Question

A three phase, 100 MVA, 25kV generator has solidly grounded neutral. The positive, negative, and the zero sequence reactances of the generator are 0.2 pu, 0.2 pu, and 0.05 pu, respectively, at the machine base quantities. If a bolted single phase to ground fault occurs at the terminal of the unloaded generator, the fault current in amperes immediately after the fault is________.

Answer 1

Concept:

In a single line to ground (LG) fault, the fault current is

\({I_n} = \frac{{3{E_a}}}{{{Z_1} + {Z_2} + {Z_0} + 3{Z_n}}}\)

Where,

Ea = terminal voltage

Z1 = positive sequence impedance

Z2 = negative sequence impedance

Z0 = zero sequence impedance

Zn = impedance in neutral grounding

The fault current can also be written as If = 3I_a1

Calculation:

Single line to ground fault, Fault current in \({I_f} = 3{I_{{a_1}}}\)

\({I_{{a_1}}} = \frac{{{E_a}}}{{{Z_1} + {Z_2} + {Z_o}}} = \frac{1}{{j0.2 + j0.2 + j0.05}}\)

= -j 2.223 Pu.

Fault current \(\left( {{I_f}} \right) = 3 \times {I_{{a_1}}}\)

= (3 × -j 2.222) =  - j 6.666 pu

Base current \( = \frac{{100 \times {{10}^6}}}{{\sqrt 3 \times 25 \times {{10}^3}}} = 2309.4\ Pu\)

Fault current I_f = 15396 A.