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EPEG 302
Advanced Electrical Machinery
Lecture 9
1
Today’s Lecture
• Three Phase Induction Motors
-
Principle of operation
Rotating Magnetic Field
Production of Torque
Equivalent circuit
Synchronous speed
Slip and its effect on rotor frequency and voltage
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Three Phase Induction Motors
• The three-phase induction motors are the most widely
used electric motors in industry for adjustable speed
drives.
• They run at essentially constant speed from no-load to
full-load.
• usually d.c. motors are preferred when large speed
variations are required.
• the 3-phase induction motors are simple, rugged, lowpriced, easy to maintain and can be manufactured with
characteristics to suit most industrial requirements.
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A typical fan motor with flow controlled by VSD
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Three Phase Induction Motors
•
Like any electric motor, a 3-phase induction motor has a stator
and a rotor.
•
The stator carries a 3-phase winding (called stator/armature
winding) while the rotor carries a short-circuited winding (called
rotor/field winding).
•
Only the stator winding is fed from 3-phase supply.
•
The rotor winding derives its voltage and power from the
externally energized stator winding through electromagnetic
induction and hence the name.
•
The induction motor may be considered to be a transformer with
a rotating secondary and it can, therefore, be described as a
“transformer type” a.c. machine in which electrical energy is
converted into mechanical energy.
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3-phase induction motor -construction
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3-phase induction motor -construction
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3-phase induction motor -construction
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Construction
• An induction motor has two main parts
i) a stationary stator
• consisting of a steel frame that supports a hollow,
cylindrical core
• core, constructed from stacked laminations,
– having a number of evenly spaced slots, providing the
space for the stator winding
Stator of IM
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Construction
ii)
a revolving rotor
• composed of punched laminations, stacked to create
a series of rotor slots, providing space for the rotor
winding
• conventional 3-phase windings made of insulated
wire (wound-rotor) » similar to the winding on the
stator=>complete set of three-phase windings exactly
as the stator. Usually Y-connected, the ends of the
three rotor wires are connected to 3 slip rings on the
rotor shaft. In this way, the rotor circuit is accessible.
• aluminum bus bars shorted together at the ends by
two aluminum rings, forming a squirrel-cage shaped
circuit (squirrel-cage)
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Construction
•
Two basic design types depending on the rotor design
– squirrel-cage induction motor
(SCIM):
– Slip ring induction motor (SRIM) wound-rotor:
Construction
1. Squirrel cage – the conductors would look like one of the exercise
wheels that squirrel or hamsters run on.
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3-phase induction motor -construction
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Construction
2. Wound rotor – have a brushes and slip ring at the end
of rotor
Notice
the slip
rings
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3-phase induction motor -construction
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3-phase induction motor –working principle
•
•
•
•
When 3-phase stator winding is energized from a 3-phase
supply, a rotating magnetic field(rmf) is set up which rotates
round the stator at synchronous speed N (= 120 f/P).
The rotating field passes through the air gap and cuts the rotor
conductors, which as yet, are stationary.
Due to the relative speed 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.
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3-phase induction motor –working principle
•
The current-carrying rotor conductors are placed in the magnetic field produced by the
stator.
•
Consequently, mechanical force acts on the rotor conductors. => F=BIL
•
The sum of the mechanical forces on all the rotor conductors produces a torque which
tends to move the rotor in the same direction as the rotating field.
•
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.
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Induction motor speed
•
At what speed will the IM run?
– Can the IM run at the synchronous speed, why?
– If rotor runs at the synchronous speed, which is the same speed of the
rotating magnetic field, then the rotor will appear stationary to the rotating
magnetic field and the rotating magnetic field will not cut the rotor. So, no
induced current will flow in the rotor and no rotor magnetic flux will be
produced so no torque is generated and the rotor speed will fall below the
synchronous speed
– When the speed falls, the rotating magnetic field will cut the rotor windings
and a torque is produced
Induction motor speed
• So, the IM will always run at a speed lower
than the synchronous speed
• The difference between the motor speed and
the synchronous speed is called the Slip (s)
nslip  nsync  nm
Where nslip= slip speed
nsync= speed of the magnetic field
nm = mechanical shaft speed of the motor
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The Slip
s
nsync  nm
nsync
Where s is the slip
Notice that : if the rotor runs at synchronous speed
s=0
if the rotor is stationary
s=1
Slip may be expressed as a percentage by multiplying the
above eq. by 100, notice that the slip is a ratio and
doesn’t have units
Induction Motors and Transformers
•
Both IM and transformer works on the principle of induced voltage
– Transformer: voltage applied to the primary windings produce an induced voltage in the
secondary windings
– Induction motor: voltage applied to the stator windings produce an induced voltage in the
rotor windings
– The difference is that, in the case of the induction motor, the secondary windings can move
– Due to the rotation of the rotor (the secondary winding of the IM), the induced voltage in it
does not have the same frequency of the stator (the primary) voltage
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Frequency
• The frequency of the voltage induced in the
rotor is given by
fr 
Pn
120
Where fr = the rotor frequency (Hz)
P = number of stator poles
n = slip speed (rpm)
P  (ns  nm )
120
P  sns

