Numerical on Synchronizing Power Coefficient

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-35

2. 2
Learning Outcomes: - (Previous 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.

3. 3
Learning Outcomes: - (Today’s Lecture_35)
 To analyse the concept of synchronizing power coefficient and synchronizing
power.
 To solve numerical on power equation of a Salient Pole Synchronous Motor.

4. 4
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
 
 
   
 
 

5. 5
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
 
       

6. 6
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

7. 7
 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 Synchronous Machines 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.

8. 8
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
 


  



     
    
   
  

9. 9
 Synchronizing torque (Ts) can be calculated as:
Where, ‘Tsy’ is the synchronizing torque coefficient.
1 1 1
. . . . ; 2 ,
60
' ' .
S
s s sy s s
s s s
NdP
T P m P where n ns Synchronous speed inrps
d
m isthenumber of phases
   
   
       

10. 10
Real Power input,
 
 
 
 
2
b
d q d
VE V 1 1
P = sinδ + - sin2δ
X 2 X X
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
 
     
     
  
  
For cylindrical pole type synchronous motor, power input/phase
So, synchronizing power coefficient/phase in Watts/electrical radian
sin
s
EV
P
X

cossy
s
dP EV
P
d X


 

11. 11
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
180
sy
s
EV
P
X


   
 
 
 
sy
s
π EV
P = 3× × cosδ
180 X
3 cos
180 2
sy
s
EV P
P
X

     sy
s
πP EV
P = 3× × cosδ
360 X
For salient pole type synchronous motor, power input/phase
So, synchronizing power coefficient/phase in Watts/electrical radian
 
 
 
 
2
d q d
EV V 1 1
P = sinδ + - sin2δ
X 2 X X
 
 
 
 
2
sy
d q d
EV 1 1
P = cosδ +V - cos2δ
X X X

12. 12
1. A 3-phase, 5000 kVA, 11 kV, 50 Hz, 1000 rpm, star connected synchronous motor operates at
full load at a power factor of 0.8 leading. The synchronous reactance in 60 % and resistance
may be neglected. Calculate the synchronizing power per mechanical degree of angular
displacement. What is the ratio of maximum to full load torque and the value of full load
torque?
3
3
3
0
0
:
11 10
/ , 6350.85
3
5000 10
, 262.43
3 11 10
0.6 14.52
14.52
262.43 36.87
, 9195.3 19.44
a
s spu base
a
s
a
b a s
Solution
Supply voltage phase V Volt
Armaturecurrent I A
V
X X Z
I
Z j
I A
Back emf E V I Z Volt

  


 

 
 
     
 
 
    

13. 13
0
max
/ ,
3 cos 593.412 .
360
1
sin sin
2
sin(90 ) 1
3.
sin sin(19.44)
31 1
114.77
2 2
s
s
g g g
s s
fl
b
gm
s s s
Synchronizing power mechanical degreedisplacement
P EV
P kW
X
EV
Torque P T P T P T
n X
T
T
VE
Maximumtorque P kNm
n n X


    


 
   
   
  
   

14. 14
1. A 20 MVA, 3-phase, star connected, 11 kV, 12 pole, 50 Hz salient pole synchronous motor has
per phase reactances of Xd = 50 Ω and Xq = 3 Ω. At full load, unity power factor and rated
voltage determine
a. Excitation voltage
b. Synchronizing power per electrical degree and the corresponding torque
c. Synchronizing power per mechanical degree and the corresponding torque
d. Active power
e. Load angle for maximum power and the corresponding power.
Eb
V jIdXd
jIqXq
Id
Iq
 
Vcos
O
Vsin
Ia

15. 15
3
6
3
1 0
0
:
11 10
( ) / , 6350.85
3
20 10
, 1049.73
3 11 10
sin
tan 26.38 ( )
cos
26.38
sin( ) 466.34
a
a q
a b
a a
d a
q a
Solution
a Supply voltage phase V Volt
Armaturecurrent I A
V I X
ve signis dueto I leading E
V I R
I I A
I I



 




 

 
 
 
      
 
 
 cos( ) 940.45
, cos( ) 8021.45b q a d d
A
Back emf E V I R I X Volt



   

16. 16
2
1
3
1 1
( ) /
3 1 1
cos 3 cos2 648.38 / . .
180
120
, 500
1 1
, 648.38 10 12
5002
2
60
syn
d q d
s
syn syn
s
b Synchronizing power eletrical degree
EV
P V kW elect deg
X X X
f
Synchronous speed inrpm N
P
Corresponding torque T P
n

 


 
     
 
 
 
    
 
 
 
2 1
2 2
383.1
( ) / /
2
3 3890.27
1
, 74298.6
2
syn syn
syn syn
s
Nm
P
c Synchronizing power mechanical degree Synchronizing power eletrical degree
P P kW
Corresponding torque T P Nm
n
 
  
 

17. 17
2
2 2 2
1 1 21
1 2
2
0
3 1 1
( ) , sin( ) sin(2 ) 20
2
( )
32 1 1
cos , ,
8 2
67.825
i g
d q d
m m m
m m m
m d q d
m
EV V
d Active power P P MW
X X X
e Load anglecorresponding tomaximum power
P P P EV V
Where P and P
P X X X
Maximum Powe
 



 
     
 
 
     
     
  
  

max 1 2, 3 sin( ) 3 sin(2 ) 33.94m m m mr P P P MW       

18. 18
Thank you