T2 has its delta winding on the bus-2 side. How does it affect zero sequence looking in from bus 2?

It isolates the line side, so bus 2 does not see the line, T1 or the generator

It passes zero sequence current to the line

2070 Asar · 4 marks

Zero sequence Thevenin impedance at bus 2

Short answer

The delta windings block zero sequence from the generator and transformer, so only the motor branch appears. Adding the motor reactance and three times the neutral reactance gives j(0.1 + 3×0.05) = j0.25 pu. Thevenin zero sequence impedance at bus 2 is j0.25 pu on 100 MVA base.

For the power system shown in figure below, draw the zero sequence network hence determine the Thevenin's equivalent zero sequence impedance as view from bus-2. [Figure: generator G 100 MVA, 13.8 kV, X0 = 0.05 pu (Y grounded) - T1 100 MVA, 13.8/138 kV, Δ/Y grounded, x = 0.10 pu - line X0 = 60 Ω - T2 100 MVA, 138/13.8 kV, Y grounded/Δ, x = 0.1 pu - bus 2 - motor M 100 MVA, 13.8 kV, X0 = 0.1 pu, neutral grounded through Xn = 0.05 pu]

The zero sequence network is drawn from the winding connections: a delta winding blocks zero sequence current from its line side, and a neutral reactance appears as $3X_n$ . Bus 2 lies between T2 (delta on the 13.8 kV side) and the motor.

Per unit values (100 MVA, 138 kV line base)

Generator: $X_0 = 0.05$ pu (Y grounded)
T1 (Δ / Y-grounded): $0.10$ pu
Line: $Z_b = 138^2/100 = 190.44\ \Omega$ , $X_0 = 60/190.44 = 0.315$ pu
T2 (Y-grounded / Δ): $0.10$ pu
Motor: $X_0 = 0.1$ pu, neutral $3X_n = 3 \times 0.05 = 0.15$ pu

Zero sequence connections

T1 (Δ on generator side, Yg on line side): the line side is connected to the reference through 0.10; the generator side is open. So the generator is isolated from the rest.
T2 (Yg on line side, Δ on bus-2 side): the line side is connected to the reference through 0.10; bus 2 side is open.
Motor: star grounded through $X_n$ , so its branch is $0.1 + 0.15 = 0.25$ pu to reference.

   G         T1          Line         T2      bus 2   M
   o    o  o----j0.315----o  o         o-------+
   |    |  |              |  |                 |
 j0.05 open j0.10      j0.10 open           j0.10
   |       |              |                    |
   |       |              |                 j0.15 (3Xn)
   |       |              |                    |
 ==+=======+==== reference (ground) ===========+==

Thevenin zero sequence impedance at bus 2

Looking into bus 2, the path through T2 is open (delta on the bus-2 side), so only the motor branch is seen:

Z_{0,th} = j(X_{0M} + 3X_n) = j(0.10 + 0.15) = j0.25\ \text{pu}

Answer: $Z_{0,th}$ at bus 2 = j0.25 pu (100 MVA base). The generator, T1, line and T2 do not contribute because of the delta windings.

✓ Answer checked by Claude Sonnet 5.5

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Zero sequence Thevenin impedance at bus 2

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