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Electrical Machines-II
6th Semester, EE and EEE
By
Dr. Binod Kumar Sahu
Associate Professor, Electrical Engg.
Siksha ‘O’ Anusandhan, Deemed to be University,
Bhubaneswar, Odisha, India
Lecture-27
2
Learning Outcomes: - (Previous Lecture_26)
 To understand the basics of Synchronous Motor.
 To know the applications of Synchronous Motors.
 To know the working principle of poly-phase Induction Motor for its
performance comparison with Synchronous Motor.
3
Learning Outcomes: - (Today’s Lecture_27)
 To understand the basics of Synchronous Motor.
 To know the working principle of poly-phase Induction Motor for its
performance comparison with Synchronous Motor.
4
R1
R2
sr
sy
sb
Stator
Flux
srI
syI
sbI
Stator
Currents
sr
sy
sb
Stator
FluxX
.
.
X
.
X
Ns Ss
R1
R2
B2
Y1 B1
Y2
Stator
Rotor
r1
r2
y1
y2
b1
b2
Working Principle of Polyphase Induction Motor: -
5
Stator Current and Stator Flux at ωt = 00
6
0
0 0
0 0
, 0 , ( ) , ( ) ,
( 120 ) 0.5 , ( 120 ) 0.5 ,
( 240 ) 0.5 ( 240 ) 0.5
sr sm sm sr sm sm
sy sm sm sy sm sm
sb sm sm sb sm sm
At t i I cos t I cos t
i I cos t I cos t
i I cos t I cos t
     
    
    
     
        
        
, 1.5s smResultant stator flux  
sr sm 
0.5sy sm 
0.5sb sm 
'
0.5S sm  1.5S sm 
0.5sb sm  
0.5sy sm  
7
Stator Current and Stator Flux at ωt = 900
8
0
0
0
, 90 ,
( ) 0,
( 120 ) 0.866 ,
( 240 ) 0.866
sr sm
sy sm sm
sb sm sm
At t
cos t
cos t
cos t

  
   
   

 
  
   
0sr 
0.866sy sm 
0.866sb sm  
Stator
Flux
0.866sb sm 
1.5S sm 
, 1.5s smResultant stator flux  
.
X
Ns
Ss R1
R2
B2
Y1 B1
Y2
Stator
Rotor
r1
r2
y1
y2
b1
b2
.
X
9
Stator Flux and
Rotor EMFs
Rotor Induced EMF: -
 Induced EMF in a winding always lags flux linking with it by 900.
10
Rotor Current: -
 Rotor current depends on the speed of the rotor or the slip at which the machine is
running. Rotor current/phase is expressed as:
 Let the rotor leakage reactance/phase at standstill is x2 = 8 Ω and rotor winding
resistance/phase is r2 = 2 Ω.
 At standstill, i.e. s = 1, rotor power factor angle is:
 So, rotor current will lag their respective phase voltages by 760. But the R-phase rotor
emf is lagging the stator flux by 900. Therefore, the R-phase rotor current will lag
resultant stator flux by 900+760 = 1660.
02
2
( ) 8
76
2
-1 -1r
r
x sx
tan tan
r

   
     
  
0
0
0 0
0 0
, 0 ,
( 166 ) 0.9701 ,
( 120 166 ) 0.275 ,
( 240 166 ) 0.6951
rr rm rm
ry rm rm
rb rm rm
At t
cos t
cos t
cos t

   
   
   

   
   
   
2
2 2
2 2( )
r
sE
I
r sx


11
Plotted at standstill with
r2 = 2 Ω and x2=8 Ω.
Stator
field
Rotor
EMF
Rotor
Field
Stator
CurrentRotor
Current
0
166
12
0
0
0 0
0 0
, 0 ,
( 166 ) 0.9701 ,
( 120 166 ) 0.275 ,
( 240 166 ) 0.6951
rr rm rm
ry rm rm
rb rm rm
At t
cos t
cos t
cos t

