Basic principles of rotating electrical machines, torque, slip, equ
THREE PHASE INDUCTION MOTORS
ELECTRICAL MACHINES
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Three Phase Induction Motor
1.1 Preliminaries
1.1.1 Introduction
Of all the ac motors, the polyphase induction motor is the one which is extensively used for
various kinds of industrial drives.
1.1.2 Advantages of Polyphase Induction Motor
i. Its cost is low and it is very reliable
ii. It has very simple and extremely ragged; almost unbreakable construction
especiallysquirrel cage type
iii. It requires minimum maintenance
iv. Requires no extra starting motor and no synchronization requred
1.1.3 Disadvantages of Polyphase Induction Motor
i. Its speed cannot be varied without sacrificing some of its efficiency
ii. Like DC shunt motor, its speed slightly decreases with increase in load
iii. Its starting torque is somewhat inferior to that of a DC shunt motor of equivalent
capacity
The transfer of energy from the stator to the rotor of an induction motor takes place entirely
inductively; with the help of a flux mutually linking the two. Hence an induction motor is
essentially a transformer with stator forming the primary and rotor forming rotating seconadary.
1.1.4 Three Phase Principle
When three-phase windings displaced in space by 120° are fed by three phae currents, they
produce a resultant magnetic flux which rotates in space as if actual magnetic poles were being
rotated mechanically. See Fig. 1 below.
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Figure 1. Three-phase system displaced by 120° electrical
1.1.5 Transformation Ratio
In a single phase transformer, the secondary values may be transferred to the primary and vice
versa. When shifting resistance or impedance from secondary to primary, it should be divided by
2
K whereas current should be multipled by K .
E2
K=
E1
Where E1∧¿ E2 are primary e.m.f and secondary e.m.f respctively.
1 R2 1 X2
R2 = 2
; X2 = 2
K K
Equivalent Impedance
1 1 1 1
R01=R1 + R2 ; X 01=X 1+ X 2 R02=R 2+ R1 ; X 02=X 2+ X 1
1.1.6 Slip
In practice, the rotor never succeeds in ‘‘catching up’’ with the stator field.
Ns
The difference between the synchronous speed N S and the actual speed N of the rotor is known
as slip.
N S−N
% slip s= x 100
NS
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Sometimes N S −N is called slip speed
Obviously actual motor speed N=N S (1−s)
Rotor Frequency
When the rotor is stationery, the frequency of rotor current is the same as the supply frequency.
But when the rotor starts revolving, then the freqiency depends on on the relative speed or on
slip speed.
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f =sf
Where f 1 is the rotor frequency and f is the stator frequency
Due to decrease in frequency of the rotor e.m.f., the rotor reactance will also decrease
∵ X r =s X 2
Where X r and X 2 are the rotor reactance under running conditions and standstil rotor reactance
per phase values respectively.
Example 1
A 3-Ø induction motor is wound for 4 poles and is supplied from 50 Hz system. Calculate (i) the
synchronous speed (ii) the rotor speed when slip is 4 % (iii) the rotor frequency when rotor runs
at 600 rpm.
Solution
i. Synchronous speed
120 f 120 x 50
NS= = =1500 rpm
P 4
ii. Rotor speed
N=N S ( 1−s ) =1500 (1−0.04 )=1440 rpm
N S−N 1500−600
iii. At 600 rpm rotor speed % slip s= x 100= x 100=60 %
NS 1500
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∴ f =0.6 x 50=30 Hz
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1.2 Eqiuvalent Circuit of the Rotor
The mechanical load on an induction motor can be represented by a non-inductive resistance of
value R2= ( 1S −1). The equivalent rotor circuit along with the load resistance R L may be drawn
as shown in Fig. 2.
Figure 2. Equivalent rotor circuit
1.3 Eqiuvalent Circuit of an Induction Motor
As in a transformer, the secondary values may be transferred to the primary and vice versa. The
eqivalent circuit of an induction motor where all values have been reffered to the primary i.e the
stator is shown in Fig. 3.
