induction motor
TRANSCRIPT
119
Induction Motor
UNIT 5 INDUCTION MOTOR
Structure
5.1 Introduction
Objectives
5.2 Construction of Induction Motor
5.2.1 Stator Parts
5.2.2 Rotor Parts
5.2.3 Other Parts
5.3 Working of 3-Phase Induction Motor
5.3.1 Slip and Effect of Slip on Rotor Parameters
5.3.2 Torque and Torque-slip Characteristic
5.3.3 Power Flow
5.3.4 Equivalent Circuit of Inductor Motor
5.4 Starting and Speed Control
5.4.1 Starters for Poly Phase Induction Motors
5.4.2 Speed Control of 3-phase Induction Motor
5.5 Measurement of Slip
5.6 Summary
5.7 Answers to SAQs
5.1 INTRODUCTION
The most commonly encountered electric motors in industry are induction motors. In this
unit, we are introducing the construction features of induction motors to explain slip ring
induction motors and squirrel cage induction motors. Next, you will study various
relations between rotor emf, rotor current and torque developed in rotor. The choice of
selection of induction motor based on torque slip characteristics has been discussed. After
that, we will discuss equivalent circuit, no-load and blocked rotor tests of induction motor.
At the end of the unit, we will study speed control of induction motor which is very useful
in industrial drives.
Objectives
After studying this unit, you should be able to
• give elementary description of various parts and types of induction motor,
• explain the principle of working of induction motor and various stator and
rotor current relations,
• describe torque-speed relations and condition for maximum torque,
• describe power stages and efficiency of motor,
• explain starting and speed control of induction motor, and
• know effect of crawling and cogging of induction motor.
120
Electrical Technology 5.2 CONSTRUCTION OF INDUCTION MOTOR
The three-phase induction motor differs constructionally from the DC machine on certain
factors: its short air-gap, absence of a commutator, simple windings (particularly the cage)
and laminated stator. It is not possible, as in the DC machine, to make the frame as a part
of the magnetic circuit. The stator core must be carried in a shell or housing which
provides a means for protecting the stator and carrying the end-covers, bearings and
terminal box.
The following parts normally constitute an induction motor (the typical constructional
variation of slip-ring type is dealt under rotor construction).
5.2.1 Stator Parts
Let us discuss, in detail, each of them separately.
Frame
The frame of an induction motor may be cast or fabricated, depending upon the size
of the motor. For miniature and small motors, the frame has a single piece of short,
hollow cylinder cast-iron casting. It is cheaper to use cast iron where losses and
efficiency is of a lesser consideration than economy and where new designs and
modifications are not to be done on the machine.
However, for medium-sized and (in particular) large induction motors, fabricated
frame structure is exclusively used. Four or more arc-shaped rolled steel pieces are
joined together to form a cylindrical shape. The outer surface may be provided with
cooling fins so as to increase the heat dissipating area without increasing the overall
diameter. The chief advantage of fabricated construction is in its application to new
designs and modifications, which can be made without reference to existing
patterns. The principal constructional difficulty is the avoidance of distortion when
the parts are welded together.
The frame gives complete support and protection to the other parts (ruggedness) and
an eye-bolt on its top is useful for transit purposes.
Stator Core
The stator core provides the space for housing for the three-phase stator windings
and also forms the path for the rotating magnetic field. They are built up of thin
sheets of thickness (called stampings or laminations) of
0.35 mm to 0.65 mm with of a special core of steel, insulated one from the other by
means of paper. The gap facing inner circumference of the plates have suitable slots
punched out, either open, semi-closed or completely closed. Normally, the stator
core has semi-closed slots where the number of slots, S, is an integral multiple of 3
times the number of poles, p for which the induction motor is designed and
constructed [S = n. (3 p)], where n = 1, 2, 3 etc.
Every part of the stator core is subject to alternate changes in polarity of the
magnetic field (due to its rotating nature) so hysteresis losses and eddy-current
losses take place. To reduce the former, about 3-5% of silicon is added to high
grade steel and to reduce the latter, a large number of thin laminations are stacked
together.
5.2.2 Rotor Parts
Rotor Core
The construction of the rotor core should be separately discussed for (a) squirrel
cage motors, and (b) slip-ring motors. Certain common features are as follows :
(a) Both types have rotors constructed to thin sheets of special core steel,
but here the thickness is larger than that of stator stampings because no
appreciable iron loss is incurred in the rotor.
121
Induction Motor (b) In small motors, the rotor core is mounted directly on the shaft, to
which it is keyed, and clamped between end-plates on the shaft and a
shrink-ring.
(c) For large motors, the rotor is built up on a fabricated spider, the cross-
section of which is shown below :
Terminal Box
Eye Bolt
Rib
Stator coreStator slotStator coilAir-gap
Longitudinal vent-slot
Shaft bore with key-way
Semi-closed slot for rotor
Frame
Stand
Figure 5.1(a) : Induction Motor Construction
(d) The rotor core surface have to be ground to obtain an accurate
air-gap.
(e) For dynamic balancing of rotor, after the winding of cage has been
done, there should be a provision in the rotor core, all along the
periphery, for small amount of weights to be attached, which finally
forms an integral part of the rotor.
Rotor Core for Squirrel Cage Induction Motor
Closed slots of either circular or rectangular shapes are provided in the
stamping. After the rotor core is stacked by the required number of
laminations, the rotor shaft is inserted through the space available at the
centre and firmly fitted by the key-way. For machines of small and medium
capacity, this assembly is then placed in a moulding machine, which forces
molten aluminium under pressure through the slots to form rotor bars, end
rings and cooling fan arrangement as an extension of end rings. This type
of die-cast rotor is simple and rugged in construction as shown in
Figure 5.1(b).
