- The document discusses voltage regulation calculations for a salient pole alternator. - It provides formulas and steps to calculate voltage regulation at lagging and leading power factors using phasor diagrams and analytical methods. - Examples are given to calculate voltage regulation for a 1000 kVA alternator at full load with 0.8 lagging pf, full load with 0.6 leading pf, and half load at unity pf.

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-13

2. 2
Learning Outcomes: - (Previous Lecture_12)
 To analyse the phasor diagram of salient pole alternator.
 To determine the voltage regulation of a salient pole alternator using
analytical method

3. 3
Learning Outcomes: - (Today’s Lecture_13)
 To solve numerical related to voltage regulation of salient pole alternator
using phasor calculation.
 To solve numerical related to voltage regulation of salient pole alternator
by analytical method.

4. 4
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
At lagging power factor: -
a d qa d qE V I R j I X j I X
    
   

5. 5
At leading power factor: -
 Case I : When armature current Ia leads the terminal voltage (V) but lags the induced
emf (E).
E
V
jIdXd
jIqXq
E/
Id
Iq



Vcos
IaRa
VsinjIaXq
Vcos q aI R d dI X
Ia IaRa
A
O
B
C
Da d qa d qE V I R j I X j I X
    
   

6. 6
At leading power factor: -
 Case I : When armature current Ia leads the terminal voltage (V) as well as the
induced emf (E).
E
V
jIdXd
jIqXq
E/
Id
Iq

 Vcos
Vsin
Vcos q aI R
Ia
IaRa
jIaX
q
IaRa
O
A
B
D
C
a d qa d qE V I R j I X j I X
    
   

7. 7
Formulae related to determination of voltage regulation of
salient pole alternator: -
 Phasor Calculation: -
The angle associated with E’ is the load angle ‘δ’.
Now internal power factor angle ψ can be determined as:
ψ = δ + φ (for lagging power factor)
= δ - φ (for leading power factor, when Ia leads ‘V’ but lags ‘E’)
= φ – δ (for leading power factor, when Ia leads both ‘V’ and ‘E’).
Direct axis current can be found using the relation Id = Ia sin ψ.
Finally, induced emf can be found using the relation:
' a aa qE V I R j I X
   
  
/
0( ) ,| | | '| | | ( )d d q d d qE E I X X or E E I X X
 
      

8. 8
V
Ia
E
 
Id
V
E

 
Id Ia
V
E



Id
Ia
Iq
Lagging Power Factor
Leading Power Factor
Ia leading ‘V’ but lagging ‘E’
Leading Power Factor
Ia leading both ‘V’ and ‘E’

9. 9
Analytical Method: -
 Internal power factor angle, ψ can be calculated using the relation:
,
, ( ' ', ' ')
, ( ' '&' ')
a
a
for laging pf
for leading pf I leading V but lagging E
for leading pf I leading both V E
  
 
 
 
 
 
cos ,
cos , ( ' ' ' ')
cos , ( ' '&' ')
q a d d
q a d d
q a d d
E V I R I X for lagging pf
V I R I X for leading pf Ialeading V but lagging E
V I R I X for leading pf Ialeading both V E



  
  
  
sin
tan
cos
sin
( ' ' ' ')
cos
sin
( ' ' ' ')
cos
a q
a a
a q
a
a a
a q
a
a a
V I X
for lagging pf
V I R
I X V
for leading pf I leaging V but laggnig E
V I R
V I X
for leading pf I leaging both V and E
V I R

















10. 10
1. For a 1000 kVA, 6.6 kV, Y-connected 3-phase salient pole alternator with Xd = 22 Ω,
Xq = 12 Ω and Ra = 4 Ω, calculate the voltage regulation (a) at full load 0.8 pf
lagging, (b) full load at 0.6 pf leading, (c) half load at unity pf.
Solution: -
a. Full load current, Ia = S/(√3 x VL) = (1000 x 103)/(√3 x 6.6 x 103) = 87.48 A.
Power factor is 0.8 lagging. So, power factor angle in 36.870.
Terminal voltage, V = 6.6 x 103/√3 = 3810.51 V
So, the load angle δ = 7.60.
Internal power factor angle, ψ = δ + φ = 7.60 + 36.870 = 44.470 .
Direct axis component of the armature current,
Id = Ia sinψ = 87.48 x sin(44.470) = 61.28 A.
0 0 0
0
' 3810.51 0 87.48 36.87 4 87.48 36.87 12
4762.14 7.6
a aa qE V I R j I X j
   
