GATE 2023 EE Q.21: solution and answer

GATE 2023 · EEQ.21 · 1 mark · MCQ

The expressions of fuel cost of two thermal generating units as a function of the respective power generation P_G1 and P_G2 are given as
F_1(P_G1) = 0.1 a P_G1^2 + 40 P_G1 + 120 Rs/hour, 0 MW ≤ P_G1 ≤ 350 MW
F_2(P_G2) = 0.2 P_G2^2 + 30 P_G2 + 100 Rs/hour, 0 MW ≤ P_G2 ≤ 300 MW
where a is a constant. For a given value of a, optimal dispatch requires the total load of 290 MW to be shared as P_G1 = 175 MW and P_G2 = 115 MW. With the load remaining unchanged, the value of a is increased by 10% and optimal dispatch is carried out. The changes in P_G1 and the total cost of generation, F (= F_1 + F_2) in Rs/hour will be as follows

(A)
P_G1 will decrease and F will increase
(B)
Both P_G1 and F will increase
(C)
P_G1 will increase and F will decrease
(D)
Both P_G1 and F will decrease

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Answer and solution

ANSWER

P_G1 will decrease and F will increase (option A).

Official answer: ASource: Official answer key, GATE 2023 EE (IIT Kanpur)

Incremental costs: $\frac{d F _{1}}{d P _{G 1}} = 0.2 a P_{G 1} + 40$ and $\frac{d F _{2}}{d P _{G 2}} = 0.4 P_{G 2} + 30$ . Both units lie inside their limits, so the optimum has equal incremental cost $λ$ and $P_{G 1} + P_{G 2} = 290 MW$ .
Original dispatch: $λ = 0.4 (115) + 30 = 76 per MWh$ . Then $0.2 a (175) + 40 = 76$ , so $35 a = 36$ and $a = \frac{36}{35} \approx 1.0286$ .
New value: $a^{'} = 1.1 a \approx 1.1314$ , so $0.2 a^{'} \approx 0.22629$ .
New optimum: $0.22629 P_{G 1} + 40 = 0.4 (290 - P_{G 1}) + 30$ , so $0.62629 P_{G 1} = 106$ and $P_{G 1} \approx 169.25 MW$ , $P_{G 2} \approx 120.75 MW$ . Hence $P_{G 1}$ decreases from $175 MW$ .
Original cost: $F_{1} = 0.1 \cdot \frac{36}{35} (175)^{2} + 40 (175) + 120 = 3150 + 7000 + 120 = 10270$ and $F_{2} = 0.2 (115)^{2} + 30 (115) + 100 = 6195$ , so $F \approx 16465 per hour$ .
New cost: $F_{1} \approx 0.11314 (169.25)^{2} + 40 (169.25) + 120 \approx 10131$ and $F_{2} = 0.2 (120.75)^{2} + 30 (120.75) + 100 \approx 6639$ , so $F \approx 16770 per hour$ , which is higher.
Check by reasoning: raising $a$ raises $F_{1}$ at every positive output, so for any fixed split of the load the cost is higher. The cheapest split under the new $a$ therefore costs more than the cheapest split under the old $a$ , and $F$ increases. Unit 1 is now dearer at the margin, so it is loaded less.
So $P_{G 1}$ decreases and $F$ increases, which is option (A).

KEY IDEA

Economic dispatch needs equal incremental cost on both units. Find a from the original dispatch, raise it by 10%, and re-solve the equal-lambda condition. A dearer unit 1 is loaded less, and the minimum total cost cannot fall when one cost curve rises.

WATCH OUT FOR

Concluding that the total cost falls because the dispatch is re-optimised. Re-optimising only limits the rise: the new minimum is still above the old one, since F_1 is higher at every output. Another slip is to compare the increase in a with a change in lambda without solving the equations.

From the GATE 2023 EE paper. Every question of this paper · Download the full paper · All 9 Dispatch & control questions, with notes

GATE 2023 EE Q.21: solution and answer

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