5.

13 Applications of equal area criterion

To illustrate the applications of equal area criterion for transient stability, different types of
disturbances on the system consisting of a finite machine connected to an infinite bus are
considered and discussed.

a) Sudden change in the mechanical input

ES X Infinite bus
OO000 00000 VI0
Fig.5.6. One line diagram
Consider a loss free
synchronous generator supplying an infinite bus through a purely
reactive transmission line as shown in Fig.5.6. The electrical
power transferred is given by
Pu= Pm Sin 8. The power angle curve of the system is as shown in Fig.5.7.
PA
A

P
P=Pm sin8
P

Fig.5.7. Power angle curve

Pis the electrical equivalent of the mechanical power
input, and So is the initial load angle
and the system is under steady state condition i.e. the machine is
running at synchronous speed,
which is represented by point a on the power angle curve.
Now, let the mechanical power be
Suddenly increased to P . Due to this increase in the shaft input, the speed increases due to
accelerating power Fa = i and the rotor speed increases. Hence, the angle ö is also
 nis
ncreased Let this new load angle be ð), which corresponds to Pi.e. at point b. Since the rotor
rcelerating and running above synchronous speed, the load angle further increases and
ac

shoots point b. When the load angle is more than 8, the output P, is more than the input
ove

the accelerating power becomes negative and the rotor decelerates and the rotor speed

but ð continues to increase to 82 where the speed again reaches the

decreases,
dec
at point c,

synchronous speed.
becomes
After this stage, the rotor swings in the reverse direction till the load angle again
increases along
80. During this reverse speed decreases along the line cb, but
swing, the rotor
The rotor speed
at a. This swinging is repeated.
the line ba and reaches synchronous speed again in the system
before coming to rest at point b, as the losses
oscillates about the point b, finally
progressively damp the amplitudes of the swing.
A2 which is equal to Aj. Hence, for the system
The system is stable because,
there is an area
is
Az A1. If the mechanical input
area A2 in A, such that =

to be stable, there should be an that,

could be increased so

there is a limit upto which this mechanical load

suddenly increased, shown in Fig.5.8.
and that limit is P" as
the system can be stable
Pt

A
Pi

8,(8,)
Fig.5.8
have a limiting value of Az in
in mechanical load, we
sudden increase the system
of dmax, upto which,
For this value of condition, & reaches a value
to Aj. Under this and there will be
further
A, which is equal mechanical load, increases öz
in the becomes unstable.
further increase Hence, the system
Can be stable. Any that Az = A1.
in A, such
available which the system
no area A2 which is
in the
mechanical input is (P"-P) upto
sudden increase the unstability of the
load results in
maximum
ence, the mechanical
in the
further increase
be stable. Any
system. (5.24)
( P=Pm Sin 81)
Om=T-8, = n
-

sin
P
b) Sudden increase in the shaft load of a synchronous motor operating on an infinite bus

(M)
Infinite bus Syn. rotor
Fig.5.9
Fig.5.9 represents the one line diagram of a power system consisting of an infinite bus
supplying power to a synchronous motor through a transmission line.
PA A A
Output of infinite bus
o r input to motor
P
Output of
P motor shaft

Fig.5.10
Fig.5.10 represents the power angle curve of a synchronous motor connected to an infinite
bus. The electrical input to the motor is P. = Pm sin 8 and P,is the shaft output under steady state
conditions. a is the initial operating point, where the input and output curves intersect each
other. is the initial operating point. When the load on the shaft is suddenly increased to P.
the final operating point isb. Instead of instantaneously setling at b, there will be overshoot till
the accelerating area Az is equal to the decelerating area A1. In other words, ô will increase to &
and oscillates between 8o and before finally settling down at 81, after oscillating about b for
sometime. If an area Az = Aj is not available in the portion A, of P-8 curve above Pline,
sufficient accelerating power is not there and oscillations will not occur and the system becomes
unstable i.e. the motor continues to decelerate and finally stops. Thus the motor will be stable

only if A2 A1. Let Pi be suddenly increased to P" such that A,= A1. Then P" represents the
maximum load that the motor shaft can carry without unstability and P"-P; represents the

sudden maximum increase in the shaft load upto which the motor will be stable. Any increase in
this sudden load causes the motor unstable and the motor decelerates and ultimately stops.

