Question

A 60 Hz generator, connected directly to an Infinite bus operating at a voltage of $1\angle 0^{\circ }p.u.,$ has a synchronous reactance of 1.35 p.u. The generator no load voltage is 1.1 pu. and its inertia constant H is 4 MJ/MVA: The generator is suddenly loaded to 60 percent of its maximum power limit. The frequency of the resulting natural oscillations of the generator rotor is

• A. 1.246 Hz
• B. 0.882 Hz
• C. 0.921 Hz
• D. 1.16 Hz

Answer 1

The correct option is B 0.882 Hz

$0.6P_{max}=P_{max}sin\delta$

$\delta =si{n}^{-1}\left(0.6\right)={36.8698}^{\circ }$

${\begin{array}{c}\frac{\partial P_e}{\partial \delta }\end{array}|}_{\delta ={36.8698}^{\circ }}=P_{max}cos\left({36.8698}^{\circ }\right)$

$=\frac{1.1\times 1}{1.35}cos\left({36.8698}^{\circ }\right)=0.6518p.u./rad$

$Frequencyofoscillations=\sqrt{\frac{(\partial P_e/\partial \delta {)}_{{36.8698}^{\circ }}}{M}}$

$Where,M=\frac{H}{\pi f}=\frac{4}{\pi \times 60}=0.0212p.u.s^2/rad$

$f_u=\sqrt{\frac{0.6518}{0.0212}}$

$f_n=5.54rad/sec=0.882Hz$