Load Sharing amongst two alternators connected in parallel

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-17

2. 2
Learning Outcomes: - (Previous Lecture_16)
 To solve numerical on load sharing.
 To analyse the parallel operation of two alternators with change in
mechanical power input:
a. At no load.
b. Under loaded condition.

3. 3
Learning Outcomes: - (Today’s Lecture_17)
 Load sharing among two alternators from speed-load characteristic.
 To solve some numerical on load sharing based on speed-load characteristics.

4. 4
Speed-load characteristics: -
 Load sharing between the two
alternators running in parallel is
controlled by speed-load characteristic
of their prime-movers.
 When the load on an alternator
increases, the electromagnetic torque
(counter torque) developed in it,
decreases the rotor speed.
 As frequency is directly proportional to
rotor speed, the increased load on the
alternator, decreases the frequency of
generation.
 Similarly increase in mechanical power
input, increases the frequency of
generation.
 Speed-Load characteristic of an
alternator is shown in Figure-2.
Frequency
Load on
Alternator
f0
fl
P
Figure 2
Speed-Load
Characteristic
Figure 1
Alternators running in
parallel
G1
G2
YB R
LOAD
Prime Mover-1
Prime Mover-2
Tm
Te

5. 5
 Initially, assume that both the alternators are sharing the load equally i.e. P1 = P2 = P/2
and the operating frequency is ‘f’. This can be clearly seen the speed-load characteristics
of the alternators. Under this condition line 1-1/ and 2-2/ represents the speed-load
characteristics of alternator-1 and alternator-2 respectively.
 At this frequency ‘f’, points a & b on the lines 1-1/ and 2-2/ indicate that the alternators
are sharing the load equally.
 When the mechanical power input to alternator-1 in increased, its speed-load
characteristic shifts upward as shown by the line 3-3/.
 Now the operating point of alternator-1 shifts to point ‘c’ at which it shares a power P1
/
which is greater than P1 and that of alternator-2 shifts to point ‘d’ at which it shares a
power P2
/ which is less than P2.
Effect of change in mechanical torque on load sharing: -

6. 6
Effect of change in mechanical
torque on load sharing
between the alternators
Load on
Alternator-1
Load on
Alternator-2
a b
c d
x y
1
2
P
P  2
2
P
P 
/
1P /
2P
//
1P //
2P
f
f1
Frequency
1
1'
2
4
2'
4'
3'
3

7. 7
 But, the total power supplied by both the alternators is P1
/ + P2
/ = P but at a frequency ‘f1’
which is greater then nominal frequency ‘f’.
 Now to bring back the operating frequency to ‘f’, the mechanical power input to alternator-2
must be decreased which shifts its speed-load characteristic (2-2/) downward as shown by the
line 4-4/.
 Now the operating point of alternator-1 shifts to point ‘x’ at which it shares a power P1
// which is
greater than P1
/ and that of alternator-2 shifts to point ‘y’ at which it shares a power P2
// which is
less than P2
/.
 Now, the operating frequency is restored to ‘f’ which is the nominal frequency. And again, the
total power supplied by the alternators is P1
// + P2
// = P .

8. 8
Numerical: -
1.Two alternators operating in parallel operating in parallel have the following data:
Alternator 1: Capacity 2 MW, frequency drops from 50 Hz at no load to 48 Hz at full load.
Alternator 2: Capacity 2 MW, frequency drops from 50.5 Hz at no load to 48.5 Hz at full
load.
Prime-mover speed regulation is linear.
a. Calculate how a total load of 3.6 MW is shared by each alternator. Also find the
operating bus frequency at this load.
b. Compute the maximum load that these two alternators can deliver without
overloading either of them.
Solution: -
Rating of alternator-1, W1 = 2 MW.
Drop in frequency of alternator-1 for a load of 2 MW = 50 – 48 = 2 Hz.
Let the alternator-1 is sharing a load of P1 MW.
So, drop in frequency of alternator-1, for a loads of P1 MW = (2/W1) x P1.
So, operating frequency of alternator-1 is, f1 = 50 – (2/W1) x P1.
Similarly operating frequency of alternator-2 is, f2 = 50.5 – (2/W2) x P2.

