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A 100 MVA, 50 Hz turbo-generator operates at no load at 3000 rpm. A load of 25 MW is suddenly applied to the machine and the steam valve to the turbine commences to open after 0.6 seconds due to the governor time-lag. Assuming the inertia constant H = 5.0 kW-s per kVA of the generator rating, the frequency to which the generated voltage drops before the steam-flow commences to meet the new load is
1. 49. Hz
2. 50.15 Hz
3. 49.24 Hz
4. 49.82 Hz

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Correct Answer - Option 3 : 49.24 Hz

Kinetic energy stored = SH

Where S is the kVA rating

H is inertia constant

Given that,

inertia constant (H) = 5.0 kW-s per kVA

kVA rating (S) = 100 MVA

The energy stored at no load = 5 ´ 100 = 500 MJ

Before the steam valves open, the energy lost by the rotor = 25 ´ 0.6 = 15 MJ

Stored kinetic energy is directly proportional to the square of the frequency.

The frequency at the end of 0.6 seconds is,

\(= 50 \times \sqrt {\frac{{500\; - \;15}}{{500}}} = 49.24\;Hz\)

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