This document summarizes key concepts about synchronous motors covered in a lecture: 1) When excitation is decreased, the synchronous motor draws reactive power from the source and operates at a lagging power factor. When excitation is increased, it supplies reactive power to the source and operates at a leading power factor. 2) Increasing or decreasing load torque does not change the active power input of a synchronous motor. It varies the load angle to maintain constant active power, causing it to operate at a lagging or leading power factor respectively. 3) Varying excitation under loaded conditions changes the back EMF and load angle while keeping the products of voltage-current and sine/cosine of angles constant, so active power

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-37

2. 2
Learning Outcomes: - (Previous Lecture_36)
 To solve numerical on power equation of a Salient Pole Synchronous Motor.
 To analyse the effect of varying excitation on a synchronous motor.

3. 3
Learning Outcomes: - (Today’s Lecture_37)
 To analyse the effect of varying excitation on a synchronous motor.
 To analyse the effect of varying load torque on a synchronous motor.
 To solve numerical on Synchronous Motor.

4. 4
Effect of varying excitation on a Synchronous Motor: -
Infinite Bus
Synchronous
Motor
XS
V = Constant
f = Constant
Ia
Eb
Field
Excitation
Mechanical
Load
Te
Tm
If

5. 5
a. No load operation: -
 Initially, assume that |Eb| = |V| and are in phase opposition in the local circuit formed by
interconnection of synchronous motor and infinite bus (i.e. load angle δ = 00 at no load).
 Expression for active and reactive power input/phase by the alternator are
( )
, , 0 0, ( ) 0, .
i i
s s
i i
s
EV V
P sin and Q V Ecos
X X
V
So At noload P and Q V E as E V
X
 

  
      
S1
N2
Axis of
Rotor Field
Axis of
Stator Field
Direction of
Movement
bE V
bE
V
f

6. 6
bE
bE
V
f aI
V
r a sE I X
'f
If the excitation is decreased, back emf ‘Eb’ becomes less
than ‘V’. So, the armature current becomes,
i.e. Armature current lags the supply voltage by 900.
So, the synchronous motor draws reactive power from the
source.
Also,
0
0 , , sin 0.
.
, ( ) .
b
i
s
b i b
s
VE
Since active power input P
X
Wehavedecreased theexcitation
V
So E V Q V E ve
X
   
     
0
90b
a
s s s
V EV E V E
I
jX jX X
 
  
    
* 0
/ , 90 0 .
, / , 0,
/ , , . . .
i a a a
i
i
Complex Power input phase S VI VI jVI
So Active Power input phase P
and Reactive Power input phase Q is ve i e reactive power is drawnbythemotor fromthe source
    



7. 7
Note: -
a. At no-load, load angle δ=00 and |Eb|=|V|, so, neither the synchronous motor draws
power from not supply power to the source. It is said to be under floating condition.
b. If the excitation is increased, the synchronous motor operates at leading power
factor and supplies reactive power to the source.
c. If the excitation is decreased, the synchronous motor operates at lagging power
factor drawing reactive power from the source.

8. 8
Waveform is plotted by taking V = 1.0 pu, Xs = 1.0 pu, by
varying the excitation voltage Eb from 0.2 pu to 1.8 pu.

9. 9
Waveform is plotted by taking V = 1.0 pu, Xs = 1.0 pu, by
varying the excitation voltage Eb from 0.2 pu to 1.8 pu.

10. 10
Waveform is plotted by taking V = 1.0 pu, Xs = 1.0 pu, by
varying the excitation voltage Eb from 0.2 pu to 1.8 pu.

11. 11
Waveform is plotted by taking V = 1.0 pu, Xs = 1.0 pu, by
varying the excitation voltage Eb from 0.2 pu to 1.8 pu.

12. 12
Effect of varying excitation under loaded condition:-
 It is known to us that, change in excitation changes the back emf ‘Eb’ but cannot
change the active power input to the synchronous motor. Active power input can only
be altered by changing the mechanical power output of the motor.
 Expression for active power and reactive power input/phase are:
 In the above power expression, V, and Xs are constant. So, change in excitation
changes the values of Eb, Ia, δ and φ, but maintains the products Ebsinδ and Iacosφ
constant, so that active power does not change before and after the change in
excitation.
 This can be easily understood from the phasor diagram.
, ( )b
a b a
s s
VE V
P sin VI cos Q V E cos VI sin
X X
       

13. 13
Eb1
VIa1
jIa1Xs
Eb2
jIa2Xs
Ia2
jIa3Xs
Eb3
Ia3
1δ
2δ 3δ
3φ 0
1φ = 02φ
a1 1
a2 2
a3 3
I cosφ
= I cosφ
= I cosφ
b1 1
b2 2
b3 3
E sinδ
= E sinδ
= E sinδ
Phasor Diagram by increasing the excitation from
its normal value, i.e. from unity power factor.
From the phasor diagram it is clear that, increase in excitation from normal value
(corresponding to unity power factor),
a. Increases the back emf, ‘Eb’.
b. Decreases the load angle, ‘δ’.
c. Increases the armature current, ‘Ia’.
d. Increases the power factor angle, ‘φ’, i.e. decreases the power factor.
e. But the products ‘Ebsinδ’ and ‘Iacosφ’ remains constant as change in excitation
cannot change the active power drawn by the motor.
f. Increases ‘Ebcosδ’, and makes the motor to supply reactive power i.e. to operate
at leading pf.
,
( )
b
a
s
b a
s
VE
P sin VI cos
X
V
Q V E cos VI sin
X
 