Figure 5.1(b) : Fabricated Rotor Spider for Induction Motor
122
Electrical Technology For machines of larger capacity, copper or brass bars are driven through the
slots manually and the ends on both sides are then welded or silver-soldered
together to form end rings. The rotor bars are slightly inclined to the shaft
axis due to the ‘skew’ provided while stacking the rotor stampings. Skewing
is done because:
(a) It helps the motor to run quietly by reducing magnetic hum.
(b) It reduces the locking tendency of the rotor.
Rotor Core for Slip Ring Induction Motor
Semi-closed slots for three-phase winding are provided in the stamping. In
cage-rotors, closed slots were used. This is the primary difference in the
stamping. Secondly, the molten aluminium or copper bars are not used.
Windings will be dealt separately.
Stator Winding
Three-phase stator winding is done on the stator core. One starting end and one
finishing end for each of the phase windings is brought out for inter-connection in Y
or ∆ fashion.
Single layer mesh windings are used for machines of smaller capacity and medium-
sized machines have double-layer lap windings. Large capacity motors employ
single layer concentric winding.
All the winding coils are placed in the slots, the slots are closed with wooden wedges
or paper insulator, which keeps the coil sides intact. The whole stator winding may
be impregnated with a thermosetting varnish and baked.
Rotor Cage or Rotor Winding
There is no rotor winding in squirrel cage motors where the short circuited
aluminium, copper and brass bars form the rotor bars. In slip ring motors, there is a
rotor winding on the rotor cage and hence are called the wound rotor motors.
The rotor is wound for the same number of poles as that of stator. The winding is
normally connected in star and the resultant three terminals are connected to three
slip-rings provided on one end of the shaft as shown below :
R
Wound Rotor
Slip Rings
Y B
Figure 5.2(a) : Slip Ring Induction Motor
When running normally, the slip-rings are short-circuited, but for starting purposes,
they are connected to a three-phase star-connected starting resistance, as shown in
Figure 5.2(b) alongside.
Rheostat
Figure 5.2(b) : Starting Resistance
123
Induction Motor In slip-ring rotors of large size, a bar winding can be used because the choice of voltage is
usually free. For smaller machines, wire-wound rotors can be used, with coil arrangements
similar to those of the stator.
5.2.3 Other Parts
Shaft and Bearing
The shaft of an induction motor is short and stiff, in order to keep as small an air-
gap as is mechanically possible. This helps in removing any significant deflection in
the rotor. Even a small deflection in the shaft, and hence rotor, would create large
irregularities in the air-gap length which would lead to production of unbalanced
magnetic pull. For small and medium capacity motors, a roller bearing may be used
at the driving end and a ball bearing at the non-driving end.
Terminal Box and End Covers
The terminal boxes have been shown in the diagram Figure 5.3 with the connection
diagram for Y or ∆ connected stator. In squirrel cage motors, the rotor bars are
short-circuited in the construction itself, so rotor terminals are not available. But in
slip ring motors, it is essential to bring out the rotor terminals as it gives possibility
of inserting an additional resistance in the rotor circuit. In the terminal box shown,
only D, E, F constitute the rotor terminals, assuming that the other ends of it have
already been connected inside the machine. Sometimes, it is desirable to bring out
ends of each of the phases of the rotor. Then, together we have (6+6), i.e. 12
terminals on the terminal box.
A1 B1 C1
C2 A2 B2
D E F
Stator terminals
Rotor terminals(in case of
slip-ring motors)
A1
A2
C2
B2
B1 C1
B2
A1 B1 C1
C2 A2 B2
Star connected stator
Delta connected stator
B1
A2 C1
C2
A1
A1
C2
B1
A2
C1
B2
Figure 5.3 : Parts of Induction Motor
Two end covers with suitable bearings provide support for the rotor assembly.
Brush Lifting and Short-Circuiting Gear
Brushes sliding on the slip rings are used for making connection of rotor windings
with the external circuit. Brush lifting and short-circuiting gear it used for operating
the brushes. This is fitted with one of the end-covers which is nearer to the slip-ring
side. Slip-rings are normally of brass or phosphor-bronze shrunk onto a cast-iron
sleeve with moulded mica insulation. The assembly is pressed onto the rotor shaft
and located either between the rotor core and the bearing, or onto the shaft
extension. This part is exclusively in slip-ring motors and not in the common type
squirrel cage motors.
5.3 WORKING OF 3-PHASE INDUCTION MOTOR
When three phase supply is connected across stator windings then a rotating flux is
produced in stator which rotates clockwise at synchronous speed giving 120
s
fN
P= .
124
Electrical Technology When the rotor of 3-phase motor is placed in air gap of stator, a emf is induced in rotor
conductors. For squirrel cage rotor, conductors are shorted together and the four terminal
wound rotor conductors are connected with additional resistance between slip rings. The
induced emf causes a current to flow in these shorted conductors which produces a flux.
Due to reaction of this flux with stator flux a torque is produced and rotor rotates in same
direction of rotating flux with a speed less than synchronous speed.
5.3.1 Slip and Effect of Slip on Rotor Parameters
The rotating field revolves with the speed of synchronism, and if the rotor conductors were
to revolve at the same speed there would not be any torque. Hence, there is a difference
between rotor and rotating field speeds. The rotor speed is less than the rotating field speed
and the difference in speed is known as the slip of motor. Generally, slip lies between zero
to five percent.
Slip (s) s r
s
N N
N
−=
where Ns is synchronous speed of rotating field in stator and Nr is speed of rotor
(motor speed)
Percent slip 100s r
s
N N
N
−= ×
or Rotor speed (1 ) sN s N= −
Three phase induction motor is a particular form of transformer which has secondary
winding rotating. In transformer, the frequencies of primary and secondary windings are
same but in induction motor the frequency of emf induced in rotor (secondary winding)
depends on slip.
