Numerical on Alternator

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-25

2. 2
Learning Outcomes: - (Previous Lecture_24)
 To solve numerical on Alternators

3. 3
Learning Outcomes: - (Today’s Lecture_25)
 To solve numerical on Alternators.

4. 4
6. A three-phase, 10 kVA, 380 V, 4-pole, 50-Hz, has synchronous impedance of (1+j15)
Ω/phase. It supplies a load of 8 kW at rated voltage and 0.8 power factor lagging.
Determine the percentage voltage regulation.
Solution: -
0 0 0
380
/ , 219.4 .
3
8000
15.19 .
3 3 380 0.8
219.4 0 15.19 36.87 (1 15) 406.98 28.19
| | | | 406.98 219.4
100 100 85.5%.
| | 219.4
a
L
a s
Terminal voltage phase V Volt
P
I A
V cos
E V I Z j
E V
Voltage Regulation
V

  
 
  
   
          
 
    

5. 5
7. A 30 kVA, 50 Hz, 4-pole alternator connected to infinite bus at rated voltage and has Xd =
1.0 pu, and Xq = 0.6 pu.
(a) For zero excitation, find the output in terms of rated power that the machine can
deliver without loosing synchronism.
(b) Under the condition of (a), estimate the current and power factor.
Solution: -
2
2
0
( ) / :
1 1
2 .
2
1 1
, 0, 2 .
2
1 1 1
, , 45 , ,
2 0.6 1
d q d
q d
a For salient polealternator expression for active power phaseis
EV V
P sin sin
X X X
V
With zeroexcitation E so P sin
X X
Now for maximum power output so P si
 


 
   
 
 
 
   
 
 
 
   
 
0
(90 ) 0.3333 .n pu

6. 6
E
V
jIdXd
jIqXq
E/
Id
Iq

 Vcos
Vsin
Vcos q aI R
Ia
IaRa
jIaX
q
IaRa
O
A
B
D
CG

jIaX
q
H
Phasor Diagram at
leading power factor

7. 7
0
/
/ 0 0
/
( ) , 0.3333 0.3333 (1) ( 1 0 )
,
45 1 0 0.6 ,
a a
a d q a aa d q a q
a
b Also P VI cos I cos asV pu
It is knowntousthat E V I R j I X j I X and E V I R j I X
E jI Excitationis zero sothealternator isoperating at leading pf
E
 

       


     
      
      
  0 0 0
/ 0 0 0
/ 0
/ 0 2 2
45 1 0 0.6 (90 )
45 (1 0.6 (90 ) 0.6 (90 ))
45 (1 0.6 0.6 )
0.6
45 (1 0.6 ) (0.6 )
1 0.6
0.6
1 0.6
a
a a
a a
-1 a
a a
a
a
a
I
E I cos j I sin
E I sin j I cos
I cos
E I sin I cos tan
I sin
I cos
I sin

 
 

 






    
        
      
 
         
 
 
 
 
0
(45 ) 1tan 

8. 8
   2 2 2 2
2 2
0.6 1 0.6
0.6 0.3333 1 0.6
1 0.6 0.3333
1.3334 (2)
0.6
, (1) (2),
1.3334 0.3333
1.3334 0.3333 1.3744
0.3333
(1),
1.3744
a a
a
a
a a
a
I cos I sin
I sin
I sin
From equations and we get
I sin I cos
I pu
Fromequation cos
 


 

  
   
 
  
  
   
  0.2425leading

9. 9
( )
( )
( 0)
, , 0, .
, ,0 .
a d q d qa d q d q a
d qd q
OR
b Voltageequationof salient polealternetor is
E V I r j I X j I X V j I X j I X as r
Again excitation E asthereis noexcitation
So thevoltageequationis V j I X j I X
Fromthe phasor dia
       

  
       

  
0
0
2 2 2 2
,
1 45
0.7071 .
1
1 45
, 1.1785 .
0.6
, 0.7071 1.1785 1.3744 .
0.7071
,
1.1785
d d d
d
q q q
q
a d q
-1 -1d
q
gram
Vcos cos
Vcos I X I pu
X
Vsin sin
Again Vsin I X I pu
X
So I I I pu
I
Internal power factor angle tan tan
I






    

    
    
   
      
0
0 0 0
0
30.96 .
, , 45 30.96 75.96
(75.96 ) 0.2425
So load power factor angle
Load power factor cos leading
  


    
 
V
jIdXd
jIqXq
E
Ia
Iq
Id
δ
δ
jIqXq
jIdXd
ψ


10. 10
8. A cylindrical rotor machine is synchronized with an infinite bus at rated voltage. Now the
steam input to the machine is increased till it begins to operate at its rated current.
Synchronous reactance of the machine is 1.2 pu.
(a) Calculate the load angle, power factor, active power and reactive power delivered
to the infinite bus.
(b) Without any change in steam input, how can this alternator be made to deliver no
reactive power to the bus? Find the excitation voltage and load angle under this
loading condition.
Solution: -
0
( ) , 1 0 .
, ' ' ' ' .
, ( ).
, ' ', '
s
a Beforethemechanical power wasincreased E V pu
Increaseinmechanical power increasestheload angle keeping theexcitationemf E constant
V
Reactive power delivered Q Ecos V
X
So increasein makesQ



 
  
 
' ( ), . . .
, 1a
ve as Ecos V i e thealernator operates at leading pf
At rated output I



 
 

