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Feeder reactor: 40 and 30 MVA, 40% reduction

Short answer

On 40 MVA base, generator reactances are 0.05 pu and 0.05333 pu in parallel, plus feeder 0.66116 pu gives 0.68696 pu. Present fault level is 58.23 MVA. To reduce by 40%, required series reactor is 0.458 pu, about 1.385 Ω.

Two generators of capacities 40MVA, 11kV and 30 MVA, 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 2 ohm. 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 40%.

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=0.68696 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} &= 0.68696\ \text{pu} \\ \text{Fault MVA} &= \frac{40}{0.68696} = 58.23\ \text{MVA} \end{aligned}

Reactor for 40 % reduction

New fault MVA =0.6×58.23=34.94= 0.6\times58.23 = 34.94 MVA.

Xeq,new=0.686960.6=1.14494 puX=1.14494−0.68696=0.45798 puX(Ω)=0.45798×3.025=1.385 Ω\begin{aligned} X_{eq,new} &= \frac{0.68696}{0.6} = 1.14494\ \text{pu} \\ X &= 1.14494 - 0.68696 = 0.45798\ \text{pu} \\ X(\Omega) &= 0.45798\times3.025 = 1.385\ \Omega \end{aligned}

Answer: Present fault level = 58.23 MVA; series reactor ≈ 0.458 pu = 1.385 Ω (fault level becomes 34.94 MVA).

✓ Answer checked by Claude Opus 5.5 (high effort)

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