Rotor emf frequency = s × frequency of applied voltage to stator = sf
The induced emf in rotor also depends on slip. When rotor is stationary the emf induced in
rotor is
2 12 1
1
Rotor emf ( )n E
E En n
= = ,
where n = turns ration = n1/n2.
where n2 and n1 are number of turns in rotor and stator windings and E1 is applied voltage
in stator. Now for moving rotor, rotor emf becomes sE2. Important thing to remember here
is that both E1 and E2 remain fixed by above relation even at different slip or load values.
The rotor resistance (R2) is constant with change in slip unless external resistance is added
or there is a change in temperature.
Rotor reactance at slip one (standstill) is X2
or X2 = 2π f L2
where f is frequency of rotor emf at standfill and L2 is inductance of rotor winding.
Now moving rotor frequency becomes sf, so
Rotor reactance = 2π sf L2 = s X2
So, rotor impedance at motion
2 2 22 2 2Z R s X= +
The rotor current, 2 22
2 2 22
2 2
sE sEI
Z R s X
= =+
125
Induction Motor and rotor power factor 2 2
22 2 2
22 2
cosR R
Z R s X
φ = =+
5.3.2 Torque and Torque-slip Characteristic
Torque
The torque developed by the rotor is proportional to the product of rotor current and
fundamental magnetic flux cutting the rotor. The total torque
T = K φ I2 cos φ2
where φ is the useful fundamental flux per pole and is proportional to rotor emf
(max.)
So T = K E2 I2 cos φ2
We know 22
2 2 22 2
sEI
R s X
=+
and cos φ2 =2
2 2 22 2
R
R s X+
so 2
2
2 2 22 2
zK s E RT
R s X=
+
For small value of slip (s X2)2 is negligible compared to
2
2R . So torque is
proportional to the slip when slip approaches unity or large values of slip, sX2 is
large compared to R2 so (R2)2 is negligible compared to (sX2)
2. Now torque is
approximately inversely proportional to the slip as shown by rectangular hyperbola.
For slip, s = 1, i.e. standsfill the torque 2
2 2
2 22 2
K R ET
R X=
+ is constant and is known as
starting torque. The approximate torque slip curve is shown in
Figure 5.4. The region where 0dT
ds> , is called stable region of operation where the
motor normally operates.
Slip Speed
Torque TorqueMax. Torque
s = 0 s = 1
Starting Torque
Figure 5.4 : Torque Slip Characteristics
Maximum Torque
By inspection of torque slip curve, we see that the maximum torque developed at
particular slip, which is derived by differentiating T with respect to slip and by
putting 0=ds
dT.
Here 22 2
2 2 22 2
K s E RT
R s X=
+
After putting 0=ds
dT, we get
2 2R s X= or 2mat
2
Rs
X=
126
Electrical Technology The maximum torque is obtained when slip is equal to ratio of rotor resistance to
rotor reactance :
22
max22
K ET
X=
and rotor power factor is .2
1
The rotor reactance should be kept as low as possible otherwise torque developed is
reduced. The maximum torque is independent of rotor resistance, but the value of
slip at which maximum torque occurs is directly proportional to rotor resistance. If
we increase the rotor resistance the value of slip, at which maximum torque occurs,
approaches to one as shown in Figure 5.5.
Slip
Speed
Torque
s = 0 s = 1
Large R2
Small R2
Medium R2
Max. TorqueR
2 is very large
Figure 5.5 : T-S Characteristic for Different Rotor Resistances
Starting Torque
We know that
Torque, 22 2
2 22 2( )
K s E RT
R s X=
+
At the time of starting, s = 1 and E2 is constant then
1 2
2 22 2
st
K RT
R X=
+
Maximum torque occurs when rotor resistance is equal to rotor reactance and the
relation between starting torque and rotor resistance becomes
1
22st
KT
R=
Tst
R2
Figure 5.6
127
Induction Motor Starting Torque in Slip Ring Motor
Three resistors are used to start a slip-ring induction motor as shown in Figure 5.8.
These resistors are variable and are varied by three armed handle which makes a
star point. At the time of starting, all resistances are in the circuit and give high
starting torque. When starting resistance is all cut out, except that in the leads, and
the switch contacts remain closed, then this is comparable with the resistance of the
rotor itself. It is necessary to short circuit the rotor with the slip rings.
Stator
3-supply
Rotor
Slip rings
Starting
Variable resistance
Figure 5.7 : Illustration for Starting Torque in Slip Ring Induction Motor
5.3.3 Power Flow
Normally, out of the power supplied to stator, a portion is wasted in stator copper loss and
another portion in stator iron loss. The remainder is supplied to rotor. Due to copper loss,
there is some voltage drop and just voltage transfer from stator to rotor. Let I1, E1 and cos
φ are phase values of current, voltage and power factor on stator side. Then power
supplied to rotor is 3 I1 E1 cosφ. If n is turn ratio of stator to rotor turns then emf induced
in rotor is n
E1 at standstill. When rotor rotates at slip s then induced voltage is n
Es 1 and
rotor current becomes n I1.
so rotor power losses = 3E2 I2 cos φ = 3s E1 I1 cos φ = s. Rotor input (i/p = input and
o/p = output)
⇒ Rotor copper loss
Rotor /s
i p=
then rotor output
= 3 E1 I1 cos φ − 3 E1 I1 cos φ . s
= 3 E1 I1 cos φ (1 − s) = Rotor input (1 – s)
so Rotor o/p
1Rotor i/p
s= −
and since Rotor copper loss
Rotor i/ps=
so s
s
−=
1o/pRotor
losscopper Rotor
Thus, Rotor i/p : rotor copper loss : rotor o/p = 1 : s : (1 − s).