Power system with a sustained line fault:

c)
Line A
Infinite
o- bus
Linc B

Fig.5.11
 Fig.5.11 re epresents a
generator connected to an
infinite bus through two lines A and B.
ines are nomal, a maximum power of
Whenthe Pmi Can be transmitted. When one of the lines
ned, the reactance of the system increases and hence the maximum
B is o
power that can
ferred decreases from Pmi Pm2
say
to
be

,Output, normal conditions

Output, fault on
A

Pi Input

Fig.5.12
is delivering power
the power angle curves, when the power system
Fig.5.12 represents is fault on the system. Pi represents the
conditions and when there
a
under normal working the input and normal
point, which is the interaction of
constant input. a is
the initial operating acceleration is zero. When
angle. The initial angular
the initial displacement
output curves. So is
curve.
below a on the fault output
directly
the fault is on, the initial operating
point drops b,to
then an
the of fault. There is
the instant of occurrence

remains at o at is accelerated
Ihe displacement angle and hence, the generator
the line ab
Pu represented by bc. At point c,
the
aCCelerating power Pa Pi
-

m o v e s along
=

The operating point becomes

the displacement angle
8 increases. decreases and
a c c e l e r a t i o n gradually
d
b to c, the and angle
z e r o . From point the infinite bus
c a U n g power is than that oft
is more
of the generator power. The
C. At this point, the speed representing retarding
at becomes negative
does so, P, displacement
O U n u e s to increase. As it when A2 A|. The angular
=

at d,
becomes zero of the generator
d, and therefore the speed
c c r e a s e s upto
point and
torque he system is
c. The system
is
still a retarding to b, through
There is d
point is S2. moves
from
point damping.
operating there is no
the and d, if
Continues to
CS
to decrease
ded and between
oscillate
b
becomes
to point finally
stable. The opei
e operating point
continues

losses and
the operating
to
diminish due
Actually the ne oscillations
angle is d.
stablished at c, when the displacement
 PA

Normal output curve

Input
7 Output curve
on fault

Om
Fig.5.13
If the initial load on the generator is increased, the operating point
is shifted above on the
normal output curve asshown in the Fig.5.13. The areas
A increases and Az will decrease. The
angle & increases to the maximum angle öm at which
Aj A2. The greatest value of Pi, the
=

generator could have without the generator going out of

fault is that value which makes stability during the existence of the
Sm Occur at the intersection of the
output curve i.e. at e, which makes A= A2. This is the input curve and the fault
critical condition in which both the
and the acceleration become zero speed
this condition occur is the
simultaneously at an angle öm. This
value of P; which makes
transient stability limit of the
system, with a sustained line fault.
If the initial
input-P, is further increased, then Az would be smaller than A and the
generator wouid reach a
point e, where the acceleration is zero with the speed above normal.
Consequently 8 would continue increase and he
to
positive. The system becomes unstable. Thus
accelerating power would again becom
limit of the system under fault using equal area criterion, the transient
conditions can be found. stability
d) Line fault with
subsequent clearing:
P

Pre-fault output curve

Post-fault output curve

P
Input
Fault output curve

Fig.5.14
 14 shows three power
Fig.5.14 shows angle curves ii) for the fault
(i) for pre-fault condition
nd (i
ondition and ii) for post-fault condition, with the faulted line disconnected from the system.

