Neutral reactance/resistance to limit L-G current

Data: 30 MVA, 13.2 kV; $X_1 = 0.30$, $X_2 = 0.40$, $X_0 = 0.05$ pu. Rated line current $= 1$ pu. Pre-fault voltage $E = 1$ pu.

$$Z_{base} = \frac{13.2^2}{30} = 5.808\ \Omega$$

With a neutral impedance $Z_n$, the zero-sequence impedance becomes $Z_0 + 3Z_n$:

$$I_f = \frac{3E}{|Z_1 + Z_2 + Z_0 + 3Z_n|}$$

Without $Z_n$: $I_f = 3/0.75 = 4$ pu, i.e. four times rated current, so it must be limited to 1 pu.

(i) Neutral reactance $X_n$

$$\begin{aligned} \frac{3}{0.30 + 0.40 + 0.05 + 3X_n} &= 1 \\ 0.75 + 3X_n &= 3 \\ X_n &= 0.75\ \text{pu} = 0.75 \times 5.808 = 4.356\ \Omega \end{aligned}$$

(ii) Neutral resistance $R_n$

The resistance is in quadrature with the reactances:

$$\begin{aligned} \frac{3}{\sqrt{(3R_n)^2 + 0.75^2}} &= 1 \\ (3R_n)^2 &= 9 - 0.5625 = 8.4375 \\ 3R_n &= 2.9047, \quad R_n = 0.9682\ \text{pu} \\ R_n &= 0.9682 \times 5.808 = 5.624\ \Omega \end{aligned}$$

Answer: (i) neutral reactance $X_n = 0.75$ pu $= 4.36\ \Omega$; (ii) neutral resistance $R_n = 0.968$ pu $= 5.62\ \Omega$.

A larger ohmic value is needed with resistance because it adds to the network reactance in quadrature, not directly.