            
  V
Ia
E
 
Id
0 0
61.28 (90 ) 61.28 82.4dI A

      

11. 11
So, the induced emf/phase or no-load emf/phase,
We can also find the no-load induced emf using the relation,
As are along the same line, i.e. no phase difference, both lead terminal
voltage ‘V’ by δ.
Voltage regulation, (|E0|-|V|)/|V| x 100 = (5374.94-3810.51)/3810.51 x 100 = 41.01 %.
b. In the second case also the alternator is running at full load.
So, the armature current, Ia = 87.48 A. (Same as in the bit (a))
Power factor is 0.6 leading. So, power factor angle in 53.130.
Terminal voltage, V = 6.6 x 103/√3 = 3810.51 V
  0 0
0
0
' 4762.14 7.6 61.28 82.4 (22 12)
5374.94 7.6
d d qE E j I X X j
V
  
         
 
 0 ' 4762.14 61.28 (22 12) 5374.94d d qE E I X X V
  
       
0 & 'E E
 
0 & 'E E
 
0 0 0
0
' 3810.51 0 87.48 53.13 4 87.48 53.13 12
3308.22 15.96
a aa qE V I R j I X j
   
          
 

12. 12
So, the load angle, δ = 15.960.
Internal power factor angle, ψ = φ - δ = 53.130 – 15.960 = 37.170.
Direct axis component of the armature current,
Id = Ia sinψ = 87.48 x sin(37.170) = 52.85 A.
So, the no-load induced emf,
Voltage regulation, (|E0|-|V|)/|V| x 100 = (2779.72-3810.51)/3810.51 x 100 = -27.05 %.
c. At half load, armature current Ia = full load current/2 = 87.48/2 = 43.74 A.
Power factor is unity. So, the power factor angle, 00.
Terminal voltage, V = 6.6 x 103/√3 = 3810.51 V
V
E

 
Id Ia
0 0
52.85 (90 ) 52.85 105.96dI A

    
  0 0
0
0
' 3308.22 15.96 52.85 105.96 (22 12)
2779.72 15.96
d d qE E j I X X j
V
  
        
 
0 0 0
0
' 3810.51 0 43.74 0 4 43.74 0 12
4019.88 7.5
a aa qE V I R j I X j
   
          
 

13. 13
So, the load angle, δ = 7.50.
Internal power factor angle, ψ = δ = 7.50.
Direct axis component of the armature current,
Id = Ia sinψ = 43.74 x sin(7.50) = 5.71 A.
Voltage regulation, (|E0|-|V|)/|V| x 100 = (4077-3810.51)/3810.51 x 100 = 7 %
V
Ia
E
 
Id
0 0
5.71 (90 ) 5.71 82.5dI A

      
  0 0
0
0
' 4019.88 7.5 5.71 82.5 (22 12)
4077 7.5
d d qE E j I X X j
V
  
         
 

14. 14
2. For a 1000 kVA, 6.6 kV, Y-connected 3-phase salient pole alternator with Xd = 22 Ω,
Xq = 12 Ω and Ra = 4 Ω, calculate the voltage regulation (a) at full load 0.8 pf
lagging, (b) full load at 0.6 pf leading, (c) half load at unity pf.
Solution: -
a. Full load current, Ia = S/(√3 x VL) = (1000 x 103)/(√3 x 6.6 x 103) = 87.48 A.
Power factor is 0.8 lagging. So, power factor angle in 36.870.
Terminal voltage, V = 6.6 x 103/√3 = 3810.51 V
So, the load angle δ = ψ – φ = 44.470 – 36.870 = 7.60.
Direct axis component of the armature current,
Id = Ia sinψ = 87.48 x sin(44.470) = 61.28 A.
Quadrature axis component of armature current,
Iq = Ia cosψ = 87.48 x cos(44.470) = 62.43 A.
V
Ia
E
 
Id0
sin 3810.51 0.6 87.48 12
tan
cos 3810.51 0.8 87.48 4
44.47
a q
a a
V I X
V I r




   
 
   
 