the initial operating point, which is the intersection of the input curve and the pre-fault
a is
So is the initial operating angle. Occurrence of the fault causes the operating point
. Tve,

fault output curve. The accelerating power ab causes the operating

from a to b on the
1o drop
the fault output curve from b to c. We assume that the circuit breakers open
t to move along
0 cleared is known as
fault is cleared. The displacement angle dc at which the fault is
tc and the
the
and the corresponding time te at which
the fault is cle ared is known as

the clearing angle

the post-fauit output
point now jumps from c to e, which is on
learing time. The operating
The generator swings between öm
upto point f, where Az Ai.
=

and moves along this curve

curve

and do and finally settles down at de. The system is stable.

f to move to the
would cause point
With the fault cleared, a higher input (or initial output)
still higher value of P; would lead to

the stability limit, would coincide with h. A

right until at
to the right is an increase in the
which would cause f to move

unstability. Another factor,

time of the fault, resulting in larger clearing angle 0c.
Clearing at which the fault
particular maximum clearing angle
there is a
any given initial load,
ror
that the system could
be stable, and
circuit breakers, so
D e Cleared by the opening of the time is known as the
de and the corresponding
g i e 1s known as the critical clearing angle breakers open beyond e
Or the circuit
beyond oe
lclearing time toe. If the fault is cleared
the system becomes unstable.
the clearing time (sum
not known directly. Instead,
Norm angle is
ally, the critical clearing from the knowledge
determine the clearing angle
of relay ume
time and time) is known. To the time of clearing. A pre-
an
fof the clearing
breaker
must be determined upto
swing curves curves is not
al tng time, the this purpose.
The use of swing
calcul curve may also
be used for criterion.
Wing area
by equal
Completely eliminated but
Cminated but is reduced to a
minimum
 WORKED EXAMPLES
5.18 A loss free generator supplies 50 MW to an infinite bus, the steady state limit of
the system being 100 MW. Determine whether the generator will remain in
synchronisnm, if the prime mover input is abruptly increased by 30 MW. Use equal
area criterion. (Kuvempu University)
Solution: P=Pm sin do. ie. 50 100 sin do. . do= 30° = 30 xonrad=0.5235 rad
180
P = Pm Sin &1, i.e. 80 = 100 sin 8
P
:. 8 = 53.13° = 53.13°xrad
180 m

= 0.9272 rad
A
P
A P,d6 S(r-P)d5 P
= (80-100sin 8) do

80 (8- 8o) -100-cos8

80 (81-8o)- 100 [-cos o1 + cos Ôo Fig.5.17
80 (0.9272-0.5235)-100[-cos 53.13° +cos 30°]
32.3 26.6 =5.7 MW-rad

om T-ô =T-0.9272 =2.2144 rad =2.2144 x180 degrees 126.87

d
Az P,dß [(P,-P)dð [(Psinð-P)ds
= =

(100sin8-80) dÑ = 100 [-cos n+ cos ol -80 [ôm-®ol

= 100 [-cos 126.87°+ cos 53.13°]-80 [2.2144 09272] =

120- 102.97
= 17.03 MW-rad

Since A2> A1, the system is stable.

5.19 A generator
with constant excitation
supplies 30 MW through a step-up
transformer and a high
voltage line to an infinite bus. If the steady state limit of
the system is 60 MW, estimate the maximum
permissible sudden increase of
generator output, if the stability is to be maintained. The resistances of the
generator, transformer and the line may be neglected. (Bangalore University
Solution: P =Pm sin ðo i.e. 30 60 sin do
 30= = 30 x rad=0.5235 rad
80 (Refer Fig.5.17)
Pm Sin d = 60 sin 8,
P =

A Pdd = J{P-P) dä
80 80

PS-oo) Pm (-cos 8 +cos ô,)-

P= Pm sin
=P sin 8 o-do) -

60 [-cos 8 + cos l
60 sin oi (O- Oo) -

60 [-cos 8 + cos
ôn] -- (1)

Ap (Pn sinð- Pi) =Pm (-cos n + cos ) -

60 sin ô, (8m-)

