Phasor diagram of salient pole 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-31

2. 2
Learning Outcomes: - (Previous Lecture_30)
 To know the advantages and disadvantages Synchronous Motor.
 To Analyse the voltage equation of a Cylindrical Pole Synchronous Motor
from its equivalent circuit.
 To understand the no-load and on-load operation of Synchronous Motor.

3. 3
Learning Outcomes: - (Today’s Lecture_31)
 To solve numerical on voltage equation and calculation of load angle of a
Cylindrical Rotor Synchronous Motor.
 To Analyse the voltage equation of a Salient Pole Synchronous Motor.
 To solve numerical on voltage equation and calculation of load angle of a
Salient Pole Synchronous Motor.

4. 4
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
fV
fI
Field CircuitArmature Circuit
bE
aIar sx
V

5. 5
Ia
IaRa
jIaxs
 

Eb
V
IaZs

b a aa s
b a s
V E I r j I x
E I Z
   
 
  
 
Lagging power factor Case-2: Armature current lags both V but leads Eb.

6. 6
Leading power factor
V
Ia


E
IaRa

jIaxs
IaZs
b a aa s
b a s
V E I r j I x
E I Z
   
 
  
 

7. 7
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
 
  

 
     

8. 8
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
 
  

 
     

9. 9
Voltage equation and phasor diagram of salient pole
synchronous motor: -
3-Phase AC
Supply
Y
B
ir
iy
ib
N1
S1
Armature
Winding
. 
.


.
S2
N2

.
.
.
.



Te
Field
Winding
R
3-Phase AC
Supply
Y
B
ir
iy
ib
S1
Armature
Winding
. 
.


.
Field
Winding
Tm
R1 R2
Y1
Y2
B2
B1
Te
S2












N2
N1
Direct
Axis
Quadrature
Axis
Cylindrical Pole Synchronous Motor
Salient Pole Synchronous Motor

10. 10
 In cylindrical pole synchronous machine air-gap between the stator and rotor is uniform
throughout. So the effect of armature is accounted for by taking a reactance called
armature reaction reactance (xar) and the synchronous reactance is
xs = xar + xl
 Whereas, in salient pole alternator synchronous machine the air-gap is minimum along the
polar axis or direct axis and maximum along the quadrature axis. So, the effect of armature
reaction along the direct axis is different as that along the quadrature axis. So, two different
armature reaction reactances namely ‘xad’ and ‘xaq’ are considered in the equivalent circuit
of the salient pole synchronous machines.
 So, the two synchronous reactances are expressed as
Xd = xad + xl ………. Direct axis synchronous reactance.
Xq = xaq + xl ………. Quadrature axis synchronous reactance.

11. 11
 Again the armature current ‘Ia’ has to be splitted into two components ‘Id’ (direct axis
component) and ‘Iq’ (quadrature axis component).
 So, the voltage equation of salient pole alternator becomes:
b a a d d q qV E I r j I X j I X
    
   

12. 12
Phasor diagram of salient pole synchronous motor
Lagging Power Factor
Case-1: - When armature current ‘Ia’ lags both terminal voltage ‘V’ and back emf ‘Eb’
   
q a d dE OI IK KL Vcos I R I X     
tan
a q
a a
Vsin I XAG AD DG
OA OB BA Vcos I R




  
 
, sin , cosd a q aWhere I I and I I  
 From the right angle triangle, OAG,

13. 13
Id
Iq
Ia
V
E
IaRa
jIdXd
jIqXq



E/
jIaXq
O
A
H
B
C
D
G

Vcos IaRa
Vsin
I
IqRa J
Vcos
KL
M

14. 14
Case-2: - When ‘Ia’ lags the terminal voltage ‘V’ but leads the back emf ‘Eb’
Id
Iq
Ia
V
E/
IaRa
jIdXd
jIqXq



E
jIaXq
O
M
G
I
IqRa
L
Vcos
K
A
B
C
D J
H
IdXd

15. 15
 From the right angle triangle, OAG,
   
q a d dE OI IK KL Vcos I R I X     
tan
a q
a a
I X VsinAG DG DA
OA OB BA Vcos I R




  
 
, sin , cosd a q aWhere I I and I I  

16. 16
Id
Iq
Ia
V
E/
IaRa
jIdXd
jIqXq



E
jIaXq
O
M
G
I
IqRa
L
Vcos
K
A
B
C
D J H
IdXd
At Leading Power Factor

17. 17
 From the right angle triangle, OAG,
   
q a d dE OI IK KL Vcos I R I X     
tan
a q
a a
Vsin I XAG AD DG
OA OB BA Vcos I R




  
 
, sin , cosd a q aWhere I I and I I  

18. 18
Numerical on determination of back emf and load angle of a salient pole
synchronous motor
1. A 20 MVA, 3-phase, 11 kV, 12-pole, 50 Hz, star connected salient pole synchronous motor has
direct and quadrature axis reactance per phase of 5 Ω and 3 Ω respectively. Determine the
excitation voltage and load angle when the machine is drawing (a) rated power at 0.5 pf lagging,
(b) half the rated power 0.8 pf leading, (c) rated power at 0.95 pf lagging.
3
6
3
0
0
11 10
, 6350.85
3
20 10
( ) , 1049.73
3 3 11 10
, 60 .
6350.85 0.5 1049.73 3
36.51
6350.85 0.5 1049.73 0
,
a
L
a q
a a
Termianl voltage V V
S
a Armaturecurrent I A
V
Power factor angle lagging
Vsin I X
tan
Vcos I R
Load angle





 

  
  

   
  
   
0 0 0
0
0
0
60 36.51 23.49
1049.73 (36.51 ) 624.6 .
1049.73 (36.51 ) 843.69 .
, 6350.85 (23.49 ) 0 624.6 5 2701.72
d a
q a
b q a d d
I I sin sin A
I I cos cos A
Soback emf E Vcos I R I X cos V
  



    
   
   
        
V
Ia
Eb




19. 19
3
6
3
0
0
11 10
, 6350.85
3
0.5 20 10
( ) , 524.86
3 3 11 10
, 36.87 .
6350.85 0.5 524.86 3
46.67
6350.85 0.5 524.86 0
a
L
a q
a a
Termianl voltage V V
S
b Armaturecurrent I A
V
Power factor angle leading
Vsin I X
tan
Vcos I R
Load an





 
 
  
  

   
  
   
0 0 0
0
0
0
, 46.67 36.87 9.8
1049.73 (36.51 ) 381.77 .
1049.73 (36.51 ) 360.2 .
, 6350.85 (9.8 ) 0 360.2 5 8167.1 .
d a
q a
b q a d d
gle
I I sin sin A
I I cos cos A
Soback emf E Vcos I R I X cos V
  



    
   
   
        
V
Ia
Eb
 


20. 20
3
6
3
0
0
11 10
, 6350.85
3
20 10
( ) , 1049.73
3 3 11 10
, 18.2 .
6350.85 0.95 1049.73 3
40.4
6350.85 0.31 1049.73 0
a
L
a q
a a
Termianl voltage V V
S
a Armaturecurrent I A
V
Power factor angle lagging
Vsin I X
tan
Vcos I R
Load ang





 

  
  

   
  
   
0 0 0
0
0
0
, 18.2 40.4 58.6
1049.73 (40.4 ) 680.16 .
1049.73 (40.4 ) 843.69 .
, 6350.85 (50.6 ) 0 680.16 5 6711.43
d a
q a
b q a d d
le
I I sin sin A
I I cos cos A
Soback emf E Vcos I R I X cos V
  



    
   
   
        
V
Ia
Eb




21. 21
Thank you