Electrical Power Systems

290 Electrical Power Systems
Solving eqn. (ii) iteratively, we get d 1 = 50º
Now
P ef = P max sind 1 = 500 sin(50°) = 383.02 MW
Initial power developed by the machine was 69.6 MW. Hence, without loss of stability, the
system can accommodate a sudden increase of
P ef – P e0 = 383.02 – 69.6 = 313.42 MW/phase
=3 ´ 313.42 = 940.3 MW (3 f) of input shaft power.
11.7 CRITICAL CLEARING ANGLE AND CRITICAL CLEARING TIME
If a fault occurs in a system, d begins to increase under the influence of positive accelerating
power, and the system will become unstable if d becomes very large. There is a critical angle
within which the fault must be cleared if the system is to remain stable and the equal–area
criterion is to be satisfied. This angle is known as the critical clearing angle. Consider the
system of ig. 11.9 operating with mechanical input P i at steady angle d 0. (P i = P e) as shown by
the point ‘a’ on the power angle diagram of ig. 11.10.
ig. 11.9: Single machine infinite bus system.
Now if a three phase short circuit occurs at the point of the outgoing radial line, the
terminal voltage goes to zero and hence the electrical power output of the generator instantly
reduces to zero, i.e., Pe = 0 and the state point drops to ‘b’. The acceleration area A1 starts to
increase while the state point moves along
bc. At time tc corresponding clearing angle dc, the fault is cleared by the opening of the line
circuit breaker. tc is called clearing time and
dc is called clearing angle. After the fault is
cleared, the system again becomes healthy
and transmits power Pe = Pmax sind, i.e., the
state point shifts to “d” on the power angle
curve. The rotor now decelerates and the
decelerating area A2 begins to increase while
the state point moves along de.
or stability, the clearing angle, dc, must
be such that area A1 = area A2. Expressing area A1 = area A2 mathematically, we have
ig. 11.10: Pe – d characteristic.
d1 Pi (dc – d0 ) = z ( Pe - Pi) dd
dc