Synchronizing Power Coefficient and numerical

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-34

2. 2
Learning Outcomes: - (Previous Lecture_33)
 To solve numerical on voltage equation to determine the back emf and load
angle of a Salient Pole Synchronous Motor.
 To analyse the power equation and power angle characteristics of a
cylindrical pole synchronous motor.

3. 3
Learning Outcomes: - (Today’s Lecture_34)
 To solve numerical on power equation of a Cylindrical Pole Synchronous
Motor.
 To analyse the power equation and power angle characteristics of a Salient
Pole Synchronous Motor.
 To analyse the concept of synchronizing power coefficient and synchronizing
power.

4. 4
2 2
cos( ) cos( ), sin( ) sin( )b b
i i
s s s s
VE VEV V
P and Q
Z Z Z Z
          
Real Power input ‘Pi’ and Reactive power input, ‘Qi’ to the cylindrical pole synchronous
motor are expressed as:
2
0 0
2 2
0 0
cos(90 ) cos(90 )
sin(90 ) sin(90 ) cos
b
i
s s
b b
s s s s
VEV
P
X X
VE VEV V
and Q Q
X X X X

 
   
      
b
s
b
s
VE
P = sinδ
X
V
Q = (V - E cosδ)
X
If the armature resistance is neglected, then, Zs = Xs and θ = 900. So, the expressions for
power input become:

5. 5
2
cos( ) cos( )b b
g
s s
VE E
P
Z Z
    
2
sin( ) sin( )b b
g
s s
VE E
Q
Z Z
    
Real Power Output (i.e. Gross Mechanical Power Developed) ‘Pm’ and Reactive power
output, ‘Qm’ of the cylindrical pole synchronous motor are expressed as:
2
0 0
cos(90 ) cos(90 )b b
g
s s
VE E
P
X X
    b
g
s
VE
P = sinδ
X
 
2 2
0 0
sin(90 ) sin(90 ) cosb b b b
g
s s s s
VE E VE E
Q
X X X X
       b
g b
s
E
Q = Vcosδ - E
X
For machines having negligible armature resistance, i.e.
0
90s a s s sZ r jX jX X    

6. 6
1. A 3-phase, 100 HP, 440 V, star connected synchronous motor has a synchronous impedance of
(0.1 + j 1) Ω/phase. Calculate the load angle, armature current, power factor and efficiency with
an excitation emf of 400 V if the excitation and torque losses are 4 kW.
0
2
440
, 254 .
3
400
, 230.94 .
3
0.1 1 1.005 84.3
, 100 746 4000 78600
3 3
cos( ) cos( )
440 400
78600 cos(84
1.005
b
s
g
b b
g
s s
Supply voltage V V
Back emf E V
Z j
Mechanical power developed P W
But mechanical power developed
VE E
P
Z Z
  
 
 
    
   
  


2
0 0
0
400
.3 ) cos(84.3 )
1.005
26.9


 
 

7. 7
0 0
0
0
254 0 230.94 26.9
, , 114.33 19
0.1 1
cos(19 ) 0.95
, 3 cos( ) 3 440 114.33 0.95 82774.6
786
, 100
b
a
s
i L a
o
i
V E
So armaturecurrent I
Z j
Operating power factor lagging
Input power tothemotro P V I W
P
Efficiency
P


 
    
    

 
        
  
00
100 94.96%
82774.6
 

8. 8
Power Equation of a salient pole Alternator: -
Eb
V
jIdXd
jIqXq
Id
Iq 

Vcos
O
Vsin
Ia

9. 9
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

  
 

   
        
From the phasor diagram we get,
So, complex power input,
sin
sin .
cos
cos
q q q
q
b
b d d d
d
V
V I X I
X
V E
V E I X I
X




  

   
cossin
* ( ) ( cos sin ) b
a q d
q d
V EV
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
sin cos sin sin cosb
q d d
E VV V
P
X X X
    
 
     
 
 
2
b
d q d
VE V 1 1
sinδ + - sin2δ
X 2 X X

10. 10
2 2 2 2
2 2
2
1 cos2 1 cos2
cos cos sin cos
2 2
1 1 1 1
cos cos2
2
b
d d q d d q
d q d q d
VEV V EV V V
Q
X X X X X X
EV V
X X X X X
 
   
 
 
         
    
         
    
    
Waveforms are plotted by taking V = 1.0 pu;
Eb = 0.98 pu; Xd = 1.0 pu and Xq = 0.6 pu.

11. 11
Condition for maximum Power in Salient Pole type Synchronous
Motor: -
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

 
 
 

 

  
   
   
  
 
   
  
 
 

12. 12
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
 
     
     
  
  

13. 13
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 synchronous motor, power input/phase
So, synchronizing power coefficient/phase in Watts/electrical radian
 For salient pole type alternator, power input/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
 
 
   
 
 

14. 14
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
 
       

15. 15
Cylindrical Pole Synchronous Motor
Salient Pole Synchronous Motor
Figures are plotted by taking V = 1.0 pu; Eb = 0.98 pu; Xs = 1.0 pu for cylindrical pole and
V = 1.0 pu; Eb = 0.98 pu; Xd = 1.0 pu and Xq = 0.6 pu for salient pole Machine

16. 16
 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 synchronous machine 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.

17. 17
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
 


  



     
    
   
  

18. 18
 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 P
d
 
  
    

19. 19
Thank you