60 [-cos (Tt -ö) + cos o]-60 sin ô,

[m -81 -8,]
60 [cos d1 + cos 8] -60 sin ®i [m -281]
60 [2 cos ô|-60sinð, (n -26) (2)
We must find the value of d} such that Az = A1. dj has to be found by trial and error
method.
Let 8 60° = 60 x rad = 1.0472 rad
180
A = 60 sin 60° (1.0472-0.5235)-60 (-cos 60° + cos 30°) = 5.2507 MW-rad
Az = 60 (2 cos 60°) - 60 sin 60° (T-2 x1.0472) = 5.5862 MW-rad

Az> Ai
For A= A,, 8, has to be increased by a small margin

Let 8 = 62° = 62 x = 1.0821 rad

180
On substitution, Aj = 5.7996 MW-rad and A=4.5574 MW-rad
Here, Az< A
Hence, , has to be reduced
and A=5.3716 MW-rad
Let d = 60.4° for which A =5.3614 MW-rad
Here A> Aj but Az A
0 may be taken as 604° = 1.0542 rad
increase of generator output
is given by
Max.permissible sudden 22.17 MW
P-P= Pm sin ô-30 60 sin
604° 30 =52.17- 30 =

a network to an intinite bus. The power

5.20 A generator supplies 50 MW
through
Determine the maximunm additional load
100 sin d.
angle diagram is given by P. =

so that the system remains stable. (Mangalore

nat can be suddenly applied,
University)
Ans: 36.95 MW
5.21 A synchronous motor is drawing 30% of the maximum steady state power from an
infinite bus. If the load on the motor is suddenly increased by 1006, would the
synchronism be lost? 1f not, what is the maximum torque angle about the ne
steady state rotor position? (Karnataka University)
Solution: Refer Fig.5.17

P=0.3 Pm and P=0.6 Pm

A= (0.6Pm -Pm sin) dÕ =

Pm 10.6 (8 -

80)-(-cos õ + cos Ón)] -(1)

A=(P sin8-0.6Pm) dÑ = P,mI-cos 8m + cos ô - 0.6 (Sm-ô)1

(2)
P= Pm sin 1.e. 0.3 Pm Pm sin o
T
.So= 17.46° =
17.46 x rad 0.3047 rad
180
P= Pm Sin 81 i.e. 0.6 Pm= Pm Sin ði

.. ôj =36.87° =36.87 x rad =

0.6435 rad
180
omT-8, = t-0.6435 = 2.498 rad
Substituting the above values in equations (1) and (2), we
get
A=0.0493 Pm MW-rad and Az 0.4873 Pm MW-rad
As
A2> A1, the system is stable.
The maximum
torque angle about the new steady state rotor
Az=A1. Substituting ô, for ôm in equation (2) and position is ö, for which
simplification, we get equating equations (1) and (2) and on
cos o+0.6
8 = 1.1368
&, has to be solved by trial and -(3)
error method.
Let &= 50° TT
= 50 xrad = 0.8726 rad
180
Substituting this value in equation (3), we
LHS =
cos 50° +0.6 get
x 0.8726 =
1.1663, which is not
Let 82= 60° Tt equal to RHS i.e. 1.1368
=
60 x = 1.0472 rad
180
Then, cos 60° +0.6 x1.0472 =1.9273 * RHS
Let &= 58° =
58
xo
180
rad =
1.0123 rad
Then cos 58° +0.6 x1.0123 =1.1373 =
RHS
8 = 58°
 state
is connected to an infinite bus and the steady
ASVnchronous generatOr o n the
torque angle is 30°. The initial load
ctability limit is 100 MW. The initial
s MW. Find out
is 50 MW. If the load is suddenly increased by 30
generator
remains stable or not. (Mysore University)
whether the system

