**Concept: **Accelerating torque = (Accelerating power, P_{a})/(Synchronous speed, ω_{s})

P_{a} is the accelerating power = fault mechanical input – fault mechanical output

**Calculation:**

Given that,

P = 4, f = 50 Hz, pf = 0.8

At fault, electrical output = 60%

MVA rating of a turbo generator is 500 MVA

**Before fault:**

Mechanical input (P_{s1}) = electrical output (P_{e1}) = 500 × 0.6 × 10^{6} = 300 MW

**During fault:**

Electrical output, P_{e2} = 0.6 × 400 × 10^{6} = 240 MW

Mechanical input, P_{s2} = 300 MW

Accelerating power, P_{a} = P_{s2} – P_{e2} = 300 – 240 = 60 MW

Now, speed \({N_s} = \frac{{120\; \times \;f}}{P} = 1500\;rpm\)

⇒ ω_{s} = 157.07 rad/sec

Accelerating torque, \({T_a} = \frac{{{P_a}}}{{{\omega _s}}} = \frac{{60\; \times \;{{10}^6}}}{{157.07}} = 0.3819\;MN - m\)