GATE 2020 EE Q.47: solution and answer

GATE 2020 · EEQ.47 · 2 marks · NAT

Bus 1 with voltage magnitude $V_{1} = 1.1$ pu is sending reactive power $Q_{12}$ towards bus 2 with voltage magnitude $V_{2} = 1$ pu through a lossless transmission line of reactance $X$ . Keeping the voltage at bus 2 fixed at 1 pu, magnitude of voltage at bus 1 is changed, so that the reactive power $Q_{12}$ sent from bus 1 is increased by 20%. Real power flow through the line under both the conditions is zero. The new value of the voltage magnitude, $V_{1}$ , in pu (rounded off to 2 decimal places), at bus 1 is ______.

GATE 2020 EE question 47: Single-line diagram: a horizontal line joining Bus 1 (left, voltage V_1) and Bus 2 (right, voltage V_2), with an arrow… — Figure from the GATE 2020 EE paper. It is part of the question.

Answer and solution

ANSWER

The new value of $V_{1}$ is about $1.12$ pu.

Official answer: 1.11 to 1.13Source: Official answer key, GATE 2020 EE (IIT Delhi)

The line is lossless, so $P_{12} = \frac{V _{1} V _{2}}{X} sin δ$ . Real power flow is zero in both conditions, so $sin δ = 0$ and $δ = 0$ in both.
The reactive power sent from bus 1 is $Q_{12} = \frac{V _{1}^{2} - V _{1} V _{2} cos δ}{X}$ . With $δ = 0$ and $V_{2} = 1$ pu this becomes $Q_{12} = \frac{V _{1} ( V _{1} - 1 )}{X}$ .
Initial case, $V_{1} = 1.1$ pu: $Q_{12} = \frac{1.1 ( 1.1 - 1 )}{X} = \frac{0.11}{X}$ .
$Q_{12}$ is increased by 20%, so the new value is $Q_{12}^{'} = 1.2 \times \frac{0.11}{X} = \frac{0.132}{X}$ . The reactance $X$ and $V_{2}$ are unchanged.
Equate the new $Q_{12}$ to the formula: $V_{1} (V_{1} - 1) = 0.132$ , which gives $V_{1}^{2} - V_{1} - 0.132 = 0$ .
Solve the quadratic: $V_{1} = \frac{1 \pm 1 + 4 ( 0.132 )}{2} = \frac{1 \pm 1.528}{2}$ , with $1.528 \approx 1.2361$ .
The negative root gives a negative voltage magnitude, which is not physical, so reject it. Then $V_{1} = \frac{1 + 1.2361}{2} \approx 1.118$ pu, which is $1.12$ pu to 2 decimal places.

KEY IDEA

With zero real power flow the angle is zero, so reactive power is $Q_{12} = V_{1} (V_{1} - V_{2}) / X$ . A 20% increase in $Q_{12}$ gives a quadratic in $V_{1}$ . Take its positive root.

WATCH OUT FOR

Using a 20% rise in voltage instead of a 20% rise in reactive power. Another is treating $Q$ as linear in $V_{1}$ . $Q$ depends on $V_{1}^{2}$ , so $V_{1}$ rises by only about 1.6%, not 20%.

From the GATE 2020 EE paper. Every question of this paper · Download the full paper · All 6 Transmission & distribution questions, with notes

GATE 2020 EE Q.47: solution and answer

Answer and solution

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