This document discusses the voltage equations and phasor diagrams of salient pole synchronous motors. It begins by stating the learning outcomes which are to analyze the voltage equation and draw the phasor diagram. It then presents the voltage equations for both cylindrical and salient pole synchronous motors, explaining how the armature current and synchronous reactances are split into direct and quadrature axis components for the salient pole case. The document proceeds to derive the phasor diagrams for three cases: when the armature current lags both voltage and back EMF, lags voltage but leads back EMF, and leads both. Key voltage equations and relationships between angles are shown for each case.

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-32

2. 2
Learning Outcomes: - (Previous 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.

3. 3
Learning Outcomes: - (Today’s Lecture_32)
 To draw and analyse the phasor diagram 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
 Voltage equation of cylindrical pole synchronous motor is
 In salient pole synchronous motor, synchronous reactance is splitted into reactances
namely direct axis synchronous reactance (Xd) and quadrature axis synchronous
reactance (Xq).
 Again the armature current ‘Ia’ is also 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
    
   
Voltage equation of salient pole synchronous motor
b a a a sV E I r j I X
   
  

5. 5
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 OI JD DM 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,
 From the right angle triangle, GHD,
a a q a
a a
DJ DJ
cos DJ I R cos I R IK
DC I R
      
 From the right angle triangle, DCJ,
q q q q
a q
I X I XGH
cos GD I X
GD GD cos


    

6. 6
b a a d d q qV E I r j I X j I X
    
   
Id
Iq
Ia
V
IaRa
jIdXd
jIqXq



jIaXq
O
A
H
B
C
D
G

Vcos IaRa
Vsin
I
IqRa J
Vcos
KL
M
Eb E/
b

7. 7
Case-2: - When ‘Ia’ lags the terminal voltage ‘V’ but leads the back emf ‘Eb’
b a a d d q qV E I r j I X j I X
    
   
Id
Iq
Ia
VIaRa
jIdXd
jIqXq



jIaXq
O
M
G
I
IqRa
L
Vcos
K
A
B
C
D J
H
IdXd EbE/
b

8. 8
 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  
 From the right angle triangle, GHD,
q q q q
a q
I X I XGH
cos GD I X
GD GD cos


    
a a q a
a a
DJ DJ
cos DJ I R cos I R IK
DC I R
      
 From the right angle triangle, DCJ,

9. 9
At Leading Power Factor
b a a d d q qV E I r j I X j I X
    
   
Id
Iq
Ia
V
IaRa
jIdXd
jIqXq



jIaXq
O
M
G
I
IqRa
L
Vcos
K
A
B
C
D J H
IdXd
Eb E/
b

10. 10
 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  
 From the right angle triangle, GHD,
q q q q
a q
I X I XGH
cos GD I X
GD GD cos


    
a a q a
a a
DJ DJ
cos DJ I R cos I R IK
DC I R
      
 From the right angle triangle, DCJ,

11. 11
Thank you