Electrical Power Systems

302 Electrical Power Systems
Solution:
H
180 f
rom eqn.(11.66), we have
rom eqn. (11.68)
a (0) =
10 .
37 . ´ 10
200 / 60
= = 3.7 × 10
180 ´ 50
–4 sec2 /degree
P a(0) = 1.0 – 0.0 = 1.0 pu
–4
= 2702.7 degree/sec 2
rom eqn. (11.69) with Dt = 0.05 sec,
rom eqn.(11.70),
rom eqn. (11.71),
Dw r (1) = 2702 .7 * 0.05 = 131.5 degree/sec.
wr (1,0) = 0 + 1315 .
2
= 67.55 degree/sec.
Dd (1) = 67.55 * 0.05 = 3.3775 degree
To complete the first iteration, we determine the initial power angle, d 0, as follows. Before
the fault
P max =
Then 4 sind 0 = 1.0
12 . ´ 10 .
030 .
= 4.0 pu
or d 0 = 14.477 degree.
With this value of the initial power angle, from eqn. (11.72), we have
or the second interval
d (1) = 14.4775 + 3.3775 = 17.855 degree.
P a (1) = 1.0 – 0.0 = 1.0
a (1) =
10 .
37 . ´ 10
–4
= 2702 degree/sec.
Dw r (2) = 2702 ´ 0.05 = 135.1 degree
w r (2,1) = w r (1) + Dw r(2)
2
= 202.65 degree/sec.
= w r (0) + Dw r (1) + Dw r(2)
2
Dd (2) = w r (2,1).Dt = 202.65 ´ 0.05 = 10.1325º