This document discusses power equations and power angle characteristics of cylindrical pole and salient pole alternators. It begins by deriving the power equations for a cylindrical pole alternator, showing that maximum power occurs at a load angle equal to the impedance angle. It then derives similar equations for a salient pole alternator, accounting for d-axis and q-axis components. The document solves several numerical problems applying the concepts. It finds maximum power outputs, load angles, and other operating parameters for different alternator configurations.

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-14

2. 2
Learning Outcomes: - (Previous 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.

3. 3
Learning Outcomes: - (Today’s Lecture_14)
 To understand the concept of active power & reactive power in a
cylindrical pole and salient pole type alternator.
 To analyse the power angle characteristic of a cylindrical pole and salient
pole type alternator.
 To solve numerical related to power equation and power angle
characteristic.
 To understand the significance of synchronizing power coefficient and
synchronizing torque coefficient.

4. 4
Power Equation of a cylindrical pole Alternator: -
Real Power, P = Real part of ‘S’, and reactive power Q = Imaginary part of ‘S’.
i.e.
Field Circuit Armature Circuit
fV
fI
Field Circuit Armature Circuit
phE
Load
aIar sx
V
0
0
0 2
0
, 0 ,
,
,
0
0
, * 0 ( )
s s
a
s s
a
s s s
Let V V
E E
Z Z
E V E V
I
Z Z
E V EV V
Complex power S V I V
Z Z Z





  



 
 
 
 
 
   
 

   
        
 
2 2
cos( ) cos( ), sin( ) sin( )
s s s s
EV V EV V
P and Q
Z Z Z Z
          

5. 5
 Condition for maximum power is
 So, the alternator delivers maximum power when the load angle is equal to impedance
angle and the maximum power is:
 If the armature resistance neglected because of its small value, then Synchronous
Impedance,
 So, expressions of active and reactive power becomes:
0
90s a s s sZ r jX jX X    
2
0 0
2 2
0 0
cos(90 ) cos(90 ) sin
sin(90 ) sin(90 ) cos
s s s
s s s s
EV V EV
P P
X X X
EV V EV V
and Q Q
X X X X
 
 
    
     
0 sin( ) 0 .
dP
d
   

     
2
cos .m
s s
EV V
P
Z Z
 

6. 6
 Note: Power angle characteristics are drawn by taking E = 1.1 pu, V = 1.0 pu, and Xs =
1.0 pu.

7. 7
Power Equation of a salient pole Alternator: -
0
, 0 , , ' ' .
, cos( ) sin( )a a a a q d
Let E E V V taking E as reference
So I I I jI I jI

  
 

    
        
E
V jIdXd
jIqXq
Id
Iq



Vcos d dI X
O
q qI X
Vsin
Ia

8. 8
From the phasor diagram we get,
So, complex power,
sin
sin .
cos
cos
q q q
q
d d d
d
V
V I X I
X
E V
V I X E I
X




  

   
sin cos
* ( ) ( cos sin )a q d
q d
V E V
S V I V I jI V jV j
X X
 
  
   
          
 
 
Real Power, P = Real part of ‘S’, and reactive power Q = Imaginary part of ‘S’.
2 2 2
1 1
sin cos sin sin cos sin sin 2
2q d d d q d
V EV V EV V
P
X X X X X X
      
 
      
 
 

9. 9
2 2 2 2
2 2
2
1 cos2 1 cos2
cos cos sin cos
2 2
1 1 1 1
cos cos2
2
d d q d d q
d q d q d
EV V V EV V V
Q
X X X X X X
EV V
X X X X X
 
   
 
 
       
    
        
    
    
Figures are plotted by taking E = 1.2 pu; V = 1.0 pu; Xd = 1.0 pu and Xq = 0.6 pu

10. 10
Condition for maximum Power in Salient Pole type Alternator: -
Active Power equation is:
Condition for maximum power is
2
1 2
1 1
sin sin 2 sin sin 2
2
m m
d q d
EV V
P P P
X X X
   
 
     
