Generator and motor currents for terminal fault

Use the internal (sub-transient) EMFs of both machines found from the pre-fault load. Base: 100 MVA, 11 kV.

Per-unit values

• Generator: $X''_g = 0.25$ pu.
• Motor: $X''_m = 0.2 \times \dfrac{100}{50} = 0.4$ pu.
• Line: $X_L = 0.05$ pu.
• Motor terminal voltage: $V_m = \dfrac{10.95}{11} = 0.9955\angle 0^\circ$ pu (reference).
• Motor input: $P = 40/100 = 0.4$ pu at 0.8 pf leading.

 Eg --j0.25--(Vt)--j0.05--(Vm)--j0.4-- Em
             |
             F (fault at generator terminals)

Pre-fault current

$$\begin{aligned} |I| &= \frac{P}{V_m\cos\phi} = \frac{0.4}{0.9955 \times 0.8} = 0.5023\ \text{pu} \\ I &= 0.5023\angle 36.87^\circ = 0.4018 + j0.3014\ \text{pu}\ \text{(leading)} \end{aligned}$$

Internal EMFs

$$\begin{aligned} V_t &= V_m + jX_LI = 0.9955 + j0.05(0.4018 + j0.3014) = 0.9804 + j0.0201\ \text{pu} \\ E''_g &= V_t + jX''_gI = 0.9804 + j0.0201 + j0.25(0.4018 + j0.3014) = 0.9050 + j0.1205 = 0.9130\angle 7.59^\circ\ \text{pu} \\ E''_m &= V_m - jX''_mI = 0.9955 - j0.4(0.4018 + j0.3014) = 1.1160 - j0.1607 = 1.1275\angle -8.20^\circ\ \text{pu} \end{aligned}$$

Currents during the fault

With the generator terminals shorted, each machine feeds the fault through its own reactance:

$$\begin{aligned} I''_g &= \frac{E''_g}{jX''_g} = \frac{0.9050 + j0.1205}{j0.25} = 0.4822 - j3.6202 = 3.6521\angle -82.41^\circ\ \text{pu} \\ I''_m &= \frac{E''_m}{j(X''_m + X_L)} = \frac{1.1160 - j0.1607}{j0.45} = -0.3572 - j2.4800 = 2.5056\angle -98.20^\circ\ \text{pu} \\ I_f &= I''_g + I''_m = 0.1250 - j6.1002 = 6.1015\angle -88.83^\circ\ \text{pu} \end{aligned}$$

Base current at 11 kV: $I_{base} = \dfrac{100 \times 10^6}{\sqrt{3} \times 11 \times 10^3} = 5248.6$ A.

Check: $I_f = V_t\left(\frac{1}{j0.25} + \frac{1}{j0.45}\right) = 0.1250 - j6.1002$ pu, the same.

Answer: generator current $\approx 19.17$ kA, motor current $\approx 13.15$ kA, total fault current $\approx 32.02$ kA.