Solaution: Refer Fig.5.17

Pm Sin Óo i.e. 50 100 sin do. S = 30° =0.5235 rad

P 80 100 sin ô1. o = 53.13° =0.9273 rad
P=Pm Sin 81 i.e.
126.87° 2.2143 rad
&m T-ôj
=
=

A (P-P, sin 8) dÕ (80-100sin8) d8

=

80 (81 -

8o) -

100 (-cos 81+ cos So)

100 (-cos 53.13° + cos 30°) = 5.6936 MW-rad
= 80 (0.9273 0.5235) -

A2 [ (P sin-P) dÕ= [ (100sin8 -80) do

= 100 (-cos8) 80 (ön-61)
80 (2.2143 -0.9273) = 17.04 MW-rad
100 [-cos 126.87° + cos 53.13°] -

ince, A2 > Aj, the system is stable

 WORKED EXAMPLES
5.6 A 60 Hz, 4 pole turbo-generator rated 500 MVA, 22 kV has an inertia constant
H 7.5 Mw-sec/MVA. Find (a) the kinetic energy stored in the rotor at
synchronous speed (b) the angular acceleration, if the electrical power developed is
400 MW, When the input less rotational losses is 7,40,000 H.P and (c) moment of
inertia (d) inertia constant M and angular acceleration. (Bangalore University)
Solution: a) Wkin HG 7.5x500 3,750 MW-S
= =
 735.5 x 10 MW544 27 MW
P 7,40,000 HP=7,40, 0X)x
MW
P.=P P,= 544.27-400144 27
GH Wn 3750 03472 MW seclelec degree
M 180f 180x 60
I80f
144 27 -415 52 elec degree/sec
Angular
acceleration
=

o :

M 0.3472
dt
120f 120x600 = I800 pm
c) Ns P 4
27nx1800
21tNs rad/ sec 60
= 188.5 rad/ sec
60

Wkin lo, . I= 2W 2x3.750x10°

kg-m 211 08 x 10 kg-
188.5
inertia constant
22 kV has a n
Hz. 4 pole turbo-generator rated 500 MVA,
A 60 which i s
5.1 The angular acceleration is 415.52
elec.degree/see

7.5 MW-sec/MVA. i n that

Find the change in 8 in electrical degrees
of 15 cycles.
constant for a period generator i s
Assume that the
the in rpm at the end of 15 cycles.
period and speed before the 15
system and has n o accelerating torque
synchronized with a large
University)
cycles period begins. (Mysore
d8
Solution: a= 415.52 elec.deg/ sec
dt
do
On integration
, do d'o=2 xx415.52 =831.04 dt
dt dt2 dt
do 28.827 V5. do
o 28.827 dt
ie
= 831.04 8,
dt dt

(15 eycles = =0.25 sec)

28.827 t
On integration, 26 =

3.6034°
.o= 12984°
8 = 14.4138 x0.25 =

N 2 0 _ 1 2 0 x 6 0 = 1,800 pm
Ns=
P 4
do 28.984x3.6034 = 103.876 ele.deg/ sec
28.827 8 =

dt
1.8129
1.8127 rad/ sec=
1.8129 rev/ sec =0.1442 rev/ sec
= 103.876 x =
4Tt
180
8.656 rpm = dN
Change in speed 0.1442 x60 =

I,808.656 rpm
1,800 +8.656
=

he speed in rpm = Ns +dN =

5.8 turbo-generator rated 20 MVA, 13.2 kV has a n inertia

constant
A S0
Hz, 4 pole rotor at
=9.0 kW-sec/KVA. Determine
the kinetic energy stored in the
if the input less the rotational
synchronous Deternmine the acceleration,
speed.
 losses is 25,000 H.P and the electrical
power developed is 15,000 kW. If the
e
acceleration computed for the generator is constant for a
period of 15 cycles.
determine the change in the torque angle in that
period and the speed at the end of
15 cycles. Assume that the
generator is synchronized with a large system and has
no
accelerating torque before the 15 cycles period begins. (Karnataka University)
Solution: Wkin= GH 20x9 180 MW-S
=