 sf e
120
fr 
Frequency
• What would be the frequency of the rotor’s induced
voltage at any speed nm?
f r  sf e
• When the rotor is blocked (s=1) , the frequency of
the induced voltage is equal to the supply frequency
• On the other hand, if the rotor runs at synchronous
speed (s = 0), the frequency will be zero
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Torque
•
Any mechanical load applied to the motor shaft will introduce
a Torque on the motor shaft. This torque is related to the
motor output power and the rotor speed
and
 load 
Pout
m
( N .m)
m 
2nm
60
(rad / s )
Horse power
•
•
•
Another unit used to measure mechanical power is the horse power
It is used to refer to the mechanical output power of the motor
Since we, as an electrical engineers, deal with watts as a unit to measure electrical
power, there is a relation between horse power and watts
1 hp  746 watts
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Example
A 208-V, 10hp, four pole, 60 Hz, Y-connected
induction motor has a full-load slip of 5
percent
1. What is the synchronous speed of this motor?
2. What is the rotor speed of this motor at rated
load?
3. What is the rotor frequency of this motor at rated
load?
4. What is the shaft torque of this motor at rated
load?
Solution
1. n
sync
2.
3.
4.

120 f e 120(60)

 1800 rpm
P
4
nm  (1  s) ns
 (1  0.05)  1800  1710 rpm
f r  sfe  0.05  60  3Hz
Pout
n
2 m
60
10 hp  746 watt / hp

 41.7 N .m
1710  2  (1/ 60)
 load 
Pout
m

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Equivalent Circuit
•
The induction motor is similar to the transformer with the
exception that its secondary windings are free to rotate
As we noticed in the transformer, it is easier if we can combine these two
circuits in one circuit but there are some difficulties
Equivalent Circuit
•
•
When the rotor is locked (or blocked), i.e. s =1, the largest
voltage and rotor frequency are induced in the rotor.
On the other side, if the rotor rotates at synchronous
speed, i.e. s = 0, the induced voltage and frequency in the
rotor will be equal to zero.
E R  sE R 0
Where ER0 is the largest value of the rotor’s induced voltage obtained
at s = 1(locked rotor)
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Equivalent Circuit
• The same is true for the frequency, i.e.
f r  sf e
• It is known that
X  L  2fL
• So, as the frequency of the induced voltage
in the rotor changes, the reactance of the
rotor circuit also changes
Where Xr0 is the rotor reactance
at the supply frequency
(at blocked rotor)
X r   r Lr  2f r Lr
 2sf e Lr
 sX r 0
Equivalent Circuit
• Then, we can draw the rotor equivalent circuit as
follows
Where ER is the induced voltage in the rotor and RR is
the rotor resistance
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Equivalent Circuit
•
•
Now we can calculate the rotor current as
IR 
ER
( RR  jX R )

