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-24

2. 2
Learning Outcomes: - (Previous Lecture_23)
 Effect of change in varying mechanical power input to an alternator
connected to an infinite bus under loaded condition.
 To solve numerical on Alternators

3. 3
Learning Outcomes: - (Today’s Lecture_24)
 To solve numerical on Alternators.

4. 4
2. A three phase star connected alternator with synchronous impedance of (0 + j 1.25) pu
delivers rated current to infinite bus-bar at 0.8 pf lagging. For the same excitation, find
the current and the power factor just before falling out of step.
Solution: -
V
E
jIaXs
δ5 = 900
Ia

0
0
0 0 0
, 1 0 .
, , 1 36.87 .
,
1 0 1 36.87 (0 1.25) 3.49 29.74
.
, 3.49
a
a s
Let thebusbar voltage V pu
So armaturecurrent I pu
Excitationemf per phase
E V I Z j pu
Withthe sameexcitationmechanical poewer input isincreased
So E pu
  
 
  
          

0
0 0
0
.
,
, 90 .
3.49 90 1 0
1.8 26.39
0 1.25
,
an
s
is maintained constant
So maximum power thealternator candevelop
before falling out of step is at load angle
E V
Newarmaturecurrent for thisconditionis I pu
Z j
So thearmaturecurr

 

   
   

1.8 cos(26.39) 0.896ent is pu and power factor is leading

5. 5
3. A three phase star connected turbo alternator, having a reactance of 10 Ω/phase, has an
armature current of 200 A at unity power factor when running on 11 kV, constant
frequency bus-bar. If the excitation is raised by 20% without changing the prime-mover
input, find the value of armature current and power factor.
Solution: -
3
0
0
0 0 0
11 10
/ , 6350.85 0
3
, 200 0 .
,
6350.85 0 200 0 (0 10) 6658.33 17.48 .
20%. ,
a
a s
Busbar voltage phase V Volt
Armaturecurrent I A
Excitationemf per phase
E V I Z j Volt
Excitationisincreased by So Newvalueof excitati


  

  
 
         
1.2 6658.33 7990
. .' ' .
/ , 6350.85 200 127010
n
a
s
onemf E Volt
Changeinexcitaioncannot changetheactive power output i e P isconstant beforeafter changeinexcitation
EV
Active power phase P VI cos sin W
X
 
  
    

6. 6
1 1 0
0
0 0
0
127010 10
, , 14.5 .
7790 6350.85
, 7990 14.5 .
7990 14.5 6350.85 0
, , 243.26 34.7
(0 10)
, , ( 34.
s
n
n
a
s
PX
So thenewload angle sin sin
E V
So E
E V
So newarmaturecurrent I A
Z j
So new power factor cos
  

 

   
       
 
   
    

  0
7 ) 0.822lagging

7. 7
4. A 3-phase, star connected alternator, rated at 11 kV, 1600 kVA, has a synchronous
reactance of 30 Ω/phase. When delivering full load current at a certain power factor, the
voltage regulation of the machine is zero. Estimate the load power factor and power
delivered by the machine.
Solution: - 3
0
3
3
0
11 10
/ , 6350.85 0
3
1600 10
, 83.98 .
3 11 10
, 0. .
0
6350.85 0 83.98 (0 30) 6350.85 0
a
a s
Busbar voltage phase V Volt
Armaturecurrent I A
Voltageregulationis zero E V It is possiblewithonlyleading pf
V I Z V
j

 

  

 
 
  
   
       
 6350.85 83.98( ( ) sin( )) (0 30) 6350.85 0
6050.85 2519.4 ( ) 2519.4 ( ) 6350.85 0
cos j j
j cos sin
 
 
     
    

8. 8
 2 2
2 2 2
2 2 2
1 0
6350.85 2519.4 2519.4 6350.85 0
6350.85 2 2519.4 6350.85 2519.4 2519.4 6350.85 ^ 2
6350.85 2519.4 6350.85
11.44
2 2519.4 6350.85
, cos(11.4
2
2 2
sin cos
sin sin cos
sin
So load power factor
 
  
 
    
      
  
      
 0
3
4 ) 0.98 .
, 3 3 11 10 83.98 0.98 1568.03L a
leading
Power delivered bythemachine P V I cos kW

      

9. 9
5. For a synchronous machine, excitation emf = 2.3 pu, terminal voltage = 1.0 pu, Xs = 1.5
pu. Find the maximum values of active and reactive power that can be delivered by the
alternator and the corresponding load angles. Also find the load angle at unity power
factor.
Solution: -
0
/ :
, 90 .
2.3 1
,max 1.5333
1.5
/ :
( ),
s
max
s
Expression for active power phaseis
EV
P sin sotheactive power will bemaximumat
X
So umumactive power P pu
Expression for reactive power phaseis
V
Q Ecos V sothereactive power will bema
X
 

 

 
  0
0
0 .
1
, (2.3 1) 0.8667
1.5
1
, 0 64.23
2.3
max
-1 -1
ximumat
So Q pu
V
For unity power factor Ecos V cos cos
E

 

  
   
        
   
Ia
jIaXs
V
E
δ
Ecosδ

10. 10
Thank you