GATE 2024 EE Q.34: solution and answer

GATE 2024 · EEQ.34 · 1 mark · NAT

The incremental cost curves of two generators (Gen A and Gen B) in a plant supplying a common load are shown in the figure. If the incremental cost of supplying the common load is Rs. 7400 per MWh, then the common load in MW is ______ (rounded off to the nearest integer).

GATE 2024 EE question 34: Graph of incremental cost (Rs./MWh, vertical axis, 5000 to 13000 in steps of 1000) against generation (MW, horizontal… — Figure from the GATE 2024 EE paper. It is part of the question.

Answer and solution

ANSWER

The common load is 35 MW. Gen A is not loaded, because its incremental cost is already 8000 per MWh at zero output, above the 7400 per MWh system value. Gen B alone supplies 35 MW.

Official answer: 35Source: Official answer key, GATE 2024 EE (IISc Bengaluru)

Read each straight line from the graph. Gen A passes through $(0 MW, 8000)$ , $(50, 9000)$ and $(100, 10000)$ , so its slope is $\frac{10000 - 8000}{100} = 20$ per MW. Hence $I C_{A} = 8000 + 20 P_{A}$ per MWh. Gen B passes through $(0, 6000)$ , $(50, 8000)$ and $(100, 10000)$ , so its slope is $\frac{10000 - 6000}{100} = 40$ per MW. Hence $I C_{B} = 6000 + 40 P_{B}$ per MWh.
For economic dispatch, every generator that is running operates at the same incremental cost $λ$ , and the outputs add up to the load: $I C_{A} = I C_{B} = λ$ and $P_{A} + P_{B} = P_{l o a d}$ .
Gen A: its incremental cost curve starts at $8000$ per MWh at $P_{A} = 0$ and only rises from there. The system incremental cost is $λ = 7400 < 8000$ , so no output of Gen A has an incremental cost as low as $λ$ . Gen A is not loaded and $P_{A} = 0$ .
Gen B: set $I C_{B} = λ$ , so $7400 = 6000 + 40 P_{B}$ . Then $P_{B} = \frac{7400 - 6000}{40} = \frac{1400}{40} = 35 MW$ . This lies on the plotted range of $0$ to $160 MW$ , so it is valid.
Common load: $P_{l o a d} = P_{A} + P_{B} = 0 + 35 = 35 MW$ .

KEY IDEA

In economic dispatch every loaded generator runs at the same incremental cost. A generator whose incremental cost at minimum output is above the system value is not loaded, so it supplies 0 MW.

WATCH OUT FOR

Extending Gen A's line back to a negative output, or setting both lines equal to 7400. That gives $P_{A} = - 30$ MW, so the total comes out as 5 MW. A generator cannot supply negative power, so Gen A must be held at 0 MW.

From the GATE 2024 EE paper. Every question of this paper · Download the full paper · All 9 Dispatch & control questions, with notes

GATE 2024 EE Q.34: solution and answer

Answer and solution

More Dispatch & control questions