sER 0
( RR  jsX R 0 )
Dividing both the numerator and denominator by s so nothing
changes we get
IR 
ER 0
RR
(  jX R 0 )
s
Where ER0 is the induced voltage and XR0 is the rotor reactance
at blocked rotor condition (s = 1)
Equivalent Circuit
• Now we can have the rotor
equivalent circuit
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Equivalent Circuit
•
Now as we managed to solve the induced voltage and different frequency
problems, we can combine the stator and rotor circuits in one equivalent circuit
Where,
a = transformation ratio
eff
X 2  a e2ff X R 0
R 2  a e2ff R R
I2 
IR
a e ff
E 1  a e ff E R 0
a e ff 
NS
NR
Power losses in Induction machines
•
Copper losses
– Copper loss in the stator (PSCL) = I12R1
– Copper loss in the rotor (PRCL) = I22R2
•
•
•
Core loss (Pcore)
Mechanical power loss due to friction and windage
How this power flow in the motor?
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Power relations
PA G
Pc o n v
1
1-s
PR C L
PAG : PRCL : Pconv
1 :
s
s : 1-s
Example
A 480-V, 60 Hz, 50-hp, three phase induction motor is
drawing 60A at 0.85 PF lagging. The stator copper losses
are 2 kW, and the rotor copper losses are 700 W. The
friction and windage losses are 600 W, the core losses are
1800 W, and the stray losses are negligible. Find the
following quantities:
1.
2.
3.
4.
The air-gap power PAG.
The power converted Pconv.
The output power Pout.
The efficiency of the motor.
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Solution
1.
Pin  3VL I L cos 
 3  480  60  0.85  42.4 kW
PAG  Pin  PSCL  Pcore
 42.4  2  1.8  38.6 kW
2.
Pconv  PAG  PRCL
 38.6 
3.
700
 37.9 kW
1000
Pout  Pconv  PF &W
 37.9 
600
 37.3 kW
1000
Solution
Pout 
4.


37.3
 50 hp
0.746
Pout
 100%
Pin
37.3
100  88%
42.4
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Example
A 460-V, 25-hp, 60 Hz, four-pole, Y-connected induction motor has the
following impedances in ohms per phase referred to the stator circuit:
R1= 0.641 R2= 0.332
X1= 1.106  X2= 0.464  XM= 26.3 
The total rotational losses are 1100 W and are assumed to be constant.
The core loss is lumped in with the rotational losses. For a rotor slip of
2.2 percent at the rated voltage and rated frequency, find the motor’s
1.
2.
3.
Speed
Stator current
Power factor
4.
Pconv and Pout
5.
ind and load
6.
Efficiency
Solution
120 f e 120  60

 1800 rpm
P
4
nm  (1  s ) nsync  (1  0.022)  1800  1760 rpm
1. n
sync

R2
0.332
 jX 2 
 j 0.464
s
0.022
2.  15.09  j 0.464  15.11.76 
1
1
Zf 

1/ jX M  1/ Z 2  j 0.038  0.0662  1.76
Z2 

1
 12.9431.1 
0.0773  31.1
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Solution
Z tot  Z stat  Z f
 0.641  j1.106  12.9431.1 
 11.72  j 7.79  14.0733.6 
4600
3
I1 

 18.88  33.6 A
Z tot 14.0733.6
3. PF  cos 33.6  0.833 lagging
4. Pin  3VL I L cos   3  460 18.88  0.833  12530 W
V
PSCL  3I12 R1  3(18.88) 2  0.641  685 W
PAG  Pin  PSCL  12530  685  11845 W
Solution
Pconv  (1  s ) PAG  (1  0.022)(11845)  11585 W
Pout  Pconv  PF &W  11585  1100  10485 W
10485
 14.1 hp
746
5.
PAG
11845
 ind 

sync 2  1800
=
6.
 load 

Pout
m

 62.8 N.m
60
10485
2  1760
 56.9 N.m
60
Pout
10485
100% 
100  83.7%
Pin
12530
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Induction Motor – Power and Torque
The output power can be found as
Pout = Pconv – PF&W – Pmisc
The induced torque or developed torque:
Example
A two-pole, 50-Hz induction motor supplies 15kW to
a
load at a speed of 2950 rpm.
1.
2.
3.
4.
What is the motor’s slip?
What is the induced torque in the motor in N.m under these conditions?
What will be the operating speed of the motor if its torque is doubled?
How much power will be supplied by the motor when the torque is doubled?
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Solution
120 f e 120  50

