GATE 2018 EE Q.29: solution and answer

GATE 2018 · EEQ.29 · 2 marks · MCQ

Consider the two bus power system network with given loads as shown in the figure. All the values shown in the figure are in per unit. The reactive power supplied by generator G1 and G2 are Q_G1 and Q_G2 respectively. The per unit values of Q_G1, Q_G2, and line reactive power loss (Q_loss) respectively are

GATE 2018 EE question 29: Single-line diagram of two buses joined by a line of reactance j0.1 pu, with the line reactive loss labelled Q_loss. — Figure from the GATE 2018 EE paper. It is part of the question.

(A)
5.00, 12.68, 2.68
(B)
6.34, 10.00, 1.34
(C)
6.34, 11.34, 2.68
(D)
5.00, 11.34, 1.34

Answer and solution

ANSWER

$Q_{G 1} = 6.34$ pu, $Q_{G 2} = 11.34$ pu and $Q_{l oss} = 2.68$ pu, which is option (C).

Official answer: CSource: Official answer key, GATE 2018 EE (IIT Guwahati)

Take per-unit values throughout. The line is a pure reactance $X = 0.1$ pu with $V_{1} = 1∠ δ$ and $V_{2} = 1∠ 0^{\circ}$ . Bus 1 has generation $20 + j Q_{G 1}$ and load $15 + j 5$ . Bus 2 has generation $15 + j Q_{G 2}$ and load $20 + j 10$ .
Real power sent into the line from bus 1 is $P_{12} = 20 - 15 = 5$ pu. At bus 2 the net real injection is $15 - 20 = - 5$ pu, so the line delivers 5 pu there. The two ends agree because the line has no resistance and so no real-power loss.
Real power over a reactance: $P_{12} = \frac{V _{1} V _{2}}{X} sin δ = \frac{1 \cdot 1}{0.1} sin δ = 5$ . Hence $sin δ = 0.5$ , $δ = 3 0^{\circ}$ and $cos δ = 0.866$ .
Reactive power leaving bus 1 into the line: $Q_{12} = \frac{V _{1}^{2} - V _{1} V _{2} cos δ}{X} = \frac{1 - 0.866}{0.1} = 1.34$ pu.
Reactive power leaving bus 2 into the line: $Q_{21} = \frac{V _{2}^{2} - V _{1} V _{2} cos δ}{X} = \frac{1 - 0.866}{0.1} = 1.34$ pu. Both ends send reactive power into the line because the two voltage magnitudes are equal.
Line reactive loss: $Q_{l oss} = Q_{12} + Q_{21} = 1.34 + 1.34 = 2.68$ pu. Check with $Q_{l oss} = X I^{2}$ : $∣ I ∣ = \frac{∣ V _{1} - V _{2} ∣}{X} = \frac{2 sin 1 5 ^{\circ}}{0.1} = 5.176$ pu, so $Q_{l oss} = 0.1 \times 26.79 = 2.68$ pu.
Reactive balance at bus 1: $Q_{G 1} = Q_{l o a d, 1} + Q_{12} = 5 + 1.34 = 6.34$ pu.
Reactive balance at bus 2: $Q_{G 2} = Q_{l o a d, 2} + Q_{21} = 10 + 1.34 = 11.34$ pu.
Therefore $(Q_{G 1}, Q_{G 2}, Q_{l oss}) = (6.34, 11.34, 2.68)$ pu, which is option (C).

KEY IDEA

A lossless reactive line carries real power $P = \frac{V _{1} V _{2}}{X} sin δ$ , which fixes $δ$ . Each end then sends $\frac{V ^{2} - V _{1} V _{2} cos δ}{X}$ of reactive power into the line, and the generators supply their bus loads plus that amount.

WATCH OUT FOR

Giving the whole line loss to one bus, or treating the line as reactive-lossless so that $Q_{l oss} = 0$ . Another slip is taking $Q_{l oss} = 1.34$ (one end's figure) instead of the sum of both ends, which gives option (B) or (D).

From the GATE 2018 EE paper. Every question of this paper · Download the full paper · All 10 Per unit & load flow questions, with notes

GATE 2018 EE Q.29: solution and answer

Answer and solution

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