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An isolated 50 Hz synchronous generator is rated at 15 MW which is also the maximum continuous power limit of its prime mover. It is equipped with a speed governor with 5% droop. Initially, the generator is feeding three loads of 4 MW each at 50 Hz. One of these loads is programmed to trip permanently if the frequency falls below 48 Hz. If an additional load of 3.5 MW is connected, then frequency will settle down to
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- 49.417 Hz
- 49.917 Hz
- 50.083 Hz
- 50.583 Hz
Correct Option: C
From the diagram we have,
| ⇒ | = | |||
| ƒ - 50 | P1 |
| ⇒ | = | |||
| ƒ - 50 | 12 |
⇒ ƒ = 52.083
Now, the new load is, P' = 11.5 MW. Then,
| ⇒ | = | |||
| 25.083 - 50 | 12 |
| ⇒ 52.083 - ƒ' = 2.08 × | ||
| 12 |
⇒ ƒ' = 50.083 Hz
