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A 50 Hz 6 pole 1000 MVA ,44 kV synchronous generator is supplying a full load at the 0.8 lagging power factor. Its output is reduced by 50% due to an electrical fault. So what will be the accelerating torque of the generator immediately after the electrical fault, ignoring the loss and considering that the primary inductive current that drives it is constant?
1. 6/π
2. 24/π
3. 12/π
4. 2/π

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Correct Answer - Option 3 : 12/π

Concept:

Accelerating power can be calculated as

Pa = P– Pe

Accelerating torque is given as

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

Where,

P= 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

P= P= 500 cosϕ = 1000 x 0.8

P= P= 800 MW

After fault Premains 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

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

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