15. 15
So, no-load induced emf,
E = Vcosδ + IqRa+ IdXd = 3810.51 x cos(7.60)+ 62.43 x 4 + 61.28 x 22
= 5374.92 V.
Voltage regulation, (|E0|-|V|)/|V| x 100 = (5374.92-3810.51)/3810.51 x 100 = 41.06 %.
b. Armature current at full load, Ia = 87.48 A.
Power factor is 0.6 leading. So, power factor angle in 53.130.
Terminal voltage, V = 6.6 x 103/√3 = 3810.51 V.
Note: - We have calculated the angle ‘ψ’ by assuming that the armature current is
leading both ‘V’ but lagging ‘E’. The negative value of ‘ψ’ indicates, Ia is leading
both ‘V’ as well as ‘E’ as shown in figure. From figure,
Load angle, δ = φ – ψ = 53.130 – 37.170 = 15.960.
Direct axis component of the armature current,
Id = Ia sinψ = 87.48 x sin(37.170) = 52.85 A.
Quadrature axis component of armature current,
Iq = Ia cosψ = 87.48 x cos(37.170) = 69.71 A.
0
sin 87.48 12 3810.51 0.8
tan 37.17
cos 3810.51 0.6 87.48 4
a q
a a
I X V
V I r

 

   
    
   
V
E

 
Id Ia

16. 16
No-load induced emf/phase,
E = Vcosδ + IqRa- IdXd = 3810.51 x cos(15.960) + 69.71 x 4 – 52.85 x 22
= 2779.77 V.
Voltage regulation, (|E0|-|V|)/|V| x 100 = (2779.77-3810.51)/3810.51 x 100 = -27.05 %.
c. Armature current at half load, Ia = 87.48/2 = 43.74 A
Power factor is unity, so the power factor angle is 00.
Load angle, δ = ψ = 7.50.
Direct axis component of the armature current,
Id = Ia sinψ = 43.74 x sin(7.50) = 5.71 A.
Quadrature axis component of armature current,
Iq = Ia cosψ = 87.48 x cos(37.170) = 43.37 A.
No-load induced emf/phase,
E = Vcosδ + IqRa+ IdXd = 3810.51 x cos(7.50) + 43.37 x 4 + 5.71 x 22
= 4077 V.
Voltage regulation, (|E0|-|V|)/|V| x 100 = (4077-3810.51)/3810.51 x 100 = -7.0 %.
0
sin 3810.51 0 43.74 12
tan 7.5
cos 3810.51 1 43.74 4
a q
a a
V I X
V I r

 

   
   
   
V
Ia
E
 
Id

17. 17
3. A salient-pole synchronous generator has the following per unit parameters : Xd =
1.0; Xq = 0.6; Ra = 0.02. If the generator is delivering rated kVA at rated voltage and
at 0.8 pf leading, compute the power angle and the excitation emf.
** In per unit calculation rated power, rated voltage and rated current are taken as 1.0 pu i.e.
100 %.
So, the terminal voltage, V = 1.0 pu, and
Load current, Ia = 1.0 pu.
So, for in phasor calculation,
Terminal voltage,
Power factor is 0.8 leading, so, the power factor angle is 36.870.
Load current,
Now,
Load angle δ = 36.870.
0
1 0V pu

 
0
1 36.87aI pu

 
0 0 0
0
' 1 0 1 36.87 0.02 1 36.87 0.6
0.82 36.87
a aa qE V I R j I X j
pu
   
          
 
V
Ia=Iq
E
 
Id = 0

18. 18
Load angle δ = 36.870.
So, internal power factor angle,
ψ = 00.
Direct axis component of ‘Ia’
Id = Iasinψ = 0.
Quadrature axis component of ‘Ia’
Iq = Iacosψ = 1.0.
Induced emf/phase
E = E/ + Id (Xd-Xq) = E/
Percentage voltage regulation
= (|E0|-|V|)/|V| x 100
= (0.82-1.0)/1.0 x 100 = -18 %.
0
0.82 36.87 pu 
E=E/
V
Id=0
Iq=Ia
 
Vcos
O
q qjI X
Vsin
a aI r
a aI r

19. 19
4. A salient-pole synchronous generator has the following per unit parameters : Xd =
1.0; Xq = 0.6; Ra = 0.02. If the generator is delivering rated kVA at rated voltage and
at 0.8 pf leading, compute the power angle and the excitation emf.
Solution: - (Analytical Method)
So, load angle, δ = φ.
Id = Ia sin ψ = 0.
Iq = Ia cos ψ = Ia = 1.0 pu.
Induced emf/phase, E = Vcosδ + Iq ra + IdXd = 1 x cos(38.67) + 1 x 0.02 = 0.82 pu.
Percentage voltage regulation
= (|E0|-|V|)/|V| x 100
= (0.82-1.0)/1.0 x 100 = -18 %.
0
sin 1 0.6 1 0.6
tan 0 .
cos 1 0.8 1 0.02
a q
a a
I X V
V I r

 

   
   
   
V
Ia=Iq
E
 
Id = 0

20. 20
Thank you