied
voltage is constant. Why?
22
22
222
XR
RkETst
From the above relation, R2 is constant and X2 is independent of‘s’. So, Tst is only depending
on E22. Therefore the starting torque cannot be altered for constant supply voltage, where
E2αV.
9. A 3-Phase, 4 pole, induction motor operates from a supply whose frequency is 50Hz.
Calculate the frequency of the rotor current at standstill and the speed at which the
magnetic field of the stator is rotating.
sff' Frequency, Current Rotor . At standstill, s=1. Therefore, f’=50Hz.
Speed of rotation of magnetic field, r.p.m 1500p
fN s
4
50120120
10. Write down the condition to get maximum torque under running condition.
The condition for maximum torque of 3 phase induction motor is, 22 sXR 11. Define Cogging.
With the equal number of stator and rotor slots, the speeds of all the harmonics produced by
the stator slotting coincide with the speed of rotor harmonics. Thus the machine refuses to
start due to magnetic locking. This is known as cogging.
Therefore the number of stator slots should not be equal to the number of rotor slots.
12. Define Crawling.
The tendency of the motor to run stably at speeds as low as one-seventh of its synchronous
speed is called crawling. The torque produced during this crawling is a harmonic induction
torque.
13. How the slip does vary with load?
The greater the load, the greater the slip.
14. What are the advantages of double cage induction motor?
i. Starting torque is high
ii. Low starting current hence suitable for direct on line starting
iii. Wide range of torque-slip characteristics can be obtained
15. What is an Induction generator? What are its limitations?
When a 3-phase induction motor made to run at a speed higher than its synchronous speed by
means of prime mover, it becomes 3-phase induction generator.
Limitations:
i. It needs reactive power from the supply network.
ii. Its output voltage and frequency cannot be controlled.
iii. It always operates at a leading powerfactor.
3
K.Rajkumar, Associate Professor/EEE, Dhanalakshmi Srinivasan Institute of Technology, Samayapuram.
PART – B
1. a. Explain the principle of operation of 3-phase induction motor and explain how the
rotating magnetic field is produced by three-phase currents. (10)
Principle of Operation:
Consider a portion of 3-phase induction motor as shown in Fig. (8.13).
The operation of the motor can be explained as follows:
i. When 3-phase stator winding is energized from a 3-phase supply, a rotating magnetic
field is set up which rotates round the stator at synchronous speed Ns(=120f/p).
ii. The rotating field passes through the air gap and cuts the rotor conductors, which as yet,
are stationary. Due to the relative motion between the rotating flux and the stationary
rotor, e.m.f’s are induced in the rotor conductors. Since the rotor circuit is short-circuited,
currents start flowing in the rotor conductors.
iii. The current-carrying rotor conductors are placed in the magnetic field produced by the
stator. Consequently, mechanical force acts on the rotor conductors. The sum of the
mechanical forces on all the rotor conductors produce a torque which tends to move the
rotor in the same direction as the rotating field.
iv. The fact that rotor is urged to follow the stator field (i.e., rotor moves in the direction of
stator field) can be explained by Lenz’s law. According to this law, the direction of rotor
currents will be such that they tend to oppose the cause producing them. Now, the cause
producing the rotor currents is the relative speed between the rotating field and the
stationary rotor conductors. Hence to reduce this
relative speed, the rotor starts running in the same
direction as that of stator field and tries to catch
it.
Rotating Magnetic Field (RMF):
Figure 1 shows three-phase windings displaced in
space by 120º, are fed by three-phase currents, displaced
in time by 120º, they produce a resultant magnetic flux,
which rotates in space as if actual magnetic poles were
being rotated mechanically.
The sinusoidal flux due to three-phase windings is
shown in Figure 2. Let the maximum value of flux due to
any one of the three phases be Φm. The resultant flux Φr, Figure 1 Three Phase Winding in motor
4
K.Rajkumar, Associate Professor/EEE, Dhanalakshmi Srinivasan Institute of Technology, Samayapuram.
at any instant, is given by the vector sum of the individual fluxes, Φ1, Φ2 and Φ3 due to three
phases. We will consider values of Φr at four instants 1/6th
time-period apart corresponding to
points marked 0, 1, 2 and 3 in Figure 2.
