Electrical Power Systems

dcr
z 1
d0
cr
zbPi-Bgdd d
dm
= C - P d
d0
b g
\ Pi-K . Pmax
sind
z0 d
dcr
zb cr
ig
d
d
dm
z dcr
2. sin - i
b g
= K Pmax d P dd
dcr
dm
dcr
dm
\ Pidd + z Pi dd
= Pmax. K1zsin ddd + Pmax K2zsin ddd d0
dcr
Power System Stability 293
\ {(d cr – d 0 ) + (d m – d cr )} P i = P max K 1 .(– cosd cr + cosd 0 ) + K 2 P max (–cosd m + cosd cr )
\ (d m – d 0)P i = K 1 P max(cosd 0 – cosd cr) + K 2P max(cosd cr – cosd m) ...(11.54)
\ Since P i = P maxsind 0 ,
\ (d m – d 0 ) sind 0 = K 1 cosd 0 – K 2 cosd m + cosd cr (K 2 – K 1 )
\ cosd cr =
1
bK2 – K1g
b g
d – d sin d + K cos d – K cosd
...(11.55)
m 0 0 2 m 1 0
rom ig. 11.13, we have
Pi = Pm sind0 = K2Pmax sindm = K2Pmax sind¢ m
\ sind0 = K2 sind¢ m = K2 sin(p – dm ) ...(11.56)
With K1, K2 and d0 specified, the critical clearing angle may be obtained from eqn. (11.55)
and eqn. (11.56)
Example 11.8: A 50Hz, synchronous generator capable of supplying 400 MW of power is
connected to a large power system and is delivering 80 MW when a three phase fault occurs at
its terminals, determine,
(a) The time in which the fault must be cleared if the maximum power angle is to be –85º.
Assume H = 7 MJ/MVA on a 100 MVA base.
(b) The critical clearing angle.
Solutions:
(a) rom eqn. (11.48),
\ sin d 0 =
P i = P max sind 0
Pmax = 400
3 MW, P 80
i =
3 MW
80I
HG 3 KJ
400I
HG 3 KJ
= 0.2
\ d 0 = 11.54º = 0.2 radian