Electrical Power Systems

294 Electrical Power Systems
and d 1 = 85º = 1.48 radian
rom eqn. (11.49),
cos d c = cos d 1 + (d 1 – d 0) sin d 0
\ cos d c = cos(1.48) + (1.48 – 0.2) sin (0.2)
\ cos d c = 0.343
\
rom eqn. (11.52),
dc = 1.22 radian.
tc =
H dc-d pfP
b g
2 0
P i (3 f) = 80 MW = 80
100
H = 7 MJ/MVA
i
b g
= 0.8 pu MW.
\ tc =
2´ 7´ 148 . -02
.
p ´ 50 ´ 0. 8
\
(b) rom eqn. (11.53)
tc = 0.377 secs = 377 ms. Ans.
dcr = cos –1 [(p – 2d0 ) sind0 – cosd0 ]
\ d cr = cos –1 [(p – 2 ´ 0.2) sin (0.2) – cos (0.2)]
d cr = cos –1 (– 0.43) = 115.46º = 2.01 radian. Ans.
Example 11.9: A syhronous genarator is connected to a large power system and supplying 0.45
pu MW of its maximum power capacity. A three phase fault occurs and the effective terminal
voltage of the generator becomes 25% of its value before the fault. When the fault is cleared,
generator is delivering 70% of the original maximum value. Determine the critical clearing
angle.
Solution:
ig. 11.14 gives the P-d characteristic.
ig. 11.14: P – d characteristic of Example 11.9.