Figure 2 Three Phase Rotating Sinusoidal Fluxes and their 120odisplaced vector
Referring to Figure 2,
a. At =0o,
mm and 2
3
2
30 321 ,
The flux Φ2 is drawn in a direction opposite to the direction
assumed positive. Therefore,
mm
o
mr 2
3
2
33
2
60
2
32 cos
b. At =60o and Point 1 in Figure 2,
02
3
2
3321 andmm ,
Therefore,
mm
o
mr 2
3
2
33
2
60
2
32 cos
It is clear that the resultant flux is again the same and rotated in
clockwise direction through an angle of 60o.
c. At =120o and Point 2 in Figure 2,
mm and 2
30
2
3321 ,
Therefore,
mm
o
mr 2
3
2
33
2
60
2
32 cos
It is found that the resultant flux is again the same and rotated in
clockwise direction through an angle of 60o.
d. At =180o and Point 3 in Figure 2,
mm and 2
3
2
30 321 ,
The resultant flux, mr 2
3 and has rotated clockwise through an additional angle 60
o.
5
K.Rajkumar, Associate Professor/EEE, Dhanalakshmi Srinivasan Institute of Technology, Samayapuram.
The important characteristics of the rotating magnetic field,
i. The resultant flux is of constant value
mr
2
3 due to any
phase.
ii. The resultant flux rotates around the stator at synchronous
speed given by p
fNs
120
b. Discuss the different power stages of an induction motor with losses. (6)
The electric power input given to the stator of an induction motor is converted into
mechanical power output at the shaft of the motor. The various losses during the energy
conversion are:
1) Fixed Losses:
a. Stator Iron Loss
b. Friction and Windage Loss
The rotor iron loss is negligible because the frequency of the rotor current is very small
under running condition.
2) Variable Losses
a. Stator Copper Loss
b. Rotor Copper Loss
The above figure shows that the electric power fed to the stator suffers losses and finally
converted into mechanical power.
The important points to be noted from the diagram are,
i. The stator input Pin= Stator output (Rotor Input) P2+ Stator I2R Loss+Stator Iron Loss
ii. Rotor input P2= Stator Output
It is because stator output is entirely transferred to the rotor through the airgap by
electromagnetic induction.
iii. Mechanical Power Developed, Pm=Pg-Rotor Copper Loss.
This mechanical power available is the gross rotor output and will produce a gross torque
Tg.
iv. Mechanical shaft output, Psh= Pm- Friction and Windage Loss.
Mechanical power available at the shaft produces a shaft torque Tsh.
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K.Rajkumar, Associate Professor/EEE, Dhanalakshmi Srinivasan Institute of Technology, Samayapuram.
2. a. Derive an expression for the torque of an induction motor and obtain the condition for
maximum torque. (8)
The torque developed is proportional to the armature current, flux per pole and the
powerfactor.
22 cosIT or 22 cosIkT
where,
I2=Rotor current at standstill
ϕ2= Angle between rotor e.m.f and rotor current
k=Constant
At standstill, the rotor e.m.f E2 is proportional to the flux ϕ.
Therefore, 222 cosIkET
Here, 2
222
2
2
2
22
XsR
Es
Z
sEI
and
22
222
2
2
2
XsR
R
Z
R
cos
Then the torque, 2
222
2
2
22
222
22
XsR
R
XsR
EskET
22
222
222
XsR
REksT
--- (1)
At the maximum torque, 02
dR
dT.
0
22
222
222
22
XsR
REks
dR
d
dR
dT
02
222
222
2222
22
22
222
XsR
RREksEksXsR
02
222
222
222
222
222
XsR
EksRXsR
02 22
22
222 RXsR
22
22
2 RXs or 22 RsX or 2
2
X
Rs max ---- (2)
Therefore the maximum torque is obtained by substituting the slip at maximum torque, smax.
222
22
22
22
2
2
222
222
2
2
22
222
222
2 XR
REk
XX
RR
REX
Rk
XsR
REksT
max
maxmax
2
22
2X
EkT max ---- (3)
From the equation (3), it is concluded that,
The maximum torque is independent of rotor resistance
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K.Rajkumar, Associate Professor/EEE, Dhanalakshmi Srinivasan Institute of Technology, Samayapuram.
