GATE 2020 EE Q.23: solution and answer

GATE 2020 · EEQ.23 · 1 mark · NAT

A single 50 Hz synchronous generator on droop control was delivering 100 MW power to a system. Due to increase in load, generator power had to be increased by 10 MW, as a result of which, system frequency dropped to 49.75 Hz. Further increase in load in the system resulted in a frequency of 49.25 Hz. At this condition, the power in MW supplied by the generator is ______ (rounded off to 2 decimal places).

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Answer and solution

ANSWER

The generator supplies 130.00 MW at 49.25 Hz.

Official answer: 123.00 to 135.00Source: Official answer key, GATE 2020 EE (IIT Delhi)

Droop control gives a straight-line frequency-power characteristic, so the generator power rises in proportion to the frequency drop: $P = P_{0} + K (f_{0} - f)$ , where $K = \frac{1}{R}$ is the slope in MW/Hz.
First operating point: $f_{0} = 50 Hz$ and $P_{0} = 100 MW$ .
Second operating point: the power rises by $10 MW$ to $110 MW$ and the frequency falls to $49.75 Hz$ , a drop of $50 - 49.75 = 0.25 Hz$ .
Slope of the characteristic: $K = \frac{Δ P}{Δ f} = \frac{10 MW}{0.25 Hz} = 40 MW/Hz$ .
At $f = 49.25 Hz$ the drop from the nominal $50 Hz$ is $50 - 49.25 = 0.75 Hz$ .
Power supplied: $P = 100 + 40 \times 0.75 = 100 + 30 = 130.00 MW$ .
Cross-check from the second point: $110 + 40 \times (49.75 - 49.25) = 110 + 20 = 130.00 MW$ , which agrees.

KEY IDEA

A droop-controlled generator follows a straight line of power against frequency. Two points fix the slope in MW/Hz, and that slope then gives the power at any other frequency.

WATCH OUT FOR

Using the 10 MW increase as if it were measured from 49.75 Hz to 49.25 Hz, or working with the 0.5 Hz step alone and forgetting the 100 MW starting power. The slope must come from the first 0.25 Hz drop, and the final power is measured from the 100 MW starting point.

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GATE 2020 EE Q.23: solution and answer

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