   
   
   

   
   
   
X
.
.
X
.
X
Ns Ss
R1
R2
B2
Y1 B1
Y2
Stator
Rotor
r1
r2
y1
y2
b1
b2
.NR
SR
X
.
Stator
Flux
Rotor
Flux
1660
.
XX
0.275ry rm 
0.6951rb rm Rotor
Flux
0.9701rr rm  
0.9701 rm
'r1.5
r
rm


 0
166
1.5R rm 
So, Resultant Rotor Flux
So, rotor resultant flux lags the stator
resultant flux by 1660.
13
 If the machine is running at a slip of 5 % i.e. s = 0.05, the rotor currents will lag their
respective phase voltages by
 So, the R-phase rotor current will lag the stator resultant flux by 900+11.310=101.310.
02
2
( ) 0.05 8
11.31
2
-1 -1r
r
x sx
tan tan
r

   
     
  
Plotted under running
condition with 5 % slip.
r2 = 2 Ω and x2=8 Ω.
Stator
field
Rotor
EMF
Rotor
Field
Stator
Current
Rotor
Current
0
101.31
14
R-phase rotor current lags stator flux by 900+11.30 = 101.30
0
0
0 0
0 0
, 0 ,
( 101.3 ) 0.1961 ,
( 120 101.3 ) 0.7511 ,
( 240 101.3 ) 0.9473
rr rm rm
ry rm rm
rb rm rm
At t
cos t
cos t
cos t

   
   
   

   
    
   
So, Resultant Rotor Flux
1.5R rm 
So, rotor resultant flux lags the stator resultant flux by
101.310.
X
.
.
X
.
X
Ns Ss
R1
R2
B2
Y1 B1
Y2
Stator
Rotor
r1
r2
y1
y2
b1
b2
NR
SR
.
X
101.30
.
X .
X
15
 If the machine is running at no load rotor speed is very close to synchronous speed i.e.
slip, s = 0, the rotor currents will lag their respective phase voltages by:
 So, the R-phase rotor current will lag the stator resultant flux by 900+00 = 900.
02
2
( ) 0.0 8
0
2
-1 -1r
r
x sx
tan tan
r

   
     
  
0
0
0 0
0 0
, 0 ,
( 90 ) 0,
( 120 90 ) 0.866 ,
( 240 90 ) 0.866
rr rm
ry rm rm
rb rm rm
At t
cos t
cos t
cos t

  
   
   

  
    
   
So, Resultant Rotor Flux
1.5R rm 
So, stator resultant flux leads the rotor resultant flux by 900.
Stator
field
Rotor
EMF
Rotor
Field
Stator
Current
Rotor
Current
0
90
16
X
.
.
X
.
X
Ns Ss
R1
R2
B2
Y1 B1
Y2
Stator
Rotor
r1
r2
y1
y2
b1
b2
NR
SR
.900
X
.X
17
 So, from no load to full load, the resultant rotor flux lags the resultant stator flux by
different angles from 900 to 1800.
 From standstill-no load-full load, both the rotor and stator resultant fluxes rotate at
synchronous speed.
 Frequency of rotor induced emf is ‘sf’. Where ‘f’ is the supply frequency.
 So, at any slip, speed of rotor magnetic field with respect to rotor is
 If the rotor is rotating at a speed ‘N’ rpm, speed of rotor field wrt stator is
 Since stator field is also rotating at Ns, the relative velocity between the stator field and the
rotor field is zero, i.e. both are stationary wrt each other under any loading conditions.
120 120r
sr s
f sf
N sN rpm
P P
  