 
 
1 2
2
2 1
2
2 1 2
2 2
1 1 2
2
2 2
1 1 21
2
0
cos 2 cos2 0
2 (2cos 1) cos 0
4 cos cos 2 0
32
cos
8
32
cos
8
m m
m m
m m m
m m m
m
m m m
m
dP
d
P P
P P
P P P
P P P
P
P P P
P

 
 
 

 

  
   
   
  
 
   
  
 
 

11. 11
So, the load angle for maximum power output can be obtained using the relation:
2 2 2
1 1 21
1 2
2
32 1 1
cos , ,
8 2
m m m
m m m
m d q d
P P P EV V
Where P and P
P X X X
 
     
     
  
  

12. 12
Numerical:-
1. A three phase, star connected alternator is delivering 20 MW and 8 MVAR to at 11
kV. The alternator has synchronous impedance of (0 + j 3) Ω/phase. Determine the
load angle and excitation of the alternator.
Solution: -
Complex power delivered by the alternator, S = (20 + j 8) MVA.
Phase voltage, V = 11 x 103/ √3 = 6350.85 Volt.
Armature current, Ia = S*/(3 x V*) = (20 - j 8) x 106/(3 x 6350.85) = A.
Induced emf/phase =
So, the load angle is δ = 22.480 and
The line value of excitation emf EL = √3 x 8236.31 = 14265.7 V
0
1130.6 21.8 
0 0 0
6350.85 0 1130.6 21.8 (0 3) 8236.31 22.48 .a sE V I Z j
  
           

13. 13
2. A three phase, star connected alternator has synchronous impedance of (1 + j 10)
Ω/phase. It is operating at a constant voltage of 6.6 kV and its field excitation is
adjusted to give an excitation voltage of 6.4 kV. Find the power output, armature
current and power factor under the condition of maximum power output.
Solution: -
For maximum power output, load angle δ = impedance angle θ = tan-1(10/1) = 84.30.
Terminal voltage/phase, V = 6.6 x 103/√3 = 3810.51 Volt.
Induced emf/phase, E = 6.4 x 103/√3 = 3695.04 Volt.
Armature current,
So, armature current, Ia = 501.25 A and power factor = cos(48.83) = 0.66 leading.
Output power, Pmax = √3VLIacos(φ) = √3 x 6.6 x 103 x 501.25 x 0.66 = 3781.84 kW.
0 0
03695.04 84.3 3810.51 0
501.25 48.83
1 10
a
s
E V
I A
Z j
 
    
   


14. 14
3. For a synchronous machine, excitation emf = 2 .3 pu , terminal voltage = 1 .0 pu, ra
= 0, Xs = 1.5 pu. Find the maximum values of active and reactive powers and the
corresponding load angles. In the reactive power diagram, indicate the reactive
power flow to or from the machine. At what load angle is the power factor unity?
Solution: -
Since, armature resistance is neglected,
Active Power,
Reactive Power,
So, active power will be maximum at δ = 900 and reactive power will be maximum at δ =
00.
Maximum active power,
Maximum reactive power,
max
2.3 1.0
sin 1.533
1.5s
EV
P pu
X


  
2
cos
s s
EV V
Q
X X
 
sin
s
EV
P
X

2
max
2.3 1.0 1.0
cos 0.866
1.5 1.5s s
EV V
Q pu
X X


    

15. 15
Phasor diagram at unity power factor
with ra = 0 is shown in the figure.
From the phasor diagram, it is clear
that under this loading condition,
cosδ = (V/E).
δ = cos-1(V/E) = cos-1(1/2.3) = 64.230.
E
V
jIaXs
 O
Ia
02.3 1
sin sin(64.23 ) 1.381 .
1.5s
EV
P pu
X


  
2
max
2.3 1.0 1.0
cos cos(64.23) 0.0022
1.5 1.5s s
EV V
Q pu
X X


     
Active power output at this load angle,
Reactive power output at this load angle,
Since the reactive power output is negative, it is drawn by the machine.

