Correct Answer - Option 3 : 12/π

__Concept:__

Accelerating power can be calculated as

Pa = Pm – Pe

Accelerating torque is given as

\({T_a} = \frac{{{P_a}}}{{{\omega _s}\;}}\)

Where,

Pm = Mechanical input power

Pe = Electrical output power

ωs = Synchronous speed in radian/sec

P = Number of poles

F = Supply frequency

Synchronous speed in rpm can be calculated as

\({N_s} = \frac{{120f}}{P}\)

Also, \({\omega _s} = \frac{{2\pi {N_s}}}{{60}}\)

__Calculation__:

Given-

f = 50 Hz, P = 6

Before fault

Pm = Pe = 500 cosϕ = 1000 x 0.8

Pm = Pe = 800 MW

After fault Pm remains same, while Pe reduces by 50%

So electrical output at the time of fault is

Pe’ = (1 – 0.5) = 0.5 x 800

Pe’= 400 MW

Now Accelerating power at the time of fault is

Pa’ = 800 – 400 = 400 MW

Now synchronous speed is

\({N_s} = \frac{{120 \times 50}}{6} = 1000\;rpm\)

\({\omega _s} = \frac{{2\pi \times 1000}}{{60}} = 104.71\;radian/sec\)

Hence accelerating torque is

\({T_a} = \frac{{400 \times {{10}^6}}}{{104.71}} = 3.82\; \times {10^6}\;N - m ={12\over\pi}\;MN - m\)