# A $30\;MVA$, 3-phase, $50\;Hz,\;13.8\;kV$, star-connected synchronous generator has positive, negative and zero sequence reactances, $15\% ,\;15\ 0 votes 47 views in General closed A \(30\;MVA$, 3-phase, $50\;Hz,\;13.8\;kV$, star-connected synchronous generator has positive, negative and zero sequence reactances, $15\% ,\;15\% \;and\;5\%$ respectively. A reactance (${X_n}$) is connected between the neutral of the generator and ground. A double line to ground fault takes place involving phases $'b'\;and\;'c'$, with a fault impedance of $j0.1\;p.u.$ The value of ${X_n}$ (in $p.u.$) that will limit the positive sequence generator current to $4270\;A$ is _________.

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We have the base current ${I_B} = \frac{{30 \times {{10}^3}}}{{\sqrt 3 \times 13.8}} = 1255.109\;A$

The positive sequence current, ${I_p} = 4270\;A$

Per unit current, ${I_{p.u.}} = \frac{{4270}}{{1255.109}} = 3.402\;p.u.$

Fault impedance = 0.1 pu

Xn = 3zf + 0.05 + 3xn = 0.35 + 3xn

$\begin{array}{l} {I_{g1}} = \frac{{{E_a}}}{{{X_1} + \left( {\left. {{X_2}} \right|} \right.\left| {\left. {{X_0}} \right)} \right.\;\;}}\\ {X_0}\; = \;3{X_n}\; + \;0.35\\ 3.402 = \frac{{1.0}}{{0.15 + \frac{{0.15 \times \left( {3{X_n} + 0.35} \right)}}{{0.15 + 3{X_n} + 0.35}}}} \end{array}$

From above equation, we get ${X_n} = 1.07\;p.u$