 3000 rpm
P
2
n n
3000  2950
s  sync m 
 0.0167 or 1.67%
nsync
3000
1. n
sync

Q no Pf W given
2.
 assume Pconv  Pload and  ind   load
 ind 
Pconv
m

15 103
 48.6 N.m
2
2950 
60
Solution
3. In the low-slip region, the torquespeed curve is linear and the induced
torque is direct proportional to slip. So,
if the torque is doubled the new slip
will be 3.33% and the motor speed will
be
nm  (1  s ) nsync  (1  0.0333)  3000  2900 rpm
4.
Pconv   ind m
 (2  48.6)  (2900 
2
)  29.5 kW
60
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Principle of operation
•
stator of three phase induction motor is supplied by three phase
alternating voltage  three phase balanced currents in the
winding.
•
current produces an mmf which is constant in magnitude and
rotates at a synchronous speed.
•
The rotating mmf produces rotating flux of constant magnitude.
•
A rotating magnetic field is one whose magnitude is constant but
whose axis of direction rotates in space.
•
The rotating field caused by the stator current induces emf in the
rotor by transformer action which produces rotor currents..
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Principle of operation (contd..)
•
The rotating field due to stator currents and the rotor currents react to
produce forces on the rotor conductors and torque.
•
The motor, thus, works on the induction principle and hence known as
induction motor.
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Rotating Magnetic Field-Analytical method
•
•
•
Figure 1 shows a three phase, two pole stator winding of an
induction motor.
The stator has slots in which coils are wound for three phases.
In Figure 1, aa’, bb’, cc’ are placed 120o apart w.r.t. each other
and the coils are supplied with three phase current.
Figure 1: Representation of three phase two pole stator winding
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Rotating Magnetic Field (contd..)
•
•
•
As shown in Figure 2 (a), three stator coils are supplied through
three phase current and its phasor representation is as shown in
Figure 2 (b).
As shown in Figure (2), current Ib lags behind Ia by 120o and
current Ic by 240o.
At time t1, Ia is maximum and Ib and Ic are half the magnitude of
that of Ia.
Figure 2: (a) Instantaneous phase current in a stator winding and
(b) Phasor diagram
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Rotating magnetic field due to three
phase current
61
Rotating magnetic filed...
•
•
consider a 2-pole, 3 phase winding .
The three phases X, Y and Z are energized from a 3-phase source and
currents in these phases are indicated as Ix, Iy and Iz
•
the fluxes produced by these currents are given by:
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Rotating magnetic field (RMF) contd...
3-phase supply produces a rotating field of constant magnitude equal to 1.5 φm.
At instant 1 the current in phase X is
zero and currents in phases Y and Z are
equal and opposite. The currents are
flowing outward in the top conductors and
inward in the bottom
conductors.
This establishes a resultant flux
towards right. The magnitude of the
resultant flux is constant and is equal to
1.5 φm
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RMF contd...
C
• φ
0
E
D
Φr = OD = 2OE = 2OC*cos30
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RMF contd....
resultant flux is
displaced 30°
clockwise from
position 1.
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RMF contd...
Φr = OD=2OE=2OC *cos 30
0
E
C
D
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RMF contd...
resultant flux is downward
i.e., it is displaced 90°
clockwise
from position 1.
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Rotating Magnetic Field (contd..)
•
•
•
•
Whenever an conductor carries a current, it produces magnetic
field that is proportional to magnitude of current.
At time t1, mmf due to a is represented by Ma due to b by Mb
and due to c by Mc.
At time t1 Mb, Mc are half the magnitude of Ma since Ib, Ic are half
the magnitude of Ia.
Figure 3 shows the resultant field produced by three phase
currents at times t1, t2 and t3
Figure 3: Rotating field produced by three phase stator currents (a) MMF
diagram (b) Flux at t1, t2 and t3
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Production of rotating field
1
69
=
70
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Φr = 3/2 * Φm
 Resultant flux is independent of
time and is a constant flux of
magnitude 1.5 times the maximum
flux per phase