9. 9
Frequency
Load on
Alternator-2
f02 = 50.5 Hz
W2 = 2
MW
fl2 = 48.5 Hz
Load on
Alternator-1
f01 = 50 Hz
fl1 = 48 Hz
W1 = 2
MW
P2P1

10. 10
Since they are operating in parallel, their operating frequency must be same.
So, f2 = f1 = f
=>50 – (2/W1) x P1 = 50.5 – (2/W2) x P2
 50 – P1 = 50.5 – P2
 P2 – P1 = 0.5
 P2 = P1+0.5 (1)
But the total load to be shared is 3.6 MW, i.e. P1 + P2 = 3.6 (2)
Solving equations (1) and (2) we have,
2 P1 = 3.1, => P1 = 1.55 MW and P2 = 3.6 – 1.55 = 2.05 MW.
So, alternator-2 is overloaded.
Operating frequency, f = f1 = f2 = 50 – (2/W1) x P1 = 50 – (2/2) x P1 = 50 – 1.55 = 48.45 Hz.
b. From the speed-load characteristics shown in the figure, it is clear that the minimum operating
frequency at which both the alternators can be operated is 48.5 Hz i.e. full load frequency of
alternator-2 as operating frequency less than 48.5 Hz, will overload the alternator-2.
So, at 48.5 Hz, alternator-2 is fully loaded sharing an output power of 2 MW.
At 48.5 Hz power shared by alternator-1 will be:
f1 = 50 – (2/W1) x P1
=>48.5 = 50 – (2/2) x P1
 P1 = 50 – 48.5 = 1.5 MW.
So, the total load the parallel combination = 2 + 1.5 = 3.5 MW.

11. 11
2. Three alternators, operating in parallel at a bus frequency of 50 Hz, share a load of 140
MW as follows:
Alternator-1: 40 MW, Alternator-2: 40 MW, Alternator-3: 60 MW.
Each alternator is rated at 100 MW. Their governor settings are so adjusted as to give the
following fall in frequency from no load to rated load:
Alternator-1: 1.25 Hz, Alternator-2: 1.5 Hz and Alternator-3: 2.0 Hz.
How will they share a load of 250 MW. Also calculate the operating frequency at this load.
Solution: -
Bus frequency when the alternators are sharing a load of 100 MW is 50 Hz.
Rating of all the alternators, W1 = W1 = W3 = 100 MW.
Fall in frequency in alternator-1 from no load to full load i.e. for 100 MW is 1.5 Hz.
So, when the alternator-1, is delivering a load of 40 MW decrease in frequency is 1.25/100 x 40
= 0.5 Hz.
So, the no-load frequency of alternator-1, f01 = 50 + 0.5 = 50.5 Hz.
Similarly, f02 = 50 + 1.5/100 x 40 = 50.6 Hz and
f03 = 50 + 2.0/100 x 60 = 51.2 Hz
Let the power shared by P1, P2, and P3 be the power shared by alternators-1, 2 and 3
respectively when they are sharing a total load of 250 MW.

12. 12
So, operating frequency of alternator-1 is:
f = 50.6 – drop in frequency when sharing a load of P1
= 50.6 – (1.25/100) x P1
 P1 = 4040 – 80 f (1)
Similarly, for alternator-1,
P2 = 3373.34 – 66.667 f (2)
And for alternator-3,
P3 = 2560 – 50 f (3)
But, P1 + P2 + P3 = 250
 4040 – 80 f + 3373.34 – 66.667 f + 2560 – 50 f = 250
 f = 49.44 Hz. (Operating Frequency)
Now, P1 = 4040 – 80 x 49.44 = 84.8 MW
P2 = 3373.34 – 66.667 x 49.44 = 77.2 MW
And P3 = 2560 – 50 x 49.44 = 88 MW.

13. 13
Thank you