(1 )sr s s sN N s N sN N    
18
Loading
Condition
Speed of
stator field
wrt stator
Speed of
rotor wrt
stator
Speed of
rotor field
wrt rotor
Speed of
rotor field
wrt stator
Relative velocity
between stator field
and rotor field
Standstill
(s = 1)
Ns N = 0 Ns Ns 0
At 5% slip Ns N = (1-
sNs) =
0.95Ns
0.05Ns 0.95Ns +
0.05Ns = Ns
0
At 0 slip Ns Ns 0 Ns 0
Relative velocity between stator, stator field, rotor and rotor field: -
For example, if the machine has 4 poles and is connected to a 50 Hz supply and running at 5 % slip,
 Speed of stator magnetic field wrt stator, Ns = (120f)/P = 1500 rpm.
 Speed of rotor wrt stator, N = (1-s) x Ns = (1-0.05) x 1500 = 1425 rpm.
 Speed of rotor field wrt rotor, Nsr = sNs = 0.05 x 1500 = 75 rpm.
 Speed of rotor field wrt stator = 1425 + 75 = 1500 rpm.
 So stator and rotor magnetic fields are stationary with respect to each other.
19
Working Principle of Synchronous Motor: -
20
Disadvantages of Synchronous Motors: -
 Synchronous motors require dc excitation which is supplied from external sources.
 These motors are not self-starting motors and need some external arrangement for its
starting and synchronizing.
 The cost per kW output is commonly higher than that of induction motors.
 Unless the incoming supply frequency is adjusted, there is no possible way to adjust the
speed.
 They cannot be started on load because its starting torque is zero.
 Slip rings and brushes are required which results in high maintenance cost.
 Synchronous motors cannot be useful for applications requiring frequent starting of
machines.
21
Synchronous Motor (Vs) Induction Motor: -
Basic Difference Synchronous Motor Induction Motor
Type of Excitation A synchronous motor is a doubly
excited machine.
An induction motor is a single Excited
machine.
Supply System Its armature winding is
energized from an AC source
and its field winding from a DC
source.
Its stator winding is energized from an
AC source.
Speed It always runs at synchronous
speed. The speed is independent
of load.
If the load increased the speed of the
induction motor decreases. It is always
less than the synchronous speed.
Starting It is not self starting. It has to be
run up to synchronous speed by
any means before it can be
synchronized to AC supply.
Induction motor has self starting
torque.
22
Basic Difference Synchronous Motor Induction Motor
Operation A synchronous motor can be
operated with lagging and
leading power by changing its
excitation.
An induction motor operates only at a
lagging power factor. At high loads the
lower factor becomes very poor.
Usage It can be used for power factor
correction in addition to
supplying torque to drive
mechanical loads.
An induction motor is used for driving
Mechanical loads only.
Efficiency It is more efficient than an
induction motor of the same
output and voltage rating.
Its efficiency is lesser than that of the
synchronous motor of the same output
and the voltage rating.
Cost A synchronous motor is costlier
than an induction motor of the
same output and voltage rating.
An induction motor is cheaper than the
Synchronous motor of the same output
and voltage rating.
Synchronous Motor (Vs) Induction Motor: -
23
Advantages of Synchronous Motors: -
 Efficiency is higher than of an induction motor of the same output and voltage rating
because there are neither losses related to slip nor the additional losses due to magnetizing
current. With synchronous motors, there is no difference of speed between air gap rotating
magnetic field and rotor. With induction motors, rotating magnetic field and rotor are not
at the same speed, so eddy losses are present and those losses introduced by the slip are
mainly responsible for reduced efficiency. In addition, with synchronous motor, the
excitation is applied directly on the rotor field winding, while with induction motor, the
power required for excitation is coming from the stator and induced on the rotor, so
additional losses due to magnetization are present with the induction motor.
24
Advantages of Synchronous Motors: -
 Synchronous motors have larger air gap. In induction motors, the electromagnetic force
induced in the rotor winding is mutually induced electromagnetic force. If the air gap is
large, then the leakage flux will increase and the mutual flux would reduce. As a
consequence, the rotor electromagnetic force and torque would be reduced. In a
synchronous motor, the magnetic flux is derived separately from the field winding at the
rotor. The electromagnetic force induced in the stator armature winding is a dynamically
induced electromagnetic force due to relative motion between the field and the conductors.
The air gap can be larger and noise and vibration are generally less than with the induction
motors.
25
Advantages of Synchronous Motors: -
 When the synchronous motor is overexcited, it generates reactive power, which improves
overall consumption power factor of the plant. Power factor will have a significant impact
on the electric utility bill costs. Electric utility companies have a minimum power factor
threshold, typically 0.9, that industrial customers must maintain in order to prevent
additional power factor charges. Synchronous motors help improve overall power factor
and may eliminate the need of power factor correction equipment, for example, capacitor
banks.
26
Thank you

