Electrical Power Systems

Power System Stability 289
or P i (d 1 – d 0 ) + P max (cosd 1 – cosd 0 ) = P i (d 1 – d 2 ) + P max (cosd 1 – cosd 2 )
...(11.44)
But Pi = Pmax sind1, which when substituted in eqn. (11.44), we get
Pmax (d1 – d0) sind1 + Pmax (cosd1– cosd0) = Pmax (d1 – d2 ) sind1 + Pmax (cosd1 – cosd2 ) ...(11.45)
Upon simplification, eqn. (11.45) becomes
(d2 – d0 ) sind1 + cosd2 – cosd0 = 0 ...(11.46)
Example 11.7: A synchronous generator, capable of developing 500MW power per phase,
operates at a power angle of 8º. By how much can the input shaft power be increased suddenly
without loss of stability? Assume that Pmax will remain constant.
Solution:
Initially, at d0 = 8º,
P e0 = P max sin d 0 = 500 sin 8º = 69.6 MW
ig. 11.8: Power angle characteristic of Example 11.7.
Let dm be the power angle to which the rotor can swing before losing synchronism. If this
angle is exceeded, Pi will again become greater than Pe and the rotor will once again be
accelrated and synchronism will be lost as shown in ig. 11.6, Threfore, the equal–area criterion
requires that eqn. (11.44) be satisfied with dm replacing d2 .
rom ig. 11.8, dm = p –d1 . Therefore, eqn. (11.46)
becomes
(p – d1 – d0 ) sind1 + cos( p – d1 ) – cosd0 = 0
\ (p – d 1 – d 0 ) sind 1 – cosd 1 – cosd 0 = 0 ...(i)
Sustituting d 0 = 8º = 0.139 radian in eqn. (i) yields
(3 – d 1) sind 1 – cosd 1 – 0.99 = 0 ...(ii)