AC Machine Stator
‘b’ phase axis
1200
1200
‘a’ phase axis
1200
‘c’ phase axis
induction motor
1
Currents int different
phases
of
AC
Machine
t
01
12
Amp
t0
t1
t2
t3
1 Cycle
induction motor
t4
time
2
MMF Due to ‘a’ phase current
1
Axis of phase a
0.8
t0
0.6
0.4
t01
0.2
Fa
0
a’
a
a’
t12
-0.2
-0.4
-0.6
t2
-0.8
-1
-90
-40
10
60
110
160
210
260
Space angle (theta) in degrees
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a
c’
Fc
b
b’
Fa
c
RMF(Rotating Magnetic Field)
1.5
F
1
Fa
0.5
Fb
a’
0
F
c’
a’
t = t1
10
113
216
Space angle () in degrees
Fc
b’
c
b
Fc
-1
-1.5
-93
Fb a
t = t 0 = t4
Fb
-0.5
t = t 0= t4
F
Fb a
c’
Fa
F b
Fc a’
t = t2
b’
c
induction motor
a
b’
b
Fc a’
t = t3
F
c
c’
Fb
4
Video of the unfolded rotating magnetic field
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RMF(Rotating Magnetic Field)
-Analogy with DC machines
The salient field structure in DC
machines is mimicked along with speed
in an AC machines by a multiphase (2 or
more) winding. The number of poles are
determined by winding distribution and is
independent of the number of phases.
The rotational speed is determined by the supply frequency and
the number of poles, such that an observer in air-gap counts same
number of poles per second, meaning the more the number of poles
the slower the machine will run and vice-versa.
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Induction Motor
•Most popular motor today in the low and medium horsepower range
•Very robust in construction
•Speed easily controllable using V/f or Field Oriented Controllers
•Have replaced DC Motors in areas where traditional DC Motors
cannot be used such as mining or explosive environments
•Of two types depending on motor construction: Squirrel Cage
or Slip Ring
•Only Disadvantage: Most of them run with a lagging power factor
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Squirrel Cage Rotor
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Slip Ring Rotor
•The rotor contains windings similar to stator.
•The connections from rotor are brought out using slip rings that
are rotating with the rotor and carbon brushes that are static.
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Torque Production in an Induction Motor
•In a conventional DC machine field is stationary and the
current carrying conductors rotate.
•We can obtain similar results if we make field structure
rotating and current carrying conductor stationary.
•In an induction motor the conventional 3-phase winding
sets up the rotating magnetic field(RMF) and the rotor
carries the current carrying conductors.
•An EMF and hence current is induced in the rotor due to
the speed difference between the RMF and the rotor,
similar to that in a DC motor.
•This current produces a torque such that the speed
difference between the RMF and rotor is reduced.
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Slip in Induction Motor
•However, this speed difference cannot become zero because that
would stop generation of the torque producing current itself.
•The parameter slip ‘s’ is a measure of this relative speed difference
120 f1
ns  n  s  
n

