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Single Line to Ground (L-G) Fault - Unloaded & Loaded

Two 11 kV, 20...

Question

Two 11 kV, 20 MVA, three-phase, star connected generators operate in parallel as shown in figure. The positive, negative and zero sequence reactance of each being respectively j0.18, j0.15, j0.10 p.u. The star point of one of the generator is isolated and that of the other is earthed through a 2 $Ω$ resistor. A single line to ground fault occurs at the terminals of one of the generators. Value of the current flowing in the grounding resistor is

A

14.72 kA

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B

3.07 kA

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C

11.52 kA

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D

8.35 kA

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Solution

The correct option is B 3.07 kA
since, both the generators are in parallal. So
$X_{1 e q} = \frac{j 0.18}{2} = j 0.09 p . u .$

$X_{2 e q} = \frac{j 0.15}{2} = j 0.075 p . u .$

But, for equivalent zero sequence reactance the neutral of one generator is not grounded. So in zero sequence network, one part will remain open circuited.

$Z_{0 e q} = j 0.1 + \frac{3 \times 2}{11^{2}} \times 20 = (0.9917 + j 0.1) p . u .$

$F a u l t c u r r e n t, I_{f} = \frac{3 E_{f}}{j (X_{1}) + j X_{2} + Z_{0 e q}}$

$= \frac{3 \times 1 ∠ 0^{\circ}}{1.0265 ∠ 14.96^{\circ}} = 2.922 ∠ - 14.96^{\circ} A$

$C u r r e n t i n g r o u n d i n g r e s i s t o r, I_{f} = 2.922 \times \frac{20}{11 \sqrt{3}} = 3.07 K A$

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