SLG fault on bus with three alternators

Base: 10 MVA, 6.6 kV (rating of each alternator).

$$I_{base} = \frac{10\times10^6}{\sqrt3 \times 6.6\times10^3} = 874.8\ \text{A}, \quad Z_{base} = \frac{6.6^2}{10} = 4.356\ \Omega$$

Each alternator: $X_1 = 0.15$, $X_2 = 0.75 \times 0.15 = 0.1125$, $X_0 = 0.30 \times 0.15 = 0.045$ pu. Pre-fault voltage 1 pu, no load.

The positive and negative networks always have all three machines in parallel:

$$X_1 = \frac{0.15}{3} = 0.05\ \text{pu}, \quad X_2 = \frac{0.1125}{3} = 0.0375\ \text{pu}$$

(i) All neutrals solidly grounded

All three zero-sequence reactances are in parallel:

$$\begin{aligned} X_0 &= \frac{0.045}{3} = 0.015\ \text{pu} \\ I_f &= \frac{3E}{j(X_1+X_2+X_0)} = \frac{3}{j(0.05+0.0375+0.015)} = \frac{3}{j0.1025} = -j29.27\ \text{pu} \\ |I_f| &= 29.27 \times 874.8 = 25\,603\ \text{A} \approx 25.6\ \text{kA} \end{aligned}$$

(ii) One neutral grounded through 0.3 Ω, others isolated

Only the grounded machine provides a zero-sequence path; its neutral resistance appears as $3R_n$:

$$\begin{aligned} R_n &= \frac{0.3}{4.356} = 0.06887\ \text{pu}, \quad 3R_n = 0.2066\ \text{pu} \\ Z_0 &= 3R_n + jX_0 = 0.2066 + j0.045\ \text{pu} \\ Z_1+Z_2+Z_0 &= 0.2066 + j(0.05 + 0.0375 + 0.045) = 0.2066 + j0.1325 \\ I_f &= \frac{3}{0.2066 + j0.1325} = 12.22\angle{-32.67^\circ}\ \text{pu} \\ |I_f| &= 12.22 \times 874.8 = 10\,692\ \text{A} \approx 10.7\ \text{kA} \end{aligned}$$

Results

Grounding only one machine through a resistance cuts the earth-fault current to about 42 % and is why generator neutrals on a common bus are normally grounded through impedance, with only one neutral grounded.