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Droop Control allows synchronous generators to run in parallel, by controlling the prime mover that drives a synchronous generator connected to an electrical grid.


Generator (1), with rated power of 500 [MW], is connected to the electricity network running at its rated frequency of 50[Hz] . A sudden load decrease of 20[MW] results in a frequency rise of 0.020[Hz] .


Generator (2) is later added to the network, with a droop of 2 [%], and running at the same rated frequency. The network now has a power frequency characteristic (lambda) of 1750 [MW/Hz].


What is, in , the rated power of Generator (2)?

Answer :

To determine the rated power of Generator (2), we can use the power frequency characteristic (lambda) and the droop value. The rated power of Generator (2) is 35 [MW].

The power frequency characteristic (lambda) represents the change in power output for a given change in frequency. In this case, the lambda value is given as 1750 [MW/Hz], which means that for every 1 [Hz] increase in frequency, the total power output of the system increases by 1750 [MW].

We know that there was a frequency rise of 0.020 [Hz] due to a sudden load decrease of 20 [MW]. Since Generator (2) has a droop of 2 [%], it means that its power output changes by 2 [%] for a 1 [Hz] change in frequency.

Using these values, we can calculate the power change caused by the load decrease:

Power change = Frequency rise * Lambda

Power change = 0.020 [Hz] * 1750 [MW/Hz]

Power change = 35 [MW]

Since Generator (2) is responsible for the power change caused by the load decrease, the rated power of Generator (2) would be equal to the power change, which is 35 [MW].

Therefore, the rated power of Generator (2) is 35 [MW].

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