GATE 2018 EE Q.21: solution and answer

GATE 2018 · EEQ.21 · 1 mark · NAT

The positive, negative and zero sequence impedances of a 125 MVA, three-phase, 15.5 kV, star-grounded, 50 Hz generator are j0.1 pu, j0.05 pu and j0.01 pu respectively on the machine rating base. The machine is unloaded and working at the rated terminal voltage. If the grounding impedance of the generator is j0.01 pu, then the magnitude of fault current for a b-phase to ground fault (in kA) is ______ (up to 2 decimal places).

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Answer and solution

ANSWER

The magnitude of the fault current for the single line-to-ground fault is about 73.52 kA.

Official answer: 73.0 to 74.0Source: Official answer key, GATE 2018 EE (IIT Guwahati)

Per-unit bases: $S_{b} = 125 MVA$ , $V_{b} = 15.5 kV$ , so $I_{b} = \frac{125 \times 1 0 ^{3}}{3 \times 15.5} A \approx 4656.1 A$ . The machine is unloaded at rated voltage, so the prefault voltage is $E = 1∠ 0^{\circ}$ pu.
A fault from one phase (here phase b) to ground is a single line-to-ground (LG) fault. For an LG fault the three sequence networks are connected in series, and the fault current is $I_{f} = 3 I_{0}$ . By symmetry the result is the same for any phase, so only the magnitude matters.
The grounding impedance carries three times the zero-sequence current, so it enters the zero-sequence network as $3 Z_{n} = 3 \times j 0.01 = j 0.03$ pu. The total zero-sequence impedance is $Z_{0}^{total} = j 0.01 + j 0.03 = j 0.04$ pu.
The sequence currents are equal: $I_{1} = I_{2} = I_{0} = \frac{E}{Z _{1} + Z _{2} + Z _{0}^{total}} = \frac{1}{j ( 0.1 + 0.05 + 0.04 )} = \frac{1}{j 0.19}$ pu.
Fault current: $∣ I_{f} ∣ = 3∣ I_{0} ∣ = \frac{3}{0.19} = 15.789$ pu.
Convert to amperes: $∣ I_{f} ∣ = 15.789 \times 4656.1 A \approx 73.52 kA$ .

KEY IDEA

For a line-to-ground fault, connect the positive, negative and zero sequence networks in series, with $3 Z_{n}$ added in the zero-sequence network. Then $I_{f} = 3 E / (Z_{1} + Z_{2} + Z_{0} + 3 Z_{n})$ , and the per-unit value is multiplied by the base current.

WATCH OUT FOR

Adding the grounding impedance $Z_{n}$ to the zero-sequence network without the factor of 3. That gives $Z_{0}^{total} = j 0.02$ , a total of $j 0.17$ pu and about 82 kA. Another slip is forgetting that $I_{f} = 3 I_{0}$ , or using the wrong base current.

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GATE 2018 EE Q.21: solution and answer

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