P= P-P, =(25,000x 0.735 15,000) x

10 MW =3.3875 MW
GH 180
M= = 0.02 MW-sec/ elec.deg
180f 180x50
From swing equation

d8P_ 3.3875 169.375 elec.deg/ -

dt2 M 0.02 sec
2.do d'o2xx
do 169.35,
On integration
dt dt2 dt

do 338.75 8+ A, where A is a constant

Att 0,8=0,=0 A =0

do 1
=338.25 8 =18.4 8,
dt do18.4 dt, do= 18.4 dt
1
On integration, 283 =184 8 9.2t or 884.36 t
t =0.3 sec, .8= 84.36 x 0.3 = 7.6°
50
To find speed

= do= 184 x V8 18.47.6 = =

50.798 elec.deg/ sec
dt

50.798 x
180
rad/sec =
50.798 x x rev/ sec
180 4T
50.798x60 4.233 rpm, 120f 120x50
rpm =
=
1,500 pm
180x4 P 4
.Rotor speed at the end of 15 cycles =
1,500 +4.233 =
1,504.233 rpm
5.9 A 50 Hz, 4 pole turbo-generator rated 100 MVA, 11 kV has an inertia constant or
8 MJ/MVA. (a) Find the stored energy in the rotor at synchronous speed (b) If the
mechanical input is suddenly raised to 80 MW for an electrical load of 50 MW,
find the rotor acceleration, neglecting mechanical and electrical losses. (c) If the
acceleration calculated is maintained constant for 10 cycles, find the change in the

torque angle and rotor speed in rpm at the end of this period. (Gulbarga

University)
 Stored
S o l u t i o n :
energy= Wkin= GH =
100 x8 =800 MJ or MW-S
b) Pa= 80- 50 = 30 MW
M= GH 800
=
0.0888 MW-secil elec. deg
180f 180x 50
d8 30
337.83 elec.deg/ sec
*d? M 0.0888

c) 2 =2x337.83,
dt
On integration, do 6756
dt

do 675 = 25.98 8. do
25.98 dt
dt

82 do = 25.98 dt, On integration, 28 25.98 t. t= 0.2 SS

0.2 S
50

8 = 12.99t, 8=168.74 = 168.74 x 0.22 =0.6749°

To find speed

=dt d = 25.98 s =25.98 v6.749 =67.49 ele.deg/ sec

= 67.49 x rad / sec = 67.49 x x rev/ sec

180 180 4t
67.49
614x 60 rpm =4.687 pm
180x4

speed at the end of 10 cycless120X 4.687 = 1,504.687 rpm

=
Rotor
4

MVA at 0.8 p.f lag, when a

5.10 A 60 Hz, 4 pole turbo-generator is delivering rated 500
fault reduces the electrical power output by 40%. Determine the accelerating

torque in Nm at the time of fault. (Kuvempu University)

P =0.6 x400 240 MW

Solution: P= 500 x 0.8 = 400 MW. P 0.6

P= Pi- P, = 400 240 = 160 Mw

2tx 120x60

= 2TUNS 4 27Tx1800188.5 rad/ sec

60 60 60
160x10 848.81 x10° Nm
Pa T, i.e. T 188.5

connected to a synchronous motor having

5.11 having H 6 MJ/MVA is
A generator =

reactances. The generator is delivering

H =4 MJ/MVA through a network of
a

when a fault occurs, which reduces the delivered

power of 1.0 p.u to the motor,
power to 0.6 p.u. Determine
the angular acceleration of the generator with respect
to the motor. (Mangalore University)
 Solution: The equivalent H is given by

Hi2 = H,H, 6X42.4 MJ/ MVA

H+H 6+4

M12 in p.uu= GH
GH 1.0x2.42 67 x 10 p.u-sec/ elec.deg
180f 180x50
d'6
2.667 x 10 2 = 1.0-0.6 =0.4 p.u
M = Pi- P i.e. dt
dt
0.4 = 1,500 ele.deg/sec= 26.18 rad/ sec
dt 2.667x10

5.12 A 2 pole, 50 Hz, turbo-generator has a rating of 60 MW at 0.85 p.f lagging. Its

rotor has a moment of inertia constant H and its angular momentum M.

