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Fault level: three generators, feeder reactor

Short answer

On 100 MVA base, G1: 0.5 pu, G2: 0.6667 pu, G3: 0.75 pu, parallel Xeq = 0.2069 pu. Fault level = 100/0.2069 = 483.33 MVA, fault current ≈ 25.37 kA. To reduce by 40% (to 290 MVA), series reactor needed = 0.1379 pu = 0.167 Ω.

The given figure shows three generators operating in parallel. Find the fault current and fault level if three phase to ground fault occurs in the outgoing feeder. What should be the value of feeder reactor to be added to reduce the fault level by 40%? [Figure: G1 20 MVA, 11 kV, X1 = 0.1 pu; G2 30 MVA, 11 kV, X2 = 0.2 pu; G3 40 MVA, 11 kV, X3 = 0.3 pu; all three connected to a common busbar (points A, B, C); an outgoing feeder from the busbar through a circuit breaker, with the fault point on the feeder beyond the C.B.]

Convert all reactances to one base, combine the three parallel generators, and use Fault MVA =MVAb/Xeq=\text{MVA}_b/X_{eq}. The feeder reactance is not given, so it is taken as zero (fault just beyond the C.B.).

Base: 100 MVA, 11 kV. Xnew=XoldMVAnewMVAoldX_{new} = X_{old}\dfrac{\text{MVA}_{new}}{\text{MVA}_{old}}; Zbase=112/100=1.21 ΩZ_{base} = 11^2/100 = 1.21\ \Omega.

 G1 20MVA   G2 30MVA   G3 40MVA
  0.1pu      0.2pu      0.3pu
   |A         |B         |C
 ==+==========+==========+=== 11 kV
              |
             C.B.
              |---- F

Reactances on 100 MVA base

GeneratorCalculationXX (pu)
G10.1×100/200.1\times100/200.5
G20.2×100/300.2\times100/300.6667
G30.3×100/400.3\times100/400.75
1Xeq=10.5+10.6667+10.75=2+1.5+1.3333=4.8333Xeq=0.20690 pu\begin{aligned} \frac{1}{X_{eq}} &= \frac{1}{0.5}+\frac{1}{0.6667}+\frac{1}{0.75} = 2+1.5+1.3333 = 4.8333 \\ X_{eq} &= 0.20690\ \text{pu} \end{aligned}

Fault level and fault current

Fault MVA=1000.2069=483.33 MVAIf=483.33×1063×11×103=25,368 A≈25.37 kA\begin{aligned} \text{Fault MVA} &= \frac{100}{0.2069} = 483.33\ \text{MVA} \\ I_f &= \frac{483.33\times10^6}{\sqrt3\times11\times10^3} = 25{,}368\ \text{A} \approx 25.37\ \text{kA} \end{aligned}

Feeder reactor for 40 % reduction

New fault MVA =0.6×483.33=290= 0.6\times483.33 = 290 MVA.

Xeq,new=100290=0.34483 puXreactor=0.34483−0.20690=0.13793 puXreactor(Ω)=0.13793×1.21=0.1669 Ω\begin{aligned} X_{eq,new} &= \frac{100}{290} = 0.34483\ \text{pu} \\ X_{reactor} &= 0.34483 - 0.20690 = 0.13793\ \text{pu} \\ X_{reactor}(\Omega) &= 0.13793\times1.21 = 0.1669\ \Omega \end{aligned}

Answer: Fault level = 483.33 MVA, fault current ≈ 25.37 kA; feeder reactor ≈ 0.167 Ω (0.1379 pu on 100 MVA) brings the level down to 290 MVA.

✓ Answer checked by Claude Opus 5.5 (high effort)

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