A 100 MW generator with a 5% governor droop responds to a -0.15 Hz frequency deviation. How much will it output?

In the context of generator operation and governor response, the concept of droop is essential. A governor droop of 5% means that for every 5% change in frequency from the nominal frequency, the generator's output changes by a corresponding amount.

When there's a frequency deviation, this droop setting indicates how much power the generator will produce or supply in response. The -0.15 Hz frequency deviation implies that the system frequency has dropped below the nominal frequency, which is typically around 60 Hz in North America; hence, a drop to 59.85 Hz is observed.

To calculate the output change, the relationship between the frequency change, the governor droop, and the generator's capacity can be applied. The formula to determine the change in output (ΔP) is given by:

[

\Delta P = \text{Droop} \times \frac{\Delta f}{f_{nominal}} \times P_{rated}

]

Substituting the known values into the formula:

• Droop = 5% = 0.05

• Δf = -0.15 Hz

• f_nominal = 60 Hz

• P_rated = 100 MW

First, find the fractional