DLG fault at bus 2, fault MVA

Base: 100 MVA; 15.75 kV (generators), 138 kV (line). Pre-fault voltage 1 pu, no load, bolted fault ($Z_f = 0$).

$$Z_{base,line} = \frac{138^2}{100} = 190.44\ \Omega, \quad X_{1,line} = \frac{25}{190.44} = 0.1313, \quad X_{0,line} = \frac{70}{190.44} = 0.3676\ \text{pu}$$

Sequence networks

Positive (emfs $E_1$, $E_2$):

 E1-j0.15-(1)-j0.1-(2)-j0.1313-(3)-j0.1-(4)-j0.2-E2
                   |
                   F

Negative: same reactances ($X_2 = X_1$), no emfs.

Zero:

• T1 is Δ (bus 1 side)/Yg (bus 2 side): bus 2 to reference through $j0.1$; G1 isolated.
• T2 is Yg (bus 3 side)/Δ (bus 4 side): bus 3 to reference through $j0.1$; G2 isolated.

 ref-j0.1-(2)-j0.3676-(3)-j0.1-ref
           |
           F      (G1 j0.05, G2 j0.1 isolated by deltas)

Thevenin impedances at bus 2

$$\begin{aligned} Z_1 = Z_2 &= j0.25\ \|\ j(0.1313 + 0.1 + 0.2) = j0.25\ \|\ j0.4313 = j0.1583\ \text{pu} \\ Z_0 &= j0.1\ \|\ j(0.3676 + 0.1) = j0.1\ \|\ j0.4676 = j0.0824\ \text{pu} \end{aligned}$$

DLG fault (b, c to ground): networks in parallel

$$\begin{aligned} Z_2\|Z_0 &= \frac{0.1583 \times 0.0824}{0.2406} = j0.0542 \\ I_{a1} &= \frac{1}{j(0.1583 + 0.0542)} = -j4.707\ \text{pu} \\ I_{a2} &= -I_{a1}\frac{Z_0}{Z_2+Z_0} = j1.612\ \text{pu} \\ I_{a0} &= -I_{a1}\frac{Z_2}{Z_2+Z_0} = j3.096\ \text{pu} \end{aligned}$$

Phase currents:

$$I_b = 7.177\angle{139.68^\circ}\ \text{pu}, \quad I_c = 7.177\angle{40.32^\circ}\ \text{pu}$$

Fault (ground) current:

$$I_f = 3I_{a0} = j9.287\ \text{pu}$$ $$I_{base} = \frac{100\times10^6}{\sqrt3\times138\times10^3} = 418.4\ \text{A} \Rightarrow I_f = 9.287 \times 418.4 = 3886\ \text{A}$$

(Each faulted line carries $7.177 \times 418.4 = 3003$ A.)

Short-circuit MVA

$$\text{SC MVA} = |I_f|_{pu} \times V_f \times \text{MVA}_{base} = 9.287 \times 1 \times 100 = 928.7\ \text{MVA}$$

Answer: $I_f \approx 9.29$ pu $= 3.89$ kA; short-circuit level ≈ 929 MVA.