Fault level: two 40 MVA generators, 2 Ω feeder

Work in per unit on a common base. Fault level $= \dfrac{\text{Base MVA}}{X_{pu}}$ and a reactor of $X$ ohms has $X_{pu} = X\cdot\dfrac{\text{MVA}_b}{\text{kV}_b^2}$.

Base: 40 MVA, 11 kV.

$$Z_{base} = \frac{11^2}{40} = 3.025\ \Omega$$

 G1 40MVA,5%   G2 40MVA,4%
     |             |
 ====+=====+=======+==== 11 kV bus
           |
          X=2 ohm feeder
           |
           F  (3-ph fault)

Fault level

$$\begin{aligned} X_{G1} &= 0.05\ \text{pu},\quad X_{G2} = 0.04\ \text{pu} \\ X_{G1}\parallel X_{G2} &= \frac{0.05\times0.04}{0.05+0.04} = 0.02222\ \text{pu} \\ X_{feeder} &= \frac{2}{3.025} = 0.66116\ \text{pu} \\ X_{eq} &= 0.02222 + 0.66116 = 0.68338\ \text{pu} \\ \text{Fault MVA} &= \frac{40}{0.68338} = 58.53\ \text{MVA} \end{aligned}$$

Fault current: $I_{base} = \dfrac{40\times10^6}{\sqrt3\times 11\times10^3} = 2099.5\ \text{A}$, so $I_f = 2099.5/0.68338 = 3072\ \text{A}$.

Reactor to reduce fault level by 40 %

New fault level $= 0.6\times58.53 = 35.12$ MVA.

$$\begin{aligned} X_{eq,new} &= \frac{40}{35.12} = \frac{0.68338}{0.6} = 1.13897\ \text{pu} \\ X_{added} &= 1.13897 - 0.68338 = 0.45559\ \text{pu} \\ X_{added}(\Omega) &= 0.45559\times3.025 = 1.378\ \Omega \end{aligned}$$

Answer: Fault level = 58.53 MVA (about 3.07 kA). A reactor of about 1.378 Ω (0.4556 pu on 40 MVA) in series with the 2 Ω feeder reduces it by 40 % to 35.12 MVA.