16. 16
3. A 3-phase, 50 Hz, 6.6 kV, star connected salient pole synchronous generator is
operating at 0.8 pf leading delivering 400 A. for this machine Xd = 4 Ω and Xq = 2.5
Ω. Calculate the maximum power that the alternator can deliver if its excitation is
maintained constant.
Solution: -
Phase voltage, V = 6.6 x 103/√3 = 3810.51 Volt.
Power factor is 0.8 leading. So the power factor angle is 36.870.
Armature current/phase Ia = 400 A.
Internal power factor angle ψ is –ve. So the armature current is leading both ‘V’ and ‘E’.
So, load angle δ = φ – |ψ| = 36.87-22.880 = 140.
So, Id = Iasinψ = 400 x sin(22.880) = 155.52 A.
Iq = Iacosψ = 400 x cos(22.880) = 368.53 A.
0
sin 400 2.5 3810.51 0.6
tan 22.88 .
cos 3810.51 0.8 400 0
a q
a a
I X V
V I r

 

   
    
   
V
E

 
Id Ia

17. 17
Excitation emf, E = Vcosδ + Iq ra – IdXd = 3810.51 x cos(140) + 0 – 152.52 x 4 = 3087.24 V.
At this load angle, power output/phase
So, total electrical power output = 3 x 2.223 = 3.669 MW.
For maximum power output, load angle can be calculated as :
2
2
0 0
1 1
sin sin 2
2
3087.24 3810.51 3810.51 1 1
sin(14 ) sin(2 14 ) 1.223 .
4 2 2.5 4
d q d
EV V
P
X X X
MW
 
 
   
 
 
  
     
 
2 2
1 1 21
1 2
2
16 1 1
cos , ,
8 2
m m m
m m m
m d q d
P P P EV V
Where P and P
P X X X
 
     
     
  
  
1
2
2
3087.24 3810.51
2.941
4
3810.51 1 1
1.089
2 2.5 4
m
m
P MW and
P MW

 
 
   
 

18. 18
2 2
1 1 21 0
2
32
cos 55.3
8
m m m
m
m
P P P
P
 
   
  
 
 
So the maximum power/phase,
1 2sin sin 2 2.941 sin(63.51) 1.089sin(2 63.51) 3.502m m m m mP P P MW       

19. 19
Synchronizing Power Coefficient: -
 The rate at which synchronous power ‘P’ varies with load angle ‘δ’ is called the
synchronizing power coefficient ‘Psy’. It is also known as stiffness of coupling, rigidity
factor or, stability factor.
 For cylindrical pole type alternator, power output/phase
So, synchronizing power coefficient/phase in Watts/electrical radian
 For salient pole type alternator, power output/phase
So, synchronizing power coefficient/phase in Watts/electrical radian
sin
s
EV
P
X

cossy
s
dP EV
P
d X


 
2
1 1
sin sin 2
2d q d
EV V
P
X X X
 
 
   
 
 
2 1 1
cos cos2sy
d q d
EV
P V
X X X
 
 
   
 
 

20. 20
So, synchronizing power coefficient (total for 3 phases) in Watts/electrical degree
 Synchronizing power coefficient (total for 3 phases) in Watts/mechanical degree
1
3 cos 3 cos
180 180
sy
s s
EV EV
P
X X

 

     
 
 
 
3 cos 3 cos
180 2 360
sy
s s
EV P P EV
P
X X
 
       

21. 21
Cylindrical rotor Alternator Salient rotor Alternator
Figures are plotted by taking E = 1.2 pu; V = 1.0 pu; Xs = 1.0 pu For cylindrical pole alternator and
E = 1.2 pu; V = 1.0 pu; Xd = 1.0 pu and Xq = 0.6 pu for salient pole alternator

