- The document discusses parallel operation of two alternators and load sharing between them. - It explains how the load current is divided between the alternators based on their impedances and induced emf values. - It also analyzes the effects of changing the mechanical power input to one alternator while they are operating in parallel. This causes a circulating current and changes how the power is shared between the alternators under both no load and loaded conditions.

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-16

2. 2
Learning Outcomes: - (Previous Lecture_15)
 To understand the need of parallel operation of alternators.
 To analyse the various conditions to synchronize an alternator with
another alternator or an infinite bus.
 To analyse the various methods of synchronization.

3. 3
Learning Outcomes: - (Today’s Lecture_16)
 To analyse the parallel operation of two alternators with change in
mechanical power input:
a. At no load.
b. Under loaded condition.
 To solve numerical on load sharing.

4. 4
Parallel Operation two Alternators: -
 After synchronizing two alternators, excitation voltages E1=E2, phase angles
δ1 = δ2 and frequencies of induced emfs f1 = f2.
 So, the resultant voltage in the local circuit forms by E1, Zs1, E2, and Zs2 is
zero i.e. in the local circuit, E1 and E2 are in phase opposition, but E1 and E2
are in phase with respect to external circuit (load terminal).
L
O
A
D
Zs1 Zs2
CI
1aI 2aI
LI
1f 2f 2 2E 1 1E 
1E 2E
External
Circuit Local
Circuit
1E
2E
 So, at no load, Ia1 = Ia2 = Ic = IL = 0.
 Under loaded condition, the total
load will be shared by both the
alternators.

5. 5
Load Sharing/Load Division: -
 
 
1 2
1 1 1 2 1 1 21 1 1
2 2 12
2 1
2
2
, (1)
, (2)
, (3)
, (3), ,
L a at L L
a a a a a at s L s L s L
a aL s L
a L
a
L s
Terminal voltage,
V I Z I I Z
But E V I Z I I Z I Z I Z Z I Z
Similarly E I Z Z I Z
E I Z
From equation wehave I
Z Z
   
       
  
 

 
    
 
 
        
 
  




6. 6
 
 
 
2
2 1
1 1 1
2
2
2 1
1 1 1
2 2
2
2
1 11
2 2
2 2
12 1 1 2
1
2
(2),a
a L
a L s L
L s
aL L
a L s
L s L s
L L
a L s
L s L s
L s L s L s s L
a
L s
Putting I inequation we get
E I Z
E I Z Z Z
Z Z
E Z I Z
E I Z Z
Z Z Z Z
Z E Z
I Z Z E
Z Z Z Z
Z Z Z Z Z Z Z Z E Z
I
Z Z

 
 
 
 

 



   

    
 
 
     
   
    
  
  
1 22
2
L s L
L s
E Z E Z
Z Z
 
 


7. 7
 
 
 
1 2 1 2
1
1 2 1 2
2 1 2 1
2
1 2 1 2
1 22 1
1 2 1 2
1 2
1 2
(4)
, (5)
(6)
(
Terminal Voltage,
Circulatingcurrent
L s
a
s s L s s
L s
a
s s L s s
s s L
t
s s L s s
C
s s
E E Z E Z
I
Z Z Z Z Z
E E Z E Z
Similarly I
Z Z Z Z Z
E Z E Z Z
V
Z Z Z Z Z
E E
I
Z Z
  

  

 

 

 
  
  
 
 
  
 
 
 
 
 
 



7)

8. 8
Numerical on Load Sharing: -
1. Two three phase star connected alternators, running in parallel supplying a load of impedance
(4+j2) Ω/phase having per phase impedances of Zs1 = (0.2+j3.2) Ω, and Zs2 = (0.25+j4.5) Ω. Their
per phase excitation emfs are 230 V with E1 leading E2 by 100. Compute current & power
shared by each alternator. Also calculate the terminal voltage.
Solution: -
 
 
0 0
1 2 1 2
1 2 1 2
0
1
1 2 1 2
2 1 2 1
0
2
1 2 1 2
230 10 , 230 0 , (0.2 3.2) , (0.25 4.5) , (4 2) .
27.59 41.5
13.77 53.41
s s L
L s
a
s s L s s
L s
a
s s L s s
Givendata
E V E V Z j Z j Z j
E E Z E Z
I A
Z Z Z Z Z
E E Z E Z
I A
Z Z Z Z Z
Load curre
  

  

            
 
  
    
 
 
  
    
 
0 0 0
1 2 27.59 41.5 13.77 53.41 40.57 37.42 .a aLnt I I I
 
          