We know Rotor o/p actual speed
1Rotor i/p synchronous speed
r
s
Ns
N= − = =
so Rotor o/p = rotor i/p × speed ssynchronou
speed actual
128
Electrical Technology But Rotor o/p =
2
60
rNT w T
π× = × 2π × Torque × Actual speed
so 60 Rotor o/p
Torque2 actual speed
×=
π ×
60 Rotor i/p
2 synchronous speed
×=
π ×
Its unit is Newton-meter.
The synchronous watt is another unit of torque which is the torque that will develop one
watt at synchronous speed.
Example 5.1
An induction motor is rated as 5 KW at 1440 r.p.m., 50 Hz, 400 V line to line. The
rotational losses due to friction and windage are 279 watts. If the maximum torque
is found at 900 rpm, determine the maximum torque developed.
Solution
Synchronous speed for 50 Hz supply nearest to 1440 r.p.m is 1500 r.p.m.
∴ Slip at full load
1500 1440 60
0.041500 1500
s−
= = =
Rotor output at full load = 5000 W
∴ Gross rotor output = 5000 + 279 = 5279 watts
96.0
04.0
04.01
04.0
s-1
s
output grossRotor
losscopper Rotor =
−==
∴ Rotor copper loss 527996.0
04.0×=
∴ Rotor input 96.0
52795279
96.0
04.05279 =×+=
Torque at full load Rotor input at full load
602 sN
= ×π
5279 60
N-m0.96 2 1500
×=
× π ×
= 0.5834 × 60 = 35.004 N-m
22 2
2 2 22 2
2max 2
2
Torque at full load
Maximum Torque
2
fl
K S E R
T R S X
T K E
X
+= =
2 2
2 2 22 2
R X S
R S X=
+
2
2
2 2 2
22
2
22
RS
X as
a SRS
X
= =+
+
129
Induction Motor
where
2
2
X
Ra = and S is the slip at full load
Slip at maximum torque max
1500 9000.4
1500Ts
−= = =
At maximum torque 2max
2
0.4T
Rs
X= = or a = 0.4
∴ Maximum torque aS
Sa
2
22 += full load torque
2 2(0.4) (0.04)
0.5834 602 0.4 0.04
+= × ×
× ×
= 2.94617 × 60 N-m = 17.666 N-m
5.3.4 Equivalent Circuit of Induction Motor
The rotor current is given by
ImpedanceRotor
emfRotor 2 =I
22
222
22
XsR
EsI
+=
If n is stator to rotor turns ratio then stator current
n
II 21 =
or 22
222
21
XsRn
EsI
+=
to produce I1 current in stator there is a requirement of voltage i.e. E1 = n E2 so that the
rotor impedance referred in stator winding
2 2 22 21 1
221 2
n R s XE EnE
II s E
n
+= = = ×
22
222
2
XsRs
n+=
( )22
2
2
22
Xns
Rn+
=
The rotor resistance s
Rn 22
can be divided as series combination of two resistance n2 R2
and .11
22
−
sRn The n
2 R2 part remains constant and represents the physical rotor
resistance referred to stator side, i.e. 2R′ . But
−1
12
2
sRn varies from zero to infinite as
s changes from unity to zero and represents the rotor output in the form of power in this
resistance. The equivalent circuit referred to stator winding is shown in Figure 5.8.
130
Electrical Technology
Ι1
Ι0
Ιm
X0
R0 =
Ιe
Ι0
Ι2n
X1
R1
n = X2
′2 N = R22 2
′R
R′2n2
2R – 11
s– 11
s
Figure 5.8
The three phase induction motor may be compared with a transformer because it involves
changing flux linkages with respect to the stator and rotor winding.
The equivalent circuit parameters can be determined by no-load test and block rotor test
which are similar to OC and SC test on transformers. This topic is not covered here.
Example 5.2
A 3-phase, 500 volt, 6 pole, 50 Hz induction motor develops 30 b.h.p. at
990 r.p.m. at a power factor of 0.86 lagging. Calculate (i) the rotor copper loss, and
(ii) the total power and the number of cycles per minute of the rotor e.m.f.
The stator losses are equal at 2000 watts. Neglect mechanical and iron losses of the
rotor.
Solution
Given Nr = 990 r.p.m., P = 6 and f = 50 Hz
Synchronous speed r.p.m. 10006
50120120=
×==
P
fNs
Slip 1000 990
0.011000
s r
s
N Ns
N
− −= = =
Rotor output = 30 h.p. = 30 × 735.5 = 22065 watts
Rotor copper loss
Rotor output 1
s
s=
−
(i) ∴ Rotor copper loss watts88.2222206501.01
01.0=×
−=
Stator output = Rotor input
= Rotor output + Rotor copper loss
= 22065 + 222.88
= 22287.88 watts
Stator input = Stator output + stator losses
= 22287.88 + 2000
= 24287.88 watts
(ii) ∴ Input power = 24.28788 kW
Rotor frequency = s × f = 0.01 × 50
= 0.5 Hz = 0.5 cycles/sec
= 0.5 × 60 cycles/minutes = 30 cycles/min
131
Induction Motor SAQ 1
A 3-phase, 400 volt, 4 pole, 50 Hz induction motor has a star connected stator and
rotor. The rotor resistance and standstill reactance are 0.2 ohm and 1.0 ohm,
respectively. The ratio of stator to rotor turns is 1.29. The full load slip is 4%.
Calculate the torque and power developed at full load. Find also the maximum
torque and speed at which it occurs.
5.4 STARTING AND SPEED CONTROL
5.4.1 Starters for Poly Phase Induction Motors
If motor is started with full voltage, the starting torque is good but very large currents, of
the order of 5-7 times the full-load current flow which causes objectionable voltage drop in
the power supply lines and hence undesirable dip in the supply line voltage. Consequently,
the operation of other equipment connected to the same supply line is affected
considerably.
If the motor is started with reduced voltage, there is no problem of high currents but it
produces an objectionable reduction in the starting torque, on account of the fact that
motor torque is proportional to the square of the applied voltage.
