Why is the zero network just two transformer reactances to ground with the line between?

Delta windings cut off G, M1 and M2, leaving only the grounded stars of T1 and T2

2071 Shrawan · 3+7 marks

L-L fault at point q with two motors

Short answer

Positive/negative Thevenin impedances at bus q: Z1=j0.0929 pu, Z2=j0.0998 pu. For the L-L fault, Ia1 = -Ia2 = -j5.19 pu. Fault currents: Ib = -8.99 pu, Ic = 8.99 pu, Ia = 0.

A single line diagram of a power system network is shown in figure below. The system data is given in the table below:

Element	X1 (pu)	X2 (pu)	X0 (pu)
G	0.1	0.12	0.05
M1	0.05	0.06	0.025
M2	0.05	0.06	0.025
T1	0.07	0.07	0.07
T2	0.08	0.08	0.08
Line	0.1	0.1	0.3

i) Draw sequence networks ii) Find fault current for a line-to-line fault on phase b and c at point q. Assume 1.0 pu pre-fault voltage throughout. [Figure: G - bus p - T1 (Δ/Y grounded) - bus q - line - bus r - T2 (Y grounded/Δ) - bus s, which feeds motors M1 (Y, ungrounded) and M2 (Y grounded)]

Assumptions: all reactances on a common base; pre-fault voltage 1 pu everywhere (no load current); bolted fault between phases b and c at q.

(i) Sequence networks

Positive (emfs Eg, Em):
 Eg-j0.1-(p)-j0.07-(q)-j0.1-(r)-j0.08-(s)-+-j0.05-Em1
                    |                     +-j0.05-Em2
                    F
Negative: same layout, G j0.12, T1 j0.07,
 line j0.1, T2 j0.08, M1 j0.06, M2 j0.06, no emfs
Zero:
 ref-j0.07-(q)-j0.3-(r)-j0.08-ref
 G (j0.05) cut off by T1 delta; M1 ungrounded,
 M2 (j0.025) cut off by T2 delta

T1 is Δ/Yg with the grounded star at q, and T2 is Yg/Δ with the grounded star at r, so the zero network is the line between two transformer reactances to ground.

(ii) L-L fault at q

Motors in parallel: positive $0.05\|0.05 = 0.025$ , negative $0.06\|0.06 = 0.03$ .

\begin{aligned} Z_1 &= j(0.1+0.07)\ \|\ j(0.1+0.08+0.025) = j0.17 \| j0.205 = j0.0929\ \text{pu} \\ Z_2 &= j(0.12+0.07)\ \|\ j(0.1+0.08+0.03) = j0.19 \| j0.21 = j0.0998\ \text{pu} \end{aligned}

Sequence currents ( $I_{a0} = 0$ ):

I_{a1} = -I_{a2} = \frac{1}{j(0.0929+0.0998)} = \frac{1}{j0.1927} = -j5.190\ \text{pu}

Fault current:

\begin{aligned} I_b &= -j\sqrt3 I_{a1} = -8.989\ \text{pu} \\ I_c &= -I_b = 8.989\ \text{pu} \end{aligned}

Answer: fault current for L-L fault at q = 8.99 pu (in phases b and c, opposite in sign; $I_a = 0$ ). The zero-sequence network does not take part in an L-L fault.

✓ Answer checked by Claude Sonnet 5.5

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L-L fault at point q with two motors

(i) Sequence networks

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