Equivalent Circuit of Synchronous Motor

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-30

2. 2
Learning Outcomes: - (Previous Lecture_29)
 To understand the basics of Synchronous Motor.
 To understand the working principle of Synchronous Motor.
 To know the various methods of starting a Synchronous Motor.

3. 3
Learning Outcomes: - (Today’s Lecture_30)
 To know the advantages and disadvantages Synchronous Motor.
 To understand the no-load and on-load operation of Synchronous Motor.
 To Analyse the voltage equation of a cylindrical pole Synchronous Motor
from its equivalent circuit.
 To solve numerical on voltage equation and calculation of load angle for
cylindrical pole Synchronous Motor .

4. 4
Synchronous Motor (Vs.) Induction Motor: -
Synchronous Motor Induction Motor
X
.
X
.
XXX
...
 
 
X
R1
Y2B2
Y1
B1
R2
R
Y
B
ir
iy
ib
Te
N1 S2 S1N2N=Ns
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
B
ir
iy
ib
R
Y
N<Ns

5. 5
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.

6. 6
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 (Contd..): -

7. 7
Note: -
 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.

8. 8
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.
 Slip rings and brushes are required which results in high maintenance cost.
 Synchronous motors cannot be useful for applications requiring frequent starting of
machines.

9. 9
Equivalent circuit of Cylindrical Rotor Synchronous Motor: -
 When the armature winding is
connected to a 3-phase supply, it
produces a rotating magnetic field
(with poles N1 & S1 as shown in
figure) which rotates at synchronous
speed.
 DC supply to the field winding also
produces a magnetic field (with poles
N2 & S2).
 Magnetic interlocking between stator and rotor poles produces an electromagnetic torque,
which makes the rotor to rotate along with stator poles.
3-Phase AC
Supply
Y
B
ir
iy
ib
N1
S1
Armature
Winding
. 
.


.
S2
N2

.
.
.
.



Te
Field
Winding
R

10. 10
 So, at steady state, stator poles and rotor poles are stationary with respect to each other.
So, the stator magnetic field can not induce an emf in the rotor/field winding.
 A part of the stator magnetic field takes part in armature reaction and the other part is the
leakage flux. So, the effect of armature magnetic field is accounted for by considering two
reactances namely, armature reaction reactance (xar) and leakage reactance (xl).
Synchronous reactance xs = xar + xl. In addition to ‘xs’ the armature winding has a
resistance of ‘ra’.
 But the synchronously rotating rotor magnetic field cuts the stator conductor (PNs times
per minute) there by, induces an emf in the stator winding (in the same manner as in an
alternator). This induced emf in the stator winding is called back emf (Eb) as it opposes
the stator voltage.

11. 11
 So the armature winding can be represented by a voltage source ‘Eb’ in series with
armature resistance ‘ra’ and synchronous reactance ‘xs’.
fV
fI
Field CircuitArmature Circuit
bE
aIar sx
V
 
 
0a a ba s
b a a s
b a a s
b a s
V I r j I x E
V E I r jx
E V I r jx
E V I Z
   
  
  
  
   
   
   
  
 Applying KVL in the armature circuit,
we have
 So, armature current,
b
a
s
V E
I
Z
 
 


12. 12
Synchronous Motor at no-load: -
 A DC motor running under no load condition, draws a small current (power) from the
supply to meet the no load losses.
 When the mechanical load on the DC motor increases, its speed decreases. The decrease in
speed reduces the back emf so that additional current is drawn by the motor from the source
to carry the increased load. So, increased mechanical load is met because of reduction in
speed.
 In induction motor, increase in mechanical load reduces the rotor speed. Reduction in rotor
speed increases the relative velocity between the rotating magnetic field and rotor
conductor.
 Increase in relative velocity increases the rotor induced emf and therefore the rotor current.
 Increase in rotor current, increases the current drawn by the motor from the 3-phase supply.

13. 13
 Similar kind of action cannot take place in a synchronous motor as it runs at a constant speed (i.e.,
synchronous speed) at steady state for all loads.
 In synchronous motor, increase in mechanical load momentarily (for a small duration) decreases
the rotor speed there by making the rotor poles to fall slightly behind the stator poles.
 The angular displacement between stator and rotor poles (by torque angle or load angle, ‘δ’) causes
the phase of back emf ‘Eb’ to change with respect to supply voltage ‘V’.
 Increase in load angle increases the armature current drawn by the motor from the 3-phase supply.
 Armature current,

 So, increase in load angle causes the armature current to increase and the increased electrical
power input further accelerates the rotor and makes it to rotate at synchronous speed but behind the
stator pole by the angle ‘δ’, (called load angle).
b
a
s
V E
I
Z
 
 

& | | .bsV Z areconstants and E is alsoconstant if theexcitationisunaltered
 

14. 14
Synchronous Motor at no-load: -
a) Ideal condition (neglecting losses): -
3-Phase AC
Supply
Y
B
ir
iy
ib
N1
S1
Armature
Winding
. 
.