The maximum torque is obtained when the rotor reactance equals its resistance.
Maximum torque varies inversely as standstill reactance. Hence, it should be kept as
small as possible.
Maximum torque varies directly as square of the applied voltage.
b. Draw and explain the slip-torque characteristics of a typical 3-phase induction motor.
Mark the starting and maximum torque regions on the diagram so drawn. (8)
The motor torque under running conditions is given by,
2
222
2
222
XsR
REksT
A curve drawn between the torque and slip, for a particular value of rotor
resistance R2 is called torque-slip characteristic. Figure shown below gives a set of torque-slip
characteristics for a slip-range from 0 to 1 for various value of rotor resistance.
Figure 3. Torque - Slip Characteristics
From the Figure 3, the following points to be noted,
i. At slip, s = 0, torque, T = 0 so that torque-slip curve starts from the origin.
ii. At normal speed, slip is small so that sX2 is negligible as compared to R2. Then the torque
will be proportional to, s/R2.
i.e, 2R
sT and finally sT when R2 is constant.
Hence torque slip curve is a straight line from zero slip to a slip that corresponds to full-
load.
iii. As slip increases beyond full-load slip, the torque increases and becomes
maximum at smax=R2/X2. This maximum torque in an induction motor is called pull-out
torque or break-down torque. Its value is at least twice the full-load value when the
motor is operated at rated voltage and frequency.
iv. As the slip increases further with increased load, then R2 becomes negligible compared to
sX2. Therefore, for larger value of slip, 22
2 Xs
sT or
sT
1 . Hence, the torque-slip
characteristic is a rectangular hyperbola. It is important to understand that the load
8
K.Rajkumar, Associate Professor/EEE, Dhanalakshmi Srinivasan Institute of Technology, Samayapuram.
increased further decrease the torque beyond the maximum torque and eventually stops at
one point.
v. It is also seen that the maximum torque does not depend on R2, but it decides the value of
slip at which the maximum torque occurs.
3. a. Describe the construction and principle of operation of a 3-Phase induction motor with
neat sketch. (10)
A 3-phase induction motor has two main parts (i) stator and (ii) rotor. The rotor
is separated from the stator by a small air-gap which ranges from 0.4 mm to 4
mm, depending on the power of the motor.
Stator:
It consists of a steel frame which encloses a hollow, cylindrical core
made up of thin laminations of silicon steel to reduce hysteresis and
eddy current losses. A number of evenly spaced slots are provided
on the inner periphery of the laminations .The insulated connected
to form a balanced 3-phase star or delta connected circuit. The 3-
phase stator winding is wound for a definite number of poles as per
requirement of speed. Greater the number of poles, lesser is the
speed of the motor and vice-versa. When 3-phase supply is given to
the stator winding, a rotating magnetic field of constant magnitude
is produced. This rotating field induces currents in the rotor by
electromagnetic induction.
Rotor:
The rotor, mounted on a shaft, is a hollow laminated core
having slots on its outer periphery. The winding placed in
these slots (called rotor winding) may be one of the
following two types: (i) Squirrel cage type (ii) Wound
type
i. Squirrel cage rotor:
It consists of a laminated cylindrical core having
parallel slots on its outer periphery. One copper or
aluminum bar is placed in each slot. All these bars are
joined at each end by metal rings called end rings as
shown in Figure 5. This forms a permanently short-
circuited winding which is permanent. The entire
construction (bars and end rings) resembles a
squirrel cage and hence the name. The rotor is not
connected electrically to the supply but has current
induced in it by transformer action from the stator.
