With T1 and T2 delta on the machine side and grounded star on the line side, what does the zero sequence network look like?

Each half-line connects to reference through its transformer; machines are isolated by the deltas

How is the pre-fault voltage at the line midpoint found?

Motor voltage plus load-current drop through T2 and half the line, giving 1.047 pu

How is the 4.033 ohm fault resistance included?

Rf as 3Rf = 0.1 pu in the series connection

2075 Chaitra · 10 marks

SLG fault at mid-line through 4.033 Ω

Short answer

On 100 MVA, 110 kV line base, pre-fault voltage 1.047 pu, Thevenin impedances Z1=j0.1624, Z2=j0.1449, Z0=j0.1562 pu and 3Rf=0.1 pu give sequence current 2.208 pu; fault current = 3 × I_a0 = 6.62 pu = 3.48 kA.

Two alternators are operating in parallel and supplying a synchronous motor which is receiving 60 MW power at 0.8 pf (lag) at 6 kV. Single line diagram for the system and its data are given below. Compute the fault current when a single line to ground fault occurs at the middle of the line through a fault resistance of 4.033 ohm. Data: G1 & G2: 11 kV, 100 MVA, xg1 = 0.20 pu, xg2 = xg0 = 0.10 pu; T1: 180 MVA, 11.5/115 kV, xT1 = 0.10 pu; T2: 170 MVA, 6.6/115 kV, xT2 = 0.10 pu; M: 6.3 kV, 160 MVA, xM1 = xM2 = 0.30 pu, xM0 = 0.10 pu; Line: xLINE1 = xLINE2 = 30.25 ohm, xLINE0 = 60.5 ohm. [Figure: G1 and G2 (star, grounded) in parallel on a bus - T1 - line - T2 - motor M (star, grounded)]

Assumptions: the figure does not mark transformer windings, so the usual arrangement is taken: T1 and T2 are Δ on the machine side and star-grounded on the 115 kV line side. Then the machine zero-sequence reactances are isolated by the deltas. Base: 100 MVA, 11 kV in the generator circuit.

Base voltages and per-unit values

\begin{aligned} V_{b,line} &= 11 \times \frac{115}{11.5} = 110\ \text{kV}, \quad Z_{b,line} = \frac{110^2}{100} = 121\ \Omega \\ V_{b,motor} &= 110 \times \frac{6.6}{115} = 6.313\ \text{kV} \end{aligned}

Element	Calculation	pu
G1 ∥ G2, $X_1$	$0.2 \\| 0.2$	0.10
G1 ∥ G2, $X_2$	$0.1 \\| 0.1$	0.05
T1	$0.1 \times \frac{100}{180}\times(\frac{11.5}{11})^2$	0.0607
Line $X_1 = X_2$	$30.25/121$	0.25
Line $X_0$	$60.5/121$	0.50
T2	$0.1 \times \frac{100}{170}\times(\frac{6.6}{6.313})^2$	0.0643
Motor $X_1 = X_2$	$0.3 \times \frac{100}{160}\times(\frac{6.3}{6.313})^2$	0.1867
Fault resistance $R_f$	$4.033/121$	0.0333

Pre-fault voltage at the middle of the line

Motor voltage $V_m = 6/6.313 = 0.9504\angle0^\circ$ pu (reference). Motor input $S = 60/0.8 = 75$ MVA $= 0.75$ pu. Half of the line is $X_{line}/2 = 0.125$ pu.

\begin{aligned} I_L &= \frac{0.75}{0.9504}\angle{-36.87^\circ} = 0.7891\angle{-36.87^\circ}\ \text{pu} \\ V_f &= V_m + I_L \cdot j(X_{T2} + X_{line}/2) = 0.9504 + 0.7891\angle{-36.87^\circ} \times j(0.0643 + 0.125) \\ &= 0.9504 + 0.7891\angle{-36.87^\circ} \times j0.1893 = 1.0469\angle{6.55^\circ}\ \text{pu} \end{aligned}

Thevenin impedances at the fault point (mid-line)

\begin{aligned} Z_1 &= j(0.10 + 0.0607 + 0.125)\ \|\ j(0.125 + 0.0643 + 0.1867) = j0.2857 \| j0.3760 = j0.1624\ \text{pu} \\ Z_2 &= j(0.05 + 0.0607 + 0.125)\ \|\ j0.3760 = j0.2357 \| j0.3760 = j0.1449\ \text{pu} \\ Z_0 &= j(0.0607 + 0.25)\ \|\ j(0.0643 + 0.25) = j0.3107 \| j0.3143 = j0.1562\ \text{pu} \end{aligned}

(Zero sequence: each half of the line, $X_0/2 = 0.25$ , goes to ground through the grounded star of its transformer.)

Fault current (sequence networks in series with $3R_f$ )

\begin{aligned} I_{a0} = I_{a1} = I_{a2} &= \frac{V_f}{Z_1+Z_2+Z_0+3R_f} = \frac{1.0469\angle{6.55^\circ}}{0.1 + j0.4635} \\ &= 2.208\angle{-71.27^\circ}\ \text{pu} \\ I_f &= 3I_{a0} = 6.624\angle{-71.27^\circ}\ \text{pu} \end{aligned}

Base current on the line:

I_{base} = \frac{100\times10^6}{\sqrt3 \times 110\times10^3} = 524.9\ \text{A}

|I_f| = 6.624 \times 524.9 = 3477\ \text{A}

Answer: SLG fault current ≈ 6.62 pu ≈ 3.48 kA (angle measured from the motor terminal voltage).

Note: $3R_f = 0.1$ pu adds resistance in series, so it both reduces the current and makes it lag by less than 90°.

✓ Answer checked by Claude Sonnet 5.5

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