Figure 3. Equivalent circuit of an induction motor
The exciting circuit may be transferred to the left, as in Fig 4 because inaccuracy involved is
negligible but the circuit and hence the calculations are simplified. This is known as the
approximate equivalent circuit of the induction motor.
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Figure 4. Approximate equivalent circuit of the induction motor per phase
Example 2
A 3-phase, star connected, 400 V, 50 Hz, 4-pole induction motor has the following per phase
parameters in ohms referred to the stator.
1 1
R1=0.15 , X 1=0.45 , R 2 =0.12 , X 2 =0.45 , X m =28.5
Compute the stator current and power factor when the motor is operated at rated voltage and
frequency with a slip of 4%.
Solution
The equivalent circuit is shown in Fig. 5
Figure 5. Equivalent circuit
R L1=R 21 ( 1S −1)=0.12( 0.041 −1)=2.88 Ω
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1 V 400/√ 3
I2 = = =67.78− j 19.36
( R 01+ R L ) + j X 01
1
( 0.15+0.12+2.88 ) + j(0.45+0.45)
V ph 400
I 0= = =− j8.1
X m √ 3 x j 28.5
Stator current
1
I 1=I 0+ I 2 =67.78−J 19.36−J 8.1
I 1=67.78−J 27.46 A
¿ I 1=73.131Cis (−22.05¿ ¿ o) A ¿
Power factor
p . f =cos φ=cos (−22¿¿ 0)=0.927 (lag)¿
1.4 Induction Motor Circle Diagram
It can be shown that the performance characteristics of an induction motor are derivable from a
circular locus. The data necessary to draw the circle diagram are found from the no-load, blocked
rotor and stator winding resistance tests.
1.4.1 No-load Test
In this test, the motor is run without any external mechanical load on it. The circuit for
deternmining the no-load parameters is shown in Fig. 6.
Figure 6. Circuit diagram for determining no-load losses of induction motor
Under this condition, the speed of the motor is very near to synchronous speed.
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The test is carried out with different values of applied voltage i.e. 20% below and above the
rated voltage.
As the motor is running light, the power factor would be less than 0.5 lagging hence the total
power input will be the difference of the two wattmeter readings i.e W 1 and W 2 .
The no-load power consists of
i. Stator copper loss
ii. Stator core loss/iron loss
iii. Friction and windage loss
The iron loss, friction and windage loss are fixed losses.
The curve drawn between the input power on no-load vs applied voltage is extrapolated to
intersect power axis at V = 0 to separate loss due to friction and windage from no-load electrical
and magnetic losses.
Figure 7. Losses separation curves
From Fig. 7, BD is friction and windage loss, DF is electrical and magnetic loss or stator copper
loss and core loss respectively.
1.4.2 Blocked-rotor Test
In this test, the rotor is locked and the rotor windings are short circuited at slip rings in the case
of wound rotor motor.
A reduced voltage is applied to the stator terminals and is so adjusted that full load current is
flows in the stator windings .
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The circuit for performing this test is the same as that for no-load test given in Fig. 6.
The relationship between short circuit current and the voltage is approximately a straight line.
The motor input on short circuit consists mainly of
i. Stator and rotor copper loss
ii. Core loss which is small
iii. Windage and friction loss is absent (rotor locked)
1.4.3 Construction of Circle Diagram
1. Draw horizontal axis OX and vertical axis OY. Here the vertical axis represents the
voltage reference.
2. With suitable scale, draw phasor OA with length corresponding to I0 at an angle Φ0 from
the vertical axis. Draw a horizontal line AB.
3. Draw OS equal to ISN at an angle ΦSC and join AS.
4. Draw the perpendicular bisector to AS to meet the horizontal line AB at C.
5. With C as centre, draw a semi-circle passing through A and S. This forms the circle
diagram which is the locus of the input current.