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Direction of resultant flux is
same as that of phase
sequence of supply.
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Slip of an induction motor
•
The rotor of the three phase induction motor cannot attain the
speed of the rotating field, because if the rotating field speed
(also called synchronous speed) were to become equal to the
rotor speed, then the relative motion of the magnetic field
between the stator and the rotor will be zero.
•
No voltage will be induced in the rotor winding and therefore
the current and the correspondingly flux or torque in the rotor
will be zero.
•
Even at no load, there will be rotational losses and therefore,
three phase induction motor will never attain the synchronous
speed even at no load condition.
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Slip of an induction motor (contd..)
•
Since three phase induction motor cannot run attain
synchronous speed, so it is said that the three phase induction
motor runs at a ‘slip’ from the synchronous speed.
•
The slip is defined as
•
In equation (6), Ns is the synchronous speed and N is the speed
of the motor (or rotor)
…………. (6)
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Frequency of rotor current and voltage
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Equivalent circuit
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Equivalent resistance
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Induced voltage
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Rotor circuit alone
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Im equivalent circuit….
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Rotor EMF and Rotor current
Rotor EMF
• When the rotor is stationary, an induction motor is equivalent to
a 3-phase transformer with secondary short-circuited.
• At the time of starting, the induced emf per phase E2 in the rotor
at the instant of starting is given as:
........... (7)
Figure 1: Per Phase Equivalent circuit of three phase induction motor
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Rotor EMF and Rotor current (contd..)
•
In equation 7, E1 is the applied voltage per phase to primary, i.e.
stator winding; N1 is the stator turns; N2 is the number of rotor
turns.
•
When the rotor starts gaining speed, the relative speed of the
rotor w.r.t. the stator flux i.e. slip is decreased.
•
Hence the induced emf in the rotor which is directly proportional
to the relative speed (i.e. slip) is also decreased and is given by
sE2
•
For slip ‘s’, the induced emf in the rotor is s times the induced
emf in the rotor at standstill.
83
Rotor EMF and Rotor current (contd..)
Rotor Current
• If R2 is the rotor resistance/phase, L2 = rotor inductance/phase.
•
At standstill:
Induced e.m.f. of rotor/phase = E2
Rotor winding resistance/phase = R2
Rotor winding reactance/phase, X2 = 2πfL2, where f is the supply
frequency.
Rotor impedance/phase,
Rotor current/phase
........... (8)
........... (9)
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Rotor EMF and Rotor current (contd..)
•
At slip ‘s’:
Induced emf of rotor/phase = sE2
Rotor winding resistance = R2
Rotor winding reactance = 2πfrL2 = s(2πfL2) = sX2
Rotor winding impedance/phase
Rotor current/phase
........... (10)
........... (11)
- The rotor current I2 lags the rotor voltage E2 by rotor power
factor angle φ2 a and is given by
........... (12)
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Rotor EMF and Rotor current (contd..)
•
Power factor of rotor current is given as
........... (13)
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Equivalent circuit referred to stator
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Power flow
88
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Losses in 3-phase Induction Motor
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Power flow in induction motor
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Power relations
Pin  3 VL I L cos   3 V ph I ph cos 
PSCL  3 I12 R1
PAG  Pin  ( PSCL  Pcore )
PRCL  3I 22 R2
Pconv  PAG  PRCL
Pout  Pconv  ( Pf  w  Pstray )
 ind 
Pconv
m
Equivalent Circuit
• We can rearrange the equivalent circuit as
follows
Actual
rotor
resistance
Resistance
equivalent to
mechanical load
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Power relations
Pin  3 VL I L cos   3 V ph I ph cos 
PSCL  3 I12 R1
PAG  Pin  ( PSCL  Pcore )  Pconv  PRCL  3I 22
R2
s
PRCL  3I 22 R2
Pconv  PAG  PRCL  3I 22
Pconv  (1  s ) PAG
R2 (1  s)
s


