Why is G1 cut off from the zero sequence network?

T1 has a delta on the generator side, which blocks zero sequence current

How is the transformer neutral reactance j0.03 included in the zero sequence network?

As 3 x j0.03 in series with the path

2079 Bhadra · 8 marks

L-G fault at bus 1 through Zf

Short answer

Thevenin impedances at bus 1 are Z1=Z2=j0.2024 pu, Z0=j0.0845 pu. With fault impedance j0.05 pu, Ia1 = 1/(j0.6393) = -j1.564 pu. Fault current I_f = 3Ia1 = 4.692 pu = 1.03 kA at 132 kV.

Figure shows a power system network. Draw positive, negative and zero sequence network. If LG fault occurs at bus 1, find fault current. Assume fault impedance Zf = 0.05 p.u.

Equipment	MVA rating	Voltage rating	X1 (p.u.)	X2 (p.u.)	X0 (p.u.)
G1	50	11 KV	0.20	0.20	0.08
G2	30	11 KV	0.25	0.25	0.1
Transformer T1	50	11/132 KV	0.1	0.1	0.1
Transformer T1 (T2)	30	11/132 KV	0.09	0.09	0.09
Line L1	45	132 KV	0.1	0.1	0.25
Line L1 (L2)	45	132 KV	0.1	0.1	0.25

[Figure: G1 (Y grounded) - T1 (Δ on generator side, Y grounded on bus 1 side) - bus 1 - two parallel lines - bus 2 - T2 (Y grounded on bus 2 side, Y grounded through j0.03 on generator side) - G2 (Y grounded); fault F at bus 1]

Assumptions: base 50 MVA, 11 kV (generators) and 132 kV (lines). The neutral reactance j0.03 of T2 is taken as already on the 50 MVA base. Pre-fault voltage 1 pu, no load.

Per-unit values on 50 MVA

Element	$X_1 = X_2$	$X_0$
G1	0.20	0.08
G2 ( $\times 50/30$ )	0.4167	0.1667
T1	0.10	0.10
T2 ( $\times 50/30$ )	0.15	0.15
Each line ( $\times 50/45$ )	0.1111	0.2778
Two lines in parallel	0.0556	0.1389

Sequence networks

Positive (negative same, no emf):
 E1-j0.2-j0.1-(1)-j0.0556-(2)-j0.15-j0.4167-E2
               F
Zero:
 ref-j0.1-(1)-j0.1389-(2)-j0.15-[3x j0.03]-j0.1667-ref
 (G1 cut off by T1 delta; T2 is Yg-Yg, so G2
  zero seq path continues through T2)

Thevenin impedances at bus 1

\begin{aligned} Z_1 = Z_2 &= j0.30\ \|\ j(0.0556 + 0.15 + 0.4167) = j0.30\ \|\ j0.6222 = j0.2024\ \text{pu} \\ Z_0 &= j0.10\ \|\ j(0.1389 + 0.15 + 0.09 + 0.1667) = j0.10\ \|\ j0.5456 = j0.0845\ \text{pu} \end{aligned}

Fault current (L-G through $Z_f = 0.05$ pu)

I_{a0} = I_{a1} = I_{a2} = \frac{V_f}{Z_1 + Z_2 + Z_0 + 3Z_f}

Taking $Z_f$ as a reactance $j0.05$ (usual in these problems), $3Z_f = j0.15$ :

\begin{aligned} I_{a0} &= \frac{1}{j(0.2024+0.2024+0.0845+0.15)} = \frac{1}{j0.6393} = -j1.564\ \text{pu} \\ I_f &= 3I_{a0} = -j4.692\ \text{pu} \end{aligned}

Base current at 132 kV:

I_{base} = \frac{50\times10^6}{\sqrt3 \times 132\times10^3} = 218.7\ \text{A}

I_f = 4.692 \times 218.7 = 1026\ \text{A}

Answer: L-G fault current at bus 1 ≈ 4.69 pu ≈ 1.03 kA.

(If $Z_f$ is taken as a pure resistance 0.05 pu, $|I_f| = 3/|0.15 + j0.4893| = 5.86$ pu.)

✓ Answer checked by Claude Sonnet 5.5

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L-G fault at bus 1 through Zf

Per-unit values on 50 MVA

Sequence networks

Thevenin impedances at bus 1

Fault current (L-G through Zf=0.05Z_f = 0.05Zf​=0.05 pu)

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