Those induction motors which employ squirrel cage
rotor are called squirrel cage induction motors. Most
of 3-phase induction motors use squirrel cage rotor
as it has a remarkably simple and robust construction
enabling it to operate in the most adverse circumstances. However, it suffers from the
disadvantage of a low starting torque. It is because the rotor bars are permanently short-circuited
Figure 4 Stator
Figure 5 Squirrel Cage Rotor
Figure 6 Slip Ring Rotor
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K.Rajkumar, Associate Professor/EEE, Dhanalakshmi Srinivasan Institute of Technology, Samayapuram.
and it is not possible to add any external resistance to the rotor circuit to have a large starting
torque.
ii. Wound rotor:
It consists of a laminated cylindrical core and carries a 3- phase winding, similar to the one
on the stator as shown in Figure 6. The rotor winding is uniformly distributed in the slots and is
usually star-connected. The open ends of the rotor winding are brought out and joined to three
insulated slip rings mounted on the rotor shaft with one brush resting on each slip ring. The three
brushes are connected to a 3-phase star-connected rheostat. At starting, external resistances are
included in the rotor circuit to give a large starting torque. These resistances are gradually
reduced to zero as the motor runs up to speed.
b. Deduce and discuss the equivalent circuit of 3-phase induction motor.(6)
The equivalent circuit of 3-phase induction motor can be developed on per phase basis by
following the same procedure as used for the development of equivalent circuit of transformer.
Let,
V1=Applied voltage per phase to the stator winding
I1 = Stator current per phase
R1 = Stator winding resistance per phase
X1 = Leakage reactance of the stator winding per phase
E1=Induced EMF per phase in stator winding
=4.44fϕT1Kw1=V1-I1(R1+jX1)
I0=No-load current per phase = Iw+Im
Im =Magnetizing current per phase
Iw= Loss component of no-load current per phase
E2= Induced EMF per phase in rotor winding at standstill
=4.44fϕT2Kw2=I2(R2+jX2)
I2 = Rotor current per phase
R2 = Rotor winding resistance per phase
X2 = Leakage reactance of the rotor winding per phase at standstill
When the motor is running at a slip, ‘s’ then the emf induced in the rotor is sE2 at slip frequency.
Hence the rotor has been represented by a resistance R2 and a variable leakage reactance sX2 (due
to slip frequency) with variable rotor voltage sE2.
Figure 7 Per phase Equivalent Circuit of 3-phase induction motor
Referring to Figure 7, the rotor current per phase is given by,
22
2
22
22
jXs
R
E
jsXR
sEI
--- (1)
10
K.Rajkumar, Associate Professor/EEE, Dhanalakshmi Srinivasan Institute of Technology, Samayapuram.
Based on the equation (1), the rotor circuit modified to represent the rotor resistance with respect
to the change in slip as R2/s. This modification is done so that the emf induced in the rotor is
independent of slip and it becomes easier to refer the rotor quantities to stator.
Figure 8 Modified Equivalent Circuit per phase
Self induced e.m.f E1 is induced in the stator winding and mutually induced e.m.f
ratio tiontransforma the is K wheresKEsEE ,'122 is induced in the rotor winding. The
secondary values may be transferred to the primary and vice versa. As in the case of transformer,
it should be remembered that when shifting impedance or resistance from secondary to primary,
it should be divided by K2 whereas current should be multiplied by K.
Rewriting the equation (1),
22
2
22
2 1
jXs
R
E
KK
I
2
222
22
KjXKs
R
E
KK
I
22
22
22
KjXKs
R
EK
K
I
''
'
22
12
jXs
R
EI
Figure 9 Per phase equivalent circuit of 3-phase induction motor referred to stator
The equivalent circuit can be further modified so that the hypothetical resistance,
s
R '2 is split
into two components:
i. '2R , the per phase rotor resistance referred to stator itself, and
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K.Rajkumar, Associate Professor/EEE, Dhanalakshmi Srinivasan Institute of Technology, Samayapuram.
ii.
s
sR
12' , the electrical equivalent of the mechanical load on the motor and is
generally called the load resistance RL or the dynamic resistance. The load resistance
RL depends upon the speed of the motor.
Figure 10 Modified Equivalent Circuit referred to stator representing rotor and the load resistances
The following approximate equivalent circuit is used as it simplifies the analysis of the induction
motor.
