Fault at receiving end of 220 kV lines

Base: 100 MVA, 11 kV (generators) and 220 kV (lines). Pre-fault voltage is taken as 1.0 pu and pre-fault load current is neglected (usual assumption for fault level calculation).

Per-unit reactances (100 MVA)

 G1-j0.15-T1-j0.10-+          +-j42 ohm-+
                   +-220 kV --+         +-- F
 G2-jX_G2-T2-jX_T2-+          +-j42 ohm-+

Thevenin reactance at F

$$\begin{aligned} X_a &= 0.15 + 0.10 = 0.25, \quad X_b = 0.1667 + 0.1067 = 0.2733 \\ X_{gen} &= \frac{0.25 \times 0.2733}{0.25 + 0.2733} = 0.1306 \\ X_{lines} &= 0.0868/2 = 0.0434 \\ X_{th} &= 0.1306 + 0.0434 = 0.1740\ \text{pu} \end{aligned}$$

Total fault current and fault level

$$\begin{aligned} I_f &= \frac{1}{0.1740} = 5.7484\ \text{pu} \\ I_{base,220} &= \frac{100 \times 10^6}{\sqrt{3} \times 220 \times 10^3} = 262.43\ \text{A} \\ I_f &= 5.7484 \times 262.43 = 1508.6\ \text{A} \\ \text{Fault level} &= \frac{100}{0.1740} = 574.84\ \text{MVA} \end{aligned}$$

Current supplied by each generator

The fault current divides between the two generator branches in inverse ratio of their reactances:

$$\begin{aligned} I_{G1} &= I_f\frac{X_b}{X_a + X_b} = 5.7484 \times \frac{0.2733}{0.5233} = 3.0023\ \text{pu} \\ I_{G2} &= I_f\frac{X_a}{X_a + X_b} = 5.7484 \times \frac{0.25}{0.5233} = 2.7460\ \text{pu} \end{aligned}$$

On the 11 kV side, $I_{base} = \dfrac{100 \times 10^6}{\sqrt{3} \times 11 \times 10^3} = 5248.6$ A:

Answer: total fault current $\approx 1509$ A at 220 kV, fault level $\approx 574.8$ MVA; G1 supplies about 15.76 kA and G2 about 14.41 kA at 11 kV.