Electrical Power Systems

Power System Stability 281
(a) ind the stored energy in the rotor at synchronous speed.
(b) If the input to the generator is suddenly raised to 60 MW for an electrical load of 50 MW,
find rotor acceleration.
(c) If the rotor acceleration calculated in part (b) is maintained for 12 cycles, find the change
in torque angle and rotor speed in rpm at the end of this period.
Solution:
(a) Stored energy = GH
G = 100 MVA, H = 10 MJ/MVA.
\ Stored energy = 100 × 10 = 1000 MJ.
(b) P a = P i – P e = 60 – 50 = 10 MW.
we know, M = GH
180 f
Now M. d d
= Pa
2
dt
\
\
2
2
5
54
2
d d
= 10 2
dt
d d 10 ´ 54
= 2
dt 5
100 ´ 10
=
180 ´ 60
= 5
54
= 108 elect-deg/Sec 2
MJ-Sec/elect deg.
\ a = acceleration = 108 elect-deg/Sec 2 Ans.
(c) 12 cycles = 12
60
= 0.2 sec.
Change in d = 1
2 a.(Dt)2 = 1
2 ´ 108 ´ (0.2) 2 elect-degree.
= 2.16 elect-degree.
Now a = 108 elect-deg/Sec 2
108
\
a = 60 ´
360º
rpm/Sec = 18 rpm/Sec.
\ Rotor speed at the end of 12 eycles
= 120 f
= 120 60
P
HG
bg
+a Dt
´
+ 18 ´ 0. 2
4
rpm
= 1803.6 rpm Ans.
I KJ