; p # of poles
s
s

p
n

s
s
where ns,s,f1 are the speeds of the RMF in RPM ,rad./sec and
supply frequency respectively
n, are the speeds of the motor in RPM and rad./sec respectively
•The angular slip frequency and the slip frequency at which voltage
is induced in the rotor is given by
N2
 2  s , f 2  sf1 , E2 s  s
E1 N1  Stator turns N2  Rotor turns
N1induction motor
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Induction Motor Example
A 100 hp, 460V, 8 pole, 60 Hz, star connected
3 phase induction motor runs at 891 rpm under full load.
Determine the synchronous speed in rpm, slip, slip
frequency (frequency of the rotor circuit),slip rpm at full load.
What is the speed of the rotor field relative to (i) rotor structure, (ii)
stator structure, (iii) stator rotating field?
Voltage induced in rotor under full load? N2/N1=0.5
Solution on Greenboard
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Induction motor Equivalent Circuit
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Relation between air-gap, gross mechanical
power and rotor copper loss
Pag : Pmech : P2  1:1  s : s
Internal efficiency =
Pmech
 1 s
Pag
Implies lower the slip higher is the induction motor efficiency
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Example problem related to the
formula shown in the previous
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Approximate Equivalent Circuit
j
•Assumes negligible magnetizing current
• Note Rc has been removed.
The sum of core losses and the windage and friction loses are treated
as constant. This is because as speed increases rotor core loss
decreases (lower f2) but windage and frictionloses increase.With
decrease of speed the converse is true. Thus the sum is constant at
any speed and is termed as rotational loss.
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IEEE Equivalent Circuit
•Assumes 30-50% magnetizing current and drop across R1+jX1 not
negligible
• As before, the sum of core losses and the windage and friction loses are
treated as constant.
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Thevnin’s equivalent of the
IEEE Equivalent Circuit
• This is done by applying Thevenin’s theorem and treating the rotor
side as load
Vth  KthV1 , Rth 
Kth2 R1 , X th
Xm
 X1 , Kth 
X m  X1
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Determining equivalent circuit parameters
j
Uses no-load test and blocked rotor tests to determine them
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Example problem related to no-load and blocked rotor test
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Performance Characteristics(1)
Pmech  Tmech mech 
I 22
R2
(1  s)  Pag (1  s);
s
 mech  (1  s) syn ; syn  4pf
1
Tmech 
Pmech
 mech

Pag
 syn
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Performance Characteristics(2)
Tmech 
Pmech
 mech
Tmech 
1
 syn

Pag
 syn

1
 syn
'
R
I 2'2 2
s
Vth2
R2'
( Rth  R2' / s) 2  ( X th  X 2' ) 2 s
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Performance Characteristics(3)
Case 1: s  small(close to zero)
R2'
Then
 Rth
s
R2'
and
 X th  X 2'
s
 Tmech 
1
 syn
Vth2
s
'
( R2 )
 Tmech  s
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Performance Characteristics(4)
Case 2: s  l arg e(close to one)
R2'
Then Rth 
 X th  X 2'
s
Vth2
1
R2'
Tmech 
 syn ( X th  X 2' ) s
 Tmech
1

s
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Performance Characteristics(5)
Combining case 1 and 2 the approximate torque speed characteristics
would look approximately like:
Tmech
Tmax
nm
ns Speed (n)
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Performance Characteristics(6)
How to obtain Tmax? By differentiating the following equation
with respect to s and equating it to zero.
Vth2
1
R2'
Tmech 
 syn ( Rth  R2' / s ) 2  ( X th  X 2' ) 2 s
One can obtain the following:
Tmax 

Vth2
1
2 syn R  R 2  ( X  X ' ) 2
th
th
th
2
1
2 syn
2
p  Vth 
1
  
( sm all R1 )
'
'
( X th  X 2 ) 2  1  ( Lth  L2 )
Vth2
Slip at maximum torque = sTmax 
R2'
[ Rth2  ( X th  X 2' ) 2 ]
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Performance Characteristics(7)
(Speed Control)
Speed control by varying
rotor resistance (vary
Tmax by varying sTmax)
(inefficient)
Speed control by varying
supply voltage and frequency
(Vth/1)
(efficient)
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Performance Characteristics(8)
Also using
Tmech 
1
 syn
and
sTmax
Vth2
R2'
( R2' / s ) 2  ( X th  X 2' ) 2 s
R2'

( X th  X 2' )
for small R1 one can write the following:
sT2max  s 2
Tmax

Tmech
2sTmax s
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Example problem based on the formula on previous to
express maximum torque and starting torque in terms of rated
torque
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Performance Characteristics(9)
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Example problem related to efficiency calculation of induction motor
based on equivalent circuit parameters
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Related to the problem
in the previous slide
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Different modes of IM operation
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Different modes of IM operation
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34
Example problem on variable frequency supply using a slip-ring
induction motor
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Speed control of SRIM with ext. resistors
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Applications of SRIM
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Wind Power applications of SRIM
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