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Fault level: two 50 MVA generators, 50% reduction

Short answer

Fault level is 500 MVA. Halving it requires doubling the existing 0.1 pu reactance to 0.2 pu; thus add a feeder reactor of 0.1 pu (0.242 Ω), giving a new fault level of 250 MVA.

For the power system network given below, calculate the fault level in MVA at the outgoing feeder for a 3 phase to ground fault on the feeder. Calculate the value of reactance to be connected in the feeder in order to reduce the fault level by 50%. Ratings of the generators are: G1 = 50 MVA, 11 kV, Xg1 = 0.15 p.u.; G2 = 50 MVA, 11 kV, Xg2 = 0.3 p.u. [Figure: G1 and G2 connected to a common busbar with a single outgoing feeder]

Generators in parallel feed the fault through the bus. Fault MVA =MVAb/Xeq=\text{MVA}_b/X_{eq}. Feeder reactance (not given) is taken as zero before adding the reactor.

Base: 50 MVA, 11 kV; Zbase=112/50=2.42 ΩZ_{base} = 11^2/50 = 2.42\ \Omega. Both generators are already on 50 MVA.

 G1 50MVA,0.15   G2 50MVA,0.3
      |               |
 =====+=======+=======+==== 11 kV
              |  feeder (+X)
              F

Fault level

Xeq=0.15×0.30.15+0.3=0.1 puFault MVA=500.1=500 MVAIf=500×1063×11×103=26.24 kA\begin{aligned} X_{eq} &= \frac{0.15\times0.3}{0.15+0.3} = 0.1\ \text{pu} \\ \text{Fault MVA} &= \frac{50}{0.1} = 500\ \text{MVA} \\ I_f &= \frac{500\times10^6}{\sqrt3\times11\times10^3} = 26.24\ \text{kA} \end{aligned}

Reactance for 50 % reduction

New fault MVA = 250 MVA, so

Xeq,new=50250=0.2 puX=0.2−0.1=0.1 puX(Ω)=0.1×2.42=0.242 Ω\begin{aligned} X_{eq,new} &= \frac{50}{250} = 0.2\ \text{pu} \\ X &= 0.2 - 0.1 = 0.1\ \text{pu} \\ X(\Omega) &= 0.1\times2.42 = 0.242\ \Omega \end{aligned}

(Halving the fault level needs the total reactance to double, so the added reactance equals the existing XeqX_{eq}.)

Answer: Fault level = 500 MVA; feeder reactance = 0.1 pu = 0.242 Ω (fault level becomes 250 MVA).

✓ Answer checked by Claude Opus 5.5 (high effort)

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