.
S2
N2

.
.
.
.



Te
Field
Winding
Auxiliary/
Pony Motor
Tm
R
R1 R2
Y1
Y2
B2
B1
S1
N2
Axis of
Rotor Field
Axis of
Stator Field
Direction of
Movement

15. 15
b) Actual condition (with losses): -
3-Phase AC
Supply
Y
B
ir
iy
ib
N1
S1
Armature
Winding
. 
.


.
Te
Field
Winding
Tm
S2
N2

.
.
.
.



0
Axis of
Stator Field
Axis of
Rotor Field
R1 R2
Y1
Y2
B2
B1
Te
S1
Axis of
Rotor Field
Axis of
Stator Field
N2
Direction of
Movement
0

16. 16
Synchronous Motor under loaded condition: -
3-Phase AC
Supply
Y
B
ir
iy
ib
N1
S1
Armature
Winding
. 
.


.
Te
Field
Winding
Tm
S2
N
2

.
.
.
. 



Axis of
Stator Field Axis of
Rotor Field
R1 R2
Y1
Y2
B2
B1
Te
S1
Axis of
Rotor Field
Axis of
Stator Field
N2
Direction of
Movement


17. 17
Note: -
 For a lossless synchronous motor with
a. At no-load, load angle ‘δ = 0’, so neither the mechanical power is delivered by nor
the electrical power is received by the synchronous motor i.e. it is in floating
condition.
b. If the mechanical load on the shaft of the motor is increased, speed of the rotor
decreases momentarily and therefore the rotor poles lags behind the stator poles by
an angle called load angle ‘δ’.
c. So, the lagging of back emf behind the supply voltage, increases the armature
current, and therefore the electrical power input from the 3-phase supply.
| | | |bV E
 


18. 18
Phasor Diagrams
 Voltage equation of synchronous motor is
 
0a a ba s
b a aa s
b a a s
V I r j I x E
V E I r j I x
V E I r jx
   
   
  
   
   
   
Lagging power factor Case-1: Armature current lags both V and Eb.
Ia
IaRa
jIaxs

  Eb
V

IaZs

19. 19
Ia
IaRa
jIaxs
 

Eb
V
IaZs

 b a a sV E I r jx
  
  
Lagging power factor Case-2: Armature current lags both V but leads Eb.

20. 20
Leading power factor
V
Ia


E
IaRa

jIaxs
IaZs
 b a a sV E I r jx
  
  

21. 21
Numerical
1. A 25 HP, 230 V, 50 Hz, 4-pole star connected synchronous motor has an armature impedance of
(0.12 + j 1.6) Ω/phase. The excitation is adjusted for a generated armature voltage per phase of
110 V. Find the (i) armature current, (ii) power factor at which the motor is operating, and (iii)
power input to the motor for a load angle of (a) 30 electrical and (b) 100 electrical.
0
0 0
0
:
230
/ , 132.8
3 3
, 132.8 0 , ' ' .
. . / , 110
( ) 3 , 110 3
132.8 0 11
( ) ,
L
b
b
b
a
s
Solution
V
Supply voltage phase V V
So V V taking V as referece
Induced emf i e back emf phase E V
a For aload angleof E V
V E
i Current drawnbythemotor I
Z


 

  
 

  
  
 
0
0
0
0 3
14.75 71.63 .
(0.12 1.6)
( ) (71.63 ) 0.32 .
( ) , 3 3 230 14.75 0.32 1880.31i L a
A
j
ii Operatating power factor of themotor cos lagging
iii Power input tothemotor P V I cos W
 
  

 
     

22. 22
0
0 0
0
:
230
/ , 132.8
3 3
, 132.8 0 , ' ' .
. . / , 110
( ) 3 , 110 10
132.8 0 1
( ) ,
L
b
b
b
a
s
Solution
V
Supply voltage phase V V
So V V taking V as referece
Induced emf i e back emf phase E V
a For aload angleof E V
V E
i Current drawnbythemotor I
Z


 

  
 

  
  
 
0
0
0
10 10
19.35 47.74 .
(0.12 1.6)
( ) (47.74 ) 0.67 .
( ) , 3 3 230 19.35 0.67 5164.7i L a
A
j
ii Operatating power factor of themotor cos lagging
iii Power input tothemotor P V I cos W
 
  

 
     

23. 23
Thank you