6. From point S, draw a vertical line SL to meet the line AB.
7. Fixing point K: For wound rotor machines where equivalent rotor resistance R2′ can be
found out: Divide SL at point K so that SK: KL = equivalent rotor resistance: stator
resistance. For squirrel cage rotor machines: Find stator copper loss using ISN and stator
winding resistance R1. Rotor copper loss = total copper loss – stator copper loss. Divide
SL at point K so that SK: KL = rotor copper loss: stator copper loss. Note: If data for
separating stator copper loss and rotor copper loss is not available then assume that stator
copper loss is equal to rotor copper loss. So divide SL at point K so that SK= KL
8. If power taken with full line voltage i.e. PSN in short circuit test would be 28,440 W
(assumed), then SM represent 28,440 W; if it measures 8.1 cm then 1cm represent
28,440/8.1 = 3,510 W. Then taking full load motor output as 14.9 kW, this measures
14,900/3,510 = 4.25 cm, line SM is extended upwards by 4.25 cm and point P is located.
(see this extension in Fig. 9)
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9. The vertical line PEFGD: PD = motor input power, PE = motor output power, EF = rotor
copper loss, FG = stator copper loss, DG = constant/fixed loss i.e. iron loss + mechanical
loss, OP = full load line current, SM = the power input on short circuit with normal
voltage applied PSN.
10.
PE
Efficiency of the machine at the operating point η=
PD
11.
Power factor of the machine=cos φ1
12.
EF
Slip s=
PF
To find the operating points corresponding to maximum power and maximum torque, draw
tangents to the circle diagram parallel to the output line and torque line respectively. The
points at which these tangents touch the circle are respectively the maximum power point
Pmax in watts and maximum torque point Tmax in synchronous watts.
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Figure 8. Circle diagram
Figure 9. Circle diagram locating point P
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Calculation of Parameters
No-load test:
Blocked rotor test:
Short circuit current with normal voltage:
Power taken with full line voltage:
1.5 Induction Motor Speed Control
In a number of industries, motors must satisfy very strict speed characteristic requirements, both
with respect to the range and smoothness of control and also with respect to economical
operation. A three phase induction motor is practically a constant speed machine like a DC shunt
motor. DC shunt motors can be made to run at any speed with good efficiency whereas in
induction motors, speed reduction is accompanied by a corresponding loss of efficiency.
Different methods by which speed control of induction motors is achieved may be grouped under
two main headings
Cotrol from stator side
Changing the applied voltage
Changing the applied frequency
Changing the number of stator poles
Control from rotor side
Rotor rheostat control
Injecting an e.m.f in the rotor circuit
1.5.1 Changing Applied Voltage
This method is the cheapest and easiest but rarely used because a large change in voltage is
required for a relatively small change in speed. This large change in voltage results in large
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change in flux density thereby seriously disturbing the magnetic conditions of the motor (emf is
directly proportional to flux in ¿ 4.44 fN ∅ ). The operation at voltages exceeding rated voltage is
restricted by magnetic saturation and consequent heating of the machine.
1.5.2 Changing the Applied Frequency
The synchronous speed (and hence the actual speed) of an induction motor can be changed by
120 f
changing the supply frequency ( N= ). This method of speed control provides wide speed
P
control range with gradual variation of the speed throughout this range. The major difficulty with
this method is how to get the variable frequency supply; the auxiliary equipment required for this
purpose results in a high first cost, increased maintenance and lowering of overall efficiency. If
an induction motor is to be operated with different frequencies with practically constant values of
efficiency, power factor and constant slip, then with the iron unsaturated, it is essential that the
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supply voltage be varied with the change in frequency such that the applied voltage per hertz
f
is maintained constant. From the emf equation, ¿ 4.44 fN ∅ , it can be understood that a change in
frequency will result in a change in flux level unless the induced emf is changed in the same
ratio. Excessive flux caused by reduction in frequency causes increase in iron loss resulting in
excessive heating of the machine.
