Fault at generator terminal feeding motor

For a bolted fault at the generator terminal, both the generator and the synchronous motor (acting as a source) feed the fault. $I_f = V_f / X_{th}$ with $V_f = 1$ pu.

Per unit values

Base: 50 MVA, 11 kV at the generator, so 132 kV on the line and 6.6 kV at the motor (voltage bases follow the transformer ratios, so no voltage correction is needed). $Z_{b,line} = 132^2/50 = 348.48\ \Omega$.

        F
 G      o   T1     Line     T2     M
(~)-0.1-+-0.025--0.0717--0.04--0.06-(~)
 |                                   |
 +====== reference (EMF = 1 pu) =====+

Fault current

Reactance on the motor side seen from F:

$$X_m = 0.025 + 0.0717 + 0.04 + 0.06 = 0.1967\ \text{pu}$$ $$\begin{aligned} X_{th} &= \frac{0.10 \times 0.1967}{0.10 + 0.1967} = 0.0663\ \text{pu} \\ I_f &= \frac{1}{0.0663} = 15.08\ \text{pu} \\ I_b &= \frac{50\times10^6}{\sqrt3 \times 11\times10^3} = 2624.3\ \text{A} \\ I_f &= 15.08 \times 2624.3 = 39\,582\ \text{A} \approx 39.6\ \text{kA} \end{aligned}$$

Contributions: generator $1/0.1 = 10$ pu (26.24 kA); motor side $1/0.1967 = 5.08$ pu (13.34 kA).

Fault MVA at the generator terminal

$$\text{Fault MVA} = \frac{S_b}{X_{th}} = \frac{50}{0.0663} = 754.1\ \text{MVA}$$

Check: $\sqrt3 \times 11 \times 39.58 = 754.1$ MVA.

Answer: Fault current ≈ 15.08 pu ≈ 39.6 kA; fault MVA ≈ 754 MVA.