Neutral reactance for 15 MVA generator

Data: 15 MVA, 13.2 kV, solidly grounded; $X_1 = 0.40$, $X_2 = 0.30$, $X_0 = 0.05$ pu. At no load $E = 1$ pu; rated line current $= 1$ pu.

Fault current without neutral reactance

$$I_f = \frac{3E}{X_1+X_2+X_0} = \frac{3}{0.75} = 4\ \text{pu}$$

This is four times rated current, so a neutral reactance $X_n$ is needed. $X_n$ carries $3I_{a0}$, so it appears as $3X_n$ in the zero-sequence network.

Required neutral reactance

$$\begin{aligned} I_f &= \frac{3E}{X_1+X_2+X_0+3X_n} \le 1 \\ 0.40 + 0.30 + 0.05 + 3X_n &= 3 \\ X_n &= \frac{3 - 0.75}{3} = 0.75\ \text{pu} \end{aligned}$$

Ohmic value

$$Z_{base} = \frac{(13.2)^2}{15} = 11.616\ \Omega$$ $$X_n = 0.75 \times 11.616 = 8.712\ \Omega$$

Answer: neutral reactance $X_n = 0.75$ pu $= 8.71\ \Omega$.

Check: with $X_n$, $I_f = 3/(0.75 + 2.25) = 1$ pu = rated current $\left(\frac{15\times10^6}{\sqrt3\times13.2\times10^3} = 656\ \text{A}\right)$.