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Fault level: two 25 MVA generators, 50% reduction

Short answer

Fault level is 250 MVA. To halve it, double the existing 0.1 pu reactance to 0.2 pu; add a feeder reactor of 0.1 pu (0.484 Ω), giving 125 MVA.

For the system shown in figure below, calculate the fault level in MVA at out going feeder for a three phase to ground fault on this feeder. Calculate the value of reactance to be connected in the feeder in order to reduce the fault level by 50%. [Figure: G1 25 MVA, 11 kV, E = 1 pu, X1 = 0.15 pu and G2 25 MVA, 11 kV, E = 1 pu, X2 = 0.3 pu connected to a common busbar; outgoing feeder through a C.B. with fault point 'F' on the feeder]

Fault MVA =MVAb/Xeq=\text{MVA}_b/X_{eq}, with E=1E = 1 pu. Feeder reactance (not given) is taken as zero before adding the reactor.

Base: 25 MVA, 11 kV; Zbase=112/25=4.84 ΩZ_{base} = 11^2/25 = 4.84\ \Omega.

 G1 25MVA,0.15   G2 25MVA,0.3
      |               |
 =====+=======+=======+==== 11 kV
              |
             C.B.--[X]-- F

Fault level

Xeq=0.15×0.30.15+0.3=0.1 puFault MVA=250.1=250 MVAIf=250×1063×11×103=13.12 kA\begin{aligned} X_{eq} &= \frac{0.15\times0.3}{0.15+0.3} = 0.1\ \text{pu} \\ \text{Fault MVA} &= \frac{25}{0.1} = 250\ \text{MVA} \\ I_f &= \frac{250\times10^6}{\sqrt3\times11\times10^3} = 13.12\ \text{kA} \end{aligned}

Reactance for 50 % reduction

New fault MVA = 125 MVA.

Xeq,new=25125=0.2 puX=0.2−0.1=0.1 puX(Ω)=0.1×4.84=0.484 Ω\begin{aligned} X_{eq,new} &= \frac{25}{125} = 0.2\ \text{pu} \\ X &= 0.2 - 0.1 = 0.1\ \text{pu} \\ X(\Omega) &= 0.1\times4.84 = 0.484\ \Omega \end{aligned}

Answer: Fault level = 250 MVA; feeder reactance = 0.1 pu = 0.484 Ω (fault level becomes 125 MVA).

✓ Answer checked by Claude Opus 5.5 (high effort)

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