(Karnataka University)
Solution: G= =
60
70.59 MVA, N,=120f120X30
2
=3,000 r.p.m.
0.85 P

(s 27TNS2Tx5000 314.16 rad/ sec

60 60

WkinIW
2
=
x8800 x 314.16=434.26 x
10=434.26 MJ
H kin 434.26 = 6.15 MJ/ MVA
G 70.59

M=
GH
70.59x6.1 -0.0482 MJ-sec/ ele. deg
180f 180x50

5.13 A 50 Hz, 4 pole turbo-generator rated 20 MVA, 11 kV has an inertia constant of

9 kW-S/KVA. Find the acceleration, if the input minus the rotational losses is
26,800 H.P and the electrical power developed is 16 MW.

Solution: P= 26,800x735.5x 10 MW =19.71 Mw

GH 20x9
M = 0.02 MW-secl elec. deg
180f 180x50

d
dt
-MRM- 19.71-16
0.02
185.5 ele.deg/ sec
5.14 A2 pole, 50 Hz, 11 kV turbo-generator has a rating of 100 MW, p.f 0.85 agging.
The rotor has a moment of inertia of 1,000 kg-m'. Caleulate H and M. (Bangalore
University)
100
Solution: G 0.8517.65MVA, Ns 120x503,000 rpm
2
l co
14 W-Ss 2TX 3000314.16 rad/ sed
1Ox= x1,000x314.16
Xl,000x3
50
os
Wkin
x10° W-S 493.48 MW-S
493.48
=

Wkin
493.48 4.194 Mw-S/ MVA
HG 117.65

GH117.65x4.194 = 0.0548 MW-sec/ ele. deg

=
180x50
M 180f below are
interconnected via a short

.furbo-alternators
with ratings given
transmission line.
0.8 lagging, I = 30,000 kg-m?
50 Hz, 60 MW,
G.:4 pole, 0.85 lagging, I=10,000 kg-m
50 Hz, 80 MW, on a base of
G: 2 pole, H of the single equivalent machine
inertia constant
Calculate the
(Gulbarga University)
200 MVA.

: Ns =
120x5= 1,500 r.p.m.
Solution: G 2 MW-S
(27tX1,500=370.11x 10° W-S=370.11
WulxIx a =x30,000 | *
60
2

G2 Ns=
120x50-3000r.p.m.
2
MW-S

10,000 x
27x3,000) x10 MW-S =493.48
Wkin2 x
60
=
370.11+493.48 = 863.59 MW-S
Wkin1t+Wkin2
is given by: Wkin=
Total kinetic energy
200 MVA base
is given by
Inertia constant H on

H=kin B63.59=4.318 MW-S/MVA

G 200 MVA and each
each rated 80
generators,
5.16 Power station-1 has
four identical rated
station-2 has 3 sets, each
while power
constant 7 MJ/MVA,
having an inertia to be regarded as a
are located close enough
Z00 MVA, 3 MJ/ MVA.
The stations constant of the
Calculate the inertia studies.
machine for stability
Single equivalent
(Mangalore University)
equivalent machine on 100 MVA base.

(80x7) x4 2,240 MJ
lution:
Power Power station-1: Wkinl =G,H, =

x3 = 1,800 MJJ
Power station-2: Wkin2=GH2 = (200x3)
4,040 MJ
otal Wkin =(2,240 +1,800) =

is given by
100 MVA base
ofthe equivalent m/c
on
**
nertia constant

H in 4040 MJ/ MVA

40.4
G 100