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Eet3082 binod kumar sahu lecture_27

  • 1. Electrical Machines-II 6th Semester, EE and EEE By Dr. Binod Kumar Sahu Associate Professor, Electrical Engg. Siksha ‘O’ Anusandhan, Deemed to be University, Bhubaneswar, Odisha, India Lecture-27
  • 2. 2 Learning Outcomes: - (Previous Lecture_26)  To understand the basics of Synchronous Motor.  To know the applications of Synchronous Motors.  To know the working principle of poly-phase Induction Motor for its performance comparison with Synchronous Motor.
  • 3. 3 Learning Outcomes: - (Today’s Lecture_27)  To understand the basics of Synchronous Motor.  To know the working principle of poly-phase Induction Motor for its performance comparison with Synchronous Motor.
  • 5. 5 Stator Current and Stator Flux at ωt = 00
  • 6. 6 0 0 0 0 0 , 0 , ( ) , ( ) , ( 120 ) 0.5 , ( 120 ) 0.5 , ( 240 ) 0.5 ( 240 ) 0.5 sr sm sm sr sm sm sy sm sm sy sm sm sb sm sm sb sm sm At t i I cos t I cos t i I cos t I cos t i I cos t I cos t                                         , 1.5s smResultant stator flux   sr sm  0.5sy sm  0.5sb sm  ' 0.5S sm  1.5S sm  0.5sb sm   0.5sy sm  
  • 7. 7 Stator Current and Stator Flux at ωt = 900
  • 8. 8 0 0 0 , 90 , ( ) 0, ( 120 ) 0.866 , ( 240 ) 0.866 sr sm sy sm sm sb sm sm At t cos t cos t cos t                       0sr  0.866sy sm  0.866sb sm   Stator Flux 0.866sb sm  1.5S sm  , 1.5s smResultant stator flux   . X Ns Ss R1 R2 B2 Y1 B1 Y2 Stator Rotor r1 r2 y1 y2 b1 b2 . X
  • 9. 9 Stator Flux and Rotor EMFs Rotor Induced EMF: -  Induced EMF in a winding always lags flux linking with it by 900.
  • 10. 10 Rotor Current: -  Rotor current depends on the speed of the rotor or the slip at which the machine is running. Rotor current/phase is expressed as:  Let the rotor leakage reactance/phase at standstill is x2 = 8 Ω and rotor winding resistance/phase is r2 = 2 Ω.  At standstill, i.e. s = 1, rotor power factor angle is:  So, rotor current will lag their respective phase voltages by 760. But the R-phase rotor emf is lagging the stator flux by 900. Therefore, the R-phase rotor current will lag resultant stator flux by 900+760 = 1660. 02 2 ( ) 8 76 2 -1 -1r r x sx tan tan r               0 0 0 0 0 0 , 0 , ( 166 ) 0.9701 , ( 120 166 ) 0.275 , ( 240 166 ) 0.6951 rr rm rm ry rm rm rb rm rm At t cos t cos t cos t                           2 2 2 2 2( ) r sE I r sx  
  • 11. 11 Plotted at standstill with r2 = 2 Ω and x2=8 Ω. Stator field Rotor EMF Rotor Field Stator CurrentRotor Current 0 166
  • 12. 12 0 0 0 0 0 0 , 0 , ( 166 ) 0.9701 , ( 120 166 ) 0.275 , ( 240 166 ) 0.6951 rr rm rm ry rm rm rb rm rm At t cos t cos t cos t                           X . . X . X Ns Ss R1 R2 B2 Y1 B1 Y2 Stator Rotor r1 r2 y1 y2 b1 b2 .NR SR X . Stator Flux Rotor Flux 1660 . XX 0.275ry rm  0.6951rb rm Rotor Flux 0.9701rr rm   0.9701 rm 'r1.5 r rm    0 166 1.5R rm  So, Resultant Rotor Flux So, rotor resultant flux lags the stator resultant flux by 1660.
  • 13. 13  If the machine is running at a slip of 5 % i.e. s = 0.05, the rotor currents will lag their respective phase voltages by  So, the R-phase rotor current will lag the stator resultant flux by 900+11.310=101.310. 02 2 ( ) 0.05 8 11.31 2 -1 -1r r x sx tan tan r               Plotted under running condition with 5 % slip. r2 = 2 Ω and x2=8 Ω. Stator field Rotor EMF Rotor Field Stator Current Rotor Current 0 101.31
