Subtransient currents for fault at motor terminals

Base: 30 MVA, 13.2 kV (the machine ratings). The motor terminal voltage is the reference. Internal EMFs are found from the pre-fault load, then each machine's contribution to the fault at the motor terminals is found separately.

Per-unit data

• $X''_g = X''_m = 0.20$ pu, line $X_L = 0.08$ pu.
• $V_f = \dfrac{12.8}{13.2} = 0.9697\angle 0^\circ$ pu.
• Motor load: $P = 20/30 = 0.6667$ pu at 0.8 pf leading.

 Eg --j0.2--+--j0.08--+--j0.2-- Em
          (gen)       (Vf) motor terminals
                       |
                       F

Pre-fault current

$$\begin{aligned} |I_L| &= \frac{0.6667}{0.9697 \times 0.8} = 0.8594\ \text{pu} \\ I_L &= 0.8594\angle 36.87^\circ = 0.6875 + j0.5156\ \text{pu} \end{aligned}$$

Internal voltages

$$\begin{aligned} E''_g &= V_f + j(X_L + X''_g)I_L = 0.9697 + j0.28(0.6875 + j0.5156) = 0.8253 + j0.1925\ \text{pu} \\ E''_m &= V_f - jX''_mI_L = 0.9697 - j0.2(0.6875 + j0.5156) = 1.0728 - j0.1375\ \text{pu} \end{aligned}$$

Sub-transient currents with the fault at the motor terminals

$$\begin{aligned} I''_g &= \frac{E''_g}{j(X''_g + X_L)} = \frac{0.8253 + j0.1925}{j0.28} = 0.6875 - j2.9476\ \text{pu} \\ I''_m &= \frac{E''_m}{jX''_m} = \frac{1.0728 - j0.1375}{j0.2} = -0.6875 - j5.3641\ \text{pu} \\ I''_f &= I''_g + I''_m = -j8.3117\ \text{pu} \end{aligned}$$

Base current: $I_{base} = \dfrac{30 \times 10^6}{\sqrt{3} \times 13.2 \times 10^3} = 1312.2$ A.

Check (Thevenin): $I''_f = V_f\left(\frac{1}{j0.28} + \frac{1}{j0.2}\right) = -j8.3117$ pu, the same.

Answer: $I''_g \approx 3972$ A, $I''_m \approx 7096$ A, $I''_f \approx 10906$ A.