Generator-reactor system fault MVA

The fault MVA is found by converting every reactance to one common MVA base, reducing the network to a single Thevenin reactance seen from F, and using $\text{Fault MVA} = \text{Base MVA}/X_{th}$ with prefault voltage 1 pu.

Step 1: Choose base and convert reactances

Base: $S_b = 20$ MVA, $V_b = 11$ kV. Use $X_{new} = X_{old}\,\dfrac{S_b}{S_{rated}}$ (same voltage, so no voltage correction).

Step 2: Reactance diagram

All generator EMFs are equal (1 pu) and their neutral is the reference, so for a 3-phase fault every source branch ends on the reference bus.

   F (G1 bus)        R1=0.1       B (G2 bus)  R2=0.1
 ---o----------------/\/\/\---------o-------/\/\/\---+
    |                               |                |
  G1 0.2                         G2 0.15          G3 0.15
    |                               |                |
 ===+===========  reference  =======+================+===

Step 3: Reduce the network as seen from F

$$\begin{aligned} X_{R2+G3} &= 0.10 + 0.15 = 0.25 \\ X_B &= 0.15 \parallel 0.25 = \frac{0.15 \times 0.25}{0.40} = 0.09375 \\ X_{R1+B} &= 0.10 + 0.09375 = 0.19375 \\ X_{th} &= 0.20 \parallel 0.19375 = \frac{0.20 \times 0.19375}{0.39375} = 0.09841\ \text{pu} \end{aligned}$$

Step 4: Fault current and fault MVA

$$\begin{aligned} I_f &= \frac{1}{X_{th}} = \frac{1}{0.09841} = 10.16\ \text{pu} \\ I_b &= \frac{S_b}{\sqrt{3}\,V_b} = \frac{20\times 10^6}{\sqrt{3}\times 11\times 10^3} = 1049.7\ \text{A} \\ I_f &= 10.16 \times 1049.7 = 10\,667\ \text{A} \approx 10.67\ \text{kA} \\ \text{Fault MVA} &= \frac{S_b}{X_{th}} = \frac{20}{0.09841} = 203.2\ \text{MVA} \end{aligned}$$

Check: $\sqrt{3} \times 11 \times 10.667 = 203.2$ MVA.

The fault MVA does not depend on the base chosen. With a 10 MVA base every pu reactance is halved, so $X_{th} = 0.04920$ pu and $10/0.04920 = 203.2$ MVA, the same result.

Answer: Fault current ≈ 10.67 kA (10.16 pu on 20 MVA base); fault MVA ≈ 203.2 MVA.