GATE 2022 EE Q.56: solution and answer

GATE 2022 · EEQ.56 · 2 marks · NAT

Two generating units rated for 250 MW and 400 MW have governor speed regulations of 6% and 6.4%, respectively, from no load to full load. Both the generating units are operating in parallel to share a load of 500 MW. Assuming free governor action, the load shared in MW, by the 250 MW generating unit is __________. (round off to nearest integer)

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Answer and solution

ANSWER

The 250 MW unit shares 200 MW.

Official answer: 188 to 192, or 198 to 202Source: Official answer key, GATE 2022 EE (IIT Kharagpur)

With free governor action both units run at the same frequency, so both see the same frequency drop $Δ f$ from the common no-load frequency $f_{0}$ .
Each unit's droop line is $\frac{Δ f}{f _{0}} = R_{i} \frac{P _{i}}{P _{i, rated}}$ , where $R_{i}$ is the per-unit regulation from no load to full load.
Unit 1 ( $250 MW$ , $R_{1} = 0.06$ ): $\frac{Δ f}{f _{0}} = 0.06 \frac{P _{1}}{250}$ . Unit 2 ( $400 MW$ , $R_{2} = 0.064$ ): $\frac{Δ f}{f _{0}} = 0.064 \frac{P _{2}}{400}$ .
Equate the two drops: $0.06 \frac{P _{1}}{250} = 0.064 \frac{P _{2}}{400}$ , i.e. $2.4 \times 1 0^{- 4} P_{1} = 1.6 \times 1 0^{- 4} P_{2}$ .
Hence $\frac{P _{1}}{P _{2}} = \frac{1.6}{2.4} = \frac{2}{3}$ .
The total load is $P_{1} + P_{2} = 500 MW$ , so $P_{1} = 500 \times \frac{2}{5} = 200 MW$ and $P_{2} = 300 MW$ .
Check: $200 < 250$ and $300 < 400$ , so neither unit is overloaded. Also $0.06 \times \frac{200}{250} = 0.048 = 0.064 \times \frac{300}{400}$ , so the drops agree.

KEY IDEA

Units in parallel under free governor action share load in proportion to rating divided by regulation, because they must have the same frequency drop. Here $P_{i} \propto P_{i, rated} / R_{i}$ .

WATCH OUT FOR

Sharing the load in proportion to the ratings alone (250:400, which gives about 192 MW). Another is using the regulation percentages directly as weights without dividing each by the unit's own rating.

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GATE 2022 EE Q.56: solution and answer

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