Methods of Starting Squirrel Cage I.M
There are basic four methods of starting the squirrel cage induction motor using
(a) Direct online starters
(b) Stator Resistor (or reactor) Starters
(c) Auto-transformer Starters
(d) Star-Delta Starters
Stop
Start
b0
b1 A
e3
UVRC
e2
O1
O2
O3
M1
M2
M3
e1
A1 C
2
C1
A2
B1
B2
S1
R
Y
B
∆-connected stator
M
Rotor
Figure 5.9 : Direct Online Starter
132
Electrical Technology Methods of Starting Slip-Ring (Wound Rotor) I.M.
Though all the above methods, except D.O.L. where the high currents may damage
the rotor windings, can also be employed for starting slip-ring motors, but it is
usually not done because the advantages of such motors can’t be fully realized. So,
the method of adding resistance to the rotor circuit is the most common method for
rotor wound I.M. starting.
D.O.L. Starters
The above Figure 5.9 shows a contactor type D.O.L. starter connected to a
motor. As soon as the push-button S1 is pressed, the contactor coil is
energized closing its contacts M1, M2 and M3 . Then, the motor windings get
full supply through back-up fuses e1, e′2, e″1and bimetallic relays O1, O2 and
O3 and the motor starts running. An auxiliary contact A in C1 retains the
contactor in closed position after the release of start switch S1. An overload
tripping device e1, working in conjunction with bimetallic relays, is placed in
series with the contactor coil, so that during sustained overload, this opens
and the motor stops automatically. For stopping the motor any time, a stop
button is provided in series with the contactor coil.
Primary Resistor Starter and Reactor Starter
This method consists of connecting the motor to the line voltage through a
series resistance in each phase. The resistors are short-circuited when the
motor accelerates to the desired speed. Sequence of operation of switches
shown in Figure 5.10 is :
(a) Initially all switches are open.
(b) Switches 1, 2 and 3 are closed simultaneously and motor starts
running with full resistances in series.
(c) Switches 4, 5 and 6 are closed when motor speed picks up and
current becomes constant. Finally switches 7, 7, and 9 are
closed to cut all resistances and motors attains final steady state
speed.
Advantages : (i) It provides closed transition starting, resulting in smooth
starting without any transition high current.
(ii) A higher p.f. than auto-transformer starters.
1R
Y
B
2
3
M
4 7
5 8
6 9
Figure 5.10 : Series Resistance Starting
Sometimes as an alternative to resistor starting, reactor starting is used. This
method is mainly used for large motors.
Auto-Transformer in the First Step Starters
In this type of starter, (A. T.) it attains the reduced voltage by means of an
auto transformer at the start. After a definite time interval (about 15 sec.),
and after the motor accelerates, it is transferred from the reduced voltage to
133
Induction Motor the full voltage in the second step. A. T. are generally provided with voltage
drops to give 40%, 60%, 75% and 100% line voltage. The starting current
and starting torque depends on the tapping selected. In the third step, the
change-over switch may be hand operated or automatic through time relay
which connects the motor finally to the line by changing over from position A
to B.
R
A
B
B
Y3phBus Bar
Auto-Transformer
V1
Stator
Rotor of sq. cage I.M.
Figure 5.11 : Simple Diagram of Auto-transformer Starter
Advantages of A.T. Starters
(i) Greater efficiency.
(ii) Taps on the transformer allow adjustment of starting torque to meet the
particular requirement.
Disadvantages of A. T. Starters
(i) It opens the circuit before the motor is connected directly to the line, thus
producing transient current and stresses.
(ii) It reduces the p.f. of the circuit.
(iii) The torque remains constant for the second step, resulting in acceleration
which is not smooth.
These disadvantages of open transition in A. T. may be overcome by the use of
Korndorfer connection, which introduces another step in starting. On the second
step, part of the A. T. remains in series with the stator windings. The third step
involves the transfer of the full-voltage without open transition.
Star-Delta Starter
It is cheaper as compared to A. T. starter. This method of starting is used for
motors designed to operate normally in delta. The six terminals from the three
phases of the stator must be available :
a, A : Terminals of phase A
b, B : Terminals of phase B
134
Electrical Technology c, C : Terminals of phase C
Commercially, the terminals are marked A1, A2; B1, B2 and C1, C2
respectively. The motor is started with TPDT switch in position 1 and
subsequently switched to position 2.
Position 1 : Starting-windings connected in Y
Position 2 : Running-windings get connected in ∆
R
Y
B
1
a
b
c
B
C
A
2
CA
a
b
c
B
B
C
A
Start Run
Y
Triple Pole Double Throw (TPDT) Switch
Figure 5.12 : Star-Delta Starter
Let VL = Line Voltage
ISP., Y = Per phase motor current during start (in Y-connection)
∴ IL, y = ISP, Y =
13 Z
VL . . .
(5.1)
[Since rotor circuit behaves almost as short circuit on starting as
2
11 0R
s
′ − =
].
If stator winding is ∆-connected, then, with direct switching the per-phase
motor current at start would be given by
ISP., ∆ =
1Z
VL . . .
(5.2)
Starting line current
, .,1
33 L
L SP
VI I
Z∆ ∆= = . . .
(5.3)
It is shown from Eq. (5.1) to (5.2) that
135
Induction Motor
., 1
,
1
1
Starting line current with star-delta starter 13
Starting line current with direct switching in delta 33
L
SL Y
LL
V
I Z
VI
Z∆
= = =
and
2
2L
Starting torque with star-delta starter 13.
Starting torque with direct switching in delta 3V
LV
= =
Example 5.3
Determine approximately the ratio of starting torque to full load torque of a three
phase induction motor with
(a) star-delta starter, and
(b) an auto-transformer starter with 50% tapping used. The short circuit current
of the motor at normal voltage is 5 times the full load current and the full load
slip is 5%.
