- The document discusses armature reaction in salient pole alternators. It explains that in salient pole alternators, the air gap is not uniform, requiring separate direct and quadrature axis armature reaction reactances (xad and xaq). - It presents the phasor diagram for a salient pole alternator under lagging and leading power factors. The diagram shows the decomposition of armature current into direct (Id) and quadrature (Iq) axis components based on the load angle δ. - Formulas are provided to calculate the induced emf, direct axis current, load angle δ, and percentage voltage regulation based on the phasor diagram analysis of the salient pole alternator.

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-11

2. 2
Learning Outcomes: - (Previous Lecture_10)
 To determine the voltage regulation using Synchronous Impedance
Method.
 To solve numerical related to determination of voltage regulation.

3. 3
Learning Outcomes: - (Today’s Lecture_11)
 To analyse the armature reaction in salient pole alternator.
 To analyse the phasor diagram of salient pole alternator.
 To determine the voltage regulation of a salient pole alternator.

4. 4
Armature Reaction in Salient Pole Alternator: -
 In cylindrical rotor type alternator, air gap between the rotor and stator is same
throughout. So the effect of armature reaction is represented by an inductive
reactance called armature reaction reactance (xar).
 But, in a salient pole alternator, air-gap is not same throughout, as the poles are
of projected type. Air-gap is minimum at the polar axis (also called direct axis)
and maximum at the inter-polar region (called quadrature axis).
 So, to analyse the effect of armature reaction we need to take two different
armature reaction reactances, one along the direct axis (xad) and the other one
along the quadrature axis (xaq).

5. 5
Cylindrical pole type Alternator Salient pole type Alternator
fNS
X
.
X
. ..
X
.
X
NaSa
a
Leakage
Flux Flux taking part in
armature reaction
fNS
 
 
X
.
X
. ..
X
.
X
Na
Sa
a
Leakage
Flux Flux taking part in
armature reaction

6. 6
 So, the total synchronous reactance along direct axis now becomes ‘Xd = xl + xad’
and that along quadrature axis becomes ‘Xq = xl + xaq’.
 Similarly, the armature current has to be splitted into two components, the direct
axis current (Id) and the quadrature axis current (Iq).
 Component ‘Id’ causes demagnetizing or magnetizing effect, and ‘Iq’ causes cross-
magnetizing effect.
 This concept of splitting the armature MMF into two components is called two
reaction theory and was proposed by Andre Blondel. So the two reaction theory is
also known as Blondel’s two reaction theory.
 So, the emf equation for a salient pole alternator can be expressed as:
0 a a d d q qE V I r j I X j I X
    
   

7. 7
Phasor diagram of salient pole type Alternator: -
At lagging power factor: -
0 a a d d q qE V I r j I X j I X
    
   
E
V
IaR
a
jIdXd
jIqXq
E/
Id
Iq



Vcos

IaR
a
Vsin

jIaX
q
Vcos q aI R d dI X
O
A
B
C
D
jIdXq
 d d qI X X

F
G H

8. 8
a a d d q qE V I R j I X j I X
    
   
In the phasor diagram, δ (between E and V) is the load angle, ψ (between E and Ia) is the
internal power factor angle and φ (between V and Ia) is the load power factor angle.
Direct axis component of the armature current,
Id = Ia sinψ
And quadrature axis component of armature current,
Iq = Ia cosψ
Voltage equation of the salient pole alternator is
Using the above expression, we cannot determine the induced emf as Id and Iq are
unknown. To determine these components we need to know the angle ‘ψ’. From the phasor
diagram, it is clear that to know the angle ψ, first of all we need to know the angle δ. Angle
‘δ’ can be determined from the phasor diagram using the relation:
' , , a aE OF FC CD whereOF V FC I r
  
    

9. 9
Portion CD can be determined from the right angle triangle CDG, by using the relation,
So,
 The angle associated with E’ is the load angle ‘δ’.
 Now internal power factor angle ψ = δ + φ .
 Direct axis current can be found using the relation Id = Ia sin ψ.
 Finally, induced emf can be found using the relation:
 Percentage voltage regulation
cos cos
cos
q q q q
a q
I X I XDG
CD CD I X
CD CD
 

      
' a aa qE V I R j I X
   
  
  0 0' ' ,| | | '| | | ( )d d q d d qE E GH E j I X X E or E E I X X
    
         
0| | | |
100
| |
E V
V
 


 

10. 10
At leading power factor: -
 Case I : When armature current Ia leads the terminal voltage (V) but lags the induced
emf (E).
0 a a d d q qE V I r j I X j I X
    
   
' a aa qE V I R j I X
   
  
  0 0' ' ,| | | '| | | ( )d d q d d qE E GH E j I X X E or E E I X X
    
         
E
V
jIdXd
jIqXq
E/
Id
Iq



Vcos
IaRa
VsinjIaXq
Vcos q aI R d dI X
Ia IaRa
A
O
B
C
D
G

H

11. 11
At leading power factor: -
 Case I : When armature current Ia leads the terminal voltage (V) as well as the
induced emf (E).
0 a a d d q qE V I r j I X j I X
    
   
' a aa qE V I R j I X
   
  
  0
0
' ,
| | | '|
| '| | | ( )
d d q
d d q
E E j I X X E or
E E GH
E I X X
   
   
 
   
E
V
jIdXd
jIqXq
E/
Id
Iq

 Vcos
Vsin
Vcos q aI R
Ia
IaRa
jIaX
q
IaRa
O
A
B
D
CG

jIaX
q
H

12. 12
Thank you