Electrical Power Systems

We know, K1 = Pmax
In ig. 11.14,
P
max
K2 = Pmax
P
max
during the fault
before the fault
after the fault
before the fault
d 0 = power angle at the time of the fault
d cr = critical power angle when fault is cleared
d m = maximum angle of swing
Area A1 = Area A2 in ig. 11.14 and rom eqn. (11.55)
cosdcr =
K
1
- K b m - gsin + K cos m - K cos
Power System Stability 295
d d0 d0 2 d 1 d0
b 2 1g
Initially, the generator is supplying 0.45 pu MW of P max Therfore,
P i = 0.45 P max = P maxsind 0
\ d 0 = 26.74 or 0.466 radian.
Now P max =
E V
g t
When the fault occurs |V t| becomes 0.25|V t|
Hence K 1 = 0.25
After the fault is cleared, with K 2 = 0.70, we have
xd
P i = K 2 P max sin d m ¢
\ d m ¢ = sin –1
L
NM
Pi
K P
2 max
O Q P = sin
\ d m ¢ = 40º or 0.698 radian.
\ d m = p – d m ¢ = 2.443 radian.
\ cosd cr =
1
b07 . - 025 . g
L
NM
045 . P
–1 max
070 . Pmax
b2443 . - 0466 . gsin b0466 . g+ 07 . cos b2443 . g
– 025 . cos 0466 .
b g = 0.29
\ d cr = 73.14º or 1.276 radian. Ans.
Example 11.10: A generator operating at 60 Hz delivers 1 pu MW power. Suddenly a three
phase fault takes place reducing the maximum power transferable to 0.40 pu MW where as
before the fault, this power was 1.80 pu MW and after the clearance of the fault, it is 1.30 pu MW.
Determine the critical clearing angle.
O Q P