SLG fault at bus 3, currents and voltages

Base: 100 MVA; 13.8 kV (generator buses 1 and 4), 115 kV (buses 2 and 3). Pre-fault voltage 1 pu, no load. From the figure: T1 is Yg (through j0.01) on the bus 1 side and Δ on the bus 2 side; T2 is Δ on the bus 3 side and Yg (through 0.02) on the bus 4 side; G1 is ungrounded, G2 solidly grounded, G3 grounded through j0.03.

(i) Sequence networks

Positive (emfs at G1, G2, G3):

 G1 j0.15 \
           (1)-j0.2-(2)-j0.3-(3)-j0.18-(4)-j0.15-G3
 G2 j0.15 /                   |
                              F

Negative: same reactances, no emfs.

Zero:

 G2: ref-j0.05-(1)        G1: open (ungrounded)
 T1: (1)-[3(j0.01)+j0.2]-ref  ;  Δ side (2): open
 line: (2)-j0.9-(3)  -> floating, no ground path
 T2: Δ side (3): open ; (4)-[0.06+j0.18]-ref
 G3: (4)-[j0.05+3(j0.03)]-ref
 -> bus 3 has NO connection to the reference

Because bus 3 sits between the Δ windings of T1 and T2, the zero-sequence impedance seen from bus 3 is infinite: $Z_0 = \infty$.

Thevenin values: $Z_1 = Z_2 = j(0.075+0.2+0.3)\ \|\ j(0.18+0.15) = j0.575\|j0.33 = j0.2097$ pu.

(ii) Phase currents

$$I_{a0} = I_{a1} = I_{a2} = \frac{1}{Z_1+Z_2+Z_0} = \frac{1}{\infty} = 0$$ $$I_a = I_b = I_c = 0\ \text{pu} = 0\ \text{A}$$

No fault current flows: the 115 kV section between the two deltas is an ungrounded system (only a small capacitive current would flow in practice).

(iii) Phase voltages at bus 3

With no current, $V_{a1} = 1$, $V_{a2} = 0$, and the condition $V_a = 0$ forces $V_{a0} = -1$ pu (the whole neutral shifts):

$$\begin{aligned} V_a &= V_{a0}+V_{a1}+V_{a2} = 0 \\ V_b &= V_{a0} + a^2V_{a1} = -1 + 1\angle{-120^\circ} = 1.732\angle{-150^\circ}\ \text{pu} \\ V_c &= V_{a0} + aV_{a1} = -1 + 1\angle{120^\circ} = 1.732\angle{150^\circ}\ \text{pu} \end{aligned}$$

Base phase voltage at bus 3 $= 115/\sqrt3 = 66.40$ kV:

• $V_a = 0$ kV
• $|V_b| = |V_c| = 1.732 \times 66.40 = 115$ kV

Results

The healthy phases rise to full line voltage, showing why ungrounded sections need insulation for line voltage to ground.