GATE 2023 EE Q.60: solution and answer

GATE 2023 · EEQ.60 · 2 marks · NAT

The two-bus power system shown in figure (i) has one alternator supplying a synchronous motor load through a Y-Δ transformer. The positive, negative and zero-sequence diagrams of the system are shown in figures (ii), (iii) and (iv), respectively. All reactances in the sequence diagrams are in p.u. For a bolted line-to-line fault (fault impedance = zero) between phases 'b' and 'c' at bus 1, neglecting all pre-fault currents, the magnitude of the fault current (from phase 'b' to 'c') in p.u. is ____________ (Round off to 2 decimal places).

GATE 2023 EE question 60: Figure (i): single-line diagram. — Figure from the GATE 2023 EE paper. It is part of the question.

Answer and solution

ANSWER

The magnitude of the line-to-line fault current is about 7.22 p.u.

Official answer: 7.1 to 7.3Source: Official answer key, GATE 2023 EE (IIT Kanpur)

Neglecting pre-fault currents, both internal emfs are $E_{g}^{''} = E_{m}^{''} = 1∠ 0^{\circ}$ p.u., so the pre-fault voltage at bus 1 is $V_{f} = 1∠ 0^{\circ}$ p.u.
Positive-sequence Thevenin impedance at bus 1 (sources shorted): the motor branch $j 0.3$ is in parallel with the path through the line and generator, $j 0.1 + j 0.1 = j 0.2$ .
$Z_{1} = \frac{j 0.3 \times j 0.2}{j 0.3 + j 0.2} = j 0.12 p.u.$
The negative-sequence network has the same topology and no sources, so $Z_{2} = Z_{1} = j 0.12$ p.u.
A line-to-line fault between $b$ and $c$ carries no zero-sequence current, so $I_{a 0} = 0$ and $Z_{0}$ is not needed. With zero fault impedance, $I_{a 1} = - I_{a 2} = \frac{V _{f}}{Z _{1} + Z _{2}} = \frac{1}{j 0.24} = - j 4.1667$ p.u.
Fault current in phase $b$ : $I_{b} = a^{2} I_{a 1} + a I_{a 2} = (a^{2} - a) I_{a 1}$ , and $a^{2} - a = - j 3$ , so $∣ I_{b} ∣ = 3 \times \frac{1}{0.24} = 7.217 \approx 7.22 p.u.$

KEY IDEA

For a line-to-line fault, only the positive and negative sequence networks are in series: $I_{a 1} = V_{f} / (Z_{1} + Z_{2})$ , with $I_{a 0} = 0$ and $∣ I_{b} ∣ = 3 ∣ I_{a 1} ∣$ . The zero-sequence network is not needed.

WATCH OUT FOR

Including the zero-sequence impedance in the line-to-line fault (it belongs to the line-to-ground and double-line-to-ground faults). Another slip is forgetting the $3$ factor, or leaving out the motor branch when finding $Z_{1}$ .

From the GATE 2023 EE paper. Every question of this paper · Download the full paper · All 8 Faults questions, with notes

GATE 2023 EE Q.60: solution and answer

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