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