PRCL
s
PRCL (1  s )
s
Pout  Pconv  ( Pf  w  Pstray )
Torque Equation
(1  s ) PAG
 ind 

m
(1  s)s
Pconv
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Power and Torque
•
From Figure 1 of lecture 12, per phase equivalent circuit of a
three phase induction motor can be as in the Figure below.
Figure 1:Per phase equivalent circuit of a three phase induction motor
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Power and Torque (contd..)
•
Figure 1 can be redrawn referring all the parameters to the
stator side as shown in Figure 2.
Figure 2: Per phase equivalent circuit of three phase induction
motor referred to the stator side.
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Power and Torque (contd..)
•
From Figure 2, current (I2) can be determined as in equation 1.
....... (1)
•
The air gap power (power transferred from stator to the rotor
through air gap) is as given by equation 2.
•
Also rotor ohmic loss is given as
•
Hence the power developed in the rotor is
....... (2)
....... (3)
....... (4)
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Power and Torque (contd..)
•
Similarly, the torque developed in the rotor is given as
....... (5)
•
Similarly the motor speed (ωm) can be written in terms of
synchronous speed (ωs) and slip (s) as in equation 6 below.
....... (6)
•
Putting the values of Pm and ωm from w=equations (4) and (6) in
equation 5 to obtain the developed torque as in equation (7)
....... (7)
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Power and Torque (contd..)
•
Similarly, substituting the value of I2 from equation (1) in
equation (7) yields
....... (8)
•
From equation (8), the slip at which maximum torque is
developed can be obtained by setting dTe/ds = 0
....... (9)
•
Substituting the value of sm from equation (9) in equation (8)
will yield the value of maximum torque (Tmax)
....... (10)
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Power and Torque (contd..)
•
From equations (9) and (10), the maximum torque (also called
pull-out torque or breakdown torque) is independent of rotor
resistance.
•
However, Sm (maximum slip) is directly proportional to the rotor
resistance.
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Squirrel cage and wound rotor
•
Squirrel cage Induction motor
Figure 3: Squirrel cage induction motor
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Squirrel cage and wound rotor (contd..)
•
-
Squirrel cage Induction motor (contd..)
Does not require slip rings, brush gear, short circuiting devices, rotor
terminals for connecting starting rheostats.
-
Star-delta starter is sufficient for starting.
-
Has slightly higher efficiency, cheaper and rugged in construction
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Squirrel cage and wound rotor (contd..)
•
-
Wound rotor (or slip ring) induction motor
Figure 4: Slip ring induction motor
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Squirrel cage and wound rotor (contd..)
•
-
Wound rotor (or slip ring) induction motor (contd..)
Requires slip rings, brush gear, short circuiting devices, rotor
terminals for connecting the starting rheostats.
-
Low efficiency, high cost and rugged in construction.
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Squirrel cage and wound rotor (contd..)
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Equivalent circuit
Fig . Per phase equivalent circuit of three phase induction motor
referred to stator
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Power relations
Pin  3VLIL cos 3VphIph cos
PSCL 3 I12R1
PAG  Pin (PSCL Pcore)  Pconv PRCL 3I2 R2
2
ER
IR 
(RR  jXR)