22. 22
 Synchronizing power coefficient (Psy), is an indication of stiffness of electromagnetic
coupling between stator and rotor magnetic field.
 Too large stiffness of coupling means, that the stator field closely follow the variation in
the rotor speed caused by a sudden disturbance in prime-mover torque.
 A sudden disturbance in the generator or motor field current, also causes the
synchronizing power to come into play, so as to maintain the synchronism.
 So, too rigid electromagnetic coupling i.e. higher stiffness causes undue mechanical
shocks, whenever the alternator is subjected to a sudden change in mechanical power
input.
 Synchronizing power coefficient is directly proportional to excitation emf (E) and
inversely proportional to Xs or Xd.
 So, overexcited alternators are more stiffer.
 Again machines having large air-gaps will have less Xs or Xd, and therefore more stiffer.
 Synchronizing power coefficient (Psy), is positive for stable operating region and
negative for unstable region.
 For smaller values of load angle, Psy value large so, the degree of stability is high.
 As δ increases, Psy decreases and therefore the degree of stability is reduced.

23. 23
Synchronizing Power: -
 The variation in synchronous power due to a small change in load angle is as called the
synchronizing power (Ps).
 When the load angle changes from δ to Δδ, the synchronizing power,
 Synchronizing power ‘Ps’ is transient in nature and comes into play whenever there is a
sudden change in steady state operating condition. Synchronizing power either flow
from or to the bus to restore steady state stability and maintain synchronism.
 The synchronizing power flows from, or to, the bus, in order to maintain the relative
velocity between interacting stator and rotor fields zero; once this is attained, the
synchronizing power vanishes.
2
cos .
1 1
cos cos2 .
s
s
d q d
EV
for cylindrial polealternator
X
dP
P
d EV
V for salient polealternator
X X X
 


  



     
    
   
  

24. 24
 Synchronizing torque (Ts) can be calculated as:
Where, ‘Tsy’ is the synchronizing torque coefficient.
1 1
. . . .s s sy
s s
dP
T P m T
d
 
  
    

25. 25
1. 5 MVA, 11 kV, 50 Hz, 4 pole, star-connected synchronous generator with
synchronous reactance 0.7 pu connected to an infinite bus. Find synchronizing power
and the corresponding torque per unit of mechanical angle displacement.
(a) at no load and
(b) at full load of 0.8 pf lagging.
Solution: -
a. At no-load, load angle, δ = 00 and induced emf/phase = terminal voltage/phase, i.e. E =
V.
Terminal voltage/phase, V = E = 11 x 103/√3 = 6350.85 Volt.
Base impedance ZB = base voltage/base current.
Base voltage is V = 6350.85 Volts.
Base current, IB = rated current = 5 x 106/(√3 x 11 x 103) = 262.43 A.
So, base Impedance ZB = 6350.85/262.43 = 24.2 Ω.
Now synchronous reactance Xs = 0.7 pu = 0.7 x 24.2 = 16.94 Ω.
Synchronizing Power per mechanical degree i.e. synchronizing power coefficient,
4 6350.85 6350.85
3 cos 3 249.33
360 360 16.94
sy
s
P EV
P kW
X
 

 
      

26. 26
Angular speed in electrical radial/sec, ω = 2πf = 314.16.
So, angular speed in mechanical radian/sec, ωm = ω/(P/2) = 314.16/2 = 157.08.
Also, ωm =2 x π x Ns/60 = 2 x π x 1500/60 = 157.08 mechanical radian/sec.
Synchronizing torque, for one degree mechanical displacement of rotor
b. At full load, armature current, Ia = 5 x 106/(√3 x 11 x 103) = 262.43 A.
Power factor is 0.8 lagging.
So, the induced emf/phase,
Synchronizing power coefficient,
31 1
249.33 10 1587.28
157.08
sy sy
m
T P Nm

     
0 0 0
( ) 6350.85 0 262.43 36.87 (0 16.94) 9694.13 21.52a a sE V I r jX j Volt
  
            
04 9694.13 6350.85
3 cos 3 cos(21.52 ) 354.058
360 360 16.94
sy
s
P EV
P kW
X
 

 
      
31 1
354.058 10 2254
157.08
sy sy
m
T P Nm

     

27. 27
Thank you