Figure 11 Approximate Equivalent Circuit of 3-phase induction motor
4. a. Write a brief note on double cage rotor induction motors. (8) (or) Explain the
advantages of double cage induction motor over single cage induction motor a double
cage induction motor. (8)
The main disadvantage of the squirrel cage induction motor is lower starting torque. But
the slip ring induction motor has the ability to increase the starting torque by inserting additional
resistance in to the rotor circuit. However such a procedure cannot be adopted for a squirrel cage
motor because its cage is permanently short-circuited. In order to provide high starting torque at
low starting current, double-cage construction is used.
Construction
Figure 12 Double Cage Induction Motor
12
K.Rajkumar, Associate Professor/EEE, Dhanalakshmi Srinivasan Institute of Technology, Samayapuram.
The outer winding consists of bars of smaller cross-section short-circuited by end rings.
Therefore, the resistance of this winding is high. Since the outer winding has relatively open
slots and a poorer flux path around its bars, it has a low inductance. Thus the resistance of the
outer squirrel-cage winding is high and its inductance is low.
The inner winding consists of bars of greater cross-section short-circuited by end rings.
Therefore, the resistance of this winding is low. Since the bars of the inner winding are
thoroughly buried in iron, it has a high inductance. Thus the resistance of the inner squirrel cage
winding is low and its inductance is high.
Operating Principle:
When a rotating magnetic field sweeps across the two windings, equal e.m.f.s are induced in
each.
At starting, the rotor frequency is the same as that of the line (i.e., 50 Hz), making the
reactance of the lower winding much higher than that of the upper winding. Because of the
high reactance of the lower winding, nearly all the rotor current flows in the high-resistance
outer cage winding. This provides the good starting characteristics of a high-resistance cage
winding. Thus the outer winding gives high starting torque at low starting current.
As the motor accelerates, the rotor frequency decreases, thereby lowering the reactance of the
inner winding, allowing it to carry a larger proportion of the total rotor current At the normal
operating speed of the motor, the rotor frequency is so low (2 to 3 Hz) that nearly all the rotor
current flows in the low-resistance inner cage winding. This results in good operating
efficiency and speed regulation.
The starting torque of this motor ranges from 200 to 250 percent of full-load torque with a
starting current of 4 to 6 times the full-load value. It is classed as a high-torque, low starting
current motor.
b. Write a brief note on induction generator. (8)
When an induction machine runs at a speed greater than its synchronous speed with the
help of prime mover, then it is called an induction generator. But it cannot function as an isolated
place without any supply, because it generates only when excitation comes from the stator side.
These generators are used in low capacity power stations, most frequently in micro hydro power
stations and wind power stations.
Figure 13 Self Excited Induction Generator
But, the external reactive source must remain permanently connected to the stator windings
responsible for the output voltage control. In interconnected applications, the synchronous grid
supplies such reactive power. In stand-alone applications, the reactive power must be supplied by
the load itself, or by a bank of capacitors connected across its terminals, or by an electronic
13
K.Rajkumar, Associate Professor/EEE, Dhanalakshmi Srinivasan Institute of Technology, Samayapuram.
inverter. When capacitors are connected to induction generator, the system is usually called a
SEIG (a self-excited induction generator).
Advantages:
It does not hunt or drop out of synchronism.
Simple and rugged construction.
Cheaper in cost
Easy maintenance
Disadvantages:
Cannot be operated independently
Can deliver only leading current
May affect the system stability
5. a. Sketch and explain the torque slip characteristics of the 3-pahse cage and slip ring
induction motors. Show the stable region in the graph. (10)
Squirrel Cage Induction Motor:
When the load is applied to the shaft of the squirrel cage induction motor, the following
reactions will be observed:
Speed decreases
Slip increases
Rotor induced e.m.f increases consequently the rotor current increases
Torque developed by the rotor increases until it is sufficient to carry the applied load
The motor continues to run at that slip which will develop the required torque for that
particular load.
The motor may be loaded continuously till pull-out torque developed, at which point the
motor will stop if more load is placed on it. Pull-out occurs at a slip well beyond the normal
range operation of the squirrel cage induction motor. Even under normal operation, however, a
suddenly applied load may require the temporary development of this maximum torque, after
which the motor will speed up to its full load value. Thus, in a manner similar to the shunt motor,
the induction motor inherently adjusts itself to the applied external torque.