9. 9
 
1 22 1
0
1 2 1 2
1 2 0
1 2
* 0 0
1 1
181.44 10.86 .
5.2 8.34
1:
3 3 181.44 10.86 27.59 41.5 14.
Terminal Voltage,
Circulatingcurrent
s s L
t
s s L s s
C
s s
t a
E Z E Z Z
V Volt
Z Z Z Z Z
E E
I A
Z Z
Complex Power shared by alternator
S V I
 

 

 
 
    
 

  

        
* 0 0
2 2
* 0 0
223 4.807
2 :
3 3 181.44 10.86 13.77 53.41 5.522 5.069
3 3 181.44 10.86 40.57 37.42 17.75 9.87
t a
t L
kW j kVAR
Complex Power shared by alternator
S V I kW j kVAR
Total Complex Power Delivered to Load
S V I kW j kVAR

         
         

10. 10
Effect of change in Mechanical Power Input: -
L
O
A
D
Zs1 Zs2
CI
1aI 2aI
LI
1f 2f 2 2E 1 1E 
Under no load condition: -
 If the mechanical power input of Alternator-1 is
increased, the acceleration of the rotor of
Alternator-1, makes emf E1 to lead E2 by an angle α
(let). Phasor diagram in the local circuit after
increasing the mechanical power input is shown in
Figure 3.
Figure 1.
Equivalent circuit.
Figure 2.
Phasor Diagram before
increasing the
mechanical power input.
Figure 3.
Phasor Diagram after
increasing the mechanical
power input.
Local
Circuit
1E
2E
Local
Circuit
1E
2E
rE
CI
1
2


11. 11
 Now, the net voltage in the local circuit is no more zero, rather it is
 This resultant emf gives rise to circulating current
 Since armature winding is highly inductive and less resistive, circulating current ‘IC’
will lag ‘Er’ by an angle close to ‘900’.
 From the phasor diagram it is clear that, ‘IC’ lags ‘E1’ by ‘φ1’ (less than 900) and
‘E2’ by ‘φ2’ (more than 900).
 So power delivered by Alternator-1 and Alternator-2 are:
 Since ‘φ1’ is less than 900 and ‘φ2’ is greater than 900, power delivered by
Alternator-1 is positive and that by Alternator-2 is negative. So, Alternator-1 is
in generating mode and Alternator-2 is in motoring mode.
1 2rE E E
  
 
1 2
r
C
s s
E
I
Z Z




1 1 1
2 2 2
cos
cos
C
C
P E I and
P E I





12. 12
 So, the electromagnetic torque developed in Alternator-1 (for which the
mechanical power input has been increased), causes retarding effect and that in
Alternator-2 produces accelerating effect.
 So there will be an automatic synchronizing action to retard the faster machine
and to accelerate the slower machine in order to maintain synchronism.
Under loaded condition: -
 Initially assume that both the Alternators have same
induced emfs and sharing the equally i.e.
 When the mechanical power to Alternator-1 is
increased, its electrical output increases. Since the total
load is constant, power delivered by Alternator-2 must
decrease i.e.
L
O
A
D
Zs1 Zs2
CI
1aI 2aI
LI
1f 2f 2 2E 1 1E 
ZL
1 1
1 2 1 2 1 2, , sin sin
2 2
L t t L
a a
s s
E V E V PI
E E I I P P
X X
 

   
      
1 1
1 2 1 2 1 1 2 2 1 2, , sin sin , . .t t
a a
s s
E V E V
E E I I P P i e
X X
   
   
     

13. 13
 Since field excitations has not been changed, the induced emfs and the terminal
voltage will remain constant. However, Alternator-1 will share more power than
Alternator-2, as the increased mechanical power to Alternator-1 makes δ1 > δ2.
Vt
Ia1=Ia2
jIa1Xs = jIa2Xs
1 2 
1 2 
E1=E2
IL
Phasor Diagram before increasing
the mechanical power input.
1 1
2 2
at s
at s
E V I X
E V I X
  
  
 
 

14. 14
Phasor Diagram after increasing
the mechanical power input.
1 1
2 2
'
'
at s
at s
E V I X
E V I X
  
  
 
 
Vt
I/
a1
jI/
a1Xs
2
E1
IL
I/
a2
E2
jI/
a2Xs
1
1
2 
Ia
IC
-IC

15. 15
 Power supplied by Alternator-1 and Alternator-2 are:
 Since Ia1> Ia2 and φ1> φ1,P1 > P2.
 So, increase in mechanical power input to Alternator-1, increases its armature
current and improves its power factor.
/
1 1 1
/
2 2 2
cos
cos
t a
t a
P V I and
P V I





16. 16
Thank you