  • 14. 14 R-phase rotor current lags stator flux by 900+11.30 = 101.30 0 0 0 0 0 0 , 0 , ( 101.3 ) 0.1961 , ( 120 101.3 ) 0.7511 , ( 240 101.3 ) 0.9473 rr rm rm ry rm rm rb rm rm At t cos t cos t cos t                            So, Resultant Rotor Flux 1.5R rm  So, rotor resultant flux lags the stator resultant flux by 101.310. X . . X . X Ns Ss R1 R2 B2 Y1 B1 Y2 Stator Rotor r1 r2 y1 y2 b1 b2 NR SR . X 101.30 . X . X
  • 15. 15  If the machine is running at no load rotor speed is very close to synchronous speed i.e. slip, s = 0, the rotor currents will lag their respective phase voltages by:  So, the R-phase rotor current will lag the stator resultant flux by 900+00 = 900. 02 2 ( ) 0.0 8 0 2 -1 -1r r x sx tan tan r               0 0 0 0 0 0 , 0 , ( 90 ) 0, ( 120 90 ) 0.866 , ( 240 90 ) 0.866 rr rm ry rm rm rb rm rm At t cos t cos t cos t                          So, Resultant Rotor Flux 1.5R rm  So, stator resultant flux leads the rotor resultant flux by 900. Stator field Rotor EMF Rotor Field Stator Current Rotor Current 0 90
  • 17. 17  So, from no load to full load, the resultant rotor flux lags the resultant stator flux by different angles from 900 to 1800.  From standstill-no load-full load, both the rotor and stator resultant fluxes rotate at synchronous speed.  Frequency of rotor induced emf is ‘sf’. Where ‘f’ is the supply frequency.  So, at any slip, speed of rotor magnetic field with respect to rotor is  If the rotor is rotating at a speed ‘N’ rpm, speed of rotor field wrt stator is  Since stator field is also rotating at Ns, the relative velocity between the stator field and the rotor field is zero, i.e. both are stationary wrt each other under any loading conditions. 120 120r sr s f sf N sN rpm P P    (1 )sr s s sN N s N sN N    
  • 18. 18 Loading Condition Speed of stator field wrt stator Speed of rotor wrt stator Speed of rotor field wrt rotor Speed of rotor field wrt stator Relative velocity between stator field and rotor field Standstill (s = 1) Ns N = 0 Ns Ns 0 At 5% slip Ns N = (1- sNs) = 0.95Ns 0.05Ns 0.95Ns + 0.05Ns = Ns 0 At 0 slip Ns Ns 0 Ns 0 Relative velocity between stator, stator field, rotor and rotor field: - For example, if the machine has 4 poles and is connected to a 50 Hz supply and running at 5 % slip,  Speed of stator magnetic field wrt stator, Ns = (120f)/P = 1500 rpm.  Speed of rotor wrt stator, N = (1-s) x Ns = (1-0.05) x 1500 = 1425 rpm.  Speed of rotor field wrt rotor, Nsr = sNs = 0.05 x 1500 = 75 rpm.  Speed of rotor field wrt stator = 1425 + 75 = 1500 rpm.  So stator and rotor magnetic fields are stationary with respect to each other.
  • 19. 19 Working Principle of Synchronous Motor: -
  • 20. 20 Disadvantages of Synchronous Motors: -  Synchronous motors require dc excitation which is supplied from external sources.  These motors are not self-starting motors and need some external arrangement for its starting and synchronizing.  The cost per kW output is commonly higher than that of induction motors.  Unless the incoming supply frequency is adjusted, there is no possible way to adjust the speed.  They cannot be started on load because its starting torque is zero.  Slip rings and brushes are required which results in high maintenance cost.  Synchronous motors cannot be useful for applications requiring frequent starting of machines.
