Eet3082 binod kumar sahu lecture_33

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-33

2
Learning Outcomes: - (Previous Lecture_32)
 To Analyse the voltage equation of a Salient Pole Synchronous Motor.
 To draw and analyse the phasor diagram of a salient pole synchronous motor.
 To derive the formulae related to determination of back emf and load angle.

3
Learning Outcomes: - (Today’s 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.

4
Formulae related to determination of back and load angle of a salient
pole Synchronous Motor
( ' ' ' ')
( ' ' ' ')
a b
a b
for lagging power factor I lagging both V and E
for lagging power factor I lagging V but leading E
for leading power factor
 
  
 


 
 
( ' ' ' ')
( ' ' ' ')
q a d d a b
b q a d d a b
q a d d
Vcos I R I X for lagging power factor I lagging both V and E
E Vcos I R I X for lagging power factor I lagging V but leading E
Vcos I R I X for leading power factor



  

  

 
( ' ' ' ')
( ' ' ' ')
a q-1
a b
a a
a q-1
a b
a a
a q-1
a a
Vsin I X
tan for lagging power factor I lagging both V and E
Vcos I R
I X Vsin
tan for lagging power factor I lagging V but leading E
Vcos I R
I X Vsin
tan for leading
Vcos I R







 
 
 
 
  
 
 
 
 
power factor











5
Numerical on determination of back emf and load angle of a salient pole
synchronous motor
1. A 20 MVA, 3-phase, 11 kV, 12-pole, 50 Hz, star connected salient pole synchronous motor has
direct and quadrature axis reactance per phase of 5 Ω and 3 Ω respectively. Determine the
excitation voltage and load angle when the machine is drawing (a) rated power at 0.5 pf lagging,
(b) half the rated power 0.8 pf leading, (c) rated power at 0.95 pf lagging.3
6
3
0
1 1
11 10
, 6350.85
3
20 10
( ) , 1049.73
3 3 11 10
, 60 .
6350.85 0.5 1049.73 3
6350.85 0.5 1049.73 0
a
L
a q
a a
Termianl voltage V V
S
a Armaturecurrent I A
V
Power factor angle lagging
Vsin I X
tan tan
Vcos I R




 

 

  
  

     
       
0
0 0 0
0
0
0
36.51
, 60 36.51 23.49
1049.73 (36.51 ) 624.6 .
1049.73 (36.51 ) 843.69 .
, 6350.85 (23.49 ) 0 624.6 5 2701.72
d a
q a
b q a d d
Load angle
I I sin sin A
I I cos cos A
Soback emf E Vcos I R I X cos V
  





    
   
   
        
V
Ia
Eb




6
3
6
3
0
1 1
11 10
, 6350.85
3
0.5 20 10
( ) , 524.86
3 3 11 10
, 36.87 .
6350.85 0.5 524.86 3
6350.85 0.5 524.86 0
a
L
a q
a a
S
b Armaturecurrent I A
V
Power factor angle leading
Vsin I X
tan tan
Vcos I R




 

 
 
  
  

    
  
    
0
0 0 0
0
0
0
46.67
, 46.67 36.87 9.8
1049.73 (36.51 ) 381.77 .
1049.73 (36.51 ) 360.2 .
, 6350.85 (9.8 ) 0 360.2 5 8167.1 .
d a
q a
b q a d d
Load angle
I I sin sin A
I I cos cos A
Soback emf E Vcos I R I X cos V
  



 
 
 
    
   
   
        
V
Ia
Eb
 


7
3
6
3
0
1 1
11 10
, 6350.85
3
20 10
( ) , 1049.73
3 3 11 10
, 18.2 .
6350.85 0.95 1049.73 3
6350.85 0.31 1049.73 0
a
L
a q
a a
S
c Armaturecurrent I A
V
Power factor angle lagging
Vsin I X
tan tan
Vcos I R




 

 

  
  

    
  
    
0
0 0 0
0
0
40.4
' ' , ' ',
, 18.2 40.4 58.6
1049.73 (40.4 ) 680.16 .
1049.73 (40.4 ) 843.69 .
, 6350.85 (
a b
d a
q a
b q a d d
Since is negative I lags V but leads E
Load angle
I I sin sin A
I I cos cos A
Soback emf E Vcos I R I X cos