Slip ring Induction motor:
Figure 14 Slip-ring Induction Motor with Variable External Resistance at Rotor Circuit
Figure 14 shows the slip-ring induction motor with a variable resistance at its rotor. If the
resistance value is zero, then the motor will act same as squirrel cage induction motor. Under
normal running condition, if the rotor resistance is doubled, the following reactions will be
observed:
The rotor current will become half of its previous condition since the rotor speed, and
hence the rotor voltage, cannot change instantaneously.
The developed torque decreases due to reduced rotor current.
14
K.Rajkumar, Associate Professor/EEE, Dhanalakshmi Srinivasan Institute of Technology, Samayapuram.
The torque will increase even when there is no change in load, the motor slows down.
Slip increases
The rotor induced voltage increases
The rotor current now increases until it is sufficient to develop enough torque to again
carry the load at a constant speed.
The torque developed at standstill is higher with increased rotor resistance. This is
explained by the fact that at standstill the rotor frequency is equal to the line frequency
and rotor resistance is maximum. The addition of rotor resistance obviously improves the
powerfactor in the circuit while reducing the rotor current. Since standstill rotor
resistance is much greater than its resistance, the powerfactor increases more rapidly than
the decrease of rotor current, and hence the developed torque is greater with the added
resistance.
Figure 15 Torque Slip Characteristics of Induction Motors
b. Describe the principle of operation of synchronous induction motor. (6)
If the rotor of a slip-ring
induction motor is fed from a DC
power, this motor can be run at a
synchronous speed. Such motors are
called synchronous induction
motors.
Consider a normal slip-ring
induction motor with a conventional
three phase rotor winding is
energized by a fixed DC excitation.
Hence the pole axes of due to DC
excitation are also fixed and do not
change like the poles due to three
phase AC excitation. These fixed
rotor poles get magnetically locked
Figure 16 Connection of Synchronous Induction Motor
Stable Region Unstable Region
Slip Ring Im
Cage Im
15
K.Rajkumar, Associate Professor/EEE, Dhanalakshmi Srinivasan Institute of Technology, Samayapuram.
with the rotating magnetic field of stator supplied from three phase AC supply. Therefore the
motor runs at a constant speed equal to synchronous speed.
Figure 15 shows the connection diagram of synchronous induction motor. The motor is
started as normal slip ring induction motor with additional resistance inserted at the rotor through
the slip ring when the switch is at the Start position. When the resistance is completely removed
from the rotor circuit, the switch is changed to Run position. Hence the exciter is connected to
the rotor winding, and the machine with DC excited rotor pulls into synchronism just like the
synchronous motor.
This motor can be stated with load as slip ring induction motor and can be operated at
constant speed and improved power factor as synchronous motor.
Advantages:
1. It starts and synchronizes itself against heavy loads
2. The exciter used can be small due to relatively smaller airgaps.
3. No separate damper winding is required.
Important Formulae
Synchronous Speed,
r.p.m p
fN s
120
Slip:
100
s
rs
N
NNSlip%
Nr-Rotor speed in r.p.m
Frequency of Rotor current, sff '
s-slip and f- Supply frequency
Full load Torque:
22
22
222
sXR
RksET f
Starting Torque:
22
22
222
XR
RkETst
Maximum Torque:
22
22
2X
kETm
16
K.Rajkumar, Associate Professor/EEE, Dhanalakshmi Srinivasan Institute of Technology, Samayapuram.
Condition for maximum torque:
2
222
X
RsorXsR mm )(
Ratio of Full load torque to maximum torque:
22
2
m
m
m
f
ss
ss
T
T
Ratio of starting torque to maximum torque:
12
2
m
m
m
st
s
s
T
T
Rotor Copper loss from Rotor Gross output and vice-versa:
s
s
output gross Rotor
Loss Copper Rotor
1
Rotor input from Rotor output and vice-versa:
sInput Rotor
Output Rotor1
Rotor Copper loss from Rotor Input and vice-versa: Input RotorsLoss Copper Rotor
Equivalent Circuit Equations:
'2101 RRR
'2101 XXX
1
12
sRRL''
0101
2jXRR
VI
L
ph
'
'
m
ph
X
VI 0
'201 III
01
1Z
VI
ph
s
RIP
'' 222 3
''223 RILoss Copper Rotor
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University Question Problems
1. An induction motor has an efficiency of 0.9 when the shaft load is 45kW. At this load, stator
ohmic loss and rotor ohmic loss each is equal to the iron loss. The mechanical loss is one-
third of the no-load losses. Neglect the ohmic losses at no-load. Calculate the slip. (8)
17
K.Rajkumar, Associate Professor/EEE, Dhanalakshmi Srinivasan Institute of Technology, Samayapuram.