Solution
(a) Star-delta starter: In star-delta starter the voltage applied to each phase at
the time of starting is 31 times the normal phase voltage.
∴ Starting current per phase scst II3
1 ==
where Isc is the short circuit current per phase, i.e. starting current per phase
without Y-∆ starter.
Since 2
2 2 22
R IT I
s sα α ,
Starting Torque
2st
stst
IT
S×
where Sst = slip at starting = 1
∴ 2st stT Iα or
2
3
scst
IT ∝ . . .
(5.4)
Also, full load torque
2st
flfl
IT
S∝ . . .
(5.5)
where Ifl and Sfl are full load current and full load slip, respectively.
Dividing Eq. (5.1) by Eq. (5.2) we get
flfl
sc
fl
st SI
I
T
T2
3
1
=
It is given that 5=fl
sc
I
I and Sfl = 0.05
∴ ( ) 417.03
25.105.05
3
1 2 ==×=fl
st
T
T
136
Electrical Technology (b) Auto-transformer starter: Let the auto-transformer tapping ratio be K. Then
the phase voltage across the motor is K times the normal voltage.
∴ Starting current scst
KII == .
At starting slip being unity, the starting torque
2st stT I∝ or 2
( )st scT KI∝
or 2 2( )st scT K I= . . .
(5.6)
Dividing Eq. (5.3) by Eq. (5.2), we get
2
2
=
st scfl
fl fl
T IK S
T I
It is given that 5.0 and 05.0,5 === KSI
Ifl
fl
sc
∴ ( ) ( )2 2
0.5 5 0.05 0.3125= × =st
fl
T
T
Example 5.4
A 6 pole, 50 Hz, three phase induction motor running on full load develops a useful
torque of 163 Newton-metres and the rotor electromotive force makes
90 complete cycles per minute. If BHP if the mechanical torque lost in friction be 14
Newton-metres, find the copper loss in the rotor windings, the input to the motor
and the efficiency. Stator losses total 750 watts.
Solution
Frequency of rotor emf 90
rps 1.5 rps60
= =
But rotor emf frequency = slip × stator emf frequency
∴ Slip 03.050
5.1==
Synchronous speed 120 120 50
1000 rpm6
s
fN
P
×= = = =
But s r
s
N NS
N
−= or
10000.03
1000
rN−=
Actual speed of motor N = 970 rpm.
BHP
9702 163
2 60 22.195 BHP746 60 746
rN Tπ× ×
π= = =
×
Gross torque produced = Useful torque + torque lost in friction
or Gross torque = 163 + 14 = 177 Newton metres
∴ Mechanical power developed 2
60
r gN Tπ=
17760
9702 ××π=
137
Induction Motor = 17979.33 watts
S
S
−=
1output grossRotor
losscopper Rotor
or
Rotor copper loss 33.1797903.01
03.0×
−=
= 556.06 watts
Rotor input = Rotor output + Rotor copper loss
= 17979.33 + 556.06 = 18535.39 watts
Motor input = Rotor input + Stator losses
= 18535.39 + 750
= 19285.39 watts
Efficiency of the motor 10039.19285
16360
9702
Input
Output×
××π
==
= 85.85%
5.4.2 Speed Control of 3-Phase Induction Motor
The stable operation of 3-phase induction motor is restricted within a small range of slips
and because of this restriction its application is limited to substantially constant speed
drives. We know that
120
s
fN
P=
Thus, we have the following methods of speed control:
(a) Frequency Variation
(b) Variation of Number of Poles
(i) using multiple windings
(ii) using consequent pole technique
(iii) pole-amplitude modulation
(c) Supply Voltage Variation
(i) Variation in Motor Parameters
(ii) Variation of rotor resistance or stator reactance
(iii) Variation of rotor reactance or both stator and rotor reactances
(d) Control of Rotor Slip Power
Frequency Variation
This method can be used for both squirrel cage and wound rotor type motors. This
method provides smooth control of speed. In frequency variation method the
variable frequency supply voltage is used. This voltage can be obtained from
(a) variable frequency alternator,
(b) rotating frequency converter,
(c) static converters and cyclo-converters.
Variation of Number of Poles
This method is generally recommended for squirrel cage induction motors and for
steady torque production where the stator and rotor poles should be equal. The
138
Electrical Technology change in the number of stator poles must be accompanied with a change in the
number of rotor poles so the pole changing can be effected by the following methods
:
(a) Using multiple windings in stator for two different speeds, one for
higher speed and another for lower speed. Change over from one speed
to another is made by a switch.
(b) Pole changing with a single stator winding also can be achieved by
using consequent pole technique. In this technique the total stator
winding coils are connected in different manner according to constant
torque, constant power and variable torque and power applications.
Supply Voltage Variation
This method is used for both squirrel cage and wound rotor motors. In this method,
there are some disadvantages like torque is reduced with the reduction in supply
voltage. This method gives smooth stepless control, but the efficiency is
comparatively lower.
Variation in Motor Parameters
The speed control can be achieved by changing motor parameters. For wound motor
we can change rotor resistance, rotor reactance and stator reactance and for squirrel
cage motors only stator reactance can be changed. By rotor resistance variation
smooth speed control can be obtained. The drawback is that large copper losses
occurs at lower speeds.
Stator reactance method is not effective at lower slips with normal rotor resistance.
Better performance can be achieved by using high resistance rotors.
Control of Rotor Slip Power
The wound rotor induction motor can be operated at lower speeds by inserting
suitable resistance in the rotor circuit. In this method slip power is dissipated in the
rotor circuit. Control or recovery of slip powers as a means of speed control of
wound motor is based on the economic use of slip power and uses a slip power
converter for this purpose. By this method, a motor can also be run at super
synchronous speeds by feeding slip power into rotor.