s
P
 RCL
s
sER0
(RR  jsXR0)
PAG *s = PRCL
Input power to rotor x s = rotor copper loss
s x [2 x π x Ns x Td] = 3 x s2 x ERO 2 x RR / RR2 +s2XR2
Torque in an induction motor
Td = [3 x ERO 2 /2πNs] * [s x RR / RR2 +s2XR2 ]
Td = [k x s x ERO 2xRR ]/ [RR2 +s2XR2 ]
Where k = 3/(2*π*Ns) = constant
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Starting torque
• At start s=12 so
Td = [3 x ERO /2πNs] * [ RR / RR2 +XR2 ]
Condition for Maximum Starting Torque
starting torque will be maximum when rotor resistance/phase is equal to standstill rotor
reactance/phase.
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Torque slip characteristics
Sm
Td = [k x s x ERO 2xRR ]/ [RR2 +s2XR2 ]
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Torque-slip characteristics.....
•
RR and XR constant Torque depends upon slip
Td = [k x s x ERO 2xRR ]/ [RR2 +s2XR2 ]
 Low slip region: At synchronous speed, s= 0 torque equals zero
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Torque-slip characteristics....
Medium slip region:
•
As slip increases (speed decrease in increase in load) term, sXR becomes
large so RR is neglected compared to sXR
Td = k x RR / sXR2
As slip increases beyond full-load slip, the torque increases and becomes
maximum at
s = R2/X2.
This maximum torque in an induction motor is called pull-out torque or breakdown torque. Its value is at least twice the full-load value when the motor is
operated at rated voltage and frequency.
Torque is inversely proportional to slip so characteristic is represented by
rectangular hyperbola.
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Torque slip characteristics contd...
 High slip region
• Torque decrease beyond the point of maximum torque
• motor slows down and eventually stops
• Overload protection should be provided to protect from overheating
The maximum torque remains the same and is independent of the value of rotor
resistance. Therefore, the addition of resistance to the rotor circuit does not
change the value of maximum torque but it only changes the value of slip at
which maximum torque occurs.
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Torque speed curve
•
•
•
•
•
•
Motor operates for value of clip between s=0 and s=Sm (value of slip for
maximum torque)
Pull out torque is 2.5 times rated full load torque
Starting torque is about 1.5 times rated full load torque.
At full load motor runs at speed N As load increase speed drops until
torque equals load torque
As soon as two torques are equal motor will run at constant but lower
speed
If torque exceeds 2.5 full load torque motor will suddenly stop
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Torque speed curve....
•
•
•
T represents the nominal full-load torque of the motor.
the starting torque (at N = 0) is 1.5 T and the maximum torque (also called
breakdown torque) is 2.5 T.
S
N
At full-load, the motor runs at a speed of N. When mechanical load increases,
motor speed decreases till the motor torque again becomes equal to the load
torque. As long as the two torques are in balance, the motor will run at constant
(but lower) speed. However, if the load torque exceeds 2.5 T, the motor will
suddenly stop.
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Torque speed characteristics
•
From equation 8 it is clear that if a motor is energized from a
fixed voltage at a constant frequency and hence the torquespeed characteristics could be determined.
•
A typical plot of developed torque as a function of slip or speed
is as shown in Figure (3).
Figure 3: Torque-speed curve of an induction motor
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Speed torque characteristic
• Motoring/ Powering mode:
• Regeneration:
• Plugging:
Torque vs speed
ωs
ωs
ωm
ωm
ωm
ωs
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Speed vs torque
Motoring mode:
• Motor rotates is same direction as
field
• As slip increases, torque increases
=>air gap flux remains constant as
stator current is not high
• At s=sm, torque is maximum, Tem
• After sm, increase in slip torque
decreases as stator current rises
(than rated current) and air gap flux
decreases
Regenerative mode
ωs
ωs
ωm
ωm
ωm
ωs
-Power fed back to from shaft into rotor circuit
-motor act as generator
-motor returns power to supply system
-torque speed similar to motoring but with negative torque
-motor speed > synchronous speed=both in same direction but with
negative slip
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Reverse plugging
ωs
ωs
ωm
ωm
ωm
ωs
-slip is greater than unity
-Motor speed opposite to field speed
-Supply is reversed while forward motoring=>field direction reversed
-Developed torque opposes the motion and acts as braking
- s>1, high motor current, low developed torque =>not recommended as
may cause excessive heating of motor
Torque speed characteristics
•
From equation 8,
....... (11)
•
•
From above equation, it is clear that if a motor is energized from
a fixed voltage at a constant frequency and hence the torquespeed characteristics could be determined.
Rearranging above equation (11) gives that the torque will be
zero when slip, s = 0 (i.e. For synchronous speed).
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Thank you
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