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Feeder reactor: 40 and 30 MVA, 50% reduction

Short answer

Existing fault level is 58.23 MVA. To reduce it by 50% to 29.11 MVA, double the total reactance from 0.687 pu to 1.374 pu by adding a series reactor of 0.687 pu (2.078 Ω).

Two generators of capacities 40MVA, 11kV and 30MVA, 11kV having their sub transient reactance of 5% and 4% respectively operating in parallel and supplying to a common load by single feeder of reactance 2Ω. Calculate the value of reactor to be connected in series with feeder so as to reduce the fault level at the end of the feeder by 50%.

Work in per unit on a common base. Fault MVA =MVAb/Xeq=\text{MVA}_b/X_{eq}; ohms to pu: Xpu=XΩ MVAb/kV2X_{pu} = X_\Omega\,\text{MVA}_b/\text{kV}^2.

Base: 40 MVA, 11 kV; Zbase=112/40=3.025 ΩZ_{base} = 11^2/40 = 3.025\ \Omega.

 G1 40MVA,5%   G2 30MVA,4%
     |             |
 ====+=====+=======+==== 11 kV
           |
          2 ohm feeder + X
           |
           F

Present fault level

XG1=0.05 pu,XG2=0.04×4030=0.05333 puXG1∥XG2=0.05×0.053330.10333=0.025806 puXfeeder=23.025=0.66116 puXeq=0.025806+0.66116=0.68696 puFault MVA=400.68696=58.23 MVA\begin{aligned} X_{G1} &= 0.05\ \text{pu},\quad X_{G2} = 0.04\times\frac{40}{30} = 0.05333\ \text{pu} \\ X_{G1}\parallel X_{G2} &= \frac{0.05\times0.05333}{0.10333} = 0.025806\ \text{pu} \\ X_{feeder} &= \frac{2}{3.025} = 0.66116\ \text{pu} \\ X_{eq} &= 0.025806 + 0.66116 = 0.68696\ \text{pu} \\ \text{Fault MVA} &= \frac{40}{0.68696} = 58.23\ \text{MVA} \end{aligned}

Reactor for 50 % reduction

New fault MVA =29.11= 29.11 MVA, so the total reactance must double:

Xeq,new=2×0.68696=1.37393 puX=1.37393−0.68696=0.68696 puX(Ω)=0.68696×3.025=2.078 Ω\begin{aligned} X_{eq,new} &= 2\times0.68696 = 1.37393\ \text{pu} \\ X &= 1.37393 - 0.68696 = 0.68696\ \text{pu} \\ X(\Omega) &= 0.68696\times3.025 = 2.078\ \Omega \end{aligned}

Answer: Present fault level = 58.23 MVA; series reactor = 0.687 pu ≈ 2.078 Ω (fault level becomes 29.11 MVA).

✓ Answer checked by Claude Opus 5.5 (high effort)

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