  



 
  
 
    
   
   
     0
50.6 ) 0 680.16 5 6711.43V   
V
Ia
Eb




8
2. A 200 kW, 3-phase, 11 kV, 10-pole, 50 Hz, star connected salient pole synchronous motor has
direct and quadrature axis reactance per phase of 1.2 pu and 0.8 pu respectively. Determine the
excitation voltage and load angle when the machine is drawing rated power input of 200 MW at
0.98 pf leading.
0
1 0
0
1 1 0
0
, 1 0 .
, (0.98) 11.48 .
, 1 11.48
1 0.199 1 0.8
45.55
1 0.98 1 0
, 45.55 14.48
a
a q
a a
Termianl voltage V pu
Power factor angle cos leading
Armaturecurrent I
Vsin I X
tan tan
Vcos I R
Load angle




  

 
 
 
 
     
          
    0 0
0
0
0
34
1 (45.55 ) 0.714 .
1 (45.55 ) 0.7 .
, 1 (34 ) 0 0.714 1.2 1.685 .
, , 1.685 11 18.535 .
d a
q a
b q a d d
b
I I sin sin pu
I I cos cos pu
Soback emf E Vcos I R I X cos pu
So linevalueof back emf E kV




   
   
        
  

9
Power Equation of a cylindrical pole Alternator: -
Real Power input, Pi = Real part of ‘Si’, and reactive power input, Qi = Imaginary part
of ‘Si’.
0
0
0 2
0
, 0 ,
,
,
0
0
, * 0 ( ) ( )
b b
s s
b b
a
s s
b b
i a
s s s
Let V V
E E
Z Z
V EV E
I
Z Z
V E VEV
Complex power input tomotor S V I V
Z Z Z





  



 
 
 
  
 
   
 

  
        
 
2 2
cos( ) cos( ), sin( ) sin( )b b
i i
s s s s
VE VEV V
P and Q
Z Z Z Z
          
fV
fI
Field CircuitArmature Circuit
bE
aIar sx
V

10
 Condition for maximum active power input is
 So, maximum power that the synchronous motor can deliver 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 ( )
b b
i
s s s
b b
b
s s s s s
VE VEV
P P
X X X
VE VEV V V
and Q Q V E cos
X X X X X
 
  
    
       
0 0
0 sin( ) 0 180 180idP
d
     

         
2 2
0
cos cos(180 ) cosb b
im
s s s s
VE VEV V
P
Z Z Z Z
    

11
0
0 2
, *
0
0
, ( ) ( )
b a
b b
a
s s
b b b
b
s s s
Complex power output of themotor S E I
V EV E
I
Z Z
V E VE E
So S E
Z Z Z



   

 
 

   
 

  
        
 
So, active power output is:
2
cos( ) cos( )b b
s s
VE E
P
Z Z
    
So, reactive power output is:
2
sin( ) sin( )b b
s s
VE E
Q
Z Z
    

12
Condition for maximum active power output is
So, maximum active power output is,
2
0
0 cos( ) cos( ) sin( ) 0 0b b
s s
VE EdP d
d d Z Z
        
 
 
             
 
2
max cos( )b b
s s
VE E
P
Z Z
 

13
Active power output is:
2
0 0
cos(90 ) cos(90 ) sinb b b
s s s
VE E VE
P
X X X
    
Reactive power output is:
2 2
0 0
sin(90 ) sin(90 ) cosb b b b
s s s s
VE E VE E
Q
X X X X
     
For machines having negligible armature resistance, i.e.
0
90s a s s sZ r jX jX X    

14
Waveforms are plotted by taking V = 1.0 pu, Eb = 0.98 pu,
Zs = 0.4 + j 2 pu and varying load angle from 00 to 1800.

17
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(84.3
1.005
b
s
b b
om
s s
Supply voltage V V
Back emf E V
Z j
Mechanical power developed W
But mechanical power developed
VE E
P
Z Z
  
 
 
    
   
  


2
0 0
0
400
) cos(84.3 )
1.005
26.9


 
 

18
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
 

Eet3082 binod kumar sahu lecture_33

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