SAQ 2
(a) The power input to a 500 volts, 50 Hz, 6 pole, three phase induction motor
running at 975 r.p.m is 40 kW. The stator losses are 1 kW. Calculate
(i) slip (ii) rotor copper loss (iii) B.H.P., and (iv) efficiency.
(b) A 4 pole, 200 V, 50 Hz, star-connected, 3-phase induction motor has primary
leakage impedance of 0.08 + j 2 Ω, and an equivalent secondary leakage
impedance at standstill of 1.2 + j 1.0 Ω per phase. The magnetizing current
may be assumed to be negligible. Find the maximum torque in synchronous
watts the motor can develop and the slip at which it occurs.
5.5 MEASUREMENT OF SLIP
There are three main methods of measuring the slip viz:
(a) by measuring the synchronous and actual speeds,
(b) by measuring the rotor frequency, and
(c) by the stroboscopic method.
139
Induction Motor The first method, depends upon the difference of two relatively large and nearly equal
quantities.
The measurement of the rotor frequency can be effected by inserting a moving-coil
ammeter, preferably when rotor frequency is very low often not exceeding one or two
cycles per second. The frequency can be determined by counting the oscillations of the
pointer. If a central zero instrument is used, the number is a complete to and fro swings
which must be counted, but in the ordinary type the pointer is pressing against the zero
stop. The percentage slip is given by
100frequency Stator
frequencyRotor ×
This method must be modified for cage − type rotors, since there are no slip rings. There is
usually a small portion of the magnetic flux going right through the centre of the rotor and
cutting the shaft, which has a small e.m.f. induced in it. This can be detected by pressing a
lead against each end of the shaft. The other ends being connected to a sensitive moving-
coil millivoltmeter. The measurement is then made in the same way as with the wound
rotor.
In the stroboscopic method a disk, with alternate black and white sectors painted on it, is
attached to the end of the motor shaft. This disk is illuminated by means of a neon lamp
operated from the same supply. If the disk is normally in a poor light, an incandescent
lamp is sufficient. Alternatively, the stroboscopic disk may be viewed through a slot cut in
another disk driven by a small synchronous motor from the same supply. Assuming the
number of black sectors to be equal to the number of poles on the motor, one sector moves
forward a pole pitch in half a cycle if there is no slip at all. Then next black sector
occupies the original position of the first. But the disk is illuminated once every half-cycle,
so that the disk appears to be stationary. Since a certain amount of slip occurs, however,
the second black sector does not quite reach the initial position of the first in half a cycle,
with the result that the disk appears to rotate slowly in a direction opposite to that of the
motor. A complete apparent revolution of the disk thus corresponds to a slip of as many
cycles as there are pairs of poles on the motor. If the apparent revolutions of the disk per
minute are counted, the frequency of slip is obtained from
Apparent rpm
Rotor frequency = × pairs of poles60
The percentage slip is then calculated as before. For testing motors with different number
of poles, a series of disks is required having different number of sectors painted on them.
Example 5.5
The power input to a 415 V, 3-phase, 6-pole, 50 Hz induction motor is 50 kW when
running at 970 r.p.m. The total stator losses are 2.0 kW and the mechanical losses
are 1.5 kW. Determine (i) the rotor copper loss (ii) the gross torque in Newton-
meters, and (iii) the rotor resistance per phase if the rotor phase current
is 110 A.
Solution
Rotor input = Stator output
= Stator input − Stator loss = 50 − 2 = 48 kW
Motor slip 970s r s
s s
N N N
N N
− −= =
The synchronous speed r.p.m. 10006
50120120=
×==
P
fNs
140
Electrical Technology Motor slip
1000 9700.03
1000s
−= =
SlipinputRotor
losscopper Rotor =
or Rotor copper loss = Slip × Rotor input
= 0.03 × 48 kW = 1.44 kW
Gross rotor output = Rotor input − Rotor copper loss
= 48 − 1.44 = 46.56 kW
Gross rotor torque Gross rotor output
2
60
rN=
π ×
346.56 10
458.366 N-m2 970
60
×= =
π ×
Rotor copper loss = 22 23I R
= 1.44 × 103 watts
2 2
1440
3 (110)R =
×
= 0.0397 Ω per phase.
SAQ 3
(a) A efficiency of a 6 pole, 3-phase, 45 kW, 50 Hz induction motor is
92 percent. The stator losses equal 102 kW. Determine (i) the slip,
(ii) the rotor copper loss, (iii) the input, and (iv) the friction and windage
losses. The motor runs at 975 r.p.m.
(b) A squirrel-cage induction motor has a ratio of rotor stand still reactance to
resistance of 4 to 1 and maximum torque is 2.25 times the normal full load
torque. Calculate (i) the full load slip, and (ii) the ratio of starting torque to
normal torque with direct-on line starting.
5.6 SUMMARY
In Section 5.2 after a brief consideration of the constructional features of induction
motors, you learnt how a revolving magnetic field is produced. Next the rotor current and
torque-speed relations were derived and we also found the condition for maximum torque.
You learnt how to calculate performance characteristics of induction motor, power stages
and efficiency. Finally, you were introduced to various starters and speed control methods.
5.7 ANSWERS TO SAQs
SAQ 1
141
Induction Motor Phase voltage Volts.
3
400=
29.1
1
nsstator tur
srotor turn==K
Standstill induced emf in the rotor per phase
2
400 1179 Volts
1.293E = × =
Rotor impedance at full load ( ) ( )222
22 SXRZ +==
or ( ) ( ) Ω==×+= 204.00416.0104.02.022
2Z
22
2
179 0.0435.1 amp.