2. A 50 HP, 6–Pole, 50 Hz, slip ring IM runs at 960 rpm on full load with a rotor current of 40
A. Allow 300 W for copper loss in S.C. and 1200 W for mechanical losses, find R 2 per
phase of the 3- phase rotor. (6)
3. An 18.65 KW, 4 pole, 50 Hz, 3 phase induction motor has friction and windage losses of
2.5% of the output. Full load slip is 4%. Find for full load (1) rotor copper loss, (2) rotor
input, (3) shaft torque and (4) the gross electromagnetic torque. (8)
4. A 6 pole, 50 Hz, 3-Phase, induction motor running on full load develops a useful torque of
160 N-m. When the rotor emf makes 120 complete cycle per minute. Calculate the shaft
power input. If the mechanical torque lost in friction and that for core loss is 10 N m,
compute
(1) The copper loss in the rotor windings.
(2) The input the motor.
(3) The efficiency.
The total stator loss in given to be 800 W. (8)
5. A 746kW, 3-phase 50Hz, 16 Pole induction motor has a rotor impedance of (0.02+j0.15) Ω
at standstill. Full load torque is obtained at 360 r.p.m. Calculate,
(i) the ratio of maximum to full load torque
(ii) The speed at maximum torque and
(iii) The rotor resistance to be added to get maximum starting torque. (12)
6. A 15kW, 400V, 50Hz, 3-Phase star connected induction motor gave the following test
results:
No-load test: 400V, 9A, 1310W
Blocked Rotor Test: 200V, 50A, 7100W
Stator and rotor ohmic losses at standstill are assumed equal.
Draw the induction motor circle diagram and calculate,
(i) Line Current
(ii) Powerfactor
(iii) Slip
(iv) Torque and efficiency at full load. (16)
7. A 4 Pole 50Hz, 7.46kW motor has, at rated voltage and frequency, at starting Torque of
160% and maximum torque of 200% of full load torque. Determine,
(i) Full load torque
(ii) Speed at maximum torque (8)
8. A 3-phase, star connected 400V, 50Hz, 4 Pole induction motor has the following per phase
parameters in ohms, referred to the stator.
R1=0.15, X1=0.45, R2'=0.12, X2'=0.45, Xm=28.5
Compute the stator current and power factor when the motor is operated at rated voltage and
frequency with Slip,s=0.04. (8)
9. A 3-phase IM has starting torque 100% and a maximum torque of 200% of full load torque.
Find slip at maximum torque. (6)
10. A 100kW (output), 3300V, 3-phase, star connected induction motor has a synchronous speed
of 1500 r.p.m. The full load slip is 1.8% and full load powerfactor 0.85. Stator copper loss =
2440W. Iron loss=3500W. Rotational losses=1200W. Calculate (1) the rotor copper loss, (2)
the line current and (3) the full load efficiency. (8)
18
K.Rajkumar, Associate Professor/EEE, Dhanalakshmi Srinivasan Institute of Technology, Samayapuram.
11. Draw the circle diagram of a 15 HP, 230V, 50Hz, 3 - phase slip ring induction motor with a
star connected stator and rotor. The winding ratio is unity. The stator resistance is 0.42 Ohm
per phase and the rotor resistance is 0.3 Ohm per phase. The following are the test readings,
No-load test: 230V, 9A, p.f=0.2143
Blocked Rotor Test: 115V, 45A, p.f=0.454
Find,
a. Starting Torque
b. Maximum Torque
c. Maximum Powerfactor
d. Slip for Maximum Torque
e. Maximum Power Output. (16)
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