Effect of varying mechanical power input to an 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-23

2. 2
Learning Outcomes: - (Previous Lecture_22)
 Effect of varying field excitation: V-Curve and inverted V-curve of an
alternator.
 Variation of reactive power (delivered/absorbed by an alternator to/from
infinite bus) with excitation.

3. 3
Learning Outcomes: - (Today’s 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

4. 4
Effect of change in varying mechanical power input to an
alternator connected to an infinite: -
Infinite BusAlternator
XS
V = Constant
f = Constant
Ia
E
Field Excitation
of Alternator
Prime Mover-1
Te
Tm
VE

 Initially assume that the alternator is
delivering an active power ‘P’ to the infinite
bus and operating:
i. With an excitation emf ‘E1’.
ii. With load angle ‘δ1’.
iii. At lagging power factor with power
factor angle ‘φ1’.
iv. With armature current ‘Ia1’.

5. 5
 Increase in mechanical power input increases the load angle there by increasing the active
power output delivered by the alternator. Expressions of active and reactive power of an
alternator are
 As the excitation is kept constant, E is constant. In the power expression, ‘V’ and ‘Xs’ are
also constant.
 So, increase in mechanical power increases the load angle ‘δ’ and ‘Iacosφ’.
 This can be easily understood from the phasor diagram:
, ( )a a
s s
EV V
P sin VI cos Q Ecos V VI sin
X X
       

6. 6
Phasor diagram with varying
mechanical power input
Ia1
Ia2
Ia3
Ia4
Ia5
jIa1Xs
jIa2Xs
jIa3Xs
jIa4Xs
jIa5
Xs
E1
E2
E3
E4
E5
δ1
δ2 δ3
δ4
1
2
5
0
3 0 
4
δ5 = 900
V

7. 7
 Increase in mechanical power input increases the active power output delivered by the
alternator.
 For different mechanical power inputs the behaviour of the alternator at different points
are as follows: -
 In lagging power factor region, if the mechanical power is increased, power factor is
improved. Whereas, in leading power factor region, increase in mechanical power input
reduces the power factor.
 After supplying the required active power demand, reactive power can be varied by
varying the field excitation.
Load Angle Induced EMF Armature Current Power Factor
δ1
E1 = E2 =
E3 = E4 = E5
Ia1 cosφ1 (lagging pf)
δ2 Ia2 cosφ2 (lagging pf)
δ3 Ia3 cosφ3 = 1 (unity pf)
δ4 Ia4 cosφ4 (leading)
δ5 Ia5 cosφ5 (leading)

8. 8
Plotted by varying the
Rotor angle from 100 to 900, and
taking E = 1.2 pu V = 1.0 pu,
Xs= 0.3 pu

9. 9
Plotted by varying the
Rotor angle from 100 to 900, and
taking E = 1.2 pu V = 1.0 pu,
Xs= 0.3 pu

10. 10
1. A three phase star connected alternator with synchronous impedance of (0 + j 5) Ω/phase
is connected to an 11 kV system. The alternator power output is found to be 10 MW and
reactive power output as 3 MVAR. Compute the armature current, load power factor,
excitation emf and load angle.
Solution: -
*
*
*
6
0
3
3
0 0
, 10 3 3
, 11
(10 3) 10
, 547.97 16.7
3 11 103
11 10
, 0 547.97 16.7 (0 5)
3
L a
L
a
L
a s
Complex Power S MW j MVAR V I
TerminalVoltage V kV
S j
Armaturecurrent I A
V
Excitationemf per phase E V I Z j


  
   

                

         0
0
0
7605.28 20.19
, 3 7605.28 13172.72 13.17
, 20.19
(16.7 ) 0.953
L
V
So linevalueof excitationemf E V kV
Load angle
Power factor cos lagging

  
   

 

11. 11
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
3.49 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
Z j
So thearmaturecurre

 

   
   

1.8 cos(26.39) 0.896nt is pu and power factor is leading

12. 12
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
 
  
    

13. 13
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 ) 0.822leading

14. 14
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
11 10
/ , 6350.85 0
3
1600 10
, 83.98 .
3 11 10
, 0
0
6350.85 ( ( ) sin( )) (0 30) 6350.85 0
6050.85 2519.4 ( ) 2
a
a s
a
Busbar voltage phase V Volt
Armaturecurrent I A
Voltageregulationis zero E V
V I Z V
I cos j j
j cos
 


 

  

 
 
  
   
      
   519.4 ( ) 6350.85 0sin   

15. 15
 2 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
6350.85 2519.4 6350.85
11.44
2 2519.4 6350.85
, cos(11.44
2
2 2
sin cos
sin sin cos
sin
So load power factor
 
  
 
    
      
  
      
 0
3
) 0.98 .
, 3 3 11 10 83.98 0.98 1568.03L a
leading
Power delivered bythemachine P V I cos kW

      

16. 16
Thank you