 
 
  

14. 14
Eb1
VIa1
jIa1Xs
Eb2
jIa2Xs
Ia2
jIa3Xs
Eb3
Ia3
1δ
2δ
3δ3φ
0
1φ = 0
2φ
a1 1
a2 2
a3 3
I cosφ
= I cosφ
= I cosφ
b1 1
b2 2
b3 3
E sinδ
= E sinδ
= E sinδ
Phasor Diagram by decreasing the excitation from its normal value, i.e. from unity
power factor.
From the phasor diagram it is clear that, decrease in excitation from normal value
(corresponding to unity power factor),
a. Decreases the back emf, ‘Eb’.
b. Increases the load angle, ‘δ’.
c. Increases the armature current, ‘Ia’.
d. Increases the power factor angle, ‘φ’, i.e. decreases the power factor.
e. But the products ‘Eb x sinδ’ and ‘Ia x cosφ’ remains constant as change
in excitation cannot change the active power drawn by the motor.
f. Makes the motor to draw reactive power, i.e. to operate at lagging pf.
,
( )
b
a
s
b a
s
VE
P sin VI cos
X
V
Q V E cos VI sin
X
 
 
 
  

15. 15

16. 16

17. 17

18. 18

19. 19

20. 20

21. 21
Effect of varying mechanical load on a Synchronous Motor: -
Infinite Bus
Synchronous
Motor
XS
V = Constant
f = Constant
Ia
Eb
Field
Excitation
Mechanical
Load
Te
Tm
If

22. 22
 Initially, assume that |Eb| = |V| and are in phase opposition in the local circuit formed by
interconnection of synchronous motor and infinite bus (i.e. load angle δ = 00 at no
load).
 Expression for active and reactive power input/phase by the alternator are
( )
, , 0 0, ( ) 0, .
i i
s s
i i
s
EV V
P sin and Q V Ecos
X X
V
So At noload P and Q V E as E V
X
 

  
      
S1
N2
Axis of
Rotor Field
Axis of
Stator Field
Direction of
Movement
bE V
bE
V
f

23. 23
 Voltage equation of the synchronous motor is:
 As δ = 0, and |E| = |V| => Ia = 0. So, the active and reactive power
received by the synchronous motor:
 So at no load, no power is delivered or, received from the infinite
bus. Therefore the synchronous motor is said to be in floating
condition.
 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.
0 0i a i aP VI cos and Q VI sin    
b a sV E j I X
  
 
 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’.
r
a
s
E
jIX

aI
V
bE


bE
V

24. 24
 Increase in load angle increases the armature current drawn by the motor from the 3-
phase supply and is given by:
 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
 

25. 25
Phasor Diagram by increasing the
mechanical load from normal
excitation, i.e. from unity power factor
From the phasor diagram it is clear that,
increase in mechanical load from normal
excitation (corresponding to unity power
factor),
a. Keeps the back emf ‘Eb’ constant.
b. Increases the load angle, ‘δ’.
c. Increases the armature current, ‘Ia’.
d. Increases the power factor angle, ‘φ’, i.e.
decreases the power factor.
e. Operates at lagging power factor.
f. Increases the active electrical power
drawn from the supply as ‘sinδ’increases.
g. Decreases the reactive power drawn by
the motor as ‘Ebcosδ’ decreases.
sin cos
( cos ) sin
b
i a
s
i b a
s
VE
P VI
X
V
Q V E VI
X
 
 
 
  
Eb1
VIa1
jIa1XsjIa2XsjIa3Xs
Ia2
1δ
2δ
3δ
4φ
0
1φ = 0
2φ
Ia3
Eb2
Eb3
Ia4
3φ
Eb4

26. 26
Eb1
V
Ia1
jIa1Xs
jIa2Xs
jIa3Xs
1δ 2δ
3δ
0
1φ = 02φ
Eb2
Eb3
3φ
Ia2
Ia3
Phasor Diagram by increasing the mechanical load from normal excitation, i.e.
from unity power factor
sin cos
( cos ) sin
b
i a
s
i b a
s
VE
P VI
X
V
Q V E VI
X
 
 
 
  

27. 27
Phasor Diagram by increasing the mechanical load from normal excitation, i.e.
from unity power factor
From the phasor diagram it is clear that, increase in mechanical load from normal excitation
(corresponding to unity power factor),
a. Keeps the back emf ‘Eb’ constant.
b. Decreases the load angle, ‘δ’.
c. Decreases the armature current, ‘Ia’.
d. Increases the power factor angle, ‘φ’, i.e. decreases the power factor.
e. Operates at leading power factor.
f. Decreases the active electrical power drawn from the supply as ‘sinδ’ decreases.
g. Increases the reactive power supplied by the motor as ‘Ebcosδ’ increases.
sin cos
( cos ) sin
b
i a
s
i b a
s
VE
P VI
X
V
Q V E VI
X
 
 
 
  

28. 28
Thank you