Why must T2 and G2 (50 MVA) values be converted?

To a common 100 MVA base: values double

Voltage ratings differ so values halve

(0.3545 in parallel with 0.45) = 0.1983 pu

In the zero-sequence network at bus 3, the G2 side path is:

Isolated by T2 delta; bus 3 grounds through T2 star plus 3Xn (0.33 pu)

2068 Chaitra · 8 marks

Equivalent Z0, Z1, Z2 at bus 3

Short answer

Positive and negative sequence impedances at bus 3: Z1 = Z2 = j0.1983 pu. Zero-sequence impedance: Z0 = j(0.1+0.0545) ∥ j0.33 = j0.105 pu (or j0.094 pu depending on neutral reactor base).

Draw positive, negative and zero sequence networks. If an unsymmetrical fault occurs at bus 3, determine equivalent Z0, Z1 and Z2. [Figure: G1 100 MVA, 11 kV, X1 = X2 = j0.2 pu, X0 = j0.05 pu (Y grounded) - T1 100 MVA, 11/220 kV, j0.1 pu (Δ / Y grounded) - parallel lines L1 (100 MVA, 220 kV, j0.1 pu) and L2 (100 MVA, 220 kV, 0.12 pu) - T2 50 MVA, 11/220 kV, j0.075 pu (Y grounded through j0.03 pu on line side / Δ) - G2 50 MVA, 11 kV, X1 = X2 = j0.15 pu, X0 = j0.03 pu (Y grounded through j0.024 pu)]

Assumptions: base 100 MVA; bus 1 = G1 terminal, bus 2 = T1 HV side, bus 3 = T2 HV side (end of the parallel lines), bus 4 = G2 terminal. Neutral reactances (j0.03 for T2, j0.024 for G2) are taken on their own 50 MVA rating. Line zero-sequence reactance is taken equal to the given value since no separate $X_0$ is given.

Per-unit values on 100 MVA

Element	$X_1 = X_2$	$X_0$
G1	0.20	0.05
T1	0.10	0.10
L1 ∥ L2	$0.1\\|0.12 = 0.0545$	0.0545
T2 ( $\times 2$ )	0.15	0.15
T2 neutral ( $\times 2$ )	–	$3\times0.06 = 0.18$
G2 ( $\times 2$ )	0.30	0.06
G2 neutral ( $\times 2$ )	–	$3\times0.048 = 0.144$

Positive sequence network

 E1-j0.2-(1)-j0.1-(2)-j0.0545-(3)-j0.15-(4)-j0.3-E2
                                |
                                F

Z_1 = j(0.2+0.1+0.0545)\ \|\ j(0.15+0.3) = j0.3545\ \|\ j0.45 = j0.1983\ \text{pu}

Negative sequence network

Same reactances, emfs shorted:

Z_2 = j0.1983\ \text{pu}

Zero sequence network

T1 Δ (G1 side)/Yg (line side): bus 2 to reference through $j0.1$ ; G1 isolated.
T2 Yg through j0.03 (line side)/Δ (G2 side): bus 3 to reference through $j(0.15+0.18) = j0.33$ ; G2 isolated.

 ref-j0.1-(2)-j0.0545-(3)-j0.33-ref
                       |
                       F
 G1 (j0.05) and G2 (j0.06+j0.144) isolated by deltas

Z_0 = j(0.1+0.0545)\ \|\ j0.33 = j0.1545\ \|\ j0.33 = j0.1053\ \text{pu}

Answer: $Z_1 = Z_2 = j0.198$ pu, $Z_0 = j0.105$ pu at bus 3.

(If the neutral reactance j0.03 is taken as already on the 100 MVA base, $Z_0 = j0.1545\|j0.24 = j0.094$ pu.)

✓ Answer checked by Claude Sonnet 5.5

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Equivalent Z0, Z1, Z2 at bus 3

Per-unit values on 100 MVA

Positive sequence network

Negative sequence network

Zero sequence network

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