1.5.3 Changing the Number of Stator Poles
120 f
The speed of an induction motor is given by the expression, N s = . Thus by changing the
P
number of poles, the speed of an induction motor can be changed. The number of pole pairs in
the stator can be changed by
Using multiple stator windings i.e by placing two or more indipendent windings
in the same stator slot, each producing a different number of poles.
By using consequent pole technique i.e by placing one or two indipendent
windings on the stator and changing the number of poles by changing the
interconnections of primary coils
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1.5.4 Rotor Rheostat Control
As the name implies, this type of control is only possible with wound rotor or slip ring induction
motors alone; not possible with squirrel cage type induction motors. The motor speed is reduced
by introducing an external resistance in the rotor circuit. Fig. 10.
Figure 10. Rotor resistance control
Wound rotor motors are usually started by connecting starting resistances in the secondary
circuit which are shorted out as the motor speeds up. If the ohmic values of these resistances are
designed for continuous operation, they can serve both as starting and speed control. The
disadvantage of this method is that with increase in rotor resistance, copper losses also increase
which decreases the operating efficiency of the motor. The external rotor resistors are also bulky
and expensive. This method is suitable where speed changes are needed for short periods only.
1.5.5 Injecting an E.M.F in the Rotor Circuit
This method is applicable to slip ring motors. The speed of an induction motor is controlled by
injecting a foreign voltage in the rotor circuit. The injected voltage must have the same freqiency
as the slip frequency. By inserting a voltage which is out of phase to the induced rotor e.m.f, it
amounts to increasng the rotor resistance whereas inserting a voltage which is in phase with the
induced rotor e.m.f is equivalent to decreasing its resistance. Therefore, by changing the phase of
the injected emf and hence the rotor resistance, the speed can be controlled.
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1.6 Electric Braking of Three Phase Induction Motors
The simplest method of stopping an induction motor or any other type of motor is to disconnect
it from the supply mains. Torque is no longer developed and the combined effect of the rotor
and the external load brings it to standstill. When rapid stopping is required, mechanical or
electrical braking is employed. Electrical braking is preffered where precise stonpping is
required. The motor is said to operate on electric braking when the direction of developed torque
is opposite to that of its rotation.
1.6.1 Plugging or Counter Current Braking
Plugging in induction motor is achieved by reversing two of the three phases which causes a
reversal of the direction of the rotating magnetic field. At the instant of switching the motor to
the plugging position, the motor runs in the opposite direction to that of the field.
1.6.2 Dynamic or Rheostatic Braking
The Rheostatic braking with a polyphase induction motor can be obtained by disconnecting the
stator winding from the AC supply and exciting it from a DC source to produce a stationery
field. The rotor winding becomes the armature winding. The source of excitation may be
provided by an independent DC source or from AC mains through a transformer-rectifier set.
The currents induced in the rotor conductors will be opposite in direction to that corresponding
to the motoring operation.
1.7 Applications of three phase Induction Motors
A three phase induction motor has got self-starting torque and no special means is
required for its starting.
Its speed falls with the increase in load and is always less than synchronous speed.
It operates on only lagging power factor which becomes very poor at light loads
The motor is used to supply mechanical load only.
Squirrel cage induction motor is suitable for constant speed industrial drives
where speed control is not required and starting torque requirement are of medium
or low value e.g. crane, hoist etc.
Wound rotor or slip-ring induction motors are used for loads requiring severe
starting conditions or for loads requiring speed control such as for driving line
shafts, lifts, pumps, generators, winding machines, crushers etc.
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Assignment 1
A 3 phase200 V, 3.73 kW, 50 Hz, star connected induction motor gave the following test
readings;
No-load test: Line voltage 200 V, line current 5 A, total input 350 W
Blocked rotor test: Line voltage 100 V, line current 26 A, total input 1700 W
The rotor copper loss at standstill is half the total copper loss.
Draw circle diagram from the test results and find
i. Full load line current
ii. Power factor at full load
iii. Slip
iv. efficiency
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