11. 11
2 2 2 2
0
0 0
0
,
( ) 1 1 (1 1.2)
0.28
2 2 1 1
73.74 .
1 74.73 1 0
, 1 36.87
1.2
, 0.8 .
, 1 1 0.8 0.8
a s
a
s
a
From leOAB
E V I X
cos
E V
E V
So I
jX j
So power factor cos leading
Active power delivered P VI cos pu and
Reactive power




 


    
  
   
 
   
   
 
    
, 0.6 .aabsorbed Q VI sin pu 
Ia
V
E
δ
jIa
Xs

Vcos
 Vsin

O A
B
C

12. 12
0
0 0
0 0
2 2
1 1 0 1 1.2
1 1 0 1.2 (90 )
1 (1 1.2 (90 )) 1.2 (90 )
1 (1 1.2 ) 1.2 )
1.2
1 (1 1.2 ) (1.2 )
1 1.2
a s
-1
OR
E V j I X
j
cos j sin
sin j cos
cos
sin cos tan
sin
Compairing themagnitudesof bo
 
 
  
  

  

  
 
       
      
      
    
 
       
 
2 2
2 2
2
2
0
:
,1 (1 1.2 ) (1.2 )
1 1 2.4 (1.2 ) (1.2 )
1 1 2.4 1.2
1.2
36.87
2.4
-1
ththe sides
So sin cos
sin sin cos
sin
sin
 
  


  
    
   
 
    
 

13. 13
0
0
*
0 0
, , 1 36.87 .
1.2 1.2 0.8
, 73.74 .
1 1.2 1 1.2 0.6
, 1 0 1 36.87 (0.8 0.6) .
, , 0.8 ,
a
-1 -1
a
So armaturecurrent I pu
cos
Load angle tan tan
sin
Complex power S V I j pu
So complex power P pu and reactive p



 
 
   
         
 
        
 
 0.6 ( )ower Q pu Absorbed from Infinitebus 

14. 14
0
2
0 0 0
22
0
2 2
, 0.8 0 .( )
, ,
1 0 0.8 0 1.2 1.3862 43.84
, , 1.3862 , 43.84
a
a s
So I pu Asthenew pf isunity
Now excitationemf
E V j I X j
So newexcitationemf E pu and newload angle

  
 
         
 
( )
.
, ,
. .
b Without changing the steaminput thereactive power supplied canbechanged by
changing the field excitation
Now withincreased excitation theactive power supply will remainconstant
and reactive power is zeroi e thealternato
2 1
2 2 1 1
2
2
.
, 0.8
0.8
1 1 0.8
0.8
a a
a
a
r isoperating at unity pf
So P P pu
I cos I cos
I
I pu
 
 
  
   
 

15. 15
9. A number of alternators are working in parallel with terminal voltage equal to the rated
value. One of the synchronous machine which has a synchronous reactance of 50 % and a
resistance of 1%, delivers 70% of its rated power output. If the emf of this alternator is
1.2 times the terminal voltage, determine the power factor of the machine at which it is
operating.
Solution: -
0
, 1 0 .
, 0.7 .
Sin ,
.
, 0.7 .
a
a
Terminal voltage V pu
Armaturecurrent I pu
cetheinduced emf is morethanthetermilal voltage let thealternator isunderexcited
and thereforethe power factor islagging
So I pu
Synchronousimpedance



 

  
, (0.01 0.5)Zs j pu 

16. 16
V
Ia
IaRa
Iaxs


E
IaZs
O
A
B
 
2
2 2 2
2 2 2
0
0 0
0
,
1.2 1 0.7 0.01 0.5
( )
0.9656
2 2 1.2 1
15.07 .
1 15.07 1 0
, 0.7 25.82
0.01 0.5
, 0.63 .
a s
a
s
From leOAB
E V I Z
cos
E V
E V
So I
jX j
So power factor cos lagging



 


   
 
  
   
 
   
    

 

17. 17
0
, 1 0 .
, 0.7 .
Sin ,
.
, 0.7 .
a
a
OR
Terminal voltage V pu
Armaturecurrent I pu
cetheinduced emf is morethanthetermilal voltage let thealternator isunderexcited
and thereforethe power factor islagging
So I pu
Synchronousimpedan



 

  
0
0
2
, (0.01 0.5)
1.2 1 0 0.7 (0.01 0.5)
1.2 1 0 0.7( ) (0.01 0.5)
1.2 (1 0.7 0.01 0.7 0.5 ) (0.7 0.5 0.7 0.01 )
1.2 (1 0.007 0.35 ) (0.35
a s
ce Zs j pu
E V I Z
j
cos jsin j
cos sin j cos sin
cos sin
 
  
    
  
  
 
 
        
       
            
      2 0.35 0.007
0.007 )
1 0.007 0.35
-1 cos sin
cos sin tan
cos sin
 
 
 
 
   
  

18. 18
2 2
2 2 2 2
2
1.2 (1 0.007 0.35 ) (0.35 0.007 )
1.2 1 0.000049 0.1225 0.014 0.0049 0.7 0.1225
0.000049 0.0049
1.44 1 0.000049 0.1225 0.7 0.014
0
cos sin cos sin
cos sin cos sin cos sin cos
sin sin cos
sin cos
   
      
  
 
     
       
 
     

 
2 2
2 2
0
0 0 0
.7 0.014 0.3175
0.014
, 0.7 0.014 0.3175
0.7
1.146 0.4535
26.97 1.146 25.82
,
-1
-1
sin cos
B
It is knowntousthat Asin Bcos A B sin tan
A
So sin tan
sin
lagging
So power factor cos
 
  




 
  
     
  
  
    
  
  
   
 0.63lagging

19. 19
Thank you