0.204
SEI
Z
×= = =
Total copper losses 2223 RI=
23 (35.1) 0.2= ×
= 739.2 watts
We know, S
S
−=
1output grossRotor
losscopper Rotor
∴ Rotor gross output ( )739.2 1 0.04
0.04
−=
= 17740 watts
= 17.74 KW
Rotor speed = (1 − 0.04) 1500
= 1440 r.p.m.
∴ Gross torque developed rotor output
2
60
r
TN
=π
17740 60
117.6 N-m2 1440
×= =
π ×
For maximum torque, R2 = S′ X2
∴ 2
2
0.20.2
1
RS
X′ = = =
Rotor e.m.f. per phase
2 179 0.2 35.8 VoltsrE S E′ ′= = × =
2 2 2 22 2 2( ) (0.2) (0.2 1)Z R s X= + = + ×
= 0.283 Ω
22
35.8126.5 amp
0.283
rEI
Z
′′ = = =
142
Electrical Technology Copper losses (Total) = 3 (126.5)2 × 0.2 = 9601 W
Rotor gross output 1 S
S
′−= ×
′ Rotor copper loss
1 0.2
9601 38404 Watts0.2
−= × =
Speed N′r at maximum torque = (1 − S′) Ns
= 0.8 × 1500
= 1200 r.p.m.
Maximum torque Tmax Rotor output
2
60
rN=
′π
38402
2 1200
60
=π ×
= 305.61 N-m
SAQ 2
(a) Input to the stator = 40 kW
Stator losses = 1 kW
Stator output = Rotor input = 40 − 1 = 39 kW
Synchronous speed corresponding to 6 poles
r.p.m. 10006
50120120=
×==
P
fNs
(i) Slip %5.2025.01000
25
1000
9751000===
−=
−=
s
s
N
NNS
(ii) Rotor copper loss = slip × rotor input
= 0.025 × 39 kW = 0.975 kW
Rotor output = Rotor input − Rotor copper loss
= 39 − 0.975 = 38.025 kW
(iii) B.H.P. 972.50746.0
025.38== BHP
(iv) %06.9540
100025.38
Input
OutputEfficiency =
×==
(b) The equivalent per phase circuit of motor is given below :
E1
0.8 Ω j2 Ω j1 Ω
1.2s
Ω
143
Induction Motor volts
3
2001 =E
sjj
EI
/2.1128.0
11 +++
=
or 3)/2.18.0(
32001 js
I++
=
( )
22
22
1)3()/2.18.0(
3200
++=
sI
The input power to the rotor circuit is given by watts/2.121 sI × which is also a
measure of the torque developed in synchronous watts.
Hence, torque developed per phase
2
22
200 1.2
3
0.8 1.2(3)
sT
s
×
=
+ +
The maximum value of the toque is obtained at a slip which makes :
0dT
ds=
2 2 2
2 2
2
22
2
1.2200 1.2 1.2 1.20.8 3 (0.8 3
3
1.21.8 (3)
d
s s ds ssdT
ds
s
− + + − + +
=
+ +
2 22
2 2
22
2
1.2200 1.2 1.2 1.2 1.20.8 3 2 0.8
3
1.21.8 (3)
s s ss s
s
− − + + + × +
=
+ +
2
2
2
22
2
200 1.2 1.2 1.2 1.20.8 3 2 0.8
3
1.21.8 (3)
s s ss
s
− + + − +
=
+ +
This is zero when
sss
2.12.18.029
2.18.0
2
+=+
+
i.e., 22
88.292.19
44.192.164.0
ssss+=+++
144
Electrical Technology or
2
1.449.64
s=
or 2 1.44
9.64s =
or max
1.44
9.64Ts s= =
giving smax T = 0.387.
Therefore maximum torque will occur at a slip of 0.387.
Substituting smax T = 0.387 in the torque expression, we get
Maximum torque per phase ( )( )
17079387.02.18.0
387.0/2.13200
2
2
=++
×=
∴ The maximum torque the motor can develop is
1707 × 3 = 5121 synchronous watts.
SAQ 3
(a) (i) Motor speed N = 975 r.p.m.
Synchronous speed r.p.m. 10006
50120120=
×==
P
fNs
Slip 025.01000
9751000=
−=
−=
s
s
N
NN
∴ s = 2.5 percent
(ii) Motor output = 45 kW
Efficiency = 92%
Motor input Output 45
kW 48.913 kWEfficiency 0.92
= = =
(iii) Rotor input = Stator output = Motor input − Stator losses
= 48.913 − 1.2 = 47.713 kW
Rotor copper loss = slip × rotor input
= 0.025 × 47.713 kW
= 1.1928 kW
(iv) Rotor output = Rotor input − Rotor copper loss
= 47.713 − 1.1928 = 46.5202 kW
Friction and windage losses = Rotor output − Net output
= 46.5202 − 45
= 1.5202 kW
(b) 1
4
resistanceRotor
standstillat reactanceRotor
2
2 ==R
X
145
Induction Motor
∴ 25.04
1
2
2 ===X
Ra
22
2
torqueMaximum
torqueload Full
Sa
aS
T
T
m
f
+==
But (given) 25.2=f
m
T
T
∴ 22
)25.0(
25.02
25.2
1
S
S
+
×= . . .
(1)
or S2 − 1.125 S + 0.0625 = 0
2
)0625.0(4)125.1(125.1 2 −±=S
2
008.1125.1 ±=S
= 0.0585, 1.0665
Here value greater than 1 is inadmissible
∴ Full load slip = 0.0585 or 5.85 percent
22)25.0(1
25.02
1
2
torqueMaximum
torqueStarting
+
×=
+==
a
a
T
T
m
s
or 0625.1
5.0=
m
s
T
T . . .
(2)
Multiplying Eq. (1) by Eq. (2), we have
0625.1
5.025.2 ×=×
m
s
f
m
T
T
T
T
or 059.1=f
s
T
T
∴ Ratio of starting torque to normal torque = 1.059.