Effect of varying excitation

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-36

2. 2
Learning Outcomes: - (Previous 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.

3. 3
Learning Outcomes: - (Today’s Lecture_36)
 To solve numerical on power equation of a Salient Pole Synchronous Motor.
 To analyse the effect of varying excitation on a synchronous motor.

4. 4
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


 
Formulae related to power equation and synchronizing power

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
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

6. 6
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

7. 7
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



   

8. 8
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
 
  
 

9. 9
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       

10. 10
Infinite BusAlternator
XS
V = Constant
f = Constant
Ia
Eb
Field
Excitation
Mechanical
Load
Te
Tm
If
Effect of varying excitation on a Synchronous Motor: -

11. 11
a. No load operation: -
 Initially, assume that |Eb| = |V| and are in phase opposition in the local circuit formed by
interconnection of synchronous motor and infinite bus (i.e. load angle δ = 00 at no load).
 Expression for active and reactive power input/phase by the alternator are
( )
, , 0 0, ( ) 0, .
i i
s s
i i
s
EV V
P sin and Q V Ecos
X X
V
So At noload P and Q V E as E V
X
 

  
      
S1
N2
Axis of
Rotor Field
Axis of
Stator Field
Direction of
Movement
bE V
bE
V
f

12. 12
 Voltage equation of the synchronous motor is:
 As δ = 0, and |E| = |V| => Ia = 0. So, the active and
reactive power received by the synchronous motor:
 So at no load, no power is delivered or, received from
the infinite bus. Therefore the synchronous motor is
said to be in floating condition.
 Now, if the excitation of motor is increased, back emf
of ‘Eb’ becomes more than bus-bar voltage ‘V’. So, the
armature current,
 So, when the motor is overexcited, it operates at leading
power factor and supplies lagging reactive power to the
source of draw leading reactive power from the source.
0 0i a i aP VI cos and Q VI sin    
b a sV E j I X
  
 
0
90a
s s s s
V E V E E V E V
I
jX jX jX X
 
    
     
r a sE jI X
aI
V
'f
bE
bE
V
f

13. 13
( )
, , 0 0,
.
, ( )
i i b
s s
i
b i b
s
EV V
P sin and Q V E
X X
So At noload P
Wehaveincreased theexcitation
V
So E V Q V E ve
X


  
  
     
So, active and reactive power input becomes:
-ve value of ‘Qi’ indicates that, the synchronous motor is supplying lagging reactive power
to the source (similar to that of a capacitor).
Also,
* 0
/ , 90 0 .
, / , 0,
/ , , . . .
i a a a
i
i
Complex Power input phase S VI VI jVI
So Active Power input phase P
and Reactive Power input phase Q is ve i e reactive power is supplied tothe sourcebythemotor
     



14. 14
Thank you