  • 21. 21 Synchronous Motor (Vs) Induction Motor: - Basic Difference Synchronous Motor Induction Motor Type of Excitation A synchronous motor is a doubly excited machine. An induction motor is a single Excited machine. Supply System Its armature winding is energized from an AC source and its field winding from a DC source. Its stator winding is energized from an AC source. Speed It always runs at synchronous speed. The speed is independent of load. If the load increased the speed of the induction motor decreases. It is always less than the synchronous speed. Starting It is not self starting. It has to be run up to synchronous speed by any means before it can be synchronized to AC supply. Induction motor has self starting torque.
  • 22. 22 Basic Difference Synchronous Motor Induction Motor Operation A synchronous motor can be operated with lagging and leading power by changing its excitation. An induction motor operates only at a lagging power factor. At high loads the lower factor becomes very poor. Usage It can be used for power factor correction in addition to supplying torque to drive mechanical loads. An induction motor is used for driving Mechanical loads only. Efficiency It is more efficient than an induction motor of the same output and voltage rating. Its efficiency is lesser than that of the synchronous motor of the same output and the voltage rating. Cost A synchronous motor is costlier than an induction motor of the same output and voltage rating. An induction motor is cheaper than the Synchronous motor of the same output and voltage rating. Synchronous Motor (Vs) Induction Motor: -
  • 23. 23 Advantages of Synchronous Motors: -  Efficiency is higher than of an induction motor of the same output and voltage rating because there are neither losses related to slip nor the additional losses due to magnetizing current. With synchronous motors, there is no difference of speed between air gap rotating magnetic field and rotor. With induction motors, rotating magnetic field and rotor are not at the same speed, so eddy losses are present and those losses introduced by the slip are mainly responsible for reduced efficiency. In addition, with synchronous motor, the excitation is applied directly on the rotor field winding, while with induction motor, the power required for excitation is coming from the stator and induced on the rotor, so additional losses due to magnetization are present with the induction motor.
  • 24. 24 Advantages of Synchronous Motors: -  Synchronous motors have larger air gap. In induction motors, the electromagnetic force induced in the rotor winding is mutually induced electromagnetic force. If the air gap is large, then the leakage flux will increase and the mutual flux would reduce. As a consequence, the rotor electromagnetic force and torque would be reduced. In a synchronous motor, the magnetic flux is derived separately from the field winding at the rotor. The electromagnetic force induced in the stator armature winding is a dynamically induced electromagnetic force due to relative motion between the field and the conductors. The air gap can be larger and noise and vibration are generally less than with the induction motors.
  • 25. 25 Advantages of Synchronous Motors: -  When the synchronous motor is overexcited, it generates reactive power, which improves overall consumption power factor of the plant. Power factor will have a significant impact on the electric utility bill costs. Electric utility companies have a minimum power factor threshold, typically 0.9, that industrial customers must maintain in order to prevent additional power factor charges. Synchronous motors help improve overall power factor and may eliminate the need of power factor correction equipment, for example, capacitor banks.