Question

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 _________.

Answer 1

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

X_n = 3z_f + 0.05 + 3x_n = 0.35 + 3x_n

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

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