Neutral reactance and L-G fault severity

For an L-G fault on a generator grounded through $X_n$, $I_f = 3E_a/[j(X_1 + X_2 + X_0 + 3X_n)]$; with solid grounding and $X_0 < X_1$ this exceeds the 3-phase current $E_a/X_1$, and the two are equal when $X_n = (X_1 - X_0)/3$.

i) Fault current for solid L-G fault on phase a

Generator sequence equations (neutral reactance appears as $3X_n$ in the zero sequence network):

$$V_{a1} = E_a - jX_1I_{a1}, \quad V_{a2} = -jX_2I_{a2}, \quad V_{a0} = -j(X_0 + 3X_n)I_{a0}$$

Boundary conditions: $I_b = I_c = 0$, $V_a = 0$.

From $I_s = A^{-1}I_p$: $I_{a0} = I_{a1} = I_{a2} = I_a/3$.

From $V_a = V_{a0} + V_{a1} + V_{a2} = 0$:

$$E_a - jI_{a1}(X_1 + X_2 + X_0 + 3X_n) = 0 \;\Rightarrow\; I_{a1} = \frac{E_a}{j(X_1 + X_2 + X_0 + 3X_n)}$$ $$I_f = I_a = 3I_{a1} = \frac{3E_a}{j(X_1 + X_2 + X_0 + 3X_n)}$$

ii) Sequence networks and interconnection

 Positive      Negative      Zero
 F1 o          F2 o          F0 o
    |jX1          |jX2          |jX0
  (Ea)            |             |j3Xn
    |             |             |
 N1 o          N2 o          N0 o

 L-G fault: networks in series
     Ia1-->
 +-(Ea)-jX1--F1--N2--jX2--F2--N0--j3Xn--jX0--F0-+
 |                                              |
 N1 --------------------------------------------+

(F1 joined to N2, F2 to N0, F0 back to N1: a single series loop carrying $I_{a1}$.)

iii) Solidly grounded: L-G more severe than 3-phase fault

With $X_n = 0$ and $X_1 = X_2$:

$$|I_{LG}| = \frac{3E_a}{2X_1 + X_0}, \qquad |I_{3\phi}| = \frac{E_a}{X_1} = \frac{3E_a}{3X_1}$$

Since $X_0 \ll X_1$, $2X_1 + X_0 < 3X_1$, so

$$|I_{LG}| > |I_{3\phi}|$$

For example, $X_1 = X_2 = 0.2$, $X_0 = 0.05$ pu: $I_{LG} = 3/0.45 = 6.67$ pu while $I_{3\phi} = 1/0.2 = 5$ pu. Hence an L-G fault at the terminals of a solidly grounded generator is more severe, and the winding bracing must be designed for it, or the neutral grounded through a reactor.

iv) Limiting neutral reactance

For equal fault currents:

$$\frac{3E_a}{2X_1 + X_0 + 3X_n} = \frac{E_a}{X_1} \;\Rightarrow\; 3X_1 = 2X_1 + X_0 + 3X_n$$ $$X_n = \frac{X_1 - X_0}{3}$$

With $X_n$ larger than this value, the L-G fault current is less than the three-phase fault current. In the example above, $X_n = (0.2 - 0.05)/3 = 0.05$ pu.