GATE EE 2019 Set 1 — Question 50

Let $X_1$, $X_2$, and $X_0$ be the positive, negative, and zero sequence reactances of the generator, respectively.
Given:
$X_1 = 0.25$ pu
$X_2 = 0.15$ pu
$X_0 = 0.05$ pu (internal zero sequence reactance of the generator)
Base MVA = 50 MVA
Base kV = 30 kVFirst, calculate the base impedance:
$Z_{base} = \frac{(Base\ kV)^2}{Base\ MVA} = \frac{(30)^2}{50} = \frac{900}{50} = 18\ \Omega$For a bolted three-phase fault at the generator terminal, the fault current is:
$I_{f,3\phi} = \frac{V_f}{X_1}$
Assuming $V_f = 1$ pu (pre-fault voltage at the fault point),
$I_{f,3\phi} = \frac{1}{0.25} = 4$ puFor a bolted line-to-ground (LG) fault at the generator terminal, the fault current is:
$I_{f,LG} = \frac{3V_f}{X_1 + X_2 + X_{0,total}}$
Where $X_{0,total}$ is the total zero sequence reactance seen from the fault point. If the generator neutral is grounded with a reactance $X_n$, then $X_{0,total} = X_0 + 3X_n$.
So, $I_{f,LG} = \frac{3V_f}{X_1 + X_2 + X_0 + 3X_n}$
Assuming $V_f = 1$ pu,
$I_{f,LG} = \frac{3}{0.25 + 0.15 + 0.05 + 3X_n} = \frac{3}{0.45 + 3X_n}$Given that the fault currents are equal, $I_{f,3\phi} = I_{f,LG}$:
$4 = \frac{3}{0.45 + 3X_n}$
$4(0.45 + 3X_n) = 3$
$1.8 + 12X_n = 3$
$12X_n = 3 - 1.8 = 1.2$
$X_n = \frac{1.2}{12} = 0.1$ puNow, convert the grounding reactance from pu to ohms:
$X_{n,ohms} = X_n \times Z_{base} = 0.1 \times 18 = 1.8\ \Omega$Note on discrepancy: The standard calculation based on symmetrical components theory yields a grounding reactance of $1.8\ \Omega$. However, the specified answer range is $0.25$ to $0.27\ \Omega$. To match the given answer, an alternative interpretation or a modification to the problem parameters would be required, which is not directly derivable from the provided information and standard formulas. Given the strict instruction to match the provided answer, we acknowledge this discrepancy. If we were to assume the answer is $0.26\ \Omega$, it would imply a different problem setup or parameters than those explicitly stated.